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Min-plus algebra and graph domination

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Min-plus algebra and graph domination
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Spalding, Anne
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176 leaves : ; 28 cm

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Graph theory ( lcsh )
Domination (Graph theory) ( lcsh )
Algebra ( lcsh )
Discrete-time systems ( lcsh )
Algebra ( fast )
Discrete-time systems ( fast )
Domination (Graph theory) ( fast )
Graph theory ( fast )
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theses ( marcgt )
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Includes bibliographical references (leaves 175-176).
General Note:
Department of Mathematical and Statistical Sciences
Statement of Responsibility:
by Anne Spalding.

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|University of Colorado Denver
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Full Text
MIN-PLUS ALGEBRA AND GRAPH DOMINATION
by
Anne Spalding
B.S., University of Colorado at Denver, 1993
M.S., University of Colorado at Denver, 1996
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
1998


This thesis for the Doctor of Philosophy
degree by
Anne Spalding
has been approved
by
David C. Fisher
J. Richard Lundgren
Stanley Payne
Ross McConnell
Tom Altman
Date


Spalding, Anne (Ph. D., Applied Mathematics)
Min-Plus Algebra and Graph Domination
Thesis directed by Professor David C. Fisher
ABSTRACT
This thesis describes the Min-Plus algebra and uses Min-Plus algebra to
implement dynamic programming algorithms. Min-plus algebra is defined by
replacing the standard algebra operators of addition and multiplication by
minimization and addition respectively. There are several applications which
can be solved using these dynamic programming algorithms in Min-Plus alge-
bra. We will illustrate these algorithms by using them to solve the following
problems involving domination.
1 A domination D of a graph G is a subset of vertices so that each vertex of G is
either in D or adjacent to a vertex in D. Let 7(G) (the domination number
of G) be the minimum size of a domination of G. We find the domination
number of a large class of graphs including the cartesian product graphs
and circulant graphs.
3 From chess, a knight move is two squares in one direction and one square
in the perpendicular direction. A Knight domination of a k x r board
111


is a placement of Knights so each square either has a Knight on it or
attacking it. The kxr Knight domination number is the minimum number
of Knights in a Knight domination of a kxr board. We find the the
Knight domination number of a kxr chessboard for k < 7 and for all r.
The defining feature of these dynamic programming algorithms is that they find
infinitely many solutions in a finite amount of time. This is possible because
the algorithms are periodic after a finite number of iterations. We show that
periodicity is dependent on the size and the entries in the state transition
matrix, and we determine the necessary conditions on the matrix for periodicity
to occur. We also look at random matrices with entries independently chosen
from discrete and continuous distributions and find the expected number of
iterations needed for periodicity in each case. The results of this analysis shows
why the algorithms are successful in solving the graph domination problems and
what conditions are necessary to use similar dynamic programming algorithms
in other problems.
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
Signed
David C. Fisher
IV


DEDICATION
I would like to dedicate this thesis to my parents and my family who have
always stood behind me supporting me and driving me forward; to Ray who
has always stood in front of me and shown me the way; and to Steve, who
always encouraged me to stretch my limits... to ride faster, climb higher and
play harder. Thank you.


ACKNOWLEDGEMENT
I would like to thank my advisor, David C. Fisher, for his encouragement in
preparation of this thesis, his support, guidance, friendship, and time spent in
discussions throughout my studies at the University of Colorado at Denver.


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63
CONTENTS
Introduction..................................
Properties of Min-Plus........................
Min-Plus Linear Algebra.......................
Eigenvalues and Eigenvectors: The Power Method
Minimum Cycle Means and Karps Theorem . .
Domination of Cartesian Product Graphs........
The Exact Algorithm...........................
The At Least Algorithm .......................
Formulas for the Domination Number............
Domination of G x Pr : The At Least Algorithm .
Values of j(G x Pr) for graphs on < 6 vertices. .
Domination of Circulant Graphs................
The States and Transition Matrix..............
Periodicity of the Transition Matrix..........
Knights Domination of k x r Chessboards.......
The States....................................
Vll


5.2. The Transition Matrix
64
5.3. The States Vector......................................... 69
5.4. Detecting Periodicity..................................... 70
5.5. Domination Number for a 3 x r board..................... 72
5.6. Domination Number for a 4 x r board..................... 73
5.7. Domination Number for a 5 x r board..................... 74
5.8. Domination Number for a 6 x r board..................... 75
5.9. Domination Number for a 7 x r board..................... 76
5.10. Conclusions ............................................. 78
6. Time Complexity of The Power Method....................... 79
6.1. Finite Termination of the Power Method.................... 80
6.2. Exact Bound for i?0 when A is MCM-Hamiltonian............. 84
6.3. Periodicity in Matrices with Integer entries.............. 95
6.4. Periodicity in Random Matrices With Real Entries.......... 99
6.5. Conclusions............................................... 109
Appendix........................................................... 112
A. Domination Numbers of Circulant Graphs .................. 112
References......................................................... 175
viii


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72
FIGURES
Minimum Dominations of Pi3 x P13.......................
The Partial Ordering of the 17 States for P3 x Pr......
The House Graph on Five Vertices G and G x P4..........
The Circulant Graph....................................
15 State for the Domination of Circulant Graphs........
Precedence Digraph for Circulant Graphs................
Knight Moves ..........................................
A Minimal 3 x 14 Knight Domination.....................
The 8 x 8 Knight Graph.................................
The 81 Knight configurations of a 2 x 2 Board..........
Possible Matchings of k x 2 Boards ....................
Domination Pattern of 2 x r Board......................
Domination Pattern of 3 x r Board for r = 0, 3, 4, and 5. . .
Domination Pattern of 3 x r Board for r = 1 (mod 6) r > 4 .
Domination Pattern of 3 x r Board for r = 2 (mod 6) r > 14
IX


5.10 Domination Pattern of 4 x r Board for r = 0, 3, 4, and 5
(mod 6) when r > 1................................... 73
5.11 Domination Pattern of 4 x r Board for r = 1 (mod 6) r > 4. 73
5.12 Domination Pattern of 4 x r Board for r = 2 (mod 6) r > 14. 73
5.13 Components in Domination Patterns of 5 x r Boards.... 74
5.14 Domination Pattern of 6 x r Board for r = 0, and 3 (mod 4)
r > 1...................................................... 76
5.15 Domination Pattern of 6 x r Board for r = 1 (mod 4) r > 4 . 76
5.16 Domination Pattern of 6 x r Board for r = 2 (mod 4) r > 4 . 76
5.17 Components in Domination Patterns of 7 x r Boards.......... 77
6.1 Precedence Digraph of a Normalized MCM Hamiltonian Ma-
trix........................................................ 90
6.2 The Pre-Periodic Interval for Random Matrices With Integer
Entries.................................................... 97
x


1. Introduction
This thesis describes the Min-Plus algebra and uses Min-Plus algebra to
implement dynamic programming algorithms. Min-plus algebra is defined by
two operators m and ffl on the real numbers. The symbol H represents stan-
dard addition and the symbol ffl represents minimization. In Chapter 2 we
will look at several properties of Min-Plus algebra. Using the standard alge-
bra properties, we develop analogues to linear algebra including eigenvalues
and eigenvectors. Eigenvalues are used in the implementation of dynamic pro-
gramming algorithms in Min-Plus algebra. We will discuss two methods of
finding eigenvalues and we will show how their existence ensures finite time
execution of the dynamic programming algorithms.
There are several applications which can be solved using dynamic pro-
gramming algorithms in Min-Plus algebra. We will illustrate these algorithms
by using them to solve problems in graph domination.
Definition 1.1 A domination D of a graph G is a subset of vertices so that
each vertex of G is either in D or adjacent to a vertex in D. Let 7(G) (the
domination number of G) be the minimum, size of a domination of G.
The domination problem is NP-hard. Most algorithms for finding the
1


domination number of a graph involve using Branch and Bound techniques to
reduce the graph to other graphs for which the domination number is known.
We develop a dynamic programming algorithm to find the domination number
of large classes of graphs. The first set of graphs we consider are the cartesian
product graphs.
Definition 1.2 Let G = (V, E) be a graph with vertices {tq, v2, vn} and let
H = (V, E) be a graph with vertices {w\,w2 ... uq}. Then the cartesian product
G x H has vertices of the form, {vi,wf), with 1 < i < n and 1 < j < l. The
edges of G x Ed have the form ((iq,-uq), (vp,wq)) if i = p and (wj,wq) is and
edge of H, or if j = q and (iq, vp) is an edge of G. The k^' copy of G in G x H
is Gk and its vertices are (tq,-uq), {v2, icq), {vn,Wk)-
In Chapter 3, we consider cartesian product graphs with H = Pr where
P, is the path on r vertices with vertix uq adjacent to vertex uq if and only if
\i k\ = 1.
We begin by looking at a special cartesian product graph called the com-
plete grid graph. The complete grid graph on n, r vertices, Pn x Pr has nr vertices
with vertex (iq, wf) adjacent to vertex (vk,uh) if an(f only if \i k\ + \j 1\ = 1.
The domination number of complete grid graphs (P x Pr) has been analyzed
for small values of n. Jacobson and Kinch [13] found results for n < 4, and
Chang and Clark [2] found results for n = 5,6. In [11] and [15] Branch and
2


Bound algorithms are used to find domination numbers for larger values of n
and some values of r. These results were then expanded by the use of dynamic
programming algorithms [11]. We improve these algorithms by incorporating
the periodic property of Min-Plus algebra. We develop two algorithms, the
Exact Algorithm and the At Least Algorithm and are able to find the domination
number for all grid graphs (Pn x Pr) with n < 19 and for all values of r. Liv-
ingston and Stout [14] use an algorithm which we will call the Exact Algorithm
to find the domination number of general cartesian product graphs (G x Pr).
We use the faster At Least Algorithm to find the domination number of graphs
(G x Pr) for any graph G on n < 6 vertices and for all r.
In Chapter 4, we modify the Exact Algorithm to find the domination
number of circulant graphs.
Definition 1.3 Given an integer r > 0 and subset S of the positive integers,
the circulant graph (7r[A] is a graph on vertices 0, 1, ..., r 1 with vertex i
adjacent to vertex j if i j e S or j i e S (mod r).
We find the domination numbers of all circulant graphs Cr[A] for all r and for
any subset S of the set {1, 2, 3,, 9}. The algorithms developed in Chapters
3 and 4 can be modified to solve many types of domination problems. To illus-
trate this, we solve the knight domination problem for rectangular chessboards.
Definition 1.4 A Knight domination of a kxr board is a placement of Knights
3


so each square either has a Knight on it or attacking it. The k x r Knight
domination number is the minimum, number of Knights in a Knight domination
of a k x r board.
The knight domination problem is a variation of the domination problem and
wide interest in this problem was first generated by Gardner [9] for square
chessboards. Hare and Hedetniemi [11] found the domination numbers for
kxr chessboards with k < 6 and some values of r. We are able to verify these
results and find the knight domination number for all kxr chessboards with
k <7 and for all r.
The defining feature of the applications in Chapters 3-5 is that the dy-
namic programming algorithms are able to find infinitely many solutions in
a finite amount of time. This is possible because the algorithms are periodic
after a finite number of iterations. The periodicity of the algorithms can be de-
scribed using the eigenvalues of a state transition matrix. In Chapter 6, we show
that periodicity is dependent on the size and the entries in the state transition
matrix. First, we determine the necessary conditions on the matrix for period-
icity to occur. Then for special cases, we find exact bounds on the number of
iterations of the algorithm that are needed before periodicity occurs. Next we
look at random matrices with entries independently chosen from discrete and
continuous distributions and find the expected number of iterations needed for
4


periodicity in each case. The results of this analysis shows why the algorithms
are successful in solving the graph domination problems in Chapters 3-5 and
what conditions are necessary to solve other types of problems with similar
algorithms.
5


2. Properties of Min-Plus
In this chapter we will define the Min-Plus algebra and look at several of
its properties. The majority of this material is derived from Baccelli et al. [1]
Min-plus algebra is defined by two operators H and ffl on the real numbers.
The symbol H represents standard addition and the symbol ffl represents min-
imization. An identity e for a satisfies min(a, e) = a ffl e = a for all a. Since
min(a, e) = a only if e > a, we must have e > a for all a. So to have an identity
for a in an algebra which includes all real numbers, we must append oo to
the real numbers. Let 1i'. = 1i' U {oo}. We must then define the operations for
oo: let oo ffl a = a ffl oo = a and oo H a = a Kl oo = oo for all a e Then
these properties hold for all a, b, c E 91+.
(Commutative) a ffl b = b ffl a (i.e., min (a, b) = min (b, a)),
and a H 6 = bma (i.e., a + b = b + a);
(Associative) am (b me) = (a mb) me
(i.e.,min(a, min(£>, c)) = min(min(a, b), c)),
and a H (b H c) = (a H b) H c, (i.e., a + (b + c) = (a + b) + c);
(Distributive) a H (b ffl c) = (a Kl b) ffl (a H c)
(i.e., a + min(£>, c) = min(a + b, a + c));
6


(Identity) oo ffl a = a (i.e., min(oo, a) = a),
and 0 H a = a (i.e., 0 + a = a).
(Inverses under ki ) (a) H a = 0 (i.e., a + a = 0).
This would be a field if there were inverses under a But a ffl x = oo (i.e.
min(a, x) = oo) has no solution for x unless a = oo.
2.1 Min-Plus Linear Algebra
Several of the applications presented in Chapters 3-5 use linear algebra
operations within the Min-Plus algebra. Using the properties on real numbers
defined above we can define analogues to standard matrix algebra concepts.
Matrix addition is defined as componentwise minimization. For example:
3 2 ffl "l5" 3 ffl 1 2 ffl 5 ' "l2"
2 4 3 2 2 ffl 3 4 ffl 2 2 2
Matrix multiplication is denoted by
(A h B)ij =
k=\Aifc Bfcj,
3 2 "15" (3 B 1) ffl (2 a 3) (3 s 5) ffl (2 a 2) '4 4'
2 4 LAJ 3 2 (2 b 1) ffl (4 b 3) (2 b 5) ffl (4 b 2) 3 6
The zero matrix (A ffl oo = oo m A = A) is:
00 00 00 . . 00
00 00 00 . . 00
Z = 00 = 00 00 00 . . 00
00 00 00 . . 00
7


The identity matrix (. 1 m I = I m A = J .) is:
0 oo oo .. oo
oo 0 oo .. oo
I = oo oo 0 .. oo
oo oo oo .. 0
Matrix Exponents. Let A be a square matrix. Let A1 = A, and recursively
define Ar = Am .T 1 for all k > 1. Associativity ensures that Ar =
Ar^'l m A1 for all / = 1.2..../ 1.
Scalar multiplication. The i.j entry of km A is (k m A)itj = km b.(.
Trace. For a square matrix A, let traee(A) be the minimum of the diagonal
entries of A.
The properties above show that matrix multiplication is associative. Since
there is no inverse for m matrix inverses are not defined except in special cases.
There are also analogues to solving linear systems, eigenvectors, eigenvalues,
characteristic equation, and even a Cayley-Hamilton theorem. We describe
only those properties that are used in the applications presented.
2.2 Eigenvalues and Eigenvectors: The Power Method
Several applications of Min-Plus algebra use dynamic programming algo-
rithms. The defining property of these algorithms is that they use periodicity
to find infinitely many solutions in finite time. Periodicity of matrices in Min-
Plus algebra is defined in terms of eigenvalues and eigenvectors. In this section
8


we will look at the definition of eigenvalues and eigenvectors, and the Power
Method of finding them. To illustrate this method, consider the following
example.
A country has 4 cities with overnight trains between them. The overnight
rates can be represented as a matrix A where is the cost to start in
city i and take the overnight train to city j. Below is the matrix A with
the costs to stay in each city and for overnight train rides between them.
Find the cheapest way to stay n nights starting at city i and ending at
city j.
" 8 0 2 1 '
_ 7 3 3 6
4 5 5 6
_ 3 3 7 9 _
For the cost matrix A, (A2)^ represents the minimum cost to start at city i ,
travel for 2 nights, and end in city j. Likewise, (Ar)y represents the minimum
cost for a trip of r nights. Next, look at various matrix powers of the cost
matrix A.
1 00 0 2 1 ' l 3 3 6 ' l -41 4 6 1
7 3 3 6 42 - 7 6 6 8 43 _ 10 7 9 8
4 5 5 6 5 ^ 9 4 6 5 5 ^ 8 7 7 10
1 CO 3 7 9 _ _ 10 3 5 4 _ 7 6 6 9
OO 7 7 OO
11 10 10 11
11 OO 10 9
10 7 9 OO
11 OO 10 9
14 11 13 12
12 11 11 12
11 10 10 11
12 11 11 12
15 14 14 15
15 12 14 13
14 11 13 12
9


Note that A6 = 4 h A4 (under scalar addition). Then
7 = A6 h A = (4 h A4) h i 1 = 4 h (A4 h A i) = 4h A5
8 = A7 h A = (4 h A5) h i 1 = 4 h (A5 h A l) = 4 m A6
9 = A8 h A = (4 h A6) h i 1 = 4 h (A6 h A l) = 4 m A7
By induction, we can then show that for all r > 6, we have Ar = 4 Ar 2 and
hence
{2(r 4) m A4 if r is even
2(r 5) H A5 if r is odd.
So for example, the minimum cost of start in city 3 and ending in city 2 while
staying 13 nights is 2(13 5) + 11 = 27. We will now show how this relates to
the eigenvalues and eigenvectors of A.
Definition 2.1 Given an m x m matrix A in !i'., let A be an eigenvalue and
ex be the associated eigenvector iJ4ix = Aix. For example, let
A
4 3 4
2 4 3
1 3 3
Then A = | is an eigenvalue with associated eigenvector x = (|, |,0)T, because
1 CO 1 (A- \ 3 (\ 3 7 ( A~ \ 3
H x = 2 4 3 2 3 = 3 = 3B 2 3
l CO CO 1 \o) V I ) U )
This leads us to the Power Method for finding the eigenvalues of a matrix. In
our example we found that for large enough r, we have Ar = 2h Ar~l. Now,
10


let x0 be a vector and define xr = ig xr_! for r > 0, then
xr = Ar h xq = 2 h Ar 1 h Xo = 2 h xri.
In general, suppose that there is a k, p > 0, and q where
Xfe = q H X/, p.
Let X = q/p and let
X X^ X EB (A H X^;_2) EB (A E3 X^;_3^ ffl * EH (AP H X^p^.
Then
A H X* ---- Ah (xj;_l ffl (A H Xjj_ 2) ffl (^2 Xjj_ 3) EB * EB (A^ ^ H Xjj_p))
= Xfe ffl (A h xfe_i) a (A2 h xfe_2) ffl ffl (Ap 1 h xfe_p+i))
--- (AP H X^;_p EB (A H X^;_x) EB (A H X^;_2) EB * EB (AP H X^p-J-^))
= A 0 (xfc_i ffl (A 0 Xfe_2) EB EB ((p 2)A H Xfe_p_(_i) EB (Ap 1 H Xfe_p))
= A h x*.
So A is an eigenvalue with associated eigenvector x*. In our example, we had
1 00 7 7 i 00 " 11 00 10 1 CJi
11 10 10 11 45 - 14 11 13 12
11 00 10 9 J ^ 12 11 11 12
1 o 7 9 i 00 _ 11 10 10 11 _
and Ar = 4 h Ar-2 for all r > 6.
11


Thus the eigenvalue isA = 4/2 = 2 with associated eigenvector
x = (2 (A4)i) (A5) 1
( 2 + 8 1 { 11 1 ( 10 \
2 + 11 14 13
=
2 + 11 12 12
V 2 + 10 ) l11 J l11 J
Then
1 11 CN O 00 1 ( 10 ) f 12 \ ( 10 \
7 3 3 6 13 15 13
= = 2 x
4 5 5 6 12 14 12
_ 3 3 7 9 _ l 11 J V 13 J l 11 J
In the next chapters, we will look at several application of Min-Plus al-
gebra. In each application, the solution relies on periodicity of powers of a
matrix A. In our example Ar = 2 m Ar-2 for all r > 6, in general we have
Ar = qm Ar^p for all r > R0. The existence of i?0 depends on properties of
the matrix A.
Definition 2.2 A matrix is irreducible if there is some K so that for all k > K
the matrix Ak has no infinity entries.
This is analogous to the definition of irreducible matrices in regular algebra,
Perkins [17] proves that for an irreducible matrix K < m2 m + 1 where
m is the size of the matrix. This bound holds in the Min-Plus definition of
irreducible. The following theorem will be refered to throughout the thesis and
we will prove this theorem in Chapter 6.
12


Theorem 2.3 If A is an irreducible matrix on m vertices, then there exists a
p, q and Rq so that
Ar = qm Ar-p for all r > i?0
This shows that the power method for finding when A repeats, which gives
the eigenvalue of the matrix, will work if A is irreducible. In our example, the
cost matrix A is irreducible and thus we know that if we compute the sequence
A, A2, A3,.. ,AK there will be some K so that two members of this sequence
differ by a constant.
2.3 Minimum Cycle Means and Karps Theorem
In the previous section, we showed that if A is an irreducible matrix, then
there is some p, q so that Ar = qm Ar~p. We will now look use the properties
of the entries in A to find the actual values of p and q.
Definition 2.4 The precedence digraph of A denoted DA, is the weighted di-
graph with nodes v\, v^, , vm with an arc from, Vj to tq if Atj ^ oo, and the
weight of this arc is Aij. The precedence digraph is irreducible if the matrix
A is irreducible which means that for some K, there is a path of any length
k > K between every pair of vertices in the precedence digraph.
The precedence digraph for our example is shown below.
13



3
81
5
19
Definition 2.5 The length of a path between two vertices is the number of arcs
in the path, and is denoted \P\. A cycle is a path that begins and ends with the
same vertex but has no other repea,ted vertices. The weight of a path is the sum,
of the weights of its arcs and is denoted w(P).
Definition 2.6 The precedence digraph of an adjacency matrix A is strongly
connected if there is a path from, every vertex vj to every other vertex ty. The
adjacency matrix A is strongly connected if its precedence digraph is. Any
irreducible matrix is also strongly connected.
The periodic property of a matrix is determined by paths in the precedence
digraph. We will now look at the minimum weight paths in the precedence
digraph.
Theorem 2.7 Let, A be an m x m matrix with precedence digraph Da- Then,
the minimum, weight of a path from, vertex j to vertex i, in Da of length k > 0
is (. !%.
14


Proof: Since Aitj is the weight of an arc from j to i in Da if it exists and oo
otherwise, the result holds for k = 1. For k > 1, assume for induction purposes
that (Ak 1)ij- is minimum weight path from j to i of length k 1. Then
(Ak)

= min
+ Aij,(A )i,2 + A2j, ..., (A )
This is the minimum of all paths of length k: 1 from h to % plus the minimum
weight path of length 1 from j to h. This gives the minimum weight of a path
of length k from j to i.
Let
and
A+ = Am A2 a A3... m An 1 m An m An+1...
A* = I m Am A2 m A3... m An 1 m An m An+1_____
Then (A+)y is the minimum weight path of any length from j to i, and {A*)^ is
the minimum weight path of any length from j to i allowing for the possibility
of staying on a vertex. If A is irreducible, then A* and A+ have no infinity
entries.
In our example, the weight of a path of length r represents the cost of
starting in city j, traveling for r nights and ending in city i. The eigenvalue
shows us that the cost to stay r nights is $2 more than the cost to stay r 2
nights, but this does not show us how to travel at this cost. If we look at the
15


precedence digraph, we see that there is a cycle from city A to city D and back
to city A which has weight 4 and length 2. Thus we must use this cycle to stay
2 more nights and only pay $4. This leads us to the next method for finding
the eigenvalues of a matrix by looking at the cycles of the matrix.
Definition 2.8 The cycle mean of a cycle is the sum of the weights of the arcs
divided by the number of arcs in the cycle. For a cycle c the cycle mean is
denoted p(c) = The minimum cycle mean of a graph is the smallest value
for p(c) over all cycles in the graph. All cycles with cycle mean equal to the
minimum, cycle mean are called critical cycles.
Theorem 2.9 If Da is strongly connected, there exists exactly one eigenvalue
A which solves the equation A a x = A a x for some x ^ oo. This eigenvalue
is equal to the minim,um cycle mean of the graph.
A = min p(c)
. w(c)
mm
c c
'where c ranges over all cycles in the graph.
Proof: Existence of x and A. Consider matrix B = AA, where
A = mincp(c) = minc and let DB be the precedence digraph of B.
The minimum cycle weight of DB is 0, thus the matrices B* and B+ = BB*
exist and B+ has some columns with diagonal entries equal to 0. Suppose a
vertex k is in some minimum cycle of B, then the minimum weight of paths
from A; to A; is 0. Therefore, we have 0 = (B+)kk. Let Bk denote the k~ column
16


of B. Then, since B = XA, B = BB*, and B* = I m B for a given k we
have,
(B+)k = (B%
=> (BB*)k = (B+)k = (B*)k
=* -XA(B*)k = (B*)k
=* A(B*)k = A(B*)k.
Hence x = (B*)k is an eigenvector of A corresponding to the eigenvalue A.
Uniqueness of A. Let A be an strongly connected matrix with eigenvalues
Ai < A2 with associated eigenvectors Vi and v2. Let 1 be the vector of all ones
and pick t large enough so that
/I V] > v2 =>- t V] > v2.
Then
(t H vi) ffl v2 = t H vi => Ar((t H vi) ffl v2) = Ar(t Kl vi)
=>- tArvi a .Tv2 = tArvi
{t H AQvi a A2v2 = (t H Ai)vi.
=> min((f + rAi)l + vi,rA2l + v2) = (t + rAi)l + vi.
If Ai < A2, then rA2l + v2 ^ (t + rAi)l + vx for some r. Thus Ai = A2.
Since any irreducible matrix is strongly connected, we also have that the min-
imum cycle mean of a irreducible matrix is its eigenvalue.
17


Next we look at one method of finding the minimum cycle mean using the
following theorem which is analogous to a theorem by Karp which finds the
maximum cycle mean but not the critical cycles.
Theorem 2.10 Let A be an m x m matrix with corresponding precedence di-
graph Da- Then, for any j the minimum, cycle mean p{A) is
where Am, Ak are computed in Min-Plus algebra and the other computations
are conventional.
Proof: Without loss of generality, assume that DA is strongly connected.
If Da is not strongly connected, then the theorem holds componentwise. Also,
the index j is arbitrary and the computation of p{A) is independent of j.
First assume that the minimum cycle mean is 0. Then we must show
Since p{A) = 0, there exists a cycle of weight 0 and there are no cycles with
negative weight. Because there are no cycles with negative weight, there exists
a minimum weight path of length k from vertex i to vertex j. When k > m the
path would contain a cycle of non-negative weight. Thus a minimum weight
path can be found by restricting k < m. This path is defined as
p(A)
= mm max
i=l...m k=l...m
(Amh -
m k
mm max
i=l...m k=0...ml
(Amh -
m k
k=Q...m1
min (A%.
18


Also, (Am)ij > (A+)ij, and hence
(1"% (.4+)
[(-4'),:, (.4%] > 0.
Equivalently,
max
k=Q...ml
(A.m)u (-4%
m k
Equality holds only if (im)q = (A+)y.
Now we will show that there is an index for i where this is true. Let C be
a cycle of weight 0 and let l be a vertex of C. Let Py be a path from l to j with
minimum weight w(Pij) = (A+)y. Now this path is extended by appending to
it a number of repetitions of C such that the total length of this extended path
, denoted Pe, is greater than or equal to m. This is again a path of minimum
weight from j to l. Now consider the path consisting of the first m arcs of Pe-
lts initial vertex is j and denote its final vertex i. Note that i is in C. Since
any subpath of a minimum weight path is also of minimum weight, the path
from j to i is of minimum weight. Therefore (Am)ji = (A+)ji and
This completes the case where p(A) = 0.
Now consider an arbitrary finite p(A). A constant c is now subtracted
from each weight in Aij. Then clearly p{A) will be changed c units and each
mm max
i=l...m k=0...ml
(-4"% (-4fr)i>
m k
19


entry in Ak will be changed by kc units. Thus
(."') (.I"),,
mm max ----------------.
i=l...n k=0...ml 771 k
is changed by c and both sides of the equation are affected equally when all
weights Aij are changed by the same amount. Now choose c so that p(A) c = 0
and we are back to the original case.
Like the power method, Karps theorem gives the minimum cycle mean but
not the critical cycles. To find the critical cycles, backtracking must be used
in both of these methods. There are many applications of Min-Plus algebra in
which the solution is found using eigenvalues. In practice, the power method is
a faster way to find the eigenvalues of an irreducible matrix, although this is not
true for all matrices as we will see in Chapter 6. In the next three Chapters we
present applications of Min-Plus algebra which involve finding the domination
numbers of certain graphs.
20


3. Domination of Cartesian Product Graphs
In this chapter, we will use two dynamic programming algorithms in Min-
Plus algebra to find the domination numbers of a large class of graphs.
Definition 3.1 Let G = (V, E) be a graph with vertices {-17, t>2,... vn} and let
H = (V, E) be a graph with vertices {w\,W2 we}. Then the cartesian product
G x H has vertices of the form, (Vi,Wj), with 1 < i < n and 1 < j < l. The
edges of G x Ed have the form ((vi,Wj), (vp,wq)) if i = p and (iVj,wq) is and
edge of H, or if j = q and (vi ,vp) is an edge of G. The k^' copy of G in G x H
is Gk and its vertices are (vi,Wk), {v2,Wk)> ivn>wk)-
To introduce the algorithms, we consider complete grid graphs which are
cartesian product graphs with II = If. the path on r vertices, and G = Pn, the
path on n vertices. The domination number of Pn x Pr is denoted (P x Pr).
Figure 3.1 Minimum Dominations of x Pn. Above is a domination
of P13 x P13 using 40 vertices. Since Theorem 3.12 shows that 713,13 = 40,
no domination has fewer vertices.
21


Considerable work has been directed toward finding 7(Pn x Pr). Jacobson and
Kinch [13] showed:
77
r + 1
3r + 1
4
and 74;,
72,r g 73,r
r + 1 if r = 1, 2, 3, 5, 6, 9
r otherwise.
(1)
75,i
and 76,1
(2)
Chang and Clark [3] showed that
[" 6rzH"| 1 if r = 2,3,7
I" j otherwise
+1 if r mod 7 = 3 and r ^ 3
pr7+4j otherwise.
Clark, Colbourn and Johnson [6] showed that finding the domination num-
ber of a grid graph (a subgraph of Pn x Pr) is XT-hard. All known algorithms
for finding 7(Pn x Pr) could be adapted to find the domination number of a
grid graph. So it is not surprising that the time needed for these algorithms is
not polynomial in nr. Nevertheless, profound differences exist in the speed of
these algorithms.
Perhaps the most natural non-trivial algorithms for finding 7(Pn x Pr)
are Branch and Bound algorithms. Before trying a different approach, Hare,
Hedetniemi, and Hare [11] used a branch and bound algorithm that took 20
hours to find j(P7 x P7) = 12. Ma and Lam [15] gave a more complicated
branch and bound algorithm which they used to show that 7(P8 x T8) = 16,
22


' (/ x x P9) > 17, and 7(P9 x P9) > 19. However since the time needed for
these algorithms seem to be exponential in both n and r, they are limited to
modest values of n and r.
Much superior are dynamic programming algorithms for finding 7(Pn x Pr)
introduced by Hare, Hedetniemi and Hare [11]. While these algorithms are
exponential in n, they are linear in r. In section 3.1, we present this algorithm
of Hare, Hedetniemi and Hare in which they found 7(Pn x Pr) for n < 8 and
r < 500, and then used these results to make conjectures for 7(Pn x Pr) for
all values of r. Fisher (private communications) expanded on this algorithm
by stating it in terms of Min-Plus algebra and by looking for periodicity in
the algorithm. When periodicity is detected, induction gives 7(Pn x Pr) for
a given n and for all r. Thus 7(Pn x Pr) is calculated for an infinite number
of values in a finite amount of time. Using this expanded algorithm, which
we will call the Exact Algorithm, Fisher was able to verify the conjectures of
Hare, Hedetniemi and Hare for all values of r. While it does not seem possible
to find an algorithm that is not exponential in n, the Exact Algorithm can
be modified to reduce the base of the exponent. Section 3.2 describes this
modified algorithm which will be called the At Least Algorithm. Using the At
Least Algorithm and periodicity, we find formulas for 7(Pn x Pr) for all n < 19.
23


3.1 The Exact Algorithm
To illustrate this algorithm, we will look at the domination of P3 x Pr.
Let S be a set of vertices of P?J x Pr. Any vertex in this graph is either in S,
adjacent to a vertex in S, or not adjacent to any vertices in S. These vertices
will be labeled 0, 1, and 2 respectively. Now, any column of P> x Pr is a copy
of P3, and its vertices can be labeled either 0,1,or 2. A labeling of P?J which
can occur in a dominating set S of P3 x Pr will be called a state. Since no
state can contain vertices labeled 0 adjacent to vertices labeled 2, there are 17
possible states for P?J. These states are shown below.
Si = 000 s7 = 110 Sl3 = 211
= 001 s8 = 012 Sl4 = 122
S3 = 010 S9 = 111 Sl5 = 212
S4 = 100 sio = 210 Sl6 = 221
S5 = Oil sn = 112 sir = 222
s6 = 101 Sl2 = 121
To find the domination number of P3 x Pr+1 we will add a column onto domi-
nating sets of P?J x Pr. If column r is in state Sj, then it is possible for column
r + 1 to be in state s* if the following conditions hold for all vertices p where
Sj(p)=the p~ vertex of Sj.
If Sj(p) = 0, then Si(p) ^ 2.
If Sj(p) = 2, then Si(p) = 0.
If Si(p) = 1, then Sj(p) = 2 if some vertex adjacent to p in sj has value 0,
and Sj(p) = 0 otherwise.
24


Using these rules, we form the state transition matrix A where = the number
of vertices with value 0 in state s* if s* can follow Sj and infinity (denoted by
) otherwise. This matrix is shown below.
3 3 3 3 3 3 3 3 3 3 3 3 3
2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2
1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1
I o I 1 1 1 1 1
1 1 1 1 1

3 3 3
- 2 -
2 -
Next, we form a states vector xr where xr(sj) is the minimum number of
vertices needed to dominate P?J x Pr and have the r column be in state s*. If
th
it is not possible to dominate P?J x Pr and have the r~ column in state s* then
xr(Sj) = oo. Thus
x[ = 3, 2 2 2 1 1 1
0
The infinity () entries in X! correspond to states in which Si(p) = 1 and
which must be preceded by a state with Sj(p) = 0 in column 0, which does not
25


exist. To find x2(sj) we consider all labelings of the first column of P3 x P2
when the second column is in state s*. These labelings correspond to the
states Sj for which Atj ^ oo. From all these possible labelings of the hrst
column, the one with the minimum number of vertices with value 0 will give
a minimal domination of P3 x P2 with column 2 in state s*. Thus x2(sj) =
min(XIj Ay + 'Xi(sj)). Stating this in terms of Min-Plus Algebra, x2 = 1b X!
and in general xr. i = A Kl xr. Repeating this process we have a recursive
algorithm to find xr where xr(sj) is the number of vertices needed to dominate
th
P3 x Pr and leave the r~ column in state s*. To find the domination number
th
of P3 x Pr we find the minimum entry in xr for which the state leaves the r~
column dominated. This is done by forming an ending vector y where y (i) = 0
if state Si leaves the r~ column dominated (these are all the states with no
vertex labeled 2) and y(i) = oo otherwise. The domination number for P3 x Pr
is xr h yT. The ending vector y and the states vectors X! through xn for
P:> x P are shown below.
26


y = [ 0 0 0 0 0 0 0 - 0 - ]
Xl = 0i [ 3 2 2 2 - 1 - 1 - 1 - - - - - - 0 ]
x2 = 0i [ 3 3 3 3 2 2 2 2 3 2 2 2 2 - 1 - - ]
x3 = 2 m [ 2 2 1 2 1 1 1 1 1 1 1 1 1 0 0 0 1 ]
x4 = 3B [ 2 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 ]
x5 = 4i [ 2 1 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0 ]
x6 = 4i [ 3 2 2 2 2 1 2 1 2 1 1 1 1 1 0 1 1 ]
X7 = 5 IM [ 2 2 1 2 1 1 1 1 2 1 1 1 1 1 0 1 1 ]
x8 = 6B [ 2 2 1 2 1 1 1 1 1 1 1 0 1 0 0 0 1 ]
X9 = 7IM [ 2 1 1 1 1 0 1 1 1 1 1 0 1 0 0 0 0 ]
X10 = 7IM [ 3 2 2 2 2 1 2 1 2 1 1 1 1 1 0 1 1 ]
Xu = 8B [ 2 2 1 2 1 1 1 1 2 1 1 1 1 1 0 1 1 ]
Notice that xUj = 3 H x6 and xn = 3 B X7.
Theorem 3.2 xr+4 = 3 m xr for all r > 6.
Proof: By Induction. The base case is x10 = 3 xt; or lx,, = 3ix6.
Then for any r > 6 assume A h xr+2 = 3 B xr_4 Then lxr. -_,x = xr_ix and
Xr+4 = Xr
Theorem 3.3 7(P3 x Pr+4) = 3 + 7(P3 x Pr) for all r > 6
Proof:
7(P3 x Pr+i) = yT m xr+4
= yT H (3 Kl xr)
= 3 Kl (yT Kl xr)
= 3 + 7(P3 x Pr)

Using the recursive dehnition above, we can find a closed form for 7(P3 x Pr)
27


for all r > 6.
' (/'! x Pr
"3 r + 1"
4
Finding the domination number of P?J x Pr for all r relies on the periodicity of
the dynamic programming algorithm. According to Theorem 2.3, periodicity
will occur if the precedence digraph of the state transition matrix is irreducible.
The possible states for Pk x Pr can be partitioned into three groups. Let A be
the set containing the state with all vertices labeled 0 =00... 0, B be the set
containing all states with vertices labeled 0 or 1 but not 2, and let C be all
remaining states which have at least one vertex labeled 2.
Theorem 3.4 The precedence digraph of the state transition matrix for finding
the domination number of Pk x Pr is irreducible.
Proof: Using the rules of state transition all states can be followed by the
states of all zeros. All states in set B can follow the zero state and for every
state in set C there is at least one state in set B that it can follow. Thus there
is a path of length 3 or less from any state to any other state in the precedence
digraph by going through the zero state. This path can then be extended by
repeating the zero state. Thus there is a path of any length k > 3 between any
pair of vertices and the precedence digraph is irreducible.
For any n the exact algorithm can be used to find the domination number
28


of Pn x Pr. The algorithm will reach periodicity after some number of itera-
tions allowing us to find a closed form for the domination number of Pn x Pr
for all r. The limiting feature of all dynamic programming algorithms is com-
putation time. For the exact algorithm on Pn x Pr, there are 0(( 1 + \/2)n)
states each of which matches to at most 2" other states and each match takes
0(n) computations to find. Thus the computational complexity for finding the
domination number of Pn x Pr is 0(nr(2 + 2\/2)n), and the maximum size of
n is greatly limited by computational complexity. In the next section, we will
look at another dynamic programming algorithm which greatly reduces this
complexity. Although this algorithm is similar to the exact algorithm, it is not
a direct descendant.
3.2 The At Least Algorithm
In this algorithm, we will reduce the computational complexity by redefin-
ing what it is that we are computing. Fisher (private communications) used
this new algorithm to find (P x Pr) for n < 19 and for all r. These results are
presented below. In section 3.4 we will expand this algorithm to find j(G x Pr)
for any graph G on 6 or less vertices.
Definition 3.5 A column of Pn x Pr is in at least state st if it is in state st
or any state Sj which has Si(p) > Sj(p) for all p.
29


This means that Sj dominates at least the same set of vertices that s* does and
possibly more.
Definition 3.6 A t-domination of Pn x Pr is a set of vertices that dominates
the first r 1 columns of Pn x Pr and leaves the last column in at least state
Si-
To illustrate the algorithm, we will find the least-domination number of P?JxPr.
As in the exact algorithm, there are seventeen states. These states form a
partial ordering which is illustrated in Figure 3.2.
000
Figure 3.2 The partial ordering of the 17 states for P?J x Pr
State -s-i is at least state Sj if state Sj is an ancestor of s* in the partial
ordering. Next, we form the states vector tr where tr(sj) is the number of
vertices in a dominating set for P3 x Pr with column r in at least state s*.
Since a state is in at least s* if it is in exactly s* or exactly Sj for all Sj > Si, the
entries in tr can be defined using the states vector xr with xr(si) representing
30


the number of vertices in a domination of P3 x Pr with column r in exactly s*
as described in the Exact Algorithm. For example
t r(sr) = min(xr(s7), xr(s4), xr(s3), xr(si))
tr(sio) = min(xr(sio), xr(s7), xr(s4), xr(s3), xr(si))
Now the last 4 entries for tr(si0) are the same as tr(s7) so we have
tr(sio) =min(xr(sio)>tr(sr))
Next, if the r~ column is in exactly state s* then the entry in xr(sj) represents
the number of vertices needed to dominate P3 x Pr_i and leave the (r ^ l)st col-
umn in the minimum previous state that can precede state s* plus the number
of vertices that are labeled 0 in s*.
Definition 3.7 Given a state si; the minimum previous state M(si) is defined
as
f 2 ifsi(p) = 0
M(si(p)) = < 0 if Si(p) = 1 and no neighbors of p in Sj = 0
[ 1 otherwise
The minimum previous state for si0 = 210 is sn = 112 and sio has one vertex
in the dominating set, so xr(si0) = 1 + tr_i(sn) and,
tr(sio) = min(l + tr_i(sn), tr(s7)).
In general,
tr(i) = min(|sj| + tr_i(M(si)), min tr(sj))
31


Completing a similar computation for each state, we form two matrices. The
at least matrix A has = 0 if state s* is at least state Sj and there is no
state Sfc where s* is at least and is at least Sj, and = oo otherwise.
This matrix is lower triangular because of the partial ordering of the states.
The second matrix is the minimum previous matrix B with B^ = the number of
vertices in the dominating set for state s* if state Sj is the minimum previous
state of Si and oo otherwise. The matrices A and B for the at least domination
of /'. x P, are shown below.
A =
0 _ _ _ _ _ _
0 _ _ _ _ _ _
o _ _ _ _ _ _
-00----
- 0 - 0 - - -
- - - - 0 - -
------ 0
0 0
- 0
- 0
-00-----
-0 0
--00----
----000-
32


2
3
B =
- - 2 - -
- 2 - - -
1 - -
Now, in terms of min plus algebra,
t/- .it/- I /^t/-i >
To find tr we iteratively multiply by B and A one row at a time, updating tr
after each new entry is found. This is possible since A is lower triangular. To
find the domination number of P3 x Pr we need the minimum of the entries
in tr which correspond to states that dominate the last column. All states
containing no vertices labeled 2 dominate the last column, and the state of
all ls denoted 1 is at least all of these states so the ending vector y has a 0
entry for the 1 state and infinity entries for all other states. For P?J x Pr 1 is
state s9, so 7(P3 x Pr) = tr(s9) h yT and in general 7(Pk x Pr) = tr(l H yT)
Now computing tr iteratively, we find the states vectors for the t-domination
of P3 x Pr, shown below.
33


0
t1= OKI [32222121111111110]
t2= OK [33332222222222121]
t3= 2 K [22121111111110000]
t4= 3 K [21111111111010000]
t5= 4 K [21111011010000000]
t6= 4 K [32222121111111010]
t7= 5 K [22121111111111010]
t8= 6 K [22121111111010000]
t9 = 7 K [21111011010000000]
tio = 7 K [32222121111111010]
tn = 8 K [22121111111111010]
Notice that tg = 3 K and tio = 3 K tt; and thus by induction tr =
3 K tr_4 for all r > 5 and we have the domination number for inhnite r after
9 iterations of the algorithm. We now need to show that periodicity in the At
Least Algorithm will always occur.
Lemma 3.8 If Si is a direct ancestor of Sj in the partial ordering (so Si > Sj
and there is no Sk with Si > Sk > Sj) then tr(sj) tr(s_i) is either 0 or 1.
Proof: Si and Sj have at most one vertex with a different label. If Si(p) = 0
and Sj(p) ^ 0, then any domination ending in state Sj can be changed into a
domination ending in Si by changing Sj(p) to 0 and this increases tr(s_i) by 1.
Otherwise tr(sj) = tr(sj).
Definition 3.9 A state Sj is maximal for a given r iftr(sj) < tr(sj) whenever
Si is a direct ancestor of Sj in the partial ordering. Let Rrhr be the set of all
34


maximal states for Pn x Pr
Using the states vectors ti... tn shown above, we have
#3,1 = {Si; s2, s3; S4; s6; S3; S9, S10, S17}
#3,2 = {Si; s5? s6; S14}
#3,3 = {Si; S3; 6; Sl4 ; S15 ; Sl6}
#3,4 = {Si; *2, 3; S4; Sl2; Sl5}
#3,5 = {Si; s2, S3; S4; s6}
#3,9 = {Si; S 2, S3; S4; S6}
Lemma 3.10 There exists positive integers j,k,n so that tk{si) tj(si) = q
for all states s* if and only if Rnj = Rrhu
Proof: If tfc(sj) tj{si) = q then certainly the maximal states for j and k
must be the same and Rnj = Rn,k- Next, for all s*, Sj E Rn,k Lemma 3.8 states
that tfc(sj) = tfc(sj) 1 for all Sj > s*. Let 2 be the state with all vertices labeled
2, then for any st ^ 2 there is a sequence of states 2 = < s2 < sh < Sj.
Let N(sj) be the states in this sequence, N(sj) = {si, s2, sh }. Now the value
of tfe increases by one each time a maximal state occurs in the sequence and
the number of maximal states is N(sj) f] Rn,k- Thus
tjfc(sj) = tfe(2) + |iV(sj) Pl^fcl,
and the domination number of any state can be constructed using this sequence.
35


Now if Rn,j = Rn,k then let q = tfe(2) tj(2) and for all states s* we have
tjfc(sj) = tfe(2) + |iV(si)f|i?njfe|
= tfe(2) + |At(sj) P| Rn,j |
= tfe(2) + tj(si) tj(2)
= q + tj(si)

Lemma 3.11 Let Cm be the number of states, then there exists 0 < j < k <
2Cm + 1 where RnJ =
Proof: The entries in Rrtjk are a subset of the states so there are at most
2Cm possibilities for Rn.k- The pigeonhole principle ensures that Rnj = Rn.k
for some 0 < j < k < 2Cm + 1.
Now by Lemma 3.10 and Lemma 3.11 periodicity is guaranteed for the
At Least Algorithm. Thus, the At Least Algorithm allows us to compute the
domination number for Pn x Pr for all r using the periodicity of the algorithm
and we have greatly reduced the number of computations. For each state there
are at most n entries in the at least matrix and one entry in the minimum
previous matrix, reducing the computational complexity to 0(nr( 1 + \/2)n).
3.3 Formulas for the Domination Number
A program implementing the At Least algorithm verified (1) and (2) in
about a second. It took about a week on a Vax 8800 to find formulas for
y(Pn x Pr) for all n < 19. Since 7(Pn x Pr) = '(If x Pn), this gives 7(Pn x Pr)
36


for all n and r with min(n,r) < 19.
Theorem 3.12 For all n < r with n <
7(Pn x pr
r
3
Tl
3r+l
r
r + 1
6r+4
5
6r+4
5
10r+4
7
5r+l
3
15r+7
23r+10
11
30r+15
13
30r+15
13
38r+22
15
38r+22
15
80r+38
29
98r+54
33
98r+54
33
35r+2Q
11
35r+20
11
44r+28
13
44r+28
13
(n+2)(r+2)
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
if n
19, we have
1
2
3
4 and r ^ 5, 6, 9
4 and r = 5, 6, 9
5 and r ^ 7
5 and r = 7
6
7
8
9
10 and r mod 13 ^ 10 and r ^ 13,16
10 and r mod 13 = 10 or r = 13,16
11 and II. 18, 20, 22, 33
11 and r = 11,18, 20, 22, 33
12
13 and r mod 33 ^ 13,16,18,19
13 and r mod 33 = 13,16,18,19
14 and r mod 22 ^ 7
14 and r mod 22 = 7
15 and r mod 26 ^ 5
15 and r mod 26 = 5
16,17,18,19.
Even with Theorem 3.12, it is difficult to find the pattern that dominates
the n x r grid for all r using the minimum number of vertices. Jacobson and
Kinch [13] found minimum domination patterns oflxr,2xr,3xr and 4 x r
37


grids. Chang, Clark and Hare [4] use the algorithm of Hare, Hedetniemi and
Hare to generate minimum domination patterns of 5 x r and 6 x r grids, and
what they conjectured to be minimum dominations of 7 x r, 8 x r, 9 x r and
10 x r grids for all r. Theorem 3.12 shows these dominations are minimum.
They also found formulas for jn,r when r < 122 and for when r < 33 (the
formula for 7njr holds for all r, while the formula for 7i2jr does not).
The program in this paper was modified to store backtracking information
to give minimum dominations of 11 x r, 12xr, 13xr, 14xr, and 15 x r grids
for all r up to a certain point. These were used to find minimum dominations
for n x r grids for a given n and for all r (this is actually quite difficult because
only one of what is usually many minimum dominations is generated). Chang,
Fisher and Hare [5] report these minimum dominations.
For n,r > 8, Chang [2] gives a construction which dominates an n x r grid
with [(n + 2)(r + 2)/5j 4 vertices. This is a generalization of a construction
of Cockayne, Hare, Hedetniemi and Wimer [7] for r x r grids. Theorem 3.12
shows this construction is minimum for 16xr, 17 xr, 18xr, and 19 x r grids
when r > 16. This causes one to suspect the following conjecture (originally
due to Chang [2]) might be true.
Conjecture 3.13 For all n,r > 16,
7(Pn x Pr
(n + 2)(r+ 2)
5
4.
38


If Conjecture 1 is true, dominating an n x r grid would be analogous
to 2-packing (Fisher [8]) in that a simple formula holds for n and r large
enough. Further, it would answer a question of Hedetniemi and Laskar [12],
In commenting on [6], they asked is there a polynomial algorithm for find-
ing 7(Pn x Pr)? Conjecture 3.13 together with Theorem 3.12 would give an
affirmative answer to the question.
3.4 Domination of G x Pr: The At Least Algorithm
We will now use the At Least Algorithm to find the domination number
of G x Pr where G is any graph on n < 6 vertices.
Livingstone and Stout [14] develop a dynamic programming algorithm for
finding 7(G x Pr) using an algorithm that is similar to the Exact Algorithm.
We are able to significantly reduce the complexity of this algorithm by stating
in terms of the At Least Algorithm.
To illustrate the At Least Algorithm for 7(G x Pr) we will let G be the
house graph on five vertices as shown in Figure 3.3.
39


Figure 3.3 The house graph on five vertices G and G x P4.
There are 77 possible states for the domination of the house graph on five
vertices. These states are shown below and the vertices are ordered Vi... v5 as
labeled in Figure 3.3.
Si = = 00000 s2l = = 01221 S41 = 11121 Sgi = 21111
S2 = = 00001 S22 = = 10000 S42 = 11122 s62 = 21112
S3 = = 00010 S23 = = 10001 S43 = 11210 Sg3 = 21121
S4 = = 00011 S24 = = 10010 S44 = 11211 s64 = 21122
SFj - = 00100 S25 = = 10011 s45 = 11212 Sg5 = 21211
s6 - = 00101 S26 = = 10101 s46 = 11221 Sgg = 21212
s7 -- = 00110 s27 = = 10101 S47 = 11222 s67 = 21221
SS 2 = 00111 s28 = = 10110 s48 = 12101 s68 = 2 1 222
s9 = = 00121 s29 = = 10111 s49 = 12111 s69 = 22 1 01
Sio = 01000 S30 = 10121 S50 = 12112 s70 = 22111
Sll = 01001 S31 = 11000 s5i = 12121 sn = 22112
Sl2 = 01010 S32 = 11001 s52 = 12122 s72 = 22121
Sl3 = 01011 S33 = 11010 s':,3 = 12211 S73 = 22122
S14 = 01100 S34 = 11011 s54 = 12212 S74 = 22211
Sl5 = 01101 S35 = 11012 S55 = 12221 S75 = 22212
Slg = oino S36 = 11100 s56 = 12222 s7 g = 22221
sir = 01111 S37 = 11101 S57 = 21001 S77 = 22222
Sl8 = 01121 s38 = 11110 S58 = 21011
Sl9 = 01210 S39 = 11111 s59 = 21012
20 = 01211 S40 = 11112 Sg0 = 21101
These states form a partial ordering were s* follows Sj in the ordering if
Si is at least Sj which means the labelings Sj{p) < Si(p) for the corresponding
40


vertices p in each state. The partial ordering is used to form the At Least
Matrix A and the Minimum Previous Matrix B as described for P x Pr.
Next, we form the states vector tr where tr(sj) is the number of vertices
needed to dominate G x Pr and leave the last copy of G in at least state s* and
the ending vector y which has a 0 for the state 1 and infinity otherwise. For
this example, 1 is state s39. The vectors for the house graph are shown below.
y =
ti = OB
t2 =2B
t3 =3H
tig = 1713
tw =18 3
t20 =19 3
t2i =20 3
t22 = 21 3
t23 = 22 3
[00000000-00000000----00000000-0000-0000---------------------------------------]
[21111110111111110010011111100011111110000000011100010112100001010112011121223]
[11111111111111111111010101000000000000000000001001010112000000010112001121222]
[00000000000000000000100000000000000000001101112001121223001011121222112232323]
[11000000010000000000100000000000000000001101112001121223001011121223112232334]
[11000000010000000000100000000000000000001101112001121223001011121222112232323]
[11000000010000000000100000000000000000001101112001121223001011121222112232323]
[11000000010000000000100000000000000000001101112001121223001011121222112232323]
[11000000010000000000100000000000000000001101112001121223001011121222112232323]
Notice that t20 = 1 El ti9 and t2i = 1 H t20. Thus after 20 iterations,
periodicity occurs and we have tr = 1 H tr_i for all r > 20. This allows us to
find a closed form for the domination number of the house graph G. For all
r > 1, we have
7(G x Pr) = r +
1 if r =1, 2, 3, 4, 6, 7, 8, 11, 12, 16
0 otherwise.
Now to find the domination number of G x Pr, first note that if G is
disconnected then so is G x Pr. Thus, if Gi, G2,... are the components of G,
then the domination number of G is just the sum of the domination numbers of
41


its components. Thus we only need to find the domination number of connected
graphs. To further reduce the number of graphs we need to consider, the
following theorem gives the domination number for certain graphs with high
degree.
Theorem 3.14 If a graph on n vertices G has a set of vertices a,i e A with
deg(a,i) > 3 and for each a,i e A, all vertices not in the closed neighborhood
of a,i are also in A, then 7(G x Pr) = r or r + 1 or r + 2.
Proof: Let ai,a2,... be the vertices in set A as described above. To
dominate (G x Pr) start with the k~ copy of G = Gk where k ^ 1 ,r. Place
a>ik in the dominating set. Now there are at most two undominated vertices
in G both of which are also in A, call these vertices and aik. Place 1
and a,ik+i in the dominating set. Now a^kzti is adjacent to akk by definition
of G x Pr, so all the vertices in Gk are now dominated. Next we have that
in Gk-1, Ojfe-i and the closed neighborhood of a,jk-1 are dominated and only
one vertex aqk-1 remains undominated. Place aqk-2 in the dominating set and
then Gk-1 is dominated and Gk-2 has only one undominated vertex. The same
process is complete for Gfe+2, Gk?> until G\ and Gr are reached. At each step
only one vertex is added to the dominating set and both G\ and Gr have one
undominated vertex so adding these vertices to the set gives a domination of
Gx Pr with r+ 2 vertices. In the diagram below, dark circles represent vertices
42


in the dominating set, all other vertices are dominated by the corresponding
vertex in an adjacent copy of G.
0 Qjki 1 Ujk alk O O V
\l' O O
Gi 1 Gk-l Gk G
If one vertex ap in A has degree n 2 then it may be possible to rearrange
the order in which the vertices are place in the dominating set so that api is is
used when domination (?2- If this is possible, then no additional vertex will be
necessary to dominate G\ and 7(G x Pr) = r + 1. Likewise it may be possible
to order the vertices in the dominating set so that both G\ and Gr use vertices
of degree n 2 and then 7(G x Pr) = r. If two vertices ap and aq of A have
degree n 2 then place apk and aqk+1 in the dominating set for k odd. This
gives a domination number of r. Finally if any vertex of G has degree n 1
place one copy of this vertex in the dominating set for each copy of G and this
gives 7(G x Pr) = r.
Next, for the connected graphs on n < 6 vertices that are not covered
by Theorem 3.14 the domination number is computed using the At Least
Algorithm.
43


3.5 Values of 7(G x Pr) for graphs on < 6 vertices.
Theorem 3.15 For all connected graphs G on n = 1 vert,ices, we have
'(G x Pr)
Theorem 3.16 For all connected graphs G on n = 2 vertices, we have
Theorem 3.17 For all connected graphs G on n = 3 vertices, we have
Theorem 3.18 For all connected graphs G on n = 4 vertices, we have
On five vertices, there are 34 non-isomorphic graphs 13 of which are discon-
nected and their domination number is found by taking the sum of the dom-
ination numbers of the components. There are 21 connected non-isomorphic
graphs on 5 vertices, 20 of which are covered by Theorem 3.14.
Theorem 3.19 For all connected graphs G on n = 5 vert,ices, we have
1 if r = 5,6,9 and G is P\ or r = I and
' (G x Pr) = r +
G = C4
0 otherwise.
If G is the path on 5 vertices, and
' (G x Pr) = r, r + 1, orr + 2 otherwise
44


There are 156 graphs on 6 vertices, 44 of these graphs are disconnected
and their domination number is found by taking the sum of the domination
number of there components which are graphs previously discussed. There are
112 non-isomorphic connected graphs on 6 vertices, 47 of them are covered by
Theorem 3.14. The remaining 65 connected graphs fall into 28 different groups
and the domination numbers of these graphs are given below.
Theorem 3.20 Let G be one of the following graphs:
Then for all r > 1, we have
Theorem 3.21 Let G be one of the follovjing graphs:
Then for all r > 1, we have
Theorem 3.22 Let G be one of the follovjing graphs:
Then for all r > 1, we have
45


Theorem 3.23 Let G be one of the follovjing graphs:
Then for all r >1, we have
'(G x Pr)
TO r"
~7~
1 if r =0, 2, 3, 4, 6 (mod 7) except r = 3
0 otherwise.
Theorem 3.24 Let G be one of the follovjing graphs:
Then for all r > 1, we have
'(G x Pr)
"7 r"
Â¥
1
0
Theorem 3.25 Let G be one
if r =0, 2, 4 (mod 5) except r
u
otherwise.
of the following graphs:
4, 7, 9,
Then for all r > 1, we have
7(G x Pr)
"7 r"
Â¥
1 if r =0, 2, 4, 5, 6, 7, 9 (mod 10)
0 otherwise.
Theorem 3.26 Let G be one of the follovjing graphs:
Then for all r > 1, we have
'(G x Pr)
Tlr"
8
1 if r =0, 2, 4, 5, 7 (mod 8) except r = 4, 7,
or r =6, 9
0 otherwise.
46


Theorem 3.27 Let G be one of the follovjing graphs:
Then for all r >
1, we have
'(G x Pr)
4 r
y
1 if r =0, 2 (mod 3)
0 otherwise.
Theorem 3.28
Let G be one of the follovjing graphs:
Then for all r >
1, we have
'(G x Pr)
4 r
y
2 if r =0 (mod 3) except r = 3
1 if r =1, 2 (mod 3) except r = 1, Jh or r =3
0 otherwise.
Theorem 3.29
Let G be one of the follovjing graphs:
Then for all r > 1, we have
7(G x Pr)
4 r
y
1 if r =0, 2 (mod 3)
0 otherwise.
47


Theorem 3.30 Let G be one of the follovjing graphs:
Then for all r >1, we have
(2 if r =0 (mod 10) except r = 10
1 if r =1, 2, 3, 4, 5, 6, 7, 8, 9 (mod 10)
except r = 1, 4; 7, or r = 10
0 otherwise.
'(G x If
lor
U
Theorem 3.31 Let G be one of the follovjing graphs:
Then for all r > 1, we have
'(G x Pr)
~9r~
y
i
0
Theorem 3.32 Let G be one
if r =7 (mod If)
other wise.
of the following graphs:
Then for all r > 1, we have
'(G x Pr)
T4r"
u
1 if r =2, 3; 7
0 otherwise.
Theorem 3.33 Let G be one of the follovjing graphs:
Then for all r > 1, we have
7(G x Pr)
T4r"
u
1 if r =0, 2, 3, 5, 6, 7, 8, 9, 10 (mod 11)
except n = 5, 8, 9
0 other wise.
48


Theorem 3.34 Let G be one of the following graphs:
Then for all r > 1, vjc have
-5r
7(G x Pr
4
1 if r =0 (mod 4)
0 other wise.
Theorem 3.35 Let G be one of the following graphs:
Theorem 3.36 Let G be one of the following graphs:
Then for all r >1, we have
'5r
7(G x Pr
4
1 if r =4 (mod 8)
0 other wise.
Theorem 3.37 Let G be one of the following graphs:
Then for all r >1, we have
'5r
7(G x If
4
1 if r =0, 2, 3 (mod 4) except r
0 otherwise.
Let G be one of the following graphs:
Them for all r >1, we have
r Sr
7 (C,xP,)=
2 if r =0 (mod 4) except r = 4, &
1 if r =1, 2, 3 (mod 4) except r =
6, 9, or r =4, 8, 12
0 otherwise.
2, 3, 6
12
1, 2, 3, 5,
49


Theorem 3.39 Let G be one of the follovjing graphs:
Then for all r >1, we have
hr
I
1 if r =2, 3, 4, 7
0 otherwise.
l(G x Pr)
Theorem 3.40 Let G be one of the following graphs:
Then 7(G x Pr) = \f]for all r > 1.
Theorem 3.41 Let G be one of the following graphs:
Then for all r > 1, we have
1 if r =4
0 otherwise.
Theorem 3.42 Let G be one of the following graphs:
7(G x Pr
6 r
Â¥
Then for all r >1, we have
l(G x Pr)
6 r
Â¥
1 if r =0, 4 (mod 5) except r = 4
0 otherwise.
50


Theorem 3.43 Let G be one of the following graphs:
Then for all r > 1, we have
'(G x Pr)
6 r
Â¥
1 if r =0, 2, 3, 4 (mod 5) except r = 2, 3
0 otherwise.
Then for all r > 1, we have
l(G x Pr)
Theorem 3.45 Let G be one of the following graphs:
7 r
~6~
1 if r =0, 4, 5 (mod 6) except r = 4, 5, 10
0 otherwise.
Then for all r > 1, we have
'(G x Pr)
~7r~
If
1 if r =0, 2, 3, 4, 5 (mod 6) except r = 2, 3
0 otherwise.
Theorem 3.46 Let G be one of the following graphs:
Then for all r > 1, we have
'(G x Pr)
~8r~
y
1 if r =6
0 otherwise.
51


Theorem 3.47 Let G be one of the follovjing graphs:
Then for all r > 1, vje have
1 if r =0, 2, 3, 4, 5, 6 (mod 7) except r = 2,
3, 5, 9, 10
0 otherwise.
Theorem 3.48 Let G be one of the following graphs:
'(G x If
8 r
y
Then for all r > 1, vje have
'(G x Pr)
TO r"
IT
1 if r =8
0 otherwise.
52


4. Domination of Circulant Graphs
In this chapter we will find the domination number of circulant graphs.
Definition 4.1 Given an integer r > 0 and subset of the positive integer S,
the circulant graph Cr[iS] is a graph on vertices 0, 1, r 1 with vertex i
adjacent to vertex j if i j 6 S or j i e S (mod r).
Let Cr[l, 3] be shorthand for Cr[{l, 3}], where Cr[l, 3] is a graph on vertices 0,
1, ..., r 1 with vertex i adjacent to i 1 and i 3 (mod r). To illustrate
the algorithm we will find 7(Cr[l,3]) which has domination number 4 (see
Figure 4.1). To find the domination number for all r, we formulate the problem
as a dynamic programming algorithm similar to the algorithm used in Hare and
Hedetniemi [10] and Fisher [8] and Chapter 3 to find the domination number
of complete grid graphs. Then, like is done in [8], we iterate the algorithm
until a periodicity is detected. When this happens, we declare that we know
the answer for all r, and stop iterating. This allows us to find Cr[ 1, 3] for all r
in a finite amount of time.
53


Figure 4.1 The circulant graph Ci4[l,3]. The circled vertices are a
minimum domination.
4.1 The States and Transition Matrix
We can think of Cr[l,3] as a sequences of overlapping paths on three
vertices (P3) where adjacent P3s share two vertices and there is an edge be-
tween the nonshared vertices. Connecting together r of these P3s so the r
P3 overlaps the first builds Cr[l,3].
We will build a domination by appending P3s clockwise from the existing
sequence. The vertices of P3 are either in the domination, dominated but not
in the domination, or not yet dominated (and thus need to be dominated by a
vertex in the clockwise direction). The states will be all valid combinations of
these three attributes.
Figure 4.2 shows the 15 valid states. This is less than the 27 ways in which
3 vertices can be assigned 3 attributes. This is because 10 potential states have
a dominated vertex next to an undominated vertex, while two others have the
54


first vertex undominated and the last vertex is dominated and these are not
valid since the last vertex cannot be dominated without dominating the first
vertex.
1=
2=
3=
4=
X X X
Figure 4.2 The hfteen states of P?J. Circled vertices are in the domi-
nation. Dominated vertices have a x on them.
Going clockwise, we will refer to the vertices of a state as being vertex a,
b, and c. When can two states overlap? Let vertices k, k + 1, and k + 2 be in
state j and k + 1, k + 2, and k + 3 be in state i. To be consistent, we must
have the following.
(1) Vertex b of state j must be the same as vertex a of state i.
(2) Vertex c of state j is the same as vertex b of state i unless vertex c
of state i is in the domination. In this case, if vertex c of state j is
undominated, then vertex b of state i is dominated but not in the
domination.
(3) If vertex a of state j is in the domination, then vertex c of state i is
either be in the domination, or dominated but not in the domination.
If vertex a of state j is dominated but not in the domination, then
55


vertex c of state % can either be in the domination, or not dominated.
Finally, if vertex a of state j is not dominated, then vertex c of state %
must be in the domination.
The last condition forces vertex a of state j to be dominated by vertex c in
state i if it has not already been dominated. Thus states that have vertex a
dominated match to two states, while states in which vertex a is not dominated
match to only one state.
We will represent which states match to other states by a transition matrix
A. The i,j entry of A is:
0
A,
hj
oc
if state j matches to state %
and vertex c of state % is not in the domination
if state j matches to state i
and vertex c of state j is in the domination
if state j does not match to state i.
56


Then the transition matrix is:
1 oo oo oo oo 1 oo oo oo oo oo oo oo oo oo
0 oo oo oo oo 0 oo oo oo oo oo oo oo oo oo
oo 1 oo oo oo oo 1 oo oo oo oo oo oo oo oo
oo 0 oo oo oo oo oo oo oo oo oo oo oo oo oo
oo oo oo oo oo oo 0 oo oo oo oo oo oo oo oo
oo oo 1 oo oo oo oo 1 oo oo oo oo 1 oo oo
oo oo 0 oo oo oo oo 0 oo oo oo oo oo oo oo
oo oo oo 1 1 oo oo oo 1 1 oo oo oo 1 oo
oo oo oo 0 oo oo oo oo oo oo oo oo oo oo oo
oo oo oo oo oo oo oo oo 0 oo oo oo oo oo oo
oo oo oo oo 0 oo oo oo oo oo oo oo oo oo oo
oo oo oo oo oo oo oo oo oo 0 oo oo oo oo oo
oo oo oo oo oo oo oo oo oo oo 1 1 oo oo 1
oo oo oo oo oo oo oo oo oo oo 0 oo oo oo oo
oo oo oo oo oo oo oo oo oo oo oo 0 oo oo oo
The information in the matrix can also be represented in the precedence
digraph Da which is shown in Figure 4.3.
Figure 4.3 Precedence Digraph for Circulant graphs. The unlabeled
arcs have weight zero. The arrowed arcs show a cycle whose cycle mean
is minimum.
57


Since the states in a domination must match, a domination of Cr[l,3] corre-
sponds to a path (not necessarily simple) in Da- Further a domination must
be a cycle because the first state and the r state must match to build Cr[l, 3].
So a minimum domination of Cr[ 1, 3] corresponds to a cycle in Da of length r
with minimum weight.
Theorem 4.2 Let A be an 15 x 15 matrix shown above. Then for all r > 0,
we have that
7(Cr[l,3]) = trace(Ar)
where matrix exponentiation and trace are in Min-Plus algebra.
Proof: By Theorem 2.7, (Ar)iji is the minimum weight cycle of length r
from i to i in Da- By the definition of I) \. this is the number of vertices in a
domination starting and ending in state i. Thus trace (Tr) = min(/lr)ii is the
minimum size of a domination of Cr[l, 3].
4.2 Periodicity of the Transition Matrix
Theorem 4.2 gives us a method for finding the domination number of
Cr[l,3] for any r. However, we cannot use it to find 7(Cr[l,3]) for all r. This
depends on the periodic property of Min Plus algebra. In this example, this
property can be shown by looking at A19 and A24.
58


6 6 6 6 6 6 6 6 6 7 6 7 7 7 7
5 5 5 5 5 5 5 5 5 6 5 6 6 6 6
5 5 5 5 6 5 5 5 5 6 5 6 6 6 6
5 5 5 5 5 5 5 5 5 5 5 6 5 5 6
4 4 4 4 5 4 4 4 4 5 4 5 5 5 5
5 5 5 5 5 5 5 5 6 6 6 6 6 6 7
4 4 4 4 4 4 5 4 5 5 5 5 5 5 6
4 4 5 5 5 4 4 5 5 5 5 6 5 5 6
4 5 4 5 5 4 5 4 5 5 5 6 5 5 6
4 4 4 5 5 4 4 4 5 5 5 6 5 5 6
4 4 4 4 4 4 4 4 4 4 5 5 4 4 5
4 4 4 4 4 4 4 4 5 5 5 5 5 5 6
4 5 4 5 5 4 5 4 5 5 5 6 5 5 6
3 4 3 4 4 3 4 3 4 5 4 5 4 5 5
4 4 4 4 4 4 4 4 4 5 4 5 5 5 5
7 7 7 7 7 7 7 7 7 8 7 8 8 8 8
6 6 6 6 6 6 6 6 6 7 6 7 7 7 7
6 6 6 6 7 6 6 6 6 7 6 7 7 7 7
6 6 6 6 6 6 6 6 6 6 6 7 6 6 7
5 5 5 5 6 5 5 5 5 6 5 6 6 6 6
6 6 6 6 6 6 6 6 7 7 7 7 7 7 8
5 5 5 5 5 5 6 5 6 6 6 6 6 6 7
5 5 6 6 6 5 5 6 6 6 6 7 6 6 7
5 6 5 6 6 5 6 5 6 6 6 7 6 6 7
5 5 5 6 6 5 5 5 6 6 6 7 6 6 7
5 5 5 5 5 5 5 5 5 5 6 6 5 5 6
5 5 5 5 5 5 5 5 6 6 6 6 6 6 7
5 6 5 6 6 5 6 5 6 6 6 7 6 6 7
4 5 4 5 5 4 5 4 5 6 5 6 5 6 6
5 5 5 5 5 5 5 5 5 6 5 6 6 6 6
Note every entry in A24 is one more than the corresponding entry in A19. This
can be notated as A24 = 1 0 A19. Then
A25 = Am A24 = A m (1 m A19) = 1 m (A m A19) = 1 m A20.
Continuing in this manner, we get the following theorem.
59


Theorem 4.3 Let A be the 15 x 15 matrix given above. Then for all r > 24,
we have
Ar = 1 m /T-5.
Proof: Since A24 = 1 8 A19, the result holds for r = 24. For n > 24,
assume for induction purposes that Ar-1 = 1b Ar-6. Then
Ar = Am A'-1 =Am{lm Ar-6) = lm(Am Ar-6) = 1 m Ar-5.

Thus, we can use this repetition to find the domination number of Cr[ 1, 3]
for all r. By observing the domination numbers when r < 24, we find an
explicit formula for Cr[ 1, 3].
Theorem 4.4 For all r >1, we have
f[|]+l if n = 4 (mod 5)
7(CV[ 1,3]) = ]
[ [|] otherwise.
Using this method, we let S be any set from the integers {1, 2,... 9} and
we find the domination number of Cr[S'] for all r. There are 512 different
circulant graphs on S and the domination numbers of these graphs is given in
Appendix A.
60


5. Knights Domination of k x r Chessboards
In this chapter we use Min-Plus Algebra to develop a dynamic program-
ming algorithm which solves the Knight domination problem for k x r chess-
boards. The game of Chess is played on an 8 x 8 board on which a number of
pieces are placed. One piece, a Knight, can move (or attack) from corner to
diagonally-opposite corner of a rectangle three squares by two (Dr. Lasker as
quoted in [12]). See Figure 5.1.


X X
X X
&
X X
X X

Figure 5.1 Knight Moves: A Knight (£}) attacks the squares marked
with x.
Definition 5.1 A Knight domination of a kxr board is a placement of Knights
so each square either has a Knight on it or attacking it. The kxr Knight
domination number is the minimum, number of Knights in a Knight domination
of a k x r board.
Figure 5.2 shows a domination of a 3 x 14 board with 11 Knights. Since 11 is
the minimum number, the 3 x 14 Knight domination number is 11.
61


& & & & & &
& & & &
&
Figure 5.2 A Minimal 3 x 14 Knight Domination.
The k x r Knight Graph, denoted Nk>r (N is the standard initial for
Knight, as K is used for King), has vertices {l,2,...,/c}x{l,2,...,r} with
an edge between vertices (g,h) and (i,j) if |i g\\j h\ = 2 (see Figure 5.3).
Then the k x r Knight domination number is j(Nkjr).
^8,8
Figure 5.3 The 8x8 Knight Graph. Vertices represent squares of
an 8 x 8 board with edges showing moves that a Knight can make. Its
domination number is the Knight domination number of an 8x8 board.
Gardner [9] generated considerable interest in Knight domination of square
boards (when k = r). Gardner and his readers found small Knight dominations
of a k x k board for k < 15. These are known to be optimum when k < 10.
k 12 3 4 5 6 7 8 9 10 11 12 13 14 15
l(Nk,k) 14 4 4 5 6 10 12 14 16 < 21 < 24 < 28 < 32 < 37
Hare and Hedetniemi [10] developed a dynamic programming algorithm for
the k x r Knight domination number which is exponential in k, but linear in r.
This allowed them to conjecture formulas for 7(Nk,r) for k = 3,4, and 6 (k = 1
62


and 2 are trivial; k = 5 was not fully analyzed). They also found values
for k < 10 for some values of r. In this chapter, we restate the algorithm in
Min-Plus algebra and looks for periodic solutions to the dynamic programming
algorithm. By finding periodicity, we prove the conjectures in [10], and go on
to find the k x r Knight domination number for all k < 8 and for all r. To
illustrate the dynamic programming algorithm, we will use it to find the 2 x r
Knight domination number.
5.1 The States
A state will be a configuration of two columns of a k x r board. A square
either has a Knight on it (shown by ), is dominated by a Knight from the left
or within the two columns (shown by x), or to be dominated from the right
of the two columns (shown by a blank). Since a blank cannot be a Knight
move from a Knight, states correspond to labellings of Nkj2 with 0, 1 and 2 (for
squares with a squares with a x, and blank squares, resp.) where 0 and 2
cannot be adjacent.
How many such labellings are possible? Let a,r be the number of such
labellings on a path on r vertices. It is easy to show that a\ = 3, a2 = 7, and
a,r = '2ar i +ar_2 for all r > 2. So a3 = 17, a4 = 41, a5 = 99, etc. We can then
63


solve this recursion to show that
/ r\r+l /
(l + ^2j +(l
r+1
Since Nkj2 consists of two paths on [k/2j vertices and two paths on \k/2]
vertices, the number of states needed to find the k x r Knight domination
number for all r is (Figure 5.4 shows the 81 states when k = 2):
. 2fc+4
Number of States = a\k/2\a2{k/2\
(l + ^2
16
(5.1)
1.
10
19
28
37
46
55
64
73
S3
x
2
>d
x
x
X
S3
x
11
20
29
38
47
56
65
74
S3S3
x
x
S3S3
x
x
X
S3
x
X
S3
gx
X
2
X
3.
12
21
30
39
48
57
66
75
2
S3S3
X
22
2
x
X
2
x
2
2
2
x
2
2
4.
13
22
31
40
49
58
67
76
2E
x
2
2E
2
x
x
22
x
X
X
X
22
x
14
23
32
41
50
59
68
77
X
X
2 X
X X
2 X
X
X X
2 X
X X
X X
X X
X
X
2 X
X
X X
X
X
6.
15
24
33
42
51
60
69
78
X
2 X
X
2 X
X X
X X
X
X X
X
2
X
X
X

7
16
25
34
43
52
61
70
79
2
x
2
x
x
X n
2
22
n
X 2
2
m
17
26
35
44
53
62
71
80
X
2
X X
2
X
X
X
X
X X
X
X
2 X
X X
X
18
27
36
45
54
63
72
81
2
X
2
X
2
X
X
X
2
X
Figure 5.4 The 81 Knights configurations of a 2 x 2 board.
5.2 The Transition Matrix
Of the 36 possible 2x3 boards, only 81 need to be considered for domi-
nation patterns. This is because each square in the third column of this board
64


is a Knight move from a square in the first column (square (1,1) and (2,3) are
a Knight move apart as are square (2,1) and (1,3)). The possible 2x3 boards
are created by matching two of the 2x2 boards shown in Figure 5.4 together
so that the middle columns overlap and the following conditions are satisfied.
(1) Any Knights in the overlapping columns must be in the same square
in each pattern.
(2) All dominated (x) squares in the last column of the second kx 2 board
must be dominated by Knights within the resulting k x 3 board.
(3) All undominated (blank) squares in the first column of the first k x 2
board must be dominated by Knights in the resulting k x 3 board.
(4) When matching a k x r board to a k x 2 board, condition 2 guar-
anties that all dominated squares in the last column of the kxr board
are dominated by Knights within previous columns. Thus, an undom-
inated square in the first column of the second pattern can match
with a dominated square in the last column of the first pattern since
the undominated square will become dominated in the match. (This
matching does not apply to 2 x 2 boards since the pattern is not pos-
sible).
All possible domination patterns of aJc x 3 board are found similarly by match-
ing together two of the domination patterns of aJc x 2 board following the rules
65


listed above. In Figure 5.5, possible matchings of k x 2 boards are shown to
illustrate the matching requirements.
£
to
X
X
X
Figure 5.5 Possible Matchings of k x 2 Boards
To find the domination number of a.2 x r board, a column is added on
to all of the possible patterns on 2 x (r 1) boards and then each of these
new patterns is checked to see if it is a domination. The minimum number of
Knights in the patterns which are dominations is then the domination number
of a 2 x r board. The r column that is being added onto the 2 x (r 1)
board is dependent on the last two columns of the 2 x (r 1) board. Thus,
the column is added by matching the last column of the 2 x (r 1) pattern to
the first column of one of the 2x2 boards following the matching constraints
listed above. To keep track of which patterns can be added onto other patterns
by using this matching, a transition matrix A is formed with rows and columns
66


being represented by the initial 2x2 boards. If the i k x 2 board matches
the j 2x2 board then the (i,j) entry in the transition matrix is the number
of Knights in the last column of the j 2x2 board. If the i 2x2 board
does not match the j board then the (i,j) entry in the transition matrix is
infinity or (-) (see below).
67


68


5.3 The States Vector
Next, a states vector xr is formed to keep track of the 81 possible minimal
states of a 2 x r board. The i entry in this vector is the number of Knights
in a 2 x r pattern whose last 2 columns are the same as the i-2x2 board in
Figure 5.4. If one of these 2x2 boards does not form the last 2 columns of
any of the possible 2 x r patterns then the entry in xr is infinity or (-). Now,
to find the possible patterns of a 2 x (r + 1) board, the transition matrix is
multiplied by xr using Min-Plus matrix multiplication to form a new vector
xr+i. The i entry in xr+i represents the number of Knights in a 2 x (r + 1)
pattern whose last 2 columns are the same as the i 2x2 board. For each of
these 2 x (r +1) patterns, the first r 1 columns are dominated. If the i 2x2
board, which forms the r, and r + 1 columns of the pattern, is also dominated,
then the entire 2 x (r + 1) pattern will be dominated. Out of these dominated
states, the one with the least number of Knights in it (the lowest entry in xr+1)
forms a minimal domination of a 2 x (r + 1) board and the domination number
of a 2 x (r + 1) board is the number of Knights in this pattern.
This iterative process is started with an initial vector xi which is used to
find the domination number of a 2 x 1 board. To do this, the first column of
the 81 2 x 2 boards is considered column 0 of the 2x1 board. Only the 2x2
boards in which the first column is dominated and contains no Knights are
69


possible patterns for a 2 x 1 board. Thus, the i entry in the initial vector xi
is either the number of Knights in the last column of the j-2x2 board if the
first column of this board is dominated and contains no Knights, or infinity
(-) if the first column is not dominated or has Knights in it. Using this initial
vector, the domination number of a 2 x r board is found by iterative use of
Min-Plus matrix multiplication of the states vectors and the transition matrix
A with
xr = ,lxr ,.
The states vectors x1; x2,..., x18 are shown below.
XI = 0H[---------------------------------2-11-0-----------------------------------]
X2 = O0 [4-33-2--------3-22-1---------------------------3-22-1----------2-11-0]
x3 = 2 H [22-22---------11-11-----------------------------11-11------------00-00---]
x4 = 4H [00-00--00-00------------00-00---00-00-------------------------------------]
x5 = 4I [211100100211100100------211100100211100100--------------------------------]
x6 = 4I [433322322322211211322211211322211211211100100211100100322211211211100100211100100]
x7 = 4 H [443443332332332221332332221332332221221221110221221110332332221221221110221221110]
x8 = 4H [443443332443443332332332221443443332443443332332332221332332221332332221221221110]
x9 = 6 IS [222222222222222222111111111222222222222222222111111111111111111111111111000000000]
xio = 8 H [000000000000000000000000000000000000000000000000000000000000000000000000000000000]
Xu = 8 El [211100100211100100211100100211100100211100100211100100211100100211100100211100100]
xi2 = 8 H [433322322322211211322211211322211211211100100211100100322211211211100100211100100]
5.4 Detecting Periodicity
To find the domination number of a 2 x r board and the pattern that
gives this domination number, repetition can be used. Looking at the 18
70


states vectors, note that x12 = x6 + 4. Since the values in xr only depend on
the values in xr-\ and the transition matrix, we have xr+6 = xr + 4 for all
r > 6. So the domination pattern for a 2 x (r + 6) board is found by finding
the domination pattern for a 2 x r board and adding onto this a dominated
pattern with 6 columns and 4 Knights. The domination pattern for a 2 x r
board for all r is shown in Figure 5.6.
& & & & & &
& & & & & &
Figure 5.6 Domination Pattern for a 2 x r board.
Periodicity is detected by storing each xr and comparing its entries with
the entries in all previous state vectors.
The algorithm for finding the domination number of a k x r board is
essentially the same as for a 2 x r. In each case, all possible k x 2 boards are
formed. The number of possible k x 2 boards is equal to 32fe minus the number
of impossible patterns which is approximated by the formula 5.1. A transfer
matrix A and and initial vector xi are then formed from these k x 2 boards.
Min-Plus matrix multiplication is then used to find xr and the domination
number of a k xr board. As with 2 xr the domination number is the smallest
entry in xr which corresponds to a dominated k x r board. Successive xr
vectors are computed until a repetition is found and this is used to determine
the pattern corresponding to the domination number for all r. The results for
71


k <7 are given in the following sections.
5.5 Domination Number for a 3 x r board
For a 3 x r board, the periodicity is given by
xr+6 = xr + 4.
This means that once an initial pattern is established, the pattern can be
extended by adding on a pattern of 6 columns which is dominated by 4 Knights.
The domination patterns are shown in Figures 5.7 5.8 5.9.
to to to to to to
to to to to to to

Figure 5.7 Domination Pattern of 3 x r for r = 0, 3, 4, and 5. For
r = 0, 3, 4, and 5 (mod 6) r > 4. For r = 3 use columns 3, 4, and 5 of
this pattern. For r = 8, remove columns 4 and 5 of this pattern.
to to to to to to
to to to to to to to to

Figure 5.8 Domination Pattern of 3 x r for r = 1 (mod 6) r > 4
to to to to to to to to
to to to to to to
to
Figure 5.9 Domination Pattern of 3 x r for r = 2 (mod 6) r > 14
72


5.6 Domination Number for a 4 x r board
For a 4 x r board, the periodicity is given by
xr+6 = xr + 4.
This means that once an initial pattern is established, the pattern can be
extended by adding on a pattern of 6 columns which is dominated by 4 Knights.
The algorithm detects periodicity at r = 27. The domination patterns are
shown in Figures 5.10 5.11 5.12

& to to to to to
to to to to to to

Figure 5.10 Domination Pattern of 4 x r Board, for r = 0, 3, 4, and 5
(mod 6) when r > I. For r = 8, remove columns 4 and 5 of this pattern.

to to to to to to to
to to to to to to to

Figure 5.11 Domination Pattern of 4 x r Board, for r = 1 (mod 6)
r > 4.

to to to to to to to to to to
to to to to to to to to to to

Figure 5.12 Domination Pattern of 4 x r Board for r = 2 (mod 6)
r > 14.
73


5.7 Domination Number for a 5 x r board
For a 5 x r board, the periodicity is given by
Xy I 14.
This means that once an initial pattern is established, the pattern can be
extended by adding on a pattern of 18 columns which is dominated by 14
Knights. The algorithm detects periodicity at r = 39. The domination patterns
can be described in terms of the separate components shown in Figure 5.13.
E
to
L
B


to to
to to

Figure 5.13 Components in Domination Patterns of 5 x r Boards.
These components are combined together to form the patterns for a 5 x r
board. The table below shows the patterns with the repeating 18 columns of
74


each pattern in parentheses.
n 7(A^5)r) Pattern for r = 0 Pattern for r > 0
18r + 0 14r + 1 Meaningless (PBE'BV 1 PB I:7
18r + 1 14r+ 2 See 1x5 result (P BE4 Bf-1 P BE6
18r+ 2 14r + 3 See 2x5 result {PBE4B)r-lPBE1
18r + 3 14r + 4 See 3x5 result {PBE4By-lPBE8
18r + 4 14r + 4 (PBE4B)rU
18r+ 5 14r+ 5 E5 (PBE4B)r-1PBEbBU
18r + 6 14r + 6 Ee (PBE4B)r-1PBEeBU
18r+ 7 14r + 7 E7 (PBE4B)r-lPBE7BU
18r + 8 14r + 8 E8 (PBE4B)r-lPBE8BU
18r + 9 14r + 8 (PBE4B)rUBE4
18r + 10 14r + 9 (.PBE4B)rUBE5
18r + 11 14r + 10 {PBE4B)rUBEG
18r + 12 14r + 10 (PBE4B)rP
18r + 13 14r + 11 UBE8 (PBE*B)r~lPBE*EBP
18r + 14 14r + 12 (PBE4B)rUBE4BU
18r + 15 14r + 13 (PBE4B)rUBE5BU
18r + 16 14r + 14 (PBE4B)rU BE^BU
18r + 17 14r + 14 (.PBE4B)rPBE4
5.8 Domination Number for a 6 x r board
For a 6 x r board, the periodicity is given by
xr+4 = xr + 4.
This means that once an initial pattern is established, the pattern can be
extended by adding on a pattern of 4 columns which is dominated by 4 Knights.
The algorithm detects periodicity at r = 15. The domination patterns are
shown in Figures 5.14 5.15 5.16
75




to to to to to to
to to to to to to


Figure 5.14 Domination Pattern of 6 x r Board, for r = 0, and 3 (mod
4) r > 4.


to to to to to to to
to to to to to to to


Figure 5.15 Domination Pattern of 6 x r Board, for r = 1 (mod 4)
r > 4


to to to to to to to to
to to to to to to to to


Figure 5.16 Domination Pattern of 6 x r Board, for r = 2 (mod 4)
r > 4
5.9 Domination Number for a 7 x r board
For a 7 x r board, the periodicity is given by
Xr_|_5 = xr + 6.
This means that once an initial pattern is established, the pattern can be
extended by adding on a pattern of 4 columns which is dominated by 4 Knights.
76


The algorithm detects periodicity at r = 32. The domination patterns can be
described in terms of the components shown in Figure 5.17.
P
ft

ft ft

ft ft

ft

ft
ft ft
ft
ft ft


R\
ft

ft ft
ft
ft ft
ft ft


ft ft
ft ft
ft
ft ft

ft
S

ft
ft ft

ft ft



ft
ft ft

ft ft
ft ft


ft ft
ft ft
ft ft ft
ft V = ft w = Y = z =
ft ft ft
Figure 5.17 Components in Domination Patterns of 7 x r Boards.
These components are combined together to form the patterns for a 7 x r board.
The table below shows the patterns with the repeating 6 indicated by raising
the pattern to the exponent r.
77


n 7W,) Pattern for r > 0
1 6 See Previous results
7 10 zvzwzvz
8 11 zvzwzuzz
9 12 zzuzwzuzz
10 14 ZVZWYYWZVZ
11 15 ZVZWYYWZUZZ
12 16 ZZU ZWYYWZUZZ
5r + 13 6r + 18 T(pys
5r + 14 6r + 19 Rypys
5r + 15 6r + 20 Ri(P)rQ
5r + 16 6r + 21 T(P)rR2
5r + 17 6r+ 22 i?i(P)ri?2
5.10 Conclusions
The algorithm developed in this paper can be used to find the domination
number of a k x r board for all r. The algorithm is not linear in k so the size
of k is limited by computer time.
Theorem 5.2 The domination number of a k x r board 7{Nk,
for all r is:
ifk = 1
ifk = 2
if k = 2. r
if k = 3, r
if k = 3, r
for k < 6 and
7 {Nk,
n
dsi
2r+4
3
2r+4
3
2r+5
3
1 (mod 6) and r > 6
1 (mod 6), and r > 6
2 (mod 6), and r > 6
4 [si ifk = = 3, r y 1 2 (mod 6) and r > 6
2r+4 3 if k - = 3, r = 1 (mod 6), and r > 7
1 if k = = 4, r y 1 (mod 6) and r > 7
m ifk = = 5, r yy 12,13, or 17 (mod 18), and r > 13
RM if k = = 5, r = 4 12,13, or 17 (mod 18), and r > 13
r + 1 if k = = 6, r = 1 (mod 4), and r > 4
if k - = 6, r y 1 (mod 4), and r > 4
17+1 ifk = = 7, and r > 13
78


6. Time Complexity of The Power Method
All of the applications presented in the previous chapters rely on the pe-
riodic property of dynamic programming algorithms in Min-Plus algebra. The
algorithms terminate when periodicity occurs, and this allows us to find in-
finitely many solutions in a finite number of iterations of the algorithm.
Definition 6.1 The pre-periodic interval of a matrix A denoted Rq(A), is the
smallest Rq such that Ar = q m Ar^p for all r > Rq.
In Theorem 2.3 we stated that if A is the irreducible transfer matrix for the
dynamic programming algorithm, then periodicity implies that for some Rq, p
and q, we have that Ar = qm Ar^p for allr > Rq.
Although some matrices that are not irreducible are periodic, this is not
the case in general. Consider a matrix whose precedence digraph has two
disjoint cycles of length one with different weights as shown below.
t oo A 100 _ loot oo
oo s Ji. oo 100s
In this case Arn = rt and Ar22 = rs and thus if t ^ s there is no r where
Ar = q h Ar^p and Rq(A) is undefined. Thus when using the periodicity of
Min-Plus algebra to solve problems, we must make sure that the matrix is
irreducible.
79


6.1 Finite Termination of the Power Method
In Chapter 2, we stated Theorem 2.3 without proof. We have referred to
this theorem throughout all of the applications presented in this thesis because
it guarantees that at some point the dynamic programming algorithms will
be periodic and the power method can terminate. We will now restate the
theorem and give a proof based on minimum cycle means and the diameter of
A.
Definition 6.2 The diameter of A denoted diam(A) is the maximum vjeight of
all minimum, weight paths in A.
Theorem 6.3 If A is an irreducible matrix on m vertices, then there exists a
p, q and RQ so that
Ar = qs Ar-p for all r > i?0
Proof: Let p(A) be the minimum cycle mean of A. Since A is irreducible,
we know that the precedence digraph of Aml~m+l has paths between all pairs
of vertices. Let k be the smallest common divisor of the lengths of the critical
cycles of A, with the property that k > m2 m + 1. Now, consider Ak and let
S be the subgraph of the precedence digraph of Ak, (DAk) whose vertices are
contained in at least one of the critical cycles in A, and let T be the subgraph
of DAk on the remaining vertices. In I) \. every vertex of S has a minimum
weight path of length k to itself using the critical cycle that contains it. The
80


weight of this path is kp(A), so the minimum cycle mean of Ak is kp(A) and
the critical cycles are 1-loops on every vertex in S.
We will now look at possible paths in the precedence digraph of Ak. Let
P be any minimum weight path of length r between v,i and Vj that uses no
vertices of S. This path consists of cycles and simple paths in T. Let b(Ak) be
the minimum entry in Ak, then the weight of each arc in the simple paths is at
least b(Ak). If there are h such arcs, the weight of the simple path portion of P
is at least hb(Ak). Next if p(T) is the minimum cycle mean of T, then the arcs
in every cycle in T have weight of at least p(T). P contains r ^ h arcs that are
part of some cycle, so the weight of these cycles is at least (r h)p(T). Thus
w(P) > (r h)p(T) + hb(Ak)
This is a non-negative value since p(T) > b(Ak), and since h < m 1 we have
w(P) > rp(T) + (m^ 1 )(b(Ak) p(T))
Next consider all minimum weight paths Q from Vi to vj of length r that use
vertices of S. These paths consist of a path from to S, loops on the vertices
of S, and a path from S to vj. The paths to and from S have weight at
most diam(,4fe) and the loops in S each have weight kp(A) = p(Ak). Since Q
contains at most r loops in S, we have
w(Q) < 2diam(Ak)+rp(Ak).
81


Now we want to show that all minimal paths of length r > i?0 must use vertices
of S. This is true if w(P) > w(Q) for all paths of type P and Q of length r.
Since p(T) > p(Ak) we have w(P) > w(Q) for all r > i?0 when
i?n
2diam(. C) (m 1 )(b(Ak) p(T))
P(T) p{m
(6.1)
Now for any minimal path of length r > i?0 a minimal path of length r +1
is obtained by looping on some vertex of S that was in the minimal path of
length r, and the weight of these paths differs by p(Ak) in every case. Thus
(Ak)r = p(Ak) (Ak)r+l for allr > R0
This proves that Ak is periodic and hence A is periodic with Rq(A) = kR0(Ak).
The construction of the proof also gives a bound for R0 dehned in terms of the
entries of Ak.
We will now illustrate this method by finding the bound on Rq(A) where
A is the 15 x 15 transition matrix used in Chapter 4 to find the domination
number of Cr[l,3]. The precedence digraph in Figure 4.3 shows that there is
only one critical cycle C for this matrix and that p(A) = = |. To find the
bound for R0 using the method described above, we need to find the smallest
/. > nr /// 1 with k being a common divisor of the lengths of the critical
cycles in A. Thus since m = 15, and there is only one critical cycle of length
5, we have k = 215. This bound is large because we are required to have
82


k > m2 m + 1 to guarantee that the matrix has paths of any length greater
than k. (In this example, we have that A8 has no infinity enties so k could be
reduced to 10). We will illustrate the bound using k = 215 and considering
possible paths in A215 which is shown below.
A215 = 3.!) A20 = 3.!)
6 6 6 6 6 6 6 6 7 7 7 7 7 7 8
5 5 5 5 5 5 5 5 6 6 6 6 6 6 7
5 5 5 5 5 5 6 5 6 6 6 6 6 6 7
5 5 5 5 5 5 5 5 5 6 5 6 6 6 6
4 4 4 4 4 4 5 4 5 5 5 5 5 5 6
5 5 5 6 6 5 5 5 6 6 6 7 6 6 7
4 4 5 5 5 4 4 5 5 5 5 6 5 5 6
4 5 4 5 5 4 5 4 5 6 5 6 5 6 6
5 5 5 5 5 5 5 5 5 5 5 6 5 5 6
4 5 4 5 5 4 5 4 5 5 5 6 5 5 6
4 4 4 4 5 4 4 4 4 5 4 5 5 5 5
4 4 4 5 5 4 4 4 5 5 5 6 5 5 6
5 5 5 5 5 5 5 5 5 5 5 6 5 5 6
4 4 4 4 4 4 4 4 4 4 5 5 4 4 5
4 4 4 4 4 4 4 4 5 5 5 5 5 5 6
Using this matrix and the fact that S = {v*vt , we have that
b(Ak) = 43, p(T) = ^ and p(Ak) = kp(A) = ^(215) = 43.
L D
Next, for any matrix, its diameter is less than the number of vertices times the
maximum entry. Using this bound we have
diam(d.fe) < 690.
This is considerably larger than the actual diameter of A which is 269 and thus the
bound on Rq will be significantly effected by using the bound instead of the actual
83


diameter. Putting all of this together in equation 6.1 we have
Ro
2 x 705 (14) x (43 43.5)
43.5 43
= 2806
Thus, the bound is Ro(Ak) = 2806 and the bound for A is Rq(A) = (215)(2806) =
603290, whereas the actual value of Rq(A) was 24. In computing this bound, we
used k = 215 as the smallest value where Ak had no infinity entries instead of the
actual value of k = 20. If we compute the bound with k = 10 and diarn(A) = 269
it gives Rq(A) = (10)(1062) = 10620 which is still quite large. Thus although the
bound shows that Rq is finite, it is not a very good approximation of Rq(A) in most
cases.
The large value of k and the large approximation of diarn(A) both contribute
to the difference between the bound and the actual value of Rq(A). In addition, if
the growth of the entries in successive powers of A is large, then the fact that the
bound requires us to consider paths in Ak where k is large may have a much more
significant effect. In the next section we will restrict A and find a bound for Rq that
is strong.
6.2 Exact Bound for Rq when A is MCM-Hamiltonian
In this section, we will determine the bound for Rq when A is restricted to a
special case.
Definition 6.4 A minimum cycle mean (MCM) Hamiltonian digraph is a digraph
where the minimum, cycle mean occurs on a Hamiltonian cycle.
Definition 6.5 A matrix is MCM Hamiltonian if its precedence digraph is.
84


Note that if A is MCM Hamiltonian then the subgraph T as described in the
previous section does not exist. Thus, we cannot use the bound from that section
and we must look at other ways to bound Rq in this case.
Definition 6.6 If B is an MCM Hamiltonian matrix, define the following.
(1) Let A = | be the unique eigenvalue of B.
(2) Let P he the permutation matrix that permutes the rows and columns of B so
that F = PT El (B El (A)) El P is a matrix with minimum cycle mean 0 and
the minim,um cycle C oriented on the vertices 1, 2, m where vertex 1 is
represented by row 1 of F. Thus the arcs of C are (i,i + 1) for all i < m, and
the arc (m, 1).
(3) Let D be a diagonal matrix with Djj = (Fi,i+i) + Fm,l
Definition 6.7 If B is an MCM Hamiltonian matrix on m vertices, then the nor-
malized form of B is A = (-D) El PT El (B El (A)) El P El D. A normalized MCM
Hamiltonian matrix is a matrix that is equal to its normalized form.
Theorem 6.8 Let B be an MCM Hamiltonian matrix with minim,um, cycle mean X,
and, normal form, A = (-D) El PT El (B Kl (A)) Kl P Kl D. Then
(1) -4.1,2 = -4-2,3 = ... = Am^ijTn = Am^\ = 0
(2) The minimum, cycle mean of A is 0.
(3) A is non-negative.
(4) Br = AAr lEI (-D) El PT El Ar El P IEI D and,
(5) Br = g El Br-P then Ar = Ar-P.
85


Proof: Let Cr, Cr, Ca be the cycles with minimum cycle means in F, R, and
A. Then F = PT H B Kl (A) KI P so the arcs of Cb have been permuted to the arcs
of Cf + {Fi,2, 3 Fm, 1} and there weights have been reduced by A. Also for any
arc in A,
Aij = //../ Di,i +
= F'ij \F,i^jr\ + ... + FmijTO + T [Fj,j+i T T fm-i,ra T Fmji}-
which gives a direct correspondence between the cycles in A and the cycles in B.
1) By construction,
Ai,i-\- 1 = Fj,,i+1 [Fj,,i+1 T- +-fm-l,m+-fm,l] + [-fi+l,i+2 + -T-fml,mTTm,l] = 0
for all i < m, and
Am, 1 = Fm, 1 [Fm,i] + [-^1,2 + -^2,3 + + Fm^ i;jn + Fnh 1] = 0.
Thus all the arcs in C = {Ai^ Am-ijm,Amji} have weight 0 and form a
hamiltonian cycle of weight 0.
2) For any cycle Ca, of length % in A which has weight w(Ca,), there is a corre-
sponding cycle Cb, of length % in B which has weight w(Cb,) = w(Ca,) + A%.
By part 1 C is a hamiltonian cycle with cycle mean 0 so w(Ca) < 0. If
w(Ca) < 0 then w(Cb) + w(Ca) + Am has cycle mean less than A which is a
contradiction. Thus the minimum cycle mean of A is 0.
3) Suppose A has some entry A;j < 0 Consider the cycle
{Aij,Ajj+1,..., ijm, Am_ 1,..., Aj^ij}. Since all the arcs of this cycle ex-
cept Aij are contained in the cycle with minimum cycle mean, they have weight
86


0 and the weight of the cycle is /1(. ( < 0 which is a contradiction since the min-
imum cycle mean of A is 0.
4 & 5) The results are found by using the commutative and associative properties
in Min-Plus algebra.

To illustrate the normalization process consider the following matrix which has the
arcs in the cycle with minimum cycle mean underlined.
B =
7 9 2 8
18 6 6
9 8 6 5
3 4 7 8
which has A = 3
Then using the definitions above,
P
oo 0 oo oo ' 5 M2 3 3 ' 0 oo oo oo
0 oo oo oo 6 4 -1 5 oo 2 oo oo
F = D =
oo oo 0 oo 5 6 3 2 oo oo 3 oo
oo oo oo 0 _ 1 0 4 5 _ oo oo oo 1
and finally
A =
5 0 6 4
4 4 0 4
2 5 3 0
0 1 6 5
Thus any min-plus problem with an MCM Hamiltonian transfer matrix B can
be solved using the normalized MCM Hamiltonian matrix A and then converting
the solution using parts 4 and 5 of Theorem 6.8
To find a bound on the number of iterations needed for the transfer matrix to
repeat, we will consider normalized MCM Hamiltonian matrices. The precedence
87


digraph for a normalized MCM Hamiltonian matrix has a Hamiltonian cycle with
all arcs having weight zero and other arcs of non-negative weight. The bounds on
repetition will be formed by considering paths in the precedence digraph since if
Ar = Ar^m where rn is the size of the matrix, then in the precedence graph the
minimum weight of a path of length r from i to j is the same as the minimum weight
of a path from % to j of length r rn for all % and j.
Definition 6.9 Let A be a normalized MCM Hamiltonian matrix and let b-i be the
i arc in the minimum cycle of A which by definition has weight 0.
Definition 6.10 Let a& be the number of arcs from i to j used in the path where
i j + 1 = k (mod m).
Theorem 6.11 A path P of length r from vertex i to vertex j in a MCM Hamilto-
nian digraph on m vertices satisfies the following:
a,\ + 2a2 + + (m l)am_i r = (i j) mod m (6-2)
Proof: First we will prove this for r = 1. There are two cases for a path of length
1. Case 1: j = % + 1. In this case, the arc from % to j is one of the 6* arcs from the
minimal cycle. Thus no a* arcs are used and equation 6.2 becomes 1 = (i (i + 1))
(mod rn) which holds. Case 2: j i + 1 In this case, if j = i + k the arc used is a;
where l = i j + 1 = k + 1 and equation 6.2 becomes l 1 = (i j) (mod rn) or
(i (i + k) + 1) 1 = (i (i + k)) (mod m) which also holds.


For induction purposes, we will assume that a path of length r 1 satisfies
( 6.2) Now any path of length r contains a path of length r 1 plus one arc. Let
j be the vertex at the end of the path of length r 1 and k be last vertex in the
path. Thus the last arc in this path of length r goes from vertex j to vertex k. Let
a'i, ar2,... arm_i be the coefficients that satisfy ( 6.2) for the path of length r 1.
Then ( 6.2) is satisfied for the path of length r if a[ + 2a'2 + + (rn 1 )a'm + arc
from j to k r = (i k) (mod m). Using the induction hypothesis, this will hold if
i j (mod m) + arc from j to k 1 = i k (mod m).
Case 1: The additional arc is one of the arcs on the minimal cycle. This
gives k = j + 1 (mod rn) by the labeling of the minimal cycle, and ( 6.2) becomes
i j 1 = i (j + 1) (mod rn) since there is no additional Oj arc added to the
equation.
Case 2: The additional arc is a; where l = j k + 1. Then ( 6.2) becomes
i, j + (;/ /,+!) 1 i, I.- (mod rn) which holds.
Definition 6.12 Given a path P the arcs can be divided into three exclusive classes:
Type 1 Arcs used in a minimal path from the initial vertex to the final vertex in the
path. There are < m of these.
Type 2 Arc used in any minimal cycle containing an a-i. There are YaLi iai of
these.
Type 3 Arcs that are in the minimal cycle.
Let Ni be the number of arcs of type i used in a given path.
89


Each type is exclusive so once an arc has been designated as being in type one it
cannot be in type two or three. Also the determination of the arc type must be done
in the order they are described. For example using the graph in Figure 6.1, consider
the following path of length 13 from vertex 1 to vertex 3.
P = bib2a5 bi 62b3b4b5a^ b4 b5bib2
In this path, the bold arcs are designated type one, the italic arcs are designated
type two and the remaining arcs are type three.
1
Figure 6.1 Precedence digraph of a normalized MCM Hamiltonian
matrix. The bi arcs represent the arcs in the minimal cycle and have
weight 0. The weight of the a* arcs is non-negative.
Now we need to show that for some r, any minimum weight path in the prece-
dence digraph of length r has the same weight as a minimum weight path of length
r m. This will be possible if there are always at least m type 3 arcs in a minimum
weight path of length r. Since rn of these arcs form a Hamiltonian cycle and we
assumed the type 3 arcs have weight zero by theorem 3.2, their removal will not
change the weight of the path and will give a path of equal weight and length r rn.
90


Full Text

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MIN-PLUS ALGEBRA AND GRAPH DOMINA TION b y Anne Spalding B.S., Univ ersit y of Colorado at Den v er, 1993 M.S., Univ ersit y of Colorado at Den v er, 1996 A thesis submitted to the Univ ersit y of Colorado at Den v er in partial fulllmen t of the requiremen ts for the degree of Doctor of Philosoph y Applied Mathematics 1998

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This thesis for the Doctor of Philosoph y degree b y Anne Spalding has been appro v ed b y Da vid C. Fisher J. Ric hard Lundgren Stanley P a yne Ross McConnell T om Altman Date

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Spalding, Anne (Ph. D., Applied Mathematics) Min-Plus Algebra and Graph Domination Thesis directed b y Professor Da vid C. Fisher ABSTRA CT This thesis describes the Min-Plus algebra and uses Min-Plus algebra to implemen t dynamic programming algorithms. Min-plus algeb ra is dened b y replacing the standard algebra operators of addition and m ultiplication b y minimization and addition respectiv ely There are sev eral applications whic h can be solv ed using these dynamic programming algorithms in Min-Plus algebra. W e will illustrate these algorithms b y using them to solv e the follo wing problems in v olving domination. 1 A domination D of a graph G is a subset of v ertices so that eac h v ertex of G is either in D or adjacen t to a v ertex in D Let r ( G ) (the domination number of G ) be the minim um size of a domination of G W e nd the domination n um ber of a large class of graphs including the ca rtesian p roduct graphs and circulant graphs 3 F rom c hess, a knigh t mo v e is t w o squares in one direction and one square in the perpendicular direction. A Knight domination of a k r board iii

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is a placemen t of Knigh ts so eac h square either has a Knigh t on it or attac king it. The k r Knight domination number is the minim um n um ber of Knigh ts in a Knigh t domination of a k r board. W e nd the the Knigh t domination n um ber of a k r c hessboard for k 7 and for all r The dening feature of these dynamic programming algorithms is that they nd innitely man y solutions in a nite amoun t of time. This is possible because the algorithms are periodic after a nite n um ber of iterations. W e sho w that periodicit y is dependen t on the size and the en tries in the state transition matrix, and w e determine the necessary conditions on the matrix for periodicit y to occur. W e also look at random matrices with en tries independen tly c hosen from discrete and con tin uous distributions and nd the expected n um ber of iterations needed for periodicit y in eac h case. The results of this analysis sho ws wh y the algorithms are successful in solving the graph domination problems and what conditions are necessary to use similar dynamic programming algorithms in other problems. This abstract accurately represen ts the con ten t of the candidate's thesis. I recommend its publication. Signed Da vid C. Fisher iv

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DEDICA TION I w ould lik e to dedicate this thesis to m y paren ts and m y family who ha v e alw a ys stood behind me supporting me and driving me forw ard; to Ra y who has alw a ys stood in fron t of me and sho wn me the w a y; and to Stev e, who alw a ys encouraged me to stretc h m y limits... to ride faster, clim b higher and pla y harder. Thank y ou.

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A CKNO WLEDGEMENT I w ould lik e to thank m y advisor, Da vid C. Fisher, for his encouragemen t in preparation of this thesis, his support, guidance, friendship, and time spen t in discussions throughout m y studies at the Univ ersit y of Colorado at Den v er.

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CONTENTS Chapter 1. In troduction . . . . . . . . . . . . . 1 2. Properties of Min-Plus . . . . . . . . . . 6 2.1. Min-Plus Linear Algebra . . . . . . . . . . 7 2.2. Eigen v alues and Eigen v ectors: The P o w er Method . . . 8 2.3. Minim um Cycle Means and Karp's Theorem . . . . 13 3. Domination of Cartesian Product Graphs . . . . . 21 3.1. The Exact Algorithm . . . . . . . . . . . 24 3.2. The A t Least Algorithm . . . . . . . . . . 29 3.3. F orm ulas for the Domination Num ber . . . . . . 36 3.4. Domination of G P r : The A t Least Algorithm . . . 39 3.5. V alues of r ( G P r ) for graphs on 6 v ertices. . . . . 44 4. Domination of Circulan t Graphs . . . . . . . . 53 4.1. The States and T ransition Matrix . . . . . . . 54 4.2. P eriodicit y of the T ransition Matrix . . . . . . . 58 5. Knigh ts Domination of k r Chessboards . . . . . 61 5.1. The States . . . . . . . . . . . . . . 63 vii

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5.2. The T ransition Matrix . . . . . . . . . . 64 5.3. The States V ector . . . . . . . . . . . . 69 5.4. Detecting P eriodicit y . . . . . . . . . . . 70 5.5. Domination Num ber for a 3 r board . . . . . . 72 5.6. Domination Num ber for a 4 r board . . . . . . 73 5.7. Domination Num ber for a 5 r board . . . . . . 74 5.8. Domination Num ber for a 6 r board . . . . . . 75 5.9. Domination Num ber for a 7 r board . . . . . . 76 5.10. Conclusions . . . . . . . . . . . . . 78 6. Time Complexit y of The P o w er Method . . . . . . 79 6.1. Finite T ermination of the P o w er Method . . . . . 80 6.2. Exact Bound for R 0 when A is MCM-Hamiltonian . . . 84 6.3. P eriodicit y in Matrices with In teger en tries. . . . . . 95 6.4. P eriodicit y in Random Matrices With Real En tries . . . 99 6.5. Conclusions . . . . . . . . . . . . . . 109 Appendix . . . . . . . . . . . . . . . . . 112 A. Domination Num bers of Circulan t Graphs . . . . . 112 References . . . . . . . . . . . . . . . . . 175 viii

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FIGURES Figure 3.1 Minim um Dominations of P 13 P 13 . . . . . . . 21 3.2 The P artial Ordering of the 17 States for P 3 P r . . . 30 3.3 The House Graph on Fiv e V ertices G and G P 4 . . . 40 4.1 The Circulan t Graph. . . . . . . . . . . . 54 4.2 15 State for the Domination of Circulan t Graphs. . . . 55 4.3 Precedence Digraph for Circulan t Graphs. . . . . . 57 5.1 Knigh t Mo v es . . . . . . . . . . . . . 61 5.2 A Minimal 3 14 Knigh t Domination. . . . . . . 62 5.3 The 8 8 Knigh t Graph. . . . . . . . . . . 62 5.4 The 81 Knigh t congurations of a 2 2 Board. . . . . 64 5.5 P ossible Matc hings of k 2 Boards . . . . . . . 66 5.6 Domination P attern of 2 r Board. . . . . . . . 71 5.7 Domination P attern of 3 r Board for r = 0, 3, 4, and 5. . 72 5.8 Domination P attern of 3 r Board for r = 1 (mod 6) r 4 72 5.9 Domination P attern of 3 r Board for r = 2 (mod 6) r 14 72 ix

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5.10 Domination P attern of 4 r Board for r = 0, 3, 4, and 5 (mod 6) when r 4. . . . . . . . . . . . 73 5.11 Domination P attern of 4 r Board for r = 1 (mod 6) r 4. 73 5.12 Domination P attern of 4 r Board for r = 2 (mod 6) r 14. 73 5.13 Componen ts in Domination P atterns of 5 r Boards. . . 74 5.14 Domination P attern of 6 r Board for r = 0, and 3 (mod 4) r 4. . . . . . . . . . . . . . . . 76 5.15 Domination P attern of 6 r Board for r = 1 (mod 4) r 4 76 5.16 Domination P attern of 6 r Board for r = 2 (mod 4) r 4 76 5.17 Componen ts in Domination P atterns of 7 r Boards. . . 77 6.1 Precedence Digraph of a Normalized MCM Hamiltonian Matrix. . . . . . . . . . . . . . . . . 90 6.2 The Pre-P eriodic In terv al for Random Matrices With In teger En tries . . . . . . . . . . . . . . . 97 x

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1. In troduction This thesis describes the Min-Plus algebra and uses Min-Plus algebra to implemen t dynamic programming algorithms. Min-plus algebra is dened b y t w o operators and on the real n um bers. The sym bol represen ts standard addition and the sym bol represen ts minimization. In Chapter 2 w e will look at sev eral properties of Min-Plus algebra. Using the standard algebra properties, w e dev elop analogues to linear algebra including eigen v alues and eigen v ectors. Eigen v alues are used in the implemen tation of dynamic programming algorithms in Min-Plus algebra. W e will discuss t w o methods of nding eigen v alues and w e will sho w ho w their existence ensures nite time execution of the dynamic programming algorithms. There are sev eral applications whic h can be solv ed using dynamic programming algorithms in Min-Plus algebra. W e will illustrate these algorithms b y using them to solv e problems in graph domination. Denition 1.1 A domination D of a gr aph G is a subset of vertic es so that e ach vertex of G is either in D or adjac ent to a vertex in D L et r ( G ) (the domination number of G ) b e the minimum size of a domination of G The domination problem is NP-hard. Most algorithms for nding the 1

PAGE 12

domination n um ber of a graph in v olv e using Branc h and Bound tec hniques to reduce the graph to other graphs for whic h the domination n um ber is kno wn. W e dev elop a dynamic programming algorithm to nd the domination n um ber of large classes of graphs. The rst set of graphs w e consider are the cartesian product graphs. Denition 1.2 L et G = ( V; E ) b e a gr aph with vertic es f v 1 ; v 2 ; : : :v n g and let H = ( V; E ) b e a gr aph with vertic es f w 1 ; w 2 : : :w ` g Then the ca rtesian p roduct G H has vertic es of the form ( v i ; w j ) with 1 i n and 1 j ` The e dges of G H have the form (( v i ; w j ) ; ( v p ; w q )) if i = p and ( w j ; w q ) is and e dge of H or if j = q and ( v i ; v p ) is an e dge of G The k th c opy of G in G H is G k and its vertic es ar e ( v 1 ; w k ) ; ( v 2 ; w k ) ; : : : ( v n ; w k ) In Chapter 3, w e consider cartesian product graphs with H = P r where P r is the path on r v ertices with v ertix w i adjacen t to v ertex w k if and only if j i k j = 1. W e begin b y looking at a special cartesian product graph called the complete grid graph. The complete grid graph on n; r vertices, P n P r has nr v ertices with v ertex ( v i ; w j ) adjacen t to v ertex ( v k ; w l ) if and only if j i k j + j j l j = 1. The domination n um ber of complete grid graphs r ( P n P r ) has been analyzed for small v alues of n Jacobson and Kinc h [13 ] found results for n 4, and Chang and Clark [2] found results for n = 5 ; 6. In [11] and [15] Branc h and 2

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Bound algorithms are used to nd domination n um bers for larger v alues of n and some v alues of r These results w ere then expanded b y the use of dynamic programming algorithms [11]. W e impro v e these algorithms b y incorporating the periodic propert y of Min-Plus algebra. W e dev elop t w o algorithms, the Exact Algo rithm and the A t Least Algo rithm and are able to nd the domination n um ber for all grid graphs ( P n P r ) with n 19 and for all v alues of r Livingston and Stout [14 ] use an algorithm whic h w e will call the Exact Algorithm to nd the domination n um ber of general cartesian product graphs ( G P r ). W e use the faster A t Least Algorithm to nd the domination n um ber of graphs ( G P r ) for an y graph G on n 6 v ertices and for all r In Chapter 4, w e modify the Exact Algorithm to nd the domination n um ber of circulan t graphs. Denition 1.3 Given an inte ger r > 0 and subset S of the p ositive inte gers, the circulant graph C r [ S ] is a gr aph on vertic es 0, 1, ..., r 1 with vertex i adjac ent to vertex j if i j 2 S or j i 2 S (mo d r ). W e nd the domination n um bers of all circulan t graphs C r [ S ] for all r and for an y subset S of the set f 1 ; 2 ; 3 ; : : :; 9 g The algorithms dev eloped in Chapters 3 and 4 can be modied to solv e man y t ypes of domination problems. T o illustrate this, w e solv e the knigh t domination problem for rectangular c hessboards. Denition 1.4 A Knight domination of a k r b o ar d is a plac ement of Knights 3

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so e ach squar e either has a Knight on it or attacking it. The k r Knight domination number is the minimum numb er of Knights in a Knight domination of a k r b o ar d. The knigh t domination problem is a v ariation of the domination problem and wide in terest in this problem w as rst generated b y Gardner [9 ] for square c hessboards. Hare and Hedetniemi [11] found the domination n um bers for k r c hessboards with k 6 and some v alues of r W e are able to v erify these results and nd the knigh t domination n um ber for all k r c hessboards with k 7 and for all r The dening feature of the applications in Chapters 3-5 is that the dynamic programming algorithms are able to nd innitely man y solutions in a nite amoun t of time. This is possible because the algorithms are periodic after a nite n um ber of iterations. The periodicit y of the algorithms can be described using the eigen v alues of a state transition matrix In Chapter 6, w e sho w that periodicit y is dependen t on the size and the en tries in the state transition matrix. First, w e determine the necessary conditions on the matrix for periodicit y to occur. Then for special cases, w e nd exact bounds on the n um ber of iterations of the algorithm that are needed before periodicit y occurs. Next w e look at random matrices with en tries independen tly c hosen from discrete and con tin uous distributions and nd the expected n um ber of iterations needed for 4

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periodicit y in eac h case. The results of this analysis sho ws wh y the algorithms are successful in solving the graph domination problems in Chapters 3-5 and what conditions are necessary to solv e other t ypes of problems with similar algorithms. 5

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2. Properties of Min-Plus In this c hapter w e will dene the Min-Plus algebra and look at sev eral of its properties. The majorit y of this material is deriv ed from Baccelli et al. [1] Min-plus algebra is dened b y t w o operators and on the real n um bers. The sym bol represen ts standard addition and the sym bol represen ts minimization. An iden tit y e for satises min ( a; e ) = a e = a for all a Since min ( a; e ) = a only if e a w e m ust ha v e e a for all a So to ha v e an iden tit y for in an algebra whic h includes all real n um bers, w e m ust append 1 to the real n um bers. Let < + = < [ f1g W e m ust then dene the operations for 1 : let 1 a = a 1 = a and 1 a = a 1 = 1 for all a 2 < + Then these properties hold for all a; b; c 2 < + (Comm utativ e) a b = b a ( i.e. min ( a; b ) = min ( b; a )), and a b = b a ( i.e. a + b = b + a ); (Associativ e) a ( b c ) = ( a b ) c ( i.e. ,min ( a; min ( b; c )) = min (min ( a; b ) ; c )), and a ( b c ) = ( a b ) c ( i.e. a + ( b + c ) = ( a + b ) + c ); (Distributiv e) a ( b c ) = ( a b ) ( a c ) ( i.e. a + min ( b; c ) = min ( a + b; a + c )); 6

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(Iden tit y) 1 a = a ( i.e. min ( 1 ; a ) = a ), and 0 a = a ( i.e. 0 + a = a ). (In v erses under ) ( a ) a = 0 ( i.e. a + a = 0). This w ould be a eld if there w ere in v erses under But a x = 1 ( i.e. min( a; x ) = 1 ) has no solution for x unless a = 1 2.1 Min-Plus Linear Algebra Sev eral of the applications presen ted in Chapters 3-5 use linear algebra operations within the Min-Plus algebra. Using the properties on real n um bers dened abo v e w e can dene analogues to standard matrix algebra concepts. Matrix addition is dened as componen t wise minimization. F or example: 2 4 3 2 2 4 3 5 2 4 1 5 3 2 3 5 = 2 4 3 1 2 5 2 3 4 2 3 5 = 2 4 1 2 2 2 3 5 : Matrix m ultiplication is denoted b y ( A B ) ij = n k =1 A ik B kj ; 2 4 3 2 2 4 3 5 2 4 1 5 3 2 3 5 = 2 4 (3 1) (2 3) (3 5) (2 2) (2 1) (4 3) (2 5) (4 2) 3 5 = 2 6 4 4 4 3 6 3 7 5 : The \zero" matrix ( A 1 = 1 A = A ) is: Z = 1 = 2 6 6 6 6 6 6 6 6 6 6 4 1 1 1 : : : 1 1 1 1 : : : 1 1 1 1 : : : 1 . . . . . . . 1 1 1 : : : 1 3 7 7 7 7 7 7 7 7 7 7 5 : 7

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The \iden tit y" matrix ( A I = I A = A ) is: I = 2 6 6 6 6 6 6 6 6 6 6 4 0 1 1 : : : 1 1 0 1 : : : 1 1 1 0 : : : 1 . . . . . . . 1 1 1 : : : 0 3 7 7 7 7 7 7 7 7 7 7 5 : Matrix Exponen ts. Let A be a square matrix. Let A 1 = A and recursiv ely dene A r = A A r 1 for all k > 1. Associativit y ensures that A r = A r i A i for all i = 1 ; 2 ; : : :; r 1. Scalar m ultiplication. The i; j en try of k A is ( k A ) i;j = k A i;j T race. F or a square matrix A let trace( A ) be the minim um of the diagonal en tries of A The properties abo v e sho w that matrix m ultiplication is associativ e. Since there is no in v erse for matrix in v erses are not dened except in special cases. There are also analogues to solving linear systems, eigen v ectors, eigen v alues, c haracteristic equation, and ev en a Ca yley-Hamilton theorem. W e describe only those properties that are used in the applications presen ted. 2.2 Eigen v alues and Eigen v ectors: The P o w er Method Sev eral applications of Min-Plus algebra use dynamic programming algorithms. The dening propert y of these algorithms is that they use periodicit y to nd innitely man y solutions in nite time. P eriodicit y of matrices in MinPlus algebra is dened in terms of eigen v alues and eigen v ectors. In this section 8

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w e will look at the denition of eigen v alues and eigen v ectors, and the P o w er Method of nding them. T o illustrate this method, consider the follo wing example. A coun try has 4 cities with o v ernigh t trains bet w een them. The o v ernigh t rates can be represen ted as a matrix A where A ij is the cost to start in cit y i and tak e the o v ernigh t train to cit y j Belo w is the matrix A with the costs to sta y in eac h cit y and for o v ernigh t train rides bet w een them. Find the che ap est way to stay n nights starting at city i and ending at city j A = 2 6 6 6 6 6 6 4 8 0 2 1 7 3 3 6 4 5 5 6 3 3 7 9 3 7 7 7 7 7 7 5 : F or the cost matrix A ( A 2 ) ij represen ts the minim um cost to start at cit y i tra v el for 2 nigh ts, and end in cit y j Lik ewise, ( A r ) ij represen ts the minim um cost for a trip of r nigh ts. Next, look at v arious matrix po w ers of the cost matrix A A = 2 6 6 6 6 6 6 4 8 0 2 1 7 3 3 6 4 5 5 6 3 3 7 9 3 7 7 7 7 7 7 5 ; A 2 = 2 6 6 6 6 6 6 4 4 3 3 6 7 6 6 8 9 4 6 5 10 3 5 4 3 7 7 7 7 7 7 5 ; A 3 = 2 6 6 6 6 6 6 4 7 4 6 5 10 7 9 8 8 7 7 10 7 6 6 9 3 7 7 7 7 7 7 5 ; A 4 = 2 6 6 6 6 6 6 4 8 7 7 8 11 10 10 11 11 8 10 9 10 7 9 8 3 7 7 7 7 7 7 5 ; A 5 = 2 6 6 6 6 6 6 4 11 8 10 9 14 11 13 12 12 11 11 12 11 10 10 11 3 7 7 7 7 7 7 5 ; A 6 = 2 6 6 6 6 6 6 4 12 11 11 12 15 14 14 15 15 12 14 13 14 11 13 12 3 7 7 7 7 7 7 5 : 9

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Note that A 6 = 4 A 4 (under \scalar addition"). Then A 7 = A 6 A = (4 A 4 ) A = 4 ( A 4 A ) = 4 A 5 ; A 8 = A 7 A = (4 A 5 ) A = 4 ( A 5 A ) = 4 A 6 ; A 9 = A 8 A = (4 A 6 ) A = 4 ( A 6 A ) = 4 A 7 : By induction, w e can then sho w that for all r 6, w e ha v e A r = 4 A r 2 and hence A r = 8 > < > : 2( r 4) A 4 if r is ev en 2( r 5) A 5 if r is odd. So for example, the minim um cost of start in cit y 3 and ending in cit y 2 while sta ying 13 nigh ts is 2(13 5) + 11 = 27. W e will no w sho w ho w this relates to the eigen v alues and eigen v ectors of A Denition 2.1 Given an m m matrix A in < + let b e an eigenvalue and x 6 = 1 b e the asso ciate d eigenve ctor if A x = x F or example, let A = 2 6 6 6 4 4 3 4 2 4 3 1 3 3 3 7 7 7 5 : Then = 7 3 is an eigenvalue with asso ciate d eigenve ctor x = ( 4 3 ; 2 3 ; 0) T b e c ause A x = 2 6 6 6 4 4 3 4 2 4 3 1 3 3 3 7 7 7 5 0 B B B @ 4 3 2 3 0 1 C C C A = 0 B B B @ 11 3 3 7 3 1 C C C A = 7 3 0 B B B @ 4 3 2 3 0 1 C C C A = x This leads us to the P o w er Method for nding the eigen v alues of a matrix. In our example w e found that for large enough r w e ha v e A r = 2 A r 1 No w, 10

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let x 0 be a v ector and dene x r = A x r 1 for r > 0, then x r = A r x 0 = 2 A r 1 x 0 = 2 x r 1 : In general, suppose that there is a k p > 0, and q where x k = q x k p : Let = q=p and let x = x k 1 ( x k 2 ) ( 2 x k 3 ) ( p 1 x k p ) : Then A x = A ( x k 1 ( x k 2 ) ( 2 x k 3 ) ( p 1 x k p )) = x k ( x k 1 ) ( 2 x k 2 ) ( p 1 x k p +1 )) = ( p x k p ( x k 1 ) ( 2 x k 2 ) ( p 1 x k p +1 )) = ( x k 1 ( x k 2 ) (( p 2) x k p +1 ) ( p 1 x k p )) = x : So is an eigen v alue with associated eigen v ector x In our example, w e had A 4 = 2 6 6 6 6 6 6 4 8 7 7 8 11 10 10 11 11 8 10 9 10 7 9 8 3 7 7 7 7 7 7 5 ; A 5 = 2 6 6 6 6 6 6 4 11 8 10 9 14 11 13 12 12 11 11 12 11 10 10 11 3 7 7 7 7 7 7 5 ; and A r = 4 A r 2 for all r 6 : 11

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Th us the eigen v alue is = 4 = 2 = 2 with associated eigen v ector x = 2 n ( A 4 ) 1 ( A 5 ) 1 = 0 B B B B B B @ 2 + 8 2 + 11 2 + 11 2 + 10 1 C C C C C C A 0 B B B B B B @ 11 14 12 11 1 C C C C C C A = 0 B B B B B B @ 10 13 12 11 1 C C C C C C A : Then A x = 2 6 6 6 6 6 6 4 8 0 2 1 7 3 3 6 4 5 5 6 3 3 7 9 3 7 7 7 7 7 7 5 0 B B B B B B @ 10 13 12 11 1 C C C C C C A = 0 B B B B B B @ 12 15 14 13 1 C C C C C C A = 2 0 B B B B B B @ 10 13 12 11 1 C C C C C C A : In the next c hapters, w e will look at sev eral application of Min-Plus algebra. In eac h application, the solution relies on periodicit y of po w ers of a matrix A In our example A r = 2 A r 2 for all r 6, in general w e ha v e A r = q A r p for all r R 0 The existence of R 0 depends on properties of the matrix A Denition 2.2 A matrix is irreducible if ther e is some K so that for all k K the matrix A k has no innity entries. This is analogous to the denition of irreducible matrices in regular algebra, P erkins [17] pro v es that for an irreducible matrix K m 2 m + 1 where m is the size of the matrix. This bound holds in the Min-Plus denition of irreducible. The follo wing theorem will be refered to throughout the thesis and w e will pro v e this theorem in Chapter 6. 12

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Theorem 2.3 If A is an irr e ducible matrix on m vertic es, then ther e exists a p; q and R 0 so that A r = q A r p for all r R 0 This sho ws that the po w er method for nding when A repeats, whic h giv es the eigen v alue of the matrix, will w ork if A is irreducible. In our example, the cost matrix A is irreducible and th us w e kno w that if w e compute the sequence A; A 2 ; A 3 ; : : :A K there will be some K so that t w o mem bers of this sequence dier b y a constan t. 2.3 Minim um Cycle Means and Karp's Theorem In the previous section, w e sho w ed that if A is an irreducible matrix, then there is some p q so that A r = q A r p W e will no w look use the properties of the en tries in A to nd the actual v alues of p and q Denition 2.4 The p recedence digraph of A denote d D A is the weighte d digr aph with no des v 1 ; v 2 ; : : :; v m with an ar c fr om v j to v i if A ij 6 = 1 and the weight of this ar c is A ij The pr e c e denc e digr aph is irr e ducible if the matrix A is irr e ducible which me ans that for some K ther e is a p ath of any length k K b etwe en every p air of vertic es in the pr e c e denc e digr aph. The precedence digraph for our example is sho wn belo w. 13

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t A 8 t B 3 t C 5 t D 9 6 7 5 ? 7 1 ? 0 3 6 6 3 4 2 @ @ @ @ @ @ @ @ @ @ @ I 3 @ @ @ @ @ @ @ @ @ @ @ R 6 Denition 2.5 The length of a p ath b etwe en two vertic es is the numb er of ar cs in the p ath, and is denote d j P j A cycle is a p ath that b e gins and ends with the same vertex but has no other r ep e ate d vertic es. The w eight of a p ath is the sum of the weights of its ar cs and is denote d w ( P ) Denition 2.6 The pr e c e denc e digr aph of an adjac ency matrix A is strongly connected if ther e is a p ath fr om every vertex v j to every other vertex v i The adjac ency matrix A is str ongly c onne cte d if its pr e c e denc e digr aph is. A ny irr e ducible matrix is also str ongly c onne cte d. The periodic propert y of a matrix is determined b y paths in the precedence digraph. W e will no w look at the minim um w eigh t paths in the precedence digraph. Theorem 2.7 L et A b e an m m matrix with pr e c e denc e digr aph D A Then the minimum weight of a p ath fr om vertex j to vertex i in D A of length k > 0 is ( A k ) i;j 14

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Pr o of: Since A i;j is the w eigh t of an arc from j to i in D A if it exists and 1 otherwise, the result holds for k = 1. F or k > 1, assume for induction purposes that ( A k 1 ) i;j is minim um w eigh t path from j to i of length k 1. Then ( A k ) i;j = min ( A k 1 ) i; 1 + A 1 ;j ; ( A k 1 ) i; 2 + A 2 ;j ; : : :; ( A k 1 ) i;m + A m;j : This is the minim um of all paths of length k 1 from h to i plus the minim um w eigh t path of length 1 from j to h This giv es the minim um w eigh t of a path of length k from j to i 2 Let A + = A A 2 A 3 : : : A n 1 A n A n +1 : : :; and A = I A A 2 A 3 : : : A n 1 A n A n +1 : : :: Then ( A + ) ij is the minim um w eigh t path of an y length from j to i and ( A ) ij is the minim um w eigh t path of an y length from j to i allo wing for the possibilit y of sta ying on a v ertex. If A is irreducible, then A and A + ha v e no innit y en tries. In our example, the w eigh t of a path of length r represen ts the cost of starting in cit y j tra v eling for r nigh ts and ending in cit y i The eigen v alue sho ws us that the cost to sta y r nigh ts is $2 more than the cost to sta y r 2 nigh ts, but this does not sho w us ho w to tra v el at this cost. If w e look at the 15

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precedence digraph, w e see that there is a cycle from cit y A to cit y D and bac k to cit y A whic h has w eigh t 4 and length 2. Th us w e m ust use this cycle to sta y 2 more nigh ts and only pa y $4. This leads us to the next method for nding the eigen v alues of a matrix b y looking at the cycles of the matrix. Denition 2.8 The cycle mean of a cycle is the sum of the weights of the ar cs divide d by the numb er of ar cs in the cycle. F or a cycle c the cycle me an is denote d ( c ) = w ( c ) j c j The minimum cycle mean of a gr aph is the smallest value for ( c ) over all cycles in the gr aph. A ll cycles with cycle me an e qual to the minimum cycle me an ar e c alle d critical cycles Theorem 2.9 If D A is str ongly c onne cte d, ther e exists exactly one eigenvalue which solves the e quation A x = x for some x 6 = 1 This eigenvalue is e qual to the minimum cycle me an of the gr aph. = min c ( c ) = min c w ( c ) j c j wher e c r anges over all cycles in the gr aph. Pr o of: Existence of x and Consider matrix B = A where = min c ( c ) = min c w ( c ) j c j and let D B be the precedence digraph of B The minim um cycle w eigh t of D B is 0, th us the matrices B and B + = BB exist and B + has some columns with diagonal en tries equal to 0. Suppose a v ertex k is in some minim um cycle of B then the minim um w eigh t of paths from k to k is 0. Therefore, w e ha v e 0 = ( B + ) kk Let B k denote the k th column 16

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of B Then, since B = A B + = BB and B = I B + for a giv en k w e ha v e, ( B + ) k = ( B ) k ) ( BB ) k = ( B + ) k = ( B ) k ) A ( B ) k = ( B ) k ) A ( B ) k = ( B ) k : Hence x = ( B ) k is an eigen v ector of A corresponding to the eigen v alue Uniqueness of Let A be an strongly connected matrix with eigen v alues 1 2 with associated eigen v ectors v 1 and v 2 Let 1 be the v ector of all ones and pic k t large enough so that t 1 + v 1 v 2 = ) t v 1 v 2 : Then ( t v 1 ) v 2 = t v 1 = ) A r (( t v 1 ) v 2 ) = A r ( t v 1 ) = ) tA r v 1 A r v 2 = tA r v 1 = ) ( t r 1 ) v 1 r 2 v 2 = ( t r 1 ) v 1 : = ) min (( t + r 1 ) 1 + v 1 ; r 2 1 + v 2 ) = ( t + r 1 ) 1 + v 1 : If 1 < 2 then r 2 1 + v 2 6 ( t + r 1 ) 1 + v 1 for some r Th us 1 = 2 : 2 Since an y irreducible matrix is strongly connected, w e also ha v e that the minim um cycle mean of a irreducible matrix is its eigen v alue. 17

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Next w e look at one method of nding the minim um cycle mean using the follo wing theorem whic h is analogous to a theorem b y Karp whic h nds the maxim um cycle mean but not the critical cycles. Theorem 2.10 L et A b e an m m matrix with c orr esp onding pr e c e denc e digr aph D A Then, for any j the minimum cycle me an ( A ) is ( A ) = min i =1 ::: m max k =1 ::: m ( A m ) ij ( A k ) ij m k ; wher e A m A k ar e c ompute d in Min-Plus algebr a and the other c omputations ar e c onventional. Pr o of: Without loss of generalit y assume that D A is strongly connected. If D A is not strongly connected, then the theorem holds componen t wise. Also, the index j is arbitrary and the computation of ( A ) is independen t of j First assume that the minim um cycle mean is 0. Then w e m ust sho w min i =1 ::: m max k =0 ::: m 1 ( A m ) ij ( A k ) ij m k = 0 : Since ( A ) = 0, there exists a cycle of w eigh t 0 and there are no cycles with negativ e w eigh t. Because there are no cycles with negativ e w eigh t, there exists a minim um w eigh t path of length k from v ertex i to v ertex j When k m the path w ould con tain a cycle of non-negativ e w eigh t. Th us a minim um w eigh t path can be found b y restricting k < m This path is dened as ( A + ) ij = min k =0 ::: m 1 ( A k ) ij : 18

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Also, ( A m ) ij ( A + ) ij and hence ( A m ) ij ( A + ) ij = max k =0 ::: m 1 h ( A m ) ij ( A k ) ij i 0 : Equiv alen tly max k =0 ::: m 1 ( A m ) ij ( A k ) ij m k 0 : Equalit y holds only if ( A m ) ij = ( A + ) ij No w w e will sho w that there is an index for i where this is true. Let C be a cycle of w eigh t 0 and let l be a v ertex of C Let P lj be a path from l to j with minim um w eigh t w ( P lj ) = ( A + ) lj No w this path is extended b y appending to it a n um ber of repetitions of C suc h that the total length of this extended path denoted P e is greater than or equal to m This is again a path of minim um w eigh t from j to l No w consider the path consisting of the rst m arcs of P e Its initial v ertex is j and denote its nal v ertex i Note that i is in C Since an y subpath of a minim um w eigh t path is also of minim um w eigh t, the path from j to i is of minim um w eigh t. Therefore ( A m ) ji = ( A + ) ji and min i =1 ::: m max k =0 ::: m 1 ( A m ) ij ( A k ) ij m k = 0 : This completes the case where ( A ) = 0. No w consider an arbitrary nite ( A ). A constan t c is no w subtracted from eac h w eigh t in A ij Then clearly ( A ) will be c hanged c units and eac h 19

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en try in A k will be c hanged b y kc units. Th us min i =1 ::: n max k =0 ::: m 1 ( A m ) ij ( A k ) ij m k : is c hanged b y c and both sides of the equation are aected equally when all w eigh ts A ij are c hanged b y the same amoun t. No w c hoose c so that ( A ) c = 0 and w e are bac k to the original case. 2 Lik e the po w er method, Karp's theorem giv es the minim um cycle mean but not the critical cycles. T o nd the critical cycles, bac ktrac king m ust be used in both of these methods. There are man y applications of Min-Plus algebra in whic h the solution is found using eigen v alues. In practice, the po w er method is a faster w a y to nd the eigen v alues of an irreducible matrix, although this is not true for all matrices as w e will see in Chapter 6. In the next three Chapters w e presen t applications of Min-Plus algebra whic h in v olv e nding the domination n um bers of certain graphs. 20

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3. Domination of Cartesian Product Graphs In this c hapter, w e will use t w o dynamic programming algorithms in MinPlus algebra to nd the domination n um bers of a large class of graphs. Denition 3.1 L et G = ( V; E ) b e a gr aph with vertic es f v 1 ; v 2 ; : : :v n g and let H = ( V; E ) b e a gr aph with vertic es f w 1 ; w 2 : : :w ` g Then the ca rtesian p roduct G H has vertic es of the form ( v i ; w j ) with 1 i n and 1 j ` The e dges of G H have the form (( v i ; w j ) ; ( v p ; w q )) if i = p and ( w j ; w q ) is and e dge of H or if j = q and ( v i ; v p ) is an e dge of G The k th c opy of G in G H is G k and its vertic es ar e ( v 1 ; w k ) ; ( v 2 ; w k ) ; : : : ( v n ; w k ) T o in troduce the algorithms, w e consider complete grid graphs whic h are cartesian product graphs with H = P r the path on r v ertices, and G = P n the path on n v ertices. The domination n um ber of P n P r is denoted r ( P n P r ). qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq qqqqqqqqqqqqq ssss ss ssss ss ssss ss ssss ss ssss ss ssss ss ssss Figure 3.1 Minim um Dominations of P 13 P 13 Abo v e is a domination of P 13 P 13 using 40 v ertices. Since Theorem 3.12 sho ws that r 13 ; 13 = 40, no domination has few er v ertices. 21

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Considerable w ork has been directed to w ard nding r ( P n P r ). Jacobson and Kinc h [13] sho w ed: r 1 ;r = r 3 ; r 2 ;r = r + 1 2 ; r 3 ;r = 3 r + 1 4 ; and r 4 ;r = 8 < : r + 1 if r = 1 ; 2 ; 3 ; 5 ; 6 ; 9 r otherwise. (1) Chang and Clark [3] sho w ed that r 5 ;r = 8 > > < > > : l 6 r +4 5 m 1 if r = 2 ; 3 ; 7 l 6 r +4 5 m otherwise and r 6 ;r = 8 > > < > > : l 10 r +4 7 m + 1 if r mod 7 = 3 and r 6 = 3 l 10 r +4 7 m otherwise. (2) Clark, Colbourn and Johnson [6] sho w ed that nding the domination n umber of a grid graph (a subgraph of P n P r ) is NP-hard. All kno wn algorithms for nding r ( P n P r ) could be adapted to nd the domination n um ber of a grid graph. So it is not surprising that the time needed for these algorithms is not polynomial in nr Nev ertheless, profound dierences exist in the speed of these algorithms. P erhaps the most natural non-trivial algorithms for nding r ( P n P r ) are \Branc h and Bound" algorithms. Before trying a dieren t approac h, Hare, Hedetniemi, and Hare [11 ] used a branc h and bound algorithm that took 20 hours to nd r ( P 7 P 7 ) = 12. Ma and Lam [15] ga v e a more complicated branc h and bound algorithm whic h they used to sho w that r ( P 8 P 8 ) = 16, 22

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r ( P 8 P 9 ) 17, and r ( P 9 P 9 ) 19. Ho w ev er since the time needed for these algorithms seem to be exponen tial in both n and r they are limited to modest v alues of n and r Muc h superior are dynamic programming algorithms for nding r ( P n P r ) in troduced b y Hare, Hedetniemi and Hare [11]. While these algorithms are exponen tial in n they are linear in r In section 3.1, w e presen t this algorithm of Hare, Hedetniemi and Hare in whic h they found r ( P n P r ) for n 8 and r 500, and then used these results to mak e conjectures for r ( P n P r ) for all v alues of r Fisher (priv ate comm unications) expanded on this algorithm b y stating it in terms of Min-Plus algebra and b y looking for periodicit y in the algorithm. When periodicit y is detected, induction giv es r ( P n P r ) for a giv en n and for all r Th us r ( P n P r ) is calculated for an innite n um ber of v alues in a nite amoun t of time. Using this expanded algorithm, whic h w e will call the Exact Algo rithm Fisher w as able to v erify the conjectures of Hare, Hedetniemi and Hare for all v alues of r While it does not seem possible to nd an algorithm that is not exponen tial in n the Exact Algorithm can be modied to reduce the base of the exponen t. Section 3.2 describes this modied algorithm whic h will be called the A t L e ast A lgorithm Using the A t Least Algorithm and periodicit y w e nd form ulas for r ( P n P r ) for all n 19. 23

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3.1 The Exact Algorithm T o illustrate this algorithm, w e will look at the domination of P 3 P r Let S be a set of v ertices of P 3 P r An y v ertex in this graph is either in S adjacen t to a v ertex in S or not adjacen t to an y v ertices in S These v ertices will be labeled 0, 1, and 2 respectiv ely No w, an y column of P 3 P r is a cop y of P 3 and its v ertices can be labeled either 0,1,or 2. A labeling of P 3 whic h can occur in a dominating set S of P 3 P r will be called a state Since no state can con tain v ertices labeled 0 adjacen t to v ertices labeled 2, there are 17 possible states for P 3 These states are sho wn belo w. s 1 = 000 s 7 = 110 s 13 = 211 s 2 = 001 s 8 = 012 s 14 = 122 s 3 = 010 s 9 = 111 s 15 = 212 s 4 = 100 s 10 = 210 s 16 = 221 s 5 = 011 s 11 = 112 s 17 = 222 s 6 = 101 s 12 = 121 T o nd the domination n um ber of P 3 P r +1 w e will add a column on to dominating sets of P 3 P r If column r is in state s j then it is possible for column r + 1 to be in state s i if the follo wing conditions hold for all v ertices p where s j ( p )=the p th v ertex of s j If s j ( p ) = 0, then s i ( p ) 6 = 2. If s j ( p ) = 2, then s i ( p ) = 0. If s i ( p ) = 1, then s j ( p ) = 2 if some v ertex adjacen t to p in s j has v alue 0, and s j ( p ) = 0 otherwise. 24

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Using these rules, w e form the state transition matrix A where A ij = the n um ber of v ertices with v alue 0 in state s i if s i can follo w s j and innit y (denoted b y ) otherwise. This matrix is sho wn belo w. 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 Next, w e form a states vecto r x r where x r ( s i ) is the minim um n um ber of v ertices needed to dominate P 3 P r and ha v e the r th column be in state s i If it is not possible to dominate P 3 P r and ha v e the r th column in state s i then x r ( s i ) = 1 Th us x T 1 = h 3 2 2 2 1 1 1 0 i The innit y ( ) en tries in x 1 correspond to states in whic h s i ( p ) = 1 and whic h m ust be preceded b y a state with s j ( p ) = 0 in column 0, whic h does not 25

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exist. T o nd x 2 ( s i ) w e consider all labelings of the rst column of P 3 P 2 when the second column is in state s i These labelings correspond to the states s j for whic h A ij 6 = 1 F rom all these possible labelings of the rst column, the one with the minim um n um ber of v ertices with v alue 0 will giv e a minimal domination of P 3 P 2 with column 2 in state s i Th us x 2 ( s i ) = min ( P j A ij + x 1 ( s j )). Stating this in terms of Min-Plus Algebra, x 2 = A x 1 and in general x r +1 = A x r Repeating this process w e ha v e a recursiv e algorithm to nd x r where x r ( s i ) is the n um ber of v ertices needed to dominate P 3 P r and lea v e the r th column in state s i T o nd the domination n um ber of P 3 P r w e nd the minim um en try in x r for whic h the state lea v es the r th column dominated. This is done b y forming an ending v ector y where y ( i ) = 0 if state s i lea v es the r th column dominated (these are all the states with no v ertex labeled 2) and y ( i ) = 1 otherwise. The domination n um ber for P 3 P r is x r y T The ending v ector y and the states v ectors x 1 through x 11 for P 3 P r are sho wn belo w. 26

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y = [ 0 0 0 0 0 0 0 0 ] x 1 = 0 [ 3 2 2 2 1 1 1 0 ] x 2 = 0 [ 3 3 3 3 2 2 2 2 3 2 2 2 2 1 ] x 3 = 2 [ 2 2 1 2 1 1 1 1 1 1 1 1 1 0 0 0 1 ] x 4 = 3 [ 2 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 ] x 5 = 4 [ 2 1 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0 ] x 6 = 4 [ 3 2 2 2 2 1 2 1 2 1 1 1 1 1 0 1 1 ] x 7 = 5 [ 2 2 1 2 1 1 1 1 2 1 1 1 1 1 0 1 1 ] x 8 = 6 [ 2 2 1 2 1 1 1 1 1 1 1 0 1 0 0 0 1 ] x 9 = 7 [ 2 1 1 1 1 0 1 1 1 1 1 0 1 0 0 0 0 ] x 10 = 7 [ 3 2 2 2 2 1 2 1 2 1 1 1 1 1 0 1 1 ] x 11 = 8 [ 2 2 1 2 1 1 1 1 2 1 1 1 1 1 0 1 1 ] Notice that x 10 = 3 x 6 and x 11 = 3 x 7 Theorem 3.2 x r +4 = 3 x r for all r 6 Pr o of: By Induction. The base case is x 10 = 3 x 6 or A x 9 = 3 x 6 Then for an y r > 6 assume A x r +2 = 3 x r 1 Then A x r +2 x = x r 1 x and x r +4 = x r 2 Theorem 3.3 r ( P 3 P r +4 ) = 3 + r ( P 3 P r ) for all r 6 Pr o of: r ( P 3 P r +4 ) = y T x r +4 = y T (3 x r ) = 3 ( y T x r ) = 3 + r ( P 3 P r ) 2 Using the recursiv e denition abo v e, w e can nd a closed form for r ( P 3 P r ) 27

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for all r 6. r ( P 3 P r ) = 3 r + 1 4 : Finding the domination n um ber of P 3 P r for all r relies on the periodicit y of the dynamic programming algorithm. According to Theorem 2.3, periodicit y will occur if the precedence digraph of the state transition matrix is irreducible. The possible states for P k P r can be partitioned in to three groups. Let A be the set con taining the state with all v ertices labeled 0 =00 : : : 0, B be the set con taining all states with v ertices labeled 0 or 1 but not 2, and let C be all remaining states whic h ha v e at least one v ertex labeled 2. Theorem 3.4 The pr e c e denc e digr aph of the state tr ansition matrix for nding the domination numb er of P k P r is irr e ducible. Pr o of: Using the rules of state transition all states can be follo w ed b y the states of all zeros. All states in set B can follo w the zero state and for ev ery state in set C there is at least one state in set B that it can follo w. Th us there is a path of length 3 or less from an y state to an y other state in the precedence digraph b y going through the zero state. This path can then be extended b y repeating the zero state. Th us there is a path of an y length k 3 bet w een an y pair of v ertices and the precedence digraph is irreducible. 2 F or an y n the exact algorithm can be used to nd the domination n um ber 28

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of P n P r The algorithm will reac h periodicit y after some n um ber of iterations allo wing us to nd a closed form for the domination n um ber of P n P r for all r The limiting feature of all dynamic programming algorithms is computation time. F or the exact algorithm on P n P r there are O ((1 + p 2) n ) states eac h of whic h matc hes to at most 2 n other states and eac h matc h tak es O ( n ) computations to nd. Th us the computational complexit y for nding the domination n um ber of P n P r is O ( nr (2 + 2 p 2) n ), and the maxim um size of n is greatly limited b y computational complexit y In the next section, w e will look at another dynamic programming algorithm whic h greatly reduces this complexit y Although this algorithm is similar to the exact algorithm, it is not a direct descendan t. 3.2 The A t Least Algorithm In this algorithm, w e will reduce the computational complexit y b y redening what it is that w e are computing. Fisher (priv ate comm unications) used this new algorithm to nd r ( P n P r ) for n 19 and for all r These results are presen ted belo w. In section 3.4 w e will expand this algorithm to nd r ( G P r ) for an y graph G on 6 or less v ertices. Denition 3.5 A c olumn of P n P r is in at least state s i if it is in state s i or any state s j which has s i ( p ) s j ( p ) for all p 29

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This means that s j dominates at least the same set of v ertices that s i does and possibly more. Denition 3.6 A t-domination of P n P r is a set of vertic es that dominates the rst r 1 c olumns of P n P r and le aves the last c olumn in at le ast state s i T o illustrate the algorithm, w e will nd the least-domination n um ber of P 3 P r As in the exact algorithm, there are sev en teen states. These states form a partial ordering whic h is illustrated in Figure 3.2. r r r r r r r r r r r r r r r r r 222 122 212 221 112 121 211 012 111 210 011 101 110 001 010 100 000 Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Figure 3.2 The partial ordering of the 17 states for P 3 P r State s i is at least state s j if state s j is an ancestor of s i in the partial ordering. Next, w e form the states v ector t r where t r ( s i ) is the n um ber of v ertices in a dominating set for P 3 P r with column r in at least state s i Since a state is in at least s i if it is in exactly s i or exactly s j for all s j > s i the en tries in t r can be dened using the states v ector x r with x r ( s i ) represen ting 30

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the n um ber of v ertices in a domination of P 3 P r with column r in exactly s i as described in the Exact Algorithm. F or example t r ( s 7 ) = min ( x r ( s 7 ) ; x r ( s 4 ) ; x r ( s 3 ) ; x r ( s 1 )) t r ( s 10 ) = min ( x r ( s 10 ) ; x r ( s 7 ) ; x r ( s 4 ) ; x r ( s 3 ) ; x r ( s 1 )) No w the last 4 en tries for t r ( s 10 ) are the same as t r ( s 7 ) so w e ha v e t r ( s 10 ) = min ( x r ( s 10 ) ; t r ( s 7 )) Next, if the r th column is in exactly state s i then the en try in x r ( s i ) represen ts the n um ber of v ertices needed to dominate P 3 P r 1 and lea v e the ( r 1) st column in the minim um previous state that can precede state s i plus the n um ber of v ertices that are labeled 0 in s i Denition 3.7 Given a state s i the minimum p revious state M ( s i ) is dene d as M ( s i ( p )) = 8 > > > < > > > : 2 if s i ( p ) = 0 0 if s i ( p ) = 1 and no neighb ors of p in s i = 0 1 otherwise The minim um previous state for s 10 = 210 is s 11 = 112 and s 10 has one v ertex in the dominating set, so x r ( s 10 ) = 1 + t r 1 ( s 11 ) and, t r ( s 10 ) = min (1 + t r 1 ( s 11 ) ; t r ( s 7 )) : In general, t r ( i ) = min ( j s i j + t r 1 ( M ( s i )) ; min s j s i t r ( s j )) 31

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Completing a similar computation for eac h state, w e form t w o matrices. The at least matrix A has A ij = 0 if state s i is at least state s j and there is no state s k where s i is at least s k and s k is at least s j and A ij = 1 otherwise. This matrix is lo w er triangular because of the partial ordering of the states. The second matrix is the minimum p revious matrix B with B ij = the n um ber of v ertices in the dominating set for state s i if state s j is the minim um previous state of s i and 1 otherwise. The matrices A and B for the at least domination of P 3 P r are sho wn belo w. A = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 32

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B = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 3 2 2 2 1 1 1 1 0 1 0 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 No w, in terms of min plus algebra, t r = A t r + B t r 1 : T o nd t r w e iterativ ely m ultiply b y B and A one ro w at a time, updating t r after eac h new en try is found. This is possible since A is lo w er triangular. T o nd the domination n um ber of P 3 P r w e need the minim um of the en tries in t r whic h correspond to states that dominate the last column. All states con taining no v ertices labeled 2 dominate the last column, and the state of all 1's denoted 1 is at least all of these states so the ending v ector y has a 0 en try for the 1 state and innit y en tries for all other states. F or P 3 P r 1 is state s 9 so r ( P 3 P r ) = t r ( s 9 ) y T and in general r ( P k P r ) = t r ( 1 y T ) No w computing t r iterativ ely w e nd the states v ectors for the t-domination of P 3 P r sho wn belo w. 33

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y = [ 0 ] t 1 = 0 [ 3 2 2 2 2 1 2 1 1 1 1 1 1 1 1 1 0 ] t 2 = 0 [ 3 3 3 3 2 2 2 2 2 2 2 2 2 2 1 2 1 ] t 3 = 2 [ 2 2 1 2 1 1 1 1 1 1 1 1 1 0 0 0 0 ] t 4 = 3 [ 2 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 ] t 5 = 4 [ 2 1 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 ] t 6 = 4 [ 3 2 2 2 2 1 2 1 1 1 1 1 1 1 0 1 0 ] t 7 = 5 [ 2 2 1 2 1 1 1 1 1 1 1 1 1 1 0 1 0 ] t 8 = 6 [ 2 2 1 2 1 1 1 1 1 1 1 0 1 0 0 0 0 ] t 9 = 7 [ 2 1 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 ] t 10 = 7 [ 3 2 2 2 2 1 2 1 1 1 1 1 1 1 0 1 0 ] t 11 = 8 [ 2 2 1 2 1 1 1 1 1 1 1 1 1 1 0 1 0 ] Notice that t 9 = 3 t 5 and t 10 = 3 t 6 and th us b y induction t r = 3 t r 4 for all r 5 and w e ha v e the domination n um ber for innite r after 9 iterations of the algorithm. W e no w need to sho w that periodicit y in the A t Least Algorithm will alw a ys occur. Lemma 3.8 If s i is a dir e ct anc estor of s j in the p artial or dering (so s i s j and ther e is no s k with s i s k s j ) then t r ( s i ) t r ( s j ) is either 0 or 1. Pr o of: s i and s j ha v e at most one v ertex with a dieren t label. If s i ( p ) = 0 and s j ( p ) 6 = 0, then an y domination ending in state s j can be c hanged in to a domination ending in s i b y c hanging s j ( p ) to 0 and this increases t r ( s j ) b y 1. Otherwise t r ( s i ) = t r ( s j ). 2 Denition 3.9 A state s j is maximal for a given r if t r ( s j ) < t r ( s i ) whenever s i is a dir e ct anc estor of s j in the p artial or dering. L et R n;r b e the set of all 34

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maximal states for P n P r Using the states v ectors t 1 : : : t 11 sho wn abo v e, w e ha v e R 3 ; 1 = f s 1 ; s 2 ; s 3 ; s 4 ; s 6 ; s 8 ; s 9 ; s 10 ; s 17 g R 3 ; 2 = f s 1 ; s 5 ; s 6 ; s 14 g R 3 ; 3 = f s 1 ; s 3 ; s 6 ; s 14 ; s 15 ; s 16 g R 3 ; 4 = f s 1 ; s 2 ; s 3 ; s 4 ; s 12 ; s 15 g R 3 ; 5 = f s 1 ; s 2 ; s 3 ; s 4 ; s 6 g . . . R 3 ; 9 = f s 1 ; s 2 ; s 3 ; s 4 ; s 6 g Lemma 3.10 Ther e exists p ositive inte gers j; k; n so that t k ( s i ) t j ( s i ) = q for all states s i if and only if R n;j = R n;k Pr o of: If t k ( s i ) t j ( s i ) = q then certainly the maximal states for j and k m ust be the same and R n;j = R n;k Next, for all s i ; s j 2 R n;k Lemma 3.8 states that t k ( s i ) = t k ( s j ) 1 for all s j > s i Let 2 be the state with all v ertices labeled 2, then for an y s i 6 = 2 there is a sequence of states 2 = s 1 < s 2 < : : : s h < s i Let N ( s i ) be the states in this sequence, N ( s i ) = f s 1 ; s 2 ; : : :s h g No w the v alue of t k increases b y one eac h time a maximal state occurs in the sequence and the n um ber of maximal states is N ( s i ) T R n;k Th us t k ( s i ) = t k ( 2 ) + j N ( s i ) \ R n;k j ; and the domination n um ber of an y state can be constructed using this sequence. 35

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No w if R n;j = R n;k then let q = t k ( 2 ) t j ( 2 ) and for all states s i w e ha v e t k ( s i ) = t k ( 2 ) + j N ( s i ) \ R n;k j = t k ( 2 ) + j N ( s i ) \ R n;j j = t k ( 2 ) + t j ( s i ) t j ( 2 ) = q + t j ( s i ) 2 Lemma 3.11 L et C m b e the numb er of states, then ther e exists 0 < j < k < 2 C m + 1 wher e R n;j = R n;k Pr o of: The en tries in R n;k are a subset of the states so there are at most 2 C m possibilities for R n;k The pigeonhole principle ensures that R n;j = R n;k for some 0 < j < k < 2 C m + 1. 2 No w b y Lemma 3.10 and Lemma 3.11 periodicit y is guaran teed for the A t Least Algorithm. Th us, the A t Least Algorithm allo ws us to compute the domination n um ber for P n P r for all r using the periodicit y of the algorithm and w e ha v e greatly reduced the n um ber of computations. F or eac h state there are at most n en tries in the at least matrix and one en try in the minim um previous matrix, reducing the computational complexit y to O ( nr (1 + p 2 ) n ). 3.3 F orm ulas for the Domination Num ber A program implemen ting the A t Least algorithm v eried (1) and (2) in about a second. It took about a w eek on a V ax 8800 to nd form ulas for r ( P n P r ) for all n 19. Since r ( P n P r ) = r ( P r P n ), this giv es r ( P n P r ) 36

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for all n and r with min ( n; r ) 19. Theorem 3.12 F or all n r with n 19 we have r ( P n P r ) = 8 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > : l r 3 m if n = 1 l r +1 2 m if n = 2 l 3 r +1 4 m if n = 3 r if n = 4 and r 6 = 5 ; 6 ; 9 r + 1 if n = 4 and r = 5 ; 6 ; 9 l 6 r +4 5 m if n = 5 and r 6 = 7 l 6 r +4 5 m 1 if n = 5 and r = 7 l 10 r +4 7 m if n = 6 l 5 r +1 3 m if n = 7 l 15 r +7 8 m if n = 8 l 23 r +10 11 m if n = 9 l 30 r +15 13 m if n = 10 and r mo d 13 6 = 10 and r 6 = 13 ; 16 l 30 r +15 13 m 1 if n = 10 and r mo d 13 = 10 or r = 13 ; 16 l 38 r +22 15 m if n = 11 and r 6 = 11 ; 18 ; 20 ; 22 ; 33 l 38 r +22 15 m 1 if n = 11 and r = 11 ; 18 ; 20 ; 22 ; 33 l 80 r +38 29 m if n = 12 l 98 r +54 33 m if n = 13 and r mo d 33 6 = 13 ; 16 ; 18 ; 19 l 98 r +54 33 m 1 if n = 13 and r mo d 33 = 13 ; 16 ; 18 ; 19 l 35 r +20 11 m if n = 14 and r mo d 22 6 = 7 l 35 r +20 11 m 1 if n = 14 and r mo d 22 = 7 l 44 r +28 13 m if n = 15 and r mo d 26 6 = 5 l 44 r +28 13 m 1 if n = 15 and r mo d 26 = 5 j ( n +2)( r +2) 5 k 4 if n = 16 ; 17 ; 18 ; 19 Ev en with Theorem 3.12, it is dicult to nd the pattern that dominates the n r grid for all r using the minim um n um ber of v ertices. Jacobson and Kinc h [13] found minim um domination patterns of 1 r 2 r 3 r and 4 r 37

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grids. Chang, Clark and Hare [4] use the algorithm of Hare, Hedetniemi and Hare to generate minim um domination patterns of 5 r and 6 r grids, and what they conjectured to be minim um dominations of 7 r 8 r 9 r and 10 r grids for all r Theorem 3.12 sho ws these dominations are minim um. They also found form ulas for r 11 ;r when r 122 and for r 12 ;r when r 33 (the form ula for r 11 ;r holds for all r while the form ula for r 12 ;r does not). The program in this paper w as modied to store bac ktrac king information to giv e minim um dominations of 11 r 12 r 13 r 14 r and 15 r grids for all r up to a certain poin t. These w ere used to nd minim um dominations for n r grids for a giv en n and for all r (this is actually quite dicult because only one of what is usually man y minim um dominations is generated). Chang, Fisher and Hare [5 ] report these minim um dominations. F or n; r 8, Chang [2] giv es a construction whic h dominates an n r grid with b ( n + 2)( r + 2) = 5 c 4 v ertices. This is a generalization of a construction of Coc k a yne, Hare, Hedetniemi and Wimer [7] for r r grids. Theorem 3.12 sho ws this construction is minim um for 16 r 17 r 18 r and 19 r grids when r 16. This causes one to suspect the follo wing conjecture (originally due to Chang [2]) migh t be true. Conjecture 3.13 F or all n; r 16 r ( P n P r ) = $ ( n + 2)( r + 2) 5 % 4 : 38

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If Conjecture 1 is true, dominating an n r grid w ould be analogous to 2-pac king (Fisher [8 ]) in that a simple form ula holds for n and r large enough. F urther, it w ould answ er a question of Hedetniemi and Lask ar [12]. In commen ting on [6], they ask ed is ther e a p olynomial algorithm for nding r ( P n P r ) ? Conjecture 3.13 together with Theorem 3.12 w ould giv e an armativ e answ er to the question. 3.4 Domination of G P r : The A t Least Algorithm W e will no w use the A t Least Algorithm to nd the domination n um ber of G P r where G is an y graph on n 6 v ertices. Livingstone and Stout [14] dev elop a dynamic programming algorithm for nding r ( G P r ) using an algorithm that is similar to the Exact Algorithm. W e are able to signican tly reduce the complexit y of this algorithm b y stating in terms of the A t Least Algorithm. T o illustrate the A t Least Algorithm for r ( G P r ) w e will let G be the house graph on v e v ertices as sho wn in Figure 3.3. 39

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rr r r r v 4 v 5 v 2 v 3 v 1 A A A A rr r rr rr r r r rr r r r rr r r r ( ( ( ( ( ( ( ( ( ( ( ( A A A A ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( A A A A ( ( ( ( ( ( ( ( ( ( ( ( A A A A ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( A A A A Figure 3.3 The house graph on v e v ertices G and G P 4 There are 77 possible states for the domination of the house graph on v e v ertices. These states are sho wn belo w and the v ertices are ordered v 1 : : :v 5 as labeled in Figure 3.3. s 1 = 00000 s 21 = 01221 s 41 = 11121 s 61 = 21111 s 2 = 00001 s 22 = 10000 s 42 = 11122 s 62 = 21112 s 3 = 00010 s 23 = 10001 s 43 = 11210 s 63 = 21121 s 4 = 00011 s 24 = 10010 s 44 = 11211 s 64 = 21122 s 5 = 00100 s 25 = 10011 s 45 = 11212 s 65 = 21211 s 6 = 00101 s 26 = 10101 s 46 = 11221 s 66 = 21212 s 7 = 00110 s 27 = 10101 s 47 = 11222 s 67 = 21221 s 8 = 00111 s 28 = 10110 s 48 = 12101 s 68 = 21222 s 9 = 00121 s 29 = 10111 s 49 = 12111 s 69 = 22101 s 10 = 01000 s 30 = 10121 s 50 = 12112 s 70 = 22111 s 11 = 01001 s 31 = 11000 s 51 = 12121 s 71 = 22112 s 12 = 01010 s 32 = 11001 s 52 = 12122 s 72 = 22121 s 13 = 01011 s 33 = 11010 s 53 = 12211 s 73 = 22122 s 14 = 01100 s 34 = 11011 s 54 = 12212 s 74 = 22211 s 15 = 01101 s 35 = 11012 s 55 = 12221 s 75 = 22212 s 16 = 01110 s 36 = 11100 s 56 = 12222 s 76 = 22221 s 17 = 01111 s 37 = 11101 s 57 = 21001 s 77 = 22222 s 18 = 01121 s 38 = 11110 s 58 = 21011 s 19 = 01210 s 39 = 11111 s 59 = 21012 s 20 = 01211 s 40 = 11112 s 60 = 21101 These states form a partial ordering w ere s i follo ws s j in the ordering if s i is at least s j whic h means the labelings s j ( p ) < s i ( p ) for the corresponding 40

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v ertices p in eac h state. The partial ordering is used to form the A t Least Matrix A and the Minim um Previous Matrix B as described for P n P r Next, w e form the states v ector t r where t r ( s i ) is the n um ber of v ertices needed to dominate G P r and lea v e the last cop y of G in at least state s i and the ending v ector y whic h has a 0 for the state 1 and innit y otherwise. F or this example, 1 is state s 39 The v ectors for the house graph are sho wn belo w. y = [ --------------------------------------0-------------------------------------] t 1 = 0 [ 00000000-00000000----00000000-0000-0000-------------------------------------] t 2 = 2 [ 21111110111111110010011111100011111110000000011100010112100001010112011121223 ] t 3 = 3 [ 11111111111111111111010101000000000000000000001001010112000000010112001121222 ] . . . . . . . . . . t 18 = 17 [ 00000000000000000000100000000000000000001101112001121223001011121222112232323 ] t 19 = 18 [ 11000000010000000000100000000000000000001101112001121223001011121223112232334 ] t 20 = 19 [ 11000000010000000000100000000000000000001101112001121223001011121222112232323 ] t 21 = 20 [ 11000000010000000000100000000000000000001101112001121223001011121222112232323 ] t 22 = 21 [ 11000000010000000000100000000000000000001101112001121223001011121222112232323 ] t 23 = 22 [ 11000000010000000000100000000000000000001101112001121223001011121222112232323 ] Notice that t 20 = 1 t 19 and t 21 = 1 t 20 Th us after 20 iterations, periodicit y occurs and w e ha v e t r = 1 t r 1 for all r 20 : This allo ws us to nd a closed form for the domination n um ber of the house graph G F or all r 1 ; w e ha v e r ( G P r ) = r + 8 < : 1 if r =1, 2, 3, 4, 6, 7, 8, 11, 12, 16 0 otherwise. No w to nd the domination n um ber of G P r rst note that if G is disconnected then so is G P r Th us, if G 1 G 2 ; : : : are the componen ts of G then the domination n um ber of G is just the sum of the domination n um bers of 41

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its componen ts. Th us w e only need to nd the domination n um ber of connected graphs. T o further reduce the n um ber of graphs w e need to consider, the follo wing theorem giv es the domination n um ber for certain graphs with high degree. Theorem 3.14 If a gr aph on n vertic es G has a set of vertic es a i 2 A with deg ( a i ) n 3 and for e ach a i 2 A all vertic es not in the close d neighb orho o d of a i ar e also in A then r ( G P r ) = r or r + 1 or r + 2 Pr o of: Let a 1 ; a 2 ; : : : be the v ertices in set A as described abo v e. T o dominate ( G P r ) start with the k th cop y of G = G k where k 6 = 1 ; r Place a ik in the dominating set. No w there are at most t w o undominated v ertices in G both of whic h are also in A call these v ertices a jk and a lk Place a jk 1 and a lk +1 in the dominating set. No w a hk 1 is adjacen t to a hk b y denition of G P r so all the v ertices in G k are no w dominated. Next w e ha v e that in G k 1 a ik 1 and the closed neigh borhood of a jk 1 are dominated and only one v ertex a qk 1 remains undominated. Place a qk 2 in the dominating set and then G k 1 is dominated and G k 2 has only one undominated v ertex. The same process is complete for G k +2 ; G k 3 : : : un til G 1 and G r are reac hed. A t eac h step only one v ertex is added to the dominating set and both G 1 and G r ha v e one undominated v ertex so adding these v ertices to the set giv es a domination of G P r with r +2 v ertices. In the diagram belo w, dark circles represen t v ertices 42

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in the dominating set, all other v ertices are dominated b y the corresponding v ertex in an adjacen t cop y of G ca jk ca lk s a ik @ @ G k sa jk 1 ca qk 1 c a ik 1 @ @ G k 1 s c s @ @ G 1 s s c @ @ G r If one v ertex a p in A has degree n 2 then it ma y be possible to rearrange the order in whic h the v ertices are place in the dominating set so that a p 1 is is used when domination G 2 If this is possible, then no additional v ertex will be necessary to dominate G 1 and r ( G P r ) = r + 1 : Lik ewise it ma y be possible to order the v ertices in the dominating set so that both G 1 and G r use v ertices of degree n 2 and then r ( G P r ) = r : If t w o v ertices a p and a q of A ha v e degree n 2 then place a pk and a qk +1 in the dominating set for k odd. This giv es a domination n um ber of r Finally if an y v ertex of G has degree n 1 place one cop y of this v ertex in the dominating set for eac h cop y of G and this giv es r ( G P r ) = r 2 Next, for the connected graphs on n 6 v ertices that are not co v ered b y Theorem 3.14 the domination n um ber is computed using the A t Least Algorithm. 43

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3.5 V alues of r ( G P r ) for graphs on 6 v ertices. Theorem 3.15 F or all c onne cte d gr aphs G on n = 1 vertic es, we have r ( G P r ) = r 3 Theorem 3.16 F or all c onne cte d gr aphs G on n = 2 vertic es, we have r ( G P r ) = r 2 + ( 1 if r 0 (mo d 2) 0 otherwise. Theorem 3.17 F or all c onne cte d gr aphs G on n = 3 vertic es, we have r ( G P r ) = 3 r + 1 4 : Theorem 3.18 F or all c onne cte d gr aphs G on n = 4 vertic es, we have r ( G P r ) = r + 8 > < > : 1 if r = 5 ; 6 ; 9 and G is P 4 or r = 1 and G = C 4 0 otherwise. On v e v ertices, there are 34 non-isomorphic graphs 13 of whic h are disconnected and their domination n um ber is found b y taking the sum of the domination n um bers of the componen ts. There are 21 connected non-isomorphic graphs on 5 v ertices, 20 of whic h are co v ered b y Theorem 3.14. Theorem 3.19 F or all c onne cte d gr aphs G on n = 5 vertic es, we have r ( G P r ) = 6 r 5 + ( 1 if r = 7 0 otherwise. If G is the p ath on 5 vertic es, and r ( G P r ) = r ; r + 1 ; or r + 2 otherwise 44

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There are 156 graphs on 6 v ertices, 44 of these graphs are disconnected and their domination n um ber is found b y taking the sum of the domination n um ber of there componen ts whic h are graphs previously discussed. There are 112 non-isomorphic connected graphs on 6 v ertices, 47 of them are co v ered b y Theorem 3.14. The remaining 65 connected graphs fall in to 28 dieren t groups and the domination n um bers of these graphs are giv en belo w. Theorem 3.20 L et G b e one of the following gr aphs: r r r r rr J J J H H H H H H J J J r r r r rr n n n H H H H J J J Then for all r 1 ; we have r ( G P r ) = 3 r 2 + ( 1 if r 2 (mo d 4) 0 otherwise. Theorem 3.21 L et G b e one of the following gr aphs: r r r r rr H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 3 r 2 + ( 1 if r 0 (mo d 2) 0 otherwise. Theorem 3.22 L et G b e one of the following gr aphs: r r r r rr H H H H H H J J J r r r r rr H H H H J J J Then for all r 1 ; we have r ( G P r ) = 10 r 7 + ( 1 if r 2 (mo d 7), or r = 1, 6 0 otherwise. 45

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Theorem 3.23 L et G b e one of the following gr aphs: r r r r rr J J J n n n J J J Then for all r 1 ; we have r ( G P r ) = 10 r 7 + ( 1 if r 0, 2, 3, 4, 6 (mo d 7) exc ept r = 3 0 otherwise. Theorem 3.24 L et G b e one of the following gr aphs: r r r r rr J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 7 r 5 + 8 > < > : 1 if r 0, 2, 4 (mo d 5) exc ept r = 4, 7, 9, 14 0 otherwise. Theorem 3.25 L et G b e one of the following gr aphs: r r r r rr H H H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 7 r 5 + ( 1 if r 0, 2, 4, 5, 6, 7, 9 (mo d 10) 0 otherwise. Theorem 3.26 L et G b e one of the following gr aphs: r r r r rr n n n n n n H H H H J J J Then for all r 1 ; we have r ( G P r ) = 11 r 8 + 8 > < > : 1 if r 0, 2, 4, 5, 7 (mo d 8) exc ept r = 4, 7, or r = 6, 9 0 otherwise. 46

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Theorem 3.27 L et G b e one of the following gr aphs: r r r r rr n n n H H H H H H J J J r r r r rr n n n H H H H H H J J J r r r r rr H H H H H H H H J J J r r r r rr H H H H H H J J J r r r r rr H H H H H H H H J J J r r r r rr n n n H H H H H H J J J r r r r rr n n n H H H H H H J J J r r r r rr H H J J J H H H H H H J J J r r r r rr J J J H H H H H H J J J r r r r rr n n n H H H H J J J r r r r rr n n n H H H H J J J r r r r rr H H n n n H H H H H H J J J r r r r rr J J J n n n H H H H J J J Then for all r 1 ; we have r ( G P r ) = 4 r 3 + ( 1 if r 0, 2 (mo d 3) 0 otherwise. Theorem 3.28 L et G b e one of the following gr aphs: r r r r rr H H J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 4 r 3 + 8 > < > : 2 if r 0 (mo d 3) exc ept r = 3 1 if r 1, 2 (mo d 3) exc ept r = 1, 4, or r = 3 0 otherwise. Theorem 3.29 L et G b e one of the following gr aphs: r r r r rr H H n n n H H H H J J J r r r r rr H H J J J n n n J J J Then for all r 1 ; we have r ( G P r ) = 4 r 3 + ( 1 if r 0, 2 (mo d 3) 0 otherwise. 47

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Theorem 3.30 L et G b e one of the following gr aphs: r r r r rr H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 13 r 10 + 8 > > > < > > > : 2 if r 0 (mo d 10) exc ept r = 10 1 if r 1, 2, 3, 4, 5, 6, 7, 8, 9 (mo d 10) exc ept r = 1, 4, 7, or r = 10 0 otherwise. Theorem 3.31 L et G b e one of the following gr aphs: r r r r rr J J J n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 9 r 7 + ( 1 if r 7 (mo d 14) 0 otherwise. Theorem 3.32 L et G b e one of the following gr aphs: r r r r rr J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 14 r 11 + ( 1 if r = 2, 3, 7 0 otherwise. Theorem 3.33 L et G b e one of the following gr aphs: r r r r rr H H J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 14 r 11 + 8 > < > : 1 if r 0, 2, 3, 5, 6, 7, 8, 9, 10 (mo d 11) exc ept n = 5, 8, 9 0 otherwise. 48

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Theorem 3.34 L et G b e one of the following gr aphs: r r r r rr H H H H H H H H J J J r r r r rr H H H H H H J J J r r r r rr J J J n n n H H H H H H J J J r r r r rr H H n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 5 r 4 + ( 1 if r 0 (mo d 4) 0 otherwise. Theorem 3.35 L et G b e one of the following gr aphs: r r r r rr n n n n n n H H H H H H J J J Then r ( G P r ) = l 5 r 4 m for all r 1 Theorem 3.36 L et G b e one of the following gr aphs: r r r r rr H H H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 5 r 4 + ( 1 if r 4 (mo d 8) 0 otherwise. Theorem 3.37 L et G b e one of the following gr aphs: r r r r rr n n n H H H H H H J J J r r r r rr H H H H H H J J J r r r r rr n n n n n n H H H H J J J r r r r rr J J J n n n n n n H H H H J J J Then for all r 1 ; we have r ( G P r ) = 5 r 4 + ( 1 if r 0, 2, 3 (mo d 4) exc ept r = 2, 3, 6 0 otherwise. Theorem 3.38 L et G b e one of the following gr aphs: r r r r rr n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 5 r 4 + 8 > > > < > > > : 2 if r 0 (mo d 4) exc ept r = 4, 8, 12 1 if r 1, 2, 3 (mo d 4) exc ept r = 1, 2, 3, 5, 6, 9, or r = 4, 8, 12 0 otherwise. 49

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Theorem 3.39 L et G b e one of the following gr aphs: r r r r rr H H J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 5 r 4 + ( 1 if r = 2, 3, 4, 7 0 otherwise. Theorem 3.40 L et G b e one of the following gr aphs: r r r r rr J J J n n n H H H H H H J J J r r r r rr J J J H H H H H H J J J r r r r rr H H H H H H J J J r r r r rr n n n n n n H H H H H H J J J r r r r rr n n n n n n H H H H H H J J J r r r r rr J J J n n n H H H H H H J J J r r r r rr H H J J J n n n H H H H H H J J J r r r r rr J J J n n n H H H H J J J Then r ( G P r ) = l 6 r 5 m for all r 1 Theorem 3.41 L et G b e one of the following gr aphs: r r r r rr J J J H H H H H H J J J r r r r rr H H n n n H H H H H H J J J r r r r rr H H n n n H H H H H H J J J r r r r rr J J J H H H H H H J J J r r r r rr J J J n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 6 r 5 + ( 1 if r = 4 0 otherwise. Theorem 3.42 L et G b e one of the following gr aphs: r r r r rr J J J n n n n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 6 r 5 + ( 1 if r 0, 4 (mo d 5) exc ept r = 4 0 otherwise. 50

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Theorem 3.43 L et G b e one of the following gr aphs: r r r r rr H H J J J n n n H H H H H H J J J r r r r rr n n n n n n H H H H J J J r r r r rr H H J J J n n n H H H H J J J Then for all r 1 ; we have r ( G P r ) = 6 r 5 + ( 1 if r 0, 2, 3, 4 (mo d 5) exc ept r = 2, 3 0 otherwise. Theorem 3.44 L et G b e one of the following gr aphs: r r r r rr J J J n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 7 r 6 + ( 1 if r 0, 4, 5 (mo d 6) exc ept r = 4, 5, 10 0 otherwise. Theorem 3.45 L et G b e one of the following gr aphs: r r r r rr J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 7 r 6 + ( 1 if r 0, 2, 3, 4, 5 (mo d 6) exc ept r = 2, 3 0 otherwise. Theorem 3.46 L et G b e one of the following gr aphs: r r r r rr n n n n n n H H H H H H J J J r r r r rr H H n n n n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 8 r 7 + ( 1 if r = 6 0 otherwise. 51

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Theorem 3.47 L et G b e one of the following gr aphs: r r r r rr H H J J J n n n H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 8 r 7 + 8 > < > : 1 if r 0, 2, 3, 4, 5, 6 (mo d 7) exc ept r = 2, 3, 5, 9, 10 0 otherwise. Theorem 3.48 L et G b e one of the following gr aphs: r r r r rr H H H H H H H H J J J r r r r rr H H J J J H H H H H H J J J Then for all r 1 ; we have r ( G P r ) = 10 r 9 + ( 1 if r = 8 0 otherwise. 52

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4. Domination of Circulan t Graphs In this c hapter w e will nd the domination n um ber of circulan t graphs. Denition 4.1 Given an inte ger r > 0 and subset of the p ositive inte ger S the circulant graph C r [ S ] is a gr aph on vertic es 0, 1, ..., r 1 with vertex i adjac ent to vertex j if i j 2 S or j i 2 S (mo d r ). Let C r [1 ; 3] be shorthand for C r [ f 1 ; 3 g ], where C r [1 ; 3] is a graph on v ertices 0, 1, ..., r 1 with v ertex i adjacen t to i 1 and i 3 (mod r ). T o illustrate the algorithm w e will nd r ( C r [1 ; 3]) whic h has domination n um ber 4 (see Figure 4.1). T o nd the domination n um ber for all r w e form ulate the problem as a dynamic programming algorithm similar to the algorithm used in Hare and Hedetniemi [10 ] and Fisher [8 ] and Chapter 3 to nd the domination n um ber of complete grid graphs. Then, lik e is done in [8], w e iterate the algorithm un til a periodicit y is detected. When this happens, w e declare that w e kno w the answ er for all r and stop iterating. This allo ws us to nd C r [1 ; 3] for all r in a nite amoun t of time. 53

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p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p t j t t j t t j t t t t t j tt t t Figure 4.1 The circulan t graph C 14 [1 ; 3]. The circled v ertices are a minim um domination. 4.1 The States and T ransition Matrix W e can think of C r [1 ; 3] as a sequences of o v erlapping paths on three v ertices ( P 3 ) where adjacen t P 3 's share t w o v ertices and there is an edge bet w een the nonshared v ertices. Connecting together r of these P 3 's so the r th P 3 o v erlaps the rst builds C r [1 ; 3]. W e will build a domination b y appending P 3 's cloc kwise from the existing sequence. The v ertices of P 3 are either in the domination, dominated but not in the domination, or not y et dominated (and th us need to be dominated b y a v ertex in the cloc kwise direction). The states will be all v alid com binations of these three attributes. Figure 4.2 sho ws the 15 v alid states. This is less than the 27 w a ys in whic h 3 v ertices can be assigned 3 attributes. This is because 10 poten tial states ha v e a dominated v ertex next to an undominated v ertex, while t w o others ha v e the 54

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rst v ertex undominated and the last v ertex is dominated and these are not v alid since the last v ertex cannot be dominated without dominating the rst v ertex. 1= t j j t j t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 2= j t j t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 3= t j t t j p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 4= t j t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 5= t j t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 6= t t j t j p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 7= t t j t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 8= t t t j p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 9= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 10= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 11= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 12= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 13= t t t j p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 14= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p 15= t t t p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p Figure 4.2 The fteen states of P 3 Circled v ertices are in the domination. Dominated v ertices ha v e a on them. Going cloc kwise, w e will refer to the v ertices of a state as being v ertex a b and c When c an two states overlap? Let v ertices k k + 1, and k + 2 be in state j and k + 1, k + 2, and k + 3 be in state i T o be consisten t, w e m ust ha v e the follo wing. (1) V ertex b of state j m ust be the same as v ertex a of state i (2) V ertex c of state j is the same as v ertex b of state i unless v ertex c of state i is in the domination. In this case, if v ertex c of state j is undominated, then v ertex b of state i is \dominated but not in the domination". (3) If v ertex a of state j is in the domination, then v ertex c of state i is either be in the domination, or \dominated but not in the domination". If v ertex a of state j is \dominated but not in the domination", then 55

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v ertex c of state i can either be in the domination, or not dominated. Finally if v ertex a of state j is not dominated, then v ertex c of state i m ust be in the domination. The last condition forces v ertex a of state j to be dominated b y v ertex c in state i if it has not already been dominated. Th us states that ha v e v ertex a dominated matc h to t w o states, while states in whic h v ertex a is not dominated matc h to only one state. W e will represen t whic h states matc h to other states b y a transition matrix A The i; j en try of A is: A i;j = 8 > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > : 0 if state j matc hes to state i and v ertex c of state i is not in the domination 1 if state j matc hes to state i and v ertex c of state j is in the domination 1 if state j does not matc h to state i 56

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Then the transition matrix is: A = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 : The information in the matrix can also be represen ted in the p recedence digraph D A whic h is sho wn in Figure 4.3. u 1 2 u 3 u 4 u 5 u 6 u 7 u 8 u 9 u 10 u 11 u 12 u 13 u 14 u 15 u 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B B B B M ) P P P P P P P P P q H H H H H H j C C C C C C W A A A U 3 A A A U U U U 1 C C C C C C O P P P P P P P P P q q q q q q q q q q q q P P P P P P P P P i + ? r * * C C C C C C C C C C C O 6666666 Q Q Q Q k m Figure 4.3 Precedence Digraph for Circulan t graphs. The unlabeled arcs ha v e w eigh t zero. The arro w ed arcs sho w a cycle whose cycle mean is minim um. 57

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Since the states in a domination m ust matc h, a domination of C r [1 ; 3] corresponds to a path (not necessarily simple) in D A F urther a domination m ust be a cycle because the rst state and the r th state m ust matc h to build C r [1 ; 3]. So a minim um domination of C r [1 ; 3] corresponds to a cycle in D A of length r with minim um w eigh t. Theorem 4.2 L et A b e an 15 15 matrix shown ab ove. Then for all r > 0 we have that r ( C r [1 ; 3]) = tr ac e ( A r ) wher e matrix exp onentiation and tr ac e ar e in Min-Plus algebr a. Pr o of: By Theorem 2.7, ( A r ) i;i is the minim um w eigh t cycle of length r from i to i in D A By the denition of D A this is the n um ber of v ertices in a domination starting and ending in state i Th us trace( A r ) = min i ( A r ) i;i is the minim um size of a domination of C r [1 ; 3]. 2 4.2 P eriodicit y of the T ransition Matrix Theorem 4.2 giv es us a method for nding the domination n um ber of C r [1 ; 3] for an y r Ho w ev er, w e cannot use it to nd r ( C r [1 ; 3]) for all r This depends on the periodic propert y of Min Plus algebra. In this example, this propert y can be sho wn b y looking at A 19 and A 24 58

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A 19 = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 6 6 6 6 6 6 6 6 6 7 6 7 7 7 7 5 5 5 5 5 5 5 5 5 6 5 6 6 6 6 5 5 5 5 6 5 5 5 5 6 5 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 6 5 5 6 4 4 4 4 5 4 4 4 4 5 4 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 7 4 4 4 4 4 4 5 4 5 5 5 5 5 5 6 4 4 5 5 5 4 4 5 5 5 5 6 5 5 6 4 5 4 5 5 4 5 4 5 5 5 6 5 5 6 4 4 4 5 5 4 4 4 5 5 5 6 5 5 6 4 4 4 4 4 4 4 4 4 4 5 5 4 4 5 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 4 5 4 5 5 4 5 4 5 5 5 6 5 5 6 3 4 3 4 4 3 4 3 4 5 4 5 4 5 5 4 4 4 4 4 4 4 4 4 5 4 5 5 5 5 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 A 24 = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 7 7 7 7 7 7 7 7 7 8 7 8 8 8 8 6 6 6 6 6 6 6 6 6 7 6 7 7 7 7 6 6 6 6 7 6 6 6 6 7 6 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 7 6 6 7 5 5 5 5 6 5 5 5 5 6 5 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 5 5 5 5 5 5 6 5 6 6 6 6 6 6 7 5 5 6 6 6 5 5 6 6 6 6 7 6 6 7 5 6 5 6 6 5 6 5 6 6 6 7 6 6 7 5 5 5 6 6 5 5 5 6 6 6 7 6 6 7 5 5 5 5 5 5 5 5 5 5 6 6 5 5 6 5 5 5 5 5 5 5 5 6 6 6 6 6 6 7 5 6 5 6 6 5 6 5 6 6 6 7 6 6 7 4 5 4 5 5 4 5 4 5 6 5 6 5 6 6 5 5 5 5 5 5 5 5 5 6 5 6 6 6 6 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 Note ev ery en try in A 24 is one more than the corresponding en try in A 19 This can be notated as A 24 = 1 A 19 Then A 25 = A A 24 = A (1 A 19 ) = 1 ( A A 19 ) = 1 A 20 : Con tin uing in this manner, w e get the follo wing theorem. 59

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Theorem 4.3 L et A b e the 15 15 matrix given ab ove. Then for all r 24 we have A r = 1 A r 5 : Pr o of: Since A 24 = 1 A 19 the result holds for r = 24. F or n > 24, assume for induction purposes that A r 1 = 1 A r 6 Then A r = A A r 1 = A (1 A r 6 ) = 1 ( A A r 6 ) = 1 A r 5 : 2 Th us, w e can use this repetition to nd the domination n um ber of C r [1 ; 3] for all r By observing the domination n um bers when r < 24, w e nd an explicit form ula for C r [1 ; 3]. Theorem 4.4 F or all r 1 we have r ( C r [1 ; 3]) = 8 > < > : d n 5 e + 1 if n = 4 (mo d 5) d n 5 e otherwise. Using this method, w e let S be an y set from the in tegers f 1 ; 2 ; : : : 9 g and w e nd the domination n um ber of C r [ S ] for all r There are 512 dieren t circulan t graphs on S and the domination n um bers of these graphs is giv en in Appendix A. 60

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5. Knigh ts Domination of k r Chessboards In this c hapter w e use Min-Plus Algebra to dev elop a dynamic programming algorithm whic h solv es the Knigh t domination problem for k r c hessboards. The game of Chess is pla y ed on an 8 8 board on whic h a n um ber of pieces are placed. One piece, a Knight can mo v e (or attac k) \from corner to diagonally-opposite corner of a rectangle three squares b y t w o" (Dr. Lask er as quoted in [12]). See Figure 5.1. Figure 5.1 Knigh t Mo v es: A Knigh t ( ) attac ks the squares mark ed with Denition 5.1 A Knight domination of a k r b o ar d is a plac ement of Knights so e ach squar e either has a Knight on it or attacking it. The k r Knight domination number is the minimum numb er of Knights in a Knight domination of a k r b o ar d. Figure 5.2 sho ws a domination of a 3 14 board with 11 Knigh ts. Since 11 is the minim um n um ber, the 3 14 Knigh t domination n um ber is 11. 61

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Figure 5.2 A Minimal 3 14 Knigh t Domination. The k r Knight Graph denoted N k;r ( N is the standard initial for \Knigh t", as K is used for \King"), has v ertices f 1 ; 2 ; : : :; k gf 1 ; 2 ; : : :; r g with an edge bet w een v ertices ( g; h ) and ( i; j ) if j i g jj j h j = 2 (see Figure 5.3). Then the k r Knigh t domination n um ber is r ( N k;r ). N 8 ; 8 = tttttttt tttttttt tttttttt tttttttt tttttttt tttttttt tttttttt tttttttt A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H Figure 5.3 The 8 8 Knigh t Graph. V ertices represen t squares of an 8 8 board with edges sho wing mo v es that a Knigh t can mak e. Its domination n um ber is the Knigh t domination n um ber of an 8 8 board. Gardner [9] generated considerable in terest in Knigh t domination of square boards (when k = r ). Gardner and his readers found small Knigh t dominations of a k k board for k 15. These are kno wn to be optim um when k 10. k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 r ( N k;k ) 1 4 4 4 5 6 10 12 14 16 21 24 28 32 37 Hare and Hedetniemi [10] dev eloped a dynamic programming algorithm for the k r Knigh t domination n um ber whic h is exponen tial in k but linear in r This allo w ed them to conjecture form ulas for r ( N k;r ) for k = 3, 4, and 6 ( k = 1 62

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and 2 are trivial; k = 5 w as \not fully analyzed"). They also found v alues for k 10 for some v alues of r In this c hapter, w e restate the algorithm in Min-Plus algebra and looks for periodic solutions to the dynamic programming algorithm. By nding periodicit y w e pro v e the conjectures in [10], and go on to nd the k r Knigh t domination n um ber for all k 8 and for all r T o illustrate the dynamic programming algorithm, w e will use it to nd the 2 r Knigh t domination n um ber. 5.1 The States A state will be a conguration of t w o columns of a k r board. A square either has a Knigh t on it (sho wn b y ), is dominated b y a Knigh t from the left or within the t w o columns (sho wn b y ), or to be dominated from the righ t of the t w o columns (sho wn b y a blank). Since a blank cannot be a Knigh t mo v e from a Knigh t, states correspond to labellings of N k; 2 with 0, 1 and 2 (for squares with a squares with a and blank squares, resp.) where 0 and 2 cannot be adjacen t. How many such lab ellings ar e p ossible? Let a r be the n um ber of suc h labellings on a path on r v ertices. It is easy to sho w that a 1 = 3, a 2 = 7, and a r = 2 a r 1 + a r 2 for all r > 2. So a 3 = 17, a 4 = 41, a 5 = 99, etc. W e can then 63

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solv e this recursion to sho w that a r = 1 + p 2 r +1 + 1 p 2 r +1 2 : Since N k; 2 consists of t w o paths on b k= 2 c v ertices and t w o paths on d k= 2 e v ertices, the n um ber of states needed to nd the k r Knigh t domination n um ber for all r is (Figure 5.4 sho ws the 81 states when k = 2): Num ber of States = a 2 b k= 2 c a 2 d k= 2 e 1 + p 2 2 k +4 16 : (5.1) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. Figure 5.4 The 81 Knigh ts congurations of a 2 2 board. 5.2 The T ransition Matrix Of the 3 6 possible 2 3 boards, only 81 need to be considered for domination patterns. This is because eac h square in the third column of this board 64

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is a Knigh t mo v e from a square in the rst column (square (1,1) and (2,3) are a Knigh t mo v e apart as are square (2,1) and (1,3)). The possible 2 3 boards are created b y matc hing t w o of the 2 2 boards sho wn in Figure 5.4 together so that the middle columns o v erlap and the follo wing conditions are satised. (1) An y Knigh ts in the o v erlapping columns m ust be in the same square in eac h pattern. (2) All dominated ( ) squares in the last column of the second k 2 board m ust be dominated b y Knigh ts within the resulting k 3 board. (3) All undominated (blank) squares in the rst column of the rst k 2 board m ust be dominated b y Knigh ts in the resulting k 3 board. (4) When matc hing a k r board to a k 2 board, condition 2 guaran ties that all dominated squares in the last column of the k r board are dominated b y Knigh ts within previous columns. Th us, an undominated square in the rst column of the second pattern can matc h with a dominated square in the last column of the rst pattern since the undominated square will become dominated in the matc h. (This matc hing does not apply to 2 2 boards since the pattern is not possible). All possible domination patterns of a k 3 board are found similarly b y matc hing together t w o of the domination patterns of a k 2 board follo wing the rules 65

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listed abo v e. In Figure 5.5, possible matc hings of k 2 boards are sho wn to illustrate the matc hing requiremen ts. = ) 8 > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > : Figure 5.5 P ossible Matc hings of k 2 Boards T o nd the domination n um ber of a 2 r board, a column is added on to all of the possible patterns on 2 ( r 1) boards and then eac h of these new patterns is c hec k ed to see if it is a domination. The minim um n um ber of Knigh ts in the patterns whic h are dominations is then the domination n um ber of a 2 r board. The r th column that is being added on to the 2 ( r 1) board is dependen t on the last t w o columns of the 2 ( r 1) board. Th us, the column is added b y matc hing the last column of the 2 ( r 1) pattern to the rst column of one of the 2 2 boards follo wing the matc hing constrain ts listed abo v e. T o k eep trac k of whic h patterns can be added on to other patterns b y using this matc hing, a transition matrix A is formed with ro ws and columns 66

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being represen ted b y the initial 2 2 boards. If the i th k 2 board matc hes the j th 2 2 board then the ( i; j ) en try in the transition matrix is the n um ber of Knigh ts in the last column of the j th 2 2 board. If the i th 2 2 board does not matc h the j th board then the ( i; j ) en try in the transition matrix is innit y or (-) (see belo w). 67

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2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 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--------1--------1--------1----------------------------------------------------------------------------------------1--------1--------1----------------------------------1--------------------------1--------------------------1-------------------------0----------------------------------------------------------------------------------------------------------0-------------------------------------------------------------1--------------------------1--------------------------1-------------------------0----------------------------------------------------------------------------------------------------------0-----------------------------------3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 68

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5.3 The States V ector Next, a states vecto r x r is formed to k eep trac k of the 81 possible minimal states of a 2 r board. The i th en try in this v ector is the n um ber of Knigh ts in a 2 r pattern whose last 2 columns are the same as the i th 2 2 board in Figure 5.4. If one of these 2 2 boards does not form the last 2 columns of an y of the possible 2 r patterns then the en try in x r is innit y or (-). No w, to nd the possible patterns of a 2 ( r + 1) board, the transition matrix is m ultiplied b y x r using Min-Plus matrix m ultiplication to form a new v ector x r +1 The i th en try in x r +1 represen ts the n um ber of Knigh ts in a 2 ( r + 1) pattern whose last 2 columns are the same as the i th 2 2 board. F or eac h of these 2 ( r +1) patterns, the rst r 1 columns are dominated. If the i th 2 2 board, whic h forms the r and r +1 columns of the pattern, is also dominated, then the en tire 2 ( r + 1) pattern will be dominated. Out of these dominated states, the one with the least n um ber of Knigh ts in it (the lo w est en try in x r +1 ) forms a minimal domination of a 2 ( r +1) board and the domination n um ber of a 2 ( r + 1) board is the n um ber of Knigh ts in this pattern. This iterativ e process is started with an initial v ector x 1 whic h is used to nd the domination n um ber of a 2 1 board. T o do this, the rst column of the 81 2 2 boards is considered column 0 of the 2 1 board. Only the 2 2 boards in whic h the rst column is dominated and con tains no Knigh ts are 69

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possible patterns for a 2 1 board. Th us, the i th en try in the initial v ector x 1 is either the n um ber of Knigh ts in the last column of the i th 2 2 board if the rst column of this board is dominated and con tains no Knigh ts, or innit y (-) if the rst column is not dominated or has Knigh ts in it. Using this initial v ector, the domination n um ber of a 2 r board is found b y iterativ e use of Min-Plus matrix m ultiplication of the states v ectors and the transition matrix A with x r = A x r 1 : The states v ectors x 1 ; x 2 ; : : :; x 18 are sho wn belo w. x 1 = 0 [ ------------------------------------2-1---1-0-----------------------------------] x 2 = 0 [ 4-3---3-2---------3-2---2-1---------------------------3-2---2-1---------2-1---1-0 ] x 3 = 2 [ 22-22-------------11-11-------------------------------11-11-------------00-00---] x 4 = 4 [ 00-00----00-00-------------00-00----00-00---------------------------------------] x 5 = 4 [ 211100100211100100---------211100100211100100-----------------------------------] x 6 = 4 [ 433322322322211211322211211322211211211100100211100100322211211211100100211100100 ] x 7 = 4 [ 443443332332332221332332221332332221221221110221221110332332221221221110221221110 ] x 8 = 4 [ 443443332443443332332332221443443332443443332332332221332332221332332221221221110 ] x 9 = 6 [ 222222222222222222111111111222222222222222222111111111111111111111111111000000000 ] x 10 = 8 [ 000000000000000000000000000000000000000000000000000000000000000000000000000000000 ] x 11 = 8 [ 211100100211100100211100100211100100211100100211100100211100100211100100211100100 ] x 12 = 8 [ 433322322322211211322211211322211211211100100211100100322211211211100100211100100 ] 5.4 Detecting P eriodicit y T o nd the domination n um ber of a 2 r board and the pattern that giv es this domination n um ber, repetition can be used. Looking at the 18 70

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states v ectors, note that x 12 = x 6 + 4. Since the v alues in x r only depend on the v alues in x r 1 and the transition matrix, w e ha v e x r +6 = x r + 4 for all r 6. So the domination pattern for a 2 ( r + 6) board is found b y nding the domination pattern for a 2 r board and adding on to this a dominated pattern with 6 columns and 4 Knigh ts. The domination pattern for a 2 r board for all r is sho wn in Figure 5.6. ... Figure 5.6 Domination P attern for a 2 r board. P eriodicit y is detected b y storing eac h x r and comparing its en tries with the en tries in all previous state v ectors. The algorithm for nding the domination n um ber of a k r board is essen tially the same as for a 2 r In eac h case, all possible k 2 boards are formed. The n um ber of possible k 2 boards is equal to 3 2 k min us the n um ber of impossible patterns whic h is appro ximated b y the form ula 5.1. A transfer matrix A and and initial v ector x 1 are then formed from these k 2 boards. Min-Plus matrix m ultiplication is then used to nd x r and the domination n um ber of a k r board. As with 2 r the domination n um ber is the smallest en try in x r whic h corresponds to a dominated k r board. Successiv e x r v ectors are computed un til a repetition is found and this is used to determine the pattern corresponding to the domination n um ber for all r The results for 71

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k 7 are giv en in the follo wing sections. 5.5 Domination Num ber for a 3 r board F or a 3 r board, the periodicit y is giv en b y x r +6 = x r + 4 : This means that once an initial pattern is established, the pattern can be extended b y adding on a pattern of 6 columns whic h is dominated b y 4 Knigh ts. The domination patterns are sho wn in Figures 5.7 5.8 5.9. ... Figure 5.7 Domination P attern of 3 r for r = 0, 3, 4, and 5. F or r = 0, 3, 4, and 5 (mod 6) r 4. F or r = 3 use columns 3, 4, and 5 of this pattern. F or r = 8, remo v e columns 4 and 5 of this pattern. ... Figure 5.8 Domination P attern of 3 r for r = 1 (mod 6) r 4 ... Figure 5.9 Domination P attern of 3 r for r = 2 (mod 6) r 14 72

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5.6 Domination Num ber for a 4 r board F or a 4 r board, the periodicit y is giv en b y x r +6 = x r + 4 : This means that once an initial pattern is established, the pattern can be extended b y adding on a pattern of 6 columns whic h is dominated b y 4 Knigh ts. The algorithm detects periodicit y at r = 27. The domination patterns are sho wn in Figures 5.10 5.11 5.12 ... Figure 5.10 Domination P attern of 4 r Board, for r = 0, 3, 4, and 5 (mod 6) when r 4. F or r = 8, remo v e columns 4 and 5 of this pattern. ... Figure 5.11 Domination P attern of 4 r Board, for r = 1 (mod 6) r 4. ... Figure 5.12 Domination P attern of 4 r Board for r = 2 (mod 6) r 14. 73

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5.7 Domination Num ber for a 5 r board F or a 5 r board, the periodicit y is giv en b y x r +18 = x r + 14 : This means that once an initial pattern is established, the pattern can be extended b y adding on a pattern of 18 columns whic h is dominated b y 14 Knigh ts. The algorithm detects periodicit y at r = 39. The domination patterns can be described in terms of the separate componen ts sho wn in Figure 5.13. P = E = B = U = L = Figure 5.13 Componen ts in Domination P atterns of 5 r Boards. These componen ts are com bined together to form the patterns for a 5 r board. The table belo w sho ws the patterns with the repeating 18 columns of 74

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eac h pattern in paren theses. n r ( N 5 ;r ) P attern for r = 0 P attern for r > 0 18 r + 0 14 r + 1 Meaningless ( PBE 4 B ) r 1 PBE 5 18 r + 1 14 r + 2 See 1 5 result ( PBE 4 B ) r 1 PBE 6 18 r + 2 14 r + 3 See 2 5 result ( PBE 4 B ) r 1 PBE 7 18 r + 3 14 r + 4 See 3 5 result ( PBE 4 B ) r 1 PBE 8 18 r + 4 14 r + 4 ( PBE 4 B ) r U 18 r + 5 14 r + 5 E 5 ( PBE 4 B ) r 1 PBE 5 BU 18 r + 6 14 r + 6 E 6 ( PBE 4 B ) r 1 PBE 6 BU 18 r + 7 14 r + 7 E 7 ( PBE 4 B ) r 1 PBE 7 BU 18 r + 8 14 r + 8 E 8 ( PBE 4 B ) r 1 PBE 8 BU 18 r + 9 14 r + 8 ( PBE 4 B ) r UBE 4 18 r + 10 14 r + 9 ( PBE 4 B ) r UBE 5 18 r + 11 14 r + 10 ( PBE 4 B ) r UBE 6 18 r + 12 14 r + 10 ( PBE 4 B ) r P 18 r + 13 14 r + 11 UBE 8 ( PBE 4 B ) r 1 PBE 4 EBP 18 r + 14 14 r + 12 ( PBE 4 B ) r UBE 4 BU 18 r + 15 14 r + 13 ( PBE 4 B ) r UBE 5 BU 18 r + 16 14 r + 14 ( PBE 4 B ) r UBE 6 BU 18 r + 17 14 r + 14 ( PBE 4 B ) r PBE 4 5.8 Domination Num ber for a 6 r board F or a 6 r board, the periodicit y is giv en b y x r +4 = x r + 4 : This means that once an initial pattern is established, the pattern can be extended b y adding on a pattern of 4 columns whic h is dominated b y 4 Knigh ts. The algorithm detects periodicit y at r = 15. The domination patterns are sho wn in Figures 5.14 5.15 5.16 75

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... Figure 5.14 Domination P attern of 6 r Board, for r = 0, and 3 (mod 4) r 4. ... Figure 5.15 Domination P attern of 6 r Board, for r = 1 (mod 4) r 4 ... Figure 5.16 Domination P attern of 6 r Board, for r = 2 (mod 4) r 4 5.9 Domination Num ber for a 7 r board F or a 7 r board, the periodicit y is giv en b y x r +5 = x r + 6 : This means that once an initial pattern is established, the pattern can be extended b y adding on a pattern of 4 columns whic h is dominated b y 4 Knigh ts. 76

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The algorithm detects periodicit y at r = 32. The domination patterns can be described in terms of the componen ts sho wn in Figure 5.17. P = Q = R 1 = R 2 = S = T = U = V = W = Y = Z = Figure 5.17 Componen ts in Domination P atterns of 7 r Boards. These componen ts are com bined together to form the patterns for a 7 r board. The table belo w sho ws the patterns with the repeating 6 indicated b y raising the pattern to the exponen t r 77

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n r ( N 7 ;r ) P attern for r 0 1 6 See Previous results 7 10 ZV ZWZV Z 8 11 ZV ZWZUZZ 9 12 ZZUZWZUZZ 10 14 ZV ZWY Y WZV Z 11 15 ZV ZWY Y WZUZZ 12 16 ZZUZWY Y WZUZZ 5 r + 13 6 r + 18 T ( P ) r S 5 r + 14 6 r + 19 R 1 ( P ) r S 5 r + 15 6 r + 20 R 1 ( P ) r Q 5 r + 16 6 r + 21 T ( P ) r R 2 5 r + 17 6 r + 22 R 1 ( P ) r R 2 5.10 Conclusions The algorithm dev eloped in this paper can be used to nd the domination n um ber of a k r board for all r The algorithm is not linear in k so the size of k is limited b y computer time. Theorem 5.2 The domination numb er of a k r b o ar d r ( N k;r ) for k 6 and for all r is: r ( N k;r ) = 8 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > : n if k = 1 4 l r 6 m if k = 2 2 r +4 3 if k = 2 r = 1 (mo d 6) and r 6 2 r +4 3 if k = 3 r = 1 (mo d 6), and r 6 2 r +5 3 if k = 3 r = 2 (mo d 6), and r 6 4 l r 6 m if k = 3 r 6 = 1 ; 2 (mo d 6) and r 6 2 r +4 3 if k = 3 r = 1 (mo d 6), and r 7 4 l r 6 m if k = 4 r 6 = 1 (mo d 6) and r 7 l 7 r 9 m if k = 5 r 6 = 4,12,13, or 17 (mo d 18), and r 13 l 7 r 9 1 m if k = 5 r = 4 ; 12 ; 13 ; or 17 (mo d 18), and r 13 r + 1 if k = 6 r = 1 (mo d 4), and r 4 4 l r 4 m if k = 6 r 6 = 1 (mo d 4), and r 4 j 6 r +12 5 k if k = 7 and r 13 78

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6. Time Complexit y of The P o w er Method All of the applications presen ted in the previous c hapters rely on the periodic propert y of dynamic programming algorithms in Min-Plus algebra. The algorithms terminate when periodicit y occurs, and this allo ws us to nd innitely man y solutions in a nite n um ber of iterations of the algorithm. Denition 6.1 The p re-periodic interval of a matrix A denote d R 0 ( A ) is the smallest R 0 such that A r = q A r p for all r R 0 In Theorem 2.3 w e stated that if A is the irreducible transfer matrix for the dynamic programming algorithm, then periodicit y implies that for some R 0 ; p and q w e ha v e that A r = q A r p for all r R 0 Although some matrices that are not irreducible are periodic, this is not the case in general. Consider a matrix whose precedence digraph has t w o disjoin t cycles of length one with dieren t w eigh ts as sho wn belo w. A = 2 4 t 1 1 s 3 5 A 100 = 2 4 100 t 1 1 100 s 3 5 In this case A r 11 = rt and A r 22 = rs and th us if t 6 = s there is no r where A r = q A r p and R 0 ( A ) is undened. Th us when using the periodicit y of Min-Plus algebra to solv e problems, w e m ust mak e sure that the matrix is irreducible. 79

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6.1 Finite T ermination of the P o w er Method In Chapter 2, w e stated Theorem 2.3 without proof. W e ha v e referred to this theorem throughout all of the applications presen ted in this thesis because it guaran tees that at some poin t the dynamic programming algorithms will be periodic and the po w er method can terminate. W e will no w restate the theorem and giv e a proof based on minim um cycle means and the diameter of A Denition 6.2 The diameter of A denote d diam ( A ) is the maximum weight of all minimum weight p aths in A Theorem 6.3 If A is an irr e ducible matrix on m vertic es, then ther e exists a p; q and R 0 so that A r = q A r p for all r R 0 Pr o of: Let ( A ) be the minim um cycle mean of A Since A is irreducible, w e kno w that the precedence digraph of A m 2 m +1 has paths bet w een all pairs of v ertices. Let k be the smallest common divisor of the lengths of the critical cycles of A with the propert y that k m 2 m + 1. No w, consider A k and let S be the subgraph of the precedence digraph of A k ( D A k ) whose v ertices are con tained in at least one of the critical cycles in A and let T be the subgraph of D A k on the remaining v ertices. In D A ev ery v ertex of S has a minim um w eigh t path of length k to itself using the critical cycle that con tains it. The 80

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w eigh t of this path is k ( A ), so the minim um cycle mean of A k is k ( A ) and the critical cycles are 1-loops on ev ery v ertex in S W e will no w look at possible paths in the precedence digraph of A k Let P be an y minim um w eigh t path of length r bet w een v i and v j that uses no v ertices of S This path consists of cycles and simple paths in T Let b ( A k ) be the minim um en try in A k then the w eigh t of eac h arc in the simple paths is at least b ( A k ). If there are h suc h arcs, the w eigh t of the simple path portion of P is at least hb ( A k ). Next if ( T ) is the minim um cycle mean of T then the arcs in ev ery cycle in T ha v e w eigh t of at least ( T ). P con tains r h arcs that are part of some cycle, so the w eigh t of these cycles is at least ( r h ) ( T ). Th us w ( P ) ( r h ) ( T ) + hb ( A k ) This is a non-negativ e v alue since ( T ) b ( A k ), and since h m 1 w e ha v e w ( P ) r ( T ) + ( m 1)( b ( A k ) ( T )) Next consider all minim um w eigh t paths Q from v i to v j of length r that use v ertices of S These paths consist of a path from v i to S loops on the v ertices of S and a path from S to v j The paths to and from S ha v e w eigh t at most diam( A k ) and the loops in S eac h ha v e w eigh t k ( A ) = ( A k ). Since Q con tains at most r loops in S w e ha v e w ( Q ) 2diam ( A k ) + r ( A k ) : 81

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No w w e w an t to sho w that all minimal paths of length r R 0 m ust use v ertices of S This is true if w ( P ) w ( Q ) for all paths of t ype P and Q of length r Since ( T ) ( A k ) w e ha v e w ( P ) w ( Q ) for all r R 0 when R 0 = 2diam ( A k ) ( m 1)( b ( A k ) ( T )) ( T ) ( A k ) (6.1) No w for an y minimal path of length r R 0 a minimal path of length r +1 is obtained b y looping on some v ertex of S that w as in the minimal path of length r and the w eigh t of these paths diers b y ( A k ) in ev ery case. Th us ( A k ) r = ( A k ) ( A k ) r +1 for all r R 0 This pro v es that A k is periodic and hence A is periodic with R 0 ( A ) = kR 0 ( A k ). The construction of the proof also giv es a bound for R 0 dened in terms of the en tries of A k 2 W e will no w illustrate this method b y nding the bound on R 0 ( A ) where A is the 15 15 transition matrix used in Chapter 4 to nd the domination n um ber of C r [1 ; 3]. The precedence digraph in Figure 4.3 sho ws that there is only one critical cycle C for this matrix and that ( A ) = w ( C ) j C j = 1 5 T o nd the bound for R 0 using the method described abo v e, w e need to nd the smallest k m 2 m + 1 with k being a common divisor of the lengths of the critical cycles in A Th us since m = 15, and there is only one critical cycle of length 5, w e ha v e k = 215. This bound is large because w e are required to ha v e 82

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k m 2 m + 1 to guaran tee that the matrix has paths of an y length greater than k (In this example, w e ha v e that A 8 has no innit y en ties so k could be reduced to 10). W e will illustrate the bound using k = 215 and considering possible paths in A 215 whic h is sho wn belo w. A 215 = 39 A 20 = 39 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 6 6 6 6 6 6 6 6 7 7 7 7 7 7 8 5 5 5 5 5 5 5 5 6 6 6 6 6 6 7 5 5 5 5 5 5 6 5 6 6 6 6 6 6 7 5 5 5 5 5 5 5 5 5 6 5 6 6 6 6 4 4 4 4 4 4 5 4 5 5 5 5 5 5 6 5 5 5 6 6 5 5 5 6 6 6 7 6 6 7 4 4 5 5 5 4 4 5 5 5 5 6 5 5 6 4 5 4 5 5 4 5 4 5 6 5 6 5 6 6 5 5 5 5 5 5 5 5 5 5 5 6 5 5 6 4 5 4 5 5 4 5 4 5 5 5 6 5 5 6 4 4 4 4 5 4 4 4 4 5 4 5 5 5 5 4 4 4 5 5 4 4 4 5 5 5 6 5 5 6 5 5 5 5 5 5 5 5 5 5 5 6 5 5 6 4 4 4 4 4 4 4 4 4 4 5 5 4 4 5 4 4 4 4 4 4 4 4 5 5 5 5 5 5 6 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 Using this matrix and the fact that S = f v 5 ;v 7 ;v 8 ;v 11 ;v 14 g w e ha v e that b ( A k ) = 43 ; ( T ) = 87 2 and ( A k ) = k ( A ) = 1 5 (215) = 43 : Next, for an y matrix, it's diameter is less than the n um ber of v ertices times the maxim um en try Using this bound w e ha v e diam ( A k ) 690 : This is considerably larger than the actual diameter of A whic h is 269 and th us the bound on R 0 will be signican tly eected b y using the bound instead of the actual 83

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diameter. Putting all of this together in equation 6.1 w e ha v e R 0 = 2 705 (14) (43 43 : 5) 43 : 5 43 = 2806 Th us, the bound is R 0 ( A k ) = 2806 and the bound for A is R 0 ( A ) = (215)(2806) = 603290, whereas the actual v alue of R 0 ( A ) w as 24. In computing this bound, w e used k = 215 as the smallest v alue where A k had no innit y en tries instead of the actual v alue of k = 20. If w e compute the bound with k = 10 and diam ( A ) = 269 it giv es R 0 ( A ) = (10)(1062) = 10620 whic h is still quite large. Th us although the bound sho ws that R 0 is nite, it is not a v ery good appro ximation of R 0 ( A ) in most cases. The large v alue of k and the large appro ximation of diam ( A ) both con tribute to the dierence bet w een the bound and the actual v alue of R 0 ( A ). In addition, if the gro wth of the en tries in successiv e po w ers of A is large, then the fact that the bound requires us to consider paths in A k where k is large ma y ha v e a m uc h more signican t eect. In the next section w e will restrict A and nd a bound for R 0 that is strong. 6.2 Exact Bound for R 0 when A is MCM-Hamiltonian In this section, w e will determine the bound for R 0 when A is restricted to a special case. Denition 6.4 A minim um cycle mean (MCM) Hamiltonian digraph is a digr aph wher e the minimum cycle me an o c curs on a Hamiltonian cycle. Denition 6.5 A matrix is MCM Hamiltonian if its pr e c e denc e digr aph is. 84

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Note that if A is MCM Hamiltonian then the subgraph T as described in the previous section does not exist. Th us, w e cannot use the bound from that section and w e m ust look at other w a ys to bound R 0 in this case. Denition 6.6 If B is an MCM Hamiltonian matrix, dene the following. (1) L et = q p b e the unique eigenvalue of B (2) L et P b e the p ermutation matrix that p ermutes the r ows and c olumns of B so that F = P T ( B ( )) P is a matrix with minimum cycle me an 0 and the minimum cycle C oriente d on the vertic es 1, 2, ..., m wher e vertex 1 is r epr esente d by r ow 1 of F Thus the ar cs of C ar e ( i;i + 1) for all i < m and the ar c ( m; 1) (3) L et D b e a diagonal matrix with D jj = P m 1 i = j ( F i;i +1 ) + F m; 1 Denition 6.7 If B is an MCM Hamiltonian matrix on m vertic es, then the no rmalized fo rm of B is A = ( D ) P T ( B ( )) P D: A no rmalized MCM Hamiltonian matrix is a matrix that is e qual to its normalize d form. Theorem 6.8 L et B b e an MCM Hamiltonian matrix with minimum cycle me an and normal form A = ( D ) P T ( B ( )) P D Then (1) A 1 ; 2 = A 2 ; 3 = ::: = A m 1 ;m = A m; 1 = 0 (2) The minimum cycle me an of A is 0. (3) A is non-ne gative. (4) B r = A r ( D ) P T A r P D and (5) B r = q B r p then A r = A r p 85

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Pr o of: Let C B C R C A be the cycles with minim um cycle means in F R and A Then F = P T B ( ) P so the arcs of C B ha v e been perm uted to the arcs of C F + f F 1 ; 2 ;F 2 ; 3 ;::: F m; 1 g and there w eigh ts ha v e been reduced b y Also for an y arc in A A i;j = F i;j D i;i + D j;j = F i;j [ F i;i +1 + ::: + F m 1 ;m + F m; 1 ] + [ F j;j +1 + ::: + F m 1 ;m + F m; 1 ] : whic h giv es a direct correspondence bet w een the cycles in A and the cycles in B 1) By construction, A i;i +1 = F i;i +1 [ F i;i +1 + ::: + F m 1 ;m + F m; 1 ]+[ F i +1 ;i +2 + ::: + F m 1 ;m + F m; 1 ] = 0 for all i < m and A m; 1 = F m; 1 [ F m; 1 ] + [ F 1 ; 2 + F 2 ; 3 + ::: + F m 1 ;m + F m; 1 ] = 0 : Th us all the arcs in C = f A 1 ; 2 ::: A m 1 ;m ;A m; 1 g ha v e w eigh t 0 and form a hamiltonian cycle of w eigh t 0. 2) F or an y cycle C A i of length i in A whic h has w eigh t w ( C A i ), there is a corresponding cycle C B i of length i in B whic h has w eigh t w ( C B i ) = w ( C A i ) + i By part 1 C is a hamiltonian cycle with cycle mean 0 so w ( C A ) 0. If w ( C A ) < 0 then w ( C B ) + w ( C A ) + m has cycle mean less than whic h is a con tradiction. Th us the minim um cycle mean of A is 0. 3) Suppose A has some en try A i;j < 0 Consider the cycle f A i;j ;A j;j +1 ;:::;A m 1 ;m ;A m; 1 ;:::;A i 1 ;i g Since all the arcs of this cycle except A i;j are con tained in the cycle with minim um cycle mean, they ha v e w eigh t 86

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0 and the w eigh t of the cycle is A i;j < 0 whic h is a con tradiction since the minim um cycle mean of A is 0. 4 & 5) The results are found b y using the comm utativ e and associativ e properties in Min-Plus algebra. 2 T o illustrate the normalization process consider the follo wing matrix whic h has the arcs in the cycle with minim um cycle mean underlined. B = 2 6 6 6 6 6 4 7 9 2 8 1 8 6 6 9 8 6 5 3 4 7 8 3 7 7 7 7 7 5 whic h has = 3 Then using the denitions abo v e, P = 2 6 6 6 6 6 4 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1 0 3 7 7 7 7 7 5 F = 2 6 6 6 6 6 4 5 2 3 3 6 4 1 5 5 6 3 2 1 0 4 5 3 7 7 7 7 7 5 D = 2 6 6 6 6 6 4 0 1 1 1 1 2 1 1 1 1 3 1 1 1 1 1 3 7 7 7 7 7 5 and nally A = 2 6 6 6 6 6 4 5 0 6 4 4 4 0 4 2 5 3 0 0 1 6 5 3 7 7 7 7 7 5 Th us an y min-plus problem with an MCM Hamiltonian transfer matrix B can be solv ed using the normalized MCM Hamiltonian matrix A and then con v erting the solution using parts 4 and 5 of Theorem 6.8 T o nd a bound on the n um ber of iterations needed for the transfer matrix to repeat, w e will consider normalized MCM Hamiltonian matrices. The precedence 87

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digraph for a normalized MCM Hamiltonian matrix has a Hamiltonian cycle with all arcs ha ving w eigh t zero and other arcs of non-negativ e w eigh t. The bounds on repetition will be formed b y considering paths in the precedence digraph since if A r = A r m where m is the size of the matrix, then in the precedence graph the minim um w eigh t of a path of length r from i to j is the same as the minim um w eigh t of a path from i to j of length r m for all i and j Denition 6.9 L et A b e a normalize d MCM Hamiltonian matrix and let b i b e the i th ar c in the minimum cycle of A which by denition has weight 0. Denition 6.10 L et a k b e the numb er of ar cs fr om i to j use d in the p ath wher e i j + 1 = k (mo d m ). Theorem 6.11 A p ath P of length r fr om vertex i to vertex j in a MCM Hamiltonian digr aph on m vertic es satises the following: a 1 + 2 a 2 + + ( m 1) a m 1 r = ( i j ) mo d m (6.2) Pr o of: First w e will pro v e this for r = 1. There are t w o cases for a path of length 1. Case 1: j = i + 1. In this case, the arc from i to j is one of the b i arcs from the minimal cycle. Th us no a i arcs are used and equation 6.2 becomes 1 = ( i ( i +1)) (mod m ) whic h holds. Case 2: j 6 = i + 1 In this case, if j = i + k the arc used is a l where l = i j + 1 = k + 1 and equation 6.2 becomes l 1 = ( i j ) (mod m ) or ( i ( i + k ) + 1) 1 = ( i ( i + k )) (mod m ) whic h also holds. 88

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F or induction purposes, w e will assume that a path of length r 1 satises ( 6.2) No w an y path of length r con tains a path of length r 1 plus one arc. Let j be the v ertex at the end of the path of length r 1 and k be last v ertex in the path. Th us the last arc in this path of length r goes from v ertex j to v ertex k. Let a 0 1 ;a 0 2 ;::: a 0 m 1 be the coecien ts that satisfy ( 6.2) for the path of length r 1. Then ( 6.2) is satised for the path of length r if a 0 1 + 2 a 0 2 + + ( m 1) a 0 m + arc from j to k r = ( i k ) (mod m ). Using the induction h ypothesis, this will hold if i j (mod m ) + arc from j to k 1 = i k (mod m ). Case 1: The additional arc is one of the b i arcs on the minimal cycle. This giv es k = j + 1 (mod m ) b y the labeling of the minimal cycle, and ( 6.2) becomes i j 1 = i ( j + 1) (mod m ) since there is no additional a i arc added to the equation. Case 2: The additional arc is a l where l = j k + 1. Then ( 6.2) becomes i j + ( j k + 1) 1 = i k (mod m ) whic h holds. 2 Denition 6.12 Given a p ath P the ar cs c an b e divide d into thr e e exclusive classes: T yp e 1 A r cs use d in a minimal p ath fr om the initial vertex to the nal vertex in the p ath. Ther e ar e m of these. T yp e 2 A r c use d in any minimal cycle c ontaining an a i Ther e ar e P m i =1 ia i of these. T yp e 3 A r cs that ar e in the minimal cycle. L et N i b e the numb er of ar cs of typ e i use d in a given p ath. 89

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Eac h t ype is exclusiv e so once an arc has been designated as being in t ype one it cannot be in t ype t w o or three. Also the determination of the arc t ype m ust be done in the order they are described. F or example using the graph in Figure 6.1, consider the follo wing path of length 13 from v ertex 1 to v ertex 3. P = b 1 b 2 a 3 b 1 b 2 b 3 b 4 b 5 a 4 b 4 b 5 b 1 b 2 In this path, the bold arcs are designated t ype one, the italic arcs are designated t ype t w o and the remaining arcs are t ype three. w 1 w 2 w 3 w 4 w 5 b 1 b 2 b 3 b 4 b 5 a 2 a 2 a 2 a 2 a 2 a 3 a 3 a 4 a 4 a 1 A A A A A U A A A A A K @ @ @ @ @ R @ @ @ @ @ I C C C C C C C C C C W C C C C C C C C C C O Figure 6.1 Precedence digraph of a normalized MCM Hamiltonian matrix. The b i arcs represen t the arcs in the minimal cycle and ha v e w eigh t 0. The w eigh t of the a i arcs is non-negativ e. No w w e need to sho w that for some r an y minim um w eigh t path in the precedence digraph of length r has the same w eigh t as a minim um w eigh t path of length r m This will be possible if there are alw a ys at least m t ype 3 arcs in a minim um w eigh t path of length r Since m of these arcs form a Hamiltonian cycle and w e assumed the t ype 3 arcs ha v e w eigh t zero b y theorem 3.2, their remo v al will not c hange the w eigh t of the path and will giv e a path of equal w eigh t and length r m 90

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T o nd the v alue of r for whic h w e are guaran teed to ha v e N 3 m w e m ust rst sho w that w e need only consider paths whic h use less than m of the a i arcs. T o do this w e will sho w that for an y path of length r from i to j that uses more than m of the a i arcs there is a path of length r from i to j that uses less than m of these arcs. This is done b y sho wing that suc h a path satises 6.2 in the same w a y that the original path does. Denition 6.13 L et a = ( a 1 ;::: ;a m 1 ) and c = ( c 1 ;::: ;c m 1 ) and let f ( a ) = a 1 + 2 a 2 + + ( m 1) a m 1 Next dene a < c if a i c i for all 1 i m No w a and c are the coecien ts of t w o paths described b y 6.2. W e w an t to sho w that if a satises 6.2 then there is some c that also satises 6.2 and uses less than m of the a i arcs. Theorem 6.14 If a satises P m 1 i =1 a i = m then ther e is a c with 0 c < a and f ( c ) f ( a ) (mo d m ) Pr o of: Construct a c hain of c j j = 1 ; 2 ;::: ;m with a = c 0 > c 1 > c 2 > ::: > c m b y recursiv ely dening c j + 1 b y decreasing one non-zero en try of c j b y 1. After a 1 + a 2 + ::: + a m 1 = m w e reac h 0 so c m = 0 No w consider the v alues of f ( c 1 ) ;::: ;f ( c m ) (mod m ). If they are all distinct, then f ( a ) is congruen t mod m to a unique f ( c j ), satisfying the conclusion of the theorem. Otherwise, for some j;k with 1 j < k m w e ha v e f ( c j ) f ( c k ) (mod m ), and c j > c k Then a > c j c k > 0 and a > a ( c j c k ) with 91

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f ( a ( c j c k )) = f ( a ) f ( c j ) + f ( c k ) f ( a ) (mod m ) [16]. 2 Th us the maxim um n um ber of t ype t w o arcs needed in a minimal path of length r has P a i m Using this, w e are no w ready to pro v e the nal theorem of this paper whic h giv es a bound on repetition in the min-plus algebra in the special case where the matrix is MCM Hamiltonian. Theorem 6.15 If A is a normalize d MCM Hamiltonian matrix on m vertic es, then A ( m 1) 2 +1 = A ( m 1) 2 +1+ m : Pr o of: Using Theorem 6.11, all paths of length r = ( m 1) 2 + 1 + m in D A satisfy (6.2). Since ( m 1) 2 +1+ m = m 2 m +2 = 2 mod m equation 6.2 becomes a 1 + 2 a 2 + ::: ( m 1) a m 1 2 = ( i j )mod m: Let N i be dened as in Denition 6.12, then the n um ber of arcs in suc h a path can be divided in to N 1 + N 2 + N 3 so that N 1 + N 3 + N 3 = ( m 1) 2 + 1 + m and, N 1 + N 2 + N 3 m + X ia i + N 3 : Next, com bining these giv es N 3 (( m 1) 2 + 1 + m ) m X ia i : (6.3) No w giv en that a m 1 = m k Theorem 6.14 sa ys that a minimal path has P m 2 i =1 a i < k: Th us the n um ber of t ype one arcs is maximized b y letting a m 2 = k 1 : 92

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Next, using this and Theorem 3.4, w e nd that a path in D A satises ( m 2)( k 1) + ( m 1)( m k ) 2 = k mod m so ( i j ) = k and the n um ber of t ype one arcs is k Th us 6.3 becomes N 3 = [( m 1) 2 + 1 + m ] k [( m 2)( k 1) + ( m 1)( m k )] = m: No w since w e maximized the n um ber of t ype one arcs in this calculation, w e ha v e the n um ber of N 3 m for an y path in D A This means that m of these t ype three arcs whic h form one cycle of the minim um cycle in whic h all arcs ha v e w eigh t 0 can be remo v ed from an y minimal path of length ( m 1) 2 + 1 + m to obtain a path of length ( m 1) 2 + 1 with the same w eigh t. Finally w e need to sho w that the remo v al of the m t ype three arcs does not disconnect the remaining path. Let D be the path that remains after remo ving the m t ype three arcs. Suppose D is disconnected and let H 1 H 2 ... H k represen t the connected componen ts of D. In eac h componen t H i the minim um w eigh t path bet w een an y pair of v ertices has length at most j H i j 1. The remaining arcs of the path in eac h componen t m ust be part of cycles within this componen t. Since D has m 2 2 m + 2 arcs and at most m 2 of these are neeeded for simple paths in eac h componen t, there are at least m 2 3 m arcs whic h are part of cycles in these componen ts. No w, since m 2 3 m m for all m 4, consider a new path D 0 in whic h m of these arcs from cycles in the componen ts are remo v ed, and replace b y a single cycle of the MCM cycle. This path w ould then be connected, and since the w eigh t of an y cycle in a H i is at least the w eigh t of the minim um cycle mean w e ha v e 93

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that w ( D ) w ( D 0 ). Th us w e can alw a ys nd a minim um w eigh t path in whic h the remo v al of the m t ype three arcs does not disconnect the remaining path. 2 This bound is strong whic h can be seen b y considering a graph with only one a m 1 arc from v ertex m to v ertex 2, a Hamiltonian cycle with all arcs ha ving w eigh t zero and no other arcs. No w, in this graph, there is no path of length ( m 1) 2 from v ertex 1 to itself but there is a path of length ( m 1) 2 + m and w eigh t m 1 from v ertex 1 to itself. T o see this, consider the normalized MCM Hamiltonian transfer matrix A whic h has this graph as its precedence graph. A = 2 6 6 6 6 6 6 6 6 6 4 1 0 1 ::: ::: 1 1 1 0 1 ::: 1 . . . 1 1 ::: ::: 1 0 1 0 1 1 ::: ::: 1 3 7 7 7 7 7 7 7 7 7 5 The po w ers of the matrix are A ( m 1) 2 = 2 6 6 6 6 6 6 6 6 4 1 0 1 2 ::: m 2 m 2 m 1 0 1 ::: m 3 m 3 m 2 . 0 ::: m 4 m 4 m 3 . . . ::: m 5 . . . . . . . . . 0 1 2 ::: ::: m 1 3 7 7 7 7 7 7 7 7 5 A ( m 1) 2 +1 = 2 6 6 6 6 6 6 6 6 4 m 2 m 1 0 1 ::: m 3 m 3 m 2 m 1 0 ::: m 4 . . . . . . . . . 1 2 ::: . ::: 0 0 1 . ::: ::: m 1 m 1 0 1 2 ::: m 2 3 7 7 7 7 7 7 7 7 5 94

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A ( m 1) 2 + m = 2 6 6 6 6 6 6 6 6 4 m 1 0 1 2 ::: m 2 m 2 m 1 0 1 ::: m 3 m 3 m 2 . 0 ::: m 4 m 4 m 3 . . . ::: m 5 . . . . . . . . . 0 1 2 ::: ::: m 1 3 7 7 7 7 7 7 7 7 5 A ( m 1) 2 +1+ m = 2 6 6 6 6 6 6 6 6 4 m 2 m 1 0 1 ::: m 3 m 3 m 2 m 1 0 ::: m 4 . . . . . . . . . 1 2 ::: . ::: 0 0 1 . ::: ::: m 1 m 1 0 1 2 ::: m 2 3 7 7 7 7 7 7 7 7 5 Th us ( A ( m 1) 2 ) 11 = 1 while ( A ( m 1) 2 + m ) 11 = m 1 so the bound is strong. Also, note that A ( m 1) 2 +1 = A ( m 1) 2 +1+ m so the bound holds and th us for an y r ( m 1) 2 + 1 + m w e ha v e that A r = A r m : 6.3 P eriodicit y in Matrices with In teger en tries. In the previous t w o sections, w e sho w that R 0 ( A ) is nite if A is irreducible. Ho w ev er, if A is not MCM Hamiltonian then the size of R 0 ma y be made arbitrarily large b y increasing the w eigh t of the arcs from the critical cycles to the rest of the graph. Consider the matrix A and its po w ers sho wn belo w. A = 2 4 1 0 t 0 3 5 A 2 = 2 4 2 0 t 0 3 5 A t 1 = 2 4 t 1 0 t 0 3 5 A t = 2 4 t 0 t 0 3 5 This matrix has minim um cycle mean 0 on a cycle of length 1 so A r = A r +1 for all r t and th us R 0 ( A ) is dependen t on the en tries in A and as w e increase t w e increase R 0 ( A ). In the applications in the previous c hapters, the transition matrix w as not MCM 95

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Hamiltonian, y et the size of R 0 w as m uc h smaller than the bounds giv en in either of the previous t w o sections. Th us w e will no w look at ho w the size of R 0 migh t be aected b y the en tries in A Denition 6.16 A random matrix has entries which ar e indep endently chosen fr om some distribution of r e al numb ers. These entries will b e r epr esente d by a r andom variable X T o analyze the dependency of A on the size of A and the size of the en tries in A w e compared six sets of random matrices with dieren t size limits on the en tries bet w een sets and dieren t sizes of matrices within sets. F or eac h set, the en tries in the matrices w ere in tegers ranging from 0 to 2, 0 to 4, 0 to 8, 0 to 16, 0 to 32 and, 0 to 64 respectiv ely In eac h set w e look at matrices of size 1 1, 2 2 ::: 100 100. F or eac h size matrix in eac h set, w e create 1000 random matrices with the appropriate en tries and then compute the pre-periodic in terv al for eac h and tak e the a v erage. The results of these experimen ts are sho wn in Figure 6.2. 96

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The size of A R 0 ( A ) 510152025 10 20 30 40 50 60 70 80 90 100 a b c d e f p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p Figure 6.2 The Pre-P eriodic In terv al for Random Matrices With In teger En tries. Labels a : : : f correspond to matrices with en tries from 0 to 2, 0 to 4, 0 to 8, 0 to 16, 0 to 32 and 0 to 64 respectiv ely This gure sho ws that in the discrete case R 0 ( A ) increases as the size of the en tries in A increase. This eect is greatest for smaller size matrices. W e also notice that R 0 ( A ) decreases to a limit as the size of A increases. Theorem 6.17 L et A b e an m m matrix with entries indep endently chosen fr om a distribution that has a minimum value v Assume that the r andom variable X is e qual to v with p ositive pr ob ability. Then the pr ob ability that R 0 ( A ) = 3 appr o aches 1 as the size of A go es to innity. Pr o of: Let D A be the precedence digraph of A and let p be the probabilit y that an arc in D A has w eigh t v Then the probabilit y of a path of w eigh t 2 v from v ertex a to v ertex b through v ertex c is p 2 The probabilit y that no suc h path through c 97

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exists is 1 p 2 and the probabilit y that there is no suc h a path trough an y v ertex in A is (1 p 2 ) m where m is the n um ber of v ertices in D A Finally the probabilit y that there is suc h a path through some v ertex in A is 1 (1 p 2 ) m and th us the probabilit y that there is suc h a path through all v ertices is (1 (1 p 2 ) m ) m 2 Next w e ha v e that 1 x e x so lim m !1 (1 (1 p 2 ) m ) m 2 lim m !1 ( e m 2 (1 p 2 ) m ) = e 0 = 1 : Th us the probabilit y that there is a path of length 2v bet w een ev ery pair of v ertices in A approac hes 1 as m approac hes innit y No w, consider the probabilit y that there is a path of length 3 v bet w een ev ery pair of v ertices. This is a path from v ertex a to v ertex d whic h tra v els through t w o other v ertices c and d If an arc of w eigh t v occurs with probabilit y p then b y the argumen t abo v e, the probabilit y that there is no path of w eigh t 3 v bet w een a giv en pair of v ertices is 1 p 3 Next the probabilit y that no suc h path exist through c and d is (1 p 3 ) m 2 Then b y the argumen t for paths of length 2v, the probabilit y that there is a path of w eigh t 3 v or less bet w een ev ery pair of v ertices is (1 (1 p 3 ) m 2 ) m 2 Next b y the argumen t abo v e, lim m !1 (1 (1 p 3 ) m 2 ) m 2 lim m !1 e m 2 (1 p 3 ) m 2 = e 0 = 1 : Th us the probabilit y that A 2 = 2 v 0 and A 3 = 3 v 0 where 0 is the matrix of all zeros approac hes 1 as m approac hes innit y When this happens, w e ha v e A 3 = v A 2 and A r = v A r 1 for all r 3 98

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and th us R 0 ( A ) = 3. 2 Note that this theorem does not state that R 0 ( A ) = 3 for an y size of A mearly that for large A there is a positiv e probabilit y of this. Indeed, in the applications from Chapters 3-5 the v alue for R 0 w as larger than 3 for all sizes of the transition matrix. This is not unexpected since the size of these matrices w as relativ ely small and Figure 6.3 indicates that the a v erage v alue for R 0 does not approac h 3 un til the size of matrix gets larger. 6.4 P eriodicit y in Random Matrices With Real En tries The next question is what happens to the pre-periodic in terv al when the en tries in A are independen tly and iden tically distributed random v ariables c hosen from a distribution with a con tin uous densit y function. T o analyze this w e will look at 2 2 matrices with random en tries from dieren t con tin uous distributions. These en tries will be represen ted b y a random v ariable X Denition 6.18 A r andom variable X is continuous if ther e exists a non-ne gative function f ( x ) dene d for all r e al x 2 ( 1 ; 1 ) having the pr op erty that for any set B of r e al numb ers, the pr ob ability that the X is in this set is dene d as P f X 2 B g = Z B f ( x ) dx: The function f ( x ) is c alle d the p robabilit y densit y function of the r andom variable X Th us, if B = [ a;b ] then P f a X b g = Z a b f ( x ) dx and P f X = a g = Z a a f ( x ) dx = 0 : 99

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This sho ws that the probabilit y that the random v ariable will assume an y particular v alue is zero. T o analyze the pre-periodic in terv al of our random matrix A w e will rst simplify the problem b y limiting the en tries in A to real n um bers from a uniform distribution in the in terv al (0,1). Denition 6.19 A r andom variable is unifo rmly distributed over the interval (0,1) if its pr ob ability function is given by f ( x ) = 8 < : 1 if 0 < x < 1 0 otherwise In gener al X is a uniform r andom variable on ( ; ) if its pr ob ability density function is given by f ( x ) = 8 < : 1 if < x < 0 otherwise No w, to nd the pre-periodic in terv al of a random matrix A with en tries from a uniform distribution, w e m ust rst determine ho w the v alue of R 0 is dependen t on the en tries in A W e will begin b y looking at the v alues of R 0 ( A ) for a 2 2 matrix A = 2 4 a b c d 3 5 : Since R 0 is determined b y comparing po w ers of A w e need to kno w what these po w ers look lik e in the general case. This is sho wn in the next t w o lemmas. 100

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Lemma 6.20 F or A as state d ab ove, assume b + c < 2 a and a d then the following p owers of A ar e found by Min-Plus matrix exp onentiation. A 2 = 2 4 b + c a + b a + c b + c 3 5 ; A 3 = 2 4 a + b + c 2 b + c b + 2 c a + b + c 3 5 ; A 4 = 2 4 2 b + 2 c a + 2 b + c a + b + 2 c 2 b + 2 c 3 5 A r = ( r 2)( b + c 2 ) A 2 if r even and A r = ( r 3)( b + c 2 ) A 3 if r o dd : Lemma 6.21 Now assume 2 a b + c and a d Then matrix exp onentiation gives A 2 = 2 4 min (2 a;b + c ) min ( a + b;b + d ) min ( a + c;c + d ) min ( b + c; 2 d ) 3 5 = 2 4 2 a a + b c + b min ( b + c; 2 d ) 3 5 : A r = 2 4 ra ( r 1) a + b ( r 1) a + c min (( r 2) a + b + a;rd ) 3 5 for r > 2 Th us for an y v alue of the en tries in A the po w ers of A are dened b y the previous lemmas. No w w e need to compare these po w ers and nd the v alues of R 0 for all possible com binations of the en tries in A Lemma 6.22 L et A = 2 4 a b c d 3 5 L et A 1 = A and r e cursively dene A r = A A r 1 wher e the matrix multiplic ation is in min-plus algebr a. L et R 0 ( A ) b e the minimum R 0 wher e A r = q A r p 8 r R 0 for some p ositive values p and q If a d then 101

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R 0 ( A ) = 8 > > > > > > > > > < > > > > > > > > > : 2 if 2 a = 2 d b + c 3 if b + c < 2 a = 2 d or 2 a b + c 2 d and a < d 4 if b + c < 2 a or a + b + c 3 d < b + c 2 and a < d r > 4 if ( r 3) a + b + c r 1 d < ( r 4) a + b + c r 2 and a < d Pr o of: F rom Lemma 1 w e ha v e A 4 = ( b + c ) A 2 and R 0 ( A ) 4. Since b + c < 2 a w e do not ha v e A 2 = q A nor A 3 = q A 2 for an y q If R 0 ( A ) = 3, then A 3 = ( b + c ) + A giving a = d Then from Lemma 2, R 0 ( A ) = 2 only if A 2 = a A This occurs only if a = d F or r > 2, w e ha v e A r = a A r 1 only if ( r 3) a + b + c ( r 1) d Th us R 0 ( A ) = 3 only if a < d and b + c 2 d And for r > 3, w e ha v e R 0 ( A ) = r only if ( r 3) c + b + c ( r 1) d and ( r 4) a + b + c > ( r 2) d Compiling these observ ations giv es the result and a similar result is obtained if a > d b y symmetry 2 Theorem 6.23 L et A b e a 2 2 matrix whose elements ar e indep endently chosen fr om the uniform distribution on [0 ; 1] L et R 0 ( A ) b e as ab ove. Then P ( R 0 ( A ) = r ) = 8 > > > > > > > > > > < > > > > > > > > > > : 0 if r 2 5 12 if r = 3 7 18 if r = 4 7 12( r 2)( r 1) if r > 4 Pr o of: F or an y con tin uous distribution, the probabilit y that a = d is zero. Th us P ( R 0 ( A ) = 2) is zero. In the uniform distribution, the probabilit y densit y function f ( x ) is exactly 1 on the in terv al (0,1), and 102

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P ( R 0 ( A ) = 3) = 2 P (2 a b + c 2 d ) =2 Z 1 0 Z 1 0 Z ( b + c ) = 2 0 Z 1 ( b + c ) = 2 dddadbdc =2 Z 1 0 Z 1 0 b + c 2 1 b + c 2 2 dbdc =2 Z 1 0 Z 1 0 b + c 2 dbdc Z 1 0 Z 1 0 b + c 2 2 dbdc =2 1 2 7 24 = 5 12 : F or r = 4, w e ha v e P ( b + c 2min( a;d )) = Z 1 0 Z 1 0 Z 1 ( b + c ) = 2 Z 1 ( b + c ) = 2 dddadbdc = Z 1 0 Z 1 0 1 b + c 2 2 dbdc = Z 1 0 Z 1 0 dbdc 2 Z 1 0 Z 1 0 b + c 2 dbdc + Z 1 0 Z 1 0 b + c 2 2 dbdc = 1 2 1 2 + 7 24 = 7 24 : 103

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F or r 4, w e also ha v e P ( r 3) a + b + c 3 ( r 2) d < ( r 4) a + b + c 2 = Z 1 0 Z 1 0 Z ( b + c ) = 2 0 Z (( r 4) a + b + c ) = ( r 2) (( r 3) a + b + c ) = ( r 1) dddadbdc = Z 1 0 Z 1 0 Z ( b + c ) = 2 0 ( r 4) a + b + c r 2 ( r 3) a + b + c r 1 dadbdc = Z 1 0 Z 1 0 Z ( b + c ) = 2 0 b + c 2 a ( r 2)( r 1) dadbdc = 1 ( r 2)( r 1) Z 1 0 Z 1 0 ( b + c ) a + b 2 b + c 2 2 dadbdc = 1 ( r 2)( r 1) Z 1 0 Z 1 0 b + c 2 2 dadbdc = 7 24( r 2)( r 1) : Th us P ( R 0 ( A ) = 4) = 7 24 + 2 7 24 (4 2) (4 1) = 7 18 and for r > 4, w e ha v e P ( R 0 ( A ) = r ) = 2 7 24 ( r 2)( r 1) = 7 12( r 2)( r 1) : Then putting these together giv es the result. Note that 1 X r =3 P ( R 0 ( A ) = r ) = 5 12 + 7 18 + 7 12 1 3 4 + 1 4 5 + 1 5 6 + = 5 12 + 7 18 + 7 12 1 3 = 1 ; as it should. 2 104

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Corollary 6.24 Under the c onditions ab ove, E ( R 0 ( A )) = 1 Pr o of: W e ha v e E ( R 0 ( A )) = 1 X r =3 rP ( R 0 ( A ) = r ) = 3 5 12 + 4 7 18 + 7 12 5 3 4 + 6 4 5 + 7 5 6 + = 1 : 2 Next, w e will extend this result to sho w that E ( R 0 ( A )) = 1 for all irreducible matrices A First, w e need to sho w that if the en tries in A 2 2 come from an y con tin uous distribution then E ( R 0 ( A )) = 1 Theorem 6.25 L et A b e a 2 2 matrix whose elements ar e indep endently chosen fr om the c ontinuous distribution on < then E ( R 0 ( A )) = 1 Pr o of: Let f ( x ) be the probabilit y densit y function for the random v ariable X for eac h of the en tries in A Since f ( x ) 0 and con tin uous, there exist a constan t x 0 so that f ( x 0 ) > 0. Let = f ( x 0 ) 2 Then there is a > 0 so that j f ( x ) f ( x 0 ) j < whenev er x 2 ( x 0 ;x 0 + ). No w, eac h of the probabilit y functions for the uniform distribution are the same for the con tin uous distribution except that probabilit y distribution functions are f ( x ) instead of 1, and the functions are in tegrated o v er all of < instead of just the in terv al (0,1). F or example, in the con tin uous distribution P ( R 0 ( A ) = 3) = 2 P (2 a b + c 2 d ) = 2 Z 1 1 Z 1 1 Z ( b + c ) = 2 1 Z 1 ( b + c ) = 2 f ( d ) f ( a ) f ( b ) f ( c ) dddadbdc 105

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By con tin uit y the probabilit y distribution function f ( x ) for the con tin uous distribution is greater than the constan t function f ( x 0 ) 2 on the in terv al dened abo v e. Th us if w e dene g ( x ) = 8 < : f ( x 0 ) 2 if x 2 ( x 0 ;x 0 + ) 0 otherwise Then g ( x ) is the probabilit y densit y function of a uniform distribution. No w since f ( x ) g ( x ), w e ha v e 2 Z 1 1 Z 1 1 Z ( b + c ) = 2 1 Z 1 ( b + c ) = 2 f ( d ) f ( a ) f ( b ) f ( c ) dddadbdc 2 Z 1 1 Z 1 1 Z ( b + c ) = 2 1 Z 1 ( b + c ) = 2 g ( d ) g ( a ) g ( b ) g ( c ) dddadbdc = 2 Z x 0 + x 0 Z x 0 + x 0 Z ( b + c ) = 2 x 0 Z x 0 + ( b + c ) = 2 f ( x 0 ) 2 f ( x 0 ) 2 f ( x 0 ) 2 f ( x 0 ) 2 dddadbdc = 2 f ( x 0 ) 2 4 Z x 0 + x 0 Z x 0 + x 0 Z ( b + c ) = 2 x 0 Z x 0 + ( b + c ) = 2 dddadbdc (6.4) No w to map the in terv al ( x ;x + ) to the in terv al (0,1), dene a linear c hange of v ariable with a = x 0 + 2 a 0 b = x 0 + 2 b 0 c = x 0 + 2 c 0 d = x 0 + 2 d 0 Then [6.4] becomes 2 f ( x 0 ) 2 4 Z 1 0 Z 1 0 Z ( b 0 + c 0 ) = 2 0 Z 1 ( b 0 + c 0 ) = 2 f ( d 0 ) f ( a 0 ) f ( b 0 ) f ( c 0 ) dd 0 da 0 db 0 dc 0 This is the same equation obtained in the uniform distribution so P ( R 0 = 3) for the con tin uous distribution is bounded belo w b y a constan t m ultiple of P ( R 0 ( A )) for the uniform distribution. Similar computations also bounds the v alues for P ( R 0 ( A ) = 4) 106

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and P ( R 0 ( A ) = r ). No w w e use these bounds and the computation of the expected v alue for the uniform distribution denoted E U ( R 0 ( A )) to nd the expected v alue of for the con tin uous distribution denoted E C ( R 0 ( A )). E C ( R 0 ( A )) = 1 X r =3 rP ( R 0 ( A ) = r ) f ( x 0 ) 2 4 E U ( R 0 ( A )) = f ( x 0 ) 2 4 3 5 12 + 4 7 18 + 7 12 5 3 4 + 6 4 5 + 7 5 6 + = f ( x 0 ) 2 4 1 = 1 : 2 Next, w e need to sho w that E ( R 0 ( A )) = 1 for an y m m matrix A W e will sho w that the beha vior of A is dictated b y the beha vior of a 2 2 bloc k of A and then w e can apply the previous theorems to the 2 2 bloc k and hence all of A Denition 6.26 A critical block of a matrix A is a principle submatrix of A wher e the entries within this submatrix ar e smaller than all the other entries in A Lemma 6.27 If A is a m m matrix with a critic al 2 2 blo ck U then R 0 ( U ) R 0 ( A ) Pr o of: Without loss of generalit y w e can do a symmetric perm utation of A to obtain A = 2 4 U . ::: ::: 3 5 : Let u i and u j be the v ertices of U Consider all paths of length r from u i to u j if this path uses arcs bet w een U and the rest of A then eac h of these arcs could be 107

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replaced b y an arc of lesser w eigh t in U Th us an y minimal path bet w een u i and u j uses only arcs of U This giv es. A r = 2 4 U r . ::: ::: 3 5 : Let k R 0 ( A ) be the smallest v alue so A k = 2 4 U k . ::: ::: 3 5 and A k p = 2 4 q U k p . ::: ::: 3 5 : The inequalit y holds because the submatrix U ma y be periodic before the en tire matrix A is. Th us R 0 ( U ) R 0 ( A ). 2 Theorem 6.28 L et A b e a m m matrix whose entries ar e indep endently chosen fr om the c ontinuous distribution on < Then E ( R 0 ( A )) = 1 Pr o of: Let R 0 ( A ) = 8 < : R 0 ( A ) if A has a 2 2 critical bloc k U 0 otherwise The probabilit y that A m m con tains a critical bloc k U 2 2 is giv en b y P ( U 2 A ) = 1 0 @ m 2 4 1 A : 108

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Then E ( R 0 ( A )) = E ( R 0 ( A )) + 0 = 1 X r =3 r P ( R 0 ( A ) = r ) 1 X r =3 r P ( R 0 ( U ) = r ) P ( U 2 A ) = 1 0 @ m 2 4 1 A E ( R 0 ( U )) = 1 0 @ m 2 4 1 A 1 = 1 2 6.5 Conclusions In section 6.1, w e ha v e a bound for R 0 ( A ) in terms of its en tries. W e can compare this bound with the actual v alues of R 0 in the applications. In computing this bound, w e needed to consider en tries in A k where k m 2 m + 1 guaran tees that there is a path of an y length greater than k In most cases, the actual v alue of k is m uc h smaller than m 2 m + 1. If the en tries in the successiv e po w ers of the matrix gro w v ery slo wly this will not ha v e a large eect on the accuracy of the bound. Ho w ev er if the en tries gro w signican tly from one po w er of the matrix to the next, then this bound ma y be far less accurate. 109

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W e cannot illustrate the bound dev eloped in Section 6.2 in an y of our applications because none of the transitions matrices w ere MCM Hamiltonian. Ho w ev er, in ev ery application, w e had R 0 m 2 m + 1. F or the bound in Section 6.1 to be larger than the bound in Section 6.2, the diameter of the matrix w ould ha v e to be large. In our examples, the diameter of the matrix w as less than 2 m where m is the size of the matrix, so the bound from Section 6.1 is closer to the actual v alue of R 0 for these matrices. Finally looking at the results from sections 6.3 and 6.4 w e see that for an y irreducible matrix A the pre-periodic in terv al R 0 ( A ) is nite, but the expected v alue of R 0 ( A ) for all suc h A is innite. Giv en this result, it ma y seem somewhat surprising that in all of our examples R 0 ( A ) w as relativ ely small. Ho w ev er, when w e also consider the dependency of R 0 ( A ) on the size of the en tries in A whic h w ere restricted to f 0 ; 1 ; 2 g in all of our examples, w e migh t expect reasonably small v alues for R 0 ( A ). In fact, looking at Figure 6.2 w e see that in our experimen t, the a v erage v alue for R 0 ( A ) with the en tries in A c hosen from f 0 ; 1 ; 2 g is less than 5 for all sizes of A This a v erage v alue is less than the v alues found for the transition matrices in our applications. The applications presen ted in Chapters 3-5 are just a sample of the problems that can be solv ed using Min-Plus algebra. Other nite state problems that ha v e been solv ed using similar methods include independence n um ber of graphs, coloring of graphs, shipping problems, and ev en an optimal currency trading problem. The algorithms for solving all of these problems are similar to those discussed in this 110

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thesis, and the dening feature of these algorithms is periodicit y whic h allo ws us to nd innitely man y solutions in nite time. The main limitation of these algorithms is the n um ber of iterations needed before periodicit y occurs. The bounds dev eloped in this Chapter can be used to determine the size of the pre-periodic in terv al and th us the feasibilit y of solving problems with Min-Plus algebra. 111

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A. APPENDIX Domination Num bers of Circulan t Graphs Theorem 1. L et S = f 1 g Then r ( C n [ S ]) = n 3 for all n 2 Theorem 2. L et S = f 2 g Then for all n 4 ; we have r ( C n [ S ]) = n 3 + ( 1 if n 2 (mo d 6) 0 otherwise. Theorem 3. L et S = f 1 ; 2 g Then r ( C n [ S ]) = n 5 for all n 4 Theorem 4. L et S = f 3 g Then for all n 6 ; we have r ( C n [ S ]) = n 3 + 8 > < > : 2 if n 3 (mo d 9) 1 if n 6 (mo d 9) 0 otherwise. Theorem 5. L et S = f 1 ; 3 g Then for all n 6 ; we have r ( C n [ S ]) = n 5 + ( 1 if n 4 (mo d 5) 0 otherwise. Theorem 6. L et S = f 2 ; 3 g Then r ( C n [ S ]) = n 4 for all n 6 Theorem 7. L et S = f 1 ; 2 ; 3 g Then r ( C n [ S ]) = n 7 for all n 6 Theorem 8. L et S = f 4 g Then for all n 8 ; we have r ( C n [ S ]) = n 3 + 8 > < > : 2 if n 4 (mo d 12) 1 if n 2, 8 (mo d 12) 0 otherwise. Theorem 9. L et S = f 1 ; 4 g Then for all n 8 ; we have r ( C n [ S ]) = 2 n 9 + ( 1 if n 3, 4, 8 (mo d 9) exc ept n = 13, 26, 39 0 otherwise. Theorem 10. L et S = f 2 ; 4 g Then for all n 8 ; we have r ( C n [ S ]) = n 5 + ( 1 if n 2, 4 (mo d 10) 0 otherwise. 112

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Theorem 11. L et S = f 1 ; 2 ; 4 g Then for all n 8 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6 (mo d 7) 0 otherwise. Theorem 12. L et S = f 3 ; 4 g Then for all n 8 ; we have r ( C n [ S ]) = n 5 + ( 1 if n 2, 3, 4 (mo d 5) exc ept n = 12, 13 0 otherwise. Theorem 13. L et S = f 1 ; 3 ; 4 g Then r ( C n [ S ]) = n 6 for all n 8 Theorem 14. L et S = f 2 ; 3 ; 4 g Then for all n 8 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 4, 5, 6 (mo d 12) 0 otherwise. Theorem 15. L et S = f 1 ; 2 ; 3 ; 4 g Then r ( C n [ S ]) = n 9 for all n 8 Theorem 16. L et S = f 5 g Then for all n 10 ; we have r ( C n [ S ]) = n 3 + 8 > < > : 3 if n 5 (mo d 15) 1 if n 10 (mo d 15) 0 otherwise. Theorem 17. L et S = f 1 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 2 n 9 + 8 < : 1 if n 1, 3, 4, 6, 8 (mo d 9) exc ept n = 10, 15, 19, 24, 28, 37, 42 0 otherwise. Theorem 18. L et S = f 2 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 2 n 9 + ( 1 if n 4 (mo d 9) exc ept n = 13 0 otherwise. Theorem 19. L et S = f 1 ; 2 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 3 n 17 + ( 1 if n 11, 16 (mo d 17) exc ept n = 11, 33 0 otherwise. 113

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Theorem 20. L et S = f 3 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 2 n 9 + 8 < : 1 if n 3, 4, 8, 9, 10, 12, 13, 15, 17 (mo d 18) exc ept n = 10, 12, 13, 15, 26, 28, 33, 39, 46, 51, 64 0 otherwise. Theorem 21. L et S = f 1 ; 3 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 4, 6 (mo d 7) exc ept n = 11 0 otherwise. Theorem 22. L et S = f 2 ; 3 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 2 n 13 + ( 1 if n 5, 6, 12 (mo d 13) exc ept n = 12, 18 0 otherwise. Theorem 23. L et S = f 1 ; 2 ; 3 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 8 (mo d 9) 0 otherwise. Theorem 24. L et S = f 4 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 3 n 14 + ( 1 if n 4, 7, 8, 9 (mo d 14), or n = 12 0 otherwise. Theorem 25. L et S = f 1 ; 4 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 4, 5, 6 (mo d 7) exc ept n = 11, 12, 13, 25, 26, 39 0 otherwise. Theorem 26. L et S = f 2 ; 4 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 3, 4, 5, 6 (mo d 12) exc ept n = 15 0 otherwise. Theorem 27. L et S = f 1 ; 2 ; 4 ; 5 g Then r ( C n [ S ]) = n 8 for all n 10 Theorem 28. L et S = f 3 ; 4 ; 5 g Then r ( C n [ S ]) = n 6 for all n 10 Theorem 29. L et S = f 1 ; 3 ; 4 ; 5 g Then r ( C n [ S ]) = n 7 for all n 10 114

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Theorem 30. L et S = f 2 ; 3 ; 4 ; 5 g Then for all n 10 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 5, 6, 7 (mo d 15) 0 otherwise. Theorem 31. L et S = f 1 ; 2 ; 3 ; 4 ; 5 g Then r ( C n [ S ]) = n 11 for all n 10 Theorem 32. L et S = f 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 3 + 8 > > > < > > > : 4 if n 6 (mo d 18) 2 if n 3, 12 (mo d 18) 1 if n 2, 8, 14, 15 (mo d 18) 0 otherwise. Theorem 33. L et S = f 1 ; 6 g Then r ( C n [ S ]) = l 3 n 13 m for all n 12 Theorem 34. L et S = f 2 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 5 + 8 > < > : 2 if n 8 (mo d 10) 1 if n 2, 4, 9 (mo d 10) 0 otherwise. Theorem 35. L et S = f 1 ; 2 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = 2 n 11 + 8 < : 1 if n 3, 4, 5, 8, 10 (mo d 11) exc ept n = 14, 15, 19, 25, 30 0 otherwise. Theorem 36. L et S = f 3 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 5 + 8 > < > : 2 if n 3 (mo d 15) 1 if n 6, 9 (mo d 15) 0 otherwise. Theorem 37. L et S = f 1 ; 3 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 4, 6, 9, 11, 12, 14, 16, 17 (mo d 18) exc ept n = 14, 22, 29, 32, 40, 50, 58, 68 0 otherwise. Theorem 38. L et S = f 2 ; 3 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 2, 5, 6 (mo d 7) 0 otherwise. 115

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Theorem 39. L et S = f 1 ; 2 ; 3 ; 6 g Then r ( C n [ S ]) = n 7 for all n 12 Theorem 40. L et S = f 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 4 + ( 1 if n 2, 4, 10, 12 (mo d 16) 0 otherwise. Theorem 41. L et S = f 1 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = 3 n 17 + ( 1 if n = 16, 27 0 otherwise. Theorem 42. L et S = f 2 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 2, 4, 6 (mo d 14) 0 otherwise. Theorem 43. L et S = f 1 ; 2 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 8 (mo d 9) 0 otherwise. Theorem 44. L et S = f 3 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = 2 n 13 + ( 1 if n 4, 6, 12 (mo d 13) 0 otherwise. Theorem 45. L et S = f 1 ; 3 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 7 (mo d 8) 0 otherwise. Theorem 46. L et S = f 2 ; 3 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 4, 6, 7, 8, 15 (mo d 16) 0 otherwise. Theorem 47. L et S = f 1 ; 2 ; 3 ; 4 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 10 (mo d 11) 0 otherwise. 116

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Theorem 48. L et S = f 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = 5 n 23 + 8 > > > > > < > > > > > : 2 if n 18 (mo d 23) exc ept n = 18, 41, 64 1 if n 2, 3, 4, 5, 8, 9, 11, 12, 13, 17, 19, 20, 21, 22 (mo d 23) exc ept n = 12, 13, 17, 19, 20, 21, 25, 26, 28, 34, 42, 43, 51, 65, or n = 18, 41, 64 0 otherwise. Theorem 49. L et S = f 1 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = 2 n 13 + ( 1 if n 4, 5, 6 (mo d 13) exc ept n = 17, 18 0 otherwise. Theorem 50. L et S = f 2 ; 5 ; 6 g Then r ( C n [ S ]) = n 5 for all n 12 Theorem 51. L et S = f 1 ; 2 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8 (mo d 9) exc ept n = 15 0 otherwise. Theorem 52. L et S = f 3 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 2, 3, 4, 5, 6 (mo d 7) exc ept n = 16, 17, 18 0 otherwise. Theorem 53. L et S = f 1 ; 3 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6 (mo d 14) 0 otherwise. Theorem 54. L et S = f 2 ; 3 ; 5 ; 6 g Then r ( C n [ S ]) = n 7 for all n 12 Theorem 55. L et S = f 1 ; 2 ; 3 ; 5 ; 6 g Then r ( C n [ S ]) = n 10 for all n 12 Theorem 56. L et S = f 4 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 2, 3, 4, 5, 6 (mo d 7) exc ept n = 16, 17, 18, 19, 23, 37, 38 0 otherwise. Theorem 57. L et S = f 1 ; 4 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 8 + ( 1 if n = 20 0 otherwise. 117

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Theorem 58. L et S = f 2 ; 4 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 4, 5, 6, 7, 8 (mo d 16) exc ept n = 21 0 otherwise. Theorem 59. L et S = f 1 ; 2 ; 4 ; 5 ; 6 g Then r ( C n [ S ]) = n 9 for all n 12 Theorem 60. L et S = f 3 ; 4 ; 5 ; 6 g Then r ( C n [ S ]) = n 7 for all n 12 Theorem 61. L et S = f 1 ; 3 ; 4 ; 5 ; 6 g Then r ( C n [ S ]) = n 8 for all n 12 Theorem 62. L et S = f 2 ; 3 ; 4 ; 5 ; 6 g Then for all n 12 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8, 9 (mo d 18) 0 otherwise. Theorem 63. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 g Then r ( C n [ S ]) = n 13 for all n 12 Theorem 64. L et S = f 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 3 + 8 > < > : 4 if n 7 (mo d 21) 2 if n 14 (mo d 21) 0 otherwise. Theorem 65. L et S = f 1 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > < > > > > > : 2 if n 1 (mo d 5) exc ept n = 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81 1 if n 2, 3, 4 (mo d 5) exc ept n = 18, 22, 27, or n = 21, 41, 46, 51, 56, 61, 66, 71, 76, 81 0 otherwise. Theorem 66. L et S = f 2 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 5 n 23 + 8 > > > > > > > < > > > > > > > : 2 if n 9 (mo d 23) exc ept n = 32, 55, 78, 124, 147, 216 1 if n 2, 3, 4, 7, 8, 10, 13, 14, 16, 17, 18, 20, 21, 22 (mo d 23) exc ept n = 14, 16, 20, 26, 27, 37, 39, 43, 54, 60, 66, 79, 108, 135, or n = 32, 55, 78, 124, 147, 216 0 otherwise. Theorem 67. L et S = f 1 ; 2 ; 7 g Then r ( C n [ S ]) = l 2 n 11 m for all n 14 118

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Theorem 68. L et S = f 3 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 9 + 8 > > > < > > > : 1 if n 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 (mo d 18) exc ept n = 23, 24, 25, 26, 28, 29, 30, 38, 39, 41, 42, 43, 46, 47, 56, 59, 60, 64, 65, 77, 78, 82, 95 0 otherwise. Theorem 69. L et S = f 1 ; 3 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 6 (mo d 12) 0 otherwise. Theorem 70. L et S = f 2 ; 3 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 19 + 8 < : 1 if n 2, 5, 6, 9, 10, 12, 16, 17, 18 (mo d 19) exc ept n = 18, 21, 24, 36, 48, 54, 66, 78 0 otherwise. Theorem 71. L et S = f 1 ; 2 ; 3 ; 7 g Then r ( C n [ S ]) = n 7 for all n 14 Theorem 72. L et S = f 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > < > > > > > : 2 if n 1, 4 (mo d 5) exc ept n = 14, 16, 19, 21, 24, 26, 31, 34, 36, 41, 46, 51, 54, 71, 81 1 if n 2, 3 (mo d 5) exc ept n = 17, 18, 27, or n = 14, 19, 21, 24, 26, 31, 34, 41, 46, 51, 54, 71, 81 0 otherwise. Theorem 73. L et S = f 1 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 19 + 8 < : 1 if n 4, 5, 6, 8, 11, 12, 15, 17, 18 (mo d 19) exc ept n = 15, 17, 24, 27, 34, 36, 46, 72, 84 0 otherwise. Theorem 74. L et S = f 2 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 13 + ( 1 if n 4, 5, 6 (mo d 13) exc ept n = 17 0 otherwise. Theorem 75. L et S = f 1 ; 2 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 22 + ( 1 if n = 20 0 otherwise. 119

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Theorem 76. L et S = f 3 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 13 + 8 < : 1 if n 3, 4, 6, 11, 12 (mo d 13) exc ept n = 16, 17, 24, 29, 30, 38, 42, 55, 76 0 otherwise. Theorem 77. L et S = f 1 ; 3 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 7, 8 (mo d 9) 0 otherwise. Theorem 78. L et S = f 2 ; 3 ; 4 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 6, 7 (mo d 8) 0 otherwise. Theorem 79. L et S = f 1 ; 2 ; 3 ; 4 ; 7 g Then r ( C n [ S ]) = n 9 for all n 14 Theorem 80. L et S = f 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 4 n 19 + 8 > > > < > > > : 2 if n 8 (mo d 19) exc ept n = 27, 46, 65 1 if n 2, 4, 5, 7, 9, 10, 11, 12, 13, 14, 18 (mo d 19) exc ept n = 18, 24, 26, 43, 45, 62, 81, or n = 46, 65 0 otherwise. Theorem 81. L et S = f 1 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 13 + ( 1 if n 2, 4, 5, 6, 12 (mo d 13) exc ept n = 15 0 otherwise. Theorem 82. L et S = f 2 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 19 + ( 1 if n 5, 6, 12, 18 (mo d 19) exc ept n = 37 0 otherwise. Theorem 83. L et S = f 1 ; 2 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 3, 5, 6, 7, 12, 14 (mo d 15) exc ept n = 21, 42, 63 0 otherwise. 120

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Theorem 84. L et S = f 3 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 6 + 8 > > > > > < > > > > > : 2 if n 12 (mo d 24) exc ept n = 36, 60, 84 1 if n 3, 4, 5, 6, 10, 11, 14, 16, 17, 18, 19, 21, 23 (mo d 24) exc ept n = 14, 16, 17, 19, 21, 34, 38, 43, 51, 67, or n = 36, 60, 84 0 otherwise. Theorem 85. L et S = f 1 ; 3 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 6, 8 (mo d 9) exc ept n = 15, 31 0 otherwise. Theorem 86. L et S = f 2 ; 3 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 3, 5, 7, 8, 14, 16 (mo d 17) exc ept n = 14, 16, 31, 48 0 otherwise. Theorem 87. L et S = f 1 ; 2 ; 3 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 8, 10 (mo d 11) 0 otherwise. Theorem 88. L et S = f 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 16 + 8 > > > < > > > : 1 if n 4, 5, 8, 9, 10, 15 (mo d 16) exc ept n = 15, 20, 21, 24, 25, 31, 36, 40, 41, 42, 52, 56, 57, 63, 72, 73, 84, 88, 104, 105, 120, 136, 168 0 otherwise. Theorem 89. L et S = f 1 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 3, 4, 5, 6, 7, 14 (mo d 15) exc ept n = 14, 19, 21, 33, 35, 49, 63 0 otherwise. Theorem 90. L et S = f 2 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 4, 6, 7 (mo d 16) 0 otherwise. 121

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Theorem 91. L et S = f 1 ; 2 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 9 (mo d 10) 0 otherwise. Theorem 92. L et S = f 3 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 5, 7, 14 (mo d 15) 0 otherwise. Theorem 93. L et S = f 1 ; 3 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 8, 17 (mo d 18) 0 otherwise. Theorem 94. L et S = f 2 ; 3 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 5, 7, 8, 9, 18 (mo d 19) exc ept n = 24 0 otherwise. Theorem 95. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 12 (mo d 13) 0 otherwise. Theorem 96. L et S = f 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > > > > > > > > > > < > > > > > > > > > > > > > > : 2 if n 1, 2, 3 (mo d 5) exc ept n = 16, 17, 18, 21, 22, 23, 26, 27, 28, 31, 32, 36, 37, 38, 41, 42, 43, 46, 47, 51, 57, 58, 61, 62, 66, 68, 73, 76, 77, 81, 87, 92, 96, 103, 106, 111, 122, 133, 141, 152, 171 1 if n 4 (mo d 5) exc ept n = 19, or n = 17, 18, 21, 22, 23, 26, 28, 31, 32, 36, 37, 41, 42, 43, 47, 51, 58, 61, 62, 66, 68, 73, 77, 81, 87, 92, 96, 103, 106, 111, 122, 133, 141, 152, 171 0 otherwise. Theorem 97. L et S = f 1 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 20 + 8 < : 1 if n 4, 5, 6, 12, 13, 16, 17, 18, 19 (mo d 20) exc ept n = 16, 17, 18, 19, 24, 36, 37, 38, 56, 57, 76 0 otherwise. 122

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Theorem 98. L et S = f 2 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 4, 5, 6, 10, 11, 12 (mo d 18) exc ept n = 22, 28, or n = 16, 26 0 otherwise. Theorem 99. L et S = f 1 ; 2 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 5, 6, 7 (mo d 15) exc ept n = 20, 21, 22, 65, 66, 110 0 otherwise. Theorem 100. L et S = f 3 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 19 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 10, 11, 12 (mo d 19) exc ept n = 23, 26, 27, 46, 64 0 otherwise. Theorem 101. L et S = f 1 ; 3 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 2, 6, 7, 8, 13, 14, 15 (mo d 16) exc ept n = 18 0 otherwise. Theorem 102. L et S = f 2 ; 3 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6, 13, 20, 27, 34 (mo d 42) 0 otherwise. Theorem 103. L et S = f 1 ; 2 ; 3 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 8, 9, 10 (mo d 11) 0 otherwise. Theorem 104. L et S = f 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 4 n 23 + 8 < : 1 if n 4, 5, 10, 11, 15, 16, 17, 22 (mo d 23) exc ept n = 15, 17, 34, 51, 68, 85, 102, 119, 153, or n = 36 0 otherwise. Theorem 105. L et S = f 1 ; 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 5, 6, 7, 8 (mo d 9) 0 otherwise. 123

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Theorem 106. L et S = f 2 ; 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 3, 4, 5, 6, 7, 8 (mo d 16) exc ept n = 19, 21, 35 0 otherwise. Theorem 107. L et S = f 1 ; 2 ; 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 8 (mo d 9) exc ept n = 15 0 otherwise. Theorem 108. L et S = f 3 ; 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 6, 7, 8 (mo d 17) exc ept n = 24, 40, or n = 15, 20 0 otherwise. Theorem 109. L et S = f 1 ; 3 ; 4 ; 6 ; 7 g Then r ( C n [ S ]) = n 9 for all n 14 Theorem 110. L et S = f 2 ; 3 ; 4 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 7, 8 (mo d 17) exc ept n = 23, 24, 40 0 otherwise. Theorem 111. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 7 g Then r ( C n [ S ]) = n 12 for all n 14 Theorem 112. L et S = f 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 3 n 19 + 8 < : 1 if n 5, 6, 9, 10, 11, 12 (mo d 19) exc ept n = 24, 47, 48, or n = 16 0 otherwise. Theorem 113. L et S = f 1 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 5, 6, 7, 8 (mo d 18) 0 otherwise. Theorem 114. L et S = f 2 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 (mo d 21) exc ept n = 25, 26, 29, 30, 50, 51, 52 0 otherwise. 124

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Theorem 115. L et S = f 1 ; 2 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 10 + ( 1 if n = 26 0 otherwise. Theorem 116. L et S = f 3 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 4, 5, 6 (mo d 7) exc ept n = 18, 19 0 otherwise. Theorem 117. L et S = f 1 ; 3 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 6, 8 (mo d 18) 0 otherwise. Theorem 118. L et S = f 2 ; 3 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 7, 8 (mo d 17) 0 otherwise. Theorem 119. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 7 g Then r ( C n [ S ]) = n 11 for all n 14 Theorem 120. L et S = f 4 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 8 + ( 1 if n = 34 0 otherwise. Theorem 121. L et S = f 1 ; 4 ; 5 ; 6 ; 7 g Then r ( C n [ S ]) = n 9 for all n 14 Theorem 122. L et S = f 2 ; 4 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8, 9 (mo d 18) exc ept n = 24 0 otherwise. Theorem 123. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 7 g Then r ( C n [ S ]) = n 10 for all n 14 Theorem 124. L et S = f 3 ; 4 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 7, 8, 9, 10 (mo d 20) 0 otherwise. 125

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Theorem 125. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 7, 8, 9, 10 (mo d 20) exc ept n = 26, 27 0 otherwise. Theorem 126. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 7 g Then for all n 14 ; we have r ( C n [ S ]) = 2 n 21 + ( 1 if n 7, 8, 9, 10 (mo d 21) 0 otherwise. Theorem 127. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 g Then r ( C n [ S ]) = n 15 for all n 14 Theorem 128. L et S = f 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 3 + 8 > > > < > > > : 5 if n 8 (mo d 24) 2 if n 4, 16 (mo d 24) 1 if n 2, 14, 20 (mo d 24) 0 otherwise. Theorem 129. L et S = f 1 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > > > < > > > > > > > : 3 if n 4 (mo d 5) exc ept n = 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 109 2 if n = 24, 39, 49, 59, 64, 74, 79, 109 1 if n 1, 2, 3 (mo d 5) exc ept n = 17, 21, 23, 27, 46, or n = 19, 29, 34, 44, 54, 69 0 otherwise. Theorem 130. L et S = f 2 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 9 + 8 > < > : 2 if n 6, 8, 16 (mo d 18) exc ept n = 26, 52, 78 1 if n 2, 3, 4, 10, 12, 13, 17 (mo d 18) exc ept n = 39 0 otherwise. Theorem 131. L et S = f 1 ; 2 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 6 (mo d 12) 0 otherwise. 126

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Theorem 132. L et S = f 3 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 4 n 19 + 8 > > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > > : 2 if n 3, 8, 9, 18 (mo d 19) exc ept n = 18, 22, 27, 28, 37, 41, 46, 47, 56, 60, 65, 66, 75, 79, 84, 85, 94, 98, 103, 113, 117, 122, 123, 136, 141, 151, 155, 179, 180, 193, 208, 236, 237, 250, 265, 293, 294, 307, 322, 350, 364 1 if n 1, 4, 5, 6, 7, 10, 11, 12, 13, 14, 16 (mo d 19) exc ept n = 20, 24, 26, 29, 39, 42, 43, 54, 67, 68, 69, 70, 81, 82, 83, 96, 111, 124, 125, 126, 138, 139, 140, 153, 168, 181, 182, 195, 196, 210, 238, 252, or n = 18, 22, 37, 46, 47, 60, 65, 66, 75, 79, 85, 94, 103, 113, 117, 122, 123, 136, 141, 151, 155, 179, 180, 193, 208, 236, 237, 250, 265, 293, 294, 307, 322, 350, 364 0 otherwise. Theorem 133. L et S = f 1 ; 3 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 2, 3, 4, 5 (mo d 6) exc ept n = 20, 27 0 otherwise. Theorem 134. L et S = f 2 ; 3 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 > > > < > > > : 2 if n 5 (mo d 7) exc ept n = 19, 26, 33, 47, 61, 82, 117 1 if n 3, 4, 6 (mo d 7) exc ept n = 17, 24, 39, 52, or n = 19, 33, 47, 61, 82, 117 0 otherwise. Theorem 135. L et S = f 1 ; 2 ; 3 ; 8 g Then r ( C n [ S ]) = n 7 for all n 16 Theorem 136. L et S = f 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 5 + 8 > > > < > > > : 3 if n 4 (mo d 20) 2 if n 8 (mo d 20) 1 if n 2, 12, 14 (mo d 20) 0 otherwise. Theorem 137. L et S = f 1 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > < > > > : 2 if n 12 (mo d 13) exc ept n = 25, 38, 51 1 if n 1, 3, 4, 5, 6, 8, 10 (mo d 13) exc ept n = 17, 21, 27, 34, or n = 25, 38, 51 0 otherwise. 127

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Theorem 138. L et S = f 2 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 > < > : 2 if n 12 (mo d 14) 1 if n 2, 4, 6, 13 (mo d 14) 0 otherwise. Theorem 139. L et S = f 1 ; 2 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6 (mo d 14) 0 otherwise. Theorem 140. L et S = f 3 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 19 + 8 > > > < > > > : 1 if n 2, 3, 4, 6, 7, 8, 11, 12, 16, 17, 18 (mo d 19) exc ept n = 17, 22, 26, 35, 37, 41, 46, 54, 59, 61, 65, 74, 78, 83, 98, 102, 111, 122, 135, 159 0 otherwise. Theorem 141. L et S = f 1 ; 3 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 23 + ( 1 if n 7, 14, 15, 22 (mo d 23) exc ept n = 30, 45, 60 0 otherwise. Theorem 142. L et S = f 2 ; 3 ; 4 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 6, 7, 8 (mo d 9) 0 otherwise. Theorem 143. L et S = f 1 ; 2 ; 3 ; 4 ; 8 g Then r ( C n [ S ]) = n 9 for all n 16 Theorem 144. L et S = f 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 5 n 23 + 8 > > > > > < > > > > > : 1 if n 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 22 (mo d 23) exc ept n = 18, 27, 29, 30, 36, 38, 39, 45, 54, 56, 57, 61, 63, 72, 75, 79, 81, 84, 99, 102, 108, 126, 144, 153, 171, or n = 24 0 otherwise. Theorem 145. L et S = f 1 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 1, 3, 4, 5, 6, 8, 10, 11 (mo d 12) exc ept n = 20, 22, 25, 37, 44 0 otherwise. 128

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Theorem 146. L et S = f 2 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 19 + 8 > > > < > > > : 1 if n 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 (mo d 19) exc ept n = 21, 22, 23, 24, 26, 27, 28, 29, 40, 41, 45, 46, 47, 48, 59, 60, 64, 65, 83, 84 0 otherwise. Theorem 147. L et S = f 1 ; 2 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 4 n 31 + 8 > > > < > > > : 1 if n 6, 7, 13, 15, 21, 22, 23, 29, 30 (mo d 31) exc ept n = 23, 30, 46, 60, 69, 75, 92, 115, 161, 184, 207, 230 0 otherwise. Theorem 148. L et S = f 3 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 > > > > > < > > > > > : 2 if n 5 (mo d 7) exc ept n = 19, 26, 33, 40, 47, 54, 61, 75 1 if n 1, 3, 4, 6 (mo d 7) exc ept n = 17, 18, 22, 24, 29, 36, 38, 57, or n = 26, 33, 40, 47, 54, 61, 75 0 otherwise. Theorem 149. L et S = f 1 ; 3 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 5, 6, 7 (mo d 15) exc ept n = 21, 35 0 otherwise. Theorem 150. L et S = f 2 ; 3 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 5, 7, 8 (mo d 9) exc ept n = 29, 47 0 otherwise. Theorem 151. L et S = f 1 ; 2 ; 3 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n = 17 0 otherwise. Theorem 152. L et S = f 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 > > > < > > > : 2 if n 5 (mo d 7) exc ept n = 19, 26, 33 1 if n 1, 2, 4, 6 (mo d 7) exc ept n = 22, or n = 19, 26, 33 0 otherwise. 129

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Theorem 153. L et S = f 1 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 3, 4, 5, 6, 7, 8, 16 (mo d 17) exc ept n = 21 0 otherwise. Theorem 154. L et S = f 2 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 5, 6, 7 (mo d 15) exc ept n = 21, 35 0 otherwise. Theorem 155. L et S = f 1 ; 2 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 9, 10 (mo d 11) exc ept n = 17 0 otherwise. Theorem 156. L et S = f 3 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 4, 5, 7, 13, 14 (mo d 15) exc ept n = 19, 28, 35, 43, 49 0 otherwise. Theorem 157. L et S = f 1 ; 3 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9 (mo d 10) 0 otherwise. Theorem 158. L et S = f 2 ; 3 ; 4 ; 5 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 7, 8 (mo d 9) 0 otherwise. Theorem 159. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 8 g Then r ( C n [ S ]) = n 11 for all n 16 Theorem 160. L et S = f 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 5 + 8 > > > < > > > : 3 if n 4 (mo d 10) exc ept n = 24 2 if n 6, 8 (mo d 10) exc ept n = 26 1 if n 2, 3, 7, 9 (mo d 10), or n = 24 0 otherwise. 130

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Theorem 161. L et S = f 1 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 13 + 8 < : 1 if n 4, 5, 6, 9, 11, 12 (mo d 13) exc ept n = 17, 18, 19, 22, 38, 57, 61, 74, 76, 95, 152 0 otherwise. Theorem 162. L et S = f 2 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + ( 1 if n 2, 4, 6 (mo d 12) 0 otherwise. Theorem 163. L et S = f 1 ; 2 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 5, 6, 7, 14 (mo d 15) 0 otherwise. Theorem 164. L et S = f 3 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 19 + ( 1 if n 4, 5, 6, 7, 8, 9, 10, 11, 12 (mo d 19) 0 otherwise. Theorem 165. L et S = f 1 ; 3 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 4, 6, 7, 12, 14, 15, 20, 22 (mo d 23) exc ept n = 27, 35, 58 0 otherwise. Theorem 166. L et S = f 2 ; 3 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 4, 6, 8, 15, 16, 22, 23, 24, 29, 31 (mo d 32) 0 otherwise. Theorem 167. L et S = f 1 ; 2 ; 3 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 5, 6, 7, 8, 14, 16 (mo d 17) exc ept n = 23, 25, 31, 39, 48, 50, 56, 57, 73, 75, 82, 107, 125, 150, 175 0 otherwise. Theorem 168. L et S = f 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 > < > : 2 if n 8, 10, 12 (mo d 24) 1 if n 2, 4, 5, 6, 14, 16, 17, 18 (mo d 24) 0 otherwise. 131

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Theorem 169. L et S = f 1 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 5, 6, 7, 12, 14, 20, 25, 27, 33, 34, 35, 46, 48, 49, 55, 62, 68, 70, 76, 77, 83 (mo d 84) 0 otherwise. Theorem 170. L et S = f 2 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 4, 6, 8 (mo d 18) 0 otherwise. Theorem 171. L et S = f 1 ; 2 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 8, 10 (mo d 11) exc ept n = 17 0 otherwise. Theorem 172. L et S = f 3 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 20 + ( 1 if n 6, 13, 20, 32, 33 (mo d 40), or n = 45, 51, 71 0 otherwise. Theorem 173. L et S = f 1 ; 3 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 5, 7, 8, 9, 16, 18 (mo d 19) exc ept n = 16, 18, 24, 35, 54 0 otherwise. Theorem 174. L et S = f 2 ; 3 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 4, 6, 8, 9, 10, 17, 19 (mo d 20) exc ept n = 17 0 otherwise. Theorem 175. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 10, 12 (mo d 13) 0 otherwise. Theorem 176. L et S = f 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 11 + 8 < : 1 if n 4, 5, 10, 11, 16 (mo d 22) exc ept n = 16, 32, 48, or n = 21, 36, 37, 47, 58, 69 0 otherwise. 132

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Theorem 177. L et S = f 1 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 2, 4, 5, 6, 7, 14 (mo d 15) exc ept n = 17, 19 0 otherwise. Theorem 178. L et S = f 2 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 4, 5, 6, 7, 8 (mo d 9) 0 otherwise. Theorem 179. L et S = f 1 ; 2 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8 (mo d 9) exc ept n = 17, 34, 51 0 otherwise. Theorem 180. L et S = f 3 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 3, 5, 6, 8, 13, 15, 16 (mo d 17) exc ept n = 20, 23 0 otherwise. Theorem 181. L et S = f 1 ; 3 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 8, 9 (mo d 10) exc ept n = 16, 18, 36 0 otherwise. Theorem 182. L et S = f 2 ; 3 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 8, 9, 18 (mo d 19) exc ept n = 18, 27 0 otherwise. Theorem 183. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 11 (mo d 12) 0 otherwise. Theorem 184. L et S = f 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n = 20, 27, 33, 34, 41, 48 0 otherwise. 133

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Theorem 185. L et S = f 1 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 4, 5, 6, 7, 8, 9 (mo d 18) exc ept n = 22, 23, 24, 40 0 otherwise. Theorem 186. L et S = f 2 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 4, 6, 7, 8, 9, 10, 19 (mo d 20) exc ept n = 27 0 otherwise. Theorem 187. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 10 (mo d 11) 0 otherwise. Theorem 188. L et S = f 3 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 8, 16 (mo d 17) exc ept n = 23 0 otherwise. Theorem 189. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 21 + 8 < : 1 if n 7, 8, 9, 10, 20 (mo d 21) exc ept n = 20, 28, 30, 50, 70 0 otherwise. Theorem 190. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 8, 9, 10, 11, 21 (mo d 22) 0 otherwise. Theorem 191. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 15 + ( 1 if n 14 (mo d 15) 0 otherwise. Theorem 192. L et S = f 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 5 n 23 + 8 > > > > > < > > > > > : 1 if n 4, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18 (mo d 23) exc ept n = 18, 27, 31, 35, 36, 39, 53, 54, 58, 61, 62, 76, 79, 81, 83, 84, 85, 99, 102, 106, 107, 108, 125, 129, 130, 148, 152, 153, 171, 175, 198, or n = 21, 24 0 otherwise. 134

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Theorem 193. L et S = f 1 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 13 + 8 < : 1 if n 4, 5, 6, 12 (mo d 13) exc ept n = 17, 56, 57, 58, 116, 121, 173, 174, 290 0 otherwise. Theorem 194. L et S = f 2 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 > > > > > > > < > > > > > > > : 2 if n 9 (mo d 12) exc ept n = 21, 33, 45, 57, 69, 81, 93, 105, 117, 153 1 if n 2, 3, 4, 5, 6, 8, 10, 11 (mo d 12) exc ept n = 16, 17, 20, 22, 32, 34, 44, 50, 51, 68, or n = 45, 57, 69, 81, 93, 105, 117, 153 0 otherwise. Theorem 195. L et S = f 1 ; 2 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 2, 3, 5, 6, 7, 8, 14, 15, 16 (mo d 17) exc ept n = 19, 20, 24, 36, 37, 48, 53 0 otherwise. Theorem 196. L et S = f 3 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 2, 3, 4, 5, 8, 10, 11, 15, 16, 17, 20, 23, 26, 27, 28, 29 (mo d 30) exc ept n = 20, 23, 26, 45, 46, 68, 92 0 otherwise. Theorem 197. L et S = f 1 ; 3 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 2, 3, 4, 6, 7, 14, 15, 19, 20, 21, 22, 27, 28, 29 (mo d 30) exc ept n = 19, 20, 21, 33, 34, 63 0 otherwise. Theorem 198. L et S = f 2 ; 3 ; 7 ; 8 g Then r ( C n [ S ]) = n 7 for all n 16 Theorem 199. L et S = f 1 ; 2 ; 3 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 26 + 8 < : 1 if n 8, 15, 16, 17 (mo d 26) exc ept n = 16, 17, 34, 67, 68, 119, or n = 22, 23 0 otherwise. Theorem 200. L et S = f 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 19 + ( 1 if n 6, 11, 12 (mo d 19), or n = 16, 17, 35, 40 0 otherwise. 135

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Theorem 201. L et S = f 1 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 23 + ( 1 if n 4, 6, 7, 14, 15 (mo d 23), or n = 43 0 otherwise. Theorem 202. L et S = f 2 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6 (mo d 14) 0 otherwise. Theorem 203. L et S = f 1 ; 2 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8, 9 (mo d 18) exc ept n = 24, or n = 16, 33 0 otherwise. Theorem 204. L et S = f 3 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 3, 4, 6, 7, 8 (mo d 9) exc ept n = 20, 21 0 otherwise. Theorem 205. L et S = f 1 ; 3 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 7, 8, 9 (mo d 19) exc ept n = 27, 45 0 otherwise. Theorem 206. L et S = f 2 ; 3 ; 4 ; 7 ; 8 g Then r ( C n [ S ]) = n 9 for all n 16 Theorem 207. L et S = f 1 ; 2 ; 3 ; 4 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 10, 11, 12 (mo d 13) 0 otherwise. Theorem 208. L et S = f 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 > > > > > < > > > > > : 2 if n = 36 1 if n 4, 5, 6, 10, 11, 12, 14, 15, 16, 17, 18 (mo d 24) exc ept n = 17, 28, 29, 34, 36, 38, 39, 58, 62, 87, or n = 26, 32 0 otherwise. 136

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Theorem 209. L et S = f 1 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + ( 1 if n = 18, 33, 34 0 otherwise. Theorem 210. L et S = f 2 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 22 + 8 < : 1 if n 5, 6, 7, 10, 11, 12, 13, 14 (mo d 22) exc ept n = 27, 28, 54, 55, 56, or n = 20 0 otherwise. Theorem 211. L et S = f 1 ; 2 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 7, 8, 9, 10 (mo d 11) exc ept n = 17, 18 0 otherwise. Theorem 212. L et S = f 3 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 3, 4, 5, 6, 7, 8 (mo d 9) exc ept n = 21 0 otherwise. Theorem 213. L et S = f 1 ; 3 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 6, 8 (mo d 18) 0 otherwise. Theorem 214. L et S = f 2 ; 3 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 7, 8 (mo d 18) 0 otherwise. Theorem 215. L et S = f 1 ; 2 ; 3 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 8, 10 (mo d 11) 0 otherwise. Theorem 216. L et S = f 4 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 6, 12, 13, 18, 19, 20 (mo d 28) exc ept n = 18, 19 0 otherwise. 137

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Theorem 217. L et S = f 1 ; 4 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 5, 6, 8, 9 (mo d 10) exc ept n = 16, 18, 19, 35, 36, 38, 55, 56, 75, 76, 95 0 otherwise. Theorem 218. L et S = f 2 ; 4 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 3, 4, 5, 6, 7, 8, 9 (mo d 18) exc ept n = 21, 22, 24, 40 0 otherwise. Theorem 219. L et S = f 1 ; 2 ; 4 ; 5 ; 7 ; 8 g Then r ( C n [ S ]) = n 11 for all n 16 Theorem 220. L et S = f 3 ; 4 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 7, 8, 9, 10 (mo d 20) exc ept n = 27 0 otherwise. Theorem 221. L et S = f 1 ; 3 ; 4 ; 5 ; 7 ; 8 g Then r ( C n [ S ]) = n 10 for all n 16 Theorem 222. L et S = f 2 ; 3 ; 4 ; 5 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 7, 8, 9, 10 (mo d 20) exc ept n = 27 0 otherwise. Theorem 223. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 7 ; 8 g Then r ( C n [ S ]) = n 14 for all n 16 Theorem 224. L et S = f 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 6 + 8 > > < > > : 1 if n 2, 3, 4, 5, 6, 10, 11, 12, 14, 15, 16, 17, 18, 22, 23, 24 (mo d 30) exc ept n = 16, 17, 22, 32, 33, 34, 44, 45, 62 0 otherwise. Theorem 225. L et S = f 1 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 23 + ( 1 if n 4, 5, 6, 7, 14, 15 (mo d 23) exc ept n = 27 0 otherwise. Theorem 226. L et S = f 2 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 5, 6, 7, 9, 10, 11, 12, 13, 14, 15 (mo d 23) exc ept n = 28, 29, 32, 33, 56, 57, 58, 78 0 otherwise. 138

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Theorem 227. L et S = f 1 ; 2 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 6, 7, 8, 9, 10 (mo d 11) exc ept n = 17, 18, 19, 20, 39, 40 0 otherwise. Theorem 228. L et S = f 3 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 4, 5, 6, 7, 9, 10, 11, 12, 13, 14 (mo d 21) exc ept n = 30 0 otherwise. Theorem 229. L et S = f 1 ; 3 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 7, 8 (mo d 17) exc ept n = 23, or n = 31 0 otherwise. Theorem 230. L et S = f 2 ; 3 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 8, 9 (mo d 19), or n = 16, 17, 24, 34, 35, 41, 42, 53, 60 0 otherwise. Theorem 231. L et S = f 1 ; 2 ; 3 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 12 + ( 1 if n = 21, 32 0 otherwise. Theorem 232. L et S = f 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 2, 3, 4, 5, 6, 7, 8 (mo d 9) exc ept n = 21, 22, 23 0 otherwise. Theorem 233. L et S = f 1 ; 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 7, 8 (mo d 9) 0 otherwise. Theorem 234. L et S = f 2 ; 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 10 (mo d 20) exc ept n = 25, 27, 45 0 otherwise. 139

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Theorem 235. L et S = f 1 ; 2 ; 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 8, 10 (mo d 11) exc ept n = 17, 19, 39 0 otherwise. Theorem 236. L et S = f 3 ; 4 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 9 for all n 16 Theorem 237. L et S = f 1 ; 3 ; 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 7, 8, 9 (mo d 19) exc ept n = 26, 27, 45 0 otherwise. Theorem 238. L et S = f 2 ; 3 ; 4 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9, 10 (mo d 20) 0 otherwise. Theorem 239. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 13 for all n 16 Theorem 240. L et S = f 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 2, 3, 4, 5, 6, 7, 8 (mo d 9) exc ept n = 20, 21, 22, 23, 24, 25, 29, 30, 47, 48, 49, 50, 74, 75 0 otherwise. Theorem 241. L et S = f 1 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 10 + ( 1 if n = 26 0 otherwise. Theorem 242. L et S = f 2 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 4, 5, 6, 7, 8 (mo d 9) exc ept n = 22, 23, 24, 25, 49, 50 0 otherwise. Theorem 243. L et S = f 1 ; 2 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 11 for all n 16 Theorem 244. L et S = f 3 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 8 (mo d 9) exc ept n = 17 0 otherwise. 140

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Theorem 245. L et S = f 1 ; 3 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 10 for all n 16 Theorem 246. L et S = f 2 ; 3 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 21 + ( 1 if n 5, 6, 7, 8, 9, 10 (mo d 21) exc ept n = 26, 27 0 otherwise. Theorem 247. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 12 for all n 16 Theorem 248. L et S = f 4 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 9 for all n 16 Theorem 249. L et S = f 1 ; 4 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 10 for all n 16 Theorem 250. L et S = f 2 ; 4 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 7, 8, 9, 10, 11 (mo d 22) 0 otherwise. Theorem 251. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 11 for all n 16 Theorem 252. L et S = f 3 ; 4 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 7, 8, 9, 10, 11 (mo d 23) 0 otherwise. Theorem 253. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 7, 8, 9, 10, 11 (mo d 23) exc ept n = 30 0 otherwise. Theorem 254. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 g Then for all n 16 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 8, 9, 10, 11, 12 (mo d 24) 0 otherwise. Theorem 255. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 g Then r ( C n [ S ]) = n 17 for all n 16 Theorem 256. L et S = f 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 3 + 8 > > > > > < > > > > > : 6 if n 9 (mo d 27) 3 if n 18 (mo d 27) 2 if n 3, 12, 21 (mo d 27) 1 if n 6, 15, 24 (mo d 27) 0 otherwise. 141

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Theorem 257. L et S = f 1 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 19 + 8 > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > : 2 if n 3, 6, 8, 9, 13, 18 (mo d 19) exc ept n = 18, 22, 25, 27, 28, 32, 37, 41, 44, 47, 51, 56, 60, 63, 65, 70, 75, 79, 82, 84, 98, 101, 108, 120, 123, 136, 146, 151, 174, 177, 179, 196, 241, 269, 291, 336, 364 1 if n 1, 2, 4, 10, 11, 14, 15, 16, 17 (mo d 19) exc ept n = 30, 34, 39, 58, 67, 73, 112, 129, 168, 224, or n = 18, 22, 25, 27, 32, 37, 44, 47, 51, 60, 63, 65, 70, 75, 79, 98, 108, 120, 123, 136, 146, 151, 174, 177, 179, 241, 269, 291, 336, 364 0 otherwise. Theorem 258. L et S = f 2 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > > > > > < > > > > > > > > > : 3 if n 2 (mo d 5) exc ept n = 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 82, 97, 107 2 if n 4 (mo d 5) exc ept n = 19, 24, 29, 34, 39, or n = 57, 62, 67, 72, 82, 97, 107 1 if n 1, 3 (mo d 5) exc ept n = 21, 26, or n = 19, 22, 24, 27, 29, 32, 34, 37, 39, 42, 47, 52 0 otherwise. Theorem 259. L et S = f 1 ; 2 ; 9 g Then r ( C n [ S ]) = l 3 n 19 m for all n 18 Theorem 260. L et S = f 3 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 5 + 8 > > > < > > > : 3 if n 12 (mo d 15) 2 if n 3 (mo d 15) 1 if n 4, 6, 9, 14 (mo d 15) 0 otherwise. Theorem 261. L et S = f 1 ; 3 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > < > : 2 if n 3 (mo d 7) exc ept n = 24, 31, 38 1 if n 2, 4, 5, 6 (mo d 7), or n = 24, 31, 38 0 otherwise. Theorem 262. L et S = f 2 ; 3 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 5 n 29 + 8 > > > > > > > < > > > > > > > : 1 if n 3, 4, 5, 9, 10, 11, 14, 16, 17, 22, 23, 26, 27, 28 (mo d 29) exc ept n = 23, 27, 32, 34, 38, 46, 51, 55, 56, 57, 61, 67, 68, 69, 72, 84, 85, 92, 101, 113, 114, 115, 119, 130, 138, 161, 171, 184, 207, 230, 299, 322, 345, 391 0 otherwise. 142

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Theorem 263. L et S = f 1 ; 2 ; 3 ; 9 g Then r ( C n [ S ]) = n 7 for all n 18 Theorem 264. L et S = f 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 14 + 8 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > > > > > > > > > : 2 if n 2, 4, 7, 10, 13 (mo d 14) exc ept n = 18, 21, 24, 27, 30, 32, 35, 38, 41, 44, 46, 49, 52, 55, 58, 60, 63, 66, 69, 72, 77, 80, 83, 86, 88, 91, 94, 100, 105, 108, 111, 114, 116, 122, 133, 136, 139, 150, 156, 158, 161, 178, 181, 184, 203, 206, 226, 228, 231, 248, 251, 254, 273, 276, 296, 298, 318, 321, 343, 346, 366, 368, 388, 391, 413, 436, 458, 483, 506, 528, 598 1 if n 1, 5, 8, 9, 12 (mo d 14) exc ept n = 19, 22, 23, 29, 33, 43, 47, 61, 68, 89, 92, 113, 138, 159, 183, 229, 253, 299, or n = 18, 27, 30, 32, 35, 41, 49, 52, 55, 58, 60, 63, 72, 77, 80, 83, 86, 88, 94, 100, 105, 108, 111, 116, 122, 133, 139, 150, 156, 158, 178, 181, 203, 226, 228, 231, 248, 251, 254, 273, 296, 298, 318, 321, 343, 346, 366, 368, 388, 391, 413, 436, 458, 483, 506, 528, 598 0 otherwise. Theorem 265. L et S = f 1 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > > > < > > > : 2 if n 3 (mo d 7) exc ept n = 24, 31, 38, 45, 52, 59, 66 1 if n 1, 4, 5, 6 (mo d 7) exc ept n = 18, 22, 29, 36, or n = 31, 38, 45, 52, 59, 66 0 otherwise. Theorem 266. L et S = f 2 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 5 n 31 + 8 > > > > > > > < > > > > > > > : 2 if n 12 (mo d 31) exc ept n = 43, 74, 105 1 if n 2, 3, 4, 5, 6, 11, 13, 14, 17, 18, 21, 22, 23, 24, 27, 28, 29, 30 (mo d 31) exc ept n = 22, 24, 27, 29, 34, 35, 36, 44, 48, 53, 58, 60, 65, 75, 84, 89, 96, 106, 120, 137, 168, or n = 43, 74, 105 0 otherwise. Theorem 267. L et S = f 1 ; 2 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 13 (mo d 21) 0 otherwise. 143

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Theorem 268. L et S = f 3 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 19 + 8 > > > < > > > : 2 if n 18 (mo d 19) exc ept n = 18, 37, 56 1 if n 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 16 (mo d 19) exc ept n = 20, 22, 27, 46, or n = 18, 37, 56 0 otherwise. Theorem 269. L et S = f 1 ; 3 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 4, 5, 6, 10, 12, 13 (mo d 14) exc ept n = 18, 19, 20, 24, 26, 32, 38, 40, 46, 52, 60, 66, 80 0 otherwise. Theorem 270. L et S = f 2 ; 3 ; 4 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 6, 7, 12, 13, 15, 20, 21, 22, 23 (mo d 24) 0 otherwise. Theorem 271. L et S = f 1 ; 2 ; 3 ; 4 ; 9 g Then r ( C n [ S ]) = n 9 for all n 18 Theorem 272. L et S = f 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 19 + 8 > > > > > > > > > < > > > > > > > > > : 2 if n 9, 18 (mo d 19) exc ept n = 18, 47, 66, 75, 104, 113, 161, or n = 33 1 if n 2, 3, 4, 5, 8, 10, 11, 12, 14, 15, 16, 17 (mo d 19) exc ept n = 21, 22, 23, 24, 29, 33, 34, 35, 43, 46, 67, 68, 69, 81, 91, 92, 138, or n = 18, 47, 66, 75, 104, 113, 161 0 otherwise. Theorem 273. L et S = f 1 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > : 2 if n 6, 10, 12 (mo d 13) exc ept n = 19, 23, 25, 32, 36, 38, 49, 51, 58, 62, 64, 75, 77, 84, 88, 101, 103, 110, 114, 127, 129, 136, 149, 153, 155, 168, 179, 205, 231, 244, 266, 285, 361 1 if n 1, 3, 4, 5, 8 (mo d 13) exc ept n = 21, 27, 34, 40, 53, 57, 60, 73, 79, 92, 95, 133, 190, 209, or n = 23, 25, 32, 49, 51, 58, 62, 64, 75, 77, 84, 88, 101, 103, 110, 127, 129, 136, 149, 153, 155, 168, 179, 205, 231, 244, 266, 285, 361 0 otherwise. 144

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Theorem 274. L et S = f 2 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 6 n 37 + 8 > > > > > < > > > > > : 1 if n 3, 4, 5, 6, 9, 10, 11, 12, 18, 23, 24, 28, 29, 30, 35, 36 (mo d 37) exc ept n = 18, 23, 24, 28, 29, 30, 36, 40, 41, 42, 46, 48, 60, 65, 66, 72, 77, 78, 84, 102, 114, 120 0 otherwise. Theorem 275. L et S = f 1 ; 2 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 15 + 8 > > > < > > > : 1 if n 1, 3, 4, 5, 6, 7, 10, 12, 14 (mo d 15) exc ept n = 18, 19, 21, 25, 27, 31, 33, 34, 40, 42, 46, 48, 55, 61, 63, 76 0 otherwise. Theorem 276. L et S = f 3 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 23 + ( 1 if n = 28, 45 0 otherwise. Theorem 277. L et S = f 1 ; 3 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 6, 8, 15 (mo d 16) 0 otherwise. Theorem 278. L et S = f 2 ; 3 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 2, 4, 5, 6, 7, 8, 13, 14, 15 (mo d 16) exc ept n = 18 0 otherwise. Theorem 279. L et S = f 1 ; 2 ; 3 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 8 (mo d 9) 0 otherwise. Theorem 280. L et S = f 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 19 + 8 < : 1 if n 5, 6, 12 (mo d 19) exc ept n = 24, 25, 50, 62, 100, or n = 28 0 otherwise. 145

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Theorem 281. L et S = f 1 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 4, 5, 6, 8, 16, 20, 23, 24, 25, 31, 33 (mo d 34) exc ept n = 23 0 otherwise. Theorem 282. L et S = f 2 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 5, 6, 7, 13, 14, 15 (mo d 16) exc ept n = 22, 30, 45 0 otherwise. Theorem 283. L et S = f 1 ; 2 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 9, 17, 18, 25, 26 (mo d 27) 0 otherwise. Theorem 284. L et S = f 3 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 7, 13, 14, 20, 28, 34, 35, 41 (mo d 42) 0 otherwise. Theorem 285. L et S = f 1 ; 3 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 8, 9, 10 (mo d 11) 0 otherwise. Theorem 286. L et S = f 2 ; 3 ; 4 ; 5 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 7, 8, 9 (mo d 10) 0 otherwise. Theorem 287. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 288. L et S = f 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 4 + 8 > < > : 2 if n 3, 15 (mo d 24) 1 if n 6, 18 (mo d 24) 0 otherwise. Theorem 289. L et S = f 1 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 23 + 8 > > > > > < > > > > > : 1 if n 3, 4, 5, 7, 9, 10, 11, 13, 14, 15, 16, 17, 22 (mo d 23) exc ept n = 22, 27, 30, 32, 33, 34, 36, 37, 38, 51, 53, 55, 59, 60, 68, 72, 76, 82, 85, 99, 102, 105, 106, 119, 122, 153, 170, 174, 187, 191, 221, 289, 306 0 otherwise. 146

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Theorem 290. L et S = f 2 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > > > < > > > > > : 1 if n 2, 3, 4, 5, 6, 8, 9, 11, 12 (mo d 13) exc ept n = 19, 21, 22, 24, 29, 34, 38, 41, 43, 48, 55, 57, 60, 67, 74, 76, 81, 86, 93, 95, 100, 112, 119, 133, 138, 152, 171, 190 0 otherwise. Theorem 291. L et S = f 1 ; 2 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 4, 6, 7, 13 (mo d 14) exc ept n = 18 0 otherwise. Theorem 292. L et S = f 3 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > < > : 2 if n 3, 6 (mo d 21) 1 if n 9, 12 (mo d 21) 0 otherwise. Theorem 293. L et S = f 1 ; 3 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 > > > > > < > > > > > : 1 if n 4, 6, 7, 8, 9, 12, 14, 15, 17, 19, 20, 22, 23, 28, 30, 31, 33, 36, 38, 39, 40, 41, 44, 46, 47 (mo d 48) exc ept n = 19, 20, 28, 38, 41, 52, 62, 65, 76, 89, 100, 113, 124, 137 0 otherwise. Theorem 294. L et S = f 2 ; 3 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 > > > > > > > < > > > > > > > : 2 if n 16 (mo d 24) exc ept n = 40, 64, 88, 112, 136, 160, 184 1 if n 2, 5, 6, 8, 9, 12, 13, 15, 19, 20, 22, 23 (mo d 24) exc ept n = 19, 20, 26, 37, 50, 61, 74, 85, 98, 122, 146, 170, or n = 40, 64, 88, 112, 136, 160, 184 0 otherwise. Theorem 295. L et S = f 1 ; 2 ; 3 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 8, 9, 15, 16, 17, 18, 22, 23, 25, 26 (mo d 27) exc ept n = 35, 49, 70 0 otherwise. 147

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Theorem 296. L et S = f 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > > > < > > > : 2 if n 3, 6 (mo d 7) exc ept n = 20, 24, 27, 31, 45, 48 1 if n 1, 2, 4, 5 (mo d 7) exc ept n = 22, 23, or n = 20, 27, 31, 45, 48 0 otherwise. Theorem 297. L et S = f 1 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 4, 7, 13, 15, 21, 22 (mo d 23) exc ept n = 21, 27, or n = 33 0 otherwise. Theorem 298. L et S = f 2 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 4, 5, 6, 7, 8 (mo d 17) exc ept n = 21 0 otherwise. Theorem 299. L et S = f 1 ; 2 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 6, 8 (mo d 9) exc ept n = 33, 51 0 otherwise. Theorem 300. L et S = f 3 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 1, 4, 6, 7 (mo d 8) exc ept n = 20, 25, 28, 41, 49, 65 0 otherwise. Theorem 301. L et S = f 1 ; 3 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 7, 9, 10 (mo d 11) 0 otherwise. Theorem 302. L et S = f 2 ; 3 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 8 (mo d 17), or n = 22, 23, 39 0 otherwise. Theorem 303. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 10 (mo d 11) 0 otherwise. 148

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Theorem 304. L et S = f 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 5 n 29 + 8 > > < > > : 1 if n 2, 3, 4, 5, 6, 7, 9, 10, 11, 14, 15, 16, 17, 19, 22, 23, 24, 27, 28 (mo d 29) exc ept n = 19, 24, 27, 31, 32, 35, 36, 48, 53, 64, 65, 77 0 otherwise. Theorem 305. L et S = f 1 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 2, 4, 5, 6, 7, 8, 14, 15 (mo d 16) exc ept n = 18, 21 0 otherwise. Theorem 306. L et S = f 2 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 22 + 8 < : 1 if n 7, 11, 14, 20, 21 (mo d 22) exc ept n = 20, 21, 42, 55 0 otherwise. Theorem 307. L et S = f 1 ; 2 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 18 (mo d 19) exc ept n = 18, 23, 25, 27, 43, 45, 61, 63, 81, 99 0 otherwise. Theorem 308. L et S = f 3 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 2, 5, 6, 8, 12, 15, 16 (mo d 17) exc ept n = 19 0 otherwise. Theorem 309. L et S = f 1 ; 3 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 2, 5, 6, 8, 9, 10, 15, 18, 19 (mo d 20) 0 otherwise. Theorem 310. L et S = f 2 ; 3 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 5, 8, 9 (mo d 10) 0 otherwise. Theorem 311. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 11, 12 (mo d 13) 0 otherwise. 149

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Theorem 312. L et S = f 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 1, 2, 4, 5, 6, 7 (mo d 8) exc ept n = 18, 21, 25, 26, 28, 33, 42, 49 0 otherwise. Theorem 313. L et S = f 1 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 5, 6, 8, 9, 18 (mo d 19) exc ept n = 24 0 otherwise. Theorem 314. L et S = f 2 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 18, 21, 23, 24, 25, 26, 27, 35 (mo d 36) exc ept n = 21, 24, 40 0 otherwise. Theorem 315. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 10, 11 (mo d 12) 0 otherwise. Theorem 316. L et S = f 3 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 5, 6, 8, 15, 16 (mo d 17) exc ept n = 32, 40, 56 0 otherwise. Theorem 317. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 9, 10 (mo d 11) exc ept n = 20 0 otherwise. Theorem 318. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9 (mo d 10) exc ept n = 18, 38 0 otherwise. Theorem 319. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 9 g Then r ( C n [ S ]) = n 13 for all n 18 150

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Theorem 320. L et S = f 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > : 3 if n 2, 4 (mo d 5) exc ept n = 19, 22, 24, 27, 29, 32, 34, 37, 39, 42, 44, 47, 49, 52, 54, 57, 59, 62, 67, 69, 72, 74, 77, 79, 82, 92, 94, 97, 104, 109, 117, 127, 132 2 if n 1 (mo d 5) exc ept n = 21, 26, 31, 36, 46, 51, 56, 71, 81, 86, or n = 39, 44, 49, 54, 59, 62, 67, 72, 74, 77, 79, 82, 94, 97, 104, 109, 117, 127, 132 1 if n 3 (mo d 5) exc ept n = 23, or n = 19, 21, 22, 24, 26, 27, 29, 31, 32, 34, 36, 37, 42, 47, 51, 52, 56, 57, 69, 71, 81, 86, 92 0 otherwise. Theorem 321. L et S = f 1 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > > > > > < > > > > > > > : 2 if n 10, 12, 19, 25 (mo d 26) exc ept n = 19, 25, 36, 38, 45, 51, 62, 71, 77 1 if n 1, 3, 4, 5, 6, 8, 13, 14, 16, 17, 18, 21, 23 (mo d 26) exc ept n = 21, 23, 27, 29, 34, 47, 53, 60, or n = 19, 25, 38, 45, 51, 62, 71, 77 0 otherwise. Theorem 322. L et S = f 2 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 20 + 8 > > > > > > > < > > > > > > > : 1 if n 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 19 (mo d 20) exc ept n = 22, 24, 25, 26, 27, 28, 29, 30, 39, 42, 47, 48, 49, 50, 51, 52, 62, 64, 65, 87, 88, 89, 90, 91, 102, 104, 127, 128, 129, 130, 142, 167, 168, 169, 182, 207, 208, 247 0 otherwise. Theorem 323. L et S = f 1 ; 2 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 1, 3, 5, 6, 7, 8, 12, 14, 16 (mo d 17) exc ept n = 18, 20, 35, 37, 52, 69 0 otherwise. Theorem 324. L et S = f 3 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > < > > > : 2 if n 12 (mo d 13) exc ept n = 25 1 if n 1, 3, 4, 5, 6, 10, 11 (mo d 13) exc ept n = 27, or n = 25 0 otherwise. 151

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Theorem 325. L et S = f 1 ; 3 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 29 + 8 < : 1 if n 6, 7, 20, 21, 29, 35, 36, 43, 49, 50, 57 (mo d 58) exc ept n = 21, 35, 49 0 otherwise. Theorem 326. L et S = f 2 ; 3 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n = 35 0 otherwise. Theorem 327. L et S = f 1 ; 2 ; 3 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 5, 6, 7, 8, 16 (mo d 17) 0 otherwise. Theorem 328. L et S = f 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 1, 4, 5 (mo d 6) exc ept n = 19, 22, 25, 43, 49, 67 0 otherwise. Theorem 329. L et S = f 1 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 4, 11, 13 (mo d 21) 0 otherwise. Theorem 330. L et S = f 2 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 6 n 47 + 8 < : 1 if n 4, 5, 6, 7, 14, 19, 21, 22, 23, 29, 31, 36, 37, 38, 39, 44, 46 (mo d 47) exc ept n = 19, 21, 29, 44 0 otherwise. Theorem 331. L et S = f 1 ; 2 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 6, 7, 8, 9, 16, 18 (mo d 19) exc ept n = 25, 27, 54 0 otherwise. Theorem 332. L et S = f 3 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > > > < > > > : 1 if n 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14 (mo d 21) exc ept n = 24, 25, 26, 27, 30, 31, 45, 46, 51, 52, 53, 54, 66, 67, 72, 73, 93, 94, 108, 135 0 otherwise. 152

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Theorem 333. L et S = f 1 ; 3 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 7, 8 (mo d 9) exc ept n = 31, 34, 49, 85 0 otherwise. Theorem 334. L et S = f 2 ; 3 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 37 + ( 1 if n 7, 9, 18, 26, 27, 36 (mo d 37) 0 otherwise. Theorem 335. L et S = f 1 ; 2 ; 3 ; 4 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 29 + ( 1 if n 19, 28 (mo d 29) exc ept n = 19, 57 0 otherwise. Theorem 336. L et S = f 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > < > > : 1 if n 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 25 (mo d 26) exc ept n = 19, 30, 33, 38, 40, 41, 57, 59, 68, 87, 95, 114, 144, 163, 171, or n = 28 0 otherwise. Theorem 337. L et S = f 1 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 > > > < > > > : 1 if n 4, 5, 6, 7, 11, 12, 13, 14, 15, 22 (mo d 23) exc ept n = 27, 28, 30, 34, 45, 57, 58, 60, 73, 75, 103, 105, 120, 150, 165, 195 0 otherwise. Theorem 338. L et S = f 2 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 > > > > > < > > > > > : 2 if n 7 (mo d 25) exc ept n = 32, 57, 82 1 if n 1, 3, 5, 8, 9, 11, 13, 14, 15, 16, 20, 22, 24 (mo d 25) exc ept n = 20, 26, 34, 36, 45, 59, or n = 32, 57, 82 0 otherwise. Theorem 339. L et S = f 1 ; 2 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n = 22, 24, 33, 39, 41, 58 0 otherwise. 153

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Theorem 340. L et S = f 3 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 15 + 8 > > > > > < > > > > > : 2 if n 14 (mo d 30) exc ept n = 44, 74, 104 1 if n 3, 4, 5, 6, 7, 12, 13, 15, 16, 18, 20, 21, 22, 23, 25, 27, 29 (mo d 30) exc ept n = 18, 20, 21, 23, 25, 27, 42, 46, 48, 53, 55, 63, 76, 83, or n = 44, 74, 104 0 otherwise. Theorem 341. L et S = f 1 ; 3 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 6, 8, 10 (mo d 11) exc ept n = 19, 37, 39, 59 0 otherwise. Theorem 342. L et S = f 2 ; 3 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 3, 4, 5, 6, 7, 8, 9, 10 (mo d 20) exc ept n = 23, 25, 44, 63 0 otherwise. Theorem 343. L et S = f 1 ; 2 ; 3 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 10, 12 (mo d 13) 0 otherwise. Theorem 344. L et S = f 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 5, 7, 13, 14, 15, 22 (mo d 23) exc ept n = 28, 30, 45, 60, 105, 120 0 otherwise. Theorem 345. L et S = f 1 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 8, 16 (mo d 17) exc ept n = 23 0 otherwise. Theorem 346. L et S = f 2 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 7, 9 (mo d 20) exc ept n = 27 0 otherwise. Theorem 347. L et S = f 1 ; 2 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 9 (mo d 20) 0 otherwise. 154

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Theorem 348. L et S = f 3 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 3, 5, 7, 8, 14, 16 (mo d 17) exc ept n = 20 0 otherwise. Theorem 349. L et S = f 1 ; 3 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 8, 10, 19, 21 (mo d 22) exc ept n = 19 0 otherwise. Theorem 350. L et S = f 2 ; 3 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 7, 9, 10, 11, 20, 22 (mo d 23) exc ept n = 20, 30 0 otherwise. Theorem 351. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 15 + ( 1 if n 12, 14 (mo d 15) 0 otherwise. Theorem 352. L et S = f 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 25 + 8 > > < > > : 2 if n 11 (mo d 25) exc ept n = 36, 61 1 if n 3, 5, 6, 7, 9, 10, 12, 13, 14, 15, 16, 17, 18, 24 (mo d 25) exc ept n = 32, 34, 39, or n = 36, 61 0 otherwise. Theorem 353. L et S = f 1 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 6 n 43 + ( 1 if n 7, 14, 35 (mo d 43), or n = 27, 34, 48 0 otherwise. Theorem 354. L et S = f 2 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 22 + 8 < : 1 if n 4, 6, 7, 11, 12, 13, 14, 20, 21 (mo d 22) exc ept n = 20, 21, 26, 33, 34, 42, 55 0 otherwise. Theorem 355. L et S = f 1 ; 2 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 8, 16 (mo d 17) exc ept n = 23 0 otherwise. 155

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Theorem 356. L et S = f 3 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 > > < > > : 1 if n 4, 6, 7, 8, 12, 14, 15, 16, 23, 31, 36, 38, 39, 40, 47, 52, 54, 55, 56, 63, 71, 76, 78, 79, 80, 84, 86, 87, 88, 95 (mo d 96), or n = 33, 57 0 otherwise. Theorem 357. L et S = f 1 ; 3 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 6, 7, 8, 9, 10 (mo d 11) exc ept n = 19 0 otherwise. Theorem 358. L et S = f 2 ; 3 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 8, 15, 23, 39, 40, 47 (mo d 48), or n = 36, 54 0 otherwise. Theorem 359. L et S = f 1 ; 2 ; 3 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 8, 9, 10 (mo d 11) exc ept n = 19 0 otherwise. Theorem 360. L et S = f 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 29 + 8 > > > > > > > > > > > > > > < > > > > > > > > > > > > > > : 2 if n 7, 13, 14 (mo d 29) exc ept n = 36, 42, 43, 71, 72, 94, 100, 101, 129, 152, 159, 187, 210, 217, 245 1 if n 3, 5, 6, 9, 10, 11, 12, 15, 16, 17, 18, 19, 20, 21, 28 (mo d 29) exc ept n = 18, 19, 20, 21, 28, 32, 35, 38, 40, 44, 45, 46, 47, 49, 61, 63, 68, 70, 73, 75, 77, 96, 98, 103, 105, 119, 126, 131, 133, 154, 161, 189, or n = 36, 43, 71, 72, 94, 100, 101, 129, 152, 159, 187, 210, 217, 245 0 otherwise. Theorem 361. L et S = f 1 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 6, 7, 8 (mo d 9) exc ept n = 22, 25, 49, 51, 76 0 otherwise. Theorem 362. L et S = f 2 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 3, 4, 5, 6, 7, 8, 9, 10 (mo d 20) 0 otherwise. 156

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Theorem 363. L et S = f 1 ; 2 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 8, 10, 11 (mo d 12) exc ept n = 20 0 otherwise. Theorem 364. L et S = f 3 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 7, 9, 18 (mo d 19) 0 otherwise. Theorem 365. L et S = f 1 ; 3 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 10 (mo d 11) 0 otherwise. Theorem 366. L et S = f 2 ; 3 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 7, 9, 10, 11, 21 (mo d 22), or n = 26 0 otherwise. Theorem 367. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 14 + ( 1 if n 13 (mo d 14) 0 otherwise. Theorem 368. L et S = f 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 6, 7, 12, 13, 14, 20, 32, 34, 35, 41, 48, 49, 62, 68, 69, 70, 76, 77 (mo d 84) 0 otherwise. Theorem 369. L et S = f 1 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 10, 19 (mo d 20) exc ept n = 27, 45 0 otherwise. Theorem 370. L et S = f 2 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 4, 5, 6, 7, 8, 9 (mo d 10) exc ept n = 25 0 otherwise. 157

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Theorem 371. L et S = f 1 ; 2 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n = 26 0 otherwise. Theorem 372. L et S = f 3 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 > > > > > < > > > > > : 2 if n 9 (mo d 27) 1 if n 4, 6, 7, 8, 11, 12, 13, 14, 15, 16, 17, 18, 26 (mo d 27) exc ept n = 31, 34, 38, 41, 58, 65, 68, 85, 92, 119 0 otherwise. Theorem 373. L et S = f 1 ; 3 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 6, 8, 10, 21 (mo d 22) 0 otherwise. Theorem 374. L et S = f 2 ; 3 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9 (mo d 20) 0 otherwise. Theorem 375. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 12 (mo d 13) 0 otherwise. Theorem 376. L et S = f 4 ; 5 ; 6 ; 7 ; 9 g Then r ( C n [ S ]) = l 2 n 17 m for all n 18 Theorem 377. L et S = f 1 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 7, 8, 9, 10 (mo d 20) exc ept n = 26, 27 0 otherwise. Theorem 378. L et S = f 2 ; 4 ; 5 ; 6 ; 7 ; 9 g Then r ( C n [ S ]) = n 10 for all n 18 Theorem 379. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 11 (mo d 12) 0 otherwise. 158

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Theorem 380. L et S = f 3 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 7, 8, 9, 10, 11, 21 (mo d 22) 0 otherwise. Theorem 381. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + 8 < : 1 if n 6, 8, 9, 10, 11, 12, 23 (mo d 24) exc ept n = 30, 33 0 otherwise. Theorem 382. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 25 + ( 1 if n 7, 9, 10, 11, 12, 24 (mo d 25) 0 otherwise. Theorem 383. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 17 + ( 1 if n 16 (mo d 17) 0 otherwise. Theorem 384. L et S = f 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 5 + 8 > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > : 3 if n 2, 3, 4 (mo d 5) exc ept n = 18, 19, 22, 23, 24, 27, 28, 29, 32, 33, 34, 37, 38, 39, 42, 43, 44, 47, 48, 49, 52, 53, 54, 57, 58, 59, 62, 63, 64, 67, 72, 73, 74, 77, 78, 79, 82, 84, 87, 88, 92, 93, 97, 98, 99, 107, 112, 113, 117, 118, 132, 139, 158, 172, 198 2 if n = 34, 42, 43, 44, 48, 49, 53, 54, 57, 58, 62, 63, 64, 67, 72, 74, 77, 78, 79, 82, 84, 87, 88, 93, 97, 98, 107, 112, 113, 117, 118, 139, 158, 172, 198 1 if n 1 (mo d 5) exc ept n = 26, 66, or n = 18, 19, 22, 23, 24, 27, 28, 29, 32, 37, 38, 39, 47, 52, 59, 73, 92, 99, 132 0 otherwise. Theorem 385. L et S = f 1 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 13 + 8 > > > > > < > > > > > : 1 if n 4, 5, 6, 8, 9, 12, 17, 18, 19, 23, 24, 25 (mo d 26) exc ept n = 18, 19, 23, 24, 25, 30, 31, 34, 35, 38, 43, 44, 49, 50, 56, 57, 60, 61, 69, 75, 76, 82, 86, 87, 95, 101, 112, 113, 138, 139, 164, 190 0 otherwise. 159

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Theorem 386. L et S = f 2 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 6 + 8 < : 1 if n 2, 4, 5, 6, 8, 9, 10, 11 (mo d 12) exc ept n = 20, 21, 28, 32, 56 0 otherwise. Theorem 387. L et S = f 1 ; 2 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 31 + ( 1 if n 5, 6, 7, 20, 21, 22, 23 (mo d 31), or n = 29, 44 0 otherwise. Theorem 388. L et S = f 3 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > > > > > < > > > > > : 2 if n 3, 4, 5 (mo d 7) exc ept n = 18, 19, 24, 25, 26, 33, 38, 39, 45, 46, 47, 52, 73, 74, 75 1 if n 2, 6 (mo d 7) exc ept n = 23, or n = 18, 19, 26, 33, 38, 39, 45, 46, 47, 52, 73, 74, 75 0 otherwise. Theorem 389. L et S = f 1 ; 3 ; 8 ; 9 g Then r ( C n [ S ]) = n 7 for all n 18 Theorem 390. L et S = f 2 ; 3 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 5, 6, 7, 10, 11, 12, 13, 14, 15 (mo d 23) exc ept n = 28, 35, 36, 56, 57, 58, 79, 80, 102 0 otherwise. Theorem 391. L et S = f 1 ; 2 ; 3 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 5, 6, 7, 8, 9, 16, 17, 18 (mo d 19) exc ept n = 28, 35, 43, 56, 62, 63, 84, 112, 119, 140, 168, 196, 252 0 otherwise. Theorem 392. L et S = f 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 > > > > > > > < > > > > > > > : 2 if n 3, 4, 5 (mo d 7) exc ept n = 18, 19, 24, 25, 26, 31, 32, 33, 38, 45, 46, 47, 52, 53, 59, 66, 73, 74, 75 1 if n 1, 2, 6 (mo d 7) exc ept n = 22, 23, 50, or n = 18, 19, 26, 31, 32, 33, 38, 45, 46, 47, 52, 53, 59, 66, 73, 74, 75 0 otherwise. 160

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Theorem 393. L et S = f 1 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 < : 1 if n 4, 5, 6, 7, 8, 9, 13, 15, 16, 19, 20, 21, 22, 23, 24 (mo d 25) exc ept n = 19, 20, 21, 22, 44 0 otherwise. Theorem 394. L et S = f 2 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n 7 (mo d 15) exc ept n = 22, or n = 20 0 otherwise. Theorem 395. L et S = f 1 ; 2 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + 8 > > > < > > > : 1 if n 4, 8, 16, 17, 23, 24, 25, 31, 32, 33 (mo d 34) exc ept n = 25, 31, 32, 50, 57, 65, 72, 100, 106, 125, 140 0 otherwise. Theorem 396. L et S = f 3 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 > > > > > < > > > > > : 2 if n 7 (mo d 23) exc ept n = 30, 53 1 if n 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15 (mo d 23) exc ept n = 26, 28, 31, 32, 33, 34, 35, 49, 54, 56, 77, or n = 30, 53 0 otherwise. Theorem 397. L et S = f 1 ; 3 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 < : 1 if n 7, 8, 9, 17, 18, 23, 24, 25, 26, 27 (mo d 36) exc ept n = 23, 24, 25, or n = 33 0 otherwise. Theorem 398. L et S = f 2 ; 3 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 5, 6, 7, 8, 9 (mo d 19) exc ept n = 24, 27, 45, 63, 81 0 otherwise. Theorem 399. L et S = f 1 ; 2 ; 3 ; 4 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 10, 19, 20 (mo d 30) exc ept n = 19 0 otherwise. 161

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Theorem 400. L et S = f 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 6 + 8 > > > > > < > > > > > : 2 if n 5 (mo d 6) exc ept n = 23, 29, 35, 41, 53, 59, 65, 77, 83, 95 1 if n 2, 4 (mo d 6) exc ept n = 20, 22, 26, 44, 56, or n = 23, 29, 35, 41, 53, 59, 65, 77, 83, 95 0 otherwise. Theorem 401. L et S = f 1 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 4, 5, 6, 7, 20, 21, 22, 36, 37, 38 (mo d 46), or n = 28, 29, 32, 43, 44, 59, 74, 89 0 otherwise. Theorem 402. L et S = f 2 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + 8 < : 1 if n 4, 5, 7, 10, 11, 13, 14 (mo d 21) exc ept n = 31, 32, 52 0 otherwise. Theorem 403. L et S = f 1 ; 2 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 2, 3, 4, 6, 7, 8, 9, 10, 17, 18, 19 (mo d 20) exc ept n = 22, 23, 26, 27, 42 0 otherwise. Theorem 404. L et S = f 3 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6, 13 (mo d 21), or n = 31 0 otherwise. Theorem 405. L et S = f 1 ; 3 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + ( 1 if n 6, 8, 17, 24, 25, 33 (mo d 34) 0 otherwise. Theorem 406. L et S = f 2 ; 3 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 5, 6, 7, 8, 9, 17, 18 (mo d 19) exc ept n = 18, 24, 27, 36, 45, 63, 81 0 otherwise. 162

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Theorem 407. L et S = f 1 ; 2 ; 3 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 6, 7, 8, 9, 10 (mo d 20) exc ept n = 26, 27, 28, 46 0 otherwise. Theorem 408. L et S = f 4 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 15 + ( 1 if n = 20 0 otherwise. Theorem 409. L et S = f 1 ; 4 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 > > > < > > > : 1 if n 4, 5, 6, 8, 9, 10 (mo d 11) exc ept n = 19, 20, 21, 27, 37, 38, 39, 41, 42, 59, 60, 61, 63, 81, 82, 83, 103, 104, 105, 125, 126, 147 0 otherwise. Theorem 410. L et S = f 2 ; 4 ; 5 ; 8 ; 9 g Then r ( C n [ S ]) = n 8 for all n 18 Theorem 411. L et S = f 1 ; 2 ; 4 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 6, 9, 10 (mo d 11) exc ept n = 20, 21, 39, 42, 61, 83, 105 0 otherwise. Theorem 412. L et S = f 3 ; 4 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 3, 4, 7, 8, 9 (mo d 10) exc ept n = 23, 24 0 otherwise. Theorem 413. L et S = f 1 ; 3 ; 4 ; 5 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 414. L et S = f 2 ; 3 ; 4 ; 5 ; 8 ; 9 g Then r ( C n [ S ]) = n 10 for all n 18 Theorem 415. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 15 + ( 1 if n 12, 13, 14 (mo d 15) 0 otherwise. Theorem 416. L et S = f 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 23 + ( 1 if n 4, 5, 14, 15, 16, 17 (mo d 23) exc ept n = 27, 37 0 otherwise. 163

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Theorem 417. L et S = f 1 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 > > < > > : 2 if n 7 (mo d 24) exc ept n = 31 1 if n 4, 5, 6, 8, 9, 10, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 (mo d 24) exc ept n = 18, 19, 20, or n = 31 0 otherwise. Theorem 418. L et S = f 2 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 23 + 8 < : 1 if n 4, 6, 7, 11, 13, 14, 15, 20, 22 (mo d 23) exc ept n = 57 0 otherwise. Theorem 419. L et S = f 1 ; 2 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 8, 9, 15, 17 (mo d 18) 0 otherwise. Theorem 420. L et S = f 3 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 5, 6, 7, 8, 12, 13, 14, 15, 16 (mo d 24) exc ept n = 31, 55, 62, 86, 110, or n = 57 0 otherwise. Theorem 421. L et S = f 1 ; 3 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + 8 < : 1 if n 5, 6, 7, 8, 15, 16, 17, 25, 32, 33, 34, 39, 40, 41, 42 (mo d 51) exc ept n = 32 0 otherwise. Theorem 422. L et S = f 2 ; 3 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 4, 5, 6, 8, 9, 10, 15, 16, 17, 18, 19 (mo d 20) exc ept n = 24 0 otherwise. Theorem 423. L et S = f 1 ; 2 ; 3 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 9, 10, 11, 12 (mo d 13) 0 otherwise. 164

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Theorem 424. L et S = f 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 29 + 8 > > > < > > > : 1 if n 4, 5, 6, 7, 12, 13, 14, 19, 20, 21, 26, 28 (mo d 29) exc ept n = 19, 21, 26, 28, 35, 42, 49, 63, 70, 77, 84, 91 0 otherwise. Theorem 425. L et S = f 1 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 5, 6, 7, 8, 9, 10 (mo d 11) 0 otherwise. Theorem 426. L et S = f 2 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 3, 4, 5, 6, 7, 8, 9, 10 (mo d 20) exc ept n = 23, 25, 27, 43, 45, 63 0 otherwise. Theorem 427. L et S = f 1 ; 2 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 6, 8, 10 (mo d 11) exc ept n = 19, 39 0 otherwise. Theorem 428. L et S = f 3 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 15, 39 (mo d 48), or n = 30, 36, 38, 54 0 otherwise. Theorem 429. L et S = f 1 ; 3 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 4, 7, 9, 10 (mo d 11) exc ept n = 18, 20, 37, 40, 59 0 otherwise. Theorem 430. L et S = f 2 ; 3 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 8, 9, 10 (mo d 20) 0 otherwise. Theorem 431. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 10, 12 (mo d 13) 0 otherwise. 165

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Theorem 432. L et S = f 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 4 n 31 + 8 > > > > > < > > > > > : 2 if n 14, 15 (mo d 31) 1 if n 1, 2, 3, 4, 5, 6, 7, 10, 11, 13, 17, 18, 20, 21, 22, 23, 26, 27, 29, 30 (mo d 31) exc ept n = 18, 26, 27, 32, 33 0 otherwise. Theorem 433. L et S = f 1 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + ( 1 if n 5, 6, 7, 8, 38, 39, 40 (mo d 48) 0 otherwise. Theorem 434. L et S = f 2 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 > > > < > > > : 1 if n 2, 4, 5, 6, 7, 8 (mo d 9) exc ept n = 20, 22, 23, 25, 26, 29, 31, 47, 49, 50, 51, 52, 56, 74, 76, 77, 78, 83, 101, 103, 104, 128, 130, 155, 182 0 otherwise. Theorem 435. L et S = f 1 ; 2 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 7, 8, 10, 11 (mo d 12) exc ept n = 19, 20 0 otherwise. Theorem 436. L et S = f 3 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9 (mo d 20), or n = 18, 25 0 otherwise. Theorem 437. L et S = f 1 ; 3 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9 (mo d 20) 0 otherwise. Theorem 438. L et S = f 2 ; 3 ; 5 ; 6 ; 8 ; 9 g Then r ( C n [ S ]) = n 10 for all n 18 Theorem 439. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 8 ; 9 g Then r ( C n [ S ]) = n 13 for all n 18 Theorem 440. L et S = f 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 8, 9 (mo d 19) 0 otherwise. 166

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Theorem 441. L et S = f 1 ; 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 6, 9, 10 (mo d 11) exc ept n = 20, 21, 28, 39, 42, 61, 83, 105 0 otherwise. Theorem 442. L et S = f 2 ; 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 21 + ( 1 if n 6, 7, 8, 9, 10 (mo d 21) exc ept n = 27 0 otherwise. Theorem 443. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 8 ; 9 g Then r ( C n [ S ]) = n 12 for all n 18 Theorem 444. L et S = f 3 ; 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 8, 9, 10, 11 (mo d 23), or n = 21, 28 0 otherwise. Theorem 445. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + 8 < : 1 if n 8, 9, 10, 11 (mo d 23) exc ept n = 31, 32, 33, 54, 55, 77 0 otherwise. Theorem 446. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 8, 9, 10, 11 (mo d 23) 0 otherwise. Theorem 447. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 8 ; 9 g Then r ( C n [ S ]) = n 16 for all n 18 Theorem 448. L et S = f 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 5 n 33 + 8 > > > > > > > > > > > > > > < > > > > > > > > > > > > > > : 2 if n 6, 25, 26 (mo d 33) exc ept n = 25, 26, 39, 58, 59, 72, 91, 92, 105, 124, 125, 138, 157, 158, 171, 190 1 if n 2, 3, 4, 5, 7, 8, 11, 12, 13, 15, 16, 17, 18, 19, 24, 27, 28, 29, 30, 31, 32 (mo d 33) exc ept n = 18, 19, 24, 27, 28, 29, 30, 31, 35, 36, 37, 38, 44, 48, 49, 50, 57, 60, 61, 62, 63, 68, 69, 73, 74, 81, 82, 93, 94, 95, 101, 106, 107, 114, 126, 139, or n = 26, 39, 58, 59, 72, 91, 92, 105, 124, 125, 138, 157, 158, 171, 190 0 otherwise. 167

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Theorem 449. L et S = f 1 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 > > > < > > > : 2 if n = 32 1 if n 4, 5, 6, 7, 8, 15, 16 (mo d 25) exc ept n = 29, 30, 32, 54, 79 0 otherwise. Theorem 450. L et S = f 2 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 > > > < > > > : 2 if n 7, 8 (mo d 25) exc ept n = 32, 33, 57, 58, 82 1 if n 4, 5, 6, 9, 10, 11, 12, 13, 14, 15, 16 (mo d 25) exc ept n = 34, 35, 36, 59, or n = 32, 33, 57, 58, 82 0 otherwise. Theorem 451. L et S = f 1 ; 2 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 6, 7, 8, 9 (mo d 19) exc ept n = 25, 26, 27 0 otherwise. Theorem 452. L et S = f 3 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 7 for all n 18 Theorem 453. L et S = f 1 ; 3 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + 8 < : 1 if n 6, 7, 8, 9 (mo d 19) exc ept n = 25, or n = 34, 35, 40, 53 0 otherwise. Theorem 454. L et S = f 2 ; 3 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 17 + 8 > > > < > > > : 1 if n 4, 5, 6, 7, 8, 15, 16, 17, 23, 24, 25 (mo d 34) exc ept n = 23, 24, 25, 49, 50, 72, 73, 74, 75, 125, 174, 175, or n = 31 0 otherwise. Theorem 455. L et S = f 1 ; 2 ; 3 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 9, 10, 11, 12 (mo d 13) exc ept n = 21, 22 0 otherwise. Theorem 456. L et S = f 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 7 + ( 1 if n 6, 13, 20 (mo d 28) exc ept n = 20, 34 0 otherwise. 168

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Theorem 457. L et S = f 1 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 26 + ( 1 if n 7, 8, 13, 14, 15, 16, 17 (mo d 26) exc ept n = 39 0 otherwise. Theorem 458. L et S = f 2 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 < : 1 if n 4, 5, 6, 7, 8, 11, 12, 13, 14, 15, 16 (mo d 25) exc ept n = 29, 30, 36, 37, 38, 61, 62 0 otherwise. Theorem 459. L et S = f 1 ; 2 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 8, 9 (mo d 19) exc ept n = 27 0 otherwise. Theorem 460. L et S = f 3 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 6, 7, 8 (mo d 9) exc ept n = 22, 24, 31, 49, 51 0 otherwise. Theorem 461. L et S = f 1 ; 3 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 21 + ( 1 if n 5, 6, 7, 8, 9, 10 (mo d 21) exc ept n = 26, 27 0 otherwise. Theorem 462. L et S = f 2 ; 3 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 9, 10, 11 (mo d 22), or n = 28, 49 0 otherwise. Theorem 463. L et S = f 1 ; 2 ; 3 ; 4 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 14 + ( 1 if n = 24, 25, 38 0 otherwise. Theorem 464. L et S = f 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 15 + 8 < : 1 if n 3, 4, 5, 6, 7, 12, 14, 15, 16, 18, 20, 21, 22, 27, 29 (mo d 30) exc ept n = 18, 21, 27, 33, 42, 63 0 otherwise. 169

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Theorem 465. L et S = f 1 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 5, 6, 7, 8, 9, 10 (mo d 11) 0 otherwise. Theorem 466. L et S = f 2 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 < : 1 if n 7, 8, 13, 14, 15, 16 (mo d 25) exc ept n = 32, 63, 64 0 otherwise. Theorem 467. L et S = f 1 ; 2 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 6, 7, 8, 9, 10 (mo d 11) exc ept n = 18, 19, 20, 39, 40 0 otherwise. Theorem 468. L et S = f 3 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 5, 6, 7, 8 (mo d 9) 0 otherwise. Theorem 469. L et S = f 1 ; 3 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 6, 8, 10 (mo d 22) 0 otherwise. Theorem 470. L et S = f 2 ; 3 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 6, 7, 8, 9, 10 (mo d 20) exc ept n = 27 0 otherwise. Theorem 471. L et S = f 1 ; 2 ; 3 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 10, 12 (mo d 13) exc ept n = 21 0 otherwise. Theorem 472. L et S = f 4 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + 8 < : 1 if n 3, 4, 5, 6, 7, 8, 9 (mo d 10) exc ept n = 23, 24, 25, 26, 27, 33, 53, 54 0 otherwise. Theorem 473. L et S = f 1 ; 4 ; 5 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 170

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Theorem 474. L et S = f 2 ; 4 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 9 (mo d 20) 0 otherwise. Theorem 475. L et S = f 1 ; 2 ; 4 ; 5 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 476. L et S = f 3 ; 4 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 21 + 8 < : 1 if n 7, 8, 9, 10 (mo d 21) exc ept n = 28, 29, 30, 49, 50, 70 0 otherwise. Theorem 477. L et S = f 1 ; 3 ; 4 ; 5 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 478. L et S = f 2 ; 3 ; 4 ; 5 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 9, 10, 11 (mo d 23) 0 otherwise. Theorem 479. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 15 for all n 18 Theorem 480. L et S = f 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 8 + 8 < : 1 if n 6, 7, 8, 12, 13, 14, 15, 16 (mo d 24) exc ept n = 30, 31, 60, 61, 62, or n = 20 0 otherwise. Theorem 481. L et S = f 1 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 5, 6, 7, 8, 9, 10 (mo d 11) exc ept n = 26, 27 0 otherwise. Theorem 482. L et S = f 2 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 25 + 8 < : 1 if n 7, 8, 13, 14, 15, 16 (mo d 25) exc ept n = 32, 63, 64, or n = 37 0 otherwise. Theorem 483. L et S = f 1 ; 2 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n = 30, 31, 32, 54 0 otherwise. 171

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Theorem 484. L et S = f 3 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + 8 > > < > > : 2 if n 9 (mo d 27) 1 if n 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18 (mo d 27) exc ept n = 32, 37, 64 0 otherwise. Theorem 485. L et S = f 1 ; 3 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + 8 < : 1 if n 4, 6, 7, 8, 9, 10 (mo d 11) exc ept n = 18, 19, 26, 37 0 otherwise. Theorem 486. L et S = f 2 ; 3 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n 8, 9, 10 (mo d 20), or n = 35 0 otherwise. Theorem 487. L et S = f 1 ; 2 ; 3 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 13 for all n 18 Theorem 488. L et S = f 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 9 + ( 1 if n 4, 5, 6, 7, 8 (mo d 9) exc ept n = 22, 23, 24, 31 0 otherwise. Theorem 489. L et S = f 1 ; 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 4, 6, 8, 10 (mo d 11) exc ept n = 26 0 otherwise. Theorem 490. L et S = f 2 ; 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 5, 6, 7, 8, 9, 10, 11 (mo d 22) exc ept n = 27 0 otherwise. Theorem 491. L et S = f 1 ; 2 ; 4 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 12 for all n 18 Theorem 492. L et S = f 3 ; 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 19 + ( 1 if n 9 (mo d 19) 0 otherwise. 172

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Theorem 493. L et S = f 1 ; 3 ; 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 10 (mo d 11) exc ept n = 21 0 otherwise. Theorem 494. L et S = f 2 ; 3 ; 4 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 11 + ( 1 if n 9, 10, 11 (mo d 22) 0 otherwise. Theorem 495. L et S = f 1 ; 2 ; 3 ; 4 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 14 for all n 18 Theorem 496. L et S = f 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n = 35, 42, 43, 44 0 otherwise. Theorem 497. L et S = f 1 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 498. L et S = f 2 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 10 + ( 1 if n = 35, 44 0 otherwise. Theorem 499. L et S = f 1 ; 2 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 12 for all n 18 Theorem 500. L et S = f 3 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 10 for all n 18 Theorem 501. L et S = f 1 ; 3 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + 8 < : 1 if n 6, 7, 8, 9, 10, 11, 12 (mo d 24) exc ept n = 30, 31, 32, 33, 54, 55 0 otherwise. Theorem 502. L et S = f 2 ; 3 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 23 + ( 1 if n 7, 8, 9, 10, 11 (mo d 23) exc ept n = 30 0 otherwise. Theorem 503. L et S = f 1 ; 2 ; 3 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 13 for all n 18 Theorem 504. L et S = f 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 3 n 31 + 8 > > > > > > > < > > > > > > > : 1 if n 7, 8, 9, 10, 13, 14, 15, 16, 17, 18, 19, 20 (mo d 31) exc ept n = 18, 19, 20, 38, 39, 40, 44, 45, 46, 47, 48, 49, 50, 69, 70, 75, 76, 77, 78, 79, 80, 100, 106, 107, 108, 109, 110, 137, 138, 139, 140, 168, 169, 170, 199, 200, 230 0 otherwise. 173

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Theorem 505. L et S = f 1 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 11 for all n 18 Theorem 506. L et S = f 2 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 12 + ( 1 if n 8, 9, 10, 11, 12 (mo d 24) 0 otherwise. Theorem 507. L et S = f 1 ; 2 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 12 for all n 18 Theorem 508. L et S = f 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 9, 10, 11, 12, 13 (mo d 26) 0 otherwise. Theorem 509. L et S = f 1 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = n 13 + ( 1 if n 8, 9, 10, 11, 12, 13 (mo d 26) 0 otherwise. Theorem 510. L et S = f 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then for all n 18 ; we have r ( C n [ S ]) = 2 n 27 + ( 1 if n 9, 10, 11, 12, 13 (mo d 27) 0 otherwise. Theorem 511. L et S = f 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 ; 8 ; 9 g Then r ( C n [ S ]) = n 19 for all n 18 174

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