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 Title:
 Extremal problems for degree sequences
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 Erbes, Catherine C. ( author )
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 Denver, CO
 Publisher:
 University of Colorado Denver
 Publication Date:
 2014
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 Potential theory (Mathematics) ( lcsh )
Ramsey numbers ( lcsh )
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 Review:
 In this thesis we present results about two extremal problems for degree sequences, the potential number (the analogue of the classical TurÌÂan number) and the potentialRamsey number.
 Thesis:
 Thesis (Ph.D.)University of Colorado Denver. Applied mathematics
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 Includes bibliographic references.
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 Department of Mathematical and Statistical Sciences
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 by Catherine C. Erbes.
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EXTREMAL PROBLEMS FOR DEGREE SEQUENCES
by
CATHERINE C. ERBES
B.A., Carleton College, 2005
M.A., Indiana University, Bloomington, 2007
A thesis submitted to the
Faculty of the Graduate School of the
University of Colorado in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
2014
This thesis for the Doctor of Philosophy degree by
Catherine C. Erbes
has been approved for the
Department of Mathematical and Statistical Sciences
by
Michael Jacobson, Chair
Michael J. Ferrara, Advisor
Ellen Gethner
Stephen Hartke
Paul Wenger
April 29, 2014
n
Erbes, Catherine C. (Ph.D., Applied Mathematics)
Extremal Problems for Degree Sequences
Thesis directed by Associate Professor Michael J. Ferrara
ABSTRACT
A sequence of nonnegative integers 7r is graphic if it is the degree sequence of
some graph G. In this case we say that G is a realization of tt. Degreesequence
analogues of many classical problems in extremal graph theory appear throughout
the literature. In this thesis we present results about two extremal problems for
degree sequences, the potential number (the analogue of the classical Turan number)
and the potentialRamsey number.
A graphic sequence 7r is potentially Hgraphic if there is a realization of 7r that
contains H as a subgraph. The potential number of a graph H, denoted a(H,n), is
the minimum even integer such that any graphic sequence of length n and sum at
least a(H,n) is potentially Rgraphic. The potential number has been determined
asymptotically for general graphs Ft, and a family V(H) of extremal sequences that
achieve this number is known.
Given nonincreasing graphic sequences = (d\,..., dn) and 7t2 = (si,..., sn),
we say that tti majorizes 1r2 if di > Si for all i, 1 < i < n. In 1970, Erdos showed
that for any Ab+ifree graph Ft, there exists an rpartite graph G such that 7r(G)
majorizes 7t(H). In 2005, Pikhurko and Taraz generalized this notion and showed
that for any graph F with chromatic number r + 1, the degree sequence of an F
free graph is, in an appropriate sense, nearly majorized by the degree sequence of
an rpartite graph. Here, we give similar results for degree sequences that are not
potentially i7graphic. In particular, we show that if tt is a graphic sequence that is
not potentially Ftgraphic, then 7r is close to being majorized by a sequence in V(H).
iii
This shows that the structure of sequences that are not potentially Hgraphic is close
to that of the extremal sequences.
Secondly, we give a stability result for the potential problem, similar to the sta
bility results of Erdos and Simonovits for the Turan problem. We say that a graph
H is crstable if every graphic sequence with sum close to a(H,n) that is not poten
tially Hgraphic can be transformed into a sequence in V(H) with o(n) additions and
subtractions. We show that, in contrast to the Turan problem, not all graphs are
(jstable. We also show that a large family of graphs are crstable.
Finally, we address a degreesequence variant of the Ramsey number, recently
introduced by Busch, et al. Given graphs Gi and G2, let rpot(Gi, G2) be the minimum
integer n such that for any graphic sequence tt of length n, either 7r is potentially G\
graphic or the complement of 7r is potentially G2graphic. We give several lower
bounds on rpot(G 1, G2), and also determine the values of rpot(Cs, Ct) and rpot(G 1, G2)
when Gi is a fixed graph of order at most four and G2 is arbitrary.
The form and content of this abstract are approved. I recommend its publication.
Approved: Michael J. Ferrara
IV
To Rachel
v
ACKNOWLEDGMENTS
First, I would like to thank my advisor Mike Ferrara, for guiding me on this
journey since day one. Thanks is also due to the members of my committee: Mike
Jacobson (without whose advice I would never have even taken graph theory), Paul
Wenger, Ellen Gethner, and Stephen Hartke, who have all helped me to become a
better mathematician. I would also like to thank Ryan Martin, who along with Paul
and Mike, helped me truly understand the necessity of a precisely written proof, no
matter how ugly it might be.
I would also like to thank the NSF Grant $4)74324 UCD GK12 Transform
ing Experiences Project and the Lynn Bateman Memorial Fellowship for funding
portions of this research.
I would not have made it to this point without the support of many people. My
fellow grad students, particularly Jenny Diemunsch and Tim Morris, my academic
siblings, whose companionship in this process (also since day one) has made it much
more fun, if not any easier; Axel Brandt, Brent Thomas, Mark Mueller, Sam Graffeo,
and Devon Sigler, who were always happy to help with math, distraction, coffee, or
drinks, whichever was most needed at the time; and Breeann Flesch and Eric Sullivan,
who served as a reminder to me that it is possible to not only graduate, but also to
find a job and succeed after grad school. My mom, dad, and brothers and sisters,
who have always believed in me, and especially my sister Anissa, whose passion for
flight and stories about learning to fly have been my greatest source of motivation.
And finally, my wife Rachel, who always knew when I needed to play video games and
when I needed to do math, and whose patience and support have meant the world to
me.
vi
TABLE OF CONTENTS
Tables....................................................................... ix
Figures ...................................................................... x
Chapter
1. Introduction............................................................... 1
1.1 Definitions and concepts ........................................... 1
1.2 An introduction to degree sequences................................. 3
2. Extremal Problems for Degree Sequences..................................... 6
2.1 An overview of extremal graph theory................................ 6
2.1.1 The Turan problem ........................................ 6
2.1.2 Ramsey theory................................................. 8
2.2 Degreesequence problems............................................ 9
2.2.1 Forcible problems versus potential problems................... 9
2.2.2 Potentially //graphic sequences ............................ 13
2.3 The potential number............................................... 17
3. The Shape of Graphic Sequences that are not Potentially //Graphic ... 23
3.1 Introduction....................................................... 23
3.2 Majorization of sequences that are not potentially //graphic .... 26
3.3 Lemmas ............................................................ 29
3.4 Proof of Theorem 3.4............................................... 30
3.5 Proof of Lemma 3.5................................................. 38
4. Stability with Respect to the Potential Number........................ 52
4.1 Graphs with degreesequence stability.............................. 54
4.2 Graphs that are not ustable....................................... 55
4.3 Technical lemmas................................................... 62
4.4 Proofs of Theorems 4.4 and 4.5..................................... 64
4.5 Implications of stability.......................................... 80
vii
5. PotentialRamsey Numbers................................................. 82
5.1 Introduction....................................................... 82
5.2 Preliminaries...................................................... 83
5.2.1 General lower bounds ...................................... 84
5.2.2 Towards an upper bound ..................................... 86
5.2.3 Potentially P4graphic sequences............................ 89
5.3 PotentialRamsey numbers for small graphs versus arbitrary graphs 91
5.4 PotentialRamsey numbers for cycles................................ 97
5.5 PotentialRamsey numbers for all graphs of order at most 4........ 110
6. Future Work............................................................. 113
6.1 Hypergraphic degree sequences..................................... 113
6.2 Further exploration of potentialRamsey numbers................... 115
References................................................................. 117
viii
TABLES
Table
4.1 Connected graphs of order at most 6 that are not ustable............ 58
4.2 Connected graphs of order at most 6 that are not weakly ustable .... 59
51 rpot(H,G) for small H and rpot(Cs, Ct).............................. 109
5.2 Small potentialRamsey numbers...................................... 110
IX
FIGURES
Figure
1.1 A 2switch.............................................................. 5
2.1 The graph (Ki U I
3.1 The shape of a graphic sequence ....................................... 23
3.2 Different ways to reduce terms in the KleitmanWang algorithm.......... 40
4.1 A graph with Va+i > 1 and 2i* V** <2a.............................. 54
U
6.1 The edgeexchange e^e ................................................ 114
V
X
1. Introduction
1.1 Definitions and concepts
A graph G consists of a vertex set V(G) and an edge set E(G), where each edge
is a set of vertices. Throughout the majority of this paper, we will restrict ourselves
to standard graphs, in which each edge contains exactly two distinct vertices. Hyper
graphs, in which an edge may contain more than two vertices (or even a single vertex),
will be discussed briefly in the last chapter. In this thesis, we will only consider finite
simple graphs, meaning that both the vertex and edge sets are finite, and there are
no repeated edges or loops (edges consisting of one vertex multiple times).
To denote edges in G, we will write xy or e = xy for an edge with endpoints x
and y, and simply e if the endpoints are not relevant. If x and y are vertices of G
and xy G E(G), we say that x is adjacent to y, and sometimes write x ~ y. When
more than one graph is under consideration, we may specify that x is adjacent to y
in G by writing x ~g y If xy E(G) we sometimes write x ^ y or x y.
The degree of a vertex x, denoted d(x) or dc(x), is the number of edges contain
ing x, or, in the case of simple graphs, the number of vertices adjacent to x. The
neighborhood of x is the set of vertices adjacent to x, and is written N(x) or Ng(x).
So, we see that do(x) = ./Vg(:e). (The subscript G is used to specify a parameter
within the graph or subgraph G. When there is only one graph being discussed or
it is understood, we do not use the subscript.) The maximum degree and minimum
degree of G are denoted A(G) and 8(G), respectively.
A graph H is a subgraph of G if V(H) C V(G) and E(H) C E(G). If H is
a subgraph of G, it is called spanning if V(H) = V(G), and induced if for each
x,y G V(H), xy G E(H) if and only if xy G E(G). If Ed is an induced subgraph
of G, we write H < G. If X C V(G), then the subgraph of G induced by X,
denoted G[X], is the graph G (V(G) \ X), which has vertex set X and edge set
{xy G E(G)  x, y G X}.
1
There are several ways to create new graphs from other graphs. One such way
is to take the complement of G, denoted G, which has V(G) = V(G) and e G E(G)
if and only if e ^ E(G). The disjoint union of graphs G and H is denoted G U H,
and has V(G U H) = V(G) U V(H) and E(G U H) = E(G) U E(H). The join of
G and H, denoted G V H has vertex set V(G) U V(H), and edge set E{G V H) =
E{G) U E{H) U {xy \ x G V(G) and y G V(H)}.
A path is a simple graph in which the vertices can be ordered V\,..., vn such that
Vi ~ vi+\ for each i G {1,..., n 1}. If, in addition, vn ~ V\, then this graph is a
cycle. We will often describe a path or cycle by its vertices, by writing v\v2 vn for
a path and vpv2 vnV\ for a cycle. Let Pn denote the path on n vertices, and Cn the
cycle on n vertices. We say that a graph is connected if there is a path between any
pair of vertices. A tree is a connected graph that contains no cycles, and we say that
a connected component of a graph is nontrivial if it has at least one edge.
A set of vertices that are pairwise adjacent is known as a clique. The complete
graph on n vertices, denoted Kn, is a clique of order n. A set of vertices that are
pairwise nonadjacent is an independent set, and the maximum order of an independent
set in G is denoted a(G). A matching is a set of disjoint edges, that is a set of edges
with no endpoints in common, and the maximum size of a matching in G is denoted
a'(G).
A graph is bipartite if the vertex set can be partitioned into two independent sets,
and kpartite if the vertex set can be partitioned into k independent sets. We will
use Kr>s to denote the complete bipartite graph with r vertices in one partite set and
s vertices in the other.
A coloring of the vertices of a graph is called proper if adjacent vertices receive
distinct colors. The chromatic number of G, denoted y(G), is the minimum k such
that G has a proper coloring using k colors.
Other terms will be defined as necessary, and for terms not defined in this disser
2
tation, see [122],
1.2 An introduction to degree sequences
The degree sequence of a graph is a list of the vertex degrees of the graph. A
sequence of nonnegative integers tt is called graphic if it is the degree sequence of
some graph G. In this case we say that G realizes tt or is a realization of tt, and we
write G = G(tt), or 7r = 7r(G). Unless otherwise noted, we will assume that all graphic
sequences are written in nonincreasing order. We will often write 7r = (d1, > d)
to indicate that each degree di appears rrii times in 7r, although we will generally
suppress the multiplicity of di if rrii = 1.
Given a graph, determining its degree sequence is a simple exercise. On the other
hand, determining when a given sequence is graphic is more difficult. The first, and
perhaps the simplest, characterization of graphic sequences is due independently to
Havel and Hakimi.
Theorem 1.1 (Havel [59], Hakimi [56]) Let 7T = (di,..., dn) be a nonincreasing
sequence of nonnegative integers. The sequence 1r is graphic if and only if the sequence
tt' = (d2 1,..., ddl+1 1, ddl+2,..., dn) is graphic.
Kleitman and Wang gave the following generalization of this theorem.
Theorem 1.2 (Kleitman and Wang [70]) Let n = (di,... ,dn) be a nonincreas
ing sequence of nonnegative integers, and let i E [n]. If 7q is the sequence defined
by
(di 1,.. * > d^ 1 j d^_i ?... j 0 li~ 1 j dii; ... ,dn) if di
(di 1,.. * > di 1 1 j d^_i 1 j  * ; d^t 1 1 j ddi+2, , dn) if di > i
then 7r is graphic if and only if 1q is graphic.
Let 7d be the sequence resulting from sorting 7q in nonincreasing order, and call
7d the residual sequence obtained by laying off di.
3
The HavelHakimi and KleitmanWang Theorems give rise to efficient algorithms
to test for graphicality. They also introduce socalled residual (sub)sequence tech
niques for analyzing graphic sequences. These techniques involve using information
that we know about the residual sequence, such as certain properties of its realiza
tions, to gain knowledge about the original sequence. For example, if the sequence tt'
is potentially //graphic (a property which will be defined in Section 2.2.2), then we
know that 7r must also be potentially //graphic.
In Chapter 3, we use a careful examination of the KleitmanWang algorithm to
create a specific realization of a sequence that meets certain criteria. In Chapter 4, we
will give some simple corollaries of the KleitmanWang algorithm that give us more
information about the residual sequences obtained from applying it.
In addition to these residualsequence characterizations of graphic sequences there
are others, many of which take the form of systems of inequalities. The bestknown
of these is due to Erdos and Gallai.
Theorem 1.3 (Erdos and Gallai [37]) A nonincreasing sequence ir = (d\,..., dn)
of nonnegative integers is graphic if and only if Y^i=\ d% even and, for all p G
{1,..., n 1},
p n
^di
i= 1 i=p\l
In [60], Sierksma and Hoogeveen give six other characterizations of graphic sequences,
and prove that they are all equivalent. Many others have given other characterizations
or sufficient conditions for graphic sequences [66, 71, 115], as well as improvements on
the ErdosGallai Criterion, in the form of a reduction of the number of inequalities
that must be checked [116, 157].
A graphic sequence may have many different realizations. A common technique
that is used to transform one realization into another is known as the edgeexchange
or 2switch. If G is a realization of tt with edges xy and uv such that xu and yv are
4
Figure 1.1: A 2switch
not edges, we can delete the edges xy and uv and replace them with the edges xu
and yv (see Figure 1.1). This maintains the degree of each vertex, so the new graph
is also a realization of tt.
Petersen [101] showed that given any pair of realizations of a graphic sequence,
one can be obtained from the other by a sequence of 2switches. This technique can
be used to prove both the HavelHakimi and KleitmanWang Theorems, as well as
many other results about degree sequences; a particularly nice example of this can be
found in [45].
An isomorphism from a graph G to a graph H is a bijection / : V(G) > V(H)
such that uv G E(G) if and only if f{u)f{v) G E(H). If there is an isomorphism from
G to H, we say that G is isomorphic to H, or that they are in the same isomorphism
class. While a 2switch always changes the structure of a labeled graph (one where
the names of the vertices are important), it may in fact preserve the isomorphism
class of the (unlabeled) graph. Recently, Barrus [4] determined sufficient conditions
for a 2switch to change the isomorphism class of a graph.
5
2. Extremal Problems for Degree Sequences
2.1 An overview of extremal graph theory
Extremal graph theory can be thought of as the study of thresholds. In particular,
we look for thresholds on graph invariants that guarantee that a graph has a certain
property. That is, given an invariant i(G), a class of graphs T, and a graph property
V, the most general formulation of an extremal graph theory problem is: What is
the minimum value m (or maximum value m!) such that every graph G E T with
i(G) > m (i(G) < m!) has property VI Those graphs in T that satisfy i(G) = m but
do not have property V are known as extremal graphs.
2.1.1 The Turan problem
One of the first and most wellstudied extremal problems is known as the Turan
problem or simply the extremal problem. It is:
Problem 2.1 Given a graph H, determine ex(H,n), the maximum number of edges
in a simple graph of order n that does not contain H as a subgraph.
In this case, the graph invariant is the number of edges, the property is contains
H as a subgraph, and the class of graphs is all simple graphs. The first answer to a
question of this type is due to Mantel, in 1907:
Theorem 2.2 (Mantel [94]) The maximum number of edges in an nvertex triangle
free simple graph is [n2/4J. That is, ex(Ks,n) = _n2/4_.
The number ex(H, n) is known as the extremal number or the Turan number for
H, as the next result, due to Turan, was the first general solution to the problem. The
graph Tra r, known as the Turan graph, is the complete rpartite graph on n vertices,
with partite sets as equal as possible. That is, if n = kr + p where 0 < p < r, then
there are p partite sets of order r + 1 and k p partite sets of order r.
Theorem 2.3 (Turans Theorem [117]) For a positive integer r, ex(Kr+i,n) =
\E(Tn>r)\, and Tn>r is the unique edgemaximal Kr+\free graph of order n.
6
Note that the Turan graph has () r(ra+pfc) = (1 1)^ + 0(n) edges.
While the exact value of the extremal function is known for very few graphs (for
some examples, see [If, 18, 36, 103, 117]), in 1966 Erdos and Simonovits [40] extended
previous work of Erdos and Stone [39] and determined ex(H, n) asymptotically for
arbitrary (nonbipartite) H.
Theorem 2.4 (The ErdosStoneSimonovits Theorem [39, 40]) If H is a graph
with chromatic number x(H) = r + 1 > 2, then
This shows that asymptotically, the Turan graphs have the proper number of edges
to be the extremal graphs for any H, not just for complete graphs. Subsequently,
Simonovits [113] and Erdos independently proved the following, sometimes known as
the First Stability Theorem.
Theorem 2.5 (Simonovits [113]) Let H be a graph with x(H) = r + 1. For every
e > 0, there exists a 5 > 0 and an n such that if n > n and G is an nvertex Hfree
graph such that
\E(G)\ > ex(H,n) 5n2,
then G can be obtained from Tn>r by changing at most tn2 edges.
With this theorem, Simonovits introduced the stability method for solving
extremal problems. Recall that a general extremal problem seeks to find the minimum
value m of a parameter (or graph invariant) i{G) over a set of objects in a certain
class T that will guarantee that every object for which i(G) > m has property V. The
first step of the stability method is to find an asymptotic solution for the extremal
problem, which usually also determines a set C of extremal objects. (If this set of
extremal objects can be shown to be the complete set, then we can find an exact
7
answer to the problem, so proving this is the ultimate goal.) The next step is to
prove a stability result, which shows that for every object G in T for which %[G) is
close to but not greater than m, the structure of G is similar to that of the objects
in C. Using the stability result, the next step is to show that every extremal object
is in fact in C. This yields the exact answer to the extremal problem.
Simonovits used this method to determine exactly the extremal number for p
disjoint copies of Kr, or pKr [113]. Since then, stability methods have been used to
attack a wide variety of extremal problems (see [7, 68, 97, 99, 102] for examples). Re
cently, stability methods have also been used to approach problems in Ramsey Theory
[53, 54, 100], and have also proven particularly helpful in studying the hypergraph
Turan problem (see [1], [96], or [98] for some examples).
2.1.2 Ramsey theory
Ramsey theory focuses on a different kind of extremal problem. In its original,
most general form, Ramseys Theorem [106] is about sets. Given a set S, we write
(j[) to denote the set of relement subsets of S. A set T C S is called ^homogeneous
if, under some coloring of the elements of (j[), all of the relement subsets of T receive
color i. If n is a positive integer, we let [n] = {1,... ,n}. With these definitions, we
can now state Ramseys Theorem.
Theorem 2.6 (Ramsey [106]) Given positive integers r andpi,... ,pk, there exists
an integer N such that every kcoloring of yields an ihomogeneous set of size
Pi for some i.
Graph theory provides a nice illustration of the case r = k = 2. In this case
the sets of order 2 are edges, and we consider a 2coloring of the edges of a complete
graph. Thus, an ihomogeneous set of size jp is a complete graph of order p^ in which
all edges have the same color. In this language, Ramseys Theorem says that given
positive integers s and t, there is an n such that any red/blue coloring of the edges
8
of Kn yields either a red Ks or a blue Kt. The number n is then called the Ramsey
number and is denoted r(s,t).
Ramsey numbers are notoriously difficult to compute. In fact, the exact value
of r(s, t) is known only for s = 3 and t G {3,... 9}, or when s = 4 and t = 4, or
s = 4 and t = 5 [105]. Asymptotically, we know the Ramsey number r(3,t) within a
constant factor:
ct2
< r(3, t) <
dt2
logi logi
where currently the best known values of the constants are c = 1/4 and d = 1 (see
[105]).
A slight relaxation of this problem is to look for monochromatic copies of graphs
other than complete graphs. Given graphs Gi,... ,Gk, the kcolor graph Ramsey
number r(Gi,..., Gk), is the minimum integer n such that any fccoloring of Kn yields
a monochromatic Gi in color i for some i. The case k = 2 is the most wellstudied,
although there are some results for k > 3 when the graphs are small or relatively
simple (see [105] for a thorough survey).
In Chapter 5 we will discuss a degreesequence variant of graph Ramsey numbers.
2.2 Degreesequence problems
2.2.1 Forcible problems versus potential problems
The goal of this dissertation is to examine degreesequence analogues of classical
extremal problems. Before we do that, however, we will discuss degreesequence prob
lems in general, beyond the simple characterization problems discussed in Chapter 1.
Recall that a graphic sequence can have many different realizations. Most degree
sequence problems deal with showing that the family of realizations of a sequence has
certain properties. Degreesequence questions can be categorized as either forcible
problems or potential problems. A forcible problem asks for conditions on a graphic
sequence that guarantee that every realization of the sequence has a certain property.
9
Potential problems, on the other hand, seek conditions which guarantee that there is
some realization of the sequence with the desired property. That is, given a graphic
sequence 7r and a graph property V, we say that 7r is forcibly Vgraphic if every
realization of 7r has property V, and potentially Vgraphic if some realization of 7r has
property V.
The extremal problems that are the focus of this dissertation are potential prob
lems, but we will give a few forcible problems here to illustrate the difference. Often,
forcible degree sequence results do not look like degree sequence results at all. For
example, consider the following theorem of Dirac:
Theorem 2.7 (Dirac [32]) If G is a simple graph with n vertices such that n > 3
and h(G) > n/2, then G is Hamiltonian.
We can rephrase this in terms of degree sequences as:
Theorem 2.8 (Theorem 2.7 rephrased) Let 7r = (d\,... ,dn) be a nonincreasing
graphic sequence with n> 3. If dn> n/2, then 1r is forcibly Hamiltonian.
Another result on forciblyHamiltonian sequences is due to Chvatal:
Theorem 2.9 (Chvatal [26]) Let G be a simple graph with vertex degrees d\ <
< dn, where n > 3. If i < n/2 implies that di > i or dn_i > n i, then G is
Hamiltonian.
While, for the property of Hamiltonicity, we only have sufficient conditions for
forcibly Pgraphic sequences, characterizations of forcibly Pgraphic sequences have
been found for many other graph properties. Chernyak, Chernyak, and Tyshke
vich characterized forcibly chordal, strongly chordal, interval, and trivially perfect
sequences in [22] and characterized the degree sequences of comparability graphs in
[118]. S.B. Rao has characterized forcibly linegraphic [109], forcibly totalgraphic,
forcibly selfcomplementary, and forcibly planar graphic sequences [110]. These last
10
three results are stated without proof in his survey of potential and forcible degree
sequence results [110], and he also developed a theory for proving characterizations
of forcibly Pgraphic sequences when P is a hereditary property [111]. A proof of the
characterization of forcibly planargraphic sequences, by Zverovich, appears in [154].
Zverovich has also characterized forcibly 3colorable [155] and forcibly 2matroidal
graphic sequences [156]. Choudum proved a characterization of forcibly outerplanar
sequences [23], and gave sufficient conditions for a sequence to be forcibly connected
[24].
The study of forcibly Pgraphic sequences has been changing in recent years.
Hammer and Simeone gave a characterization of split graphs in terms of their degree
sequences [57]. There is also a characterization of such graphs in terms of forbidden
subgraphs (c.f. [5]). Inspired by this, Barrus, Kumbhat, and Hartke [5] investigated
degreesequence forcing sets, that is, a set of graphs T such that if any realization
of a graphic sequence it is Pfree, then every realization of that sequence is Pfree.
Bauer et al. gave necessary and sufficient conditions for a sequence to be forcibly
ftough [3], and sufficient conditions for a sequence to be forcibly fcfactorable [2],
The conditions for itoughness are a direct extension of the conditions in Theorem
2.9, both of which are shown to be best monotone, a property which we will discuss
below. These kinds of results, which move beyond simple characterizations, show
that the study of forcible degree sequence problems is far from complete.
To understand what is meant by a best monotone theorem, we first need a defini
tion. Given two sequences it = (x\,... xn) and it' = (iji,..., yn), both in either non
increasing or nondecreasing order, we say that it majorizes it' if Xi > yi for each i, 1 <
i < n. Theorem 2.9 is best possible in the sense that if a sequence fails the condition
at some position i, then it is majorized by a sequence that has a nonHamiltonian
realization. In particular, it is majorized by it = (f, (n i l)n~2\ (n 1)*), which
is uniquely realized by the graph (Ah U Kn_2i) V Ah (see Figure 2.1). Every graphic
11
Figure 2.1: The graph (Ki U Kn_2i) V fQ
sequence that majorizes this one, however, does satisfy the conditions of Theorem
2.9, and hence is forcibly Hamiltonian. This is what is meant by a best monotone
theorem; in some sense it is the strongest possible degreesequence condition that
guarantees Hamiltonicity. As mentioned above, best monotone theorems for several
other graph properties have been found, and this is an active area of research.
Now we turn our attention to potential degreesequence problems. If we look
simply at graph properties, there are several results on potentially Pgraphic se
quences. There are characterizations of potentially Pgraphic sequences for each of
the following properties: fcedgeconnected [34], fcconnected [120], fcfactorable [72]
and connected fcfactorable (if the sequence is already potentially fcfactorable) [108],
and selfcomplementary [30]. In fact, selfcomplementary is the only property for
which there is a characterization of both forcibly and potentially Pgraphic sequences.
There are also potential versions of classical graph theory problems. We present
just a few examples before turning to the one that is the focus of this dissertation.
Our first example is a potential version of Hadwigers conjecture, a longstanding
open problem in graph theory.
Conjecture 2.10 (Hadwigers Conjecture [55]) If x(G) > k, then G contains a
Kkrninor.
12
A slight relaxation of Hadwigers Conjecture is Hajos Conjecture, which says that a
graph with chromatic number at least k must have a AVsubdivision. Hajos Con
jecture has been shown to be false for almost all graphs (see [95] p. 4556 for a nice
proof). However, Dvorak and Mohar [33] recently proved that the potential version
of Hajos Conjecture is true. That is, they showed that if y(7r) is the maximum
value of y(G) over all realizations G of 7r, then for any graphic sequence tt, there is a
realization of tt that contains a subdivided complete graph of order x(7r).
Another example of a potential version of a classical problem is a variant of the
Graph Minor Theorem of Robertson and Seymour [112], which states that in every
infinite set of graphs, there is a pair of graphs such that one is a minor of the other.
This is not true, however, if minor is replaced by induced subgraph. For the
potential version, S.B. Rao [111] conjectured that in any infinite set of graphs, there
is a pair of graphs, say G and H, such that ir(G) has a realization containing H as
an induced subgraph. This was recently proved by Chudnovsky and Seymour [25].
There are also known results for potential versions of the ErdosSos conjecture
on trees [137], the SauerSpencer graph packing problem [10], and a conjecture of
Bollobas and Scott about spanning bipartite graphs [58].
2.2.2 Potentially Hgraphic sequences
The property that we we are chiefly concerned with is that of having a subgraph
isomorphic to a graph H. If some realization of tt has this property, we say that
7r is potentially Hgraphic. The study of this property will ultimately lead us to a
potential version of the Turan problem, but first we give some useful results about
potentially Hgraphic sequences.
When a sequence is potentially Hgraphic, it is often nice to know something
about the realizations that contain H. The next two theorems, which we will use
many times, tell us that we can find realizations of potentially Hgraphic sequences
that have nice properties. The first shows that we can find our subgraph H in a
13
realization where H is on the vertices of highest degree.
Theorem 2.11 (Gould, Jacobson, Lehel [50]) If it = (d\,, dn) is potentially
Hgmphic, then there exists a realization G of tt such that the vertices of H have the
V(H)\ largest degrees ofn.
If H is a complete split graph, Kr\J Kt, then we can draw a stronger conclusion.
Theorem 2.12 (Yin [127]) A nonincreasing sequence ir = (d\,..., dn) is poten
tially Kr V Ktgraphic if and only if there is a realization of ir containing a copy of
Kr V Kt such that the vertices of the complete graph of order r have degrees di,... ,dr
and the vertices of the independent set of order t have degrees dr+i,..., dr+t.
Let H be a graph with degree sequence ir(H). If tt(H) = (si,... ,Sk), then we
say 7r = (d\,... ,dn) is degree sufficient for H if di > Si for each i, 1 < i < k. If
both sequences are nonincreasing, then it is clear that being degree sufficient for H is
a necessary condition for a sequence to be potentially Hgraphic. However, it is not
sufficient; for example, the sequence (2, 2, 2, 2, 2) is degree sufficient for id3, but the
unique realization of this sequence is C$, which does not contain K3.
Because they will be needed in Chapter 5, we present here some characterizations
of potentially Hgraphic sequences for small H. These results serve to illustrate the
point that degreesufficiency for H is rarely enough to guarantee that a sequence has
a realization containing H] there are usually exceptional families of sequences that
prevent the obvious necessary condition from being sufficient.
Theorem 2.13 (Luo [92]) Let 7r = (d\, d2,..., dn) be a graphic sequence with n >
3. Then ir is potentially K^graphic if and only ifd3 > 2 except for two cases: ir = (24)
and 7r = (25).
Theorem 2.14 (Luo [92]) Let 7r = (di,d2, , cln) be a graphic sequence, where the
di are in nonincreasing order. Then 7r is potentially C^graphic if and only if the
14
following conditions hold:
1. d4 > 2
2. d\ = n 1 implies d2 > 3
3. Ifn = 5,6, then ir ^ (2n).
Theorem 2.15 (Luo [92]) T graphic sequence ir = (d\, d2, , dn) is potentially
C^graphic if and only if n satisfies the following conditions:
1. 4 > 2 and tt ^ (2ra) for n = 6, 7.
R For i = 1,2, di = n i implies d^ > 3.
3. If 7r = (cii, g?2, 2fc, lra_fc_2); Â£/jen eh + eh < n + k 2.
The following characterization of potentially Fhgraphic sequences is due to Rao
[107], although his paper was never published. Kezdy and Lehel [69] gave a proof using
network flows, and Yin gave a constructive proof in [128]. For a graphic sequence 7r,
we let
Theorem 2.16 (Rao [107]) Let n > k and tt = (eh,..., dn) be a nonincreasing se
quence of nonnegative integers, ir is potentially Kkgraphic if and only if the following
conditions hold:
(i) dk> k 1,
(ii)
(Hi) For any s and t with 0 < s < k and 0 < t < n k 1,
S t
di + dk+i Y (s + t) (s + t 1)
i 1 i 1
k n
+ min{s + t, di k + 1 + s} + min{s + t, di}.
i=s+l i=k\t\1
15
This theorem is difficult to use in practice, as it involves evaluating a large set
of inequalities. The next theorem gives simple sufficient conditions for a sequence to
be potentially Ahgraphic. The simplicity of these conditions make them much more
practical to use, as we demonstrate several times throughout this dissertation.
Theorem 2.17 (Yin and Li [135]) Let 7r = (di,..., dn) be a nonincreasing graphic
sequence and let k be a positive integer.
(a) If dk > k 1 and di > 2(k 1) i for 1 < i < k 2, then ir is potentially
Kkgraphic.
(b) If dk > k 1 and d2k > k 2, then ir is potentially Kkgraphic.
As can be seen from the previous theorems, characterizations of potentially 17
graphic sequences are often highly technical. However, there are such characteri
zations for a large number of graphs and graph families. In particular, potentially
CVgraphic sequences were characterized for various k in [13, 92, 125, 126] and [145].
For k < Â£, graphic sequences with a realization containing cycles of each length be
tween k and I, inclusive, are known as potentially fcC)graphic. These have been
characterized for k = 3 and Â£ = 4, 5, 6, for k = 4 and Â£ = 5, and for k = 5 and Â£ = 6
[21, 132], Characterizations of Ahgraphic and Kr Hgraphic sequences for small r
have been studied by many different authors. Among these are characterizations for
Ah [93]; Ah e [41, 75]; Ah e, Ah, and Ah [148]; Ah H where H is one of C.4
[63], P4, P3 U Ah, Ah, Ah,3, 2Ah [62], Z4 = Ah P3 (the paw) [80]; P5 and Y4, where
I4 is the tree with degree sequence (3, 2,1,1,1) [64]; Ah C6 and Ah 2C3 = Ah,3
[61], Ah A3 [150], and Ah C5 [124], For larger complete graphs, we have char
acterizations for Kr+i e [146]; Ah+1 P3 and Kr+4 2Ah [121] (Ah 2Ah was
done in [91]). A characterization of potentially Ah,4 + egraphic sequences is given in
[80], and more generally for potentially Ah,* + egraphic sequences in [16]. Finally, we
have some characterizations for multipartite graphs, in particular for Ah,3 and Ah,4
16
[131]; Ki;i)S for s = 4,5 [151] and s = 6 [153]; and finally for K^iyS when s > 2 and
n > 3s + 1 [130]. Similar to the case for complete graphs, there is a characterization
for a sequence to have a realization containing a complete split graph using a set
of inequalities similar to Theorem 2.16 in [127], and simpler sufficient conditions are
given in [129].
There is, however, another way to guarantee that a graphic sequence has a re
alization containing a specific subgraph. For that, we turn at last to the potential
number.
2.3 The potential number
The degree sequence analogue of the Turan problem was introduced by Erdos,
Jacobson, and Lehel in 1991 [38].
Problem 2.18 Determine cr(H, n), the minimum even integer such that every nterm
graphic sequence ir with a(n) > a(H,n) is potentially Hgraphic.
We refer to a(H,n) as the potential number or potential function of H. When
they proposed the problem, Erdos, Jacobson, and Lehel conjectured that a(Kk,n) =
(k 2)('2n k + 1) + 2, based on the graphic sequence ((n l)k~2, (k 2)n~k+2) which
is uniquely realized by Kk2 ViWfc+2 They proved this conjecture for k = 3. Gould,
Jacobson, and Lehel [50] and Li and Song [86] proved the case k = 4 independently,
and then Li and Song proved it for k = 5 [87] two years later. Finally, Li, Song, and
Luo settled the conjecture for k > 6 and n > (k) + 3 [88].
Clearly, if a simple characterization of potentially //graphic sequences is known
for a given H (that is, one that gives simple conditions on the terms of the sequence
and perhaps excludes exceptional families), then determining the potential number for
H is not difficult. Thus, those graphs for which such characterizations of potentially
//graphic sequences are known also have known potential number. Often we do not
have any characterization of potentially Hgraphic sequences, or such characteriza
17
tions are very complex, such as Theorem 2.16. Thus, it is sometimes easier to simply
determine the potential number than it is to determine a simple characterization.
Aside from those listed above, the potential number is known for the following
graphs and graph families. Note that in most cases n is assumed to be sufficiently
large.
pKz and C4 [50]
The fan graph, F2m+i = Kx V P2m+i1 [19]
Pk and Ck, where Ck is a cycle with k chords incident to a vertex on the cycle
[139]
Single cycles: C5, Ce [73]; C\ [76]
Sets of cycles: 3Ci for / = 4,5,6 and n > l [84]; 3Cj for 3 < l < 8 and n > l,
and 3C9 for n > 12 [85]; 3Q for all l and n sufficiently large [89]; for l > 7
and 3 < k < l.
Complete multipartite graphs: KTyS for r = s = 3,4 [134], for r > s > 3
[141], for r = 2 [142]; K1A>2 [74]; idU;3 [77]; Kw for n > t + 4 [143, 14];
Kri,r2,...,ri,2,s for s > 3 and n > 2s2 + 8s + 3(r3 + ri) +4 [149]; Kri_>ri
s>r>r/>>ri>0, r>3 [136]; Kfs, which is the complete tpartite graph
with s vertices in each part (in fact it is shown that <7(AT*, n) = cr(Kj + KStS, n))
[15]
Friendship graphs, Fk, consisting of k triangles sharing a single vertex [44]
Generalized friendship graphs, Ft r^, which are formed from k copies of Kt that
overlap in a set of r vertices [133]
Disjoint union of cliques [43]
18
Small graphs: K5 C4 [78]; AT5 P4 and K5 P5, see [80]; AT5 e [144, 65]
Kr+1 e [135]
Kr+1 (&:P3 U tK2) for n > 4r + 10, where r, k, and f satisfy r + 1 > 3A: + 2t,
k +1 > 2, k > 1, and t > 0 [82]
Kr+i pK2 for r > 2, 1 < p < L^J and n > 3r + 3 [138]
. Kr+i I<3 [147]
Kr+i P3 and Kr+1 F, where F is A^free and contains some tree on 4 vertices
[SI]
Ar+1 A4, Ar+1 (Ah e), and Ah+1 Z4 (recall Z3 = K4 P3), and Kr+4 Z,
where Z is Crfree and contains Z4 [79]
Ah+i U, where U is C4free, Z4free, and contains K3 U P4 [83]
Graphs with independence number 2 [47] (note that this result subsumes many
of the previous results)
After many years of having only results for single graphs or small graph classes
similar to those above, Ferrara, LeSaulnier, Moffatt, and Wenger [46] took a large
step forward by determining the potential number asymptotically for all graphs H.
Their result uses some ideas from [47], where the potential number was determined
for graphs with independence number 2. Our work in Chapters 3 and 4 is based on
this result, so we will describe it next in detail.
Let H be a graph on k vertices with at least one nontrivial connected component.
For each i E {o'(P) + 1,..., k}, define
Vi(tf) = min{A(F) : F < H, \V(F) \ = 1} ,
19
where F < H denotes that F is an induced subgraph of H. Let n be sufficiently
large, and consider the sequence
nt(H, n) = ((n l)k~\ (k i + V,(tf) l)n~k+i).
This sequence is graphic provided that n k + i and Vi(H) 1 are not both odd.
If they are both odd, then reduce the last term of the sequence by 1. For each
i & {ol(H) + 1,,k}, the sequence 7Ti(H, n) is realized by a graph G that consists of
a clique on k i vertices joined to a (nearly) (Vi(H) l)regular graph on n k + i
vertices. This graph does not contain H, because any ^vertex subgraph of G must
use i vertices from the (Vi(H) l)regular graph; as such, it has an ?'vertex induced
subgraph with maximum degree Vi(H) 1. However, F[ has no such subgraph, so Ft
is not contained in G. Thus, iu(H, n) is not potentially Hgraphic, and consequently,
a(H, n) > maxj n))).
Let Oi{H) = 2(k i) + 'Vi(H) 1. This is the leading coefficient of
Since we are concerned with the asymptotic behavior of a(H,n), we are chiefly con
cerned with the maximum value of Ui{H). Note that the maximum value of Ui{H)
can be determined by finding the minimum value of 2i Vi(H). Thus, there may
be more than one value of i for which ai(Ft) achieves its maximum value. Hence, we
define i*(H) to be the smallest index i e {ot(H) + 1,..., k} that maximizes Ui{H).
We will often write just i* instead of i*(H), when the context is clear.
The main result of [46] states that 7Ti*(H,n) determines a(H,n) asymptotically
for all Ft, which can be viewed as an ErdosStoneSimonovitstype theorem for the
potential problem.
Theorem 2.19 (Ferrara, LeSaulnier, Moffatt and Wenger [46]) If Ft is a graph
20
and n is a positive integer, then
cr(H, n) =
We define V(H) to be the set of sequences 7Ti(H,n) that achieve the maximum
value of the leading coefficient. That is, V(H) = {7Ti(H,n) : Ui(H) = Note
that n does not play a role in the definition of V(H), because the coefficient of n
in a(7Ti(H,n)) is what puts the sequence into this set, not the actual sum of the
sequence.
Example 1. Consider H = Kk Z\, where Z\ = id4 P3. Clearly a(H) = 3, so we
need to compute Vj(i7) for i e {4,..., k}. The subgraph induced by the four vertices
involved in the copy of Z\ has maximum degree 2, so V4(i7) = 2. For i > 5, every
subgraph on i vertices has a dominating vertex, so Vi(H) = i 1 for each i > 5. The
minimum value of 2i Vi(H) is 6, and this is achieved by both i = 4 and i = 5. We
have a{H) = a5(H) = 2k 7, so cr(H, n) ~ (2k 7)n. In [79], Lai showed that
So we see that the asymptotic result matches the exact result.
It is interesting to note that there are two sequences in V(H)] they are 7r4(i7, n) =
((n l)fc4, (k 3)n~fc+4) and n5(H,n) = ((nl)k~5, (k 2)n~k+5) (where the last term
of each may be decreased by 1, depending on parity). However, only the sequence
((n l)fc4, (k 3)fc+4) is given as an extremal sequence in [79], since a(T4(i7, n)) =
ct(tt5(H, n)) 2. Thus, the method of determining extremal sequences by computing
Vj(i7) may actually yield more information than previous methods.
(k 2) (2n k + 1) 3(n k + 1) + 1 if n = k mod 2
(k 2)(2n k + 1) 3(n k + 1) + 2 if n ^ k mod 2.
21
Chapters 3 and 4 explore V(H) and graphic sequences that are close to being in
V(H) in more detail. In Chapter 3, we will explore the structure of sequences that are
not potentially //graphic, in terms of majorization of degree sequences. In Chapter
4, we prove a stability result for the potential number, and discuss what this result
means for our study of the potential function.
22
3. The Shape of Graphic Sequences that are not Potentially //Graphic
3.1 Introduction
In this chapter, we study the structure of degree sequences that are not poten
tially //graphic. As the realizations of a graphic sequence may have a great deal of
structural variety, it is perhaps more appropriate to say that we examine the shape
of these sequences. One way to think about the shape of a graphic sequence is to
draw a picture of it; that is, given a sequence tt = (d\,..., dn), we draw rectangles
of height di,..., dn and line them up, similar to a Ferrers diagram, as in Figure 3.1.
This gives us a graphical representation of the sequence that allows us to discern,
at a glance, the relative differences in the values and multiplicities of terms in the
sequence.
Given (not necessarily graphic) sequences S\ = (aq,..., xn) and S2 = (jq,..., yn),
we say S i majorizes S2 and write S i y S2 if ay > ty for all/, 1 < i < n. In terms of
pictures, if Si majorizes S2, then the drawing of S2 fits beneath the drawing of S\.
Our study of the shape of graphic sequences is based on this idea of majorization,
and what must be done to a sequence to ensure it will be majorized by another
sequence. Since sequences that are not potentially //graphic are the degreesequence
analogue of Hfree graphs, our inspiration comes from several results on Hfree graphs
from the extremal literature.
Figure 3.1: The shape of a graphic sequence
23
In [35], Erdos showed the following.
Theorem 3.1 If G is a Kr+\free graph of order n, then there exists an nvertex
rpartite graph F such that ir(F) F tt(G).
We give the proof here for completeness.
Proof. We proceed by induction on r. When r = 1, the result is trivial, so
suppose that r > 1 and the result is true for all smaller values of r. Let G be
a Kr+ ifree graph with V(G) = {v\,...,n}, such that V\ has degree A(G) and
N(vi) = {r2, , E\(G)+i} Let H be the subgraph of G induced by N(vi). Clearly,
Ft is Krfree, so there is an (r l)partite graph F' with vertex set N(vi) such that
dF>(vi) > dn{vi) for all i with 2 < i < A(G) + 1. Create a new graph F on V(G)
by joining every vertex in {t>i, ua(g)+2, Wn} to the vertices of F'. This graph is
rpartite, and since ir(F') F tt(Ft) and each vertex in V(G) \ N(vi) has degree at
least A(G), n(F) F n(G).
Since the sum of the degree sequence of a graph is twice the number of edges in
the graph, this result says that for any Ab+ifree graph G, there is an rpartite graph
with at least as many edges as G. Observing that the Turan graph Tn>r is the unique
edgemaximal rpartite graph of order n, Turans Theorem (Theorem 2.3) follows as
a corollary of Theorem 3.1.
Having established a result on the degree sequences of fdr+1free graphs, we next
discuss a similar result for Ftfree graphs for general Ft. To do this, we need to use
another type of majorization, introduced by Pikhurko and Taraz in [104], Given
positive integers m and k, define Dkrn(Si) to be the sequence
(xk to, ... ,xk m, xk+i m,... ,xn m).
k times
We say that S2 (k,m)majorizes S\ if S2 F Dk,m{S 1).
24
Theorem 3.2 (Pikhurko and Taraz [104]) Let H be a graph with chromatic num
ber x(H) = r + 1 > 2. For any e > 0 and n > no(e, H), the degree sequence of an
Hfree graph G of order n is (en,en)majorized by the degree sequence of some r
partite graph of order n.
It was noted in [104] that both the operation of leveling off the first k terms
of 7r(G) and the operation of reducing all of the terms in 7t(G) by m are necessary.
For example, consider F = Kt,t The graph Kt1 V Kn_t+i is Ffree and has degree
sequence ((n l)t_1, it l)n~t+1). However, y;(F) = 2, so we have r = 1, and in this
case an rpartite graph must be the empty graph with degree sequence (0ra). Clearly,
we need both operations to reduce ((n l)t_1, (t F)n~t+l) to a sequence of zeroes.
The operations used to create Fm en(7r(G)) in Theorem 3.2 reduce
most e2n2, so Theorem 3.2 says that reducing the degree sum of G by o(n2) results
in a graph whose degree sequence is majorized by the degree sequence of an rpartite
graph. This implies the ErdosStoneSimonovits Theorem (Theorem 2.4), just as
Theorem 3.1 implies Turans Theorem.
That the degree sequences of rpartite graphs appear as the bounding class in
Theorems 3.1 and 3.2 is unsurprising given the central role played by the Turan
graph in the extremal literature. As shown in Chapter 2, the sequences that form the
bounding class for the potential problem are
^(H, n) = ((n l)k~\ (ki + V,(H) l)n~k+*),
where Fd is a graph of order k. It is our goal to examine the structure of degree
sequences that are not potentially Hgraphic in a manner similar to Theorems 3.1
and 3.2, by comparing them to ni(H, n).
25
3.2 Majorization of sequences that are not potentially Hgraphic
Throughout the remainder of this chapter, unless otherwise noted we will assume
that all sequences have minimum term at least 1. Given two nterm graphic sequences
7Ti = (di,, dn) and 7r2, and nonnegative integers oq, a2, and b with oq < a2, we say
that 7T1 is ([i, a2], b)close to 7r2 if there is a (not necessarily graphic) sequence 7r[
with 7r2 >: 7r( such that 7r^ can be obtained from 7Ti via the following two steps:
1. Create the sequence
Si (dl dd^ 1 da2 , ; d(j2 + l ; * ; dffj .
&2 Q1 + 1 times
2. Create 71j from Ai by subtracting a tota/ of at most b from the terms of Si.
We will refer to step (1) as leveling off terms oq to a2 of tv\ and the procedure in
step (2) as editing the sequence Si.
In contrast to the idea of (k, m)majorization, ([op, a2], 6)closeness leaves the
first oq 1 terms unchanged, and after the leveling off step allows for variable editing,
provided that the total amount of editing in step (2) is at most b.
As an example, consider the sequences 7Ti = (195,145,105, 55) and 7r2 = (192, 818).
For the leveling off step, we can reduce terms 3 through 11 of 7Ti to the value of
the IIth term to get the sequence Si = (192,1013,55). The sequence Si is still not
majorized by 7r2, so we edit by subtracting 2 from each term that is equal to 10; this
is a total editing of 26, and results in tv[ = (192, 813, 55), which is majorized by tt2
Thus, 7T1 is ([3,11], 26)close to 712
We show that if a sequence is not potentially fCgraphic, then it is close (in the
above sense) to being majorized by one of the sequences 7?i(H,n). Our first result
concerns sequences which fail to be potentially fCgraphic simply by failing to be
degree sufficient for H.
26
Theorem 3.3 Let H be a graph with degree sequence tt(H) = (hi,..., hk), and let
7r = (di,... ,cln) be a graphic sequence that is not degree sufficient for H. Further,
let j be the largest integer for which dkj+i < hkj+i If j > a(H) + 1, then ir is
majorized by 7Tj(H,n). If j < a(H) + 1, then ir is ([k a(H),k j + l],0)close to
Ka(H)+i(H, n).
Proof. First note that for each i > a(H) + 1, hki+i < k i + Vi(H). Otherwise,
every zvertex induced subgraph of H has maximum degree greater than Vi(H),
contradicting the definition of Vi(FI).
Similarly, if i < a(H) + 1, then hki+i
a(H) vertices have degree at most k a(H), contradicting the fact that there are at
least a(H) vertices in H with degree at most k a(H).
Recall that j is the largest integer for which dkj+i < hkj+i First suppose that
j > a(H) + 1. In this case, we show that tt is majorized by 7Tj(H,n). Clearly, the
first k j terms of 7r are majorized by the first k j terms of 7fj(H, n). As dkj+i <
hk~j+1 < k j + Vj(H), the remaining terms of tt are at most k j + Vj(H) 1.
Thus, 7r is majorized by 7Vj(H, n).
Now suppose that j < a(H) + 1. Here we show that tt is ([k a(H), k j + 1], 0)
close to 7ra(H)+i(H,n). We know that dkj+i < hkj+i < k a(H). Since
k a(H) + Va(ij)+1 2 > k a(H) 1, reducing terms dk~a(H) through dkj
of 7r to dkj+i results in a sequence that is majorized by Tra(H)+i(H, n).
The number of terms leveled off in Theorem 3.3 is best possible in light of the
following example. Let H = Kkr\tKr, where r is at least 2, and for 1 < j < a(H) +1
let
27
where n is sufficiently large. If the sum of Hj is even, then 7tj is graphic; if the sum
is odd, then reducing the last term by 1 yields a graphic sequence. Clearly 7r, is not
degree sufficient for H.
Note that the (k j + l)st term of 7Tj is the first place that degree sufficiency for
H fails. Since k r 1 < k 2, the last n k + j terms of 7iq are termwise dominated
by the last n k+j terms of 7ra(H)+i(H, n). However, 7iq has k j terms equal to
of which only the first k a(H) 1 are dominated by ira(H)+i(H, n). Therefore,
we need to reduce terms dka(H), , duj of 7iq to dkj+i = k r 1, and each of
these reductions is on the order of n. This yields a sequence that is majorized by
Tra(H)+i(H, ti), but reducing any smaller number of terms would not suffice.
The case where 7r is degree sufficient for H seems to be much more technical,
and requires both the leveling off and editing operations outlined above. Recall that
i* = i*(H) is the smallest i in {a(H) + 1,..., k} that minimizes 2i Vj(df).
Theorem 3.4 Let H be a graph of order k vnth at least one nontrivial component
and let 1r be an nterm graphic sequence that is degree sufficient for H. If 7r is not
potentially Hgraphic, then tt is ([k i* + 1, k\, (6a + 3)k2 + a3k)close to7Ti*(H,n).
To show the sharpness of Theorem 3.4, consider H = I+k Let
7Tfc = (n 1, (2k 5)2fc3, ffi2fc+2) .
The ErdosGallai criteria show that 7Tfc is graphic; clearly irk is degree sufficient for
Kk. Observe that ttu is potentially /C^graphic if and only if the sequence 7r'k =
((2k 6)2fc_3), obtained by performing the HavelHakimi algorithm, is potentially
fdfc_igraphic. However, the complement of any realization of n'k is a 2regular graph,
so the maximum size of a clique in any realization of n'k is at most k 2. Hence 7
is not potentially fdfcgraphic.
28
Since 2i Vi(Kk) = z + 1 for each i e {2,..., k}, we have i* = 2. Note that
K2{Kk, n) = ((n l)k~2, (k 2)n~k+3). As k i* + 1 = k 1, leveling off terms k 1
and k of tt'k does not change the sequence. However, each entry from k 1 through
2k 2 is larger than k 2, so we need to reduce each of these entries by k 3, for a
total of k(k 3) editing. Thus we perform a total of 0(ak2) editing.
This shows that Theorem 3.4 is in some sense best possible up to the coefficient
of ak2. However, since the amount of editing given in the theorem is (6a + 3)k2 + a3k,
the leading term of this expression is not k2 when a is much larger than k1^2. Thus,
in this case we do not yet have any information about the sharpness of this result.
3.3 Lemmas
In addition to some of the results on degree sequences presented in Chapters 1
and 2, we will need the following results for the proof of Theorem 3.4. First, we have
a lemma that is central to the proof of Theorem 3.4, and is likely also of independent
interest. As the proof of this result is quite technical, we postpone it until Section
3.5.
Lemma 3.5 Let r and k be positive integers with r < k, and let ir = (d\,... ,dn) be
a nonincreasing graphic sequence. Suppose that dk_r dk > r(k + 2). If there are at
least r(k + r + 1) terms among dk+1,..., dn with values in {k r,... ,k 1}, then
7r has a realization containing the graph Kk_rr with vertices of degree di,... ,dkr
forming the partite set of order k r.
We also use the next lemma, which gives a bound on the length of a sequence
with fixed maximum term that is not potentially Hgraphic.
Lemma 3.6 Let H be a graph with tt(H) = (hi,..., hk) and let tt = (d\,..., dn) be
a graphic sequence with d\ < M such that there are terms dit,... ,dik of tt satisfying
dij > hj for 1 < j < k. If tt has at least 2M2 + k positive terms, then there is a
realization G of tt with a copy of H that lies on vertices of degree d^,... ,dik.
29
Proof. We may assume that dij = dj for all j and also that n > 2M2 + k and dn> 1.
First note that if M = 1, then H must be a set of disjoint edges and isolated vertices,
and 7r is potentially fTgraphic as long as n > k. We therefore assume M > 2.
Let V(H) = {u\,... ,Uk}, with the vertices in nonincreasing order by degree. In
a realization G of tt, let S = {tq,... ,vk} be the vertices with the k highest degrees
(in order) and let Hs be the graph with vertex set S and V{Vj Â£ E(HS) if and only if
Ui'Uj Â£ E(H). If all of the edges of Hs are in G, then Hs is a subgraph of G that is
isomorphic to H.
Assume now that G is a realization of tt that maximizes \E(HS) C\E(G)\, but this
quantity is less than \E(HS)\. Thus, there exist vi}Vj Â£ V(G) such that ^ E{G)
but ViVj Â£ E(Hs). Since tt is degree sufficient for H, it follows that Vi and vj must
each have a neighbor, say cq and cij, respectively, such that Via^Vjaj ^ E(Hs) but
Viai,Vjdj Â£ E(G). Note that possibly di = dj.
Since the maximum degree in G is M, there are at most M2 + 1 vertices at dis
tance at most 2 from cq, and at most M2 + 1 vertices at distance at most 2 from
dj. Since di and dj have distinct neighbors in S, there are at most k 2 vertices
in S that are distance at least 3 from both di and dj. Therefore, there is a vertex
w in V(G)\S that is distance at least 3 from both a* and dj. Let i be a neighbor
of w; consequently x is not adjacent to or dj, and xw E(HS). Exchanging the
edges Vidi,Vjdj, and wx for the nonedges ViVj,diW, and djX yields a realization G' of
7T such that \E(Hs)r\E(G')\ > \E(Hs)C\E(G)\, contradicting the maximality of G.
3.4 Proof of Theorem 3.4
For the proof of Theorem 3.4, we actually prove a more technical result that
follows below. First we define some terminology that is used in the proof.
Given a graphic sequence 7r that is degree sufficient for Kr V Kk_r, we create
a sequence 7tw called the want sequence of tt for Kr V Kk~r Begin by Ending a
30
realization G of tt on the vertices {ig,..., vn} with d(vi) = di that maximizes the sum
of (a) the number of edges amongst Vi,... ,vr and (b) the number of edges joining
{ui,... ,vr} and {wr+i, ,vk}. Let Gr = G[vr+i,... ,vn], and let ti = (uy+i,... ,wn)
be the degree sequence of Gr, indexed so that ug = dGr(vi).
For each rg with i < r, we want rg to be adjacent to each of the vertices in the
set Si = {tg,..., vk} \ {vi}. Since tt is degree sufficient for Kr V Kk_r, we see that for
each nonneighbor of rg in Si, there is a neighbor of rg in {vk+i,... ,vn}, and each of
these neighbors is distinct. Let Wi be a subset of NGr(vi) fl {vk+i, ,vn} that has
size k 1 dsi(vi). Let W be the multiset Uf=1VFj.
To create the want sequence from 7r, we make the following modifications. Each
time the vertex vy appears in W, add 1 to entry wy of 7t For each j with r + 1 <
j < k, subtract r d{vl^,tVry(Vj) from Wj. The sequence that results from these
modifications is the want sequence, ttw. Note that the largest value that can be
subtracted from any entry is r, and the only entries that might be reduced are those
with index at most k. Since 7r is degree sufficient for Kr V Kkr, no entry of 7tw is
negative and at most r terms of 7tw are 0. The largest value that will be added to
any entry of tt is at most r, and only terms with index at least k + 1 are increased,
so the largest entry of 7rw is at most the maximum of wr+1 and wk+1 + r.
If 7tw is graphic, we can find a realization of 7r that contains Kr V Kk_r. To do
this, take the union of the complete split graph Kr V Kk_r on the vertices {ig,..., vk}
(with the clique on the vertex set {v\..., vr}) and a realization of 7tw on the vertices
{vr+\,..., vn}. Then join each vertex belonging to the clique of the complete split
graph (that is, 7g such that i < r) to the vertices in N(vi) fl {ufc+i,... ,vn}\Wi. This
graph has degree sequence 7r, so we have a realization of 7r that contains the desired
complete split graph.
We will prove the following, more specific result than that stated in Theorem 3.4.
31
Theorem 3.7 Let H be a fixed graph of order k with at least one nontrivial com
ponent. If 7r is a graphic sequence of length n that is degree sufficient for H but not
potentially Hgraphic, then ir is
{[k i* + 1, k\, k2 + ki* + 2 + (6k2 + (i* V*. {H))2k + V;* {H)){i* V** (H) 2))close
to 7n).
Since
2i*  < 2(a(H) + 1) VQ(ff)+i(tf) < 2a(H) + 1,
and i* > a(H) + 1, we see that i* Vi*(H) < a(H). Thus,
k2 + ki* + 2+(6fc2 + (** Vt*(H))2k + Vi*(H))(i? Vi*(H) 2)
< k2 + ki* + 2 + (6fc2 + a2k + Vi*{H))a
< 6ak2 + a3k + 3A:2 + 2.
Hence Theorem 3.4 (which claims that (6o + 3)k2 + a3k editing suffices) follows di
rectly from Theorem 3.7.
Proof. Let it = (di,, dn) and label the vertices of H with {v\,..., Vk} such that
d(vi) > d(vj) when i < j. To simplify notation, we will write a for a(H) and let
t = i* Vi*(H). Let f(H) = 6k2 + {t)2k + VifiH). With this notation, we prove
that 7r is {{k i* + 1, k\, k2 + ki* + 2 + f(H)(Â£* 2))close to 7h*(H, n).
If either dk > 2k 3 or d^k > k 2, then by Theorem 2.17, tt is potentially fdfc
graphic. As this would imply that 7r is potentially Hgraphic, we assume henceforth
that dk < 2k 4 and, if n > 2k, that d2fc < k 3.
32
We will break the proof into several cases. In Cases 13, we will show that,
after reducing the value of terms dki*+i, , dki to dk, we only require at most
k2 + ki* + 2 + f(H)(Â£* 2) editing. In Case 4, we show that 7r is in fact potentially
//graphic, so no editing is required.
Case 1: n < 2k.
In this case, after leveling off terms dki*+i through dk1, we need to reduce at
most k + i* terms of the sequence. Each of those terms is reduced by at most k+Â£* 3,
so the total amount of editing is at most k2 + ki* + (Â£* 3)(k + ?'*), which is less than
k2 + ki* + 2 + f(H)(Â£* 2).
Case 2: 2k
Reducing the terms dki*+1,..., dk1 of tt to dk creates a sequence where at most
the first k i* terms may be greater than 2k 4. Now in this sequence, each of the
terms from position k i* + 1 to position 2k 1 is at most 2k 4, so they must be
reduced by at most k + Â£* 3. Each term from position 2k to the end of the sequence
is at most k 3, so must be reduced by at most Â£* 2. Thus the amount of editing
required for this sequence is at most
(k+i* l)(k+Â£* 3) + (n2k + l)(Â£* 2) < (Â£*2)(f(H)k+i*) + (kl)(k+i*l).
Case 3: n > f(H) and df(H) < k Â£* 1.
Here, we will edit only the terms up to df(H)i, since all subsequent terms are at
most k Â£* 1. We again level off terms dki*+i through dk1, making them equal
to dk Each of the terms from position k i* + 1 to position 2k 1 must be reduced
by at most k + Â£* 3, and each term from position 2k to position f(H) 1 must be
33
reduced by at most Â£* 2. The total amount of editing required is at most
(k+i* l)(k+t 3) + (f(H)2k)(Â£* 2) < {t2)(f(H)k+i*) + (kl)(k+i*l).
Case 4: n > f(H) and df(u) > k Â£* 1.
Now we show that tt is potentially //graphic. If Â£* = 1, then df(H) > k 1, and
since /(//) > 2k, this means d'2k > k 1. Thus by part (b) of Theorem 2.17, 7r is
potentially Kkgraphic. We assume henceforth that Â£* >2.
Let t be such that dt> k 1 but dt+\ < k 1. We then have two cases.
Case 4a: t < k Â£*.
In this case, we wish to show that tt has a realization that contains the complete
split graph Kt V Kk_t. First note that tt is degree sufficient for such a graph because
dt > k 1 and dk > k Â£* > t. Let 7r be the want sequence of tt for Kt V Kk_t.
Since every entry of tt is at most k 2 (because dt+\ < k 1), and t is also at most
k 1, the largest entry of Trf is less than 2k.
Zverovich and Zverovich [157] showed that a sequence with maximum term r
and minimum term s is graphic as long as the length of the sequence is at least
. Since nf has length n t and up to t terms may be 0, 7r is graphic if
n 2t > (k + l)2. Since t < k Â£*, this is true if n > k2 + + 1 27*. The
observation that n > f(H) > 6k2 shows that this is true, and 7r is graphic.
Now observe that if Trf is graphic, then tt is potentially (Kt V iW^graphic. If
t > k a(H), then this complete split graph contains a copy of H, and we are done.
So we assume that t < k a(H). Let Fj denote an /vertex induced subgraph of H
that achieves A (id) = Vi(H). If tt has a realization containing Kt\/ Kk_t with a copy
of Fk_t on the vertices in the independent set, then tt is potentially //graphic.
34
By Theorem 2.12, we know that a realization G of 7r can be found that contains
Kt V Kkt such that the t vertices of degree k 1 are on the t highestdegree ver
tices of G, and the k t vertices of degree t are on the next k t highestdegree
vertices of G. Delete the t vertices of highest degree in G, and let tt' = (d[,..., d!n_t)
be the degree sequence of the resulting subgraph of G. It follows that tt' satis
fies d[ < k 2 and d'^H^_t > k Â£* t > 1. Since A(Fk_t) = Vkt(H) and
d'k_t > d'^Hj_t > k Â£* > Vkt(H), if follows that tt' is degree sufficient for Fk_t.
Applying Lemma 3.6 with H = Fk_t and M = k 2, we see that tt' is potentially
Ffc_tgraphic as long as at least 2[k 2)2 + [k t) terms of F are positive. Since
/(H) t> 6k2, this is true. Thus, there is a realization of F that contains a copy of
Fk_t on the vertices of highest degree; overlapping this with the vertices vt+\,..., vk
of G, we get a realization of tt containing KtV Fk_t, which implies that tt is potentially
//graphic.
Case 4b: t > k Â£*.
First, suppose dkt*dk > Â£*{k+2). This implies that dke* > Â£*{k+2)+dk. Since
t > 2, it follows that dk_i* > 3k. We claim that this implies that tt is potentially
//graphic.
Since f{H) > Â£*(k+Â£* + 1) and df^H) A kÂ£*, Lemma 3.5 yields a realization G of
tt containing the complete bipartite graph Kk_i*^*, where the vertices {iq,..., vk_i*}
form the partite set of order k Â£*. Let S be the set of vertices in the copy of
Kk_t*ti*, let S' = {iq,... ,vke*}, and let R = V(G) \ S. If the vertices of S' induce a
complete graph, then G contains Kke* V /Q*; consequently G contains a copy of H
since k Â£* > k a. Suppose there are vertices rq and Vj in S' such that ViVj ^ E(G).
Since dsivi) < k 2 and d(vi) > 3k 2, we know that Vi has at least 2k neighbors in
R. Similarly, Vj has at least 2k neighbors in R. If each neighbor of iq in R is adjacent
to each neighbor of Vj in R, then each of these vertices in R has degree at least 2k 1.
35
Since Vi has at least 2k neighbors in R, it follows that G has at least 2k vertices with
degree at least 2k 1, contradicting the assumption that dk < 2k 4. Thus there are
vertices x and y in R such that ViX, Vjy E E(G), and xy ^ E(G). Hence we can replace
the edges v^x and Vjy with the nonedges xy and ViVj to obtain a new realization of tt.
Iteratively performing this process for each nonadjacent pair of vertices in S' yields a
realization of tr in which S' induces a complete graph. Consequently tt is potentially
//graphic.
Finally, we must consider the case where dke* < dk + Â£*(k + 2) < 2k + Â£*{k + 2).
Observe that tt is degree sufficient for Kk_f* VKg* since t > k G and /(//) > k. Let
tt2 = (gi,. , gn') be the want sequence of tt for Kk_g* V Kg* where n' = n (k Â£*).
Since constructing the want sequence increases each term by at most k Â£*, the facts
that dkg*+1 < 2k + Â£*{k + 2), dk < 2k 3, and < k 1 imply that 7r has the
following properties:
gt < 3k + (k + l)t for 1 < < t,
gi < 3k Â£* for G + 1 < i < Â£* + k, and
gi < 2k  G for G + k + 1 < i < n.
Claim 3.1 The sequence tis graphic.
Proof of Claim 3.1. Note that m! = n (k Â£*) > 6k2 + (Â£*)2k.
Tripathi and Vijay [116] showed that the ErdosGallai criteria (Theorem 1.3) need
only be checked for certain values of p: it suffices to check all p < s, where s is the
largest integer for which ds > s 1, or to check only those values of p for which dp is
strictly greater than dp+\. We will use the ErdosGallai criteria and this observation
to show that irf is graphic.
Since g^ <2 k Â£* for large enough i, we only need to check the inequalities for
indices up to 2k. We can write the right side of the ErdosGallai inequality (Inequality
36
(1.1)) as
r n'
p(pi)+ Yp+ Y 9i>
i=p\1 i=r\1
where r > p+1 is the largest index such that gr > p but gr+i < p. This then simplifies
to
nf
p(r 1) + 'Y/ di > p(r 1) + n' r
i=r\1
= r(p 1) p + n'
> (j> + l)(p 1) p + n'
= p2 p 1 + n'
So we need to show that n' + p2 p 1 > gi for each p < 2k.
First suppose p < l*. Then
v
Y9i
i= 1
Since n1 > 6k2 + (t*)2k, the desired inequality holds.
If t + 1 < p < t + k, then
v
Ygi M*k + (t)2(k + 1) + (p t)(3k t) < 2tk + (t)2(k + l) + 3k2.
i=1
For p in this range,
n' +p2 p 1 > 6k2 + (r)2fc + (r + l)2 1 =6k2 + )2(fc + 1) + 2t,
so the inequality holds.
37
Finally, if Â£* + k + 1 < p < 2k, then
p
^gt< 2tk + (t)2(k + l) + 3k2 + {ptk)(2kt) < 5k2 + (t)2k + 2(t)2tk.
i= 1
Now, n! + p2 p 1 > 6k2 + (t)2k + 4fc2, so the ErdosGallai inequality is satisfied.
Thus Claim 1 is proved.
Now that we have shown that irf is graphic, we can use a realization of 7rf to
create a realization of tt containing a copy of Kki* V KÂ£*. Since H C Kke* V Ke*,
this implies that tt is potentially //graphic.
3.5 Proof of Lemma 3.5
Idea of the proof: The proof of Lemma 3.5 is based on a careful analysis of
repeated applications of the KleitmanWang algorithm (Theorem 1.2). Observe that
when laying off a term ck from a graphic sequence, the di terms of highest degree,
aside from di, are each reduced by 1. If there are many terms of the same value
that will be reduced, the order in which these reductions occur does not matter. In
particular, provided we reduce the correct number of terms, we may reduce any of
the terms equal to d^ and will get the same residual sequence. This fact is the key
to constructing a realization of tt that contains Kkr,r, as referenced in the statement
of Lemma 3.5.
To see this more clearly, consider a sequence tt = (d\,..., dig), where d5 = de =
= dV2, and dVi = 7. If we lay off dig in the standard way, then we reduce terms di
through dj by one, and then reorder the sequence to make it nonincreasing. This gives
us the residual sequence tt = (di, d2, dg, d^, dg, dg, dio, dn, di2, d
(see Figures 3.2a and 3.2b). However, since dg = = du, we could reduce any three
of these terms and the resulting sequence would be the same except in the order of the
38
terms. So, we could reduce terms d\ through d, d$, dg, and dw, as in Figure 3.2c, and
get the residual sequence tt' = (di,..., cU, d&, dj, dg, du, du, de, dg, dw, du,..., dw),
but the values of the terms are the same as in the first case. Alternatively, we could
reduce the first four terms of 7r and the last three terms of the constant subsequence
(that is, dw, dn, and du), and then we would not have to reorder the sequence at
all, as in Figure 3.2e. Figures 3.2b, 3.2d, and 3.2e show that while the order of the
terms in the sequence changes depending on which terms we reduce, the values of
those terms are the same, so as far as the sequences are concerned, it does not matter
which terms we reduce.
Flowever, if each term di is associated with a particular vertex Vi in a realization
of 7r, it does matter which terms are reduced as we perform the KleitmanWang
algorithm. Thus, we will perform the algorithm but carefully keep track of which
terms we reduce and the order of the terms in the residual sequence. Sometimes we
will reduce terms in the standard way prescribed by Theorem 1.2, but sometimes we
will wish to change the order of the sequence as little as possible, and will reduce
terms starting with the end of a constant subsequence as in Figure 3.2e. This will
allow us to build a realization of 7r with the desired properties.
The KleitmanWang algorithm provides a means by which the desired realization
can be constructed on the vertex set V = {tq,... ,vn} so that the vertices Vj have
degree dj for j = 1,..., n. The vertex Vj is associated with the jth term in 7r. When dj
is laid off, the resulting sequence, tt' is a graphic sequence. We can use it to construct
a graph on V \ {vj} with (the reordered) 7v' as its degree sequence. The vertex Vj is
then added, adjacent to the first dj members of {vi,... ,Vji,Vj+i,... ,vn}. In this
way, when we lay off a term dj of tt, we will say that the vertices associated with
the terms that are reduced are assigned to the neighborhood of Vj. Repeating this
process, we will create a realization of 7r containing Kkr,r
39
(a) Lay off di3 by reducing terms d\
through dj
(c) Lay off di3 by reducing terms d\
through c?4, de, dg, and dio
(e) Lay off d\s by reducing terms d\
through c?4, dio, dn, and du
(b) The sequence that results from the
reduction in 3.2a
(d) The sequence that results from the
reduction in 3.2c
(f) The sequence that results from the re
duction in 3.2e
Figure 3.2: Different ways to reduce terms in the KleitmanWang algorithm
40
The problem with this procedure is that applying it more than once requires that
each of the degree sequences must be reordered, which makes keeping track of the
vertices that are assigned to a particular neighborhood difficult.
For clarity, we will often abuse terminology and say we lay off vertex Vj to mean
we lay off the term of 7r whose value corresponds to the degree of Vj. This makes
sense when we think about laying off a term dj of 7r as assigning a set of vertices
to the neighborhood of Vj. We will lay off at most r(k + r + 1) vertices with the
aim of obtaining just r of them whose neighborhood contains {tq,... ,tq_r}. Our
parameters are chosen just for this purpose. The entries in {dk+i, , dn} that have
value in {k r,... ,k 1} will be the candidates for entries to lay off. Because
dkr dk > r(k + 2), we can guarantee that, for each of the degree sequences that
result from the laying off procedure, the entries that correspond to tq,... ,Vk~r will
always stay within the first k 1 entries.
Terms and definitions: Now we proceed to prove that the procedure outlined
above does indeed produce the graph we want. We will lay off entries of 7r corre
sponding to vertices vai,va2,..., vap,..., where vap will be determined at step p. Let
Vo = V and for p = 1,2,..., let Vp = Vp_i va,p. The neighborhood we assign to
vap, which we will call Np, is a subset of Vp. We will call the process of laying off
vai, , Var(h+2) the Layingoff Algorithm.
For p = 0,1,2,..., we define dp(vi) to be the remaining degree of Vi af
ter vai,...,va are laid off. That is, for every vertex tq, d0(vi) = di and for
p = 1,2,..., r(k + 2), we have dp{Vi) := di {Aj: 1 < j < p and ry G Nj}\. Iter
To determine which vertex vap to lay off, for p = 1,2,..., we define Sp_ 1 C Tp_i to be
atively,
the set of all vertices w G for which clp_i(w) E {k r,... ,k 1}. Then choose
41
vap to be a vertex in Sp1 for which dp\{vap) is minimum. Let ip\ = dpi(vap); this
is the number of vertices that will be assigned to Np. Note that the neighborhood of
vap may not consist solely of the vertices in Np. In particular, if dp\{vap) < do(vap),
then vav is in Np< for some p' < p. Thus, the neighborhood of vap in our hnal graph
contains vap, although vap/ is not in Np.
The natural ordering on V = V0 is simply (iq,... ,vn). This corresponds to the
nonincreasing order of tt. We say that Vi naturally precedes Vj if i < j, and will write
Vi oc Vj. We will define ttp to be the sequence given by each dp(vi), for all q G Vp,
that is nonincreasing and, when equality holds, to obey the natural ordering. That
is, dp{Vi) precedes dp(vj) in ttp if either (a) dp(vi) > dp(vj), or (b) dp(vi) = dp(vj)
and i < j. This is simply the degree sequence obtained from 7r by p iterations of the
KleitmanWang algorithm; thus, ttp is graphic.
Observe that in defining ttp, we have prescribed the order of the terms based on
the vertices with which they are associated. This is because we need to keep track of
not only the remaining degree of a vertex but also the position of that vertex in ttp.
To make this precise, let rp be a function from {1,..., \VP\} > Vp in which rp(j) is
the vertex in the jth position in the order defined by tvp, and let Tp be the sequence
tp(1), , Tp(np). Thus, Tp is simply the sequence of vertices of Vp, ordered according
to the position of their remaining degree in ttp. A subsequence rp(bi),..., rp(bm) of Tp
is consistent if rp(6j) oc rp(bj) for all bi < bj. In essence, this means that all vertices in
the subsequence are in order by index, from lowest to highest. We say that Tp itself
is consistent if rp(1),..., rp(n p) is consistent.
In the KleitmanWang algorithm, when the term di is laid off it is first removed
from the sequence; then the first di terms of the resulting sequence are each reduced
by one. To incorporate this into the Layingoff Algorithm, we define 7rp_i to be 7rp_i
with the term associated with vap removed. Then, rp_i and Tp_i are the corresponding
order function and sequence of vertices.
42
Finding the neighborhoods Np: Now we can describe our modification of
the KleitmanWang algorithm more precisely. At step p of the Layingoff Algorithm,
we choose Np in the following way:
1. If Tp_i is consistent, then simply let Np be the first Â£p\ vertices in Tp\.
2. If Tp_i is not consistent but dp_i(rp_i(tp_i)) > dp_i(rp_i(Â£pi + 1)), then we
again let Np consist of the first tp\ vertices in Tp_i.
3. If Tp_i is not consistent but dp_i(Tp_i(Tp_i)) = dp_i(rp_i(fp_i + 1)), then Np
consists of all vertices w Â£ Vp_\ for which dp_i(u>) > dp_i(fp_i(Â£p_i)), and the
vertices x with the highest index for which dp_i(:r) = dp_i(rp_i(Tp_i)).
In words, what we do is identify the vertices with largest dp values and reduce their
values by 1. If Tp_i is not consistent and we cannot reduce all of those with the same
value, we reduce those with largest index (i.e., those that come later in the ordering,
as in Figure 3.2e). When Tp_i is consistent, we still take the first tp_\ vertices, even
if all of those with the same value are not reduced. We will say that Np is good if
{ui,... Vkr} C Np. The existence of at least r vertices among {vk+i,... ,vn} such
that laying oh each gives a good Np will yield the Kk_r r we seek.
An example: Let us do a small example to illustrate the way the Layingoff
Algorithm works.
Begin with the graphic sequence 7r = (9, 9, 9, 9, 8, 8, 7, 7, 7, 7, 4,4, 4, 4) =
(d(vi),..., d(vu)). For the purposes of this example, we will only identify the neigh
borhoods of the vertices with degree 4.
Step 1 Since the original ordering of vertices is consistent, we assign the neighbor
hood of W14 to be Ni = {v\, v2, v3, n4}. The new sequence is = (86, 74,43),
and since di(u4) > d\(v5), there is no reordering of vertices and 7\ is consistent.
Step 2 Since T\ is consistent, we can simply assign the set N2 = {v\, v2, iq, w4} to the
43
neighborhood of 1*13. Now 7t2 = (8, 8, 7s, 42). However, the vertices are no longer
in their original order; the sequence T2 is: v3,Vg,Vi,V2,v3,V4, Vj, Vs,Vg,Vio,Vn,Vi2.
Step 3 Since T2 is not consistent, we must consider d2(r2(4)). Since d2(r2(4)) =
<^2 (^2 (5)), we cannot simply assign the four highestdegree vertices to N3. We
begin with N3 = {us,^}, the two highestdegree vertices. Then we need two
more vertices, so we take the two vertices of degree d2(r2(4)) = 7 that have
the highest index, that is vg and v\g. So N3 = {v3) v&, Vg, uio} This leaves
7r3 = (7s, 62, 4), and T3 is consistent.
Step 4 Since T3 is consistent, we let iV4 = {iq, u2, v3, u4}. Then 7r4 = (74,66).
Observe that at each step 7q is exactly the sequence we would get after i iterations
of the KleitmanWang algorithm if a term of value 4 is laid off each time.
Proof that the Layingoff Algorithm gives r good neighborhoods: Now
we will show that this process does create r vertices among {vk+i, that have
good neighborhoods. We begin with several claims that develop useful properties
of the Layingoff Algorithm, in particular the key observation that Â£p > Â£p\ for all
p < rk. Then, we show that at each iteration of the algorithm, the sequence Tp has
a certain structure that allows us to easily count the number of iterations needed to
find r good Nps.
Claim 1. If p < r(k + 1) and i < j, then dp(vj) < dp(vi) + 1.
Proof of Claim 1. If dp(vj) > dp(vi) + 2 then, since d0{vi) > d0{vj), there exists a
p' such that dp/i(vj) = dpi\(vi), ry G Np/ and Vj Np/, and there also exists a p"
such that dpni{vj) = dp"\(vi) + 1, tq G N^/ and Vj ^ Npn. But such a p" cannot
exist because if dpi(vj) > dp"\(vi), then ty G Npn implies Vj is also in Npn. This
contradiction proves Claim 1.
44
Claim 2. If p < r(k + 1) and j < k r, then dp\{vj) > k. In addition, if
Spi fl iVp / 0, then Np is good.
Proof of Claim 2. If Vj is not laid off, then dj decreases by at most 1 at each step and
so dp_i(vj) > dj (p 1). Because dkr > dk + r(k + 2), we have the following:
dpi(vj) > dj (p 1) > dkr ip 1) > dk + r(fc + 2) (p 1) > dk + r.
The conditions on the sequence force dk > k r, giving dp_i(u,) > dk + r > k. As
a result, if Sp\ fl Np ^ 0, then Np must contain every vertex with remaining degree
greater than dPivap) < k 1. This includes all of {tq,... ,Vkr} and so Np must be
good.
Let gp denote the number of good neighborhoods Np< with p' < p. We may assume
that gp < r 1 for all p < rfk + 1). Otherwise, we would have r good neighborhoods,
hence our copy of Kkr,r In particular, by Claim 2 we can assume that there are at
most r 1 values of p for which Sp_ i fl Np ^ 0.
Claim 3. If p < rfk + 1), then fl Np\ < r 1 and (Apil > r(fc + r + 1)
gp~iir 1) ip 1) > 2r. In addition, every v G Np has dpi(n) at least as large as
the least value of dpi among members of Sp\.
Proof of Claim 3. Consider the vertex vap. It has degree at most k 1 when it is
laid off. By Claim 2, there are at least k r vertices Vj with dp_i(uj) > k and so
ISpi fl Np\ < ik 1) ik r) = r 1. Because a vertex will only leave the set Sp
if it has been laid off or assigned to the neighborhoods of enough other vertices that
45
its remaining degree is too low,
\SP1 > r(k + r +1)
p1
n N,}
3 = 1
(p 1) > r(fc + r + l) gpi(r 1) (p 1).
Since gp_\ < r 1, we have Ap_i > r{k + r + 1) (r l)2 (p 1) > 2r. If we
include the vertices {v\,... ,Vkr}, there are a total of at least k vertices w for which
dpi{w) is at least the minimum value of dp_i among the members of Sp_This
proves Claim 3.
Claim 4. If Â£p < Â£p1 for some p < rk, then at most r more iterations of the Laying
off Algorithm will create the desired copy of Kkr,r
Proof of Claim f. By definition, Â£p_\ = clpi{vap) and Â£p = dP(vap+1) Since va,p was
chosen to minimize dv\ among Sv1, we know that dpi{va +1) > Â£pi > k r.
Since dp{vap+1) > dpi(vap+1) 1, we get Â£p = dp{vap+1) > Â£Pi 1. Thus,
Â£p = Â£pil. This means that dpi(vap+1) = Â£p\ and vap+1 e Np. Since vap+1 e Sp1
as well, Claim 2 gives that Np is good.
Further, as is also the minimum remaining degree of any vertex in Sp_i,
Claim 3 gives that every vertex w in Np has dp_i{w) > Â£v\. Since va +1 G Np, we
conclude that dp_i(rp_i(Tp_i)) = Â£p\. Since dp_i(ufc_r) > dp_i(uap+1), there are at
most Â£p_i 1 vertices w with dp_1{w) > dp_i(ufc_r).
So, if we can show that Â£p/ > Â£p\ 1 for all p' such that p
each Np/ is good and we have the desired Kk~r,r in at most r more steps. From Claim
3, there are at most r 2 vertices in fl Np that have remaining degree larger than
dpi(vap+1) = Â£p1. Also from Claim 3, Ap_i > r{k + r + 1) gpi(r 1) (p 1).
46
Thus, there are at least
r(k + r + 1) gpi{r 1) (p 1) (r 2) > r(r gp_i)
vertices of remaining degree equal to lp\ in Sp\. Since Claim 3 gives that ,fy_i fl
Np>  < r 1, for all p' > p, each of the next rgp iterations of the Layingoff Algorithm
will remove at most r vertices from Sp1 which have remaining degree equal to lp\.
Hence there is always a vertex in Sp\ with degree equal to tv\. Thus, no vertex
with degree ip\ 1 will be placed into Np/, and Claim 4 is proved.
We can thus assume that Â£p > Â£p\ for all p < rk.
Now we are prepared to examine the structure of the sequence Tp. Claim 5
below is the main observation, that even when the Layingoff Algorithm results in
an inconsistent sequence, the sequence that results is of a very specific form. Thus,
the Layingoff Algorithm ensures that the number of iterations between consistent
sequences is less than k.
To show this, we say the sequence Tp is of proper form if there is a partition of
Vp into four ordered sets tp1\ tp2\ tp3^ and Tp4'1 (where the order is inherited from rp)
such that dp is constant on each of rp^ and rp^ and, when i < j and Vi,Vj G Vp, Vj
precedes ty if and only if Vj G r^ and Vi G rp3\ By Claim 1, we know that in this
case dp{vj) < dp(vi) + 1. Note that this allows for rp2^ and tp0> to be empty, in which
.(3)
case Tp is consistent.
We will abuse notation to let U represent the concatenation of ordered sets;
that is, Tp U Tp ^ is also an ordered set, where the elements of Tp'1 precede those of
tp\ and within each set the original order is maintained. Thus, if Tp is of proper
form, both Tp1'1 U Tp2'1 and Tpa> U t^1 are consistent. For a sequence Tp that is of proper
(3), Ah
form, the inconsistency of Tp is
ri2) U r^3)
. A consistent sequence has inconsistency
zero.
47
Claim 5. For all p E {0,..., rk 1}, Tp is of proper form. If Tp_i is consistent or has
inconsistency at least k, then Np is good. If Tp is inconsistent, then Utp3')  < tp\.
If Tp_ i has positive inconsistency, then either
Tp is consistent (and Np is good),
Tp = Tp_i and Np is good, or
Tp has inconsistency strictly less than the inconsistency of Tp_i.
Proof of Claim 5. We will prove the claim by induction on p.
If p = 0, then Tp = T0 is consistent. Moreover Np+i is good because it is simply
the first Â£p entries of Tp, which must contain Vi,... Vkr hi fact, this is true for any
consistent Tp and this will be our base case for the induction.
We assume the statement of the claim is true for T0,..., Tp_i.
Case 1: Tp_i is consistent.
The set Np is good because it is simply the first lp\ entries of Tp_i, which must
contain wi,..., Wfc_r. If Tp is consistent, then it is, by definition, of proper form.
If Tp is not consistent, then dp_i(rp_i(f!p_i)) = dp_i(rp_i(Â£p_i + 1)). We can
partition Tp_i into where rpl\L contains all vertices with remaining
degree exactly dp_i(rp_i(fp_i)) and Tp^i contains those with lower remaining degree.
We can further partition into and where contains all vertices
of that are included in Np, and consists of those that are not.
Now Tp can be partitioned into
(i)
a(D
1 pl
(2)
'p 1
(3)
f(4)Li
'p 1 )
and
(4)
y(4 )R
'pl >
48
and it is of proper form. Clearly Irp1"1 U rp3') = lv\.
Observe that if Tp_i is consistent and Tp is not, then {v\,... ,Vkr} is contained
in rp^ U rp^ U Tp3'1.
Case 2: Tp_i is not consistent.
Recall that ip_\ = Np\, the number of vertices in Vp that are reduced by one when
a vertex of Tp_i is laid off. The effect of the Layingoff Algorithm on Tp depends on
the value of ip\.
Note that tp\ < fpi\ is not possible because Claim 4 allows us to assume that
ip1 > ip2 Since ip2 > rpi\ U Tp3\  > rpl\, this is a contradiction.
With this information, we can show that if the inconsistency of Tp_i is at least
k, then Np is good. The largest k entries of Tp_i are in rpi\ U rp^\ U rp3\ and
Np contains all of Because there are r(k + r + 1) vertices eligible to be laid
off from {vk+i,... ,un}, and weve laid off at most rk, the vertices Vkr, . ,Vk will
not be laid off. If Vk is in r^, then its value is at most c4 1, and if Vk is in
Tp_i, then its value is at most dk. But the degree of each of V\,... ,Vk~r is at least
dkr ~ hr > dk + r(k + 2) kr > dk So, each of V\,..., Vk~r are in rpi\ and will be
in Np as long as the inconsistency is at least k.
Case 2a: fpi\ < ip\ < If^ U
In this case, we can partition Tp2}1 into two pieces: and rp^. The members
(i)
42)
of Tp^f are reduced when vap is laid off, but those of fpj^ are not.
After reordering, we obtain the following:
(2 )L
r(i) =y(b
p pi)
(2) = a(2 )L
p 1 )
r(3) =y(3)
p 1 p 1)
(4) =f(2 )R f(4)
p p 1 u p1
49
Moreover, Â£p_i > Â£p_2 > fpi\ U Tp3\ = \tp^ U tp3^\. In addition, the inconsistency of
Tp is tp2^ U Tp311 = \t^1i U fp^l, which is strictly less than the inconsistency of Tp_i
because rp^i is a strict subset of
To proceed through the next cases, we must partition rpi}l into two pieces:
and The members of Tp^ have the same remaining degree as those in rp3}1, and
those of r^_i have smaller remaining degree (either or both of these may be empty).
Case 2b: Ir^ U f^l < Â£pi < U  + \rpA}i\.
In this case, the values of r^}l U rp2}1 as well as some of are reduced. Since
the members of U (and the unreduced values of now have the same
value, reordering results in Tp being a consistent sequence.
Case 2c: U \ + r^f  < Â£pi < U U U t^\.
In this case, we can partition into two pieces: rp^ and rp^. The members
i(3)
(4)Ls
of Tp3}^ are reduced but those of are not.
p
i(3 )l
After reordering, we obtain the following:
r =r(1) Ur(3)L
Tp 'p 1 U 'p 1 >
T
T,
(1)
0
(2)
V
(3)
T,
(2)
p 1)
T
P
(4)
T,
(3 )R
p 1 ;
P
T,
(4)
P~ I'
Moreover, Â£p_i > Â£p_2 > rp1}l U Tp3\ = Tp1'1 U tp3'1 . In addition, the inconsistency of
Tp is rp2^ U Tp3'11 = \t^1i U Tp3_}^\, which is strictly less than the inconsistency of Tp_i
because is a strict subset of t
Case 2d: U T{p\ U U t^\ < Â£p_x.
In this case, no rearranging is necessary: the order of the vertices in Tp is the
same as the order in Tp_i.
50
Because the only vertices out of order are in tp_i U Np will contain all
of the first lp_\ vertices. Since ip_\ > k r, the neighborhood Np must contain
{ui,... ,Vkr} and thus be good.
This concludes the proof of Claim 5.
Given Claim 5, the proof of Lemma 3.5 follows easily. There can be at most k 1
neighborhoods that are not good between consecutive good neighborhoods. So after
(r 1 )k + 1 iterations of the procedure, there will be r good neighborhoods, giving
us the desired realization.
51
4. Stability with Respect to the Potential Number
In Chapter 3, we showed that graphic sequences that are not potentially H
graphic are close to being majorized by a sequence from V(H). In this chapter, we
will show that for some graphs H, if the sequence also has the property that its
sum is close to the potential number, then it is actually very close to being one of
the sequences in V(H), not just to being majorized by one. However, this actually
depends largely on the structure of H, so it is not true for all graphs.
Our goal is to prove a stability result akin to the result of Simonovits (The
orem 2.5) for the Turan problem, so we examine this theorem again in a slightly
different light. We can rephrase Theorem 2.5 in terms of edit distance, which will
help to motivate our definition of distance between graphic sequences. Given graphs
G and G' on the same labeled vertex set, the edit distance between G and G', denoted
dist(G, G'), is \E(G)AE(G')\. With this terminology, we restate Theorem 2.5:
Theorem 4.1 (Simonovits [113]) Let H be a graph with x(H) = r + 1. For every
e > 0, there exists a 5 > 0 and an ne such that if n > n and G is an nvertex Hfree
graph such that
E(G)\ > ex(H,n) 5n2,
Then dist(G,Tn>r) < en2.
In order to define a stability concept for graphic sequences, we first need to define
a measure of how far apart two graphic sequences are. The concept of ([a\, a2],b)
closeness defined in Chapter 3 is not fine enough for our purposes, so instead we
will use the standard f1 norm. For graphic sequences = (aq,... ,xn) and 7r2 =
(t/i,..., ym) where m < n, we let 7Ti 7T21 = YTj=i\xj ~ Vj I) where we define
ym+i, .. .yn to be 0 if m 7^ n.
This is in keeping with our motivation. Indeed, if G is a graph such that
dist(G,Trar) < en, then we would like to have 7t(G) 7r(Trar) be small as well.
We know that, assuming appropriate divisibility conditions, 7r(Tn>r) = ((^n)). Let
52
Vl be the set of vertices in G that have degree at least Lp^n and be the set of
vertices in G with degree less than r^n. Then
vÂ£Vs ^
< den.
This justifies our use of the t1 norm to define a distance between graphic sequences.
We can now precisely define what it means for a graph to be stable with respect
to the potential number.
Definition 4.2 A graph H is stable with respect to the potential number, or
estable, if for any e > 0, there exists an no = n(e, H) and 5 > 0 such that for
any graphic sequence ir of length n > no that is not potentially Hgraphic and that
satisfies
a(H, n) 5n,
there is some it' G V(H) such that 7r ir'\\ < en.
We also dehne the following weaker version of ustability:
Definition 4.3 A graph H is weakly 0, there exists an
n0 = n(e, H) and 5 > 0 such that for any graphic sequence ir of length n > n0 that is
degree sufficient for H but not potentially Hgraphic and satisfies
o{ir) > a(H, n) 8n,
there is some it' G V(H) such that 7r ir'\\ < en.
The difference between these conditions is that ustability does not require that
a sequence be degree sufficient for H. There are graphs that are weakly
53
Figure 4.1: A graph with Va+i > 1 and 2i* Vi* < 2a
not crstable; complete graphs are one example, which will be discussed in more detail
in Section 4.2.
4.1 Graphs with degreesequence stability
Recall that for a graph H of order k the sequences in V(H) are determined by
the value 2i Vi(H) for i E {ot{H) + 1,..., k}. Let i*(H) be the smallest index
i E {a + 1,..., k} such that 2i Vi is minimized. (When it is understood, we will
suppress the argument H in our use of parameters like a, Vi, and i*.) We have
2i* Vi* < 2(a + 1) Va+i < 2a + 1.
If Va+1(H) > 1, then 2(a + 1) Va+i < 2a(H), so 2i* Vi* < 2a(H). Thus, if
2i* Vi* = 2a(H) + 1, we must have Va+i(H) = 1. In this case, we know that
there is a set of a + 1 vertices in H that induce a graph consisting of a matching and
isolated vertices.
It is worth noting that 2i* Vi* < 2a(H) does not necessarily mean that
Va+i(H) > 1. For example, if H is the graph in Figure 4.1, we have a(H) = 4
with V5(H) = 1 and Ve(H) = 5. Thus, 2i* Vi* = 2(6) 5 = 7 < 2a(H).
Our main theorem states that there is a large class of graphs that are crstable.
Theorem 4.4 If H is a graph such that 2i* Vi*(H) < 2a(H), then H is crstable.
54
On the other hand, if 2i* Vi* (H) = 2a(H) + l (and thus Va+i(H) = 1), whether
H is crstable depends more strongly on the structure of H, as can be seen in the next
That is, it is K\>x U K^y with an edge joining the vertices of degree greater than 1.
Theorem 4.5 If H is a graph of order k such that
(a) 2i*Vi.(H) = 2a(H) + l,
(b) H has a set X of a(H) + 1 vertices such that H[X) has one edge, and
(c) HC Kka(H)~2 V Sbltb2 for some b\ and b2 with b\ + 62 = 01(H),
then H is crstable.
These results imply that complete split graphs, complete bipartite graphs, friend
ship graphs, and odd cycles, among many others, are crstable. We will say that H is
Type 1 if 2i* Vi*(H) < 2a(H), and Type 2 if 2i* Vi*(H) = 201(H) + 1. Of those
graphs just listed, the complete split graphs and complete bipartite graphs are Type
1, and the odd cycles and friendship graphs are Type 2.
4.2 Graphs that are not astable
The hypotheses of Theorem 4.5 suggest that graphs that do not satisfy all of
these conditions may not be crstable. This is in fact the case, at least if H satisfies
condition (a) but not condition (c), as we see in the next theorem.
Theorem 4.6 If H is a graph such that 2i* Vi*(H) = 2a(H) + 1, and H (Z
Kka(H)2 V Sbltb2 for anV bi and b2 with b\ + b2 = ot(H), then H is not crstable.
Proof. Consider the sequence
theorem. Let Sx,y be the double star with central vertices of degree x + 1 and y + 1.
55
This is the degree sequence of the graph Kg V Sn^i and this graph is in fact the
only realization of pe. Note that cr(pe) = 2{Â£ + 1 )n {Â£ + 2){Â£ + 1).
If H is Type 2, then a(H, n) = 2(kal)n+o(n). Thus a(pka2) > cr(H, n)5n
for any 8, provided n is large enough. The sequence pka2 is not potentially H
graphic, for if H C Kk_a_2 V Sn(ka 2) n (fc ex 2) j then at least a + 2 vertices must
2 2
come from the set of vertices that induce a double star; since at most a of these
vertices can be independent in H, at most a of the vertices of degree k a 1 may
be used. This implies that H is a subgraph of Kk_a(H)_2 V Skl^2 for some b\ and b2
with b\ + b2 = a, contradicting our hypothesis.
It remains to show that \\pka2 tt > eri for every tt E V(H) and some choice
of e. Since 2i* Vi* = 2a + 1, we know that for each 7Tj(H,n) G V(H), we have
2j Vj(H) = 2a + 1. We also know that Ha+\(H) E V(H). Recall that yfy (TT, n) =
{{n l)k~j, (k j + Vj l)n~k+j). When j = a + 1, we have 7ra+i(H, n) = ((n
1 (ka  so
II Pk
a2
7ra+1(H,n)\\ =n k + a.
For any j > a + 1 with n) E V(H), we have
pfc2 TTjiH^W = (j a 2 )(n 1 (k j + Vj 1))
+ (n k + a)(k j + V j 1 (k a 1))
2 n(j a 1) + (j a) (a + j V j 2k) + 4.
Hence, pk.
a2
7ij(H, 77.)  > ea for each 7ij(H, n) E V(H) and any e < 1.
56
Which graphs satisfy the conditions of Theorem 4.6? In Tables 4.1 and 4.2, we
present all graphs of order at most 6 that satisfy the hypotheses of this theorem, and
hence are not ustable. The graphs in Table 4.2 have the additional property that
they are not weakly ustable, because the sequence pka2 is degree sufficient for each
of these graphs, demonstrating that the conditions for weak astability cannot be met
in these cases.
We can generalize many of the graphs in Tables 4.1 and 4.2 to fold larger families
of graphs that are not ustable.
Claim 4.1 Let H be a graph of order k that is Type 2 and satisfies H Kka(H)2 V
5,6i,62 for any bi and 62 with b\ + 62 = ot{H). If a(H) < ^44, then for all p > 1, the
graph H V Kp is not astable.
Proof. By Theorem 4.6, H itself is not crstable. Let Hp = H V Kp, we will use
Theorem 4.6 to show that Hp is not crstable either. First note that a(Hp) = a(H).
To ensure that Hp is Type 2, we need to show that 2i*{Hp) Vi*{Hp) = 2a(H) + 1.
We already know that 2i*(H) Vi*(H) = 2a(H) + 1, so we need to check that
2i Vi(Hp) > 2a(H) + 1 for each i e {a(H),...,k + p}. For i < k, we have
Vi(Hp) = VfiH), because for these values of i, any induced subgraph of H is also an
induced subgraph of Hp. When i > k, we have VfiHp) = i 1, because there will
be i k vertices of degree i 1 in any set of i vertices. Thus, for i > k, we have
2i Vi(Hp) = i + 1, which is smallest when i = k + 1. Since a(H) < ^44; we know
(k + 1) + 1 > 2a(H) + 1, so Hp is Type 2.
We also need to show that Hp Kk+pa(H)2 V for any b\ and &2 with
b\ + 62 = ot(H). This is clear, because if Hp were such a subgraph, then H would
be a subgraph of Kka2VSbltb2 Thus, Theorem 4.6 implies that Hp is not crstable.
57
Table 4.1: Connected graphs of order at most 6 that are not ustable
58
Table 4.2: Connected graphs of order at most 6 that are not weakly erstable
In particular, this result shows that the complete graph Kk and the graphs Kk
P3 = Kk_4 V Z\ and Kk_5 V P5 are not erstable. For H = Kk or H = Kk P3,
there is a set X of a + 1 vertices such that H[X) contains exactly one edge, showing
that we cannot weaken the hypotheses of Theorem 4.5. For H = Kk_5 V P5, there is
no such set. However, there is a set of a + 1 vertices that induce a matching of size
two and isolated vertices. We have not yet shown anything about the ustability of
such graphs, so only know that they are not ustable if they fall under the hypotheses
of Theorem 4.6. For further insight into these graphs, it is worth pointing out that
graphs H4, H9, H10, H13, H14, and H16 from Table 4.1 and graphs F2 through F7
from Table 4.2 each have two edges in any matching induced by a vertex set of order
a 4 1.
We can make other generalizations from the graphs in Tables 4.1 and 4.2. For
example, there are several graphs consisting of a complete graph with one or more
pendant vertices. For a graph G, let cu(G), called the clique number of G, be the
order of the largest clique in G. Let Kp(mi,... ,mp) denote the complete graph on
vertex set {iq,... ,vp} with rrij leaves incident to vertex Vj for each j, 1 < j < p.
59
This graph has order k = p + Y^=i mi and, if any rrti = 0, the independence number
is a = 1 + Y^=i mi (if110 mi = 0) then the independence number is simply Y^=i mi)
Claim 4.2 The graph Kp(0, 0,1, 2,... ,p 2) is not astable.
Proof. First note that k a 2 = p 3 for any graph of this form. Let G(pka2)
be the unique realization of pka2, which has clique number k a = p 1. Since
Kp(rrii,... ,mp) contains a clique of order p, we see that Kp(mi,... ,mp) is not a
subgraph of G(pfc__2).
For j E {1,... ,k a;}, every set of a + j vertices in Kp( 0, 0,1,.. .p 2) induces
a subgraph of maximum degree at least 2j 1. In fact, the subgraph induced by the
set of all of the pendant vertices as well as vertices ii,v2, .Vj from the clique has
maximum degree exactly 2j 1. Thus, Va+j = 2j 1 for each j, and in particular
this means that the graph is Type 2. Theorem 4.6 then implies that it is not ustable.
The graph Kp(0,0,1,... ,p 2) has an interesting property, namely that 2i
Vi(Kp(0, 0,1,... ,p 2)) = 2a; + 1 for each i E {a + 1,... k}, which means that the
sequence 7Ti(Kp(0, 0,1,... ,p 2), n) is in V(KP(0, 0,1,... ,p 2)) for each L That is,
there are k a sequences in V(KP(0, 0,1,... p 2)), and none of them is within en of
Pka2 Any subgraph of Kp(0, 0,1,... ,p 2) that contains Kp has 2T Vj* = 2a + l,
so is also not ustable. However, these subgraphs may not satisfy Va+J = 2j 1 for
each j and do not have the same property.
It is interesting to note that Kp( 1,1,..., 1) is Type 2 but satisfies the conditions
of Theorem 4.5, so is ustable, and Kp(2, 2,..., 2) is Type 1, which means it is also
crstable. Thus, relatively small differences in the structure of a graph can change
whether the graph is crstable.
60
The previous example points out another class of graphs that cannot be astable.
Since u{G{pf)) = Â£ + 2, if a graph that is Type 2 has clique number greater than
k a, it cannot be a subgraph of G(pka2) Since the clique number of a graph of
order k cannot be larger than k a + 1, this means that any graph H that is Type
2 and has u(H) = k a(H) + 1 is not astable.
We have shown that there are graphs that are not weakly astable are there
graphs that are weakly erstable? In fact, complete graphs are weakly crstable even
though they are not crstable.
Theorem 4.7 The complete graph Kk is weakly astable for all k > 3.
Proof. To see this, consider Theorem 2.17. If a sequence 7r = (di,..., dn) is degree
sufficient for Kk, then dk > k 1. Theorem 2.17 says that if in addition e?2fc > k 2,
or if di > 2(k 1) i for each i with 1 < i < k 2, then 7r is potentially Kkgraphic.
Thus if 7r is not potentially AVgraphic, we must have c^fc < k 3, and thus dj < k 3
for each j > 2k, as well as having some i with i < k 2 such that di < 2{k 1) i.
Thus, the graphic sequence with the largest sum that is degree sufficient for Kk but
not potentially AVgraphic is n = ((n l)k~3, (k l)k+2, (k 3)n2fc+1), which has
sum (2n 2k)(k 3) + (k \){k + 2). As discussed in Chapter 2, the potential number
for Kk is a(Kk, n) = (k 2)(2n k +1) + 2. Thus, cr(Kk, n) er(7r) = 2n 4fc + 2. This
shows that among graphic sequences that are degree sufficient for Kk, but not poten
tially Arfcgraphic, there are none that satisfy the condition that cr(7r) > a(Kk, n) 5n
for 5 < 2. Thus, taking 5=1 trivially satisfies the conditions for weak erstability.
Before closing this section, we point out some facts about the erstability of cycles.
All cycles are Type 2. Odd cycles are crstable, because they satisfy the hypotheses
of Theorem 4.5. The cycle Cq is not astable, because it is not a subgraph of G(p 1) =
G(pka2), and it is not weakly astable because p\ is degree sufficient for CÂ§. For a
61
cycle of length 2p with p > 3, however, we have that C12P C Kv_2 V for some x,y
with x + y = p. This does not imply that C2P is ustable, though, because while it is
Type 2, it has at least two edges in every graph of maximum degree one induced by
p + 1 vertices, so does not satisfy the hypotheses of Theorem 4.5. Investigating the
astability of these and other Type 2 graphs with this property (containing a set of
a + 1 vertices that induces a matching with at least two edges) is the next step in
our research.
Now we will turn our attention to the proofs of Theorems 4.4 and 4.5.
4.3 Technical lemmas
For a graph G, let V^(G) denote the family of subgraphs of G obtained by
deleting exactly t vertices from G. This is the family of induced subgraphs of G with
order C(G) t. We say that a graphic sequence tt is potentially 2?^(G)graphic if
there is a realization of tt containing any graph in V^\G).
To prove Theorems 4.4 and 4.5, we will need the following useful consequences of
Theorem 1.2, the KleitmanWang algorithm.
Corollary 4.8 Let iTj be the sequence obtained from tr = (d\,... ,cln) by laying off
the term dj. Then:
1. There is a realization of tt in which the vertex of degree dj is adjacent to the dj
vertices of highest degree other than itself.
2. If TTj is potentially Hgraphic, then ir is potentially Hgraphic.
3. If G is a graph with degree sequence ir and v is a vertex in G, then if ir(G v)
is potentially Hgraphic, then tt is potentially Hgraphic.
4 If tt = (n 1 ,d2,... ,dn), then tt is potentially Hgraphic if and only if tt\ =
(d2 1,..., dn 1) is potentially P^1)(H)graphic.
62
Part 1 of Corollary 4.8 guarantees the existence of a realization of 7r in which the
vertex of maximum degree, d\, is adjacent to the next d\ vertices of highest degree.
Following [48], we call such a realization a canonical realization of tt.
Lemma 4.9 If H is a graph of order k and t < k cx(H), then there is an F G
V^(H) such that a(H,n) 2tn > a(F,n).
Proof. Since t < ka(H), we know cx(H) < kt. Thus, there is a graph F G T>^(H)
with a(F) = a(H), for we could simply take a subgraph of H of order k t that
contains a maximum independent set of H. For each j G {o(T) + 1,..., k t}, we
have Vj(F) < V j(H), because every jvertex subgraph F' of F is a jvertex subgraph
of H. This means that for each such j,
2j V3(F) > 2j Vj(H) > 2i* 
since 2i*(H) Vi*(H) is the minimum value of 2j Vj(H) over the set {a(H) +
1 In particular, we see that 2 i*(F) Vi*(F) > 2 i*(H) Recall from
Theorem 2.19 that a(H,n) = (2(k i*(H)) + Vi*(H) 1 )n + o(n). We compute
a(F, n) = (2(k t) 2i*(F) + V,* (F) 1 )n + o(n)
= (2k (2 i*(F) Vj* (F)) 1 )n 2 tn + o(n)
< (2k (2i*(H) Vi* (H)) 1 )n 2tn + o(n)
= cr(H, n) 2tn.
Our proof of Theorem 4.4 follows very closely the proof of Theorem 2.19 in [46].
As such, we will need the following result from that paper.
63
Theorem 4.10 (The Bounded Max Degree Theorem [46]) Let H be a graph
of order k and ir = (di,..., dn) be a nonincreasing graphic sequence with n sufficiently
large satisfying the following:
1. tt is degree sufficient for H, and
2. dn> k a(H).
There exists a function f = f(a(H),k) such that if d\ < n f(a(H),k), then ir is
potentially Hgraphic.
For our purposes, it is useful to know that the function / is given by
f(a(H),k)
k a(H)
2
1)
+ a(H) 1
+ Ak2 + k + 1
4.4 Proofs of Theorems 4.4 and 4.5
Theorem 4.4 is an immediate corollary of the following lemma. Theorem 4.5 will
follow with a little more work; in particular when 2i* Vi*(H) = 2a.(H) + 1, then
Lemma 4.11 may result in a realization of tt containing Kk_a_i V KatyH)+1. In this
case we must do further analysis to show that H is ustable.
Lemma 4.11 Let H be a graph of order k and let e > 0 be given. There exists an
no = n(e, H) and 5 <  such that any graphic sequence it of length n > no with
<7(tt) > a(H,n) 5n, either
1. tt is potentially Hgraphic;
2. 117r tt'W < en for some tt' G V(H); or
3. tt is potentially {Kk_a^H)_i V Ka(H)+i)graphic.
64
Our goal is to create a realization of 7r containing a supergraph of H, or show that
117r 7T711 < en for some tt' E V(H). This will be done in three stages. Stage 1 describes
an algorithm that iteratively builds a realization of tt with a desired structure, while
keeping track of changes made so that we can discover more about the terms of the
sequence as we proceed. Stage 2 uses this realization and the information gained from
the algorithm to show that in many cases, this realization actually does contain H (if
H is Type 1), or it contains (Kka(H)i V Ka(H)+i) (if H is Type 2). Finally, Stage
3 shows that in those cases where we do not have one of these desirable realizations,
we can show that tt is close (in our metric) to one of the sequences in V(H).
Proof of Lemma 4.11. Let tt be a graphic sequence of length n, where we
assume n is sufficiently large. Suppose also that for some 8 < (2(k i*) +
1 8)n.
Stage 1: In this stage, we describe the algorithm that reduces tt to a resid
ual sequence 7iy, assuming at each step that the sequence obtained does not satisfy
the conditions of the Bounded Max Degree Theorem (Theorem 4.10) for a certain
complete split graph.
Define
{0 if H is Type 1
1 if H is Type 2.
We initialize the algorithm by applying the KleitmanWang algorithm (Theorem 1.2)
to tt to obtain a sequence ttq with minimum degree at least k i* + Vi*~1~'5. Since
8 < 1, we do this by iteratively laying off terms of value at most k i* + Vf*~2,
beginning with the smallest such term. After laying off the first term, we get a
65
sequence t\\ with n 1 terms such that
(2[k i*) + Vj* 1 5)n (2(k i*) + V** 1) + 1
=(2(k ?:*) + V,* 1 5)(n 1) + (1 5).
As we continue to lay off terms of low value, we obtain a sequence 7r, of length n j
with
(2(k ~ i*) +  1 5)(n j) + j(l 5).
If j(l 5) > 28(n j), then
(2(fc **) + Vi* l)(n j) + 5(n j) > a(H,nj),
implying that ttj is potentially Hgraphic. By Part 2 of Corollary 4.8, this means that
7T is also potentially //graphic. Thus, laying off at most terms must result in a
sequence with the desired minimum degree. Call this sequence 7dh Note the following
properties of 7r^: it is not potentially //graphic, it has length n0 > n( 1 j^), and
its smallest term is at least k i* + > k a bH.
After this initialization, we perform the following steps to create sequences
7b1),... ,7for some l < k a(H) bn The purpose of each step is to reduce
the maximum term and increase the minimum term of each successive sequence. For
each t > 0, let 7r^ = (df[ \ ..., dn}) be the sequence that results from the tth itera
tion of the algorithm, and let Rt be a canonical realization of on the vertex set
{vf\ ..., Vn}} such that d(v^) = dtp.
Starting with t = 0, the algorithm proceeds as follows:
(i) Remove the nonneighbors of the vertex vP from Rt to obtain a graph Rt. Let
7?^ = 7T(Rt) Note that in Rt, the vertex vp1 is a dominating vertex; since Rt
was a canonical realization of 7r^, this means that if vertex was removed
66
from Rt, then vertex v^ is also removed if p > j.
(ii) Lay off the largest term of 7r^, and call the resulting sequence This is the
degree sequence of the neighborhood of vf'* in Rt.
(iii) As in the initialization step, apply Theorem 1.2 to laying off terms of
smallest value until we have a sequence with minimum term at least
k_2f_V1L _l + (2 + t)S
Let 7r^+1^ be the sequence that results from this step.
(iv) Terminate if df+1'> < nt+1 (ffc/2i) (^2 1) or t + 1 = A; a{H) bn Otherwise,
return to Step (i).
When the algorithm terminates, let 7= 7r(t+1h The remainder of the proof
deals with determining properties of 7and its realizations.
Let us first examine the effects of the algorithm. Step (i) yields a graph with a
dominating vertex, whose degree sequence has largest term nt 1, where nt is the
length of 7f(t). Since the algorithm stops when is too small, the number of
vertices removed is at most 1)> and Theorem 2.17 (sufficient conditions
for a graphic sequence to be potentially iv^graphic) gives us a bound on the size of
those terms. Laying off the largest term of 7?^) in Step (ii) removes a dominating
vertex from the graph. Finally, in Step (iii), we lay off terms to obtain a graphic
sequence with minimum degree at least k (i + 1) a bn
Before moving on to Stage 2, we state and prove several claims which, along with
the discussion in the previous paragraph, will help us determine if we can achieve the
desired realization of 7r or show that it is close to an extremal sequence.
Claim 4.3 If tt^ is potentially (H)graphic, then 7r is potentially Hgmphic.
Additionally, there is a realization G of n such that G contains K1 V G^, where G^
67
is a realization of .
Proof of Claim 4.3. First we will show that for each t with 0 < t < Â£, if
is potentially 'D('t\H)gr&phic, then 7r is potentially iAgraphic. The proof is by
induction on t. If is potentially ^(i^graphic, then by repeated applications of
Part 2 of Corollary 4.8, tt is potentially iAgraphic. Assume that the statement is true
for some t < Â£, and suppose 7r^+b is potentially 2At+1)(i4)graphic. Again, repeated
applications of Part 2 of Corollary 4.8 show that is potentially graphic.
Then, as 7f'A is created from 7f^) by the removal of a dominating vertex, we see that
7is potentially graphic. This is an application of Part 4 of Corollary 4.8.
Finally, Part 3 of Corollary 4.8 implies that ir^ is potentially V^^H)graphic, and
the induction hypothesis shows that 7r is potentially iAgraphic.
Now we will show that for each t <Â£, tv has a realization containing Kt V G^\
where is a realization of 7CA We proceed by induction on t. Let G^ be a real
ization of 7Ch By repeated applications of Part 2 of Corollary 4.8, 7r is potentially
G^graphic, establishing the base case of our induction. Suppose the statement is
true for some t. with 0 < t. < Â£, and let G^+b be a realization of 7r^+1h Again, we use
Part 2 of Corollary 4.8 several times to show that 7?^) is potentially G^+1^graphic.
Part 4 of Corollary 4.8 then shows that 7?^) is potentially G^t+1Agraphic, and we
know that moreover, since we have added a dominating vertex to the graph, there is
a realization of tf^d that contains Ad V G^+1b To get from 7to 7f^, we removed
the nonneighbors of a vertex from a canonical realization Rt of 7r^d Doing this
does not change the graph induced by NRt(vi'). Thus we may assume that G*A is
a realization of ir^ in which G^+b C AG(i> (rf); that is, we can assume that is
the vertex acting as Ad in the copy of Ad V G^+1h Now the induction hypothesis
gives us a realization of 7r that contains Kt V G^, and we see that we can view G*A
as Ad V G^+b with additional vertices that are not adjacent to This gives us a
realization of tt containing Ad+i V G^+b, as desired. Thus, there is a realization of 7r
68
that contains TQ V .
The construction of the realization of 7r containing Ki V shows that after the
f th step of the algorithm, there is a realization of tt in which t vertices are adjacent
to all but a fraction of the vertices in the graph. The exact value of this fraction will
be determined after the next claim, as we need some of the facts developed in that
claim in order to calculate it.
Claim 4.4 (2(k i*) + V^* (H) 1 (1 +t)8 2t)nt, and at most
iterations of the KleitmanWang algorithm are needed at each implementation of Step
(Hi).
Proof of Claim f.f. First we will prove the lower bound on a(n). Again the proof
is by induction on t. The claim holds for by hypothesis. Suppose t > 0 and
a(ir^) > (2(k i*) + V**(H) 1 (1 +1)8 2t)nt. We will show that the inequality
holds for 7r^+1\ Let M = 2(k 2)(fc/2])(^2 !) Since d^ > nt (fc/2])(^2 1)>
creating 7?^) from entails removing at most (fc/2) (^2 1) vertices, each of which
has degree at most k 2 (by Theorem 2.17). Thus, <7(7?^)) > <7(77^) M. We have
assumed that n is sufficiently large, so in particular, since nt increases with n, we
may assume M < 8nt, which means that
(t(tt('*')) >(2(k i*) + Vi*(H) 1 (t + 1)Â£ 2t)nt 8nt
=(2(k  i*) + Vi*(H) 1 (1 + (t + 1 ))8 2t)nt
>(2(k  i*) + Vi*(H) 1 (1 + (t + 1 ))8 2t)nt,
69
where rit is the length of 7f^h Creating 7from 7fh) requires laying off a term of
value rit 1, so <7(7?^) = <7(7?^) 2(nt 1). We now have
(2(k i*) + Vi*(H) 1 (1 + (t + 1))Â£ 2t)nt 2(nt 1)
>(2(/c 2*) + 1 (1 + {t + 1))$ 2{t + l))rq
>(2(/c 2*) + 1 (1 + (t + 1))$ 2{t + l))7q,
where nt is the length of 7r^. Finally, the last step of our algorithm involves laying
off terms from that have value less than k 2l ~Vi* 1+b+2b (f, + 1). The first
iteration of this process yields a new sequence 7^ with
> cr(7r(t)) [2(k i*) + Vt*(H) 1 (1 + (t + 1))Â£ 2{t + 1)]
> [2(k i*) + Vt*(H) 1 (1 + (t + 1))
Repeating this j times yields the sequence 7fj*') with
*{%?) > [2(k 1*) + V,*(H) 1 (1 + {t + 1))
= [2(fc 2*) + Vi*(H) 1 (1 + (f + 1))$ 2(t + 1 )\ntj,
where nt,j is the length of ir^. Since 7rit+1) = for some j, we have established the
lower bound on
If we analyze this final step more carefully, we get the second half of the claim. We
have seen that a(w> (2(fc 2*) + V* (77) 1 (1 + (f +1))Â£ 2(t+l))ht. When we
lay off a term of 7whose value is too small, we subtract 2(k z*) + Vi* 2 2(t + 1)
from <7(7?^), so that, if we call the resulting sequence Trf\ we see that <7(7?^) is at
70
least
[(2(k i*)+Vi.{H) 1 (f + 2))5 2{t + 1 )]nt
[2(fc**) + Vil2(i + l)] + l
= [(2(k i*) + Vi*{H) 1 ~(t + 2))8 2it + 1 )](nt 1)
+ (1 (t + 2)5).
After repeating this j times, we get a sequence with sum at least
[(2(fc f) + V,* (//) 1 (t + 2))5 2{t + 1)] (th j) + (1 (t + 2)5)j.
Now, if criirf1) > a(H,nt j) 2{t + l)(nt j), then by Lemma 4.9, cr(p^) >
happen if (1 (t + 2)5) j > (t + 3)6(nt j), which means that if j > nt >
then 7is potentially 2?(t+1)(i7)graphiC. Since 7T(t+1) = 7fjt} for some j, this in turn
means 7dt+1) is potentially Z^t+1) (i7)graphiC. Finally, Claim 4.3 shows that in this
case, 7T is potentially //graphic. Thus, at most ^f^~nt iterations of the Kleitman
Wang algorithm result in the desired sequence 7r^+1h
We would like to know how many terms have been removed from 7r in the creation
of 7That is, what is the value of n np. At each iteration of the algorithm, the
vertex vf1 misses at most (fe/2) (^2 1) vertices, so in Step (i) at most this many
vertices are removed. In Step (ii), only one vertex is removed. Then, Claim 4.4 shows
that at most ^3^gSn terms are laid off in Step (iii). Since we do the algorithm at
most k a 1 times, and do the initialization step once, this means that
 < (A a 1) ( ( ^)
71
Claim 4.3 shows that there is a realization of 7r that contains Kg V G^e\ where
GC)
is a realization of ir^. The Â£ vertices in the clique are adjacent to every vertex
in G^, so they are only nonadjacent to vertices that have been removed or laid off
through the course of the algorithm. Thus, they are not adjacent to at most n ng
vertices; that is, the expression in Equation (4.1) gives an upper bound on how many
vertices are not adjacent to the clique in this realization.
Stage 2: In Stage 2, we analyze the realizations of 7r^ with the goal of showing
that some realization contains a supergraph of H. Recall that in Stage 1, our algo
rithm stops either when we have iterated enough times, or when the sequence tt^
fails the maximum degree condition of Theorem 4.10.
If the algorithm stops because we have iterated k a(H) bn times, then the
graph Kg V G^ contains Kk_a^H)_bH V Ka(H)+bH, which is a supergraph of H when H
is Type 1, so in this case tt is potentially Hgraphic. If H is Type 2, then we are left
with a realization of tt containing Kka(H)i V AG(ir)+i This suffices to prove Parts
1 and 3 of the lemma.
Therefore we assume that the algorithm stops when t+l=Â£
and Sf* < ng (fc/2)(fc2 !) Now has maximum term df\ which is less than
rig (fc/2])(^2 1) and minimum term dnj, which is at least k a(H) bn Â£
If 7r^ is also degree sufficient for Sg = Kk_a_bll_g V Ka+bll, then the Bounded Max
Degree Theorem (Theorem 4.10) implies that is potentially S^graphic, because
the value (fc/2]) (^2 1) was chosen so that (^ ) (fc2 1) > f(a + bn, k Â£), where
/ is the function given by Theorem 4.10. Since Kg\J Sg = Kk_a_blI V Ka+bn, when
H is Type 1 this shows that tt is potentially Hgraphic. If H is Type 2, then there
is a realization of 7r containing JGai V Ka+\. Again, this suffices to prove Parts 1
and 3 of the lemma.
We can therefore assume that 7r^ is not degree sufficient for Sg. Let p = max{j :
df'* > k Â£ 1}. Since 7r^ is not degree sufficient for Sg but the minimum term of
72
is at least k a(H) bn Â£, we know that p < k ct(H) bH Â£ Thus, 7is
degree sufficient for Kp V Kk_Â£_p.
For j E {a + 1,..., k}, let Fj be a jvertex induced subgraph of H such that
A(Fj) = Vj(H). We will show that if is degree sufficient for Kp V Fk_Â£_p, then 7r
is potentially //graphic. Unfortunately, we cannot simply apply the Bounded Max
Degree Theorem for this graph, because we do not know what a(Fk_Â£_p) is. Instead
we will use the BMDT to show that we have a realization containing Kp V Kk_Â£_p,
and then create a copy of Fk_Â£_p on the vertices of the independent set.
To begin, note that 7r^ satisfies the conditions of the Bounded Max Degree
Theorem (Theorem 4.10) for Kp V Kk_Â£_p, since the minimum term of 7is at
least k a(H) bn Â£ > p = \V{KP V Kk_Â£_p)\ a(Kp V Kk_Â£_p) and dp <
nt (fc/2l) (k2 1) < nÂ£ f[k i p, k Â£).
Let GÂ£ be a realization of 7on the vertex set {ui,... ,vne} that contains Kp V
KkÂ£_p on the vertices {v\)... ,Vkt} such that the first p vertices induce a clique.
Such a realization exists by Lemma 2.12. Delete the vertices of the clique from GÂ£ to
get the graph G'Â£, and let p = it(G'Â£), with the order of the vertices maintained. We
want to show that there is a realization of p that contains a copy of Fk_Â£_p on the
vertex set {vp+i,..., vk_Â£}. Note that since the minimum term of 7r^ is greater than
p, the minimum term of p is at least 1.
To construct the desired realization, place a copy of Fk_Â£_p on the vertices
vp+i,... ,vk_Â£. Since 7r^ is degree sufficient for Kp V Fk_Â£_p, we know p is degree
sufficient for Fk_Â£_p. Thus, the degrees of the vertices in {vp+\,... ,vk_Â£} prescribed
by p are at least as large as the degrees of these vertices in Fk_Â£_p. If, after doing this,
any vertex in {vp+i,..., vkÂ£} does not yet have the degree prescribed by p, join that
vertex to distinct vertices among the remaining nÂ£ [k ) vertices. The sequence
p' obtained by subtracting the degrees of the vertices in the graph constructed so far
from the corresponding terms of p has at least nÂ£ (k ) (k l p){k Â£ 3)
73
positive terms and maximum term at most k Â£ 2. By the ErdosGallai criteria
(Theorem 1.3), the sequence // is graphic. Combining a realization of p! with the
constructed graph gives us the desired realization of p, showing that F^ is potentially
(Kp V Fkip)graphic. Since Kp V Fkep is a supergraph of a graph in X>F (H), Claim
4.3 implies that tt is potentially //graphic.
Stage 3: We have shown that in many cases, we can construct a realization
of 7r that contains either a supergraph of H or Kk_a_i V Ka+i. This gives us the
conclusions in Parts 1 and 3 of the lemma. Now we will show that in the remaining
case, that is when F^ is not degree sufficient for Kp V Fk_Â£_p, we have 7r 7r/1 < tn
for some F G V(H), establishing Part 2.
To do this, we first create a sequence rj from F^ by adding Â£ vertices to a real
ization of F^ and taking the degree sequence of the resulting graph. Then we will
compute
h~v\\,
\\q TTkÂ£p{H, rn + Â£)\\, and
7rkeP(H,ne + Â£) 7rk_e_p(H,n)\\.
After showing that 7Tke~p(H, n) G V(H), we will use the triangle inequality to show
that 117r 7Tkip(H,n)\\ < tn.
We begin by noting that if F^ is not degree sufficient for Kp V Fk_Â£_pi then it
has the following properties:
dff* < ng 1 for j < p.
Sp < k Â£ 2 for p + l < j < k Â£ 1.
< Vkip + p 1 for j > k Â£.
74
We thus have the following upper bound on
= (2p + Vkep ~ l)nt + (k Â£ l)(k Â£ 1 VkeP 2p)
= (2p + VkeP ~ 1 )ne + o(n).
We also know, by Claim 4.4, that
(2(fc f) + V7; 1 (Â£ + 1)4 2Â£)nf
> (2(fc Â£) (2(fc Â£p)~ VkeP) l(Â£+ l)5)m
= (2p + Vkep 1 4(Â£ + 1 ))riÂ£.
Let rj be the degree sequence of Ah V G^\ where is a realization of 7r^h Then
a(q) =
> (2p + Vkep ~ 1 S(Â£ + l))ne + 2Â£m + Â£2 Â£.
Now consider the sequence 'kkepi.H, n^+Â£). We need to show that ifkeP(H, riÂ£+
Â£) G V(H). The sum of this sequence is
cr(7rkep(H, ne + Â£)) = (2p + VkeP ~ 1 )ne p2 pV keP + 2Â£ng + Â£2 Â£
> o(rj).
After simplifying, we see that the coefficient of ne + Â£ in cr(irkep(H, ni + Â£)) is
2{Â£ + p) + Vkep ~ 1 If 7rkeP(H, ne + A) is not in V(H), then
(2(^ + p) + Vfc_,_p l)(rq + Â£) < (2(fc T) V,* 1 )(ne + Â£). (4.2)
75
However, we also know that
a(7ckep(H,ne + Â£)) > a(rj) > {2(k i*) V;* 1 (Â£ + l)5)ne + 2Â£ne + Â£2 Â£. (4.3)
Inequalities (4.2) and (4.3) together imply that
(2(k i*)   1 (Â£+l)5)ne + 2Â£ne+Â£2 Â£< (2(k i*) V,* 1 )(ne + t),
which is not true for sufficiently large rq. Thus, irkÂ£p(H, rq + Â£) E V(H).
In order to calculate ?] TvkÂ£p{H,nÂ£ + Â£), note that the only terms of q =
(q,..., ant+Â£) that are greater than those of TTkÂ£P{H, nÂ£ +1) are those aj for which
p + Â£ + l < j < k 1. Thus, we can bound the distance between these sequences
by taking the absolute difference of their sums and adding two times the difference
of the terms in that range. Since the largest terms of rj in that range are at most
k 2, and the corresponding terms of kkÂ£P equal VkÂ£P + p + Â£ 1, we add at
most 2(k Â£ p 1 VkÂ£P)(k Â£ p 1) to the difference of sums. This yields
\\V ~ 7TkÂ£P{H, nÂ£ + Â£)\\ < a(7rk_Â£_p(H, ne + Â£)) a(q) + 2k2
< 5{Â£ + l)riÂ£ + 2k2.
Now we calculate 7r q\\] this is where we use all of the information we gained
in Stage 1 about the structure of 7r. Recall from Equation (4.1) that at most
Â£
1) + 1 +
(3 + k)6 '
rn
1 + 5
25
TT4
n
terms are removed from tt to create 7r^h Of these, the terms that were laid off in
Step (ii) because they were the first term of some 7have high degree, and those
that were removed during Steps (i) and (iii) of the algorithm have very low degree.
76
Theorem 2.17 implies that those terms removed in Steps (i) and (iii) are in fact at
most k 3, and the hypotheses of the algorithm imply that the terms removed in
Step (ii) are at least ne (p,/2iK^2 !) The terms of rj that give the degrees of
vertices in the clique in Ah V G^ correspond to the terms of 7r that were removed
in Step (ii). So the difference between these terms in the two sequences is at most
("fc/2])(^2 !) The terms of 7r that were removed in Steps (i) and (iii) correspond
to zeros in rj. Thus, the difference between these terms is at most k 3. We see that
since ne < nt < n and l < k a(H) 1, we have
,(2 + ka(H))6
ivr 7711 < ( n
1 + 6
\k/2]
(k2 1) (k 3)(k a(H) 1)
k
\k/2]
(k2 l)(k a(H) 1) +
26
TTI
n
(k3 + 2)6
< n
~ 1+5
\k/21
k\
Finally, we know that 7TkeP{H, nÂ£ + Â£) irkep{H, n)  = (n (ne + l))(2(k
i*) + Vi* 1), and by Equation (4.1), this is at most
1 ) + k
(3 + k)6
1 + 6
n
<
k
\k/2]
kA + 2k}
, (3 + k)6
1 + 6
n.
(2 (k
}*) + Vi* 1)
Thus, we have
17T nkeP(H,n)\\ < \\tt rj  +  rj Trkep(H, ne + Â£) 
+ H^FfcÂ£p(H> + Â£) TÂ£kÂ£p(Hj Tl)
(k3 + 2)6
< 6(Â£ + 1 )n + k2 + x + n + 2y^k/2] ^ + 2k
k
2(3 + k)6
1 + 6
n
< ( 6(k a) +
(.k3 + 2)6
1 + 6
2/;2 (3 + k)s
1 + 6
k
U + 2[\k/2l]k4 + k2'
77
Thus, as 5 approaches 0 this expression is less than tn.
From the proof of Lemma 4.11, we see that when H is Type 1, then we only reach
the conclusions in Parts 1 and 2 of the lemma, thus implying that H is ustable. Thus,
Theorem 4.4 is an immediate corollary. When H is Type 2, however, we may end
up in the case where Lemma 4.11 can only guarantee a realization of 7r containing
Kka(H)i V KafyH)+1 Now, to prove Theorem 4.5, we analyze what happens in this
case.
Proof of Theorem 4.5. Let H be a graph that satisfies the hypotheses of
Theorem 4.5. Let e > 0 be given, and let tt = (di,..., dn) be a graphic sequence of
length n that is not potentially fTgraphic and such that a(?r) > (2(k ?'*) + Vi*
1 5)n, where 5 < By Lemma 4.11, either 7r 7r71 < tn for some tt' E V(H), or
tt is potentially (I
Since the former case implies that H is ustable, we may assume the latter case
holds. By Theorem 2.12, we may assume that there is a realization G of tt on the
vertex set {rq,..., vn} such that d(vi) = di for 1 < i < n, the vertices v\,..., Vka{H)i
induce the clique and the vertices Vk~a(H),..., vk form the independent set in the
complete split graph. Let S = {vk_a(H), ,vk}. If S is not an independent set in
G, then {rq,..., vk} induces a supergraph of H, a contradiction. We may therefore
assume that S is an independent set in G.
If every vertex in S has degree kal, then since 7r is nonincreasing, dj < ka 1
for each j > k a. Since na+i(H, n) = ((n (&; a we see
that each term of 7r is less than or equal to the corresponding term of 7fa+i (H,n).
Thus, 117r na+i(H,n)\\ < a(na+i(H, n)) a{7r) < Sn.
If at least two vertices in S, say u and v, have degree at least k a in G, then we
can do an edge exchange to create a supergraph of H, implying that tt is potentially
Hgraphic. Without loss of generality, suppose d(u) > d{v) > k a. Since S is an
independent set in G, there must be vertices
78
that u ~ a\ and v ~ 02 It is possible that a\ = a2. By Theorem 2.17, we may assume
that each vertex in V(G) {ui,..., Vk} has degree at most 2k 4. Thus, there are at
most (2fc)2 + l vertices at distance at most 2 from ci\, and at most (2fc)2 + l vertices at
distance at most 2 from a2. Since n is sufficiently large, there is a vertex w that is at
distance at least 3 from both a\ and a2. Since a{ji) > a(H, n) 8n, that vertex must
have positive degree in G. Let r be a neighbor of w; note that x is not adjacent to
ai or a2, but it is possible that x G {v\}..., Vk} \ {it, v}. We can exchange the edges
itai, na2, and wx for the nonedges uv, wa\, and xa2 to get a realization of 7r that
contains (Kkai V Ka+i) + e, where e is the edge uv in the independent set. Since
H has a set of a + 1 vertices that induces one edge, this shows that 7r is potentially
//graphic.
We are left with the case where exactly one vertex in S has degree greater than
k a 1. We may assume that this vertex is Vk~a, so dk~a > k a but dj < k a1
for all j > k a + 1. For j G {1,..., k a 1}, let Wj be the set of neighbors of
Vka that are not adjacent to Vj. Recall that H C /dfc_Q,_2 V Sbltb2 fr some b\ and &2.
Suppose b\ > 62. If dka > (k a 1)(&2 + 1), then for some p G {1,..., k a 1}
we have \ WP\ > 62. Let x G W,p. Then we can exchange the edges xVka and vpVka+i
for the nonedges xvp and VkaPka+i We can do a similar edge exchange for &2
vertices in Wp and the vertices Vk~a+2, ,Vka+b2 until Vk~a is adjacent to each of
Vka+1, Vka+b2, while vp is no longer adjacent to these vertices. Thus, we create
a realization of 7r where vp and Vka are the centers of a double star Sbltb2 that is
joined to a complete graph on the vertices {v\,... ,Vka1} \ {vp}, which means that
7r is potentially //graphic.
So dka < (k a1)(62 + 1). In this case we can show that 7r 7tq+i(/7, n)  < eri.
Since Tra+i(H, n) = ((n 1 )k~a~1,(k a 1 )ra(fc1)); we see that na+i(H,n)
79
majorizes 7r except at the (k a)th term. Thus,
Ik 7ra+i(H,n)\\ < (
< 5n + (k a 1)62
As 8 becomes smaller, we see that this is less than en, so H is ustable.
4.5 Implications of stability
Once we know that a graph H is ustable, the next step in the stability method is
to use this fact to show that the sequences in V(H) are the only extremal sequences. If
the full set of extremal sequences for a graph is known, then we also know the precise
value of the potential number for that graph. This is one of the main goals in proving
a stability result for an extremal problem. However, in our case the property of not
being erstable is also very interesting, because it draws attention to the existence of
other extremal sequences for such graphs.
We showed in Section 4.2 that if H is a graph of order k and independence
number a that is not a subgraph of Kk_a_2 V SXyV for some x, y, with x + y = a,
then H is not erstable. The reason such a graph is not astable is that the sequence
Pi = ((n 1 )e, , (f1 + l)ra^2), with l = k a 2, is not potentially Hgraphic,
although its sum is arbitrarily close to the potential number for large n. This implies
that the sequences in V(H) are not the only extremal sequences for these graphs.
As an example, consider the complete graph on k vertices, Kk. We have shown
that Kk is not erstable, but it is weakly erstable. This implies that the sequence
7T2(H,n) = ((n 1 )k~2,(k 2)n~k+2), which is the only extremal sequence noted
in the literature, is not the only extremal sequence for Kk. In fact, the sequence
pk~3 = {{n l)fc3, (ra~2+3)2 {k 2)n~k~5) has the same sum as 7r2(f4, n), and is also
not potentially AVgraphic, as shown in Section 4.2. Thus, this sequence also belongs
in the set of extremal sequences for Kk.
80
The fact that Kk is weakly crstable is reflected in the observation that pk3 is
not degree sufficient for Kk, and in fact if there are other extremal sequences for Kk,
they must also fail to be degree sufficient. If a graph H is weakly erstable but not
crstable, any extremal sequence that is not in V(H) must not be degree sufficient
for H. If a graph is not weakly erstable, then any extremal sequences in addition to
those of V(H) may or may not be degree sufficient for the graph. A natural question
to ask is whether the sequence pka2 is the only additional extremal sequence for
graphs that are not crstable, or if others can be found.
Currently we know very little about graphs that are Type 2 but have a set of a +1
vertices that induce a matching of size at least 2. We have shown that if such a graph
is not a subgraph of Kka2 V Sx
results on the crstability of such graphs. Determining more about such graphs is the
next step in our study of crstability.
81
5. PotentialRamsey Numbers
5.1 Introduction
As discussed in Chapter 2, the fccolor graph Ramsey number r(Gi,..., Gk),
where G\,... ,Gk are graphs, is the minimum integer n such that any fcedgecoloring
of Kn yields a monochromatic Gi in color i for some i. The potential version of
this problem was introduced by Busch, Ferrara, Hartke and Jacobson in [9], and
is defined first for two colors (we will discuss a multicolor version in Chapter 6).
Given graphs Gi and G2 and a graphic sequence tt = (d\,... ,cln), we write 7r >
(Gi,G2) if either tt is potentially Gigraphic or 7f is potentially G2graphic, where
7f = (n 1 d\, n 1 d2, , n 1 dn) is the complementary degree sequence of
tt. The potentialRamsey number of G\ and G2, denoted rpot(Gi, G2), is the minimum
integer n such that if tt is a graphic sequence of length at least n, then tt > (Gi, G2).
We can easily show that rpot(Gi,G2) < r(Gi,G2), because a realization of a
graphic sequence of length n can be thought of as giving a 2edgecoloring of Kn.
That is, a realization of tt determines the red edges in Kn, and the complementary
realization of tt determines the blue edges. Thus, if every red/blue coloring of the
edges of Kn produces a red Gi or blue G2, then every graphic sequence of length
n is either potentially Gigraphic or its complement is potentially G2graphic. This
bound is sharp in some cases; in particular we have the following result from [9].
Lemma 5.1 (Busch et al. [9]) Letr = r(Gi,G2) and letG be a graph of order r1
such that Gi ^ G and G2 ^ G. If tt(G) is unigraphic, then r(Gi,G2) = rpot(Gi,G2).
Busch et al. used this and a result of Gerencser and Gyarfas [49] to show that
rpot{Ps, Pt) = r(Ps, Pt) = s + LfJ 1.
Determining the Ramsey number r(Ks, Kt) is one of the foremost open problems
in combinatorics. A wellknown lower bound on r(Kt,Kt) is r(Kt,Kt) > ^t2tP,
proven by Spencer in [114]. However, the potentialRamsey number is comparatively
simple; in fact, Busch et al. determined rpot(Ks, Kt) for all s,t, thus showing that
82
rpot{Kt, Kt) is linear in t (and that the simple bound in Lemma 5.1 can be very far
from sharp).
Theorem 5.2 (Busch et al. [9]) For s > t > 3, rpot(Ks, Kt) = 2s +t 4, except
when s = t = 3, in which case rpot(K3, K3) = 6.
They also determined the potentialRamsey numbers rpot(Cs, Kt) and rpot(Ps, Kt).
Theorem 5.3 (Busch et al. [9]) If s > 3 and _yj > t > 2, then rpot(Cs, Kt) =
s + t 2. If s > 4 and t > _yj > 2, then rpot(Cs, Kt) = 2t 2 + ~~.
Theorem 5.4 (Busch et al. [9]) For s > 6 and t >3,
In this chapter, we present results on rpot(H, G), where H is one of K2, 2K2, K3, P3,
or P4, and G is an arbitrary graph of order t, as well as rpat(H, G) for all H and G of
order at most 4. We will also state and prove the value of rpot(Cs, Ct).
5.2 Preliminaries
We will assume from now on that G is a graph of order t. If H is a nontrivial
graph, then it is easy to see that rpot(H, G) > t, for if tt = (0t_1), tt (H, G). It is
also easy to see that rpot(H,tKi) = rpot(tKi, H) = t. We can thus assume that all
graphs have at least one nontrivial component. We begin with a simple observation
about the potential Ramsey number of graphs with isolated vertices.
Proposition 5.5 Suppose G = G\ U K\ and rpot(H,Gi) = r. If r < R(G), then
rpot{H,G) = \V(G)\, and if r > \V(G)\, then rpot(H,G) = r.
Proof. Let n = R(G). First suppose r < n, and let tt be a graphic sequence of
length n. If tt is not potentially Pgraphic, then we know that F has a realization R
s + t 2 if t < LyJ
2f2+LU if t> LfJ.
83
containing G\ as a subgraph. Since P(Gi) < n, there is a vertex in R that is not
used by the copy of G\. Adding this vertex gives us a copy of G in R, so tt > (H, G).
Since rpot(H,G) > F(G), the conclusion holds.
If r > n, let tt be a graphic sequence of length r. If tt is not potentially 17graphic,
then again 7f has a realization R containing Gi, and since r > n, there is a vertex
in R that is not used by the copy of Gi, so there is in fact a copy of G in R. Thus,
7r > (H, G). Since rpot(H, Gi) = r, we know that rpot(H, G) > r, so rpot(H, G) = r.
We may henceforth assume that G and H do not have any isolated vertices.
5.2.1 General lower bounds
We now determine several lower bounds on rpot(H, G).
Lemma 5.6 If H is a graph of order t that is not a subgraph of Ki t_i and G has no
isolated vertices, then rpot(H, G) > t + 1.
Proof. Consider tt = (t 1, lt_1). This sequence is uniquely realized by the star
Kiji, so 7r is not potentially Hgraphic. The complement of 7r is uniquely realized
by Kt1 + Ki, which can contain no graph on t vertices unless that graph has at least
one isolate. So rpot(H, G) > t + 1.
The maximum size of a matching in the complement of G, a'(G), turns out to
play an important role in determining rpot(H, G). To simplify notation, we will often
write a' in place of cx'(G) when G is understood.
Lemma 5.7 If G is a graph of order t > 3 and H is a graph with a connected
component of order at least 3, then rpot(H,G) > max{2(t a'(G) 1) + l,t}.
Proof. We have already seen that rpot(H, G) > t. Thus we may assume 2{t a'
1) + 1 > t. Let 7T = (I2!*"'1)). This degree sequence is uniquely realized by a
matching of size t a' 1, so it does not contain a copy of any graph H with a
84
component of order at least 3. Any graph on t vertices in the complement of this
realization must use at least a' + 1 pairs of vertices that are endpoints of one of the
edges, which requires a matching of size at least a' + 1 in G. Thus, tt (H, G) and
f'poti.H, G) > 2(t af 1) + 1.
Recall that if the degree sequence of G is (x\,... xt) and tt = (d\,..., dn), then
7r is degree sufficient for G if di > Xi for each i, 1 < i < k. A simple necessary
condition for 7r to be potentially Ggraphic is that tt be degree sufficient for G. Thus,
sequences that fail to be degree sufficient for H and G often play a role in determining
rpot(H,G). In particular, the minimum degree of G becomes important when a'(G)
is large and the lower bound provided by Lemma 5.7 is simply t. When this happens,
the following lemma may be used.
Lemma 5.8 If G is a graph of order t > 5 such that 8(G) > "^] and H is a graph
with three or more vertices of degree at least 2, then rpot(H, G) >t + 2.
Proof. Suppose first that t is odd, so 8(G) > Consider the sequence 7Ti =
((t + 1 d)2,l*1), which is graphic by the ErdosGallai criteria (Theorem 1.3).
Clearly, 7Ti is not potentially iTgraphic because it is not degree sufficient for H.
On the other hand, 7fi is not degree sufficient for G, so ni (H, G).
If t is even, consider the sequence 7r2 = (t + 2 8, t + 1 8, lt_1), which has
7f2 = ((t l)t1, <5 1,8 2). Since 8 > the ErdosGallai criteria again show
that 7r2 is graphic. However, 7r2 is not degree sufficient for H, so is not potentially
Hgraphic, while 7f2 is not degree sufficient for any graph of order t and minimum
degree 8. The result follows.
85
5.2.2 Towards an upper bound
We will often want to show that a graph H is a subgraph of the complement of
another graph, G. To do this, we will look for a mapping / with the property that
if f{u) f(v), then u ooH v. If there is such a mapping, and it is injective, then
HCG.
As we will discuss, when H is P3, P4, or K3, sequences that are not potentially
//graphic must end with a string of ls or 0s. Thus, these sequences are realized
by graphs containing connected components with at least three vertices, and a set of
disjoint edges and vertices. The following results will therefore help us to determine
upper bounds on the potential Ramsey number.
Lemma 5.9 If G is a graph of order t > 3 and R is a graph of order at least
max{2(t a'(G) 1) + l,t} such that A(R) < 1, then G is a subgraph of R.
Proof. First suppose a1 < tj2, so that max{2(f of 1) + 1, t} = 2t 2a! 1. If
\R\ = 2(t of 1) + 1, then there is at least one isolated vertex in R, and at most
t, a' 1 (disjoint) edges. By taking each isolated vertex and at most one endpoint
of each edge, we can find t of independent vertices in R. To find the remaining of
vertices needed for G, we must pick both endpoints of at most of edges in R. We
can then identify the edges of R with a matching in G, and this gives a mapping of
G into R, showing that G C R.
If /? > 2(t a1 1) + 1, there are again at least t a1 independent vertices in
R; choosing the largest independent set in R, at most a' more vertices are needed to
create a copy of G. This means at most a' edges of R must be used, and this can
also be identified with a matching in G as before.
If of = t/2, then max{2(f a1 1) + 1, t} = t. In this case, if \R\ = t, then there
are at most t/2 disjoint edges in R, and the edges of a matching in G can be identified
with these edges and any isolated vertices in R to see that G C R. If \R\ > t, it is
86
easy to see that G C R.
We now have the following easy corollary.
Corollary 5.10 Let G be graph of order t > 3, and R a graph of order at least
2(t a'(G) l) + l such that R = RiUR2, where R\ is the maximal subgraph of R with
A(i?i) < 1. Suppose there is a subgraph G2 of G such that G2 C R2, and let Gi be the
subgraph of G induced by V(G)\V(G2). If\Ri\ > max{2(Gi afGi) 1) +1, Gi},
then G C R.
Proof. Lemma 5.9 shows that if f?i > max{2(Gi a!{Gi) 1) + 1, Gi}, then
Gi C i?j. Since there are no edges between R1 and R2 in R, this implies that G C R.
Lemma 5.11 Let G and R be as given in Corollary 5.10, and suppose \R2\ = q + a
where a = a{R2), and e[ = \R\ (2[t a' 1) + 1). If
(i) a' > L^J and \R\ t > q or
(ii) a' < L^J and q' q + a > 0,
then G C R.
Proof. Choose a set Vi of t a vertices of G so that a'(G[Vi]) is maximized, and
let Gi = G[Vi]. We claim that a'(Gi) = min{ Clearly, this is an upper
bound on the size of a'(Gi). Let M be a maximum matching in G. To maximize
a'(Gi), we begin by selecting for V\ only those vertices that are saturated by M. If
t a > 2a1, then Vi contains all vertices saturated by M, so a'(Gi) > a'. Otherwise,
we can choose up to t a vertices that are saturated by M so that we have at least
L^J edges of M included. In this case, a'{Gf) = L^J
87
Suppose that condition (i) holds. Then i?i = \R\ (q+a) > q+t (q+a) = t a.
Since a' > _^J, we have a'(G\) = which implies that max{2((f a) a'(Gi)
1) + 1 ,t a} = t a, so Corollary 5.10 shows that G C R.
If condition (ii) holds, then a'(Gi) = a'(G), and
\R\  = \R\ (q + a) = q' + 2t 2a'(G) 1 q a
= q' q + a + (2(t a) 2a'(G) 1) > 2(t a) 2a'(G\) 1.
Thus, Corollary 5.10 again shows that G C R.
Finally we present a lemma which can be used when a sequence is not degree
sufficient for a graph with at least three vertices of degree at least 2.
Lemma 5.12 Let G be a graph of order t, and letir = (d\,... ,dn) be a nonincreasing
graphic sequence such that d^ < 1 and n > 2{t a'(G) 1) + 1. If a'(G) < then
IF is potentially Ggraphic.
Proof. Since cfe < 1, 7r = (d\,d2, l^n~2~Â£\0Â£). Every realization of this sequence is a
graph R = R1UR2, where A(f?i) < 1 and R2 is either empty (that is, R\ = R), a star,
two disjoint stars, or a double star. If there is any realization of 7r where A(f?i) = 0,
then there are at least t independent vertices in R and thus G C R (if is potentially
Ggraphic).
If A(Ri) = 1 in every realization of 7r, we can apply Lemma 5.11. Clearly q' > 0
and, since R2 is a star, two stars, or a double star, a > 2 and q < 2. This im
plies that q + a > 0, so part (ii) of the lemma holds. Since a' < ^ implies
2(t a' 1) + 1 > t + 2, part (i) of the lemma also holds. Thus, G C R.
88
5.2.3 Potentially P4graphic sequences
Before we can address potentialRamsey numbers for small graphs, we will prove
the following characterization of potentially P4graphic sequences. This proof serves
as an illustration of common degreesequence techniques such as the use of 2switches
and residual sequences, as discussed in Chapter 1.
Proposition 5.13 If 7r = (di,...dn) is a nonincreasing graphic sequence of length
n > 4, then ir is potentially P^graphic if and only if cl2 > 2 and d4 > 1.
A key step in the proof is the next lemma.
Lemma 5.14 If tt = (d\, d2,..., dm) is a nonincreasing graphic sequence such that
7Ti = (g?2 1, ^3 1, , d^+i 1, d(i1+2,..., dm) is potentially Pngraphic, then n is
potentially Pn+igraphic.
Proof. Let H be a realization of 7Ti containing P, a path of order n. Label the
vertices of P with w\,..., wn, such that Wi ~ wi+4 for 1 < i < n 1. Add a vertex v
of degree d\ to H by making v adjacent to the vertices of degree d2 1, , d^+i 1,
and call this new graph H'. Note that H1 is a realization of tt. If v ~ w\ or v ~ wn,
then there is clearly a Pn+\ in H1. Also note that if v has consecutive neighbors on P,
then by detouring through v we get a Pn+4, so we can assume this does not happen.
Suppose v has two neighbors, x\ and X2, that are not on P. Then v,x\, and
X2 are each not adjacent to w\ or u?2, or we get a longer path. Exchange the edges
vx\ and uqu>2 with the nonedges vw2 and aquq to get a realization of tt in which
X2VW2W3 wn is a path of order n + 1.
Thus we may assume that v has all but at most one of its neighbors on P. Let
Wi and Wj be two neighbors of v on P. If Wi 00 Wj, then we can exchange the edges
WiiWi and vwj with the nonedges vwi\ and WiWj to get v ~ uy_i, which creates a
path of order n + 1. Thus, all neighbors of v on P must be adjacent. This means that
if Wi is adjacent to v, then Wi has at least d\ 2 neighbors on P that are neighbors
89
of v, two neighbors on P that are not neighbors of v, and one neighbor not on P (v
itself) for a degree of at least d\ + \. But every vertex in P[ has degree at most d\, so
this is not possible. Thus, 7r is Pn+igraphic.
Now we can prove the desired characterization. Note that the sequence 7r =
(di,..., dn) is potentially P3graphic if and only if d\ > 2.
Proof of Proposition 5.13. If 7r is P4graphic, then clearly the conditions must
hold.
So assume 7r = (d\, d2,..., dn) with d2 > 2 and d4 > 1. If d2 > 3, then 7Ti (as
dehned in Lemma 5.14) is potentially P3graphic, so Lemma 5.14 shows that 7r is
potentially P4graphic.
If d2 = 2, let V\ and v2 be vertices of degree d\ and d2, respectively, in a realization
p[ of 7r. Let x and y be the neighbors of v2. Suppose that x = v\. If v\ ~ y, then we
have a 3cycle v\v2y in H. Since d4 > 1, there is a fourth vertex w in H and either
w ~ v\, w ~ y, or there is another edge, say wz, in H. In the first two cases, wv\v2y
or wyv\V2 is a P4. In the second case we can exchange the edges wz and V\V2 with
nonedges wv\ and zv2 to get the path zv2yv4.
If V\ 00 y, then since d\> 2 there must be another vertex w that is adjacent to
Vi, and we again get the path wv\v2y.
If v\ 00 v2, but V\ and v2 have a common neighbor, say x, then v\xv2y is a P4.
If they do not have any common neighbors, then there are at least two vertices w,z
adjacent to V\. liw^x (or y) then v\wxv2 is a P4. If not, then we can exchange the
edges v\w and v2x with the nonedges v\v2 and wx to get the path zv\v2y. Thus, 7r is
potentially P4graphic.
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EXTREMALPROBLEMSFORDEGREESEQUENCES by CATHERINEC.ERBES B.A.,CarletonCollege,2005 M.A.,IndianaUniversity,Bloomington,2007 Athesissubmittedtothe FacultyoftheGraduateSchoolofthe UniversityofColoradoinpartialfulllment oftherequirementsforthedegreeof DoctorofPhilosophy AppliedMathematics 2014
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ThisthesisfortheDoctorofPhilosophydegreeby CatherineC.Erbes hasbeenapprovedforthe DepartmentofMathematicalandStatisticalSciences by MichaelJacobson,Chair MichaelJ.Ferrara,Advisor EllenGethner StephenHartke PaulWenger April29,2014 ii
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Erbes,CatherineC.Ph.D.,AppliedMathematics ExtremalProblemsforDegreeSequences ThesisdirectedbyAssociateProfessorMichaelJ.Ferrara ABSTRACT Asequenceofnonnegativeintegers is graphic ifitisthedegreesequenceof somegraph G .Inthiscasewesaythat G isa realization of .Degreesequence analoguesofmanyclassicalproblemsinextremalgraphtheoryappearthroughout theliterature.Inthisthesiswepresentresultsabouttwoextremalproblemsfor degreesequences,thepotentialnumbertheanalogueoftheclassicalTurannumber andthepotentialRamseynumber. Agraphicsequence is potentially H graphic ifthereisarealizationof that contains H asasubgraph.Thepotentialnumberofagraph H ,denoted H;n ,is theminimumevenintegersuchthatanygraphicsequenceoflength n andsumat least H;n ispotentially H graphic.Thepotentialnumberhasbeendetermined asymptoticallyforgeneralgraphs H ,andafamily P H ofextremalsequencesthat achievethisnumberisknown. Givennonincreasinggraphicsequences 1 = d 1 ;:::;d n and 2 = s 1 ;:::;s n wesaythat 1 majorizes 2 if d i s i forall i ,1 i n .In1970,Erd}osshowed thatforany K r +1 freegraph H ,thereexistsan r partitegraph G suchthat G majorizes H .In2005,PikhurkoandTarazgeneralizedthisnotionandshowed thatforanygraph F withchromaticnumber r +1,thedegreesequenceofan F freegraphis,inanappropriatesense,nearlymajorizedbythedegreesequenceof an r partitegraph.Here,wegivesimilarresultsfordegreesequencesthatarenot potentially H graphic.Inparticular,weshowthatif isagraphicsequencethatis notpotentially H graphic,then isclosetobeingmajorizedbyasequencein P H iii
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Thisshowsthatthestructureofsequencesthatarenotpotentially H graphicisclose tothatoftheextremalsequences. Secondly,wegiveastabilityresultforthepotentialproblem,similartothestabilityresultsofErd}osandSimonovitsfortheTuranproblem.Wesaythatagraph H is stableifeverygraphicsequencewithsumcloseto H;n thatisnotpotentially H graphiccanbetransformedintoasequencein P H with o n additionsand subtractions.Weshowthat,incontrasttotheTuranproblem,notallgraphsare stable.Wealsoshowthatalargefamilyofgraphsare stable. Finally,weaddressadegreesequencevariantoftheRamseynumber,recently introducedbyBusch,etal.Givengraphs G 1 and G 2 ,let r pot G 1 ;G 2 betheminimum integer n suchthatforanygraphicsequence oflength n ,either ispotentially G 1 graphicorthecomplementof ispotentially G 2 graphic.Wegiveseverallower boundson r pot G 1 ;G 2 ,andalsodeterminethevaluesof r pot C s ;C t and r pot G 1 ;G 2 when G 1 isaxedgraphoforderatmostfourand G 2 isarbitrary. Theformandcontentofthisabstractareapproved.Irecommenditspublication. Approved:MichaelJ.Ferrara iv
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ToRachel v
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ACKNOWLEDGMENTS First,IwouldliketothankmyadvisorMikeFerrara,forguidingmeonthis journeysincedayone.Thanksisalsoduetothemembersofmycommittee:Mike JacobsonwithoutwhoseadviceIwouldneverhaveeventakengraphtheory,Paul Wenger,EllenGethner,andStephenHartke,whohaveallhelpedmetobecomea bettermathematician.IwouldalsoliketothankRyanMartin,whoalongwithPaul andMike,helpedmetrulyunderstandthenecessityofapreciselywrittenproof,no matterhowuglyitmightbe. IwouldalsoliketothanktheNSFGrant#074324UCDGK12TransformingExperiencesProject"andtheLynnBatemanMemorialFellowshipforfunding portionsofthisresearch. Iwouldnothavemadeittothispointwithoutthesupportofmanypeople.My fellowgradstudents,particularlyJennyDiemunschandTimMorris,myacademic siblings,whosecompanionshipinthisprocessalsosincedayonehasmadeitmuch morefun,ifnotanyeasier;AxelBrandt,BrentThomas,MarkMueller,SamGraeo, andDevonSigler,whowerealwayshappytohelpwithmath,distraction,coee,or drinks,whicheverwasmostneededatthetime;andBreeannFleschandEricSullivan, whoservedasaremindertomethatitispossibletonotonlygraduate,butalsoto ndajobandsucceedaftergradschool.Mymom,dad,andbrothersandsisters, whohavealwaysbelievedinme,andespeciallymysisterAnissa,whosepassionfor ightandstoriesaboutlearningtoyhavebeenmygreatestsourceofmotivation. Andnally,mywifeRachel,whoalwaysknewwhenIneededtoplayvideogamesand whenIneededtodomath,andwhosepatienceandsupporthavemeanttheworldto me. vi
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TABLEOFCONTENTS Tables........................................ix Figures.......................................x Chapter 1.Introduction...................................1 1.1Denitionsandconcepts........................1 1.2Anintroductiontodegreesequences..................3 2.ExtremalProblemsforDegreeSequences...................6 2.1Anoverviewofextremalgraphtheory.................6 2.1.1TheTuranproblem.......................6 2.1.2Ramseytheory..........................8 2.2Degreesequenceproblems........................9 2.2.1Forcibleproblemsversuspotentialproblems..........9 2.2.2Potentially H graphicsequences................13 2.3Thepotentialnumber..........................17 3.TheShapeofGraphicSequencesthatarenotPotentially H Graphic...23 3.1Introduction...............................23 3.2Majorizationofsequencesthatarenotpotentially H graphic....26 3.3Lemmas.................................29 3.4ProofofTheorem3.4..........................30 3.5ProofofLemma3.5...........................38 4.StabilitywithRespecttothePotentialNumber...............52 4.1Graphswithdegreesequencestability.................54 4.2Graphsthatarenot stable......................55 4.3Technicallemmas............................62 4.4ProofsofTheorems4.4and4.5.....................64 4.5Implicationsofstability.........................80 vii
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5.PotentialRamseyNumbers...........................82 5.1Introduction...............................82 5.2Preliminaries...............................83 5.2.1Generallowerbounds......................84 5.2.2Towardsanupperbound....................86 5.2.3Potentially P 4 graphicsequences................89 5.3PotentialRamseynumbersforsmallgraphsversusarbitrarygraphs91 5.4PotentialRamseynumbersforcycles..................97 5.5PotentialRamseynumbersforallgraphsoforderatmost4.....110 6.FutureWork...................................113 6.1Hypergraphicdegreesequences.....................113 6.2FurtherexplorationofpotentialRamseynumbers...........115 References ......................................117 viii
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TABLES Table 4.1Connectedgraphsoforderatmost6thatarenot stable........58 4.2Connectedgraphsoforderatmost6thatarenotweakly stable....59 5.1 r pot H;G forsmall H and r pot C s ;C t ...................109 5.2SmallpotentialRamseynumbers......................110 ix
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FIGURES Figure 1.1A2switch...................................5 2.1Thegraph K i [ K n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 i K i ........................12 3.1Theshapeofagraphicsequence......................23 3.2DierentwaystoreducetermsintheKleitmanWangalgorithm.....40 4.1Agraphwith r +1 > 1and2 i )222(r i 2 ................54 6.1Theedgeexchange e u v e 0 ..........................114 x
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1.Introduction 1.1Denitionsandconcepts A graph G consistsofa vertexset V G andan edgeset E G ,whereeachedge isasetofvertices.Throughoutthemajorityofthispaper,wewillrestrictourselves tostandardgraphs,inwhicheachedgecontainsexactlytwodistinctvertices.Hypergraphs,inwhichanedgemaycontainmorethantwoverticesorevenasinglevertex, willbediscussedbrieyinthelastchapter.Inthisthesis,wewillonlyconsidernite simplegraphs,meaningthatboththevertexandedgesetsarenite,andthereare norepeatededgesorloopsedgesconsistingofonevertexmultipletimes. Todenoteedgesin G ,wewillwrite xy or e = xy foranedgewithendpoints x and y ,andsimply e iftheendpointsarenotrelevant.If x and y areverticesof G and xy 2 E G ,wesaythat x is adjacent to y ,andsometimeswrite x y .When morethanonegraphisunderconsideration,wemayspecifythat x isadjacentto y in G bywriting x G y .If xy 62 E G wesometimeswrite x 6 y or x 6 G y The degree ofavertex x ,denoted d x or d G x ,isthenumberofedgescontaining x ,or,inthecaseofsimplegraphs,thenumberofverticesadjacentto x .The neighborhood of x isthesetofverticesadjacentto x ,andiswritten N x or N G x So,weseethat d G x = j N G x j .Thesubscript G isusedtospecifyaparameter withinthegraphorsubgraph G .Whenthereisonlyonegraphbeingdiscussedor itisunderstood,wedonotusethesubscript.Themaximumdegreeandminimum degreeof G aredenoted G and G ,respectively. Agraph H isa subgraph of G if V H V G and E H E G .If H is asubgraphof G ,itiscalled spanning if V H = V G ,and induced ifforeach x;y 2 V H xy 2 E H ifandonlyif xy 2 E G .If H isaninducedsubgraph of G ,wewrite H G .If X V G ,thenthesubgraphof G inducedby X denoted G [ X ],isthegraph G )]TJ/F15 11.9552 Tf 12.496 0 Td [( V G n X ,whichhasvertexset X andedgeset f xy 2 E G j x;y 2 X g 1
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Thereareseveralwaystocreatenewgraphsfromothergraphs.Onesuchway istotakethe complement of G ,denoted G ,whichhas V G = V G and e 2 E G ifandonlyif e= 2 E G .The disjointunion ofgraphs G and H isdenoted G [ H andhas V G [ H = V G [ V H and E G [ H = E G [ E H .The join of G and H ,denoted G H hasvertexset V G [ V H ,andedgeset E G H = E G [ E H [f xy j x 2 V G and y 2 V H g A path isasimplegraphinwhichtheverticescanbeordered v 1 ;:::;v n suchthat v i v i +1 foreach i 2f 1 ;:::;n )]TJ/F15 11.9552 Tf 12.405 0 Td [(1 g .If,inaddition, v n v 1 ,thenthisgraphisa cycle .Wewilloftendescribeapathorcyclebyitsvertices,bywriting v 1 v 2 v n for apathand v 1 v 2 v n v 1 foracycle.Let P n denotethepathon n vertices,and C n the cycleon n vertices.Wesaythatagraphis connected ifthereisapathbetweenany pairofvertices.A tree isaconnectedgraphthatcontainsnocycles,andwesaythat aconnectedcomponentofagraphis nontrivial ifithasatleastoneedge. Asetofverticesthatarepairwiseadjacentisknownasa clique .Thecomplete graphon n vertices,denoted K n ,isacliqueoforder n .Asetofverticesthatare pairwisenonadjacentisan independentset ,andthemaximumorderofanindependent setin G isdenoted G .A matching isasetofdisjointedges,thatisasetofedges withnoendpointsincommon,andthemaximumsizeofamatchingin G isdenoted 0 G Agraphis bipartite ifthevertexsetcanbepartitionedintotwoindependentsets, and k partite ifthevertexsetcanbepartitionedinto k independentsets.Wewill use K r;s todenotethe completebipartitegraph with r verticesinonepartitesetand s verticesintheother. Acoloringoftheverticesofagraphiscalled proper ifadjacentverticesreceive distinctcolors.The chromaticnumber of G ,denoted G ,istheminimum k such that G hasapropercoloringusing k colors. Othertermswillbedenedasnecessary,andfortermsnotdenedinthisdisser2
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tation,see[122]. 1.2Anintroductiontodegreesequences The degreesequence ofagraphisalistofthevertexdegreesofthegraph.A sequenceofnonnegativeintegers iscalled graphic ifitisthedegreesequenceof somegraph G .Inthiscasewesaythat G realizes orisa realization of ,andwe write G = G ,or = G .Unlessotherwisenoted,wewillassumethatallgraphic sequencesarewritteninnonincreasingorder.Wewilloftenwrite = d m 1 1 ;:::;d m t t toindicatethateachdegree d i appears m i timesin ,althoughwewillgenerally suppressthemultiplicityof d i if m i =1. Givenagraph,determiningitsdegreesequenceisasimpleexercise.Ontheother hand,determiningwhenagivensequenceisgraphicismoredicult.Therst,and perhapsthesimplest,characterizationofgraphicsequencesisdueindependentlyto HavelandHakimi. Theorem1.1Havel[59],Hakimi[56] Let = d 1 ;:::;d n beanonincreasing sequenceofnonnegativeintegers.Thesequence isgraphicifandonlyifthesequence 0 = d 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ;:::;d d 1 +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ;d d 1 +2 ;:::;d n isgraphic. KleitmanandWanggavethefollowinggeneralizationofthistheorem. Theorem1.2KleitmanandWang[70] Let = d 1 ;:::;d n beanonincreasingsequenceofnonnegativeintegers,andlet i 2 [ n ] .If i isthesequencedened by i = 8 > > < > > : d 1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ;:::;d d i )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 ;d d i +1 ;:::;d i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ;d i +1 ;:::;d n if d i
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TheHavelHakimiandKleitmanWangTheoremsgiverisetoecientalgorithms totestforgraphicality.Theyalsointroducesocalledresidualsubsequencetechniquesforanalyzinggraphicsequences.Thesetechniquesinvolveusinginformation thatweknowabouttheresidualsequence,suchascertainpropertiesofitsrealizations,togainknowledgeabouttheoriginalsequence.Forexample,ifthesequence 0 ispotentially H graphicapropertywhichwillbedenedinSection2.2.2,thenwe knowthat mustalsobepotentially H graphic. InChapter3,weuseacarefulexaminationoftheKleitmanWangalgorithmto createaspecicrealizationofasequencethatmeetscertaincriteria.InChapter4,we willgivesomesimplecorollariesoftheKleitmanWangalgorithmthatgiveusmore informationabouttheresidualsequencesobtainedfromapplyingit. Inadditiontotheseresidualsequencecharacterizationsofgraphicsequencesthere areothers,manyofwhichtaketheformofsystemsofinequalities.Thebestknown oftheseisduetoErd}osandGallai. Theorem1.3Erd}osandGallai[37] Anonincreasingsequence = d 1 ;:::;d n ofnonnegativeintegersisgraphicifandonlyif P n i =1 d i isevenand,forall p 2 f 1 ;:::;n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 g p X i =1 d i p p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ n X i = p +1 min f d i ;p g : .1 In[60],SierksmaandHoogeveengivesixothercharacterizationsofgraphicsequences, andprovethattheyareallequivalent.Manyothershavegivenothercharacterizations orsucientconditionsforgraphicsequences[66,71,115],aswellasimprovementson theErd}osGallaiCriterion,intheformofareductionofthenumberofinequalities thatmustbechecked[116,157]. Agraphicsequencemayhavemanydierentrealizations.Acommontechnique thatisusedtotransformonerealizationintoanotherisknownasthe edgeexchange or 2switch .If G isarealizationof withedges xy and uv suchthat xu and yv are 4
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Figure1.1: A2switch notedges,wecandeletetheedges xy and uv andreplacethemwiththeedges xu and yv seeFigure1.1.Thismaintainsthedegreeofeachvertex,sothenewgraph isalsoarealizationof Petersen[101]showedthatgivenanypairofrealizationsofagraphicsequence, onecanbeobtainedfromtheotherbyasequenceof2switches.Thistechniquecan beusedtoproveboththeHavelHakimiandKleitmanWangTheorems,aswellas manyotherresultsaboutdegreesequences;aparticularlyniceexampleofthiscanbe foundin[45]. An isomorphism fromagraph G toagraph H isabijection f : V G V H suchthat uv 2 E G ifandonlyif f u f v 2 E H .Ifthereisanisomorphismfrom G to H ,wesaythat G is isomorphicto H ,orthattheyareinthesame isomorphism class .Whilea2switchalwayschangesthestructureofalabeledgraphonewhere thenamesoftheverticesareimportant,itmayinfactpreservetheisomorphism classoftheunlabeledgraph.Recently,Barrus[4]determinedsucientconditions fora2switchtochangetheisomorphismclassofagraph. 5
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2.ExtremalProblemsforDegreeSequences 2.1Anoverviewofextremalgraphtheory Extremalgraphtheorycanbethoughtofasthestudyofthresholds.Inparticular, welookforthresholdsongraphinvariantsthatguaranteethatagraphhasacertain property.Thatis,givenaninvariant i G ,aclassofgraphs F ,andagraphproperty P ,themostgeneralformulationofanextremalgraphtheoryproblemis:Whatis theminimumvalue m ormaximumvalue m 0 suchthateverygraph G 2F with i G >m i G
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NotethattheTurangraphhas )]TJ/F20 7.9701 Tf 5.479 4.379 Td [(n 2 )]TJ/F20 7.9701 Tf 13.15 5.698 Td [(r n + p )]TJ/F20 7.9701 Tf 6.586 0 Td [(k 2 = )]TJ/F17 7.9701 Tf 13.151 4.707 Td [(1 r n 2 2 + O n edges. Whiletheexactvalueoftheextremalfunctionisknownforveryfewgraphsfor someexamples,see[11,18,36,103,117],in1966Erd}osandSimonovits[40]extended previousworkofErd}osandStone[39]anddeterminedex H;n asymptoticallyfor arbitrarynonbipartite H Theorem2.4TheErd}osStoneSimonovitsTheorem[39,40] If H isagraph withchromaticnumber H = r +1 2 ,then ex H;n = 1 )]TJ/F15 11.9552 Tf 13.151 8.088 Td [(1 r n 2 + o n 2 : Thisshowsthatasymptotically,theTurangraphshavethepropernumberofedges tobetheextremalgraphsforany H ,notjustforcompletegraphs.Subsequently, Simonovits[113]andErd}osindependentlyprovedthefollowing,sometimesknownas theFirstStabilityTheorem. Theorem2.5Simonovits[113] Let H beagraphwith H = r +1 .Forevery > 0 ,thereexistsa > 0 andan n suchthatif n>n and G isan n vertex H free graphsuchthat j E G j ex H;n )]TJ/F19 11.9552 Tf 11.955 0 Td [(n 2 ; then G canbeobtainedfrom T n;r bychangingatmost n 2 edges. Withthistheorem,Simonovitsintroducedthestabilitymethod"forsolving extremalproblems.Recallthatageneralextremalproblemseekstondtheminimum value m ofaparameterorgraphinvariant i G overasetofobjectsinacertain class F thatwillguaranteethateveryobjectforwhich i G >m hasproperty P .The rststepofthestabilitymethodistondanasymptoticsolutionfortheextremal problem,whichusuallyalsodeterminesaset C ofextremalobjects.Ifthissetof extremalobjectscanbeshowntobethecompleteset,thenwecanndanexact 7
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answertotheproblem,soprovingthisistheultimategoal.Thenextstepisto proveastabilityresult,whichshowsthatforeveryobject G in F forwhich i G is closetobutnotgreaterthan m ,thestructureof G issimilartothatoftheobjects in C .Usingthestabilityresult,thenextstepistoshowthateveryextremalobject isinfactin C .Thisyieldstheexactanswertotheextremalproblem. Simonovitsusedthismethodtodetermineexactlytheextremalnumberfor p disjointcopiesof K r ,or pK r [113].Sincethen,stabilitymethodshavebeenusedto attackawidevarietyofextremalproblemssee[7,68,97,99,102]forexamples.Recently,stabilitymethodshavealsobeenusedtoapproachproblemsinRamseyTheory [53,54,100],andhavealsoprovenparticularlyhelpfulinstudyingthehypergraph Turanproblemsee[1],[96],or[98]forsomeexamples. 2.1.2Ramseytheory Ramseytheoryfocusesonadierentkindofextremalproblem.Initsoriginal, mostgeneralform,Ramsey'sTheorem[106]isaboutsets.Givenaset S ,wewrite )]TJ/F20 7.9701 Tf 5.479 4.379 Td [(S r todenotethesetof r elementsubsetsof S .Aset T S iscalled i homogeneous if,undersomecoloringoftheelementsof )]TJ/F20 7.9701 Tf 5.479 4.379 Td [(S r ,allofthe r elementsubsetsof T receive color i .If n isapositiveinteger,welet[ n ]= f 1 ;:::;n g .Withthesedenitions,we cannowstateRamsey'sTheorem. Theorem2.6Ramsey[106] Givenpositiveintegers r and p 1 ;:::;p k ,thereexists aninteger N suchthatevery k coloringof )]TJ/F17 7.9701 Tf 5.479 4.379 Td [([ N ] r yieldsan i homogeneoussetofsize p i forsome i Graphtheoryprovidesaniceillustrationofthecase r = k =2.Inthiscase thesetsoforder2areedges,andweconsidera2coloringoftheedgesofacomplete graph.Thus,an i homogeneoussetofsize p i isacompletegraphoforder p i inwhich alledgeshavethesamecolor.Inthislanguage,Ramsey'sTheoremsaysthatgiven positiveintegers s and t ,thereisan n suchthatanyred/bluecoloringoftheedges 8
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of K n yieldseitherared K s orablue K t .Thenumber n isthencalledthe Ramsey number andisdenoted r s;t Ramseynumbersarenotoriouslydiculttocompute.Infact,theexactvalue of r s;t isknownonlyfor s =3and t 2f 3 ;:::; 9 g ,orwhen s =4and t =4,or s =4and t =5[105].Asymptotically,weknowtheRamseynumber r ;t withina constantfactor: ct 2 log t r ;t c 0 t 2 log t ; wherecurrentlythebestknownvaluesoftheconstantsare c =1 = 4and c 0 =1see [105]. Aslightrelaxationofthisproblemistolookformonochromaticcopiesofgraphs otherthancompletegraphs.Givengraphs G 1 ;:::;G k ,the k colorgraphRamsey number r G 1 ;:::;G k ,istheminimuminteger n suchthatany k coloringof K n yields amonochromatic G i incolor i forsome i .Thecase k =2isthemostwellstudied, althoughtherearesomeresultsfor k 3whenthegraphsaresmallorrelatively simplesee[105]forathoroughsurvey. InChapter5wewilldiscussadegreesequencevariantofgraphRamseynumbers. 2.2Degreesequenceproblems 2.2.1Forcibleproblemsversuspotentialproblems Thegoalofthisdissertationistoexaminedegreesequenceanaloguesofclassical extremalproblems.Beforewedothat,however,wewilldiscussdegreesequenceproblemsingeneral,beyondthesimplecharacterizationproblemsdiscussedinChapter1. Recallthatagraphicsequencecanhavemanydierentrealizations.Mostdegreesequenceproblemsdealwithshowingthatthefamilyofrealizationsofasequencehas certainproperties.Degreesequencequestionscanbecategorizedaseither forcible problems or potentialproblems .Aforcibleproblemasksforconditionsonagraphic sequencethatguaranteethat every realizationofthesequencehasacertainproperty. 9
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Potentialproblems,ontheotherhand,seekconditionswhichguaranteethatthereis some realizationofthesequencewiththedesiredproperty.Thatis,givenagraphic sequence andagraphproperty P ,wesaythat is forcibly P graphic ifevery realizationof hasproperty P ,and potentially P graphic ifsomerealizationof has property P Theextremalproblemsthatarethefocusofthisdissertationarepotentialproblems,butwewillgiveafewforcibleproblemsheretoillustratethedierence.Often, forcibledegreesequenceresultsdonotlooklikedegreesequenceresultsatall.For example,considerthefollowingtheoremofDirac: Theorem2.7Dirac[32] If G isasimplegraphwith n verticessuchthat n 3 and G n= 2 ,then G isHamiltonian. Wecanrephrasethisintermsofdegreesequencesas: Theorem2.8Theorem2.7rephrased Let = d 1 ;:::;d n beanonincreasing graphicsequencewith n 3 .If d n n= 2 ,then isforciblyHamiltonian. AnotherresultonforciblyHamiltoniansequencesisduetoChvatal: Theorem2.9Chvatal[26] Let G beasimplegraphwithvertexdegrees d 1 d n ,where n 3 .If ii or d n )]TJ/F20 7.9701 Tf 6.586 0 Td [(i n )]TJ/F19 11.9552 Tf 12.518 0 Td [(i ,then G is Hamiltonian. While,forthepropertyofHamiltonicity,weonlyhavesucientconditionsfor forcibly P graphicsequences,characterizationsofforcibly P graphicsequenceshave beenfoundformanyothergraphproperties.Chernyak,Chernyak,andTyshkevichcharacterizedforciblychordal,stronglychordal,interval,andtriviallyperfect sequencesin[22]andcharacterizedthedegreesequencesofcomparabilitygraphsin [118].S.B.Raohascharacterizedforciblylinegraphic[109],forciblytotalgraphic, forciblyselfcomplementary,andforciblyplanargraphicsequences[110].Theselast 10
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threeresultsarestatedwithoutproofinhissurveyofpotentialandforcibledegreesequenceresults[110],andhealsodevelopedatheoryforprovingcharacterizations offorcibly P graphicsequenceswhen P isahereditaryproperty[111].Aproofofthe characterizationofforciblyplanargraphicsequences,byZverovich,appearsin[154]. Zverovichhasalsocharacterizedforcibly3colorable[155]andforcibly2matroidal graphicsequences[156].Choudumprovedacharacterizationofforciblyouterplanar sequences[23],andgavesucientconditionsforasequencetobeforciblyconnected [24]. Thestudyofforcibly P graphicsequenceshasbeenchanginginrecentyears. HammerandSimeonegaveacharacterizationofsplitgraphsintermsoftheirdegree sequences[57].Thereisalsoacharacterizationofsuchgraphsintermsofforbidden subgraphsc.f.[5].Inspiredbythis,Barrus,Kumbhat,andHartke[5]investigated degreesequenceforcingsets,thatis,asetofgraphs F suchthatifanyrealization ofagraphicsequence is F free,theneveryrealizationofthatsequenceis F free. Baueretal.gavenecessaryandsucientconditionsforasequencetobeforcibly t tough[3],andsucientconditionsforasequencetobeforcibly k factorable[2]. Theconditionsfor t toughnessareadirectextensionoftheconditionsinTheorem 2.9,bothofwhichareshowntobebestmonotone,apropertywhichwewilldiscuss below.Thesekindsofresults,whichmovebeyondsimplecharacterizations,show thatthestudyofforcibledegreesequenceproblemsisfarfromcomplete. Tounderstandwhatismeantbyabestmonotonetheorem,werstneedadenition.Giventwosequences = x 1 ;:::;x n and 0 = y 1 ;:::;y n ,bothineithernonincreasingornondecreasingorder,wesaythat majorizes 0 if x i y i foreach i; 1 i n .Theorem2.9isbestpossibleinthesensethatifasequencefailsthecondition atsomeposition i ,thenitismajorizedbyasequencethathasanonHamiltonian realization.Inparticular,itismajorizedby = i i ; n )]TJ/F19 11.9552 Tf 12.082 0 Td [(i )]TJ/F15 11.9552 Tf 12.082 0 Td [(1 n )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 i ; n )]TJ/F15 11.9552 Tf 12.082 0 Td [(1 i ,which isuniquelyrealizedbythegraph K i [ K n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 i K i seeFigure2.1.Everygraphic 11
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Figure2.1: Thegraph K i [ K n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 i K i sequencethatmajorizesthisone,however,doessatisfytheconditionsofTheorem 2.9,andhenceisforciblyHamiltonian.Thisiswhatismeantbyabestmonotone theorem;insomesenseitisthestrongestpossibledegreesequenceconditionthat guaranteesHamiltonicity.Asmentionedabove,bestmonotonetheoremsforseveral othergraphpropertieshavebeenfound,andthisisanactiveareaofresearch. Nowweturnourattentiontopotentialdegreesequenceproblems.Ifwelook simplyatgraphproperties,thereareseveralresultsonpotentially P graphicsequences.Therearecharacterizationsofpotentially P graphicsequencesforeachof thefollowingproperties: k edgeconnected[34], k connected[120], k factorable[72] andconnected k factorableifthesequenceisalreadypotentially k factorable[108], andselfcomplementary[30].Infact,selfcomplementary"istheonlypropertyfor whichthereisacharacterizationofbothforciblyandpotentially P graphicsequences. Therearealsopotentialversionsofclassicalgraphtheoryproblems.Wepresent justafewexamplesbeforeturningtotheonethatisthefocusofthisdissertation. OurrstexampleisapotentialversionofHadwiger'sconjecture,alongstanding openproblemingraphtheory. Conjecture2.10Hadwiger'sConjecture[55] If G k ,then G containsa K k minor. 12
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AslightrelaxationofHadwiger'sConjectureisHajos'Conjecture,whichsaysthata graphwithchromaticnumberatleast k musthavea K k subdivision.Hajos'Conjecturehasbeenshowntobefalseforalmostallgraphssee[95]p.4556foranice proof.However,DvorakandMohar[33]recentlyprovedthatthepotentialversion ofHajos'Conjectureistrue.Thatis,theyshowedthatif isthemaximum valueof G overallrealizations G of ,thenforanygraphicsequence ,thereisa realizationof thatcontainsasubdividedcompletegraphoforder Anotherexampleofapotentialversionofaclassicalproblemisavariantofthe GraphMinorTheoremofRobertsonandSeymour[112],whichstatesthatinevery innitesetofgraphs,thereisapairofgraphssuchthatoneisaminoroftheother. Thisisnottrue,however,ifminor"isreplacedbyinducedsubgraph."Forthe potentialversion,S.B.Rao[111]conjecturedthatinanyinnitesetofgraphs,there isapairofgraphs,say G and H ,suchthat G hasarealizationcontaining H as aninducedsubgraph.ThiswasrecentlyprovedbyChudnovskyandSeymour[25]. TherearealsoknownresultsforpotentialversionsoftheErd}osSosconjecture ontrees[137],theSauerSpencergraphpackingproblem[10],andaconjectureof BollobasandScottaboutspanningbipartitegraphs[58]. 2.2.2Potentially H graphicsequences Thepropertythatwewearechieyconcernedwithisthatofhavingasubgraph isomorphictoagraph H .Ifsomerealizationof hasthisproperty,wesaythat is potentially H graphic .Thestudyofthispropertywillultimatelyleadustoa potentialversionoftheTuranproblem,butrstwegivesomeusefulresultsabout potentially H graphicsequences. Whenasequenceispotentially H graphic,itisoftennicetoknowsomething abouttherealizationsthatcontain H .Thenexttwotheorems,whichwewilluse manytimes,tellusthatwecanndrealizationsofpotentially H graphicsequences thathaveniceproperties.Therstshowsthatwecanndoursubgraph H ina 13
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realizationwhere H isontheverticesofhighestdegree. Theorem2.11Gould,Jacobson,Lehel[50] If = d 1 ;:::;d n ispotentially H graphic,thenthereexistsarealization G of suchthattheverticesof H havethe j V H j largestdegreesof If H isacompletesplitgraph, K r K t ,thenwecandrawastrongerconclusion. Theorem2.12Yin[127] Anonincreasingsequence = d 1 ;:::;d n ispotentially K r K t graphicifandonlyifthereisarealizationof containingacopyof K r K t suchthattheverticesofthecompletegraphoforder r havedegrees d 1 ;:::;d r andtheverticesoftheindependentsetoforder t havedegrees d r +1 ;:::;d r + t Let H beagraphwithdegreesequence H .If H = s 1 ;:::;s k ,thenwe say = d 1 ;:::;d n is degreesucientfor H if d i s i foreach i; 1 i k .If bothsequencesarenonincreasing,thenitisclearthatbeingdegreesucientfor H is anecessaryconditionforasequencetobepotentially H graphic.However,itisnot sucient;forexample,thesequence ; 2 ; 2 ; 2 ; 2isdegreesucientfor K 3 ,butthe uniquerealizationofthissequenceis C 5 ,whichdoesnotcontain K 3 BecausetheywillbeneededinChapter5,wepresentheresomecharacterizations ofpotentially H graphicsequencesforsmall H .Theseresultsservetoillustratethe pointthatdegreesuciencyfor H israrelyenoughtoguaranteethatasequencehas arealizationcontaining H ;thereareusuallyexceptionalfamiliesofsequencesthat preventtheobviousnecessaryconditionfrombeingsucient. Theorem2.13Luo[92] Let = d 1 ;d 2 ;:::;d n beagraphicsequencewith n 3 .Then ispotentially K 3 graphicifandonlyif d 3 2 exceptfortwocases: = 4 and = 5 Theorem2.14Luo[92] Let = d 1 ;d 2 ;:::;d n beagraphicsequence,wherethe d i areinnonincreasingorder.Then ispotentially C 4 graphicifandonlyifthe 14
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followingconditionshold: 1. d 4 2 2. d 1 = n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 implies d 2 3 3.If n =5 ; 6 ,then 6 = n Theorem2.15Luo[92] Agraphicsequence = d 1 ;d 2 ;:::;d n ispotentially C 5 graphicifandonlyif satisesthefollowingconditions: 1. d 5 2 and 6 = n for n =6 ; 7 2.For i =1 ; 2 d 1 = n )]TJ/F19 11.9552 Tf 11.955 0 Td [(i implies d 4 )]TJ/F20 7.9701 Tf 6.586 0 Td [(i 3 3.If = d 1 ;d 2 ; 2 k ; 1 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 ,then d 1 + d 2 n + k )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 Thefollowingcharacterizationofpotentially K k graphicsequencesisduetoRao [107],althoughhispaperwasneverpublished.KezdyandLehel[69]gaveaproofusing networkows,andYingaveaconstructiveproofin[128].Foragraphicsequence welet denotethesumofthetermsof Theorem2.16Rao[107] Let n k and = d 1 ;:::;d n beanonincreasingsequenceofnonnegativeintegers. ispotentially K k graphicifandonlyifthefollowing conditionshold: i d k k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ii iseven, iiiForany s and t with 0 s k and 0 t n )]TJ/F19 11.9552 Tf 11.956 0 Td [(k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 s X i =1 d i + t X i =1 d k + i s + t s + t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 + k X i = s +1 min f s + t;d i )]TJ/F19 11.9552 Tf 11.955 0 Td [(k +1+ s g + n X i = k + t +1 min f s + t;d i g : 15
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Thistheoremisdiculttouseinpractice,asitinvolvesevaluatingalargeset ofinequalities.Thenexttheoremgivessimplesucientconditionsforasequenceto bepotentially K k graphic.Thesimplicityoftheseconditionsmakethemmuchmore practicaltouse,aswedemonstrateseveraltimesthroughoutthisdissertation. Theorem2.17YinandLi[135] Let = d 1 ;:::;d n beanonincreasinggraphic sequenceandlet k beapositiveinteger. aIf d k k )]TJ/F15 11.9552 Tf 12.487 0 Td [(1 and d i 2 k )]TJ/F15 11.9552 Tf 12.486 0 Td [(1 )]TJ/F19 11.9552 Tf 12.486 0 Td [(i for 1 i k )]TJ/F15 11.9552 Tf 12.487 0 Td [(2 ,then ispotentially K k graphic. bIf d k k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 and d 2 k k )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 ,then ispotentially K k graphic. Ascanbeseenfromtheprevioustheorems,characterizationsofpotentially H graphicsequencesareoftenhighlytechnical.However,therearesuchcharacterizationsforalargenumberofgraphsandgraphfamilies.Inparticular,potentially C k graphicsequenceswerecharacterizedforvarious k in[13,92,125,126]and[145]. For k<` ,graphicsequenceswitharealizationcontainingcyclesofeachlengthbetween k and ` ,inclusive,areknownaspotentially k C l graphic.Thesehavebeen characterizedfor k =3and ` =4 ; 5 ; 6,for k =4and ` =5,andfor k =5and ` =6 [21,132].Characterizationsof K r graphicand K r )]TJ/F19 11.9552 Tf 11.69 0 Td [(H graphicsequencesforsmall r havebeenstudiedbymanydierentauthors.Amongthesearecharacterizationsfor K 4 [93]; K 4 )]TJ/F19 11.9552 Tf 12.481 0 Td [(e [41,75]; K 5 )]TJ/F19 11.9552 Tf 12.481 0 Td [(e K 5 ,and K 6 [148]; K 5 )]TJ/F19 11.9552 Tf 12.481 0 Td [(H where H isoneof C 4 [63], P 4 ;P 3 [ K 2 ;K 3 ;K 1 ; 3 ; 2 K 2 [62], Z 1 = K 4 )]TJ/F19 11.9552 Tf 12.223 0 Td [(P 3 thepaw[80]; P 5 and Y 4 ,where Y 4 isthetreewithdegreesequence ; 2 ; 1 ; 1 ; 1[64]; K 6 )]TJ/F19 11.9552 Tf 12.213 0 Td [(C 6 and K 6 )]TJ/F15 11.9552 Tf 12.212 0 Td [(2 C 3 = K 3 ; 3 [61], K 6 )]TJ/F19 11.9552 Tf 12.401 0 Td [(K 3 [150],and K 6 )]TJ/F19 11.9552 Tf 12.4 0 Td [(C 5 [124].Forlargercompletegraphs,wehavecharacterizationsfor K r +1 )]TJ/F19 11.9552 Tf 12.735 0 Td [(e [146]; K r +1 )]TJ/F19 11.9552 Tf 12.736 0 Td [(P 3 and K r +1 )]TJ/F15 11.9552 Tf 12.735 0 Td [(2 K 2 [121] K 6 )]TJ/F15 11.9552 Tf 12.735 0 Td [(2 K 2 was donein[91].Acharacterizationofpotentially K 1 ; 4 + e graphicsequencesisgivenin [80],andmoregenerallyforpotentially K 1 ;t + e graphicsequencesin[16].Finally,we havesomecharacterizationsformultipartitegraphs,inparticularfor K 2 ; 3 and K 2 ; 4 16
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[131]; K 1 ; 1 ;s for s =4 ; 5[151]and s =6[153];andnallyfor K 1 ; 1 ;s when s 2and n 3 s +1[130].Similartothecaseforcompletegraphs,thereisacharacterization forasequencetohavearealizationcontainingacompletesplitgraphusingaset ofinequalitiessimilartoTheorem2.16in[127],andsimplersucientconditionsare givenin[129]. Thereis,however,anotherwaytoguaranteethatagraphicsequencehasarealizationcontainingaspecicsubgraph.Forthat,weturnatlasttothepotential number. 2.3Thepotentialnumber ThedegreesequenceanalogueoftheTuranproblemwasintroducedbyErd}os, Jacobson,andLehelin1991[38]. Problem2.18 Determine H;n ,theminimumevenintegersuchthatevery n term graphicsequence with H;n ispotentially H graphic. Wereferto H;n asthe potentialnumber or potentialfunction of H .When theyproposedtheproblem,Erd}os,Jacobson,andLehelconjecturedthat K k ;n = k )]TJ/F15 11.9552 Tf 10.781 0 Td [(2 n )]TJ/F19 11.9552 Tf 10.781 0 Td [(k +1+2,basedonthegraphicsequence n )]TJ/F15 11.9552 Tf 10.781 0 Td [(1 k )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 ; k )]TJ/F15 11.9552 Tf 10.781 0 Td [(2 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(k +2 which isuniquelyrealizedby K k )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 K n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k +2 .Theyprovedthisconjecturefor k =3.Gould, Jacobson,andLehel[50]andLiandSong[86]provedthecase k =4independently, andthenLiandSongproveditfor k =5[87]twoyearslater.Finally,Li,Song,and Luosettledtheconjecturefor k 6and n )]TJ/F20 7.9701 Tf 5.479 4.379 Td [(k 2 +3[88]. Clearly,ifasimplecharacterizationofpotentially H graphicsequencesisknown foragiven H thatis,onethatgivessimpleconditionsonthetermsofthesequence andperhapsexcludesexceptionalfamilies,thendeterminingthepotentialnumberfor H isnotdicult.Thus,thosegraphsforwhichsuchcharacterizationsofpotentially H graphicsequencesareknownalsohaveknownpotentialnumber.Oftenwedonot haveanycharacterizationofpotentially H graphicsequences,orsuchcharacteriza17
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tionsareverycomplex,suchasTheorem2.16.Thus,itissometimeseasiertosimply determinethepotentialnumberthanitistodetermineasimplecharacterization. Asidefromthoselistedabove,thepotentialnumberisknownforthefollowing graphsandgraphfamilies.Notethatinmostcases n isassumedtobesuciently large. pK 2 and C 4 [50] Thefangraph, F 2 m + i = K 1 P 2 m + i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [19] P k and C k ,where C k isacyclewith k chordsincidenttoavertexonthecycle [139] Singlecycles: C 5 C 6 [73]; C k [76] Setsofcycles: 3 C l for l =4 ; 5 ; 6and n l [84]; 3 C l for3 l 8and n l and 3 C 9 for n 12[85]; 3 C l forall l and n sucientlylarge[89]; k C l for l 7 and3 k l Completemultipartitegraphs: K r;s for r = s =3 ; 4[134],for r s 3 [141],for r =2[142]; K 1 ; 1 ; 2 [74]; K 1 ; 1 ; 3 [77]; K 1 ; 1 ;t for n t +4[143,14]; K r 1 ;r 2 ;:::;r l ; 2 ;s for s 3and n 2 s 2 +8 s +3 r 1 + r l +4[149]; K r 1 ;r 2 ;:::;r l ;r;s for s r r l r 1 0, r 3[136]; K t s ,whichisthecomplete t partitegraph with s verticesineachpartinfactitisshownthat K t s ;n = K j + K s;s ;n [15] Friendshipgraphs, F k ,consistingof k trianglessharingasinglevertex[44] Generalizedfriendshipgraphs, F t;r;k ,whichareformedfrom k copiesof K t that overlapinasetof r vertices[133] Disjointunionofcliques[43] 18
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Smallgraphs: K 5 )]TJ/F19 11.9552 Tf 11.955 0 Td [(C 4 [78]; K 5 )]TJ/F19 11.9552 Tf 11.955 0 Td [(P 4 and K 5 )]TJ/F19 11.9552 Tf 11.955 0 Td [(P 5 ,see[80]; K 5 )]TJ/F19 11.9552 Tf 11.956 0 Td [(e [144,65] K r +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e [135] K r +1 )]TJ/F15 11.9552 Tf 12.148 0 Td [( kP 3 [ tK 2 for n 4 r +10,where r;k ,and t satisfy r +1 3 k +2 t k + t 2, k 1,and t 0[82] K r +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(pK 2 for r 2,1 p b r +1 2 c and n 3 r +3[138] K r +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(K 3 [147] K r +1 )]TJ/F19 11.9552 Tf 10.198 0 Td [(P 3 and K r +1 )]TJ/F19 11.9552 Tf 10.197 0 Td [(F ,where F is K 3 freeandcontainssometreeon4vertices [81] K r +1 )]TJ/F19 11.9552 Tf 10.52 0 Td [(K 4 K r +1 )]TJ/F15 11.9552 Tf 10.519 0 Td [( K 4 )]TJ/F19 11.9552 Tf 10.519 0 Td [(e ,and K r +1 )]TJ/F19 11.9552 Tf 10.519 0 Td [(Z 1 recall Z 1 = K 4 )]TJ/F19 11.9552 Tf 10.519 0 Td [(P 3 ,and K r +1 )]TJ/F19 11.9552 Tf 10.52 0 Td [(Z where Z is C 4 freeandcontains Z 1 [79] K r +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(U ,where U is C 4 free, Z 1 free,andcontains K 3 [ P 4 [83] Graphswithindependencenumber2[47]notethatthisresultsubsumesmany ofthepreviousresults Aftermanyyearsofhavingonlyresultsforsinglegraphsorsmallgraphclasses similartothoseabove,Ferrara,LeSaulnier,Moatt,andWenger[46]tookalarge stepforwardbydeterminingthepotentialnumberasymptoticallyforallgraphs H Theirresultusessomeideasfrom[47],wherethepotentialnumberwasdetermined forgraphswithindependencenumber2.OurworkinChapters3and4isbasedon thisresult,sowewilldescribeitnextindetail. Let H beagraphon k verticeswithatleastonenontrivialconnectedcomponent. Foreach i 2f H +1 ;:::;k g ,dene r i H =min f F : F H; j V F j = i g ; 19
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where F H denotesthat F isaninducedsubgraphof H .Let n besuciently large,andconsiderthesequence e i H;n = n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k )]TJ/F20 7.9701 Tf 6.587 0 Td [(i ; k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k + i : Thissequenceisgraphicprovidedthat n )]TJ/F19 11.9552 Tf 12.331 0 Td [(k + i and r i H )]TJ/F15 11.9552 Tf 12.331 0 Td [(1arenotbothodd. Iftheyarebothodd,thenreducethelasttermofthesequenceby1.Foreach i 2f H +1 ;:::;k g ,thesequence e i H;n isrealizedbyagraph G thatconsistsof acliqueon k )]TJ/F19 11.9552 Tf 11.64 0 Td [(i verticesjoinedtoanearly r i H )]TJ/F15 11.9552 Tf 11.639 0 Td [(1regulargraphon n )]TJ/F19 11.9552 Tf 11.64 0 Td [(k + i vertices.Thisgraphdoesnotcontain H ,becauseany k vertexsubgraphof G must use i verticesfromthe r i H )]TJ/F15 11.9552 Tf 11.367 0 Td [(1regulargraph;assuch,ithasan i vertexinduced subgraphwithmaximumdegree r i H )]TJ/F15 11.9552 Tf 11.141 0 Td [(1.However, H hasnosuchsubgraph,so H isnotcontainedin G .Thus, e i H;n isnotpotentially H graphic,andconsequently, H;n max i e i H;n Let e i H =2 k )]TJ/F19 11.9552 Tf 11.985 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.985 0 Td [(1.Thisistheleadingcoecientof e i H;n Sinceweareconcernedwiththeasymptoticbehaviorof H;n ,wearechieyconcernedwiththemaximumvalueof e i H .Notethatthemaximumvalueof e i H canbedeterminedbyndingtheminimumvalueof2 i )265(r i H .Thus,theremay bemorethanonevalueof i forwhich e i H achievesitsmaximumvalue.Hence,we dene i H tobethesmallestindex i 2f H +1 ;:::;k g thatmaximizes e i H Wewilloftenwritejust i insteadof i H ,whenthecontextisclear. Themainresultof[46]statesthat e i H;n determines H;n asymptotically forall H ,whichcanbeviewedasanErd}osStoneSimonovitstypetheoremforthe potentialproblem. Theorem2.19Ferrara,LeSaulnier,MoattandWenger[46] If H isagraph 20
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and n isapositiveinteger,then H;n = e i H n + o n : Wedene P H tobethesetofsequences e i H;n thatachievethemaximum valueoftheleadingcoecient.Thatis, P H = f e i H;n : e i H = e i H g .Note that n doesnotplayaroleinthedenitionof P H ,becausethecoecientof n in e i H;n iswhatputsthesequenceintothisset,nottheactualsumofthe sequence. Example1. Consider H = K k )]TJ/F19 11.9552 Tf 11.203 0 Td [(Z 1 ,where Z 1 = K 4 )]TJ/F19 11.9552 Tf 11.203 0 Td [(P 3 .Clearly H =3,sowe needtocompute r i H for i 2f 4 ;:::;k g .Thesubgraphinducedbythefourvertices involvedinthecopyof Z 1 hasmaximumdegree2,so r 4 H =2.For i 5,every subgraphon i verticeshasadominatingvertex,so r i H = i )]TJ/F15 11.9552 Tf 11.129 0 Td [(1foreach i 5.The minimumvalueof2 i )212(r i H is6,andthisisachievedbyboth i =4and i =5.We have e 4 H = e 5 H =2 k )]TJ/F15 11.9552 Tf 11.955 0 Td [(7,so H;n k )]TJ/F15 11.9552 Tf 11.955 0 Td [(7 n .In[79],Laishowedthat H;n = 8 > > < > > : k )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(3 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k +1+1if n k mod2 k )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(3 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k +1+2if n 6 k mod2 : Soweseethattheasymptoticresultmatchestheexactresult. Itisinterestingtonotethattherearetwosequencesin P H ;theyare e 4 H;n = n )]TJ/F15 11.9552 Tf 9.861 0 Td [(1 k )]TJ/F17 7.9701 Tf 6.587 0 Td [(4 ; k )]TJ/F15 11.9552 Tf 9.861 0 Td [(3 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k +4 and e 5 H;n = n )]TJ/F15 11.9552 Tf 9.861 0 Td [(1 k )]TJ/F17 7.9701 Tf 6.586 0 Td [(5 ; k )]TJ/F15 11.9552 Tf 9.862 0 Td [(2 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(k +5 wherethelastterm ofeachmaybedecreasedby1,dependingonparity.However,onlythesequence n )]TJ/F15 11.9552 Tf 10.659 0 Td [(1 k )]TJ/F17 7.9701 Tf 6.587 0 Td [(4 ; k )]TJ/F15 11.9552 Tf 10.66 0 Td [(3 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(k +4 isgivenasanextremalsequencein[79],since e 4 H;n = e 5 H;n )]TJ/F15 11.9552 Tf 11.682 0 Td [(2.Thus,themethodofdeterminingextremalsequencesbycomputing r i H mayactuallyyieldmoreinformationthanpreviousmethods. 21
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Chapters3and4explore P H andgraphicsequencesthatareclosetobeingin P H inmoredetail.InChapter3,wewillexplorethestructureofsequencesthatare notpotentially H graphic,intermsofmajorizationofdegreesequences.InChapter 4,weproveastabilityresultforthepotentialnumber,anddiscusswhatthisresult meansforourstudyofthepotentialfunction. 22
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3.TheShapeofGraphicSequencesthatarenotPotentially H Graphic 3.1Introduction Inthischapter,westudythestructureofdegreesequencesthatarenotpotentially H graphic.Astherealizationsofagraphicsequencemayhaveagreatdealof structuralvariety,itisperhapsmoreappropriatetosaythatweexaminetheshape" ofthesesequences.Onewaytothinkabouttheshapeofagraphicsequenceisto drawapictureofit;thatis,givenasequence = d 1 ;:::;d n ,wedrawrectangles ofheight d 1 ;:::;d n andlinethemup,similartoaFerrersdiagram,asinFigure3.1. Thisgivesusagraphicalrepresentationofthesequencethatallowsustodiscern, ataglance,therelativedierencesinthevaluesandmultiplicitiesoftermsinthe sequence. Givennotnecessarilygraphicsequences S 1 = x 1 ;:::;x n and S 2 = y 1 ;:::;y n wesay S 1 majorizes S 2 andwrite S 1 S 2 if x i y i forall i ,1 i n .Intermsof pictures,if S 1 majorizes S 2 ,thenthedrawingof S 2 tsbeneath"thedrawingof S 1 Ourstudyoftheshape"ofgraphicsequencesisbasedonthisideaofmajorization, andwhatmustbedonetoasequencetoensureitwillbemajorizedbyanother sequence.Sincesequencesthatarenotpotentially H graphicarethedegreesequence analogueof H freegraphs,ourinspirationcomesfromseveralresultson H freegraphs fromtheextremalliterature. Figure3.1: Theshapeofagraphicsequence 23
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In[35],Erd}osshowedthefollowing. Theorem3.1 If G isa K r +1 freegraphoforder n ,thenthereexistsan n vertex r partitegraph F suchthat F G Wegivetheproofhereforcompleteness. Proof. Weproceedbyinductionon r .When r =1,theresultistrivial,so supposethat r> 1andtheresultistrueforallsmallervaluesof r .Let G be a K r +1 freegraphwith V G = f v 1 ;:::;v n g ,suchthat v 1 hasdegree G and N v 1 = f v 2 ;:::;v G +1 g .Let H bethesubgraphof G inducedby N v 1 .Clearly, H is K r free,sothereisan r )]TJ/F15 11.9552 Tf 11.976 0 Td [(1partitegraph F 0 withvertexset N v 1 suchthat d F 0 v i d H v i forall i with2 i G +1.Createanewgraph F on V G byjoiningeveryvertexin f v 1 ;v G +2 ;:::;v n g totheverticesof F 0 .Thisgraphis r partite,andsince F 0 H andeachvertexin V G n N v 1 hasdegreeat least G F G Sincethesumofthedegreesequenceofagraphistwicethenumberofedgesin thegraph,thisresultsaysthatforany K r +1 freegraph G ,thereisan r partitegraph withatleastasmanyedgesas G .ObservingthattheTurangraph T n;r istheunique edgemaximal r partitegraphoforder n ,Turan'sTheoremTheorem2.3followsas acorollaryofTheorem3.1. Havingestablishedaresultonthedegreesequencesof K r +1 freegraphs,wenext discussasimilarresultfor H freegraphsforgeneral H .Todothis,weneedtouse anothertypeofmajorization,introducedbyPikhurkoandTarazin[104].Given positiveintegers m and k ,dene D k;m S 1 tobethesequence x k )]TJ/F19 11.9552 Tf 11.956 0 Td [(m;:::;x k )]TJ/F19 11.9552 Tf 11.955 0 Td [(m  {z } k times ;x k +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(m;:::;x n )]TJ/F19 11.9552 Tf 11.955 0 Td [(m : Wesaythat S 2 k;m majorizes S 1 if S 2 D k;m S 1 24
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Theorem3.2PikhurkoandTaraz[104] Let H beagraphwithchromaticnumber H = r +1 2 : Forany > 0 and n n 0 ;H ,thedegreesequenceofan H freegraph G oforder n is n;n majorizedbythedegreesequenceofsome r partitegraphoforder n Itwasnotedin[104]thatboththeoperationoflevelingo"therst k terms of G andtheoperationofreducingallofthetermsin G by m arenecessary. Forexample,consider F = K t;t .Thegraph K t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 K n )]TJ/F20 7.9701 Tf 6.587 0 Td [(t +1 is F freeandhasdegree sequence n )]TJ/F15 11.9552 Tf 11.564 0 Td [(1 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ; t )]TJ/F15 11.9552 Tf 11.565 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(t +1 .However, F =2,sowehave r =1,andinthis casean r partitegraphmustbetheemptygraphwithdegreesequence n .Clearly, weneedbothoperationstoreduce n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ; t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(t +1 toasequenceofzeroes. Theoperationsusedtocreate D n;n G inTheorem3.2reduce G byat most 2 n 2 ,soTheorem3.2saysthatreducingthedegreesumof G by o n 2 results inagraphwhosedegreesequenceismajorizedbythedegreesequenceofan r partite graph.ThisimpliestheErd}osStoneSimonovitsTheoremTheorem2.4,justas Theorem3.1impliesTuran'sTheorem. Thatthedegreesequencesof r partitegraphsappearastheboundingclassin Theorems3.1and3.2isunsurprisinggiventhecentralroleplayedbytheTuran graphintheextremalliterature.AsshowninChapter2,thesequencesthatformthe boundingclassforthepotentialproblemare e i H;n = n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k )]TJ/F20 7.9701 Tf 6.587 0 Td [(i ; k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k + i ; where H isagraphoforder k .Itisourgoaltoexaminethestructureofdegree sequencesthatarenotpotentially H graphicinamannersimilartoTheorems3.1 and3.2,bycomparingthemto e i H;n 25
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3.2Majorizationofsequencesthatarenotpotentially H graphic Throughouttheremainderofthischapter,unlessotherwisenotedwewillassume thatallsequenceshaveminimumtermatleast1.Giventwo n termgraphicsequences 1 = d 1 ;:::;d n and 2 ,andnonnegativeintegers a 1 ;a 2 ,and b with a 1 a 2 ,wesay that 1 is[ a 1 ;a 2 ] ;b close to 2 ifthereisanotnecessarilygraphicsequence 0 1 with 2 0 1 suchthat 0 1 canbeobtainedfrom 1 viathefollowingtwosteps: 1.Createthesequence S 1 = d 1 ;:::;d a 1 )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ;d a 2 ;:::;d a 2  {z } a 2 )]TJ/F20 7.9701 Tf 6.586 0 Td [(a 1 +1times ;d a 2 +1 ;:::;d n : 2.Create 0 1 from S 1 bysubtractinga total ofatmost b fromthetermsof S 1 Wewillrefertostepas levelingo terms a 1 to a 2 of 1 andtheprocedurein stepas editing thesequence S 1 Incontrasttotheideaof k;m majorization,[ a 1 ;a 2 ] ;b closenessleavesthe rst a 1 )]TJ/F15 11.9552 Tf 10.671 0 Td [(1termsunchanged,andafterthelevelingostepallowsforvariableediting, providedthatthetotalamountofeditinginstepisatmost b Asanexample,considerthesequences 1 = 5 ; 14 5 ; 10 5 ; 5 5 and 2 = 2 ; 8 18 Forthelevelingostep,wecanreduceterms3through11of 1 tothevalueof the11 th termtogetthesequence S 1 = 2 ; 10 13 ; 5 5 .Thesequence S 1 isstillnot majorizedby 2 ,soweeditbysubtracting2fromeachtermthatisequalto10;this isatotaleditingof26,andresultsin 0 1 = 2 ; 8 13 ; 5 5 ,whichismajorizedby 2 Thus, 1 is[3 ; 11] ; 26closeto 2 Weshowthatifasequenceisnotpotentially H graphic,thenitiscloseinthe abovesensetobeingmajorizedbyoneofthesequences e i H;n .Ourrstresult concernssequenceswhichfailtobepotentially H graphicsimplybyfailingtobe degreesucientfor H 26
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Theorem3.3 Let H beagraphwithdegreesequence H = h 1 ;:::;h k ,andlet = d 1 ;:::;d n beagraphicsequencethatisnotdegreesucientfor H .Further, let j bethelargestintegerforwhich d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(j +1
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where n issucientlylarge.Ifthesumof j iseven,then j isgraphic;ifthesum isodd,thenreducingthelasttermby1yieldsagraphicsequence.Clearly j isnot degreesucientfor H Notethatthe k )]TJ/F19 11.9552 Tf 11.614 0 Td [(j +1 st termof j istherstplacethatdegreesuciencyfor H fails.Since k )]TJ/F19 11.9552 Tf 10.851 0 Td [(r )]TJ/F15 11.9552 Tf 10.851 0 Td [(1 k )]TJ/F15 11.9552 Tf 10.851 0 Td [(2,thelast n )]TJ/F19 11.9552 Tf 10.851 0 Td [(k + j termsof j aretermwisedominated bythelast n )]TJ/F19 11.9552 Tf 12.225 0 Td [(k + j termsof e H +1 H;n .However, j has k )]TJ/F19 11.9552 Tf 12.225 0 Td [(j termsequalto n k +1 ,ofwhichonlytherst k )]TJ/F19 11.9552 Tf 11.305 0 Td [( H )]TJ/F15 11.9552 Tf 11.306 0 Td [(1aredominatedby e H +1 H;n .Therefore, weneedtoreduceterms d k )]TJ/F20 7.9701 Tf 6.587 0 Td [( H ;:::;d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(j of j to d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(j +1 = k )]TJ/F19 11.9552 Tf 12.35 0 Td [(r )]TJ/F15 11.9552 Tf 12.35 0 Td [(1,andeachof thesereductionsisontheorderof n .Thisyieldsasequencethatismajorizedby e H +1 H;n ,butreducinganysmallernumberoftermswouldnotsuce. Thecasewhere isdegreesucientfor H seemstobemuchmoretechnical, andrequiresboththelevelingoandeditingoperationsoutlinedabove.Recallthat i = i H isthesmallest i in f H +1 ;:::;k g thatminimizes2 i )222(r i H Theorem3.4 Let H beagraphoforder k withatleastonenontrivialcomponent andlet bean n termgraphicsequencethatisdegreesucientfor H .If isnot potentially H graphic,then is [ k )]TJ/F19 11.9552 Tf 11.25 0 Td [(i +1 ;k ] ; +3 k 2 + 3 k closeto e i H;n ToshowthesharpnessofTheorem3.4,consider H = K k .Let k = )]TJ/F19 11.9552 Tf 5.479 9.684 Td [(n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ; k )]TJ/F15 11.9552 Tf 11.956 0 Td [(5 2 k )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 ; 1 n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 k +2 : TheErd}osGallaicriteriashowthat k isgraphic;clearly k isdegreesucientfor K k .Observethat k ispotentially K k graphicifandonlyifthesequence 0 k = )]TJ/F15 11.9552 Tf 5.48 9.684 Td [( k )]TJ/F15 11.9552 Tf 11.955 0 Td [(6 2 k )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 ,obtainedbyperformingtheHavelHakimialgorithm,ispotentially K k )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 graphic.However,thecomplementofanyrealizationof 0 k isa2regulargraph, sothemaximumsizeofacliqueinanyrealizationof 0 k isatmost k )]TJ/F15 11.9552 Tf 12.135 0 Td [(2.Hence k isnotpotentially K k graphic. 28
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Since2 i )273(r i K k = i +1foreach i 2f 2 ;:::;k g ,wehave i =2.Notethat e 2 K k ;n = n )]TJ/F15 11.9552 Tf 11.528 0 Td [(1 k )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 ; k )]TJ/F15 11.9552 Tf 11.528 0 Td [(2 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(k +3 .As k )]TJ/F19 11.9552 Tf 11.528 0 Td [(i +1= k )]TJ/F15 11.9552 Tf 11.529 0 Td [(1,levelingoterms k )]TJ/F15 11.9552 Tf 11.528 0 Td [(1 and k of 0 k doesnotchangethesequence.However,eachentryfrom k )]TJ/F15 11.9552 Tf 12.133 0 Td [(1through 2 k )]TJ/F15 11.9552 Tf 11.772 0 Td [(2islargerthan k )]TJ/F15 11.9552 Tf 11.773 0 Td [(2,soweneedtoreduceeachoftheseentriesby k )]TJ/F15 11.9552 Tf 11.772 0 Td [(3,fora totalof k k )]TJ/F15 11.9552 Tf 11.955 0 Td [(3editing.Thusweperformatotalof O k 2 editing. ThisshowsthatTheorem3.4isinsomesensebestpossibleuptothecoecient of k 2 .However,sincetheamountofeditinggiveninthetheoremis +3 k 2 + 3 k theleadingtermofthisexpressionisnot k 2 when ismuchlargerthan k 1 = 2 .Thus, inthiscasewedonotyethaveanyinformationaboutthesharpnessofthisresult. 3.3Lemmas InadditiontosomeoftheresultsondegreesequencespresentedinChapters1 and2,wewillneedthefollowingresultsfortheproofofTheorem3.4.First,wehave alemmathatiscentraltotheproofofTheorem3.4,andislikelyalsoofindependent interest.Astheproofofthisresultisquitetechnical,wepostponeituntilSection 3.5. Lemma3.5 Let r and k bepositiveintegerswith r
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Proof. Wemayassumethat d i j = d j forall j andalsothat n 2 M 2 + k and d n 1. Firstnotethatif M =1,then H mustbeasetofdisjointedgesandisolatedvertices, and ispotentially H graphicaslongas n k .Wethereforeassume M 2. Let V H = f u 1 ;:::;u k g ,withtheverticesinnonincreasingorderbydegree.In arealization G of ,let S = f v 1 ;:::;v k g betheverticeswiththe k highestdegrees inorderandlet H S bethegraphwithvertexset S and v i v j 2 E H S ifandonlyif u i u j 2 E H .Ifalloftheedgesof H S arein G ,then H S isasubgraphof G thatis isomorphicto H Assumenowthat G isarealizationof thatmaximizes j E H S E G j ,butthis quantityislessthan j E H S j .Thus,thereexist v i ;v j 2 V G suchthat v i v j = 2 E G but v i v j 2 E H S .Since isdegreesucientfor H ,itfollowsthat v i and v j must eachhaveaneighbor,say a i and a j ,respectively,suchthat v i a i ;v j a j = 2 E H S but v i a i ;v j a j 2 E G .Notethatpossibly a i = a j Sincethemaximumdegreein G is M ,thereareatmost M 2 +1verticesatdistanceatmost2from a i ,andatmost M 2 +1verticesatdistanceatmost2from a j .Since a i and a j havedistinctneighborsin S ,thereareatmost k )]TJ/F15 11.9552 Tf 12.663 0 Td [(2vertices in S thataredistanceatleast3fromboth a i and a j .Therefore,thereisavertex w in V G n S thatisdistanceatleast3fromboth a i and a j .Let x beaneighbor of w ;consequently x isnotadjacentto a i or a j ,and xw= 2 E H S .Exchangingthe edges v i a i ;v j a j ,and wx forthenonedges v i v j ;a i w ,and a j x yieldsarealization G 0 of suchthat j E H S E G 0 j > j E H S E G j ,contradictingthemaximalityof G 3.4ProofofTheorem3.4 FortheproofofTheorem3.4,weactuallyproveamoretechnicalresultthat followsbelow.Firstwedenesometerminologythatisusedintheproof. Givenagraphicsequence thatisdegreesucientfor K r K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r ,wecreate asequence w ,calledthe wantsequenceof for K r K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r .Beginbyndinga 30
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realization G of onthevertices f v 1 ;:::;v n g with d v i = d i thatmaximizesthesum ofathenumberofedgesamongst v 1 ;:::;v r andbthenumberofedgesjoining f v 1 ;:::;v r g and f v r +1 ;:::;v k g .Let G r = G [ v r +1 ;:::;v n ],andlet w 0 = w r +1 ;:::;w n bethedegreesequenceof G r ,indexedsothat w i = d G r v i Foreach v i with i r ,wewant v i tobeadjacenttoeachoftheverticesinthe set S i = f v 1 ;:::;v k gnf v i g .Since isdegreesucientfor K r K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r ,weseethatfor eachnonneighborof v i in S i ,thereisaneighborof v i in f v k +1 ;:::;v n g ,andeachof theseneighborsisdistinct.Let W i beasubsetof N G r v i f v k +1 ;:::;v n g thathas size k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.956 0 Td [(d S i v i .Let W bethemultiset [ k i =1 W i Tocreatethewantsequencefrom w 0 ,wemakethefollowingmodications.Each timethevertex v y appearsin W ,add1toentry w y of w 0 .Foreach j with r +1 j k ,subtract r )]TJ/F19 11.9552 Tf 13.009 0 Td [(d f v 1 ;:::;v r g v j from w j .Thesequencethatresultsfromthese modicationsisthewantsequence, w .Notethatthelargestvaluethatcanbe subtractedfromanyentryis r ,andtheonlyentriesthatmightbereducedarethose withindexatmost k .Since isdegreesucientfor K r K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r ,noentryof w is negativeandatmost r termsof w are0.Thelargestvaluethatwillbeaddedto anyentryof w 0 isatmost r ,andonlytermswithindexatleast k +1areincreased, sothelargestentryof w isatmostthemaximumof w r +1 and w k +1 + r If w isgraphic,wecanndarealizationof thatcontains K r K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r .Todo this,taketheunionofthecompletesplitgraph K r K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r onthevertices f v 1 ;:::;v k g withthecliqueonthevertexset f v 1 :::;v r g andarealizationof w onthevertices f v r +1 ;:::;v n g .Thenjoineachvertexbelongingtothecliqueofthecompletesplit graphthatis, v i suchthat i r totheverticesin N v i f v k +1 ;:::;v n gn W i .This graphhasdegreesequence ,sowehavearealizationof thatcontainsthedesired completesplitgraph. Wewillprovethefollowing,morespecicresultthanthatstatedinTheorem3.4. 31
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Theorem3.7 Let H beaxedgraphoforder k withatleastonenontrivialcomponent.If isagraphicsequenceoflength n thatisdegreesucientfor H butnot potentially H graphic,then is [ k )]TJ/F19 11.9552 Tf 11.021 0 Td [(i +1 ;k ] ;k 2 + ki +2+ k 2 + i )144(r i H 2 k + r i H i )144(r i H )]TJ/F15 11.9552 Tf 11.022 0 Td [(2 close to e i H;n Since 2 i )222(r i H 2 H +1 )222(r H +1 H 2 H +1 ; and i H +1,weseethat i )222(r i H H .Thus, k 2 + ki +2+ k 2 + i )222(r i H 2 k + r i H i )222(r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 k 2 + ki +2+ k 2 + 2 k + r i H 6 k 2 + 3 k +3 k 2 +2 : HenceTheorem3.4whichclaimsthat +3 k 2 + 3 k editingsucesfollowsdirectlyfromTheorem3.7. Proof. Let = d 1 ;:::;d n andlabeltheverticesof H with f v 1 ;:::;v k g suchthat d v i d v j when i
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Wewillbreaktheproofintoseveralcases.InCases13,wewillshowthat, afterreducingthevalueofterms d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(i +1 ;:::;d k )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 to d k ,weonlyrequireatmost k 2 + ki +2+ f H ` )]TJ/F15 11.9552 Tf 12 0 Td [(2editing.InCase4,weshowthat isinfactpotentially H graphic,sonoeditingisrequired. Case1: n< 2 k Inthiscase,afterlevelingoterms d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(i +1 through d k )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,weneedtoreduceat most k + i termsofthesequence.Eachofthosetermsisreducedbyatmost k + ` )]TJ/F15 11.9552 Tf 10.122 0 Td [(3, sothetotalamountofeditingisatmost k 2 + ki + ` )]TJ/F15 11.9552 Tf 11.194 0 Td [(3 k + i ,whichislessthan k 2 + ki +2+ f H ` )]TJ/F15 11.9552 Tf 11.955 0 Td [(2. Case2: 2 k n
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reducedbyatmost ` )]TJ/F15 11.9552 Tf 11.955 0 Td [(2.Thetotalamountofeditingrequiredisatmost k + i )]TJ/F15 11.9552 Tf 10.098 0 Td [(1 k + ` )]TJ/F15 11.9552 Tf 10.098 0 Td [(3+ f H )]TJ/F15 11.9552 Tf 10.098 0 Td [(2 k ` )]TJ/F15 11.9552 Tf 10.099 0 Td [(2 ` )]TJ/F15 11.9552 Tf 10.098 0 Td [(2 f H )]TJ/F19 11.9552 Tf 10.099 0 Td [(k + i + k )]TJ/F15 11.9552 Tf 10.098 0 Td [(1 k + i )]TJ/F15 11.9552 Tf 10.098 0 Td [(1 : Case4: n f H and d f H >k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F15 11.9552 Tf 11.955 0 Td [(1. Nowweshowthat ispotentially H graphic.If ` =1,then d f H k )]TJ/F15 11.9552 Tf 11.887 0 Td [(1,and since f H 2 k ,thismeans d 2 k k )]TJ/F15 11.9552 Tf 12.396 0 Td [(1.ThusbypartbofTheorem2.17, is potentially K k graphic.Weassumehenceforththat ` 2. Let t besuchthat d t k )]TJ/F15 11.9552 Tf 11.956 0 Td [(1but d t +1 t .Let w 1 bethewantsequenceof for K t K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(t Sinceeveryentryof w 0 isatmost k )]TJ/F15 11.9552 Tf 11.556 0 Td [(2because d t +1 6 k 2 showsthatthisistrue,and w 1 isgraphic. Nowobservethatif w 1 isgraphic,then ispotentially K t K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t graphic.If t k )]TJ/F19 11.9552 Tf 11.638 0 Td [( H ,thenthiscompletesplitgraphcontainsacopyof H ,andwearedone. Soweassumethat t
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ByTheorem2.12,weknowthatarealization G of canbefoundthatcontains K t K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t suchthatthe t verticesofdegree k )]TJ/F15 11.9552 Tf 12.458 0 Td [(1areonthe t highestdegreeverticesof G ,andthe k )]TJ/F19 11.9552 Tf 12.667 0 Td [(t verticesofdegree t areonthenext k )]TJ/F19 11.9552 Tf 12.667 0 Td [(t highestdegree verticesof G .Deletethe t verticesofhighestdegreein G ,andlet 0 = d 0 1 ;:::;d 0 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(t bethedegreesequenceoftheresultingsubgraphof G .Itfollowsthat 0 satises d 0 1 k )]TJ/F15 11.9552 Tf 13.162 0 Td [(2and d 0 f H )]TJ/F20 7.9701 Tf 6.587 0 Td [(t k )]TJ/F19 11.9552 Tf 13.162 0 Td [(` )]TJ/F19 11.9552 Tf 13.162 0 Td [(t 1.Since F k )]TJ/F20 7.9701 Tf 6.586 0 Td [(t = r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t H and d 0 k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t d 0 f H )]TJ/F20 7.9701 Tf 6.587 0 Td [(t k )]TJ/F19 11.9552 Tf 12.463 0 Td [(` r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(t H ,itfollowsthat 0 isdegreesucientfor F k )]TJ/F20 7.9701 Tf 6.586 0 Td [(t ApplyingLemma3.6with H = F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t and M = k )]TJ/F15 11.9552 Tf 12.427 0 Td [(2,weseethat 0 ispotentially F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t graphicaslongasatleast2 k )]TJ/F15 11.9552 Tf 12.529 0 Td [(2 2 + k )]TJ/F19 11.9552 Tf 12.529 0 Td [(t termsof 0 arepositive.Since f H )]TJ/F19 11.9552 Tf 11.491 0 Td [(t 6 k 2 ,thisistrue.Thus,thereisarealizationof 0 thatcontainsacopyof F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(t ontheverticesofhighestdegree;overlappingthiswiththevertices v t +1 ;:::;v k of G ,wegetarealizationof containing K t F k )]TJ/F20 7.9701 Tf 6.586 0 Td [(t ,whichimpliesthat ispotentially H graphic. Case4b: t k )]TJ/F19 11.9552 Tf 11.956 0 Td [(` First,suppose d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F19 11.9552 Tf 9.59 0 Td [(d k ` k +2.Thisimpliesthat d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` ` k +2+ d k .Since ` 2,itfollowsthat d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` 3 k .Weclaimthatthisimpliesthat ispotentially H graphic. Since f H >` k + ` +1and d f H k )]TJ/F19 11.9552 Tf 10.054 0 Td [(` ,Lemma3.5yieldsarealization G of containingthecompletebipartitegraph K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` ;` ,wherethevertices f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` g formthepartitesetoforder k )]TJ/F19 11.9552 Tf 12.995 0 Td [(` .Let S bethesetofverticesinthecopyof K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` ;` ,let S 0 = f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` g ,andlet R = V G n S .Iftheverticesof S 0 inducea completegraph,then G contains K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` K ` ;consequently G containsacopyof H since k )]TJ/F19 11.9552 Tf 10.717 0 Td [(` k )]TJ/F19 11.9552 Tf 10.717 0 Td [( .Supposetherearevertices v i and v j in S 0 suchthat v i v j = 2 E G Since d S v i k )]TJ/F15 11.9552 Tf 11.372 0 Td [(2and d v i 3 k )]TJ/F15 11.9552 Tf 11.372 0 Td [(2,weknowthat v i hasatleast2 k neighborsin R .Similarly, v j hasatleast2 k neighborsin R .Ifeachneighborof v i in R isadjacent toeachneighborof v j in R ,theneachoftheseverticesin R hasdegreeatleast2 k )]TJ/F15 11.9552 Tf 10.637 0 Td [(1. 35
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Since v i hasatleast2 k neighborsin R ,itfollowsthat G hasatleast2 k verticeswith degreeatleast2 k )]TJ/F15 11.9552 Tf 11.029 0 Td [(1,contradictingtheassumptionthat d k 2 k )]TJ/F15 11.9552 Tf 11.03 0 Td [(4.Thusthereare vertices x and y in R suchthat v i x;v j y 2 E G ,and xy= 2 E G .Hencewecanreplace theedges v i x and v j y withthenonedges xy and v i v j toobtainanewrealizationof Iterativelyperformingthisprocessforeachnonadjacentpairofverticesin S 0 yieldsa realizationof inwhich S 0 inducesacompletegraph.Consequently ispotentially H graphic. Finally,wemustconsiderthecasewhere d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` k .Let w 2 = g 1 ;:::;g n 0 bethewantsequenceof for K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` K ` where n 0 = n )]TJ/F15 11.9552 Tf 11.379 0 Td [( k )]TJ/F19 11.9552 Tf 11.38 0 Td [(` Sinceconstructingthewantsequenceincreaseseachtermbyatmost k )]TJ/F19 11.9552 Tf 11.082 0 Td [(` ,thefacts that d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` +1 < 2 k + ` k +2, d k < 2 k )]TJ/F15 11.9552 Tf 12.233 0 Td [(3,and d 2 k 6 k 2 + ` 2 k TripathiandVijay[116]showedthattheErd}osGallaicriteriaTheorem1.3need onlybecheckedforcertainvaluesof p :itsucestocheckall p s ,where s isthe largestintegerforwhich d s s )]TJ/F15 11.9552 Tf 11.559 0 Td [(1,ortocheckonlythosevaluesof p forwhich d p is strictlygreaterthan d p +1 .WewillusetheErd}osGallaicriteriaandthisobservation toshowthat w 2 isgraphic. Since g i < 2 k )]TJ/F19 11.9552 Tf 12.178 0 Td [(` forlargeenough i ,weonlyneedtochecktheinequalitiesfor indicesupto2 k .WecanwritetherightsideoftheErd}osGallaiinequalityInequality 36
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.1as p p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ r X i = p +1 p + n 0 X i = r +1 g i ; where r p +1isthelargestindexsuchthat g r p but g r +1
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Finally,if ` + k +1 p 2 k ,then p X i =1 g i 2 ` k + ` 2 k +1+3 k 2 + p )]TJ/F19 11.9552 Tf 11.203 0 Td [(` )]TJ/F19 11.9552 Tf 11.203 0 Td [(k k )]TJ/F19 11.9552 Tf 11.203 0 Td [(` 5 k 2 + ` 2 k +2 ` 2 )]TJ/F19 11.9552 Tf 11.203 0 Td [(` k: Now, n 0 + p 2 )]TJ/F19 11.9552 Tf 11.771 0 Td [(p )]TJ/F15 11.9552 Tf 11.772 0 Td [(1 > 6 k 2 + ` 2 k +4 k 2 ,sotheErd}osGallaiinequalityissatised. ThusClaim1isproved. Nowthatwehaveshownthat w 2 isgraphic,wecanusearealizationof w 2 to createarealizationof containingacopyof K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` K ` .Since H K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` K ` thisimpliesthat ispotentially H graphic. 3.5ProofofLemma3.5 Ideaoftheproof: TheproofofLemma3.5isbasedonacarefulanalysisof repeatedapplicationsoftheKleitmanWangalgorithmTheorem1.2.Observethat whenlayingoaterm d i fromagraphicsequence,the d i termsofhighestdegree, asidefrom d i ,areeachreducedby1.Iftherearemanytermsofthesamevalue thatwillbereduced,theorderinwhichthesereductionsoccurdoesnotmatter.In particular,providedwereducethecorrectnumberofterms,wemayreduceanyof thetermsequalto d d i andwillgetthesameresidualsequence.Thisfactisthekey toconstructingarealizationof thatcontains K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r;r ,asreferencedinthestatement ofLemma3.5. Toseethismoreclearly,considerasequence = d 1 ;:::;d 18 ,where d 5 = d 6 = = d 12 ,and d 13 =7.Ifwelayo d 13 inthestandardway,thenwereduceterms d 1 through d 7 byone,andthenreorderthesequencetomakeitnonincreasing.Thisgives ustheresidualsequence 0 = d 1 ;d 2 ;d 3 ;d 4 ;d 8 ;d 9 ;d 10 ;d 11 ;d 12 ;d 5 ;d 6 ;d 7 ;d 14 ;:::;d 18 seeFigures3.2aand3.2b.However,since d 5 = = d 12 ,wecouldreduceanythree ofthesetermsandtheresultingsequencewouldbethesame exceptintheorderofthe 38
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terms .So,wecouldreduceterms d 1 through d 4 d 6 ;d 9 ,and d 10 ,asinFigure3.2c,and gettheresidualsequence 0 = d 1 ;:::;d 4 ;d 5 ;d 7 ;d 8 ;d 11 ;d 12 ;d 6 ;d 9 ;d 10 ;d 14 ;:::;d 18 butthevaluesofthetermsarethesameasintherstcase.Alternatively,wecould reducetherstfourtermsof andthelastthreetermsoftheconstantsubsequence thatis, d 10 ;d 11 ,and d 12 ,andthenwewouldnothavetoreorderthesequenceat all,asinFigure3.2e.Figures3.2b,3.2d,and3.2eshowthatwhiletheorderofthe termsinthesequencechangesdependingonwhichtermswereduce,the values of thosetermsarethesame,soasfarasthesequencesareconcerned,itdoesnotmatter whichtermswereduce. However,ifeachterm d i isassociatedwithaparticularvertex v i inarealization of ,itdoesmatterwhichtermsarereducedasweperformtheKleitmanWang algorithm.Thus,wewillperformthealgorithmbutcarefullykeeptrackofwhich termswereduceandtheorderofthetermsintheresidualsequence.Sometimeswe willreducetermsinthestandardwayprescribedbyTheorem1.2,butsometimeswe willwishtochangetheorderofthesequenceaslittleaspossible,andwillreduce termsstartingwiththeendofaconstantsubsequenceasinFigure3.2e.Thiswill allowustobuildarealizationof withthedesiredproperties. TheKleitmanWangalgorithmprovidesameansbywhichthedesiredrealization canbeconstructedonthevertexset V = f v 1 ;:::;v n g sothatthevertices v j have degree d j for j =1 ;:::;n .Thevertex v j isassociatedwiththe j thtermin .When d j islaido,theresultingsequence, 0 ,isagraphicsequence.Wecanuseittoconstruct agraphon V nf v j g withthereordered 0 asitsdegreesequence.Thevertex v j is thenadded,adjacenttotherst d j membersof f v 1 ;:::;v j )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ;v j +1 ;:::;v n g .Inthis way,whenwelayoaterm d j of ,wewillsaythattheverticesassociatedwith thetermsthatarereducedareassignedtotheneighborhoodof v j .Repeatingthis process,wewillcreatearealizationof containing K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r;r 39
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aLayo d 13 byreducingterms d 1 through d 7 bThesequencethatresultsfromthe reductionin3.2a cLayo d 13 byreducingterms d 1 through d 4 d 6 d 9 ,and d 10 dThesequencethatresultsfromthe reductionin3.2c eLayo d 13 byreducingterms d 1 through d 4 d 10 d 11 ,and d 12 fThesequencethatresultsfromthereductionin3.2e Figure3.2: DierentwaystoreducetermsintheKleitmanWangalgorithm 40
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Theproblemwiththisprocedureisthatapplyingitmorethanoncerequiresthat eachofthedegreesequencesmustbereordered,whichmakeskeepingtrackofthe verticesthatareassignedtoaparticularneighborhooddicult. Forclarity,wewilloftenabuseterminologyandsaywelayovertex v j tomean welayothetermof whosevaluecorrespondstothedegreeof v j .Thismakes sensewhenwethinkaboutlayingoaterm d j of asassigningasetofvertices totheneighborhoodof v j .Wewilllayoatmost r k + r +1verticeswiththe aimofobtainingjust r ofthemwhoseneighborhoodcontains f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r g .Our parametersarechosenjustforthispurpose.Theentriesin f d k +1 ;:::;d n g thathave valuein f k )]TJ/F19 11.9552 Tf 13.028 0 Td [(r;:::;k )]TJ/F15 11.9552 Tf 13.028 0 Td [(1 g willbethecandidatesforentriestolayo.Because d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r )]TJ/F19 11.9552 Tf 12.345 0 Td [(d k r k +2,wecanguaranteethat,foreachofthedegreesequencesthat resultfromthelayingoprocedure,theentriesthatcorrespondto v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r will alwaysstaywithintherst k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1entries. Termsanddenitions: Nowweproceedtoprovethattheprocedureoutlined abovedoesindeedproducethegraphwewant.Wewilllayoentriesof correspondingtovertices v a 1 ;v a 2 ;:::;v a p ;::: ,where v a p willbedeterminedatstep p .Let V 0 = V andfor p =1 ; 2 ;::: ,let V p = V p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 )]TJ/F19 11.9552 Tf 12.571 0 Td [(v a p .Theneighborhoodweassignto v a p ,whichwewillcall N p ,isasubsetof V p .Wewillcalltheprocessoflayingo v a 1 ;:::;v a r k +2 theLayingoAlgorithm. For p =0 ; 1 ; 2 ;::: ,wedene ^ d p v i tobethe remainingdegree of v i after v a 1 ;:::;v a p arelaido.Thatis,foreveryvertex v i ^ d 0 v i = d i andfor p =1 ; 2 ;:::;r k +2,wehave ^ d p v i := d i )285(jf N j :1 j p and v i 2 N j gj .Iteratively, ^ d p v i = 8 > > < > > : ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v i if v i = 2 N p ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v i )]TJ/F15 11.9552 Tf 11.956 0 Td [(1if v i 2 N p : Todeterminewhichvertex v a p tolayo,for p =1 ; 2 ;::: ,wedene S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 V p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 tobe thesetofallvertices w 2 V p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 forwhich ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w 2f k )]TJ/F19 11.9552 Tf 12.055 0 Td [(r;:::;k )]TJ/F15 11.9552 Tf 12.054 0 Td [(1 g .Thenchoose 41
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v a p tobeavertexin S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 forwhich ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v a p isminimum.Let ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 = ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v a p ;this isthenumberofverticesthatwillbeassignedto N p .Notethattheneighborhoodof v a p maynotconsistsolelyoftheverticesin N p .Inparticular,if ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v a p < ^ d 0 v a p then v a p isin N p 0 forsome p 0 ^ d p v j ,orb ^ d p v i = ^ d p v j and i
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Findingtheneighborhoods" N p : Nowwecandescribeourmodicationof theKleitmanWangalgorithmmoreprecisely.Atstep p oftheLayingoAlgorithm, wechoose N p inthefollowingway: 1.If T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isconsistent,thensimplylet N p betherst ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 verticesin ^ T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2.If T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isnotconsistentbut ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 > ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 +1,thenwe againlet N p consistoftherst ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 verticesin T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 3.If T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isnotconsistentbut ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 = ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 +1,then N p consistsofallvertices w 2 V p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 forwhich ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w > ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,andthe vertices x withthehighestindexforwhich ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 x = ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 Inwords,whatwedoisidentifytheverticeswithlargest ^ d p valuesandreducetheir valuesby1.If T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isnotconsistentandwecannotreduceallofthosewiththesame value,wereducethosewithlargestindexi.e.,thosethatcomelaterintheordering, asinFigure3.2e.When T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isconsistent,westilltaketherst ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 vertices,even ifallofthosewiththesamevaluearenotreduced.Wewillsaythat N p is good if f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r g N p .Theexistenceofatleast r verticesamong f v k +1 ;:::;v n g such thatlayingoeachgivesagood N p willyieldthe K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r;r weseek. Anexample: LetusdoasmallexampletoillustratethewaytheLayingo Algorithmworks. Beginwiththegraphicsequence = ; 9 ; 9 ; 9 ; 8 ; 8 ; 7 ; 7 ; 7 ; 7 ; 4 ; 4 ; 4 ; 4= d v 1 ;:::;d v 14 .Forthepurposesofthisexample,wewillonlyidentifytheneighborhoodsoftheverticeswithdegree4. Step1 Sincetheoriginalorderingofverticesisconsistent,weassigntheneighborhoodof v 14 tobe N 1 = f v 1 ;v 2 ;v 3 ;v 4 g .Thenewsequenceis 1 = 6 ; 7 4 ; 4 3 andsince ^ d 1 v 4 ^ d 1 v 5 ,thereisnoreorderingofverticesand T 1 isconsistent. Step2 Since T 1 isconsistent,wecansimplyassigntheset N 2 = f v 1 ;v 2 ;v 3 ;v 4 g tothe 43
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neighborhoodof v 13 .Now 2 = ; 8 ; 7 8 ; 4 2 .However,theverticesarenolonger intheiroriginalorder;thesequence T 2 is: v 5 ;v 6 ;v 1 ;v 2 ;v 3 ;v 4 ;v 7 ;v 8 ;v 9 ;v 10 ;v 11 ;v 12 Step3 Since T 2 isnotconsistent,wemustconsider ^ d 2 ^ 2 .Since ^ d 2 ^ 2 = ^ d 2 ^ 2 ,wecannotsimplyassignthefourhighestdegreeverticesto N 3 .We beginwith N 3 = f v 5 ;v 6 g ,thetwohighestdegreevertices.Thenweneedtwo morevertices,sowetakethetwoverticesofdegree ^ d 2 ^ 2 =7thathave thehighestindex,thatis v 9 and v 10 .So N 3 = f v 5 ;v 6 ;v 9 ;v 10 g .Thisleaves 3 = 8 ; 6 2 ; 4,and T 3 isconsistent. Step4 Since T 3 isconsistent,welet N 4 = f v 1 ;v 2 ;v 3 ;v 4 g .Then 4 = 4 ; 6 6 Observethatateachstep i isexactlythesequencewewouldgetafter i iterations oftheKleitmanWangalgorithmifatermofvalue4islaidoeachtime. ProofthattheLayingoAlgorithmgives r goodneighborhoods: Now wewillshowthatthisprocessdoescreate r verticesamong f v k +1 ;:::;v n g thathave goodneighborhoods.Webeginwithseveralclaimsthatdevelopusefulproperties oftheLayingoAlgorithm,inparticularthekeyobservationthat ` p ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 forall p rk .Then,weshowthatateachiterationofthealgorithm,thesequence T p has acertainstructurethatallowsustoeasilycountthenumberofiterationsneededto nd r good N p s. Claim1. If p r k +1and i ^ d p 00 )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v i ,then v i 2 N p 00 implies v j isalsoin N p 00 .This contradictionprovesClaim1. 44
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Claim2. If p r k +1and j k )]TJ/F19 11.9552 Tf 13.207 0 Td [(r ,then ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v j k .Inaddition,if S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N p 6 = ; ,then N p isgood. ProofofClaim2. If v j isnotlaido,then d j decreasesbyatmost1ateachstepand so ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v j d j )]TJ/F15 11.9552 Tf 11.955 0 Td [( p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1.Because d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r d k + r k +2,wehavethefollowing: ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v j d j )]TJ/F15 11.9552 Tf 11.955 0 Td [( p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 d k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r )]TJ/F15 11.9552 Tf 11.955 0 Td [( p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 d k + r k +2 )]TJ/F15 11.9552 Tf 11.956 0 Td [( p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 d k + r: Theconditionsonthesequenceforce d k k )]TJ/F19 11.9552 Tf 12.256 0 Td [(r ,giving ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v j d k + r k .As aresult,if S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N p 6 = ; ,then N p mustcontaineveryvertexwithremainingdegree greaterthan ^ d p v a p k )]TJ/F15 11.9552 Tf 12.222 0 Td [(1.Thisincludesallof f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r g andso N p mustbe good. Let g p denotethenumberofgoodneighborhoods N p 0 with p 0 p .Wemayassume that g p r )]TJ/F15 11.9552 Tf 11.057 0 Td [(1forall p r k +1.Otherwise,wewouldhave r goodneighborhoods, henceourcopyof K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r;r .Inparticular,byClaim2wecanassumethatthereareat most r )]TJ/F15 11.9552 Tf 11.955 0 Td [(1valuesof p forwhich S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N p 6 = ; Claim3. If p r k +1,then j S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 N p j r )]TJ/F15 11.9552 Tf 12.558 0 Td [(1and j S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j >r k + r +1 )]TJ/F19 11.9552 Tf 422.702 23.98 Td [(g p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 r )]TJ/F15 11.9552 Tf 11.889 0 Td [(1 )]TJ/F15 11.9552 Tf 11.89 0 Td [( p )]TJ/F15 11.9552 Tf 11.889 0 Td [(1 2 r .Inaddition,every v 2 N p has ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v atleastaslargeas theleastvalueof ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 amongmembersof S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ProofofClaim3. Considerthevertex v a p .Ithasdegreeatmost k )]TJ/F15 11.9552 Tf 12.457 0 Td [(1whenitis laido.ByClaim2,thereareatleast k )]TJ/F19 11.9552 Tf 12.488 0 Td [(r vertices v j with ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v j k andso j S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N p j k )]TJ/F15 11.9552 Tf 12.104 0 Td [(1 )]TJ/F15 11.9552 Tf 12.104 0 Td [( k )]TJ/F19 11.9552 Tf 12.103 0 Td [(r = r )]TJ/F15 11.9552 Tf 12.104 0 Td [(1.Becauseavertexwillonlyleavetheset S p ifithasbeenlaidoorassignedtotheneighborhoodsofenoughotherverticesthat 45
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itsremainingdegreeistoolow, j S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j r k + r +1 )]TJ/F25 11.9552 Tf 10.789 20.922 Td [( p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ j =1 f S j )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N j g )]TJ/F15 11.9552 Tf 10.789 0 Td [( p )]TJ/F15 11.9552 Tf 10.789 0 Td [(1 r k + r +1 )]TJ/F19 11.9552 Tf 10.788 0 Td [(g p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 r )]TJ/F15 11.9552 Tf 10.788 0 Td [(1 )]TJ/F15 11.9552 Tf 10.788 0 Td [( p )]TJ/F15 11.9552 Tf 10.788 0 Td [(1 : Since g p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 r )]TJ/F15 11.9552 Tf 12.309 0 Td [(1,wehave j S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j r k + r +1 )]TJ/F15 11.9552 Tf 12.309 0 Td [( r )]TJ/F15 11.9552 Tf 12.31 0 Td [(1 2 )]TJ/F15 11.9552 Tf 12.31 0 Td [( p )]TJ/F15 11.9552 Tf 12.309 0 Td [(1 2 r .Ifwe includethevertices f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r g ,thereareatotalofatleast k vertices w forwhich ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w isatleasttheminimumvalueof ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 amongthemembersof S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .This provesClaim3. Claim4. If ` p <` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 forsome p rk ,thenatmost r moreiterationsoftheLayingoAlgorithmwillcreatethedesiredcopyof K k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r;r ProofofClaim4. Bydenition, ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 = ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v a p and ` p = ^ d p v a p +1 .Since v a p was chosentominimize ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 among S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,weknowthat ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v a p +1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(r Since ^ d p v a p +1 ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v a p +1 )]TJ/F15 11.9552 Tf 13.216 0 Td [(1,weget ` p = ^ d p v a p +1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 )]TJ/F15 11.9552 Tf 13.216 0 Td [(1.Thus, ` p = ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 )]TJ/F15 11.9552 Tf 11.518 0 Td [(1.Thismeansthat ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v a p +1 = ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and v a p +1 2 N p .Since v a p +1 2 S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 aswell,Claim2givesthat N p isgood. Further,as ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isalsotheminimumremainingdegreeofanyvertexin S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 Claim3givesthateveryvertex w in N p has ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Since v a p +1 2 N p ,we concludethat ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 = ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Since ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r > ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v a p +1 ,thereareat most ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1vertices w with ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 w ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r So,ifwecanshowthat ` p 0 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 )]TJ/F15 11.9552 Tf 10.324 0 Td [(1forall p 0 suchthat p p 0 p + r )]TJ/F19 11.9552 Tf 10.324 0 Td [(g p ,then each N p 0 isgoodandwehavethedesired K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r;r inatmost r moresteps.FromClaim 3,thereareatmost r )]TJ/F15 11.9552 Tf 10.704 0 Td [(2verticesin S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 N p thathaveremainingdegreelargerthan ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v a p +1 = ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .AlsofromClaim3, j S p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j >r k + r +1 )]TJ/F19 11.9552 Tf 11.945 0 Td [(g p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 r )]TJ/F15 11.9552 Tf 11.945 0 Td [(1 )]TJ/F15 11.9552 Tf 11.945 0 Td [( p )]TJ/F15 11.9552 Tf 11.945 0 Td [(1. 46
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Thus,thereareatleast r k + r +1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(g p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 r )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( r )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 >r r )]TJ/F19 11.9552 Tf 11.956 0 Td [(g p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 verticesofremainingdegreeequalto ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 in S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .SinceClaim3givesthat j S p 0 )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 N p 0 j r )]TJ/F15 11.9552 Tf 9.715 0 Td [(1,forall p 0 p ,eachofthenext r )]TJ/F19 11.9552 Tf 9.716 0 Td [(g p iterationsoftheLayingoAlgorithm willremoveatmost r verticesfrom S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 whichhaveremainingdegreeequalto ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Hencethereisalwaysavertexin S p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 withdegreeequalto ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .Thus,novertex withdegree ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1willbeplacedinto N p 0 ,andClaim4isproved. Wecanthusassumethat ` p ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 forall p rk Nowwearepreparedtoexaminethestructureofthesequence T p .Claim5 belowisthemainobservation,thatevenwhentheLayingoAlgorithmresultsin aninconsistentsequence,thesequencethatresultsisofaveryspecicform.Thus, theLayingoAlgorithmensuresthatthenumberofiterationsbetweenconsistent sequencesislessthan k Toshowthis,wesaythesequence T p isof properform ifthereisapartitionof V p intofourorderedsets p p p and p wheretheorderisinheritedfrom p suchthat ^ d p isconstantoneachof p and p and,when i
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Claim5. Forall p 2f 0 ;:::;rk )]TJ/F15 11.9552 Tf 10.77 0 Td [(1 g T p isofproperform.If T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isconsistentorhas inconsistencyatleast k ,then N p isgood.If T p isinconsistent,then j p [ p j ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 If T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 haspositiveinconsistency,theneither T p isconsistentand N p isgood, T p = ^ T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and N p isgood,or T p hasinconsistencystrictlylessthantheinconsistencyof T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ProofofClaim5. Wewillprovetheclaimbyinductionon p If p =0,then T p = T 0 isconsistent.Moreover N p +1 isgoodbecauseitissimply therst ` p entriesof ^ T p ,whichmustcontain v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r .Infact,thisistrueforany consistent T p andthiswillbeourbasecasefortheinduction. Weassumethestatementoftheclaimistruefor T 0 ;:::;T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Case1: T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isconsistent. Theset N p isgoodbecauseitissimplytherst ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 entriesof ^ T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,whichmust contain v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r .If T p isconsistent,thenitis,bydenition,ofproperform. If T p isnotconsistent,then ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 = ^ d p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 +1.Wecan partition ^ T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 into^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,where^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 containsallverticeswithremaining degreeexactly ^ d p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 containsthosewithlowerremainingdegree. Wecanfurtherpartition^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 into^ L 1 p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and^ L 2 p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,where^ L 1 p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 containsallvertices of^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 thatareincludedin N p ,and^ L 2 p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 consistsofthosethatarenot. Now T p canbepartitionedinto p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 p =^ L 2 p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 p =^ L 1 p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; and p =^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; 48
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anditisofproperform.Clearly j p [ p j = ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 Observethatif T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isconsistentand T p isnot,then f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r g iscontained in p [ p [ p Case2: T p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isnotconsistent. Recallthat ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 = j N p j ,thenumberofverticesin V p thatarereducedbyonewhen avertexof T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 islaido.TheeectoftheLayingoAlgorithmon T p dependson thevalueof ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Notethat ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j isnotpossiblebecauseClaim4allowsustoassumethat ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 .Since ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 j p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j > j ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j ,thisisacontradiction. Withthisinformation,wecanshowthatiftheinconsistencyof T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 isatleast k ,then N p isgood.Thelargest k entriesof T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arein p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and N p containsallof^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .Becausethereare r k + r +1verticeseligibletobelaid ofrom f v k +1 ;:::;v n g ,andwe'velaidoatmost rk ,thevertices v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r ;:::;v k will notbelaido.If v k isin^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,thenitsvalueisatmost d k )]TJ/F15 11.9552 Tf 12.96 0 Td [(1,andif v k isin ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,thenitsvalueisatmost d k .Butthedegreeofeachof v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r isatleast d k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r )]TJ/F19 11.9552 Tf 12.026 0 Td [(kr d k + r k +2 )]TJ/F19 11.9552 Tf 12.026 0 Td [(kr>d k .So,eachof v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.587 0 Td [(r arein^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 andwillbe in N p aslongastheinconsistencyisatleast k Case2a: j ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j <` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 < j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j Inthiscase,wecanpartition^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 intotwopieces:^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and^ R p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .Themembers of^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arereducedwhen v a p islaido,butthoseof^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arenot. Afterreordering,weobtainthefollowing: p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 : 49
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Moreover, ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j = j ^ p [ ^ p j .Inaddition,theinconsistencyof T p is j p [ p j = j ^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j ,whichisstrictlylessthantheinconsistencyof T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 because^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isastrictsubsetof^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Toproceedthroughthenextcases,wemustpartition^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 intotwopieces:^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Themembersof^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 havethesameremainingdegreeasthosein^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,and thoseof^ R p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 havesmallerremainingdegreeeitherorbothofthesemaybeempty. Case2b: j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j + j ^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j Inthiscase,thevaluesof^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 aswellassomeof^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arereduced.Since themembersof^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 andtheunreducedvaluesof^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 nowhavethesame value,reorderingresultsin T p beingaconsistentsequence. Case2c: j ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j + j ^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j <` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 < j ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ L p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 j Inthiscase,wecanpartition^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 intotwopieces:^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Themembers of^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arereducedbutthoseof^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 arenot. Afterreordering,weobtainthefollowing: p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; p =^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 : Moreover, ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j = j ^ p [ ^ p j .Inaddition,theinconsistencyof T p is j p [ p j = j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ R p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j ,whichisstrictlylessthantheinconsistencyof T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 because^ R p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isastrictsubsetof^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Case2d: j ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ L p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 j ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 Inthiscase,norearrangingisnecessary:theorderoftheverticesin T p isthe sameastheorderin ^ T p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 50
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Becausetheonlyverticesoutoforderarein^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ ^ p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 N p willcontainall oftherst ` p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 vertices.Since ` p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 k )]TJ/F19 11.9552 Tf 12.868 0 Td [(r ,theneighborhood N p mustcontain f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [(r g andthusbegood. ThisconcludestheproofofClaim5. GivenClaim5,theproofofLemma3.5followseasily.Therecanbeatmost k )]TJ/F15 11.9552 Tf 10.825 0 Td [(1 neighborhoodsthatarenotgoodbetweenconsecutivegoodneighborhoods.Soafter r )]TJ/F15 11.9552 Tf 12.18 0 Td [(1 k +1iterationsoftheprocedure,therewillbe r goodneighborhoods,giving usthedesiredrealization. 51
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4.StabilitywithRespecttothePotentialNumber InChapter3,weshowedthatgraphicsequencesthatarenotpotentially H graphicareclosetobeingmajorizedbyasequencefrom P H .Inthischapter,we willshowthatforsomegraphs H ,ifthesequencealsohasthepropertythatits sumisclosetothepotentialnumber,thenitisactuallyveryclosetobeingoneof thesequencesin P H ,notjusttobeingmajorizedbyone.However,thisactually dependslargelyonthestructureof H ,soitisnottrueforallgraphs. OurgoalistoproveastabilityresultakintotheresultofSimonovitsTheorem2.5fortheTuranproblem,soweexaminethistheoremagaininaslightly dierentlight.WecanrephraseTheorem2.5intermsofeditdistance,whichwill helptomotivateourdenitionofdistancebetweengraphicsequences.Givengraphs G and G 0 onthesamelabeledvertexset,the editdistance between G and G 0 ,denoted dist G;G 0 ,is j E G 4 E G 0 j .Withthisterminology,werestateTheorem2.5: Theorem4.1Simonovits[113] Let H beagraphwith H = r +1 .Forevery > 0 ,thereexistsa > 0 andan n suchthatif n>n and G isan n vertex H free graphsuchthat j E G j ex H;n )]TJ/F19 11.9552 Tf 11.955 0 Td [(n 2 ; Thendist G;T n;r
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V L bethesetofverticesin G thathavedegreeatleast r )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 r n and V S bethesetof verticesin G withdegreelessthan r )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 r n .Then k G )]TJ/F19 11.9552 Tf 11.955 0 Td [( T n;r k = X v 2 V L d v )]TJ/F19 11.9552 Tf 13.151 8.088 Td [(r )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 r n + X v 2 V S r )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 r n )]TJ/F19 11.9552 Tf 11.955 0 Td [(d v < 4 n: Thisjustiesouruseofthe ` 1 normtodeneadistancebetweengraphicsequences. Wecannowpreciselydenewhatitmeansforagraphtobestablewithrespect tothepotentialnumber. Denition4.2 Agraph H is stablewithrespecttothepotentialnumber ,or stable ,ifforany > 0 ,thereexistsan n 0 = n ;H and > 0 suchthatfor anygraphicsequence oflength n n 0 thatisnotpotentially H graphicandthat satises H;n )]TJ/F19 11.9552 Tf 11.956 0 Td [(n; thereissome 0 2P H suchthat k )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 k 0 ,thereexistsan n 0 = n ;H and > 0 suchthatforanygraphicsequence oflength n n 0 thatis degreesucientfor H butnotpotentially H graphicandsatises H;n )]TJ/F19 11.9552 Tf 11.956 0 Td [(n; thereissome 0 2P H suchthat k )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 k
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Figure4.1: Agraphwith r +1 > 1and2 i )222(r i 2 not stable;completegraphsareoneexample,whichwillbediscussedinmoredetail inSection4.2. 4.1Graphswithdegreesequencestability Recallthatforagraph H oforder k thesequencesin P H aredeterminedby thevalue2 i )288(r i H for i 2f H +1 ;:::;k g .Let i H bethesmallestindex i 2f +1 ;:::;k g suchthat2 i )247(r i isminimized.Whenitisunderstood,wewill suppresstheargument H inouruseofparameterslike ; r i ,and i .Wehave 2 i )222(r i 2 +1 )222(r +1 2 +1 : If r +1 H > 1,then2 +1 )269(r +1 2 H ,so2 i )269(r i 2 H .Thus,if 2 i )285(r i =2 H +1,wemusthave r +1 H =1.Inthiscase,weknowthat thereisasetof +1verticesin H thatinduceagraphconsistingofamatchingand isolatedvertices. Itisworthnotingthat2 i )352(r i 2 H doesnotnecessarilymeanthat r +1 H > 1.Forexample,if H isthegraphinFigure4.1,wehave H =4 with r 5 H =1and r 6 H =5.Thus,2 i )222(r i =2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(5=7 < 2 H Ourmaintheoremstatesthatthereisalargeclassofgraphsthatare stable. Theorem4.4 If H isagraphsuchthat 2 i )220(r i H 2 H ,then H is stable. 54
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Ontheotherhand,if2 i )65(r i H =2 H +1andthus r +1 H =1,whether H is stabledependsmorestronglyonthestructureof H ,ascanbeseeninthenext theorem.Let S x;y bethedoublestarwithcentralverticesofdegree x +1and y +1. Thatis,itis K 1 ;x [ K 1 ;y withanedgejoiningtheverticesofdegreegreaterthan1. Theorem4.5 If H isagraphoforder k suchthat a 2 i )222(r i H =2 H +1 b H hasaset X of H +1 verticessuchthat H [ X ] hasoneedge,and c H K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 S b 1 ;b 2 forsome b 1 and b 2 with b 1 + b 2 = H then H is stable. Theseresultsimplythatcompletesplitgraphs,completebipartitegraphs,friendshipgraphs,andoddcycles,amongmanyothers,are stable.Wewillsaythat H is Type1 if2 i )226(r i H 2 H ,and Type2 if2 i )226(r i H =2 H +1.Ofthose graphsjustlisted,thecompletesplitgraphsandcompletebipartitegraphsareType 1,andtheoddcyclesandfriendshipgraphsareType2. 4.2Graphsthatarenot stable ThehypothesesofTheorem4.5suggestthatgraphsthatdonotsatisfyallof theseconditionsmaynotbe stable.Thisisinfactthecase,atleastif H satises conditionabutnotconditionc,asweseeinthenexttheorem. Theorem4.6 If H isagraphsuchthat 2 i )328(r i H =2 H +1 ,and H 6 K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 S b 1 ;b 2 forany b 1 and b 2 with b 1 + b 2 = H ,then H isnot stable. Proof. Considerthesequence ` = n )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 ` ; n )]TJ/F19 11.9552 Tf 11.955 0 Td [(` 2 2 ; ` +1 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 : 55
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Thisisthedegreesequenceofthegraph K ` S n )]TJ/F21 5.9776 Tf 5.756 0 Td [(` 2 ; n )]TJ/F21 5.9776 Tf 5.756 0 Td [(` 2 ,andthisgraphisinfactthe onlyrealizationof ` .Notethat ` =2 ` +1 n )]TJ/F15 11.9552 Tf 11.956 0 Td [( ` +2 ` +1. If H isType2,then H;n =2 k )]TJ/F19 11.9552 Tf 9.548 0 Td [( )]TJ/F15 11.9552 Tf 9.548 0 Td [(1 n + o n .Thus k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 H;n )]TJ/F19 11.9552 Tf 9.547 0 Td [(n forany ,provided n islargeenough.Thesequence k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 isnotpotentially H graphic,forif H K k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 S n )]TJ/F18 5.9776 Tf 5.756 0 Td [( k )]TJ/F21 5.9776 Tf 5.756 0 Td [( )]TJ/F18 5.9776 Tf 5.756 0 Td [(2 2 ; n )]TJ/F18 5.9776 Tf 5.756 0 Td [( k )]TJ/F21 5.9776 Tf 5.756 0 Td [( )]TJ/F18 5.9776 Tf 5.757 0 Td [(2 2 ,thenatleast +2verticesmust comefromthesetofverticesthatinduceadoublestar;sinceatmost ofthese verticescanbeindependentin H ,atmost oftheverticesofdegree k )]TJ/F19 11.9552 Tf 11.957 0 Td [( )]TJ/F15 11.9552 Tf 11.958 0 Td [(1may beused.Thisimpliesthat H isasubgraphof K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 S b 1 ;b 2 forsome b 1 and b 2 with b 1 + b 2 = ,contradictingourhypothesis. Itremainstoshowthat k k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 )]TJ/F19 11.9552 Tf 11.992 0 Td [( k >n forevery 2P H andsomechoice of .Since2 i )273(r i =2 +1,weknowthatforeach e j H;n 2P H ,wehave 2 j )232(r j H =2 +1.Wealsoknowthat e +1 H 2P H .Recallthat e j H;n = n )]TJ/F15 11.9552 Tf 12.325 0 Td [(1 k )]TJ/F20 7.9701 Tf 6.587 0 Td [(j ; k )]TJ/F19 11.9552 Tf 12.325 0 Td [(j + r j )]TJ/F15 11.9552 Tf 12.325 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.586 0 Td [(k + j .When j = +1,wehave e +1 H;n = n )]TJ/F15 11.9552 Tf 422.702 23.98 Td [(1 k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; k )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(k + +1 ,so k k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 )]TJ/F25 11.9552 Tf 12.169 0 Td [(e +1 H;n k = n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k + : Forany j> +1with e j H;n 2P H ,wehave k k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 )]TJ/F25 11.9552 Tf 12.169 0 Td [(e j H;n k = j )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.955 0 Td [(j + r j )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 +2 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k + +2 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.956 0 Td [(j + r j )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 + n )]TJ/F19 11.9552 Tf 11.955 0 Td [(k + k )]TJ/F19 11.9552 Tf 11.955 0 Td [(j + r j )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 =2 n j )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ j )]TJ/F19 11.9552 Tf 11.955 0 Td [( + j )222(r j )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 k +4 : Hence, k k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 )]TJ/F25 11.9552 Tf 12.169 0 Td [(e j H;n k >n foreach e j H;n 2P H andany < 1. 56
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WhichgraphssatisfytheconditionsofTheorem4.6?InTables4.1and4.2,we presentallgraphsoforderatmost6thatsatisfythehypothesesofthistheorem,and hencearenot stable.ThegraphsinTable4.2havetheadditionalpropertythat theyarenot weakly stable,becausethesequence k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 isdegreesucientforeach ofthesegraphs,demonstratingthattheconditionsforweak stabilitycannotbemet inthesecases. WecangeneralizemanyofthegraphsinTables4.1and4.2tondlargerfamilies ofgraphsthatarenot stable. Claim4.1 Let H beagraphoforder k thatisType2andsatises H 6 K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 S b 1 ;b 2 forany b 1 and b 2 with b 1 + b 2 = H .If H k +1 2 ,thenforall p 1 ,the graph H K p isnot stable. Proof. ByTheorem4.6, H itselfisnot stable.Let H p = H K p ;wewilluse Theorem4.6toshowthat H p isnot stableeither.Firstnotethat H p = H Toensurethat H p isType2,weneedtoshowthat2 i H p )222(r i H p =2 H +1. Wealreadyknowthat2 i H )312(r i H =2 H +1,soweneedtocheckthat 2 i )317(r i H p 2 H +1foreach i 2f H ;:::;k + p g .For i k ,wehave r i H p = r i H ,becauseforthesevaluesof i ,anyinducedsubgraphof H isalsoan inducedsubgraphof H p .When i>k ,wehave r i H p = i )]TJ/F15 11.9552 Tf 12.393 0 Td [(1,becausetherewill be i )]TJ/F19 11.9552 Tf 12.507 0 Td [(k verticesofdegree i )]TJ/F15 11.9552 Tf 12.507 0 Td [(1inanysetof i vertices.Thus,for i>k ,wehave 2 i )233(r i H p = i +1,whichissmallestwhen i = k +1.Since H k +1 2 ,weknow k +1+1 2 H +1,so H p isType2. Wealsoneedtoshowthat H p 6 K k + p )]TJ/F20 7.9701 Tf 6.587 0 Td [( H )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 S b 1 ;b 2 forany b 1 and b 2 with b 1 + b 2 = H .Thisisclear,becauseif H p weresuchasubgraph,then H would beasubgraphof K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 S b 1 ;b 2 .Thus,Theorem4.6impliesthat H p isnot stable. 57
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Table4.1: Connectedgraphsoforderatmost6thatarenot stable H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11 H12 H13 H14 H15 H16 H17 H18 H19 H20 H21 H22 H23 58
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Table4.2: Connectedgraphsoforderatmost6thatarenotweakly stable F1 F2 F3 F4 F5 F6 F7 F8 Inparticular,thisresultshowsthatthecompletegraph K k andthegraphs K k )]TJ/F19 11.9552 Tf 422.701 23.98 Td [(P 3 = K k )]TJ/F17 7.9701 Tf 6.587 0 Td [(4 Z 1 and K k )]TJ/F17 7.9701 Tf 6.586 0 Td [(5 P 5 arenot stable.For H = K k or H = K k )]TJ/F19 11.9552 Tf 12.711 0 Td [(P 3 thereisaset X of +1verticessuchthat H [ X ]containsexactlyoneedge,showing thatwecannotweakenthehypothesesofTheorem4.5.For H = K k )]TJ/F17 7.9701 Tf 6.587 0 Td [(5 P 5 ,thereis nosuchset.However,thereisasetof +1verticesthatinduceamatchingofsize twoandisolatedvertices.Wehavenotyetshownanythingaboutthe stabilityof suchgraphs,soonlyknowthattheyarenot stableiftheyfallunderthehypotheses ofTheorem4.6.Forfurtherinsightintothesegraphs,itisworthpointingoutthat graphsH4,H9,H10,H13,H14,andH16fromTable4.1andgraphsF2throughF7 fromTable4.2eachhavetwoedgesinanymatchinginducedbyavertexsetoforder +1. WecanmakeothergeneralizationsfromthegraphsinTables4.1and4.2.For example,thereareseveralgraphsconsistingofacompletegraphwithoneormore pendantvertices.Foragraph G ,let G ,calledthe cliquenumber of G ,bethe orderofthelargestcliquein G .Let K p m 1 ;:::;m p denotethecompletegraphon vertexset f v 1 ;:::;v p g with m j leavesincidenttovertex v j foreach j; 1 j p 59
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Thisgraphhasorder k = p + P p i =1 m i and,ifany m i =0,theindependencenumber is =1+ P p i =1 m i ifno m i =0,thentheindependencenumberissimply P p i =1 m i Claim4.2 Thegraph K p ; 0 ; 1 ; 2 ;:::;p )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 isnot stable. Proof. Firstnotethat k )]TJ/F19 11.9552 Tf 11.974 0 Td [( )]TJ/F15 11.9552 Tf 11.974 0 Td [(2= p )]TJ/F15 11.9552 Tf 11.974 0 Td [(3foranygraphofthisform.Let G k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 betheuniquerealizationof k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 ,whichhascliquenumber k )]TJ/F19 11.9552 Tf 12.302 0 Td [( = p )]TJ/F15 11.9552 Tf 12.302 0 Td [(1.Since K p m 1 ;:::;m p containsacliqueoforder p ,weseethat K p m 1 ;:::;m p isnota subgraphof G k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 For j 2f 1 ;:::;k )]TJ/F19 11.9552 Tf 11.934 0 Td [( g ,everysetof + j verticesin K p ; 0 ; 1 ;:::p )]TJ/F15 11.9552 Tf 11.934 0 Td [(2induces asubgraphofmaximumdegreeatleast2 j )]TJ/F15 11.9552 Tf 11.611 0 Td [(1.Infact,thesubgraphinducedbythe setofallofthependantverticesaswellasvertices v 1 ;v 2 ;:::v j fromthecliquehas maximumdegreeexactly2 j )]TJ/F15 11.9552 Tf 12.17 0 Td [(1.Thus, r + j =2 j )]TJ/F15 11.9552 Tf 12.171 0 Td [(1foreach j ,andinparticular thismeansthatthegraphisType2.Theorem4.6thenimpliesthatitisnot stable. Thegraph K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 12.752 0 Td [(2hasaninterestingproperty,namelythat2 i )]TJ 422.702 23.98 Td [(r i K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 12.093 0 Td [(2=2 +1foreach i 2f +1 ;:::k g ,whichmeansthatthe sequence e i K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 11.283 0 Td [(2 ;n isin P K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 11.283 0 Td [(2foreach i .Thatis, thereare k )]TJ/F19 11.9552 Tf 10.597 0 Td [( sequencesin P K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 10.597 0 Td [(2,andnoneofthemiswithin n of k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 .Anysubgraphof K p ; 0 ; 1 ;:::;p )]TJ/F15 11.9552 Tf 10.147 0 Td [(2thatcontains K p has2 i )71(r i =2 +1, soisalsonot stable.However,thesesubgraphsmaynotsatisfy r + j =2 j )]TJ/F15 11.9552 Tf 12.02 0 Td [(1for each j anddonothavethesameproperty. Itisinterestingtonotethat K p ; 1 ;:::; 1isType2butsatisestheconditions ofTheorem4.5,sois stable,and K p ; 2 ;:::; 2isType1,whichmeansitisalso stable.Thus,relativelysmalldierencesinthestructureofagraphcanchange whetherthegraphis stable. 60
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Thepreviousexamplepointsoutanotherclassofgraphsthatcannotbe stable. Since G ` = ` +2,ifagraphthatisType2hascliquenumbergreaterthan k )]TJ/F19 11.9552 Tf 12.067 0 Td [( ,itcannotbeasubgraphof G k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 .Sincethecliquenumberofagraphof order k cannotbelargerthan k )]TJ/F19 11.9552 Tf 12.065 0 Td [( +1,thismeansthatanygraph H thatisType 2andhas H = k )]TJ/F19 11.9552 Tf 11.955 0 Td [( H +1isnot stable. Wehaveshownthattherearegraphsthatarenotweakly stable{arethere graphsthat are weakly stable?Infact,completegraphsareweakly stableeven thoughtheyarenot stable. Theorem4.7 Thecompletegraph K k isweakly stableforall k 3 Proof. Toseethis,considerTheorem2.17.Ifasequence = d 1 ;:::;d n isdegreesucientfor K k ,then d k k )]TJ/F15 11.9552 Tf 11.639 0 Td [(1.Theorem2.17saysthatifinaddition d 2 k k )]TJ/F15 11.9552 Tf 11.639 0 Td [(2, orif d i 2 k )]TJ/F15 11.9552 Tf 11.472 0 Td [(1 )]TJ/F19 11.9552 Tf 11.472 0 Td [(i foreach i with1 i k )]TJ/F15 11.9552 Tf 11.472 0 Td [(2,then ispotentially K k graphic. Thusif isnotpotentially K k graphic,wemusthave d 2 k
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cycleoflength2 p with p> 3,however,wehavethat C 2 p K p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 S x;y forsome x;y with x + y = p .Thisdoesnotimplythat C 2 p is stable,though,becausewhileitis Type2,ithasatleasttwoedgesineverygraphofmaximumdegreeoneinducedby p +1vertices,sodoesnotsatisfythehypothesesofTheorem4.5.Investigatingthe stabilityoftheseandotherType2graphswiththispropertycontainingasetof +1verticesthatinducesamatchingwithatleasttwoedgesisthenextstepin ourresearch. NowwewillturnourattentiontotheproofsofTheorems4.4and4.5. 4.3Technicallemmas Foragraph G ,let D t G denotethefamilyofsubgraphsof G obtainedby deletingexactly t verticesfrom G .Thisisthefamilyofinducedsubgraphsof G with order j V G j)]TJ/F19 11.9552 Tf 18.12 0 Td [(t .Wesaythatagraphicsequence ispotentially D t G graphicif thereisarealizationof containinganygraphin D t G ToproveTheorems4.4and4.5,wewillneedthefollowingusefulconsequencesof Theorem1.2,theKleitmanWangalgorithm. Corollary4.8 Let j bethesequenceobtainedfrom = d 1 ;:::;d n bylayingo theterm d j .Then: 1.Thereisarealizationof inwhichthevertexofdegree d j isadjacenttothe d j verticesofhighestdegreeotherthanitself. 2.If j ispotentially H graphic,then ispotentially H graphic. 3.If G isagraphwithdegreesequence and v isavertexin G ,thenif G )]TJ/F19 11.9552 Tf 11.801 0 Td [(v ispotentially H graphic,then ispotentially H graphic. 4.If = n )]TJ/F15 11.9552 Tf 12.448 0 Td [(1 ;d 2 ;:::;d n ,then ispotentially H graphicifandonlyif 1 = d 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ;:::;d n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ispotentially D H graphic. 62
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Part1ofCorollary4.8guaranteestheexistenceofarealizationof inwhichthe vertexofmaximumdegree, d 1 ,isadjacenttothenext d 1 verticesofhighestdegree. Following[48],wecallsucharealizationa canonicalrealization of Lemma4.9 If H isagraphoforder k and t
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Theorem4.10TheBoundedMaxDegreeTheorem[46] Let H beagraph oforder k and = d 1 ;:::;d n beanonincreasinggraphicsequencewith n suciently largesatisfyingthefollowing: 1. isdegreesucientfor H ,and 2. d n k )]TJ/F19 11.9552 Tf 11.955 0 Td [( H Thereexistsafunction f = f H ;k suchthatif d 1 0 begiven.Thereexistsan n 0 = n ;H and < 1 k suchthatanygraphicsequence oflength n n 0 with H;n )]TJ/F19 11.9552 Tf 11.955 0 Td [(n ,either 1. ispotentially H graphic; 2. k )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 k
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Ourgoalistocreatearealizationof containingasupergraphof H ,orshowthat k )]TJ/F19 11.9552 Tf 9.998 0 Td [( 0 k > < > > : 0if H isType1 1if H isType2 : WeinitializethealgorithmbyapplyingtheKleitmanWangalgorithmTheorem1.2 to toobtainasequence 0 withminimumdegreeatleast k )]TJ/F19 11.9552 Tf 12.247 0 Td [(i + r i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 )]TJ/F20 7.9701 Tf 6.586 0 Td [( 2 .Since < 1,wedothisbyiterativelylayingotermsofvalueatmost k )]TJ/F19 11.9552 Tf 12.799 0 Td [(i + r i )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 2 beginningwiththesmallestsuchterm.Afterlayingotherstterm,wegeta 65
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sequence 1 with n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1termssuchthat 1 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F19 11.9552 Tf 11.955 0 Td [( n )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 = k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.956 0 Td [( n )]TJ/F15 11.9552 Tf 11.956 0 Td [(1+ )]TJ/F19 11.9552 Tf 11.955 0 Td [( : Aswecontinuetolayotermsoflowvalue,weobtainasequence j oflength n )]TJ/F19 11.9552 Tf 11.691 0 Td [(j with j k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.955 0 Td [( n )]TJ/F19 11.9552 Tf 11.955 0 Td [(j + j )]TJ/F19 11.9552 Tf 11.956 0 Td [( : If j )]TJ/F19 11.9552 Tf 11.956 0 Td [( > 2 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(j ,then 0 t k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n )]TJ/F19 11.9552 Tf 11.955 0 Td [(j + n )]TJ/F19 11.9552 Tf 11.955 0 Td [(j H;n )]TJ/F19 11.9552 Tf 11.955 0 Td [(j ; implyingthat j ispotentially H graphic.ByPart2ofCorollary4.8,thismeansthat isalsopotentially H graphic.Thus,layingoatmost 2 1+ n termsmustresultina sequencewiththedesiredminimumdegree.Callthissequence .Notethefollowing propertiesof :itisnotpotentially H graphic,ithaslength n 0 n )]TJ/F17 7.9701 Tf 16.232 4.708 Td [(2 1+ ,and itssmallesttermisatleast k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 )]TJ/F20 7.9701 Tf 6.587 0 Td [( 2 k )]TJ/F19 11.9552 Tf 11.956 0 Td [( )]TJ/F19 11.9552 Tf 11.955 0 Td [(b H Afterthisinitialization,weperformthefollowingstepstocreatesequences ;:::; ` forsome `
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from R t ,thenvertex v t p isalsoremovedif p>j iiLayothelargesttermof b t ,andcalltheresultingsequence q t .Thisisthe degreesequenceoftheneighborhoodof v t 1 in b R t iiiAsintheinitializationstep,applyTheorem1.2to q t ,layingotermsof smallestvalueuntilwehaveasequencewithminimumtermatleast k )]TJ/F15 11.9552 Tf 13.15 8.088 Td [(2 i )222(r i 2 )]TJ/F15 11.9552 Tf 13.151 8.088 Td [(1++ t 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +1 : Let t +1 bethesequencethatresultsfromthisstep. ivTerminateif d t +1 1
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isarealizationof ` ProofofClaim4.3. Firstwewillshowthatforeach t with0 t ` ,if t ispotentially D t H graphic,then ispotentially H graphic.Theproofisby inductionon t .If ispotentially D H graphic,thenbyrepeatedapplicationsof Part2ofCorollary4.8, ispotentially H graphic.Assumethatthestatementistrue forsome t<` ,andsuppose t +1 ispotentially D t +1 H graphic.Again,repeated applicationsofPart2ofCorollary4.8showthat q t ispotentially D t +1 H graphic. Then,as q t iscreatedfrom b t bytheremovalofadominatingvertex,weseethat b t ispotentially D t H graphic.ThisisanapplicationofPart4ofCorollary4.8. Finally,Part3ofCorollary4.8impliesthat t ispotentially D t H graphic,and theinductionhypothesisshowsthat ispotentially H graphic. Nowwewillshowthatforeach t ` hasarealizationcontaining K t G t where G t isarealizationof t .Weproceedbyinductionon t .Let G bearealizationof .ByrepeatedapplicationsofPart2ofCorollary4.8, ispotentially G graphic,establishingthebasecaseofourinduction.Supposethestatementis trueforsome t with0 t<` ,andlet G t +1 bearealizationof t +1 .Again,weuse Part2ofCorollary4.8severaltimestoshowthat q t ispotentially G t +1 graphic. Part4ofCorollary4.8thenshowsthat b t ispotentially G t +1 graphic,andwe knowthatmoreover,sincewehaveaddedadominatingvertextothegraph,thereis arealizationof b t thatcontains K 1 G t +1 .Togetfrom t to b t ,weremoved thenonneighborsofavertex v t 1 fromacanonicalrealization R t of t .Doingthis doesnotchangethegraphinducedby N R t v t 1 .Thuswemayassumethat G t is arealizationof t inwhich G t +1 N G t v t 1 ;thatis,wecanassumethat v t 1 is thevertexactingas K 1 inthecopyof K 1 G t +1 .Nowtheinductionhypothesis givesusarealizationof thatcontains K t G t ,andweseethatwecanview G t as K 1 G t +1 withadditionalverticesthatarenotadjacentto v t 1 .Thisgivesusa realizationof containing K t +1 G t +1 ,asdesired.Thus,thereisarealizationof 68
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thatcontains K ` G ` Theconstructionoftherealizationof containing K ` G ` showsthatafterthe t 'thstepofthealgorithm,thereisarealizationof inwhich t verticesareadjacent toallbutafractionoftheverticesinthegraph.Theexactvalueofthisfractionwill bedeterminedafterthenextclaim,asweneedsomeofthefactsdevelopedinthat claiminordertocalculateit. Claim4.4 t k )]TJ/F19 11.9552 Tf 11.56 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.56 0 Td [(1 )]TJ/F15 11.9552 Tf 11.561 0 Td [(+ t )]TJ/F15 11.9552 Tf 11.56 0 Td [(2 t n t ,andatmost t +3 1+ n t iterationsoftheKleitmanWangalgorithmareneededateachimplementationofStep iii. ProofofClaim4.4. Firstwewillprovethelowerboundon t .Againtheproof isbyinductionon t .Theclaimholdsfor byhypothesis.Suppose t 0and t k )]TJ/F19 11.9552 Tf 11.392 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.391 0 Td [(1 )]TJ/F15 11.9552 Tf 11.392 0 Td [(+ t )]TJ/F15 11.9552 Tf 11.392 0 Td [(2 t n t .Wewillshowthattheinequality holdsfor t +1 .Let M =2 k )]TJ/F15 11.9552 Tf 12.311 0 Td [(2 )]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 12.312 0 Td [(1.Since d t 1 >n t )]TJ/F25 11.9552 Tf 12.312 9.684 Td [()]TJ/F20 7.9701 Tf 13.478 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 12.312 0 Td [(1, creating b t from t entailsremovingatmost )]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.392 0 Td [(1vertices,eachofwhich hasdegreeatmost k )]TJ/F15 11.9552 Tf 12.028 0 Td [(2byTheorem2.17.Thus, b t t )]TJ/F19 11.9552 Tf 12.028 0 Td [(M .Wehave assumedthat n issucientlylarge,soinparticular,since n t increaseswith n ,we mayassume M n t ,whichmeansthat b t k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.956 0 Td [( t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t n t )]TJ/F19 11.9552 Tf 11.955 0 Td [(n t = k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t n t k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t b n t ; 69
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where b n t isthelengthof b t .Creating q t from b t requireslayingoatermof value b n t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1,so q t = b t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 b n t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1.Wenowhave q t k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 t b n t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 b n t )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1 b n t k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1 q n t ; where q n t isthelengthof q t .Finally,thelaststepofouralgorithminvolveslaying otermsfrom q t thathavevaluelessthan k )]TJ/F17 7.9701 Tf 12.633 5.458 Td [(2 i r i 2 )]TJ/F17 7.9701 Tf 12.633 5.699 Td [(1+ t +2 2 )]TJ/F15 11.9552 Tf 11.438 0 Td [( t +1.Therst iterationofthisprocessyieldsanewsequence q t 1 with q t 1 q t )]TJ/F15 11.9552 Tf 11.955 0 Td [([2 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1] [2 k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1] q n t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 : Repeatingthis j timesyieldsthesequence q t j with q t j [2 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1] q n t )]TJ/F19 11.9552 Tf 11.955 0 Td [(j =[2 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 t +1] q n t;j ; where q n t;j isthelengthof q t j .Since t +1 = q t j forsome j ,wehaveestablishedthe lowerboundon t +1 Ifweanalyzethisnalstepmorecarefully,wegetthesecondhalfoftheclaim.We haveseenthat q t k )]TJ/F19 11.9552 Tf 10.573 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 10.574 0 Td [(1 )]TJ/F15 11.9552 Tf 10.573 0 Td [(+ t +1 )]TJ/F15 11.9552 Tf 10.573 0 Td [(2 t +1 q n t .Whenwe layoatermof q t whosevalueistoosmall,wesubtract2 k )]TJ/F19 11.9552 Tf 10.641 0 Td [(i + r i )]TJ/F15 11.9552 Tf 10.64 0 Td [(2 )]TJ/F15 11.9552 Tf 10.64 0 Td [(2 t +1 from q t ,sothat,ifwecalltheresultingsequence q t 1 ,weseethat q t 1 isat 70
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least [ k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +2 )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 t +1] q n t )]TJ/F15 11.9552 Tf 11.955 0 Td [([2 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1]+1 =[ k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 t +1] q n t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 + )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +2 : Afterrepeatingthis j times,wegetasequence q t j withsumatleast [ k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i + r i H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +2 )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 t +1] q n t )]TJ/F19 11.9552 Tf 11.955 0 Td [(j + )]TJ/F15 11.9552 Tf 11.955 0 Td [( t +2 j: Now,if q t j H; q n t )]TJ/F19 11.9552 Tf 12.926 0 Td [(j )]TJ/F15 11.9552 Tf 12.927 0 Td [(2 t +1 q n t )]TJ/F19 11.9552 Tf 12.927 0 Td [(j ,thenbyLemma4.9, q t j F; q n t )]TJ/F19 11.9552 Tf 10.146 0 Td [(j forsome F 2D t +1 H ,so q t j ispotentially D t +1 H graphic.Thiswill happenif )]TJ/F15 11.9552 Tf 10.419 0 Td [( t +2 j t +3 q n t )]TJ/F19 11.9552 Tf 10.418 0 Td [(j ,whichmeansthatif j t +3 1+ n t t +3 1+ q n t then q t j ispotentially D t +1 H graphic.Since t +1 = q t j forsome j ,thisinturn means t +1 ispotentially D t +1 H graphic.Finally,Claim4.3showsthatinthis case, ispotentially H graphic.Thus,atmost t +3 1+ n t iterationsoftheKleitmanWangalgorithmresultinthedesiredsequence t +1 Wewouldliketoknowhowmanytermshavebeenremovedfrom inthecreation of ` .Thatis,whatisthevalueof n )]TJ/F19 11.9552 Tf 12.055 0 Td [(n ` ?Ateachiterationofthealgorithm,the vertex v t 1 missesatmost )]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 12.43 0 Td [(1vertices,soinStepiatmostthismany verticesareremoved.InStepii,onlyonevertexisremoved.Then,Claim4.4shows thatatmost + k 1+ n termsarelaidoinStepiii.Sincewedothealgorithmat most k )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.956 0 Td [(1times,anddotheinitializationsteponce,thismeansthat n )]TJ/F19 11.9552 Tf 11.955 0 Td [(n ` k )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1+ + k 1+ n + 2 1+ n: .1 71
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Claim4.3showsthatthereisarealizationof thatcontains K ` G ` ,where G ` isarealizationof ` .The ` verticesinthecliqueareadjacenttoeveryvertex in G ` ,sotheyareonlynonadjacenttoverticesthathavebeenremovedorlaido throughthecourseofthealgorithm.Thus,theyarenotadjacenttoatmost n )]TJ/F19 11.9552 Tf 12.117 0 Td [(n ` vertices;thatis,theexpressioninEquation.1givesanupperboundonhowmany verticesarenotadjacenttothecliqueinthisrealization. Stage2: InStage2,weanalyzetherealizationsof ` withthegoalofshowing thatsomerealizationcontainsasupergraphof H .RecallthatinStage1,ouralgorithmstopseitherwhenwehaveiteratedenough"times,orwhenthesequence t failsthemaximumdegreeconditionofTheorem4.10. Ifthealgorithmstopsbecausewehaveiterated k )]TJ/F19 11.9552 Tf 12.33 0 Td [( H )]TJ/F19 11.9552 Tf 12.331 0 Td [(b H times,thenthe graph K ` G ` contains K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F20 7.9701 Tf 6.587 0 Td [(b H K H + b H ,whichisasupergraphof H when H isType1,sointhiscase ispotentially H graphic.If H isType2,thenweareleft witharealizationof containing K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 K H +1 .ThissucestoproveParts 1and3ofthelemma. Thereforeweassumethatthealgorithmstopswhen t +1= `f + b H ;k )]TJ/F19 11.9552 Tf 11.791 0 Td [(` ,where f isthefunctiongivenbyTheorem4.10.Since K ` S ` = K k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F20 7.9701 Tf 6.586 0 Td [(b H K + b H ,when H isType1thisshowsthat ispotentially H graphic.If H isType2,thenthere isarealizationof containing K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 K +1 .Again,thissucestoproveParts1 and3ofthelemma. Wecanthereforeassumethat ` isnotdegreesucientfor S ` .Let p =max f j : d ` j k )]TJ/F19 11.9552 Tf 12.083 0 Td [(` )]TJ/F15 11.9552 Tf 12.083 0 Td [(1 g .Since ` isnotdegreesucientfor S ` buttheminimumtermof 72
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` isatleast k )]TJ/F19 11.9552 Tf 11.557 0 Td [( H )]TJ/F19 11.9552 Tf 11.557 0 Td [(b H )]TJ/F19 11.9552 Tf 11.556 0 Td [(` ,weknowthat pp = j V K p K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p j)]TJ/F19 11.9552 Tf 20.003 0 Td [( K p K k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p and d ` 1 < n ` )]TJ/F25 11.9552 Tf 11.955 9.684 Td [()]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1
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positivetermsandmaximumtermatmost k )]TJ/F19 11.9552 Tf 12.386 0 Td [(` )]TJ/F15 11.9552 Tf 12.386 0 Td [(2.BytheErd}osGallaicriteria Theorem1.3,thesequence 0 isgraphic.Combiningarealizationof 0 withthe constructedgraphgivesusthedesiredrealizationof ,showingthat ` ispotentially K p F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p graphic.Since K p F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p isasupergraphofagraphin D ` H ,Claim 4.3impliesthat ispotentially H graphic. Stage3: Wehaveshownthatinmanycases,wecanconstructarealization of thatcontainseitherasupergraphof H or K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 K +1 .Thisgivesusthe conclusionsinParts1and3ofthelemma.Nowwewillshowthatintheremaining case,thatiswhen ` isnotdegreesucientfor K p F k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p ,wehave k )]TJ/F19 11.9552 Tf 11.441 0 Td [( 0 k
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Wethushavethefollowingupperboundon ` : ` p n ` )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F19 11.9552 Tf 11.955 0 Td [(p k )]TJ/F19 11.9552 Tf 11.956 0 Td [(` )]TJ/F15 11.9552 Tf 11.955 0 Td [(2+ n ` )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.956 0 Td [(` )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p + p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 = p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` + k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )222(r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 p = p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` + o n : Wealsoknow,byClaim4.4,that ` k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( ` +1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 ` n ` k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F15 11.9552 Tf 11.955 0 Td [( k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F19 11.9552 Tf 11.955 0 Td [(p )222(r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( ` +1 n ` = p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.955 0 Td [( ` +1 n ` : Let bethedegreesequenceof K ` G ` ,where G ` isarealizationof ` .Then = ` +2 `n ` + ` 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(` p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.955 0 Td [( ` +1 n ` +2 `n ` + ` 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(`: Nowconsiderthesequence e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` .Weneedtoshowthat e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` 2P H .Thesumofthissequenceis e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` = p + r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` )]TJ/F19 11.9552 Tf 11.955 0 Td [(p 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(p r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p +2 `n ` + ` 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(` : Aftersimplifying,weseethatthecoecientof n ` + ` in e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` is 2 ` + p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1.If e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` isnotin P H ,then ` + p + r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` + ` < k )]TJ/F19 11.9552 Tf 11.956 0 Td [(i )222(r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` + ` : .2 75
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However,wealsoknowthat e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` k )]TJ/F19 11.9552 Tf 11.022 0 Td [(i )144(r i )]TJ/F15 11.9552 Tf 11.022 0 Td [(1 )]TJ/F15 11.9552 Tf 11.022 0 Td [( ` +1 n ` +2 `n ` + ` 2 )]TJ/F19 11.9552 Tf 11.022 0 Td [(`: .3 Inequalities.2and.3togetherimplythat k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i )222(r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F15 11.9552 Tf 11.955 0 Td [( ` +1 n ` +2 `n ` + ` 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(` k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i )222(r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 n ` + ` ; whichisnottrueforsucientlylarge n ` .Thus, e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` 2P H Inordertocalculate k )]TJ/F25 11.9552 Tf 12.785 0 Td [(e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` k ,notethattheonlytermsof = a 1 ;:::;a n ` + ` thataregreaterthanthoseof e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` arethose a j forwhich p + ` +1 j k )]TJ/F15 11.9552 Tf 12.443 0 Td [(1.Thus,wecanboundthedistancebetweenthesesequences bytakingtheabsolutedierenceoftheirsumsandaddingtwotimesthedierence ofthetermsinthatrange.Sincethelargesttermsof inthatrangeareatmost k )]TJ/F15 11.9552 Tf 12.47 0 Td [(2,andthecorrespondingtermsof e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p equal r k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p + p + ` )]TJ/F15 11.9552 Tf 12.47 0 Td [(1,weaddat most2 k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F19 11.9552 Tf 11.955 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )222(r k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p k )]TJ/F19 11.9552 Tf 11.955 0 Td [(` )]TJ/F19 11.9552 Tf 11.955 0 Td [(p )]TJ/F15 11.9552 Tf 11.955 0 Td [(1tothedierenceofsums.Thisyields k )]TJ/F25 11.9552 Tf 12.169 0 Td [(e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` k e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` )]TJ/F19 11.9552 Tf 11.955 0 Td [( +2 k 2 ` +1 n ` +2 k 2 : Nowwecalculate k )]TJ/F19 11.9552 Tf 12.055 0 Td [( k ;thisiswhereweusealloftheinformationwegained inStage1aboutthestructureof .RecallfromEquation.1thatatmost ` k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1+ + k 1+ n + 2 1+ n termsareremovedfrom tocreate ` .Ofthese,thetermsthatwerelaidoin Stepiibecausetheywerethersttermofsome t havehighdegree,andthose thatwereremovedduringStepsiandiiiofthealgorithmhaveverylowdegree. 76
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Theorem2.17impliesthatthosetermsremovedinStepsiandiiiareinfactat most k )]TJ/F15 11.9552 Tf 12.544 0 Td [(3,andthehypothesesofthealgorithmimplythatthetermsremovedin Stepiiareatleast n ` )]TJ/F25 11.9552 Tf 12.614 9.683 Td [()]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 12.614 0 Td [(1.Thetermsof thatgivethedegreesof verticesinthecliquein K ` G ` correspondtothetermsof thatwereremoved inStepii.Sothedierencebetweenthesetermsinthetwosequencesisatmost )]TJ/F20 7.9701 Tf 13.477 4.379 Td [(k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 12.33 0 Td [(1.Thetermsof thatwereremovedinStepsiandiiicorrespond tozerosin .Thus,thedierencebetweenthesetermsisatmost k )]TJ/F15 11.9552 Tf 11.532 0 Td [(3.Weseethat since n ` n t n and ` k )]TJ/F19 11.9552 Tf 11.956 0 Td [( H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1,wehave k )]TJ/F19 11.9552 Tf 11.955 0 Td [( k + k )]TJ/F19 11.9552 Tf 11.955 0 Td [( H 1+ n + k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k )]TJ/F15 11.9552 Tf 11.955 0 Td [(3 k )]TJ/F19 11.9552 Tf 11.955 0 Td [( H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 + k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k )]TJ/F19 11.9552 Tf 11.955 0 Td [( H )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ 2 1+ n k 3 +2 1+ n + k d k= 2 e k 4 : Finally,weknowthat k e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n ` + ` )]TJ/F25 11.9552 Tf 12.088 0 Td [(e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.586 0 Td [(p H;n k = n )]TJ/F15 11.9552 Tf 11.875 0 Td [( n ` + ` k )]TJ/F19 11.9552 Tf 422.702 23.981 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1,andbyEquation.1,thisisatmost k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k d k= 2 e k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+ k + k 1+ n k )]TJ/F19 11.9552 Tf 11.955 0 Td [(i + r i )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 k d k= 2 e k 4 +2 k 2 + k 1+ n: Thus,wehave k )]TJ/F25 11.9552 Tf 12.169 0 Td [(e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n kk )]TJ/F19 11.9552 Tf 11.955 0 Td [( k + k )]TJ/F25 11.9552 Tf 12.168 0 Td [(e k )]TJ/F20 7.9701 Tf 6.586 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` k + k e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n ` + ` )]TJ/F25 11.9552 Tf 12.169 0 Td [(e k )]TJ/F20 7.9701 Tf 6.587 0 Td [(` )]TJ/F20 7.9701 Tf 6.587 0 Td [(p H;n k ` +1 n + k 2 + k 3 +2 1+ n +2 k d k= 2 e k 4 +2 k 2 + k 1+ n k )]TJ/F19 11.9552 Tf 11.955 0 Td [( + k 3 +2 1+ +2 k 2 + k 1+ n +2 k d k= 2 e k 4 + k 2 : 77
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Thus,as approaches0thisexpressionislessthan n FromtheproofofLemma4.11,weseethatwhen H isType1,thenweonlyreach theconclusionsinParts1and2ofthelemma,thusimplyingthat H is stable.Thus, Theorem4.4isanimmediatecorollary.When H isType2,however,wemayend upinthecasewhereLemma4.11canonlyguaranteearealizationof containing K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( H )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 K H +1 .Now,toproveTheorem4.5,weanalyzewhathappensinthis case. ProofofTheorem4.5. Let H beagraphthatsatisesthehypothesesof Theorem4.5.Let > 0begiven,andlet = d 1 ;:::;d n beagraphicsequenceof length n thatisnotpotentially H graphicandsuchthat k )]TJ/F19 11.9552 Tf 12.17 0 Td [(i + r i )]TJ/F15 11.9552 Tf 422.702 23.98 Td [(1 )]TJ/F19 11.9552 Tf 12.009 0 Td [( n ,where < 1 k .ByLemma4.11,either k )]TJ/F19 11.9552 Tf 12.009 0 Td [( 0 k
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that u a 1 and v a 2 .Itispossiblethat a 1 = a 2 .ByTheorem2.17,wemayassume thateachvertexin V G )149(f v 1 ;:::;v k g hasdegreeatmost2 k )]TJ/F15 11.9552 Tf 11.079 0 Td [(4.Thus,thereareat most k 2 +1verticesatdistanceatmost2from a 1 ,andatmost k 2 +1verticesat distanceatmost2from a 2 .Since n issucientlylarge,thereisavertex w thatisat distanceatleast3fromboth a 1 and a 2 .Since H;n )]TJ/F19 11.9552 Tf 11.247 0 Td [(n ,thatvertexmust havepositivedegreein G .Let x beaneighborof w ;notethat x isnotadjacentto a 1 or a 2 ,butitispossiblethat x 2f v 1 ;:::;v k gnf u;v g .Wecanexchangetheedges ua 1 va 2 ,and wx forthenonedges uv wa 1 ,and xa 2 togetarealizationof that contains K k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 K +1 + e ,where e istheedge uv intheindependentset.Since H hasasetof +1verticesthatinducesoneedge,thisshowsthat ispotentially H graphic. Weareleftwiththecasewhereexactlyonevertexin S hasdegreegreaterthan k )]TJ/F19 11.9552 Tf 10.667 0 Td [( )]TJ/F15 11.9552 Tf 10.668 0 Td [(1.Wemayassumethatthisvertexis v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( ,so d k )]TJ/F20 7.9701 Tf 6.587 0 Td [( k )]TJ/F19 11.9552 Tf 10.667 0 Td [( but d j k )]TJ/F19 11.9552 Tf 10.668 0 Td [( )]TJ/F15 11.9552 Tf 10.668 0 Td [(1 forall j k )]TJ/F19 11.9552 Tf 12.253 0 Td [( +1.For j 2f 1 ;:::;k )]TJ/F19 11.9552 Tf 12.252 0 Td [( )]TJ/F15 11.9552 Tf 12.252 0 Td [(1 g ,let W j bethesetofneighborsof v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( thatarenotadjacentto v j .Recallthat H K k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 S b 1 ;b 2 forsome b 1 and b 2 Suppose b 1 b 2 .If d k )]TJ/F20 7.9701 Tf 6.587 0 Td [( k )]TJ/F19 11.9552 Tf 11.916 0 Td [( )]TJ/F15 11.9552 Tf 11.916 0 Td [(1 b 2 +1,thenforsome p 2f 1 ;:::;k )]TJ/F19 11.9552 Tf 11.916 0 Td [( )]TJ/F15 11.9552 Tf 11.916 0 Td [(1 g wehave j W p j b 2 .Let x 2 W p .Thenwecanexchangetheedges xv k )]TJ/F20 7.9701 Tf 6.586 0 Td [( and v p v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( +1 forthenonedges xv p and v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( v k )]TJ/F20 7.9701 Tf 6.587 0 Td [( +1 .Wecandoasimilaredgeexchangefor b 2 verticesin W p andthevertices v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( +2 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( + b 2 until v k )]TJ/F20 7.9701 Tf 6.587 0 Td [( isadjacenttoeachof v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( +1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( + b 2 ,while v p isnolongeradjacenttothesevertices.Thus,wecreate arealizationof where v p and v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( arethecentersofadoublestar S b 1 ;b 2 thatis joinedtoacompletegraphonthevertices f v 1 ;:::;v k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 gnf v p g ,whichmeansthat ispotentially H graphic. So d k )]TJ/F20 7.9701 Tf 6.586 0 Td [( k )]TJ/F19 11.9552 Tf 9.96 0 Td [( )]TJ/F15 11.9552 Tf 9.96 0 Td [(1 b 2 +1.Inthiscasewecanshowthat k )]TJ/F25 11.9552 Tf 10.174 0 Td [(e +1 H;n k
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majorizes exceptatthe k )]TJ/F19 11.9552 Tf 11.955 0 Td [( th term.Thus, k )]TJ/F25 11.9552 Tf 12.169 0 Td [(e +1 H;n k < e +1 H;n )]TJ/F19 11.9552 Tf 11.955 0 Td [( + d k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F15 11.9552 Tf 11.956 0 Td [( k )]TJ/F19 11.9552 Tf 11.955 0 Td [( )]TJ/F15 11.9552 Tf 11.955 0 Td [(1
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Thefactthat K k isweakly stableisreectedintheobservationthat k )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 is notdegreesucientfor K k ,andinfactifthereareotherextremalsequencesfor K k theymustalsofailtobedegreesucient.Ifagraph H isweakly stablebutnot stable,anyextremalsequencethatisnotin P H mustnotbedegreesucient for H .Ifagraphisnotweakly stable,thenanyextremalsequencesinadditionto thoseof P H mayormaynotbedegreesucientforthegraph.Anaturalquestion toaskiswhetherthesequence k )]TJ/F20 7.9701 Tf 6.587 0 Td [( )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 istheonlyadditionalextremalsequencefor graphsthatarenot stable,orifotherscanbefound. CurrentlyweknowverylittleaboutgraphsthatareType2buthaveasetof +1 verticesthatinduceamatchingofsizeatleast2.Wehaveshownthatifsuchagraph isnotasubgraphof K k )]TJ/F20 7.9701 Tf 6.586 0 Td [( )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 S x;y ,thenitisnot stable,butwehavenopositive resultsonthe stabilityofsuchgraphs.Determiningmoreaboutsuchgraphsisthe nextstepinourstudyof stability. 81
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5.PotentialRamseyNumbers 5.1Introduction AsdiscussedinChapter2,the k colorgraphRamseynumber r G 1 ;:::;G k where G 1 ;:::;G k aregraphs,istheminimuminteger n suchthatany k edgecoloring of K n yieldsamonochromatic G i incolor i forsome i .Thepotentialversionof thisproblemwasintroducedbyBusch,Ferrara,HartkeandJacobsonin[9],and isdenedrstfortwocolorswewilldiscussamulticolorversioninChapter6. Givengraphs G 1 and G 2 andagraphicsequence = d 1 ;:::;d n ,wewrite G 1 ;G 2 ifeither ispotentially G 1 graphicor ispotentially G 2 graphic,where = n )]TJ/F15 11.9552 Tf 11.75 0 Td [(1 )]TJ/F19 11.9552 Tf 11.75 0 Td [(d 1 ;n )]TJ/F15 11.9552 Tf 11.749 0 Td [(1 )]TJ/F19 11.9552 Tf 11.75 0 Td [(d 2 ;:::;n )]TJ/F15 11.9552 Tf 11.75 0 Td [(1 )]TJ/F19 11.9552 Tf 11.75 0 Td [(d n isthecomplementarydegreesequenceof .The potentialRamseynumber of G 1 and G 2 ,denoted r pot G 1 ;G 2 ,istheminimum integer n suchthatif isagraphicsequenceoflengthatleast n ,then G 1 ;G 2 Wecaneasilyshowthat r pot G 1 ;G 2 r G 1 ;G 2 ,becausearealizationofa graphicsequenceoflength n canbethoughtofasgivinga2edgecoloringof K n Thatis,arealizationof determinestherededgesin K n ,andthecomplementary realizationof determinestheblueedges.Thus,ifeveryred/bluecoloringofthe edgesof K n producesared G 1 orblue G 2 ,theneverygraphicsequenceoflength n iseitherpotentially G 1 graphicoritscomplementispotentially G 2 graphic.This boundissharpinsomecases;inparticularwehavethefollowingresultfrom[9]. Lemma5.1Buschetal.[9] Let r = r G 1 ;G 2 andlet G beagraphoforder r )]TJ/F15 11.9552 Tf 9.39 0 Td [(1 suchthat G 1 G and G 2 G .If G isunigraphic,then r G 1 ;G 2 = r pot G 1 ;G 2 Buschetal.usedthisandaresultofGerencserandGyarfas[49]toshowthat r pot P s ;P t = r P s ;P t = s + b t 2 c)]TJ/F15 11.9552 Tf 19.925 0 Td [(1. DeterminingtheRamseynumber r K s ;K t isoneoftheforemostopenproblems incombinatorics.Awellknownlowerboundon r K t ;K t is r K t ;K t p 2 e t 2 t= 2 provenbySpencerin[114].However,thepotentialRamseynumberiscomparatively simple;infact,Buschetal.determined r pot K s ;K t forall s;t ,thusshowingthat 82
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r pot K t ;K t islinearin t andthatthesimpleboundinLemma5.1canbeveryfar fromsharp. Theorem5.2Buschetal.[9] For s t 3 r pot K s ;K t =2 s + t )]TJ/F15 11.9552 Tf 12.2 0 Td [(4 ,except when s = t =3 ,inwhichcase r pot K 3 ;K 3 =6 TheyalsodeterminedthepotentialRamseynumbers r pot C s ;K t and r pot P s ;K t Theorem5.3Buschetal.[9] If s 3 and b 2 s 3 c t 2 ,then r pot C s ;K t = s + t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 .If s 4 and t> b 2 s 3 c 2 ,then r pot C s ;K t =2 t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2+ d s 3 e Theorem5.4Buschetal.[9] For s 6 and t 3 r pot P s ;K t = 8 > > < > > : s + t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2if t b 2 s 3 c 2 t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2+ b s 3 c if t> b 2 s 3 c : Inthischapter,wepresentresultson r pot H;G ,where H isoneof K 2 ; 2 K 2 ;K 3 ;P 3 or P 4 ,and G isanarbitrarygraphoforder t ,aswellas r pot H;G forall H and G of orderatmost4.Wewillalsostateandprovethevalueof r pot C s ;C t 5.2Preliminaries Wewillassumefromnowonthat G isagraphoforder t .If H isanontrivial graph,thenitiseasytoseethat r pot H;G t ,forif = t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 6! H;G .Itis alsoeasytoseethat r pot H;tK 1 = r pot tK 1 ;H = t .Wecanthusassumethatall graphshaveatleastonenontrivialcomponent.Webeginwithasimpleobservation aboutthepotentialRamseynumberofgraphswithisolatedvertices. Proposition5.5 Suppose G = G 1 [ K 1 and r pot H;G 1 = r .If r j V G j ,then r pot H;G = j V G j ,andif r> j V G j ,then r pot H;G = r Proof. Let n = j V G j .Firstsuppose r n ,andlet beagraphicsequenceof length n .If isnotpotentially H graphic,thenweknowthat hasarealization R 83
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containing G 1 asasubgraph.Since j V G 1 j n ,let beagraphicsequenceoflength r .If isnotpotentially H graphic, thenagain hasarealization R containing G 1 ,andsince r>n ,thereisavertex in R thatisnotusedbythecopyof G 1 ,sothereisinfactacopyof G in R .Thus, H;G .Since r pot H;G 1 = r ,weknowthat r pot H;G r ,so r pot H;G = r Wemayhenceforthassumethat G and H donothaveanyisolatedvertices. 5.2.1Generallowerbounds Wenowdetermineseverallowerboundson r pot H;G Lemma5.6 If H isagraphoforder t thatisnotasubgraphof K 1 ;t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and G hasno isolatedvertices,then r pot H;G t +1 Proof. Consider = t )]TJ/F15 11.9552 Tf 12.596 0 Td [(1 ; 1 t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Thissequenceisuniquelyrealizedbythestar K 1 ;t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,so isnotpotentially H graphic.Thecomplementof isuniquelyrealized by K t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 + K 1 ,whichcancontainnographon t verticesunlessthatgraphhasatleast oneisolate.So r pot H;G t +1. Themaximumsizeofamatchinginthecomplementof G 0 G ,turnsoutto playanimportantroleindetermining r pot H;G .Tosimplifynotation,wewilloften write 0 inplaceof 0 G when G isunderstood. Lemma5.7 If G isagraphoforder t 3 and H isagraphwithaconnected componentoforderatleast3,then r pot H;G max f 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 G )]TJ/F15 11.9552 Tf 11.956 0 Td [(1+1 ;t g Proof. Wehavealreadyseenthat r pot H;G t .Thuswemayassume2 t )]TJETq1 0 0 1 517.474 126.781 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 517.474 119.96 Td [( 0 )]TJ/F15 11.9552 Tf 422.702 23.98 Td [(1+1 t .Let = 2 t )]TJETq1 0 0 1 254.941 105.005 cm[]0 d 0 J 0.359 w 0 0 m 5.439 0 l SQBT/F20 7.9701 Tf 254.941 100.319 Td [( 0 )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Thisdegreesequenceisuniquelyrealizedbya matchingofsize t )]TJETq1 0 0 1 218.308 78.821 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 218.308 72 Td [( 0 )]TJ/F15 11.9552 Tf 12.721 0 Td [(1,soitdoesnotcontainacopyofanygraph H witha 84
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componentoforderatleast3.Anygraphon t verticesinthecomplementofthis realizationmustuseatleast 0 +1pairsofverticesthatareendpointsofoneofthe edges,whichrequiresamatchingofsizeatleast 0 +1in G .Thus, 6! H;G and r pot H;G 2 t )]TJETq1 0 0 1 203.926 642.925 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 203.926 636.104 Td [( 0 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1. Recallthatifthedegreesequenceof G is x 1 ;:::;x t and = d 1 ;:::;d n ,then isdegreesucientfor G if d i x i foreach i; 1 i k .Asimplenecessary conditionfor tobepotentially G graphicisthat bedegreesucientfor G .Thus, sequencesthatfailtobedegreesucientfor H and G oftenplayaroleindetermining r pot H;G .Inparticular,theminimumdegreeof G becomesimportantwhen 0 G islargeandthelowerboundprovidedbyLemma5.7issimply t .Whenthishappens, thefollowinglemmamaybeused. Lemma5.8 If G isagraphoforder t 5 suchthat G d t +1 2 e and H isagraph withthreeormoreverticesofdegreeatleast2,then r pot H;G t +2 Proof. Supposerstthat t isodd,so G t +1 2 .Considerthesequence 1 = t +1 )]TJ/F19 11.9552 Tf 13.285 0 Td [( 2 ; 1 t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,whichisgraphicbytheErd}osGallaicriteriaTheorem1.3. Clearly, 1 isnotpotentially H graphicbecauseitisnotdegreesucientfor H Ontheotherhand, 1 isnotdegreesucientfor G ,so 1 6! H;G If t iseven,considerthesequence 2 = t +2 )]TJ/F19 11.9552 Tf 12.804 0 Td [(;t +1 )]TJ/F19 11.9552 Tf 12.804 0 Td [(; 1 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,whichhas 2 = t )]TJ/F15 11.9552 Tf 12.487 0 Td [(1 t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ; )]TJ/F15 11.9552 Tf 12.487 0 Td [(1 ; )]TJ/F15 11.9552 Tf 12.487 0 Td [(2.Since t +2 2 ,theErd}osGallaicriteriaagainshow that 2 isgraphic.However, 2 isnotdegreesucientfor H ,soisnotpotentially H graphic,while 2 isnotdegreesucientforanygraphoforder t andminimum degree .Theresultfollows. 85
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5.2.2Towardsanupperbound Wewilloftenwanttoshowthatagraph H isasubgraphofthecomplementof anothergraph, G .Todothis,wewilllookforamapping f withthepropertythat if f u G f v ,then u H v .Ifthereissuchamapping,anditisinjective,then H G Aswewilldiscuss,when H is P 3 P 4 ,or K 3 ,sequencesthatarenotpotentially H graphicmustendwithastringof1'sor0's.Thus,thesesequencesarerealized bygraphscontainingconnectedcomponentswithatleastthreevertices,andasetof disjointedgesandvertices.Thefollowingresultswillthereforehelpustodetermine upperboundsonthepotentialRamseynumber. Lemma5.9 If G isagraphoforder t 3 and R isagraphoforderatleast max f 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 G )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 ;t g suchthat R 1 ,then G isasubgraphof R Proof. Firstsuppose 0 2 t )]TJETq1 0 0 1 201.686 241.062 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 201.686 234.24 Td [( 0 )]TJ/F15 11.9552 Tf 12.044 0 Td [(1+1,thereareagainatleast t )]TJETq1 0 0 1 406.963 241.062 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 406.963 234.24 Td [( 0 independentverticesin R ;choosingthelargestindependentsetin R ,atmost 0 moreverticesareneededto createacopyof G .Thismeansatmost 0 edgesof R mustbeused,andthiscan alsobeidentiedwithamatchingin G asbefore. If 0 = t= 2,thenmax f 2 t )]TJETq1 0 0 1 271.714 145.141 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 271.714 138.32 Td [( 0 )]TJ/F15 11.9552 Tf 11.54 0 Td [(1+1 ;t g = t .Inthiscase,if j R j = t ,thenthere areatmost t= 2disjointedgesin R ,andtheedgesofamatchingin G canbeidentied withtheseedgesandanyisolatedverticesin R toseethat G R .If j R j >t ,itis 86
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easytoseethat G R Wenowhavethefollowingeasycorollary. Corollary5.10 Let G begraphoforder t 3 ,and R agraphoforderatleast 2 t )]TJ/F19 11.9552 Tf 10.087 0 Td [( 0 G )]TJ/F15 11.9552 Tf 10.087 0 Td [(1+1 suchthat R = R 1 [ R 2 ,where R 1 isthemaximalsubgraphof R with R 1 1 .Supposethereisasubgraph G 2 of G suchthat G 2 R 2 ,andlet G 1 bethe subgraphof G inducedby V G n V G 2 .If j R 1 j max f 2 j G 1 j)]TJ/F19 11.9552 Tf 14.85 0 Td [( 0 G 1 )]TJ/F15 11.9552 Tf 10.414 0 Td [(1+1 ; j G 1 jg then G R Proof. Lemma5.9showsthatif j R 1 j max f 2 j G 1 j)]TJ/F19 11.9552 Tf 18.75 0 Td [( 0 G 1 )]TJ/F15 11.9552 Tf 12.364 0 Td [(1+1 ; j G 1 jg ,then G 1 R 1 .Sincetherearenoedgesbetween R 1 and R 2 in R ,thisimpliesthat G R Lemma5.11 Let G and R beasgiveninCorollary5.10,andsuppose j R 2 j = q + a where a = R 2 ,and q 0 = j R j)]TJ/F15 11.9552 Tf 17.933 0 Td [( t )]TJETq1 0 0 1 311.892 371.244 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 311.892 364.423 Td [( 0 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 .If i 0 > b t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c and j R j)]TJ/F19 11.9552 Tf 17.932 0 Td [(t q ,or ii 0 b t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c and q 0 )]TJ/F19 11.9552 Tf 11.955 0 Td [(q + a 0 then G R Proof. Chooseaset V 1 of t )]TJ/F19 11.9552 Tf 12.399 0 Td [(a verticesof G sothat 0 G [ V 1 ]ismaximized,and let G 1 = G [ V 1 ].Weclaimthat 0 G 1 =min fb t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c ; 0 g .Clearly,thisisanupper boundonthesizeof 0 G 1 .Let M beamaximummatchingin G .Tomaximize 0 G 1 ,webeginbyselectingfor V 1 onlythoseverticesthataresaturatedby M .If t )]TJ/F19 11.9552 Tf 11.355 0 Td [(a 2 0 ,then V 1 containsallverticessaturatedby M ,so 0 G 1 0 .Otherwise, wecanchooseupto t )]TJ/F19 11.9552 Tf 12.04 0 Td [(a verticesthataresaturatedby M sothatwehaveatleast b t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c edgesof M included.Inthiscase, 0 G 1 = b t )]TJ/F20 7.9701 Tf 6.586 0 Td [(a 2 c 87
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Supposethatconditioniholds.Then j R 1 j = j R j)]TJ/F15 11.9552 Tf 13.975 0 Td [( q + a q + t )]TJ/F15 11.9552 Tf 9.977 0 Td [( q + a = t )]TJ/F19 11.9552 Tf 9.976 0 Td [(a Since 0 > b t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c ,wehave 0 G 1 = b t )]TJ/F20 7.9701 Tf 6.586 0 Td [(a 2 c whichimpliesthatmax f 2 t )]TJ/F19 11.9552 Tf 10.349 0 Td [(a )]TJ/F19 11.9552 Tf 10.349 0 Td [( 0 G 1 )]TJ/F15 11.9552 Tf 422.702 23.98 Td [(1+1 ;t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a g = t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a ,soCorollary5.10showsthat G R Ifconditioniiholds,then 0 G 1 = 0 G ,and j R 1 j = j R j)]TJ/F15 11.9552 Tf 17.933 0 Td [( q + a = q 0 +2 t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 0 G )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 )]TJ/F19 11.9552 Tf 11.956 0 Td [(q )]TJ/F19 11.9552 Tf 11.956 0 Td [(a = q 0 )]TJ/F19 11.9552 Tf 11.955 0 Td [(q + a + t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 0 G )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 0 G 1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 : Thus,Corollary5.10againshowsthat G R Finallywepresentalemmawhichcanbeusedwhenasequenceisnotdegree sucientforagraphwithatleastthreeverticesofdegreeatleast2. Lemma5.12 Let G beagraphoforder t ,andlet = d 1 ;:::;d n beanonincreasing graphicsequencesuchthat d 3 1 and n 2 t )]TJ/F19 11.9552 Tf 11.526 0 Td [( 0 G )]TJ/F15 11.9552 Tf 11.525 0 Td [(1+1 .If 0 G t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 ,then ispotentially G graphic. Proof. Since d 3 1, = d 1 ;d 2 ; 1 n )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 )]TJ/F20 7.9701 Tf 6.586 0 Td [(` ; 0 ` .Everyrealizationofthissequenceisa graph R = R 1 [ R 2 ,where R 1 1and R 2 iseitheremptythatis, R 1 = R ,astar, twodisjointstars,oradoublestar.Ifthereisanyrealizationof where R 1 =0, thenthereareatleast t independentverticesin R andthus G R ispotentially G graphic. If R 1 =1ineveryrealizationof ,wecanapplyLemma5.11.Clearly q 0 0 and,since R 2 isastar,twostars,oradoublestar, a 2and q 2.Thisimpliesthat )]TJ/F19 11.9552 Tf 9.299 0 Td [(q + a 0,sopartiiofthelemmaholds.Since 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 implies 2 t )]TJETq1 0 0 1 137.244 150.761 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 137.244 143.94 Td [( 0 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 t +2,partiofthelemmaalsoholds.Thus, G R 88
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5.2.3Potentially P 4 graphicsequences BeforewecanaddresspotentialRamseynumbersforsmallgraphs,wewillprove thefollowingcharacterizationofpotentially P 4 graphicsequences.Thisproofserves asanillustrationofcommondegreesequencetechniquessuchastheuseof2switches andresidualsequences,asdiscussedinChapter1. Proposition5.13 If = d 1 ;:::d n isanonincreasinggraphicsequenceoflength n 4 ,then ispotentially P 4 graphicifandonlyif d 2 2 and d 4 1 Akeystepintheproofisthenextlemma. Lemma5.14 If = d 1 ;d 2 ;:::;d m isanonincreasinggraphicsequencesuchthat 1 = d 2 )]TJ/F15 11.9552 Tf 12.096 0 Td [(1 ;d 3 )]TJ/F15 11.9552 Tf 12.096 0 Td [(1 ;:::;d d 1 +1 )]TJ/F15 11.9552 Tf 12.096 0 Td [(1 ;d d 1 +2 ;:::;d m ispotentially P n graphic,then is potentially P n +1 graphic. Proof. Let H bearealizationof 1 containing P ,apathoforder n .Labelthe verticesof P with w 1 ;:::;w n ,suchthat w i w i +1 for1 i n )]TJ/F15 11.9552 Tf 11.665 0 Td [(1.Addavertex v ofdegree d 1 to H bymaking v adjacenttotheverticesofdegree d 2 )]TJ/F15 11.9552 Tf 11.094 0 Td [(1 ;:::;d d 1 +1 )]TJ/F15 11.9552 Tf 11.094 0 Td [(1, andcallthisnewgraph H 0 .Notethat H 0 isarealizationof .If v w 1 or v w n thenthereisclearlya P n +1 in H 0 .Alsonotethatif v hasconsecutiveneighborson P thenbydetouringthrough v wegeta P n +1 ,sowecanassumethisdoesnothappen. Suppose v hastwoneighbors, x 1 and x 2 ,thatarenoton P .Then v;x 1 ,and x 2 areeachnotadjacentto w 1 or w 2 ,orwegetalongerpath.Exchangetheedges vx 1 and w 1 w 2 withthenonedges vw 2 and x 1 w 1 togetarealizationof inwhich x 2 vw 2 w 3 w n isapathoforder n +1. Thuswemayassumethat v hasallbutatmostoneofitsneighborson P .Let w i and w j betwoneighborsof v on P .If w i w j ,thenwecanexchangetheedges w i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w i and vw j withthenonedges vw i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and w i w j toget v w i )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,whichcreatesa pathoforder n +1.Thus,allneighborsof v on P mustbeadjacent.Thismeansthat if w i isadjacentto v ,then w i hasatleast d 1 )]TJ/F15 11.9552 Tf 12.088 0 Td [(2neighborson P thatareneighbors 89
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of v ,twoneighborson P thatarenotneighborsof v ,andoneneighbornoton P v itselfforadegreeofatleast d 1 +1.Buteveryvertexin H hasdegreeatmost d 1 ,so thisisnotpossible.Thus, is P n +1 graphic. Nowwecanprovethedesiredcharacterization.Notethatthesequence = d 1 ;:::;d n ispotentially P 3 graphicifandonlyif d 1 2. ProofofProposition5.13. If is P 4 graphic,thenclearlytheconditionsmust hold. Soassume = d 1 ;d 2 ;:::;d n with d 2 2and d 4 1.If d 2 3,then 1 as denedinLemma5.14ispotentially P 3 graphic,soLemma5.14showsthat is potentially P 4 graphic. If d 2 =2,let v 1 and v 2 beverticesofdegree d 1 and d 2 ,respectively,inarealization H of .Let x and y betheneighborsof v 2 .Supposethat x = v 1 .If v 1 y ,thenwe havea3cycle v 1 v 2 y in H .Since d 4 1,thereisafourthvertex w in H andeither w v 1 w y ,orthereisanotheredge,say wz ,in H .Inthersttwocases, wv 1 v 2 y or wyv 1 v 2 isa P 4 .Inthesecondcasewecanexchangetheedges wz and v 1 v 2 with nonedges wv 1 and zv 2 togetthepath zv 2 yv 1 If v 1 y ,thensince d 1 2theremustbeanothervertex w thatisadjacentto v 1 ,andweagaingetthepath wv 1 v 2 y If v 1 v 2 ,but v 1 and v 2 haveacommonneighbor,say x ,then v 1 xv 2 y isa P 4 Iftheydonothaveanycommonneighbors,thenthereareatleasttwovertices w;z adjacentto v 1 .If w x or y then v 1 wxv 2 isa P 4 .Ifnot,thenwecanexchangethe edges v 1 w and v 2 x withthenonedges v 1 v 2 and wx togetthepath zv 1 v 2 y .Thus, is potentially P 4 graphic. 90
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5.3PotentialRamseynumbersforsmallgraphsversusarbitrarygraphs Wearenowpreparedtodetermine r pot H;G forsmall H Proposition5.15 If G isanygraphoforder t 2 ,then r pot K 2 ;G = t Proof. Weknowthat r pot K 2 ;G t .Toseethatthisisenough,let beanygraphic sequenceoflength t .Ifnorealizationof contains K 2 ,then = t ,andtheunique realizationof is K t ,whichcontainsanygraphon t vertices.So r pot K 2 ;G = t Giventhisresult,wewillnolongerconsider K 2 inthedeterminationofthe followingpotentialRamseynumbers.Thatis,wewillingeneralassumethatallour graphshaveorderatleast3. Proposition5.16 If G isagraphoforder t 3 ,then r pot K 2 ;G = t +1 unless G = K t ,where r pot K 2 ;K t = t +2 Proof. If isagraphicsequenceoflength n thatisnotpotentially2 K 2 graphic,then isoneofthefollowing: ; 1 ; 0 n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 ; 2 ; 2 ; 0 n )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 ,or ; 1 ; 1 ; 0 n )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 ,whicharethe degreesequencesof K 2 [ n )]TJ/F15 11.9552 Tf 11.127 0 Td [(2 K 1 K 3 [ n )]TJ/F15 11.9552 Tf 11.127 0 Td [(3 K 1 ,and P 3 [ n )]TJ/F15 11.9552 Tf 11.127 0 Td [(3 K 1 ,respectively. Thecomplementsofeachofthesegraphsare K n )]TJ/F19 11.9552 Tf 12.446 0 Td [(e K n )]TJ/F19 11.9552 Tf 12.447 0 Td [(K 3 = K n )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 K 3 ,and K n )]TJ/F19 11.9552 Tf 12.425 0 Td [(P 3 ,respectively.If G = K t ,then G isasubgraphof K n )]TJ/F19 11.9552 Tf 12.425 0 Td [(K 3 or K n )]TJ/F19 11.9552 Tf 12.425 0 Td [(P 3 if andonlyif n t +2,anditisasubgraphof K n )]TJ/F19 11.9552 Tf 11.69 0 Td [(e ifandonlyif n t +1.Thusif n t +2,then K 2 ;K t If G isnotacompletegraphoforder t ,then 0 G 1.Thus, G isasubgraph of K n )]TJ/F19 11.9552 Tf 11.845 0 Td [(K 3 or K n )]TJ/F19 11.9552 Tf 11.845 0 Td [(P 3 aslongas n t +1,and G isasubgraphof K n )]TJ/F19 11.9552 Tf 11.845 0 Td [(e if n t Thismeansthatif n t +1,then K 2 ;G Itiseasytoseethatagraphicsequence = d 1 ;:::;d n ispotentially P 3 graphic ifandonlyif d 1 2.Wewillusethisinthenextresult. 91
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Proposition5.17 If G isagraphoforder t 3 ,then r pot P 3 ;G = 8 > > < > > : 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 G )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 if 0 G 2 t )]TJETq1 0 0 1 474.521 484.438 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 474.521 477.617 Td [( 0 )]TJ/F15 11.9552 Tf 11.764 0 Td [(1+1if andonlyif 0 G = t= 2,theresultfollows. Intheabovecases,thatisfor H = K 2 ; 2 K 2 ,and P 3 ,thepotentialRamseynumber r pot H;G isequaltothegraphRamseynumber r H;G forall G see[29]forthe graphRamseynumbers. Theorem5.18 Let G beagraphoforder t 3 .Then r pot P 4 ;G = 8 > > < > > : 2 t )]TJETq1 0 0 1 304.222 285.293 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 304.222 278.472 Td [( 0 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 if 0 G t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 t +1 if 0 G > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 : Proof. Let beagraphicsequencethatisnotpotentially P 4 graphic.Webeginby assumingthat 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 and haslength2 t )]TJETq1 0 0 1 341.038 194.76 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 341.038 187.939 Td [( 0 )]TJ/F15 11.9552 Tf 10.254 0 Td [(1+1.Lemma5.7givesthedesired lowerboundon r pot H;G ,soweneedonlyprovethat ispotentially G graphic. Since isnotpotentially P 4 graphic,Proposition5.13showsthateither d 2 =2 and d 4 =0,or d 2 1.Intherstcase, isuniquelyrealizedby K 3 [ t )]TJ/F15 11.9552 Tf 10.402 0 Td [(2 0 )]TJ/F15 11.9552 Tf 10.403 0 Td [(4 K 1 Thecomplementofthisgraphis K 2 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 0 )]TJ/F17 7.9701 Tf 6.586 0 Td [(4 3 K 1 ,whichcontainsa K t .Thus, is 92
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potentially G graphicforanygraphoforder t .If d 2 1,then d 3 1andLemma 5.12showsthat ispotentially G graphic. Nowsuppose 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 2 and haslength t +1.If 0 > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 ,then2 t )]TJETq1 0 0 1 516.811 666.906 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 516.811 660.085 Td [( 0 )]TJ/F15 11.9552 Tf 422.701 23.981 Td [(1+1 t +1 r pot P 4 ;G ,byLemma5.6.Lemma5.9showsthatif d 1 1,then ispotentially G graphic.If d 2 1,andthereissomerealizationof withno componentsisomorphicto K 2 ,thenthereareatleast t independentverticesinthis realizationandwearedone.Ontheotherhand,if d 2 1andtherearecomponents ofordertwoineveryrealizationof ,thenwecanapplyLemma5.11.Wehave q =1, a 2,and 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 2 b t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a 2 c .Theobservationthat j R j)]TJ/F19 11.9552 Tf 17.933 0 Td [(t =1completestheproof. Otherwise d 2 =2and d 4 =0,sotheuniquerealizationof is R = K 3 + t )]TJ/F15 11.9552 Tf 10.791 0 Td [(2 K 1 whichmeans R =3 K 1 K t )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 .Since 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 2 ,therearetwononadjacentvertices in G ofdegreeatmost t )]TJ/F15 11.9552 Tf 12.211 0 Td [(2.Identifythesetwoverticeswithtwooftheverticesof degree t )]TJ/F15 11.9552 Tf 11.221 0 Td [(2in R ,andidentifytheremainingverticesof G withtheverticesofdegree t +2in R .Thismappingshowsthat G R ,so P 4 ;G and r pot P 4 ;G = t +1. Finally,weaddress r pot K 3 ;G .Werestatethecharacterizationofpotentially K 3 graphicsequencesfromChapter2forreference. Theorem5.19Luo[92] Let = d 1 ;d 2 ;:::;d n beagraphicsequencewith n 3 .Then ispotentially K 3 graphicifandonlyif d 3 2 exceptfortwocases: = 4 and = 5 Firstwewillconcernourselveswiththegeneralcasewhere t 5,andthenwe willprovearesultforsmallergraphs. 93
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Theorem5.20 If G isagraphoforder t 5 ,then r pot K 3 ;G = 8 > > > > > > < > > > > > > : 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 G )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 if 0 G t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 t +2 if 0 G > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 and G d t +1 2 e t +1 if 0 G > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 and G < d t +1 2 e : Proof. Since t 5,Lemma5.6showsthatweonlyneedtoconsidersequencesof lengthatleast6.Lemmas5.6,5.7,and5.8giveustheappropriatelowerboundsin eachcase.Wethusneedtoshowthattheselowerboundsarealsoupperbounds. Firstsuppose 0 t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 andlet beagraphicsequenceoflength2 t )]TJETq1 0 0 1 482.632 504.287 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 482.632 497.466 Td [( 0 )]TJ/F15 11.9552 Tf 12.395 0 Td [(1+1 thatisnotpotentially K 3 graphic.ByTheorem5.19,thethirdtermin mustbeat most1,andLemma5.12showsthat K 3 ;G Nowconsiderthecase 0 > t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 .If G d t +1 2 e ,let = d 1 ;:::;d t +2 bea K 3 freegraphicsequence.Since t +2 7,wemusthave d 3 1.Thismeansthatevery realization R of hastheform R 1 [ R 2 ,where R 1 1and R 2 iseitherempty with R 1 = R ,astar,twodisjointstars,oradoublestar.Ifthereisarealizationin which R 1 =0,thenthereareatleast t independentverticesin R ,so R contains K t and G R If R 1 =1foranyrealizationof ,thenwecanapplyLemma5.11with q 0 1, a 2,and q 2.Since a 2,weareinthecasewhere 0 > b t )]TJ/F20 7.9701 Tf 6.586 0 Td [(a 2 c .Since j R j = t +2 and q 2,wehave G R .Thisconcludesthecasewhere G d t +1 2 e If 0 > t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 and G < d t +1 2 e ,then t +1isalowerboundon r pot K 3 ;G .Let = d 1 ;:::;d t +1 bea K 3 freegraphicsequence.If d 1 =1,thenLemma5.9implies that K 3 ;G .If d 1 > 1and d 2 1,thenwecanagainuseLemma5.11with q =1and a 2.Since 0 > t )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 2 ,weonlyneedtosatisfypartiofthelemma,and since haslength t +1, j R j)]TJ/F19 11.9552 Tf 17.933 0 Td [(t =1= q foranyrealization R of 94
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Finally,considerthecasewhere = d 1 ;d 2 ; 1 r ; 0 t )]TJ/F20 7.9701 Tf 6.587 0 Td [(r )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and d 1 and d 2 areatleast 2.BytheErd}osGallaicriteriaTheorem1.3,wemusthave d 1 + d 2 r +2,and d 1 + d 2 musthavethesameparityas r .If d 1 + d 2 = r +2,thentheuniquerealization of isadoublestarandisolatedvertices.Let u 1 and u 2 betheverticesofdegree d 1 and d 2 ,respectively,inarealization R of .Obviously u 1 u 2 ,and u 2 hasatmost b r 2 cb t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 c neighborsotherthan u 1 .Since G < d t +1 2 e ,thereisavertex w in G withatleast d t 2 e)]TJ/F15 11.9552 Tf 21.169 0 Td [(1= b t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 c nonneighborsin G .Let : V G V R send w to u 2 and d 2 )]TJ/F15 11.9552 Tf 11.973 0 Td [(1nonneighborsof w to N R u 2 nf u 1 g andmapsallotherverticesto N R u 1 andtheisolatedverticesof R arbitrarily.Thismappingshowsthat G R If d 1 + d 2 r ,thenthereisarealizationof consistingoftwodisjointstars, disjointedges,andisolatedvertices.Let R = R 1 [ R 2 besucharealizationof where R 2 isthesubgraphconsistingofthetwostars,and R 1 istheremainderof thegraph.WewanttoapplyCorollary5.10toshowthat G R .Todothis, wewanttondsubgraphs G 1 and G 2 of G suchthat G 2 hasorder d 1 + d 2 = a andisasubgraphof R 2 ,and 0 G 1 t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 .Ifthiscanbedone,wewillhave j R 1 j = j G 1 j = t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a max f 2 t )]TJ/F19 11.9552 Tf 11.956 0 Td [(a )]TJ/F19 11.9552 Tf 11.955 0 Td [( 0 G 1 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ;t )]TJ/F19 11.9552 Tf 11.955 0 Td [(a g Tothisend,let w beavertexofminimumdegreein G suchthat w hasatleast t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 d 2 nonneighborsin G ,let M beamaximumantimatchingin G ,andlet be ahomomorphismthatsends w to u 2 and d 2 nonneighborsof w totheneighborhood of u 2 .Thisusestheendpointsofatmost d 2 edgesof M .Since 0 t )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 2 ,thisleaves atleast t )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(d 2 = t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 )]TJ/F15 11.9552 Tf 11.956 0 Td [(2 d 2 2 t )]TJ/F19 11.9552 Tf 11.955 0 Td [(d 1 )]TJ/F19 11.9552 Tf 11.956 0 Td [(d 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 2 = t )]TJ/F19 11.9552 Tf 11.956 0 Td [(a )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 edgesof M wherebothendpointshavenotyetbeenused.Sincetheverticesinthe neighborhoodof u 1 areindependent,wecanchooseedgesof M tomaptoedgesof R 1 ,andthenchoosetheverticesthatmapto N R u 1 tobeanyverticesthatareleft afterthischoice.Thisgivesus G 1 with 0 G 1 t )]TJ/F20 7.9701 Tf 6.587 0 Td [(a )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 ,soCorollary5.10givesus ourresult. 95
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Finally,weneedtoconsider K 3 versusgraphsoforderlessthan5.Graphsof order3anddisconnectedgraphsoforder4areaddressedbyPropositions5.5,5.15, 5.16and5.17. Proposition5.21 ThefollowingarepotentialRamseynumbersfor K 3 versusconnectedgraphsoforder4: r pot K 3 ;K 4 =7 r pot K 3 ;P 4 =5 r pot K 3 ;C 4 =6 r pot K 3 ;K 1 ; 3 =6 r pot K 3 ;K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e =6 r pot K 3 ;K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(P 3 =6 Proof. Thevaluesof r pot K 3 ;K 4 and r pot K 3 ;P 4 followfromTheorem5.2and Theorem5.18,respectively. Fortheothers,let = d 1 ;:::;d 6 beagraphicsequencethatisnotpotentially K 3 graphic.If d 1 = d 6 =1,then isrealizedbyamatchingofsize3and isrealized by K 6 )]TJ/F15 11.9552 Tf 12.48 0 Td [(3 K 2 ,whichcontainsacopyof K 4 )]TJ/F19 11.9552 Tf 12.48 0 Td [(e .If d 1 > 1,then d 3 =1,andthere areatleast4independentverticesinanyrealizationof .Thusanyrealizationof contains K 4 andif G isanynoncompletegraphoforder4, r pot K 3 ;G 6. Toseethatthisisalsoalowerboundon r pot K 3 ;G ,consider = 5 .Since = 5 6! K 3 ;G if G isanyof C 4 K 1 ; 3 K 4 )]TJ/F19 11.9552 Tf 10.047 0 Td [(e ,or K 4 )]TJ/F19 11.9552 Tf 10.047 0 Td [(P 3 .Thus r pot K 3 ;G 6 ineachofthesecases,andwegetourresult. 96
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5.4PotentialRamseynumbersforcycles Inthissection,wewillestablish r pot C s ;C t forallvaluesof t and s with t s 3. Forthesakeofcomparison,hereiswhatisknownabouttheRamseynumberfor cycles: Theorem5.22KarolyiandRosta[67] Let 3 s t beintegers.Then r C s ;C t = 8 > > > > > > > > > > < > > > > > > > > > > : 6 if s = t =3 or 4 t + s= 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 if s;t areeven max f t + s= 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ; 2 s )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 g if t isodd ;s iseven 2 t )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 otherwise : Ourresultsbeginwithsomespecialcases.When s =3,wehavealreadycomputed r pot C 3 ;G forall G .Inparticular,thisyields r pot C 3 ;C 3 = r pot C 3 ;C 4 =6and r pot C 3 ;C t = t +1for t 5. When s =4,wehave: Proposition5.23 For t 5 r pot C 4 ;C t = t +1 .When t =4 r pot C 4 ;C 4 =6 Proof. Todetermine r pot C 4 ;C t ,for t =4and t 6,wecanuseLemma5.1.Chvatal andHarary[28]showedthat r C 4 ;C 4 =6,and C 5 isaselfcomplementarygraphof order5thatdoesnotcontain C 4 .Since C 5 =2 5 hasonlyonerealization,wesee that r pot C 4 ;C 4 isalso6. ChartrandandSchuster[12]showedthat r C 4 ;C t = t +1for t 6,with K 1 ;t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 asthecriticalgraph.Thedegreesequenceof K 1 ;t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 hasnootherrealizations,and thegraphdoesnotcontain C 4 .Thecomplementof K 1 ;t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 is K t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 [ K 1 ,whichdoes notcontain C t .SoweagainuseLemma5.1tosee r pot C 4 ;C t = t +1. Unfortunately, r C 4 ;C 5 =7[42],andwewishtoshowthat r pot C 4 ;C 5 =6,so wecannolongeruseLemma5.1.Forthelowerbound,weconsider K 1 ; 4 = ; 1 4 97
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whichisnotpotentially C 4 graphicandwhosecomplementisnotpotentially C 5 graphic.Thus,weknow r pot C 4 ;C 5 6. Let beanygraphicsequenceoflength6thatisnotpotentially C 4 graphic.By Theorem2.14,wehavethefollowingpossibilitiesfor : d 4 1.Inthiscase, isoneofthefollowingsequences: d 1 ;d 2 ;d 3 ; 1 3 d 1 ;d 2 ;d 3 ; 1 2 ; 0, d 1 ;d 2 ;d 3 ; 1 ; 0 2 ,or d 1 ;d 2 ;d 3 ; 0 3 .Inorderforeachofthese tobegraphic,wemusthave d 2 3bytheErd}osGallaiCriterion.For eachsequence, is: 3 ; 5 )]TJ/F19 11.9552 Tf 12.758 0 Td [(d 3 ; 5 )]TJ/F19 11.9552 Tf 12.759 0 Td [(d 2 ; 5 )]TJ/F19 11.9552 Tf 12.758 0 Td [(d 1 ; 4 2 ; 5 )]TJ/F19 11.9552 Tf 12.759 0 Td [(d 3 ; 5 )]TJ/F19 11.9552 Tf 12.758 0 Td [(d 2 ; 5 )]TJ/F19 11.9552 Tf 12.758 0 Td [(d 1 2 ; 4 ; 5 )]TJ/F19 11.9552 Tf 10.895 0 Td [(d 3 ; 5 )]TJ/F19 11.9552 Tf 10.895 0 Td [(d 2 ; 5 )]TJ/F19 11.9552 Tf 10.894 0 Td [(d 1 ,or 3 ; 5 )]TJ/F19 11.9552 Tf 10.895 0 Td [(d 3 ; 5 )]TJ/F19 11.9552 Tf 10.894 0 Td [(d 2 ; 5 )]TJ/F19 11.9552 Tf 10.895 0 Td [(d 1 ,respectively.Analyzing eachsequence,weseethat satisesconditionsthroughofTheorem 2.15,sotheyareallpotentially C 5 graphic. d 1 =5and d 2 2.Ifthishappensandwearenotintherstcase,then =5 ; 2 4 ; 1and =4 ; 3 4 ; 0.Again satisestheconditionsofTheorem 2.15,soitispotentially C 5 graphic. = 6 .Inthiscase, = 6 ,whichdoeshavearealizationcontaining C 5 so C 4 ;C 5 Thus, r pot C 4 ;C 5 =6. Theproofofthemaintheoreminthissectionusesastandardtechniqueinthe degreesequenceliteraturecalledtheedgeexchange.Thisisageneralizationofthe 2switchdescribedinChapter1.An alternatingcircuit inagraph G isacircuit C withedges e 1 e 2 e 2 p suchthatforeven i ,theedge e i 2 E G andforodd i ,the edge e i = 2 E G .Exchangingtheedgesforthenonedgesthatis,removingtheedges from G andaddingthenonedgesto G yieldsagraphwiththesamedegreesequence as G .Thisallowsustochangerealizationsofadegreesequencetoobtaindesirable properties,butgivesusalittlemorefreedomthanthestandard2switch. 98
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Theorem5.24 For t s 3 ,thepotentialRamseynumberforcyclesis r pot C s ;C t = t + b s )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 2 c ,exceptfor r pot C 4 ;C 4 =6 r pot C 3 ;C 3 =6 ,and r pot C 3 ;C 4 =6 Proof. Giventheabovepropositions,wemayassume s 5.Let k = b s )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 c .Tosee that r pot C s ;C t t + k ,considerthedegreesequence k + t )]TJ/F15 11.9552 Tf 11.832 0 Td [(2 k ;k t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,whichhas theuniquerealization G = K k K t )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .Anysubgraphoforder s in G mustuseat least s )]TJ/F19 11.9552 Tf 10.466 0 Td [(k verticesfromtheindependentset,forcingtheretobeanindependentsetof order s )]TJ/F19 11.9552 Tf 10.701 0 Td [(k inthegraph.However, C s = b s 2 c
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Let d =max x 2 V G n V C d C x .Supposerstthat d =0.Ifthereexistsanedge xy in G n C ,then v 1 v 3 v 2 xyv 2 v 1 isanalternatingcircuitin G .Exchangingtheedges inthiscircuitcreatesarealizationinwhich v 1 v 3 v 4 v s +1 isan s cycle.Ifthereis nosuchedge,then G n C isasetofisolatedvertices.Let f w 1 ;w 2 ;:::;w t + k )]TJ/F17 7.9701 Tf 6.587 0 Td [( s +1 g = V G n V C .Sincetheverticesthatarenoton C havenoneighborson C ,wehave a t cyclein G .If s iseven,onesuchcycleis w 1 w 2 w t )]TJ/F20 7.9701 Tf 6.587 0 Td [(k )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 v 2 v 4 v 2 k +2 v 1 w 1 ,andif s isodd,apossiblecycleis w 1 w 2 w t )]TJ/F20 7.9701 Tf 6.586 0 Td [(k )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 v 2 v 4 v 2 k +2 v 3 w 1 Nowsuppose d> 0.If xv i 2 E G ,then xv i +3 = 2 E G becausethenwewould havean s cycle v i xv i +3 v i +4 v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Thus,theremustbeanindex j suchthat xv j 2 E G and xv j +1 = 2 E G .Wecanthenexchangetheedges xv j and v j +1 v j +2 withthe nonedges xv j +1 and v j v j +2 ,togetarealizationof containingan s cycle. Case2: Somerealizationof containsa C s +2 Let C = v 1 v 2 v s +2 bean s +2cycleinarealization G of .ByCase1,wecan assumethat G doesnotcontaina C s +1 .Then,foreachindex i ,weagainhavethat v i v i +2 and v i v i +3 arenotedgesin G .Ifthereisan x 2 V G n V C suchthat xv i and xv i +3 arebothedgesin G ,then v i xv i +3 v i +4 v i )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isan s +1cyclein G .Soeither thereisan x 2 V G n V C andanindex j suchthat xv j 2 E G and xv j +1 = 2 E G orno x 2 V G n V C hasneighborsin C .Intherstcase,wecanexchangeedges xv j and v j +1 v j +2 withnonedges xv j +1 and v j v j +2 togetan s +1cyclein G Inthesecondcase,ifthereisanedge xy in G n C ,then v 1 v 3 v 2 xyv 2 v 1 isanalternatingcircuitin G ,andexchangingtheedgesinthiscircuityieldsan s +1cycle v 1 v 3 v 4 v s +2 .Iftherearenoedgesin G n C ,thenwecannda t cyclein G aswe didinCase1:If s iseven,wehave w 1 w 2 w t )]TJ/F20 7.9701 Tf 6.587 0 Td [(k )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 v 2 v 4 v 2 k +2 v 2 k +4 w 1 ,andif s is odd,wehave w 1 w 2 w t )]TJ/F20 7.9701 Tf 6.586 0 Td [(k )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 v 2 v 4 v 2 k +2 v 1 w 1 Case3: Somerealizationof containsa C p ,where p s +3. Choosearealization G of sothat p ismaximal,andassumebyCase2that G containsno C s +2 .Wemayassume v 1 v s = 2 E G ,becausethatwouldcreatean 100
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s cycle.If v p v s +1 2 E G ,thenwehavethe s +2cycle v 1 v 2 v s +1 v p ,sowecan assumethatthisedgeisalsonotin G .Thus,wecanexchangetheedges v p v 1 and v s v s +1 forthenonedges v 1 v s and v s +1 v p togetarealizationof withan s cycle. Case4: Everyrealizationof hascircumferenceatmost s )]TJ/F15 11.9552 Tf 11.955 0 Td [(1. Let G bearealizationof containingalongestcycle C = v 1 v p ,where p s )]TJ/F15 11.9552 Tf 9.809 0 Td [(1 andismaximumoverallrealizationsof .Let H bethesubgraphof G inducedby V G n V C .Thenexttwoclaims,whichdevelopthestructureof H ,aresimilarto thosein[9]. Claim5.1 H hasnocycles. Suppose x 1 x ` isacyclein H .If x i isadjacentto v j ,thenwemusthave v j +1 x i = 2 E G and v j +2 x i +1 = 2 E G ,forintherstcase,wegetthecycle v 1 v j x i v j +1 v p andinthesecondcasewegetthecycle v 1 v j x i x i +1 v j +2 v p .Eachoftheseisa cycleoflength p +1in G ,contradictingourhypothesis.Thus,wecanexchangethe edges v j v j +1 and x i x i +1 forthenonedges x i v j +1 and x i +1 v j +1 togetarealizationof withacycleoflength p + ` ,contradictingourmaximalchoiceof G and p .Ifno vertexonacyclein H hasaneighboron C ,thenwecandothesameedgeexchange. Thus H mustnotcontainanycycles. Claim5.2 If H > 1 ,thentheonlynontrivialcomponentof H isastar. Firstsupposethatthereareatleasttwovertices, x and y ,in H with d H x > 1 and d H y > 1.Ifsuch x and y cannotbefoundinthesamecomponentof H ,then theyareeachthecenterofastar,sotherearevertices x 0 and y 0 suchthat xx 0 and yy 0 areedgesof H but xy and x 0 y 0 arenotedges.Wecanthenexchangetheedgeswith thenonedgessothat x and y areinthesamecomponent,andareinfactadjacent. Thuswecanassumethat xy 2 E G Since H hasnocycles,therearevertices x 0 2 N H x nf y g and y 0 2 N H y nf x g suchthat x 0 6 y 0 and x 0 6 = y 0 .Since C ismaximalin G ,theremustbeavertex v i on 101
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C suchthat x 6 v i and y 6 v i +1 .Thus, v i xx 0 y 0 yv i +1 v i isanalternatingcircuitin G ; exchangingtheedgesandnonedgesonthiscycleyieldsacycleoflength p +2in G contradictingourmaximalchoiceof G Thismeansthatthereisatmostonevertexofdegreegreaterthan1in H .Suppose thatthereareatleasttwonontrivialcomponentsin H .Oneofthesecomponents mustbeasingleedge,say yy 0 .Let x bethevertexofdegreegreaterthan1.If x misses consecutiveverticeson C ,say v i and v i +1 ,then v i xx 0 x 00 xv i +1 v i isanalternatingcircuit in G ,where x 0 and x 00 areneighborsof x in H .Exchangingedgesonthisalternating circuityieldsacycleoflength p +1,againcontradictingthechoiceof G .Clearly, x is notadjacenttoconsecutiveverticeson C ,elsewewouldalreadyhavealongercycle. Sowemayassume,withoutlossofgenerality,that x 6 v 1 but xv 2 and xv p areedges in G .Now,if y 6 v 2 ,then v 1 v 2 yy 0 x 0 xv 1 isanalternatingcircuitin G ,andexchanging theedgesonthisalternatingcircuitallowsustoextend C byusingthepath v 1 xv 2 Similarly,if y 0 6 v p ,thenexchangingedgesonthealternatingcircuit yy 0 v p v 1 xx 0 y allowsustoextend C byusingthepath v p xv 1 .Thus,wemusthave y v 2 and y 0 v p .However,thisallowsustoextend C byreplacingthepath v p v 1 v 2 withthe path v p y 0 yv 2 ,onceagaincontradictingourchoiceof G .Thus,wecannothavetwo nontrivialcomponentsin H Claim5.3 If H > 1 ,then C s ;C t Thepreviousclaimsshowthatif H > 1,thentheonlynontrivialcomponent of H isastar.Let x bethesinglevertexofdegreegreaterthan1in H ,andlet y 1 ;:::;y ` betheneighborsof x in H .If x v i forany v i 2 C ,then v i +1 y j for 1 j ` ,and v i +1 x ,becauseotherwisewewouldhavealongercyclein G .If x hastwoconsecutivenonneighborsin C ,say v i and v i +1 ,then v i v i +1 xy 1 y 2 xv 1 isan alternatingcircuitin G ;exchangingtheedgesandnonedgesofthiscircuitcreatesa realizationof withalongercycle.Thus, x mustbeadjacenttopreciselyeveryother 102
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vertexon C ,whichimpliesthat p iseven. Suppose,withoutlossofgenerality,that x isadjacentto v i foreveryodd i .If z isanisolatedvertexin H ,then z v i +1 foranyodd i ,becausethiscreatesthe alternatingcircuit xv i +1 zxy 1 y 2 x ,whoseedgeswecanexchangetocreatearealization of withalongercycle.Thus,if i iseventhen v i hasnoneighborsin H .Also, v i v i +2 foranyeven i ,becausesuchanedgegivesacycle v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 xv i +1 v i v i +2 v i +3 v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 in G oflength p +1.Nowwecanndacycleoflength t + k )]TJ/F20 7.9701 Tf 12.611 5.256 Td [(p 2 in G :if z 1 ;z 2 ;:::z m aretheisolatesin H ,let C 0 = v 2 xv 4 y 1 y ` z 1 z m v 6 v 8 v p .Since k )]TJ/F20 7.9701 Tf 13.272 5.256 Td [(p 2 0,itis possiblethat C 0 islongerthannecessary;wecanshortenittolength t byremoving verticesfromthestarorisolates.Thisestablishestheclaim. Nowweassume H 1.If j H j t ,thenwecannda t cyclein G bytaking anysetof t verticesfrom H ,andwearedone.Ifnot,then j H j k since j H j = t + k )]TJ/F19 11.9552 Tf 12.362 0 Td [(p ,whichfurtherimpliesthat j H j t )256(d s )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 eb s )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 2 c +1 d s )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 2 ed p 2 e .Ourgoalistonda t cyclein G .Wewilldothisbyndingacyclein G thatalternatesbetweenverticesof C andverticesof H untilituses d p 2 e verticesof C .Since j H j + l p 2 m = t + k )]TJ/F19 11.9552 Tf 11.955 0 Td [(p + l p 2 m = t + k )]TJ/F25 11.9552 Tf 11.956 13.271 Td [(j p 2 k = t + s )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 )]TJ/F25 11.9552 Tf 11.956 13.27 Td [(j p 2 k t addingfurtherverticesfrom H willresultinacycleoflength t Webeginbyexaminingthestructureof H inrelationto C .Assume C hasan orientationinthedirectionofincreasingindices.Firstnotethatno w 2 V H has twoconsecutiveneighborson C ,forthiswouldcontradictthemaximalityof p .Call avertex v i 2 C surrounded iftherearevertices w 1 ;w 2 2 H suchthat w 1 v i and 103
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w 2 v i 1 .Notethat w 1 w 2 ispossible.Inthiscase,wecanexchangeedges w 2 v i )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and w 1 v i withnonedges w 2 v i and w 1 v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 togetalongercycleusingthepath v i w 2 v i +1 .Sotherecanbenosurroundedverticeson C Ifsomevertex w 2 V H hasneighbors v i and v j on C ,then v i )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v j )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,because suchanedgewouldcreatethecycle v 1 C + v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v j )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 C )]TJ/F19 11.9552 Tf 7.084 4.338 Td [(v i wv j C + v 1 ,where C )]TJ/F15 11.9552 Tf 12.314 4.338 Td [(means wefollow C againstitsorientationand C + meanswefollow C withitsorientation. Thiscyclehaslengthgreaterthan p ,sothiscontradictsourchoiceof G .Similarly,if v i +1 v j +1 ,thecycle v 1 C + v i xv j C )]TJ/F19 11.9552 Tf 7.085 4.339 Td [(v i +1 v j +1 C + v 1 isacycleoflength p +1. Amonganythreeconsecutiveverticeson C ,everypairofverticesin H hasa commonnonneighbor.Suppose z 1 and z 2 areanypossiblyadjacentverticesin H andconsider,withoutlossofgenerality, v 1 ;v 2 ,and v 3 on C .If z 1 isadjacenttoboth v 1 and v 3 ,then z 2 v 2 ,becausethen v 2 issurrounded.If z 1 missesapairofconsecutive verticesamong v 1 v 2 ; and v 3 ,thensince z 2 cannotbeadjacenttoconsecutivevertices, z 2 mustmissoneofthese.If z 1 v 2 ,then z 2 cannotbeadjacenttoboth v 1 and v 3 or v 2 issurrounded,soagaintheyhaveacommonnonneighbor. Let xy beanedgeof H .Ifforsome v i 2 C wehave x v i ,then y isnotadjacent toeither v i +1 or v i +2 ,becauseeitherofthesewouldcreatealongercyclein G .If x v i while y v i ,wecanexchangeedges v i v i +1 and xy withthenonedges xv i +1 and yv i togetalongercyclein G .Soforeveryedge xy in H x v i implies y v i andthisimpliesthattheendpointsofeachedgein H areadjacenttoatmostevery thirdvertexon C Suppose xy 2 E H suchthat x;y v i ,and z isanisolatein H .If z v i +1 then z v i ,andexchangingedges yv i and zv i +1 withthenonedges yv i +1 and zv i createsalongercyclegoingthrough xy .So z v i +1 .Similarly, z v i )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 Suppose x 1 y 1 and x 2 y 2 areedgesin H ,andsuppose x 1 ;y 1 v i forsome i .Then x 2 ;y 2 v i +1 ,becauseinthiscaseexchangingtheedges x 1 y 1 and x 2 y 2 withthe nonedges x 1 x 2 and y 1 y 2 wouldcreatealongercyclein G .Similarly, x 2 ;y 2 v i +2 104
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If x 2 ;y 2 v i ,thenagainexchanging x 1 y 1 and x 2 y 2 for x 1 x 2 and y 1 y 2 willgiveus theedge x 1 x 2 withonly x 1 v i ,whichwehaveshownleadstoalongercyclein G So x 2 and y 2 mustalsobeadjacentto v i ;thisistrueforanyedgein H ,soevery nonisolatedvertexin H hasthesameneighborhoodon C Withthisstructure,wecannowcreateacycle C 0 in G oflength j H j + d p 2 e .Let C = v 1 v 2 v p .Suppose H =0,andlet V H = f w 1 ;:::;w j H j g .Withoutlossof generality,wemayassume w 1 and w 2 aresuchthat v 1 w 1 and v 1 w 2 .Webegin C 0 with w 1 v 1 w 2 ,andaddto C 0 bypickingverticesfrom C greedilyaccordingtothe orderof C andalternatingwithverticesfrom H alwaysendingwithavertexin H Supposewehavebuilt C 0 uptothepointwherewehave C 0 = w 1 v 1 w 2 v m )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w l so v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 w l isanonedgeof G .Let H l = H )222(f w 1 ;:::;w l g Ifthereareatleasttwoverticesin H l ,weproceedasfollows.If w l v m andsome vertex w l +1 of H l isalsonotadjacentto v m ,thenweadd v m w l +1 to C 0 .Supposethat w l v m buteveryvertexof H l isadjacentto v m .Inthiscase,if w l isalsonotadjacent to w m +1 ,thenweadd w l v m +1 w l +1 to C 0 ,sincenovertexin H l canbeadjacentto v m +1 .Ifontheotherhand w l isadjacentto v m +1 ,thennoneoftheverticesin H l are adjacentto v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 .Here,wewouldreplace w l in C 0 withsome z 1 from H l ,sothatwe have v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 z 1 v m +1 z 2 toextend C 0 ,where z 2 isanothervertexfrom H l If w l v m ,let z 1 and z 2 beverticesof H l .Thevertices w l and z i where i is either1or2musthaveacommonnonneighborintheset f v m )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ;v m ;v m +1 g ,andsince w l v m ,weknowthat w l missesboth v m )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and v m +1 .Ifsome z i misses v m +1 ,then wecontinue C 0 withthepath v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 w l v m +1 z i .Ifeveryvertexof H l isadjacentto v m +1 thentheseverticesmustmissboth v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and v m .Inthiscase,insteadofusing v m )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w l in C 0 ,wecontinuethecyclewith v m )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 z 1 v m z 2 v m +2 w l Thus,aslongasthereareatleasttwoverticesremainingin H l ,wecancontinue building C 0 inthisway.Thisconstructionensuresthatatleasteveryothervertexof C isincludedin C 0 {weneveromittwoconsecutiveverticesof C from C 0 .Also,if 105
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v j w isthelatestedgeaddedto C 0 ,noneoftheverticesin f v j +1 ;:::;v p g arein C 0 Nowsuppose j H j > d p 2 e andwehavebuilt C 0 with d p 2 e verticesfrom H and d p 2 e)]TJ/F15 11.9552 Tf 16.039 0 Td [(1 verticesfrom C .Weneedtoaddonemorevertexof C to C 0 ,andthenwecancollect t )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 d p= 2 e verticesof H tonishthecycle. Let v q bethelastvertexof C thatwasaddedto C 0 ,andlet b = d p 2 e sothelast edgeof C 0 sofaris v q w b .Sinceatleasteveryothervertexof C mustbeusedin C 0 weseethat q 2 l p 2 m )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 )]TJ/F15 11.9552 Tf 11.956 0 Td [(1= 8 > > < > > : p )]TJ/F15 11.9552 Tf 11.955 0 Td [(3if p iseven p )]TJ/F15 11.9552 Tf 11.955 0 Td [(2if p isodd. If p iseven,orif p isoddand q p )]TJ/F15 11.9552 Tf 12.435 0 Td [(3,thenthereareatleastthreeconsecutive verticesof C thataren'tusedin C 0 yet;thus, w b andanyothervertexin H must haveacommonnonneighboramongthesethreevertices.Addingthisnonneighbor to C 0 givesus d p 2 e verticesof C andalastvertexin H .Fromhere,therestof H can beinsertedinto C 0 until C 0 isacycleoflength t in G Theremainingcasetoconsiderisif p isoddand q = p )]TJ/F15 11.9552 Tf 12.109 0 Td [(2.Ifthishappens,we musthaveskippedeverypossiblevertexof C ;thatis,every v i with i evenisnotused in C 0 ,andevery v j with j oddisusedin C 0 ,upto j = q = p )]TJ/F15 11.9552 Tf 11.955 0 Td [(2. Nowweseethatthelastedgein C 0 sofaris v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 w b .Thereisatleastonevertex, say z ,unusedin H .If z and w b haveacommonnonneighborin f v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ;v p g ,weuse thatnonneighboronapathfrom w b to z andnishthecyclewithverticesfrom H Otherwise,if w b v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 ,andsome z 2 H b isnotadjacentto v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 ,thenwecontinue C 0 with w b v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 z ,andnishwithverticesof H .Ifevery z 2 H b isadjacentto v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(3 thennoneofthemisadjacentto v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 .Inthiscase,insteadofusing v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 w b in C 0 weuse v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 z .Thuswemayassume w b v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 .Ifsome z 2 H b isadjacentto v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 then v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 mustbeanonneighborofboth w b and z ,sowecanadd w b v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 z to C 0 ,and nishthecyclewithverticesof H .Finallywehavethecasewhere z v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 and w b and z havenocommonnonneighborsin f v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ;v p g .Wemayassumethat w b v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 106
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and z v p becauseifweinsteadhave w b v p and z v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 ,wecan2switchthese edgestoendinthiscase.Since v p )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 and v p arethensuccessorsofneighborsof w b theyarenotadjacent.Thus,wemaynish C 0 bytakingthepath v p )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 v p w b z andthen continuingwithverticesof H Nowwemustconsiderthecasewhere j H j = d p 2 e .Thisimpliesthat p = s )]TJ/F15 11.9552 Tf 12.056 0 Td [(1, s iseven,and s = t .Thecreationof C 0 proceedsasaboveuntilwehaveadded d p 2 e)]TJ/F15 11.9552 Tf 19.014 0 Td [(1 verticesof H and d p 2 e)]TJ/F15 11.9552 Tf 18.448 0 Td [(2verticesof C to C 0 .Again,sinceatleasteveryothervertex of C isaddedto C 0 ,theindex, q ,ofthelastvertexaddedto C 0 mustbeatmost 2 d p 2 e)]TJ/F15 11.9552 Tf 20.609 0 Td [(2 )]TJ/F15 11.9552 Tf 12.297 0 Td [(1= p )]TJ/F15 11.9552 Tf 12.297 0 Td [(4.Nowwehave C 0 = w 1 v 1 v q w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 ,where b = d p 2 e ,andwe needtoaddtwomoreverticesof C to C 0 ,whileendingbackat w 1 If w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and w b bothmiss v q +1 ,thenwecontinue C 0 with w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v q +1 w b .Then,there arethreeconsecutiveunusedvertices v q +2 ;v q +3 ,and v q +4 amongwhich w b and w 1 musthaveacommonnonneighbor.Addingthisvertexto C 0 andreturningto w 1 givesusthedesired t cycle.Ifboth w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and w b miss v q +2 ,and w b v q +1 ,thenwe add w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +2 w b to C 0 .Now,if w b and w 1 haveacommonnonneighbor,say v 0 ,in f v q +3 ;:::;v p g ,thenwenish C 0 with w b v 0 w 1 .Ifnot,weknowthat q +4= p andwe mayassumethat w b v q +3 and w 1 v q +4 .Since w b v q +1 ,weknowthat v q +2 is notadjacentto v q +4 ,sowenish C 0 with v q +2 v q +4 w b w 1 Next,suppose w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 and w b miss v q +2 but w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +1 .If w b v q ,theninstead ofusing v q w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 in C 0 ,wetake w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(2 v q w b v q +2 w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 .Weneedonemorevertexof C to nish C 0 ;if w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and w 1 haveacommonnonneighbor, v 0 ,in f v q +3 ;:::;v p g ,thenwe nish C 0 with w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v 0 w 1 .Ifnot,weknowthat q +4= p andwemayassumethat w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +3 and w 1 v q +4 .Since w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 isadjacenttoboth v q +1 and v q +3 ,weknow that v q +2 v q +4 .Thus,wecanend C 0 with v q +2 v q +4 w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 w 1 Wemaythereforeassumethat w b v q ,and C 0 continuesfrom w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 with w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +2 w b .Nowweneedonemorevertexfrom C .If w b and w 1 haveacommonnonneighborin f v q +3 ;:::;v p g ,thenweusethisvertextonish C 0 .Ifnot,we 107
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mayonceagainassumethat w b v q +3 and w 1 v q +4 .Since w b v q ,weseethat v q +1 v q +4 .Exchangetheedges w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +1 and w 1 v q +4 withthenonedges w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 w 1 and v q +1 v q +4 .Thisyieldsthecycle v 1 C + v q w b v q +3 C )]TJ/F19 11.9552 Tf 7.084 4.338 Td [(v q +1 v q +4 v 1 oflength p +1in G acontradiction.Thismeanswemusthavethecommonnonneighborweneededto nishthecycle. Finally,suppose w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 and w b havenocommonnonneighborin f v q +1 v q +2 g .We mayassumethat w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +1 and w b v q +2 orwecan2switchtomakethis happen.If w 1 v q +1 ,thenwemayend C 0 with v q w b )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 v q +3 w b v q +1 w 1 .If w 1 v q +2 thenweend C 0 with v q w b v q +3 w b )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 v q +2 w 1 Nowweconsiderthecasewhere H =1.Let x 1 y 1 ;:::;x r y r betheedgesof H andlet w 1 ;:::;w ` betheisolatedverticesof H .Asbefore,wewillndacycle, C 0 ,oflengthatleast t in G byndingacyclethatuses d p 2 e verticesfrom C andthe remainingverticesfrom H Let C = v 1 v 2 v p .Wemayassumethat v 1 issuchthat x 1 v 1 and x 1 v 2 andwebegin C 0 with x 1 v 1 y 1 v 2 x 2 .Ifthereisonlyoneedgein H ,thenin f v 2 ;:::;v p g theremustbeavertexthatisnotadjacenttoboth y 1 and w 1 ,soweusethelowestindexedsuchvertextoreturnto H ,thencontinueasinthe H =0case.If p< 5, wealreadyhave d p 2 e verticesof C andwecannish C 0 byusingtheotherverticesof H .If p =5,thensinceedgesof H canbeadjacenttoatmosteverythirdvertexon C ,theremustbesome v j x 2 ,andweadd x 2 v j y 2 ,thennish C 0 withverticesof H If p 6,thenagainsinceedgesof H areadjacenttoatmosteverythirdvertex on C ,either v 3 or v 4 mustbenonadjacentto x 2 .Weadd x 2 v j y 2 v j +1 or x 2 v j y 2 v j +2 where j =3or4andweusethesecondoptionif x 2 v j +1 ,tocontinue C 0 .We continuethisprocess,ateachstepaddingtherstvertexof C 0 thatisavailable,until wehaveobtained d p 2 e verticesfrom C 0 andendedwithavertexfrom H If2 r d p 2 e ,weendthisportionoftheprocesswith v m y j in C 0 forsome m
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verticesof H to C 0 untilwehaveacycleoflength t ,andwearedone.If2 r< d p 2 e thensince j H jd p 2 e ,theremustbeisolatedverticesin H ,andwehave C 0 that endswith v m y r atthispoint.Sincewemissedatmosteverythirdvertexof C ,there mustbeatleastthreeunusedconsecutiveverticeson C ,andinthissettheremust beavertex v j thatisnotadjacenttoboth y r and w 1 .Wecontinue C 0 with y r v j w 1 andnowproceedwithaddingverticesto C 0 asinthecasewhere H =0.Since fewerverticesof C canbemissedwhenweareusingedgesof H atmosteverythird vertex,ratherthanatmosteveryother,wewillbeabletondenoughverticesof C tocompletethecycleasdesired. Thisgivesusa t cyclein G ,so C s ;C t ,andwehaveourresult. Table5.1summarizesthegeneralresultsgiveninthischapter.Inthetable, H is axedgraphand G isagraphoforder t with t j V H j .Thepropertiesof G that aectthevalueof r pot H;G aregiveninthecolumnunder G Table5.1: r pot H;G forsmall H and r pot C s ;C t H G r pot H;G K 2 G t 2 K 2 K t t +2 G 6 = K t t +1 P 3 0 G t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 t +1 K 3 K 4 7 P 4 5 C 4 ;K 1 ; 3 ;K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e;K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(P 3 6 0 G t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 2 t )]TJETq1 0 0 1 424.467 161.022 cm[]0 d 0 J 0.478 w 0 0 m 7.522 0 l SQBT/F19 11.9552 Tf 424.467 154.2 Td [( 0 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+1 0 G > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 and G d t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 2 e t +2 0 G > t )]TJ/F17 7.9701 Tf 6.587 0 Td [(3 2 and G < d t )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 2 e t +1 C 4 C 4 6 C s ;s 4 C t ;t s;t 6 =4 t + b s )]TJ/F17 7.9701 Tf 6.587 0 Td [(1 2 c 109
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5.5PotentialRamseynumbersforallgraphsoforderatmost4 InspiredbytheworkofChvatalandHararyin[29],wegiveinTable5.2atable ofallpotentialRamseynumbers r pot H;G where H and G areeachoneoftheten graphsoforder4thatdonothaveisolatedvertices. Table5.2: SmallpotentialRamseynumbers K 2 P 3 2 K 2 K 3 P 4 K 1 ; 3 C 4 K 1 ; 3 + e K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e K 4 K 2 2 3 4 3 4 4 4 4 4 4 P 3 3 4 5 4 5 4 5 5 7 2 K 2 5 5 5 5 5 5 5 6 K 3 6 5 6 6 6 6 7 P 4 5 5 5 5 5 7 K 1 ; 3 6 6 6 7 8 C 4 6 6 7 8 K 1 ; 3 + e 6 7 8 K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e 7 8 K 4 8 Theresultsinthispaperyieldtherstverowsofthistable,aswellasthe valuefor r pot C 4 ;C 4 .Theorems5.2and5.3giveus r pot K 4 ;K 4 and r pot C 4 ;K 4 respectively.Todeterminetheothervaluesonthistable,wemakeuseofthefollowing characterizationsofpotentially H graphicsequencesforsmall H Observation5.25 Anonincreasinggraphicsequence = d 1 ;:::;d n with n 4 is potentially K 1 ; 3 graphicifandonlyif d 1 3 Theorem5.26Chen,Li[16] Anonincreasinggraphicsequence = d 1 ;:::;d n with n 4 ispotentially K 1 ; 3 + e graphicifandonlyif d 1 3 and d 3 2 Theorem5.27Eschen[41] Let = d 1 ;:::;d n beanonincreasinggraphicsequenceoflength n 4 .Then ispotentially K 4 )]TJ/F19 11.9552 Tf 12.759 0 Td [(e graphicifandonlyifthe followingconditionshold: 110
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a d 2 3 b d 4 2 ,and cif n =5 ; 6 ,then 6 = 2 ; 2 n )]TJ/F17 7.9701 Tf 6.586 0 Td [(2 and 6 = 6 Theorem5.28Luo[93] Let = d 1 ;:::;d n beanonincreasinggraphicsequence oflength n 4 .Then ispotentially K 4 graphicifandonlyif d 4 3 and 6 = n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 ; 3 s ; 1 n )]TJ/F20 7.9701 Tf 6.587 0 Td [(s )]TJ/F17 7.9701 Tf 6.586 0 Td [(1 foreach s =4 ; 5 exceptthefollowingsequences: n =5 : ; 3 4 4 ; 2 n =6 : 6 2 ; 3 4 ; 3 4 ; 2 6 5 ; 1 4 ; 2 2 n =7 : 7 3 ; 3 4 ; 3 6 ; 3 5 ; 1 6 ; 2 5 ; 2 ; 1 n =8 : 7 ; 1 6 ; 1 2 Considerthesequence = 5 .Thissequenceisnotpotentially H graphic foranyof C 4 K 1 ; 3 K 1 ; 3 + e K 4 )]TJ/F19 11.9552 Tf 12.454 0 Td [(e ,and K 4 ,anditisselfcomplementary.Thus, r pot H;G 6if H and G areanyofthegraphsjustlisted.Recallthat r pot H;G r H;G foranygraphs H and G .Since r K 1 ; 3 ;K 1 ; 3 =6and r K 1 ; 3 ;C 4 =6[29]we seethat r pot K 1 ; 3 ;K 1 ; 3 =6and r pot K 1 ; 3 ;C 4 =6aswell. Similarly,thesequence = 6 isnotpotentially H graphicforanyofthese graphs,anditscomplement, = 6 isnotpotentially K 4 )]TJ/F19 11.9552 Tf 12.897 0 Td [(e graphic.Thus, r pot H;K 4 )]TJ/F19 11.9552 Tf 12.519 0 Td [(e 7foranyofthese H .Since r H;K 4 )]TJ/F19 11.9552 Tf 12.52 0 Td [(e =7for H = K 1 ; 3 ;C 4 and K 1 ; 3 + e ,weseethatthepotentialRamseynumberineachofthesecasesis7. Toseethat r pot K 4 )]TJ/F19 11.9552 Tf 12.675 0 Td [(e;K 4 )]TJ/F19 11.9552 Tf 12.675 0 Td [(e =7,considerthenonincreasinggraphicsequence = d 1 ;:::;d 7 .ByTheorem5.27,wemayassumethateither d 2 2or d 2 3and d 4 1.If d 2 2,thenin = d 1 ;:::; d 7 whichisrearrangedtobenonincreasing wemusthave d 6 4,whichmeans ispotentially K 4 )]TJ/F19 11.9552 Tf 12.905 0 Td [(e graphic.If d 2 3 111
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and d 4 1,then d 4 5,so isalsopotentially K 4 )]TJ/F19 11.9552 Tf 12.176 0 Td [(e graphic,andwehaveour conclusion. Nowwedetermine r pot K 1 ; 3 + e;H for H = K 1 ; 3 ;C 4 ,and K 1 ; 3 + e .If = d 1 ;:::;d 6 isanonincreasinggraphicsequencethatisnotpotentially K 1 ; 3 + e graphic,thenbyTheorem5.26,either d 1 2or d 3 1.If d 1 2,thenin we seethat d 6 3,andif d 3 1,then d 4 4.Ineithercase,thesequence satises theconditionstobepotentially H graphicforeach H underconsideration,sothe potentialRamseynumberineachcaseis6. Finallyweconsider r pot H;K 4 for H = K 1 ; 3 ;K 1 ; 3 + e ,and K 4 )]TJ/F19 11.9552 Tf 11.229 0 Td [(e .Foreachsuch H ,thesequence = 7 isnotpotentially H graphic,anditscomplement, = 7 isnotpotentially K 4 graphic.Thus, r pot H;K 4 8ineachcase.For n =8,we seethat = d 1 ;:::;d n ispotentially K 4 graphicunless d 4 2or isoneofthe exceptionslistedinTheorem5.28.If d 4 2,then d 5 5,andthisisenoughto guaranteethat ispotentially H graphicforeach H weareconsidering.If isone oftheexceptionalsequences,then isoneof 3 ; 4 4 ; 0, 2 ; 4 5 ; 0, ; 4 7 ,or 2 ; 4 6 eachofwhichisalsopotentially H graphicforthegivengraphs.Thus r pot K 4 ;H =8 for H = K 1 ; 3 ;K 1 ; 3 + e ,and K 4 )]TJ/F19 11.9552 Tf 11.955 0 Td [(e ,completingthetable. 112
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6.FutureWork InChapter4,wediscussedopenproblemsintheareaof stabilityofgraphs.In thischapter,wegiveproblemsrelatedtootherresultspresentedinthisdissertation. 6.1Hypergraphicdegreesequences Ahypergraphisageneralizationofthegraphswehavediscussedsofar,inwhich edgesarenolongerrestrictedtocontainingonlytwovertices;thatis,ahypergraph H consistsofavertexset V H andanedgeset E H whereanedgecanbeanysubsetof V H .Ahypergraphiscalled k uniform orisa k graph ifeveryedgecontainsexactly k vertices,and simple iftherearenorepeatededgesorloopsedgeswhichcontain avertexmorethanonce.Thus,thedegreeofavertexinasimplehypergraph isthenumberofedgesthatcontainit,aswithstandardgraphs.Asequence of nonnegativeintegersis k graphic ifthereisasimple k uniformhypergraphwith as itsdegreesequence. Incontrasttotheeldofgraphicsequences,wherethereisasignicantbody ofliterature,muchlessisknownabout k graphicsequences.Infact,thefollowing theoremofDewdney[31]istheonlycharacterizationof k graphicsequencescurrently known. Theorem6.1Dewdney1975 Let = d 1 ;:::;d n beanonincreasingsequence ofnonnegativeintegers. is k graphicifandonlyifthereexistsanonincreasing sequence 0 = d 0 2 ;:::;d 0 n ofnonnegativeintegerssuchthat 1. 0 is k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 graphic, 2. n X i =2 d 0 i = k )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 d 1 ,and 3. 00 = d 2 )]TJ/F19 11.9552 Tf 11.955 0 Td [(d 0 2 ;d 3 )]TJ/F19 11.9552 Tf 11.955 0 Td [(d 0 3 ;:::;d n )]TJ/F19 11.9552 Tf 11.955 0 Td [(d 0 n is k graphic. RecentworkbythisauthorandBehrens,etal.[6]hasproducedsomenewdirections fromwhichtoapproachproblemsrelatingto k graphicsequences. 113
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Figure6.1: Theedgeexchange e u v e 0 Inparticular,weshowedthatthereisafamilyofedgeexchanges,similartothe 2switchfor2graphs,thatcanbeusedtomovebetweenrealizationsofa k graphic sequence.Let e and e 0 bedistinctedgesina k graph G ,andchoosevertices u 2 e n e 0 and v 2 e 0 n e .Theoperation e u v e 0 deletestheedges e and e 0 andaddstheedges e )]TJ/F19 11.9552 Tf 10.502 0 Td [(u + v and e 0 )]TJ/F19 11.9552 Tf 10.502 0 Td [(v + u where e )]TJ/F19 11.9552 Tf 10.502 0 Td [(u + v denotestheset e )101(f u g [f v g ;seeFigure6.1. SimilartotheresultofPetersenthatshowedthat2switchessucetotransformany realizationofagraphicsequenceintoanyother,weshowedthattheseedgeexchanges sucetomovebetween k realizationsof k graphicsequences.However,inperforming theseedgeexchanges,itissometimesnecessarytocreaterealizationsofthesequence thatarenotsimple k graphs{theyhaverepeatededges.Thisraisesthefollowing problem: Problem6.2 Determinetheminimalcollection Q ofedgeexchangessuchthatfor any k graphicsequence ,thereisasequenceofexchangesfrom Q thatsuceto movebetweenrealizationsof usingonlysimple k graphs. Recently,GuandLai[52]provedacharacterizationofpotentially r edgeconnected k graphicsequences.Togetherwithotherresultsin[6],thisshowsthatdespitethe dicultyindeterminingasimplecharacterizationof k graphicsequences,manyother questionsabout k graphicsequencescanbeanswered. OnesuchquestionrelatestoatheoremofKundu. 114
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Theorem6.3Kundu[72] Let = d 1 ;:::;d n beagraphicsequence.Thereis arealizationof thatcontainsa c factorifandonlyif 0 = d 1 )]TJ/F19 11.9552 Tf 12.351 0 Td [(c;:::;d n )]TJ/F19 11.9552 Tf 12.351 0 Td [(c is alsographic. Theproofofthistheoremusesonly2switches.Assuch,itseemspossiblethatwith thefamilyofedgeexchangesdescribedherewemaybeabletoproveasimilarresult ontheexistenceofrealizationsof k graphicsequenceswith c factors. IhopetostudyproblemssuchasProblem6.2andKundu'stheoremfor k graphic sequences,aswellasotherpotentialproblemsforhypergraphicsequences,inthe future. 6.2FurtherexplorationofpotentialRamseynumbers ThestudyofpotentialRamseynumbersisstillinitsinfancy.Assuch,thereare manyquestionsthatcanbeaskedaboutthesenumbers. OneniceresultingraphRamseytheoryisthatif T isan n vertextree,then r T;K t = n )]TJ/F15 11.9552 Tf 11.516 0 Td [(1 t )]TJ/F15 11.9552 Tf 11.517 0 Td [(1+1,whichwasprovedbyChvatalin1977[27].Iwouldlike todetermine r pot T;K t foran n vertextree T .Manyofourresultson r pot H;G for smallxed H dependontheexistenceofasimplecharacterizationofthepotentially H graphicsequences,whichgiveussimpleboundsontheexactvaluesofsomeofthe termsofasequence.Whileitiseasytoseewhenagraphicsequenceisthedegree sequenceofatree,thecharacterizationdoesnotsaymuchaboutthevaluesofspecic terms.Thus,exploring r pot T;K t mayrequiresomeadditionaltechniques. Inthesamevein,determining r pot mG;nH forgraphs G and H ,where mG means m disjointcopiesof G ,maybemuchmoredicultthandetermining r pot G;H Burr,Erd}os,andSpencer[8]gaveupperandlowerboundsfor r nG;nH thatdepend onlyon n andtheordersandindependencenumbersof G and H .Ifsimilarlysimple resultscouldbefoundfor r pot nG;nH oreven r pot mG;nH ,thiswouldadvanceour understandingofthepotentialRamseynumber. 115
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WhilethepotentialRamseynumberandRamseynumberareveryfarapartin somecasessuchascliques,ithasbeenshownthatmanypairsofgraphshaveRamsey numbersthatarelinearintheorderofthegraphs.SinceweknowthatpotentialRamseynumbersarelinear,itmaybeworthwhiletodeterminethepotentialRamsey numbersforsuchpairsofgraphs,tocomparethesevaluestotheRamseynumbers. AsindicatedinChapter5,wecanalsodenethepotentialRamseynumberfor morethan2colors.Todenethis,wemustdiscusswhatitmeansforgraphicsequencestopack.Givengraphicsequences 1 ;:::; k with i = d i 1 ;:::;d i n ,we saythat 1 ;:::; k pack ifthereexistedgedisjointgraphs G 1 ;:::;G k withvertexset f v 1 ;:::;v n g suchthat d G i v j = d i j foreach i ,and d G 1 [[ G k v j = P n i =1 d i j .Note thatthesequences 1 ;:::; k neednotbeinnonincreasingorder,andinfacttheorder ofthetermsofthesequencemaynotbealtered. Nowwedene r pot G 1 ;:::;G k tobetheleastinteger n suchthatforanycollectionofgraphicsequences 1 ;:::; k ,eachoflength n ,thattermwisesumto n )]TJ/F15 11.9552 Tf 12.537 0 Td [(1 andpack,thereexistedgedisjointgraphs F 1 ;:::;F k thatrealizethispackingandfor some i ,thegraph F i contains G i asasubgraph. Thenaturalrstquestiontoaskinthisveinis,whatis r pot K 3 ;K 3 ;K 3 ?GreenwoodandGleasonshowedthat r K 3 ;K 3 ;K 3 =17[51],butthisistheonlyclassical completegraphmulticolorRamseynumberknown.Giventhat2colorpotentialRamseynumbersareperhapseasiertocomputethan2colorRamseynumbers,but degreesequencepackingproblemsareverydicult,itwouldbeinterestingtoseeif moreprogresscanbemadeinthemulticolorpotentialRamseycase. 116
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