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CONDITIONAL COLORING by Mark R. Dillon B.S., Purdue University, 1982, 1985 M.S., Purdue University, 1984 A thesis submitted to the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Doctor of Philosophy Applied Mathematics 1998
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This thesis for the Doctor of Philosophy degree by Mark R. Dillon has been approved by Kathryn L. Fraughnaugh David C. Fisher J. Richard Lundgren Tom Russell Tom Altman Date
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Dillon, Mark R. (Ph.D., Applied Mathematics) Conditional Coloring Thesis directed by Associate Professor Kathryn L. Fraughnaugh ABSTRACT The conditional chromatic number X(G, P) of a graph G with respect to a graphical property P is the minimum number of colors needed to color the vertices of G such that each color class induces a subgraph of G with property P. When P is the property that a graph contains no subgraph isomorphic to a graph F, we write X(G, ,F). The conditional chromatic number of a graph has been studied by various authors since 1968. We focus on two conditional chromatic numbers, specifically X( G, ,Cj) and X( G, ,pj), where Cj is a cycle of length j for some fixed j 2: 3 and Pj is a path of length j1 for some fixed j 2: 2. We find X(G, ,Cj) for graphs missing at most j1 edges and X(G, ,pj) for graphs missing at most 2j 5 edges. To accomplish this, we characterize all Hamiltonian graphs of order n with at least (n 1) edges and all graphs with no Hamiltonian paths with at least (2n5) edges. We determine both conditional chromatic numbers for all graphs with acyclic complements. We also determine a lower bound on X(G, ,pj) in terms of the size of G. Finally, we show the problem of determining if X(G, ,P3 ) k, for some k 2: 0, is NPcomplete. This abstract accurately represents the content of the candidate's thesis. I rec ommend its publication. Signed ____________ Kathryn L. Fraughnaugh lll
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DEDICATION To my wife, Wendy Tweten Dillon, whose love, understanding, patience, and encouragement made it possible to complete this dissertation.
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ACKNOWLEDGEMENTS I would like to express my deep gratitude to my advisor, Professor Kathryn Fraughnaugh for her tireless guidance, patience, inspiration and encouragement during the course of my research and throughout the preparation of this disser tation. My sincere thanks to Professor Richard Lundgren and Professor David Fisher who were always there to answer questions. I would also like to thank my mother and father, Pat and Russell Dillon who always gave me the support I needed and my brothers and sister, Steve, Dave, Scott, Laura and Paul Dillon for their constant encouragement. I also thank Mr. John Drubert, my sixth grade teacher, who was the first person to interest me in the field of mathematics and my close friends Doug Mckissack, Arthur Shulman and Kevin Williams for their constant encouragement. Finally, I received, and gratefully acknowledge financial assistance from Hughes Aircraft Company without whose support, this thesis would never have been com pleted.
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Contents 1 Introduction to coloring and its applications 1.1 1.2 1.3 Overview of Thesis Results Definitions and Notation Conditional Coloring .. 2 Basic Results for x(G, Pj) and x(G, Cj) 3 NPcomplete 4 Determining the conditional chromatic number for graphs with acyclic complements 4.1 4.2 Determining x(G, Cj) when G is acyclic Determining x(G, Pj) when G is acyclic 4.3 The difference between the Cjand Pjchromatic numbers of a 1 4 5 9 16 22 29 29 35 graph. . . . . . . . . . . . . . . . . 37 5 Determining x(G, Cj) 39 5.1 Preliminaries . . 39 Vl
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5.2 Determining x(G, 'Cj) when e(G) (j1) 46 6 Determining x(G, 'Pj) 58 6.1 Determining which graphs of large size have a Hamiltonian path 58 6.2 Graphs missing complete subgraphs or a star 6.3 Determining x(G, ,pj) when e(G) is large .. 6.4 Determining bounds on x(G, ,pj) given the number of edges in a 73 75 graph. . . . . . . . . . . . . . . . . 79 A Appendix 84 B References 95 Vll
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1 Introduction to coloring and its applications Let G be a graph. A vertex coloring of G is an assignment of colors to its vertices so that no two adjacent vertices receive the same color. The chromatic number X( G) is the minimum number of colors needed to color G. An equivalent definition of the chromatic number is the minimum integer k such that there is a partition of the vertices into k sets so that the subgraph induced by each set is an independent set. Vertex coloring has a wide variety of applications. One such application is the following. In the United States Government, there are congressional committees with members of Congress serving on multiple committees. When assigning meet ing times for these committees, one must not schedule simultaneous meetings for two committees that have a common member. A schedule solution is found by de termining the minimum number of time slots required for the committees to meet. We can model this problem by constructing a graph G whose vertices represent committees. We draw an edge between two vertices if the committees represented by these vertices have a common member. Determining the minimum number of time slots required for the committees to meet is equivalent to determining X(G). For further details, see Roberts [47] or Bodin and Friedman [8]. Another application is the channel assignment problem. Radio stations in a region are to be assigned transmitting frequencies so that radio stations which are geographically close, say 50 miles, receive different frequency assignments. The problem of assigning frequencies is a graph coloring problem. Let each vertex 1
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represent a radio station and draw an edge between two vertices if the radio sta tions represented by those vertices are within 50 miles. The number of frequencies required so that the radio stations do not interfere with each other is the vertex chromatic number of this graph. For more information regarding the channel as signment problem, see Cozzens and Roberts [20], Hale [28], Opsut and Roberts [43], or Pennotti [46]. The last application of vertex coloring we will discuss is the classic map coloring problem. This is a much studied problem and is discussed in most graph theory textbooks. For example, see Bondy and Murty [10], Harary [29], Chartrand and Lesniak [17], or Roberts [47]. Given a map with various countries, we would like to color the countries in such a way that we use the fewest number of colors, and if two countries share a common border, they receive a different color. This problem can be translated into a graph coloring problem by building a graph with each vertex representing a country and drawing an edge between two vertices if the countries represented by those vertices have a common border. To determine the minimum number of colors required to color the map, we find X( G). This graph has the property that it is planar. The Four Color Theorem states that all planar graphs can be colored using at most four colors. Now, suppose we relax the condition that there be no edges in each color class and allow certain configurations of edges. For example, in the committee scheduling application, we could allow each committee to have a time conflict with at most one other committee with a common member. In this case, each color class can contain isolated vertices and edges. We will call this problem the relaxed committee scheduling problem. 2
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This is an example of a generalization for graph coloring called conditional graph coloring. We define the conditional chromatic number X(G, P) of G with respect to a graph theoretical property P to be the minimum integer k such that there is a partition of the vertices into k sets so that the subgraph induced by each set has the property P. Another application for conditional coloring is the circuit manufacturing prob lem. A designer draws an electrical circuit to be manufactured. Several circuit boards sandwiched one on top of the other may be required to build the entire circuit since, for this circuit to work properly, each circuit board must be built without intersecting edges. Now, the manufacturer would like to determine the minimum number of required circuit boards. The problem can be modeled as a conditional graph coloring problem. Let each vertex in our graph represent a junc tion in the circuit. Draw an edge between two vertices in the graph if the junctions represented by those vertices are connected in the circuit design. Determining the minimum number of circuit boards required is equivalent to determining X(G, P), where P is the property that each color class be drawn with no intersecting edges, i.e., is planar. This particular conditional chromatic number is also known as the vertex thickness of a graph. For more information regarding vertex thickness, see Beineke and White [7] or Cimikowski [18]. For other electrical circuit problems, see Hutchinson [37] or Garey and Johnson [26]. We will study two types of conditional chromatic numbers. The first permits no paths of order j in a color class, and the second permits no cycles of order j in a color class. The relaxed committee scheduling problem is an example of conditional coloring with the property that no color class contains P3 Studying 3
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these two numbers may g1ve insight into and/ or bounds on other conditional chromatic numbers or the original chromatic number. It is often the case that the understanding of a generalization leads to a better understanding of the original concept. 1.1 Overview of Thesis Results This thesis will present some new results about the conditional chromatic number X(G, 'Cj) with respect to the property of having no cycles of a fixed length j and some new results about the conditional chromatic number X(G, ,pj) with respect to the property of having no paths of a fixed length j 1. The first chapter will provide some basic definitions from graph theory and review the existing literature regarding X(G, ,Cj) and X(G, ,pi) The second chapter will present and prove some basic results regarding X ( G, ,Cj) and X(G, ,pj) Some of these results will be needed in later chapters. The third chapter will show that the problem of determining if a graph can be ,P3colored using k colors is NPcomplete. The fourth chapter will answer the question: given any graph with acyclic complement, what is X(G, ,Cj) and X(G, ,pi)? A construction which illustrates that X(G, ,pj)X(G, 'Cj) for any positive integer a is also provided. The fifth chapter will determine X( G, ,Cj) for all graphs of large size. Dargen and Fraughnaugh [21] characterized X(G, ,Cj) when a graph is missing at most j2 edges. In this chapter, we extend this result to graphs missing at most j1 edges. To accomplish this, we characterize all Hamiltonian graphs of order n with at least (n1) edges. 4
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The sixth chapter will determine X(G, ,pj) for all graphs of large size and determine an upper bound on the size of G given a bound for X(G, ,pj) To determine X(G, ,pj) for all graphs of large size, we will characterize all graphs of order n with no Hamiltonian paths having at least (2n5) edges. 1.2 Definitions and Notation A graph G consists of a finite nonempty set V = V (G) of vertices and a collection E = E( G) of distinct pairs of vertices, called edges. Throughout this paper, let G be a graph. The size or number of edges of G is denoted by e = e( G), and the order or number of vertices of G is denoted by n = n( G). A graph is simple if there is at most one edge between any distinct pair of vertices and there are no loops. For this paper, we will assume that all graphs are simple. If u and v are vertices of G, we write the edge joining u and vas uv and call u and v neighbors. The open neighborhood N ( u) of a vertex u is the set of neighbors of u, and the closed neighborhood N [ u] of u is N ( u) U { u}. The degree de ( v) of a vertex is defined to be the number of edges in G incident with that vertex. When it clear to which graph we are referring, we may write d( v). The minimum degree J(G) is the minimum degree among all vertices of G while the maximum degree is the maximum degree among all vertices of G. If J(G) = r, then all vertices have the same degree and G is regular of degree r or rregular. A 3regular graph is a cubic graph. If a vertex is not incident to any edge, then this vertex is an isolated vertex. A subgraph H of G is a graph having all of its vertices and edges in G, and we write H G. For any set S of vertices in G, the induced subgraph (S) is the 5
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maximal subgraph of G with vertex set S. If S is a nonempty set of edges in G, then (S) is the subgraph whose edge set is S and whose vertex set is the set of ends of edges in S. A graph is complete if every pair of vertices is joined by an edge. The complete graph on n vertices is Kn. The graph on n vertices with no edges is InThe graph with distinct vertices v1 ... Vn and edges v1v2 v2v3 ... VnlVn is a path Pn. We say the vertices v1 and Vn are connected by the path PnWe sometimes call this path a ( v1 vn)path. The length of a path is the number of edges in it. The distance d( u, v) from u to v in G is the length of a shortest path from u to v. The diameter of G is the maximum distance between any two vertices of G. If u = v and there are at least three vertices on the path, then this path is a cycle. The graph which consists of a cycle on j vertices is Cj, and a graph which contains no cycles is acyclic. A graph G is Hamiltonian if it has a cycle containing all the vertices of G. A graph G has a Hamiltonian path if it has a path containing every vertex of G. A graph G is Hamiltonian connected if for every pair u and v of distinct vertices of G, there exists a Hamiltonian ( u, v )path. A graph is connected if for any two vertices u and v, there is a path from u to v. A maximal connected subgraph of G is called a component of G. If G is not connected, then we say G is disconnected. The connectivity !);(G) of G is the minimum number of vertices whose removal results in a disconnected or trivial graph. If the connectivity of G is !);, we say G is !);connected. A connected acyclic graph is a tree, and a forest is a graph each of whose components is a tree. In a tree T, we can make T a rooted tree by designating any vertex as the root vertex. Once we choose a root u, the level of a vertex v is d( u, v). The root is the 6
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only vertex at level 0. Further, all adjacent vertices differ by exactly one level, and each vertex at level i + 1 is adjacent to exactly one vertex at level i. The maximum level is the height hu (T) of the tree. The complement G of G also has V(G) as its vertex set, but two vertices are adjacent in G if and only if they are not adjacent in G. A clique of G is a complete subgraph of G. The clique number w( G) of G is the maximum order among all cliques of G. An independent set of vertices of G is a set V such that xy :. E for all x, y E I. An independent set of edges of G has no two of its edges incident, and such a set is a matching. Next, we will discuss some operations defined on graphs. Let GI and G 2 be two graphs with disjoint vertex sets. The union G 1 + G 2 of GI and G 2 has V(G 1 + G 2 ) = V(GI) U V(G 2 ) and E(G1 + G 2 ) = E(G1 ) U E(G2). In general, mG is the pairwise vertex disjoint union of m copies of G. The join GI V G 2 of two graphs GI and G 2 has V(G 1 V G 2 ) = V(GI) U V(G 2 ) and E(G1 V G 2 ) = E(GI) U E(G2 ) U {uv: u E V(GI) and v E V(G 2 )}. The cartesian product GI x G 2 of two graphs GI and G 2 has V(GI x G 2 ) = {(u, v): u E V(Gt) and v E V(G 2 )}, and (u1 vi) and (u2 v 2 ) are adjacent when ever either uiu2 E E(GI) and VI = v2 or ui = u 2 and viv2 E E(G2). The strong product G 1 oG 2 of two graphs GI and G 2 has V( G 1 oG 2 ) = {( u, v) : u E V( G 1 ) and v E V(G 2 )}, and (u 1 v 1 ) and (u 2 v2 ) are adjacent whenever either u 1 u 2 E E(GI) and vi = v2, or ui = u2 and viv2 E E(G2), or uiu2 E E(GI) and viv2 E E(G2). The conjunction GI 1\ G 2 of two graphs GI and G 2 has V ( GI 1\ G 2 ) = { ( u, v) : u E V(Gt) and v E V(G 2 )}, and (u 1 v 1 ) and (u 2 v2 ) are adjacent whenever u1u2 E E(GI) and VIV2 E E(G2). 7
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A vertex coloring is an assignment of labels to the vertices of G so that any two adjacent vertices receive different labels. We think of each distinct label as a color and call each set of vertices assigned a fixed color a color class. The vertex chromatic number X( G) of G is the minimum k for which G has a kcoloring, i.e., one using k colors. A graph is kcolorable if we can color it using k colors. An edge coloring is an assignment of labels (or colors) to the edges of G so that any two incident edges receive a different label. The edge chromatic number of a graph is defined to be the minimum number of colors needed to edge color G. If we use the term "chromatic number" or "coloring" in this paper, we mean vertex chromatic number or vertex coloring. A graph is said to be embedded in a surface S when it is drawn on S so that no two edges intersect. A graph is planar if it can be embedded in the plane; a plane graph has already been embedded in the plane. We will refer to the regions defined by a plane graph as its faces. The Four Color Theorem states that every planar graph is 4colorable. For a proof of this result, see Appel and Haken [5]. A bipartite graph G is a graph whose vertex set V can be partitioned into two subsets V1 and V2 such that every edge of G joins V1 and 1;2. The bipartition number b( G) of a graph G is given by b(G) = max{ e(B) : B G and B is bipartite}. For other basic graph theory definitions and terminology, the reader is referred to Harary [29]. 8
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1.3 Conditional Coloring An equivalent definition of vertex coloring is a partition of the vertex set so that the subgraph induced by each set of this partition is an independent set. Stated differently, each color class contains no path on two vertices. This led several authors [15], [16], [35] to a more general concept of vertex colorings. Let P be any graph theoretic property. For example, P could be the property that a graph contains a clique of a certain order, that a graph does not contain a cycle of order 6, that a graph does not contain an induced cycle of order 6, or the maximum degree of a graph is 5. We define a Pcoloring of a graph to be an assignment of colors to its vertices so that the subgraph induced by each color class satisfies the property P. The ? conditional chromatic number X( G, P) of G (or briefly Pchromatic number) is the minimum k for which G has a Pcoloring with k colors. When Pis the property that a graph consists entirely of isolated vertices, the ?chromatic number is the usual chromatic number. When P is the property that a graph contains no subgraph (not necessarily induced) isomorphic to a graph F, we write X(G, F) for the ?chromatic number and refer to a Pcoloring as a ,p coloring and the ?chromatic number as the Fchromatic number. If X( G, F) ::; k, then we say G is ,p kcolorable. When P is the property that a graph contains no induced subgraph isomorphic to a graph F, we write X(G, F!) for the ?chromatic number and refer to a Pcoloring as a F!coloring and the ?chromatic number as the F!chromatic number. The concept of partitioning the vertex set of a graph so that the subgraph induced by the vertices in each set in the partition has the property P seems to 9
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have been independently discovered by several authors around 1968 (see [15], [16], [35]) and again by several authors around 1985 (see [3], [4], [12], [14], [23], [30], [42]). Mathematicians have studied various conditional chromatic numbers since 1968. In 1968, Chartrand, Kronk, and Wall [16] studied a conditional coloring number called the arboricity of a graph where color classes are acyclic. Hedetniemi [35] proved that the arboricity of a planar graph is at most three. In 1968, Chartrand, Geller and Hedetniemi [15] proved several results about X(G, ,pj) For example, if the diameter of G is d, then X(G, ,pj) ::; dj + 3 for j 2. Further, for every j 2, the authors constructed a planar graph G such that X(G, ,pj) = 4. This proved that there is no stronger result than the Four Color Theorem for X(G, ,pj) for j 2. Also, in 1968, Sachs and Schauble [48] proved that given j 2 and K k 1, there exists a graph G with X(G, ,Kj) = k and a ,Kj Kcoloring of G containing at least k color classes isomorphic to Kjl In 1969, Kramer and Kramer [39] discussed X(G, P), where Pis the property that the subgraph induced by each color class has minimum degree j for some fixed j 0. In 1970, Lick and White [41] published a paper with results for the same conditional chromatic number. In 1975, Cook [19] studied X(G, ,K1,j) for In 1977, Harary and Kainen [34] discussed X(G, ,K3 ) for planar graphs. Also, LesniakFoster and Straight [40] published results for X(G, P), where P is the property that each color class induces a complete graph or a graph with no edges. Sampathkumar, Prabha, Neeralagi, and Venkatachalam [50] published results for 10
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X(G, ,pj) for j 2. In 1978, Beineke and White [7] discussed the thickness of a graph. The concept of thickness is equivalent to X(G, P), where P is the property that each color class is planar. The authors determined the thickness of complete and complete bipartite graphs. In 1985, Harary [30] published a paper providing an overview of conditional coloring. This paper discussed the conditional vertex and edge chromatic num bers for several different properties and has become the standard for conditional coloring terminology. In 1985, Andrews and Jacobson [3] studied X(G, where the maximum vertex degree in each color class is at most t. The authors related X(G, to X( G) by proving that X(G, X(G)/(t + 1). Furthermore, they showed X(G, n2 j(tn + n2 2e), where nand e are the order and size of G. Harary and Fraughnaugh [31] also published results for X(G, in 1985. Also in 1985, Mynhardt and Broere [42] discussed conditional coloring with respect to the property that each color class is a disjoint union of complete sub graphs. They also studied the problem of finding a graph subject to certain restric tions for which the conditional chromatic number is arbitrarily large. Mynhardt and Broere [12] also studied conditional coloring with respect to the property that each color class has no induced subgraph isomorphic to a graph F. In 1986, Harary and Fraughnaugh [32] generalized the concept of bipartition number to that of a conditional bipartition number. Specifically, given a property P, a graph is conditionally bipartite with respect to P if V (G) is the disjoint union of sets X and Y where the induced subgraphs (X) and (Y) both have property 11
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P. The conditional bipartition number b(G,P) is max{e(B): B G and B is conditionally bipartite with respect toP}. The authors studied b(G, P) for several minimum and maximum degree properties. Also in 1986, Domke, Laskar, Hedetniemi, and Peters [23] discussed X(G, P), where Pis the property that the subgraph induced by each color class is a complete rpartite graph for any r and X(G, Q), where Q is the property that each color class is a disjoint union of complete subgraphs. In 1987, Andrews and Jacobson [4] published a paper with results for X(G, Also in 1987, Brown and Corneil [14] studied conditional coloring with respect to the property that each color class has no induced subgraph isomorphic to a set of graphs. In 1989, Akiyama, Era, Gervacio, and Watanabe [1] discussed the kpath chro matic number. The kpath chromatic number X( G, Pk) of G is the smallest number c of distinct colors with which V (G) can be colored such that each connected com ponent of (Vi) is a path of order at most k, 1 ::; i ::; c. The authors proved that X(G, Pk) ::; I for rregular graphs. Since the 2path chromatic number is equivalent to our ,P3chromatic number, we immediately get that X(G, ,P3 ) ::; 2 for cubic graphs. Also in 1989, Albertson, Jamison, Hedetniemi, and Locke [2] discussed conditional coloring with respect to the property that each color class is a disjoint union of complete subgraphs. In 1990, Baldi [6] published results for X(G, ,pj) for j 2. In 1991, Harary and Hsu [33] provided theorems relating the conditional chro matic number of the Cartesian product, join, strong product, and conjunction of two graphs to the conditional chromatic number of the original pair of graphs. 12
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In 1992, Borodin [11] discussed the kcyclic chromatic number. A coloring of the vertices of a planar graph is kcyclic if whenever two vertices lie in the boundary of the same face of size at most k, their colors are different. There are also conditional coloring papers published in 1992 where a color class is not permitted to contain some induced subgraph. For example, Brown and Corneil [13], discussed uniquely ,H! kcolorable graphs where His any graph. A graph G is uniquely kcolorable if G is kcolorable and there is only one kcoloring (up to a permutation of colors). In fact, Brown and Corneil conjectured that for all graphs of order at least two and for all nonnegative integers k, there exist uniquely H! kcolorable graphs. So far, they have shown this result whenever G is 2connected or G is 2connected. In 1992, Johns and Saba [38] considered X(G, Pj) They proved that given integers l 1 and j 2, there exists a graph G such that X(G, Pj) = l, namely K(jl)l Also, there exists a graph such that X(G, Pj)X(G, Pj+t) = l, namely Kj(i+l)t In 1993, Dargen and Fraughnaugh [21] determined X(G, Cj) for graphs miss ing up to j2 edges and Sampathkumar [49] discussed X(G, P), where Pis the property that the subgraph induced by each color class has independence j for some fixed j 0. Also in 1993, Hutchinson [37] applied thickness results to testing printed circuit boards for electrical shorts. In 1995, Cimikowski [18] presented some heuristics for the graph thickness problem, i.e., decomposing a graph into the minimum number of planar sub graphs. The heuristics are based on some algorithms for finding a maximal planar subgraph of a nonplanar graph. He also proved that T(G) :::; + 3/2J, 13
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where T(G) is the thickness of G and e is the size of G. For a survey of different properties studied, who studied them, and when they were studied, see Table 1.1. Table 1.1. Conditional Coloring Properties studied Color Class Property Year Reference disconnected or trivial 1968 Hedetniemi [35] 1970 Hedetniemi [36] acyclic 1968 Chartrand, Kronk, Wall [16] 1968 Hedetniemi [35] has no K3 1977 Harary, Kainen [34] has no Kj for some fixed j 1968 Sachs, Schauble [48] has no Pj for some fixed j 1968 Chartrand, Geller, Hedetniemi [15] 1977 Sampathkumar, Prabha, Neeralagi, Venkatachalam [50] 1990 Baldi [6] 1992 Johns, Saba [38] complete or a graph without edges 1977 LesniakFoster, Straight [40] has maximum degree j for some 1985 Harary, Fraughnaugh [31] fixed j 1985 Andrews, Jacobson [3] 1986 Harary, Fraughnaugh [32] 1987 Andrews, Jacobson [4] complete rpartite graph for any r 1986 Domke, Laskar, Hedetniemi, Peters [23] has no induced subgraph 1985 Broere, Mynhardt [12] isomorphic to graph F 1985 Mynhardt, Broere [42] 1987 Brown, Corneil [14] has no induced subgraph isomor1987 Brown, Corneil [14] phic to any graph F in a set of graphs F disjoint union of complete 1985 Mynhardt, Broere [42] subgraphs 1986 Domke, Laskar, Hedetniemi, Peters [23] 1989 Albertson, Jamison, Hedetniemi, Locke [2] contains no K1,j 1975 Cook [19] minimum degree j for fixed j 1969 Kramer, Kramer [39] 1970 Lick, White [41] 14
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Table 1.1. Conditional Coloring Properties studied (cont.) Color Class Property Year Reference disjoint union of paths of order at 1989 Akiyama, Era, Gervacio, Watanabe most k [1] j independent for fixed j 1993 Sampathkumar [49] has no Cj for fixed j 1993 Dargen, Fraughnaugh [21] planar 1978 Beineke, White [7] 1993 Hutchinson [37] 1995 Cimikowski [18] kcyclic 1992 Borodin [11] contains no induced subgraph isomor1992 Brown, Corneil [13] phic to H survey of properties 1985 Harary [30] In this thesis, we concentrate primarily on two conditional chromatic numbers, X(G, ,Cj) where Cj is a cycle of order j 2: 3 and X(G, ,pj) where Pj is a path of order j 2: 2. 15
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2 Basic Results for x(G, Pj) and x(G, Cj) This chapter provides the necessary background for the topics presented in Chap ters 3, 4, 5, and 6. We start by proving some basic relationships for X(G, Cj) and X(G, ,pj) The following straightforward result has been known since 1968 [15]. Theorem 2.1 Let G be a graph of order n. If j 2: 2, then X(G, ,pj) I Proof. Let n = a(j 1) +r, where 0 r < j 1. Color G as follows: create a color classes of size j 1 and one color class of size r if r > 0. Hence X(G, ,pj) I D Corollary 2.1 If j 2: 2, then X(Kn, 'Pj) = I Proof. Since a complete subgraph of order at least j contains Pj, each color class of Kn can contain at most j 1 vertices. Therefore, X(Kn, ,pj) 2: I and equality follows from Theorem 2.1. D The next theorem provides the most basic relationship between X(G, ,Cj) and X(G, ,pj) Theorem 2.2 If j 2: 3 and G is a graph, then X(G, ,Cj) X(G, ,pj) Proof. Let C be any minimum ,Prcoloring of G. Since a color class that contains no Pj contains no Cj, the coloring C is also a ,Crcoloring of G. So, D 16
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Now that we have an upper bound for X(G, ,pj) and a relationship between X(G, ,Cj) and X(G, ,pj), we can find an upper bound for X(G, ,Cj) Corollary 2.2 If j 3 and G is a graph of order n, then X(G, 'Cj) < I Further, X(Kn, 'Cj) = I Proof. By Theorem 2.2 and Theorem 2.1, we get X(G, 'Cj) ::; X(G, ,pj) ::; I Let G = Kn. A subgraph induced by j or more vertices of Kn contains Cj, and therefore cannot be a color class of KnHence, X(Kn, 'Cj) D The next theorem relates two different conditional coloring numbers. Theorem 2.3 If j k 2, then X(G, ,pj) ::; X(G, ,pk) Proof. Let C be any minimum ,pkcoloring of G. Since a color class that contains no Pk contains no Pj for k :=::; j, the coloring C is also a ,Prcoloring of D One might ask if the above type of theorem is true for X(G, ,Cj) In fact, there is no relationship in general in either direction. For example, X(C4 ,C4 ) = 2 and X(C4 ,C3 ) = 1, which implies that X(G, ,C4 ) 1:_ X(G, ,C3 ) for at least one graph. Further, X(C3 ,C4 ) = 1 and X(C3 ,C3 ) = 2, which implies that X(G,,04 ) X(G,,03 ) for at least one graph. The next theorem provides a relationship between the conditional chromatic numbers of a graph and its subgraphs. Theorem 2.4 Let G be a graph with H G. If j > 2, then X(H, ,pj) < X(G, ,pj) If j 3, then X(H, 'Cj) ::; X(G, 'Cj) 17
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Proof. Let C be any minimum 'Cj('Pj)coloring of G. Since H G, the coloring C is also a ,Cj('Pj)coloring of H. So, X(H, ,Cj) ::::; X(G, ,Cj) and D Now that we have results for subgraphs for X(G, ,Cj) and X(G, ,pj), we can derive an elementary lower bound for X(G, ,pj) and X(G, 'Cj) Theorem 2.5 If G is a graph and j 2: 3, then X(G, ,pi) 2: X(G, ,Cj) 2: Proof. First of all, Kw(G) G. By Corollary 2.2, X(Kw(G), ,Cj) = By Theorem 2.4, we get X(G, ,Cj) 2: X(Kw(G), ,Cj) = By Theorem 2.2, we have X(G, ,pi) 2: X(G, 'Ci) D This bound is attained when G = Kn, and therefore the bound is tight. In order to derive some upper bounds on the order of a color class of some graph G, we need some well known results regarding Hamiltonian graphs, Hamiltonian connected graphs, and graphs which have a Hamiltonian path. First, we state Ore's Theorem, Dirac's Theorem, and another theorem which appears as Corollary 4.6 in Bondy and Murty [10]. Theorem 2.6 (Ore [44}) If G is a graph of order n 2: 3 such that for all distinct nonadjacent vertices u and v, d(u) + d(v) 2: n, then G is Hamiltonian. Theorem 2. 7 (Dirac {22}) If G is a graph of order n 2: 3 and each vertex has degree at least then G is Hamiltonian. Theorem 2.8 (Ore [45}, Bondy {9}) If G is a graph of order n 2: 3 and e(G) 2: then G is Hamiltonian. Moreover, the only nonHamiltonian graphs with n vertices and edges are K1 V (K1 + Kn2) and K2 V I3. 18
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The following theorem is also well known. Parts ( i) and (iii) appear as an exercise in West [51] and part (ii) is simply a restatement of the first sentence in Theorem 2.8. Theorem 2.9 Let G be a graph of order n. If n 2 and e( G) ( n 4), then G is Hamiltonian connected. ( i) Ifn 3 and e(G) (n3), then G is Hamiltonian. (ii) If n 2 and e(G) (n2), then G has a Hamiltonian path. (iii) Proof. ( i). (by induction on n). If n 3, then ( i) is vacuously true. If n = 4, then G = K4 and a complete graph is Hamiltonian connected. If n = 5, then G E { K5 K5 e} where e is an edge. These two graphs are Hamiltonian connected. If n = 6, then G E {K6 K6 e, K6 E(2K2), K6 E(P3)} = Q where e is an edge. Each G E g is Hamiltonian connected. So assume ( i) holds for appropriate graphs on n 1 vertices. Let G be a graph on n 7 vertices and e(G) (n4). We will show G is Hamiltonian connected. Let {u,v} V(G) with u #v. If d(u) n2, then e(Gu) e(G)(n2) (n4)(n2) = (n5) and, by the induction hypothesis, Gu is Hamiltonian connected. Now e(G) (n4) implies that 6(G) 3 for we would need to remove at least n3 edges from a complete graph in order for 6(G) = 2. Since d(u) 3, we can choose z E N(u){v} and add uz to a Hamiltonian (z,v)path in Gu to form a Hamiltonian (u,v)path in G. If u is adjacent to every other vertex in G (i.e., d(u) = n 1), then e(Gu) = e(G)(n1) (n4)(n1) = (n4) and, by Theorem 2.8, 19
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Gu is Hamiltonian. Break an edge involving v (say vw) on the Hamiltonian cycle in Gu and add the edge wu to obtain a Hamiltonian (u,v)path in G. In either case, G has a Hamiltonian (u,v)path. Therefore, G is Hamiltonian connected. (ii). For a proof of Ore's Theorem, see Roberts [47]. (iii). If G is complete, then G is Hamiltonian. So, assume G is not complete. Since e(G) (n2), we get n 3. Let uv be any missing edge of G. Consider G + uv. Now, e(G + uv) = e(G) + 1 (n3). By (ii), the graph G + e has a Hamiltonian cycle. Therefore, G has a Hamiltonian path. D We can see that all three statements in the previous theorem are exact since K2 V (K1 + Kn_3 ) has (n 3) edges and is not Hamiltonian connected, K1 V (K1 +Kn_2 ) and K2 V !3 (n2) edges and are not Hamiltonian (in fact, Theorem 2.8 points out that these are the only such graphs), and K1 + Knl has ( n 1) edges and contains no Hamiltonian path. Lastly, we state a sufficient condition on the size of a graph to obtain cycles of all orders. Theorem 2.10 IJG is a graph with order n 3 and e(G) (n3), then Ck G for all k = 3, 4 ... n. Proof. (by induction on n). If n = 3, then G = K3 and the result follows immediately. Assume that n 4 and that any graph of order n 1 with at least (n4) edges contains cj for j = 3, 4, ... 'n1. Let G be a graph of order n with e(G) (n3). By Theorem 2.9, we get Cn G. If G is complete, the result follows immediately. If G is not complete, then b(G) :=::; n2. Choose a vertex v E V(G) of minimum degree in G. Consider Gv. Then e(Gv) = e(G)b(G) (n3)(n2) = 20
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(n4). By the induction hypothesis, Gv contains C3 C4 ... Cn_1 as subgraphs. Therefore, G contains the same C3 ... Cn_1 as subgraphs. D Now that we have some basic results, we discuss the computational complexity of the following problem: Given a graph G and a positive integer k, can G be ,P3colored using k colors? In the next chapter, we show that this problem is NPcomplete. 21
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3 NPcomplete A polynomial time algorithm is an algorithm whose worstcase running time is O(nk), where the input to the algorithm is of cardinality n and k is some con stant. A problem II is defined to be a binary relation on a set I of problem instances and a set S of problem solutions. For example, consider the problem SHORTESTPATH of finding a shortest path between two given vertices in G. An instance for SHORTESTPATH is a triple consisting of a graph and two ver tices. A solution is a sequence of vertices in the graph G with perhaps the empty sequence denoting that no path exists. The problem SHORTESTPATH itself is a relation that associates each instance of a graph and two vertices with a shortest path in the graph that connects the two vertices. For simplicity, the theory of NP completeness restricts itself to decision problems, those having a yes/no solution. In this case, we can view an abstract decision problem as a function that maps the instance set I to the solution set {0, 1 }. For example, a decision problem related to SHORTESTPATH is as follows: Given a graph G, two vertices {u,v} V(G), and a positive integer k, does there exist a path of length at most k? There are decision problems which can be solved in polynomial time and those which require superpolynomial time. A polynomial time solvable problem is one which can be solved using a deterministic polynomial time algorithm. The complexity class NP is the class of problems where given a "yes" solution, we can verify this solution in polynomial time. A basic idea in the theory of NP completeness is that of a polynomial transformation. Let lh and lh denote two 22
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decision problems. We say that there is a polynomial transformation from II1 to II2, written IIl ex II2, if the following two conditions hold: (a) There exists a function F transforming any instance I of II1 to an instance F(I) of II 2 such that the answer to I with respect to II1 is "yes" if and only if the answer to F(I) is "yes" with respect to II2 (b) There exists an polynomial time algorithm to compute F(I). A decision problem II is NPcomplete if II E N P and for every problem II' E N P, II' ex II. The kCOLORING problem is stated as follows: given a graph G = (V, E) and integer 3:::; k:::; lVI, is G kcolorable? We know from Garey and Johnson [27] that the kCOLORING problem is NPcomplete for k 2: 3. The goal of this chapter is to determine how difficult it is to solve the problem of conditionally coloring a graph. Observe that the ,p2 kCOLORING problem is the usual kCOLORING problem. Does relaxing the condition for each color class being an independent set to each color class containing no Pj for j 2: 3 change the computational complexity of the colorability problem? We will show the answer is "no" when j = 3. The ,p3 kCOLORING problem is stated as follows: given a graph G = (V, E) and integer 3 :::; k :::; (Note: By Theorem 2.1, a graph can always be ,P3colored with at most flfll colors), does there exist a kpartition of the vertex set so that each subgraph induced by a set in the partition does not contain P3 as a subgraph (not necessarily induced)? Theorem 3.1 The ,p3 kCOLORING problem II is NPcomplete fork 2: 3. Proof. Let G = (V, E) be a graph of order n, and C a ,p3 kcoloring of G. The following is a polynomial time algorithm which can be used to check that C is a 23
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valid .P3coloring of G. Examine all triples of distinct vertices in each color class and determine whether the graph induced by each triple contains P3 Determining if a triple contains P3 is 0(1). Since each set in the partition can contain at most n vertices, there are O(n3 ) triples. This algorithm is of order n3 and this shows that II E NP. Next, we will show that a known NPcomplete problem can be transformed to II. Construct a graph F(G) from G = (V, E) as follows: for each v E V, let F(G) contain two vertices VI and v2 We say that VI and v2 are associated with the vertex v. In F(G), join v1 and v2 to the vertices associated with each neighbor of v and join VI to v2 Observe that this transformation is polynomial. See Figure 3.1 for the transformation of G = C4 We will show that X( G) k implies that X(F(G), .P3 ) k. Assume X( G) X W D u v Figure 3.1. The construction of F(C4). k. Let H = F(G) and C beakcoloring of G. Color Has follows: For each vertex v E V, assign C(v) to the two vertices associated with v in H. This coloring uses at most k colors to color H. To see that this coloring is a valid .P3coloring of H, 24
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suppose, by way of contradiction, that a color class of H contains P3 = uvw. By construction, ab E E(H) if and only if either a and bare associated with the same vertex in G or there exists cd E E( G) such that a is associated with c and b is associated with d. Therefore, if two adjacent vertices in H are assigned the same color, then they must be associated with the same vertex in G since G has a valid kcoloring. Thus, uv E E(H) implies that there exists c E V(G) such that u and v are associated with c. Similarly, vw E E(H) and v being a vertex associated with c imply that v and w are associated with c. This is a contradiction since, by construction, there are exactly two vertices in H associated with a vertex in G. Therefore, X(H, P3 ) ::; k. Next, we will show that X(H, ,P3 ) :=::; k implies that X( G) :=::; k. Let d: V(H) + {1, 2, ... k} be a ,p3 kcoloring of H. We color the vertices of G using Algorithm 3.1 presented in Figure 3.1. Let c: V(G)+ {1,2, ... ,k} be the function that assigns colors to the vertices of G based on Algorithm 3.1. Note 1: Observe that Algorithm 3.1 assigns a color to z E V(G) from one of the colors assigned to the two vertices in H associated with z. Nate 2: For every v E V (G) with associated vertices VI and v2 there is at most one wE NH(vi, v2 ) colored c(v). Suppose by way of contradiction that there exist x,y E V(H) such that {x,y} NH(vi,v2 ) and d(x) = d(y) = c(v). By Note 1, we know that either VI or v2 is colored c(v), say VI But XVIY forms P3 in H, which is a contradiction. Each time through the loop (Lines 722), a vertex in G is assigned a color in either Line 9 or Line 17. Since the number of vertices in a graph is finite, all the vertices of G are assigned a color. We will show that c is a valid kcoloring of G. 25
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Input: Graphs G and H, and a .P3coloring d: V(H) + {1, 2, ... k} of H. Output: A kcoloring c: V(G)+ {1, 2, ... k} of G. 1. For every z E V(G) with associated vertices z1 and z2 in H. 2. If d(z1 ) = d(z2), then let c(z) = d(z1). 3. If d(z1 ) tf_ d(N(z1)) and z is uncolored, then let c(z) = d(z1). 4. If d(z2 ) tf_ d(N(z2)) and z is uncolored, then let c(z) = d(z2). 5. If G has uncolored vertices, then 6. Let previous_color = 1 and z be an uncolored vertex in G. 7. Repeat steps 822 until all vertices of G are colored. 8. If previous_color =I d(z1), then 9. let c(z) = d(z1), 10. let previous_color = d(z1), and w be the unique vertex in V(G) with which the vertex in NH(z1 ) with color d(z1 ) is associated. 11. If w is uncolored, then 12. let z = w. 13. else if there are uncolored vertices in G, then 14. let previous_color = 1 and z be an uncolored vertex in G. 16. else (previous_color = d(z1)) 17. let c(z) = d(z2), 18. let previous_color = d(z2), and w be the unique vertex in V(G) with which the vertex in NH(z2 ) with color d(z2 ) is associated. 19 If w is uncolored, then 20. let z = w. 21. else if there are uncolored vertices in G, then 22. let previous_color = 1 and z be an uncolored vertex in G. 23. Output c and stop. Figure 3.1. Algorithm 3.1 26
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Let v E V(G) with c(v) =a, u E N(v), VI and v2 be the vertices in H associated with v, and u1 and u2 be the vertices in H associated with u. We will show that c(u) 1 a. Assume v was assigned color a in Line 2 of Algorithm 3.1. Now, u cannot be assigned color a or else the subgraph induced by the color class in H containing vi, v2 and one of the vertices associated with u would contain P3 Thus, all vertices assigned a color in Line 2 of Algorithm 3.1 receive a valid color. Assume v was assigned color a in Line 3 of Algorithm 3.1, that is, d(vi) =a and the neighbors of v1 (which include u1 and u2 ) are not colored a. Since u is assigned its color from one of the colors assigned to ui or u2 (Note 1), we have c( u) 1 a. Thus, all vertices colored in Line 3 receive a valid color. A similar argument proves that every vertex colored in Line 4 receives a valid color. We have just shown that if u or v were assigned a color in Lines 2, 3, or 4, then c( u) 1 c( v). So assume u and v were assigned colors in Lines 9 or 17 of Algorithm 3.1. Since only one vertex is colored at at time, let's assume that u was already colored when v is colored. Suppose that c( u) = c( v) = a. Assume u was assigned color a in Line 9 of Algorithm 3.1. In the loop (Lines 722) with z = u, in Line 9 we have c(u) = d(ui) = a and we let previous_color = a (Line 10). Since we assumed that c( v) = a, either v1 or v2 is colored a (Note 1). Further, since there is at most one neighbor of ui colored a (Note 2), we must have z = v (Line 12), i.e., vis the next vertex to be colored in the loop (Lines 722). When we go through the loop to color v, either it gets colored in Line 9 or 17. If v is colored in Line 9, then since previous_color = a, we know that the color of c( vi) 1 a (Line 8) and c( v) 1 a 27
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(Line 9). So v must be colored in Line 17 and we know that d(v1 ) =a (Line 16) and c(v) = d(v2 ) =a (Line 17). Now, d(v2 ) =/=a or else v would have been colored in Line 2. Thus c( v) =!= a, which contradicts the assumption that c( v) = a. By applying the previous argument replacing x1 with x2 and starting with z = x in Line 17, we again reach a contradiction. Thus, for every adjacent u, v E V(G), we have c(u) =!= c(v), which implies that Algorithm 3.1 produces a kcoloring of G. We have transformed a known NP complete problem to our problem. Therefore, the P3 kCOLORING problem is NPcomplete. D This construction for H does not help to prove that the ,pj kcolorability problem is NPcomplete for j 2: 4. We have tried other constructions for H for the ,pj kcolorability problem with no success. Suggested areas for future research include determining whether the ,pj kcolorability problem, the Cj kcolorability problem, and the Kj kcolorability problem are NPcomplete for j 2: 3. Now, we turn our attention to finding X(G, Pj) and X(G, Cj) for graphs with acyclic complements. 28
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4 Determining the conditional chromatic number for graphs with acyclic complements In this chapter we will show that X(G, .Cj) and X(G, .Pj) may have different values for many graphs. First, we will examine the values of X(G, .Cj) for graphs whose complements are acyclic and then perform a similar analysis for X(G, .Pj) Finally, the main theorem of this chapter shows that the difference between X(G, .Cj) and X(G, .Pj) can be made arbitrarily large in a particular family of graphs. 4.1 Determining x(G, Cj) when G is acyclic To determine X(G, .Cj) for j 3, it is necessary to determine how large any color class can be in a minimum Crcoloring of G. If j = 3, then we will see the largest color class must be of size 4 or less and if j 4, then the largest color class must be of size j or less. Therefore, we determine X(G, .C3 ) and X(G, .Cj) for j 4 in two separate theorems. We first address the size of the largest possible color class for X(G, .C3 ) with the following lemma. Lemma 4.1 If G is a graph of order n 5 and G is acyclic, then G contains C3 as a subgraph. Proof. Let G be a graph of order n 5 with acyclic complement. Let H be a subgraph of G with five vertices. We may assume fi is a tree (otherwise we remove edges from H until fi is a tree). If fi has a vertex v such that dfl(v) 3, then since fi contains no cycles, N fl ( v) is an independent set in fi of size at least 29
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3. If every vertex in fl has degree at most 2, then fl = P5 which clearly contains an independent set of size 3. In either case, the vertices from the independent set in fl forms C3 in H. Since H G, the graph G contains C3 D The following theorem evaluates X(G, .C3 ) when G is acyclic. Theorem 4.1 Let G be a graph with acyclic complement. Let m be the number of edges in a maximum matching of G. If m 0, then X(G, .C3 ) = I n;m l Proof. Let A be a color class in a .C3coloring of G. By Lemma 4.1, IAI :::; 4. Let a be the number of color classes of size 4 and b the number of color classes of size 3. The only C3free graphs of order 4 whose complements are acyclic are P4 and C4 Each of these graphs is missing two independent edges. Therefore, the subgraph induced by each color class of size 4 must be missing two independent edges. The only C3free graphs of order 3 whose complements are acyclic are P3 and K2 + K1 Each of these graphs is missing an edge. Therefore, the subgraph induced by each color class of size 3 must be missing an edge. Since color classes are disjoint, we can form a matching M1 in G with a pair of edges from each color class of size 4 and one from each color class of size 3. Thus IM11 = 2a + b, and since m is the size of a maximum matching, we get 2a + b = IMll :::=; m. Thus, X( G, .C3) n4g3b l +a+ b = n;ab l n;m l To construct a coloring, let M be a maximum matching in G. We form as many color classes of size 4 as possible by using the endpoints of two edges in M in each, then based on the parity of m, zero (if m is even) or one (if m is odd) color class of size 3 using the endpoints of one edge in M and an additional vertex, and finally as many color classes of size 2 as possible. With this construction, the 30
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graph induced by a color class of size 4 is either C4 or P4 Further, the graph induced by a color class of size 3 is either P3 or K2+K1 Each of these graphs is C3free and are valid color classes. If m is even, there are !J} color classes of size 4, and the rest are of size at most 2. With this coloring, we get X(G,C3 ) :=:; + !J]= ln2ml If m is odd and n > 2m, there are m;1 color classes of size 4 and one color class of size 3. The rest are of size at most 2. With this coloring, we get X(G, C3 ) :=:; I n4(r)3l + m;1 + 1 = 1n 2ml If m is odd and n = 2m, there are m;1 color classes of size 4 and one color class of size 2. With this coloring, X(G, C3 ) :=:; m;1 + 1 = mtl = I r; l = I n2m l D Next we will complete the study of graphs whose complements are acyclic by determining X(G, Cj) for j 4. Again, we need to determine how large any color class can be in a minimum Crcoloring of a graph. Recall that Theorem 2.10 guarantees cycles of all orders if a graph is missing at most j3 edges. The following theorem shows that we still have long cycles for some graphs missing up ton1 edges. Lemma 4.2 Let G be a graph of order n 5 with acyclic complement. Then Cn1 Furthermore, if K1,n2 'l:. G, then G is Hamiltonian. Proof. We may assume G = T where T is a tree and K1,n_2 T only if K1,n_2 G. (Otherwise, remove edges from G until we get a graph whose complement is a tree. Since n 5, this can be done without creating K1,n_2 in the complement.) Choose a root vertex v0 of T so that the height of the tree hv0 (T) is maximum. 31
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Since n 5, the diameter of the tree T is at least 2 and therefore, hvo (T) 2. Further, dr ( v0 ) = 1 and there is exactly one vertex VI at level 1. Let T E be the subgraph of G induced by the vertices on even levels ofT, and T0 be the subgraph of G induced by the vertices on odd levels ofT. Since TE and T0 are complete graphs, every induced subgraph of TE or T0 is Hamiltonian connected. Further, all vertices on level i are adjacent to all vertices on levels i 2, i 3, ... 1, 0 in G. In each of the following cases, we will show CnI G and either KI,n2 Cn Case 1. hv0 (T) 4. Let Vi be a vertex on the tree at level i for i = 2, 3, 4. Now, a Hamiltonian (v0,v4)path from TE v2 together with v4vi together with a Hamiltonian (vi,v3)path from To together with v3v0 forms CnI G. Further, a Hamiltonian (v0,v4)path from TE together with v4vi together with a Hamiltonian (vi,v3)path from To together with v3v0 forms Cn Case 2. hv0 (T) = 3. Let WI, w2 ... be the vertices on level 2 and XI, x2 ... the vertices on level 3. Case 2A. The number of vertices in level 2 is 1. Since n 5, there are at least two vertices XI and x2 on level 3. Then CnI is formed in G by using a Hamiltonian (xi,x2)path in T0 together with xiv0x2 In this case, we have KI,n2 T induced by the vertices on levels 1, 2, and 3. Case 2B. The number of vertices in level 2 is at least 2. Since hv0 (T) = 3, there is at least one vertex XI on level 3. Assume without loss of generality that WIXI E E(T). If level 3 has only one vertex, then a Hamiltonian (v0,w2)path from TE to gether with w2xivo forms CnI G. Further, KI,n2 T is induced by the 32
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vertices on levels 0, 1, and 2. Assume level 3 has at least two vertices XI, x2 (and WIXI E E(T)). If x2wi E E(T), then form CnI G by using a Hamiltonian (v0,w2)path in TE together with w2x2 together with a Hamiltonian (x2,xi)path in T0 VI together with XI v0 A Hamiltonian cycle is formed in G by using a Hamiltonian (v0,w2)path in TE together with w2x2 together with a Hamiltonian (x2,xi)path in T0 together with Otherwise, without loss of generality assume x2w2 E E(T), and we can form CnI G by using a Hamiltonian (vo,wi)path in TE together with WIX2 together with a Hamiltonian (x2,xi)path in T0 VI together with xiv0 A Hamiltonian cycle is formed in G by using a Hamiltonian (vo,wi)path in TE together with wix2 together with a Hamiltonian (x2,xi)path in T0 together with XI v0 Case 3. hv0(T) = 2. Then G = KI,nI and TE = KnI which contains CnI D We use the above result to get an upper bound on the size of a color class in a minimum Crcoloring of G. We complete our analysis of graphs whose complements are acyclic with the following theorem which determines X(G, .Cj) when j 4. Theorem 4.2 Let G be a graph of order n with acyclic complement. Let m 0 be the maximum number of pairwise vertex disjoint copies of KI,j2 in G. If j 4, then X(G, .Cj) =max (I l, I I l). Proof. Let A be a color class in a Crcoloring of G. Notice that (A) is acyclic since G is acyclic. If IAI j + 1, then by Lemma 4.2, we get cj (A), a contradiction. Hence IAI j, and it follows that X(G, .Cj) I I l 33
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Suppose IAI = j. We will show K 1 ,j_ 2 (A). Suppose j = 4. Every graph of order 4 with acyclic complement contains either C4 or K1 ,2 Since A is a color class, we must have K 1 2 (A). If j 5, then by Lemma 4.2, we have K1,j_ 2 (A). Let a be the number of color classes of size j. Since color classes are vertex disjoint, we get a:::; m. Therefore, X(G,Cj) +a= lj::::r1 and X(G,,Cj) IIl). To show the inequality in the other direction, we produce a minimum 'Cr coloring. Consider {U1 U 2 ... Um}, where each Ui is a K 1 ,j_ 2 in G and the U/s are pairwise vertex disjoint. LetS= V(G)V(Ui) Now, n = (j1)m + s. We consider two cases: m ::S: s and 0 ::S: s < m. If m :::; s, then n mj, which implies that I 1 I I 1 Fori = 1, ... m, let V(Ui) together with a vertex from S form a color class of size j (the subgraph induced by each such color class is Crfree since it contains a vertex of degree at most one). We partition the remaining sm vertices into as many color classes of size j 1 as possible. Clearly each such subgraph is Crfree. Thus, we have X(G,'Ci) ::S: m+ = = If 0:::; s < m, then n < mj, which implies that I I 1 I l Let r = I m;s 1 and U = {U1, U2, ... Umr }. Further, let T = S u V(Ui)). For i = 1, ... mr, let V(Ui) together with a vertex from T form a color class of size j. Note that m r = m I m.t 1 = l mi7+s J = li J, and there are sufficient elements in T to form li J color classes since ITI = s+r(j 1) = s+ I m;s 1 (j 1) s + m;s (j 1) = I. If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Crfree, we will show that lEI < j. 34
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Now, lEI ITIly j s+r(j1)(mr) rj(ms) ms < (.+1)j(ms) J J If B = 0, then there were no remaining vertices after forming mr color classes of size j. Therefore, j divides n and X ( G, Cj) l I J = I I l If B =!= 0, then j does not divide nand we get X(G, Cj) li J + 1 = I I l In either case, we have produced a Crcoloring with I I l colors. D Now that we have determined X(G, Cj) for graphs with acyclic complements, the next section gives X(G, Pj) for this same class of graphs. 4.2 Determining x(G, Pj) when G is acyclic The following lemma determines the longest path in a graph whose complement is acyclic. This lemma will be used in the main theorem of this section which determines X(G, Pj) when G is acyclic. Lemma 4.3 If G is a graph of order n 2: 2 with acyclic complement, then Pnl G. Furthermore, if G =/= K1,n_1 then G has a Hamiltonian path. Proof. If n = 2, the result is trivial. If n = 3, then G is missing at most two edges and P2 Further, if G =/= K1 2 then G = P3 or G = K3 and both contain 35
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where e is an edge. All graphs contain P4 except K4 E(K1 3), which contains P3 So, assume n 2: 5. By Lemma 4.2, the graph G contains Cn_1 as a subgraph and therefore Pn_1 G. If G #K1,n_1 then b(G) 2: 1 and the vertex in G not on Cn1 can be added to Cn1 to form Pn D The following theorem determines X(G, ,pj) for graphs whose complements are acyclic. Theorem 4.3 Let G be a graph of order n with acyclic complement. Let m 2: 0 be the maximum number of pairwise vertex disjoint copies of K1,j_1 in G. If j 2: 2, Proof. Let A be a color class in a ,Prcoloring of G. Notice that (A) is acyclic since G is acyclic. If IAI > j, then by Lemma 4.3, we get Pj (A), which contradicts the assumption that A is a color class. Therefore, IAI :::; j. Suppose IAI = j. Now, (A) is acyclic and is Prfree. Therefore, by Lemma 4.3, we get K1,j_1 = (A). Let a be the number of color classes of size j. Since color classes are vertex disjoint, we get a:::; m. Therefore, X(G, ,pj) 2: +a= r > r n_ml I J1 I J1 To show the inequality in the other direction, we produce a minimum ,pr coloring. Consider {U1 U2 ... Um}, where each Ui is a K1,j_1 in G and the U/s are pairwise vertex disjoint. For each i = 1, ... m, form a color class of size j using V(Ui) (the subgraph induced by each such color class is Prfree since it contains a vertex of degree zero). With the remaining vertices, form as many color classes of size j 1 as possible. With this coloring, we get X( G, ,pj) :::; Therefore, X(G, ,pj) = I j:=_7l 36 r +m = r n_ml I J1 I J1 D
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The following corollary will be used several times in upcoming chapters and is stated here for reference. Corollary 4.1 If G is a graph with (E(G)) = K1,m and j 2, then ifO m < j 1, if j 1 m n 1. Proof. If m < j1, then G contains no copy of K 1 ,j_ 1 and by Theorem 4.3, X(G,,Pj) = If m j 1, then G contains exactly one K1,j1 and by Theorem 4.3, X(G,,Pj) = l;::::il D Now, we can state the main theorem of this chapter. 4.3 The difference between the .Cjand .Pjchromatic numbers of a graph Theorem 4.4 If 0, j 4 and n a(j1)j, then there exists a connected graph G of order n such that X(G, ,pj)X(G, 'Cj) Proof. This is a proof by construction. Let G be the disjoint union of Ina(j1)2 and a(j 1) pairwise vertex disjoint copies of K 1 ,j_2 See Figure 4.1 for the construction of G when j = 5, a = 1 and n = 24. In this case, G consists of I8 and four copies of K 1 ,3 Further, X(G, ,C5 ) = 5 and X(G, ,P5 ) = 6. Observe that G contains a(j 1) pairwise vertex disjoint copies of K 1 ,j_2 and no copyofK1 ,j_ 1 ByTheorem4.3, X(G,,Pj) = Sincen a(j1)j, Theorem 4.2 implies that X(G, 'Cj) = Therefore, X(G, ,pj) X(G, 'Cj) I = I I +a= a. D Since we have shown these two conditional chromatic numbers may be different, we will address each separately in the following two chapters. 37
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Figure 4.1. The construction of G for j = 5, a= 1 and n = 24 Some possible directions for further research include answering the following questions. What happens if we remove the restriction that the complement is acyclic and examine all graphs missing n 1 edges or examine graphs whose complements are bipartite? Will the difference increase and, if so, by how much? Also, are there families of graphs where X( G, .Pj)X( G, .Cj) is O(nk) fork 1? The following questions also remain open. For the set of graphs of a fixed order n, how much can these two conditional chromatic numbers differ if we remove half of the edges from the complete graph? And, in general, given the size of G, what are upper and lower bounds for X(G, .Pj) X(G, .Cj)? Which graphs attain these bounds? 38
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5 Determining x(G, 'Cj) Next we address the problem: given a graph of order n withe edges, what are the bounds on the conditional chromatic number? In 1995, Dargen and Fraughnaugh published the following theorem. Theorem 5.1 (Dargen and Fraughnaugh, [21}) If e(G) (j2), then ifG = Kl,j2 or G = K3 and j = 5 otherwise. We can deduce from this theorem that we need to remove at least j2 edges from a complete graph to obtain a graph whose ,Crconditional chromatic number is I j=i l A natural question one may ask is how many edges do we need to remove so that X( G, ,Cj) = I ;=n? We will show that this number can be obtained by removing only two edges from a complete graph when j = 3. In this chapter, we will determine which graphs are Hamiltonian given that the size of their complement is at most n 1. Using this information, the final theorem of this chapter will expand the knowledge of X(G, 'Cj) to the class of graphs missing exactly j1 edges. 5.1 Preliminaries To prove the final theorem, we will first address some special cases. We will soon see that the special cases for the final theorem are graphs missing a star of a particular order and graphs missing pairwise vertex disjoint complete subgraphs. First, we will address graphs missing a star of a particular order. 39
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Theorem 5.2 Let G be a graph of order n. If j 3 and e( G) (2j 6) and K1,j_2 G, then X(G,Cj) = fj::::n. Proof. Let X be a set of vertices which induce K1,j_2 G. Since K1,j_2 G and e(G) 2j6, there is exactly one vertex v0 of degree at least j2 in G. Therefore, v0 EX. Let A be a color class in a Crcoloring of G. Suppose v0 If IAI j, then (A) would be missing at most j4 edges and by Theorem 2.10, we get Cj (A). Therefore, for A to be a color class, we must have IAI j1. Suppose v0 E A, and consider Av0 If lAv01 j, then by the same argument, we get Cj (Av0). Therefore, lAv01 j1, which implies IAI j. Thus, there can be at most one color class of size j (one containing v0), and X(G,Cj) fj::::fl + 1 = fj::::il To show the inequality in the other direction, form one color class B of size j by using the vertices in X together with one vertex from G X and form as many other color classes as possible of size j 1 with the remaining vertices in GX. Now (B) is Crfree since v0 is adjacent to at most one other vertex in (B). Therefore, X(G,Cj) fj::::fl +1 = fj::::il D Next, we would like to determine X(G, Cj) when G is missing a set of mutu ally disjoint complete graphs of a specific order. To accomplish this, we use the following two lemmas. Lemma 5.1 Let G be a graph of order n 3 such that G = Vf=1 Si where Si = Ir; (ri 1,p 1) for each i. Then either G is Hamiltonian or maxi{ri} I Proof. Let v E V(G). Then v E Si for some i and d(v) = (n1)(ri1) = nri Therefore, since each vertex in G is in some Si, we get 6(G) = nmaxi{ri}. If 40
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maxi{ri} < ln!ll then b(G) > nln!l1 = ln;1J. Thus, b(G) nl2 and by Dirac's Theorem (Theorem 2.7), G is Hamiltonian. D Lemma 5.2 Let G be a graph of order n 4 such that G = Vf=1 Si where Si = Ir; (1 :::; ri :::; 1 p 1) for each i. Then either G contains an ( n1) cycle or there exist r 1 and r 2 such that r1 = r2 = 1 Proof. Assume r 1:::; and ri < when i =j;l. Let v E S1 and G' = Gv. Then G' = Ir1I V (Vf=2 Si) and maxi2:2{ri, r1 1} < l By Lemma 5.1, G' is Hamiltonian, and therefore G contains an (n1)cycle. D The following theorem determines X(G, 'Cj) when G is missing a set of mutually disjoint complete graphs of a specific order. Theorem 5.3 Let G be a graph of order n. If j 3 and (E(G)) = mK lil, then X(G, ,Ci) = { I j=71 if j is odd max (I j=71 I y l) if j is even Proof. Observe that (E(G)) = mK lj!ll if and only if G = Vi=l 2 Si where si = Iril fori= 1,2, ... ,m and si = I1 fori= m+ 1,m+2, ... ,nIL m. Assume j is odd. Then I j!11 = j!1 Let A be a color class in a ,Crcoloring of G. Suppose that IAI > j + 1. Let B A such that lEI = j + 2. Then (B) = V7=1 Iq; where 1 :::; ql :::; (j + 1)12, 1 :::; q2 :::; (j + 1) I 2, and 1 :::; qi :::; (j 1) I 2 for 3 :::; i :::; k since I B I = j + 2 implies that (B) can contain at most two copies of By removing one vertex from each 2 ofiq1 andiq2 formC B such that ICI =j. Now, (C)= Iq1IVIq21V(V7=3Iq;), 41
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and maxi{qi} :::; j;1 By Lemma 5.1, (C) is Hamiltonian. This contradicts the assumption that (A) is Crfree. So IAI :::; j + 1. If IAI = j + 1, then by Lemma 5.2, either 2Jill (A) or (A) contains a 2 jcycle. Because (A) cannot contain Cj, we must have 2Jill (A). Now, if 2 IAI = j, then by Lemma 5.1, we have Jill (A). Let a be the number of color 2 classes of size j + 1 and b be the number of color classes of size j in a Cr coloring of G. Since color classes are vertex disjoint, we get 2a + b :::; m. Now, X(G .c ) > I n(j+l)ajbl +a+ b = I n2abl > I n.ml 1 I 11 I 11 I 11 To show the inequality in the other direction, we split the proof into three cases. Recall that G = Vi=l Si where Si = Jill for = 1, 2, ... m and 2 si = Jl fori= m + 1, m + 2, ... 'n( i) m. Case 1. m is even. Color G as follows: For 1 :::; l :::; m/2, form a color class of size j + 1 by using all the vertices from 821_1 and 821 The subgraph induced by each of these color classes is isomorphic to M = I ill V I ill, which is bipartite. 2 2 Since bipartite graphs only contain even cycles, M is Crfree. With the remaining vertices, form as many other color classes of size j 1 as possible. With this coloring, we get X(G,Cj):::; + = Case 2. m is odd and n m + j;t. Color G as follows: For 1 :::; l :::; m21 form a color class of size j + 1 by using all the vertices from 821_1 and 821 Again, since the subgraph induced by each of these color classes is bipartite, it is Crfree. Observe that n m (Jt1 ) +j;l guarantees that there are at least j vertices not yet assigned to color classes. Form one color class C of size j using all the vertices from Sm together vertices from V(Si) Now, (C)= Jill V Ki=.l and 2 2 the largest cycle we can form is cj1 obtained by alternately traversing between 42
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the vertices of K :L::::1_ and I i.! until we use all the vertices in K :L::::1_. Therefore, (C) 2 2 2 is Crfree. With the remaining vertices, form as many other color classes of size j1 as possible. Then X(G, Cj) :::; I + m;l + 1 = I j=_r;l Case 3. m is odd and n < m ( + j;l. This means that 0 :::; nm ( i) < j;l. We will show that nm is not divisible by j 1. Suppose that nm = z(j 1) for some nonnegative integer z. Then 0 < nm (j; 1 ) j1 <2 iff 0 < (j + 1) j1 z(j1) + mm 2 < 2 iff 0 < 2z(j 1) + 2m mj m < j 1 iff 0 < 2z(j1) m(j1) < j1 iff 0 < (2zm)(j1) < j1 iff 0 < 2zm < 1 iff m 2z. However, this contradicts the assumption that m is odd. Therefore, nm is not divisible by j 1. This implies that I = I j=71 Now back to the proof. Form a new graph F by taking G and adding back in all the missing edges in 81 Now F is missing m 1 pairwise vertex disjoint copies of K d b k' C 1 S X(F C)
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Let A be a color class in a Crcoloring of G. Suppose that IAI > j. Let B A such that lEI = j + 1. Then (B) = V7=1 Iq; where 1 q1 (j + 2)/2, and 1 qi j /2 for 2 i k since lEI = j + 1 implies that (B) can contain at most one copy of I ill By removing one vertex from Iqu form C B such that 2 ICI = j. Now, (C)= Iq1I V (V7=2Iq;), and maxi{qi} By Lemma 5.1, (C) is Hamiltonian. This contradicts the assumption that (A) is Crfree. So IAI j and X(G, .Cj) I I l If IAI = j, then since (A) is Crfree, by Lemma 5.1 we have Jill (A). Let a 2 be the number of color classes of size j in a Crcoloring of G. Since color classes are vertex disjoint, we get a m. Therefore, X( G, Cj) I 1 + m = I j:=.7l Hence, X(G, .Cj) (I I 1, I j:=.71) To show the inequality in the other direction, we produce a minimum Cr coloring. Let U = V(G) U:1 V(Si) Now, n = + s. We consider two cases: ym s and 0 s < i{!:m. If ym s, then n mj and lj:=.71 II1 For 1 i m, form a color class of size j by using all the vertices from Si together with y vertices from U not yet assigned to a color class. Since ym s, U contains a sufficient number of vertices to form these m color classes. Now, all the subgraphs induced by these color classes are isomorphic to C = I ill V K .i.::::.1. and the largest cycle we can form is 2 2 Cj_2 obtained by alternately traversing between the vertices of K .i.::::.1. and I ill until 2 2 we use all the vertices inK .i.::::.1.. Therefore, Cis Crfree. We partition the remaining 2 s j;2m vertices into as many color classes of size j 1 as possible. Thus, we h X(G C) < ls(i?)ml = lmj(i?)m+sml = I(J..)m+sml = 1nm1 ave , J m + 1 1 1 1 JJJJ44
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If 0 :::; s < ym, then n < mj and I I l I j=_7l Let r = I (2? )ms l and T = U u (U:mr+ 1 V ( Si)).. For 1 :::; i :::; mr, form a color class of size j by using all the vertices from Si together with y vertices from T not yet assigned to a color class. Again these subgraphs are Crfree since their longest cycle is Ci_2 Note that mr = = lmj(if)m+sj = l(ijm+sj = liJ, and there are sufficient elements in T to form li J color classes since ITI = s+r ( t.) = s+ (it2) s+ (it2) = (i;2) (i;2) liJ Again, all the subgraphs induced by these color classes are isomorphic to C = I ill V K .i.::::.1., 2 2 and we have already shown that Cis Crfree. If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Crfree, we will show that IBI < j. Now, lEI ITI ly j (j; 2 ) j + 2 (j2) s + r(2)(mr) 2 s + rjm (j; 2 ) < s+ ((S')jms +l)jme;2) J. If B = 0, then there were no remaining vertices after forming m r color classes of size j. Therefore, j divides n and X(G, Cj) :::; li J = I I l If B 10, then j does not divide nand we get X(G, Cj) :::; li J + 1 = I I l In either case, we have produced a Crcoloring with I I l colors. D 45
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5.2 Determining x(G, .Cj) when e(G) > (j1) To determine the ,Crchromatic number when j 8 for graphs given the size of the complement is at most j1, we first determine which graphs of order n 8 are Hamiltonian given the size of their complement is at most n 1. Theorem 5.4 Let G be a graph of order n 8 with e( G) :::; n 1. Then either Kl,n2 G or G is Hamiltonian. Proof. Assume K1,n_2 c:_ G. If G = Kn, then G is obviously Hamiltonian. So assume G is not complete and let u and v be a pair of nonadjacent vertices. Since K1,n_2 c:_ G, we must have d(u) 2 and d(v) 2. Let H = G{ u, v }. We break the proof into cases based on the number of edges in G incident to u or v. Case 1. Suppose that the number of edges in G incident to u or v is n1. Since e(G) = n1, His complete. If IN(u) n N(v)l = 0, then let x, y E N(u) and a, bE N(v) be distinct vertices. These vertices exist since d(u) 2 and d(v) 2. Now xuybva together with a Hamiltonian (a,x)path from H{y, b} forms CnSo assume IN(u) n N(v)l 1. If IN(u) n N(v)l = 1, then let w, x, and y be distinct vertices in H such that w E N(u) n N(v), x E N(u) andy E N(v). Then xuwvy together with an (y,x)path from H{ w} forms Cn. So assume IN ( u) n N ( v) I 2. Since there are n 2 edges missing from { u, v} into H out of a possible 2(n2), there must be n2 edges from { u, v} into H. Since n 8, there are at least six edges from { u, v} into H. Therefore, we can assume d( u) or d( v) is at least 3. Without loss of generality, assume d( u) 3. Let w, x, y E N(u), and w, x E N(v). Then xvwuy together with a Hamiltonian 46
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(y,x)path from H{ w} forms enTherefore, we can assume that every such pair of vertices can be incident to no more than n 2 edges in G. Case 2. Suppose that the number of edges in G incident to u or v is exactly n 2. Since there are n 3 missing edges from { u, v} into H, H is missing at most one edge, and there are 2(n2) (n3) = n1 edges from {u,v} into H. Suppose IN ( u) n N ( v) I = 0. Then in order to have n1 edges from { u, v} into Hand IN(u) n N(v)l = 0, the subgraph H needs to have at least n1 vertices. But H only contains n 2 vertices, and we reach a contradiction. Therefore, we can assume IN(u) nN(v)l 1. Let wE N(u) nN(v). Consider J = G{ u, v, w }. Then J is a graph of order n3 missing at most one edge. Since n 8, we have e(J) (n;3 ) 1 (n;3 ) (n7) and J is Hamiltonian connected by Theorem 2.9. If N(u) =/= N(v), then since d(u) 2, d(v) 2,and n 8, there exist distinct vertices x, y E V(J) such that x E N(u) andy E N(v). Now xuwvy together with a Hamiltonian (x,y)path in J forms enSo we can assume that N(u) = N(v). Recall that there are n1 edges from {u,v} into H. Therefore, d(u) = d(v) = n;1 The integer n must be odd in order for n2l to be an integer. Therefore, n 9 and d(u) = d(v) 4. This implies that there exist distinct vertices x, y E V(J) such that x, y E N(u) n N(v), x =/= w, and y =/= w. Then xuwvy together with a Hamiltonian (x,y)path in J forms en Thus, we may assume that for every nonadjacent pair of vertices u and v, the number of edges in G incident to u or v is at most n 3. 47
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Case 3. Suppose that the number of edges in G incident to every nonadjacent pair of of vertices u, v is at most n 3. There are at most n4 edges missing from { u, v} into H. Therefore there are at least 2(n2)(n4) = n edges from {u,v} into H. So, d(u) + d(v) 2: n and since u and v are arbitrary nonadjacent vertices, by Ore's Theorem (Theorem 2.6) G is Hamiltonian. D Now, armed with Theorems 4.1, 4.2, 5.2, 5.3, and 5.4, we can proceed with determining X(G, Cj) when e(G) 2: (j 1). We begin by determining X(G, Cj) when j 2: 8. Theorem 5.5 If j 2: 8 and e(G) 2: (j1), then { I j::::i l if Kl,j2 G X(G,Cj) = I otherwise. Proof. Suppose K1,j_2 G. We have j 1 2j 6 since j 2: 8. Therefore, e(G) = (j 1) 2: (2j6), and by Theorem 5.2, we have X(G, Cj) = lj=il Assume K1,j_2 C1 G. Let A be a color class in a Crcoloring of G. We will show that IAI < j, which implies that X( G, Cj) 2: I and equality follows from Corollary 2.2. Suppose that IAI 2: j. Let B A such that lEI = j. Now K1,j_2 C1 G implies that K1,j_2 C1 (B). By Theorem 5.4, (B) is Hamiltonian, a contradiction. D We wish to extend the above result for all j 2: 3. To accomplish this, we need to determine which graphs of order n are Hamiltonian given the size of their complement is at most n 1. 48
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Theorem 5.6 Let G be a graph of order n 3 with e( G) ::; n 1. Then G is Hamiltonian except in the following cases: K1,n2 G (i) K3 G and n = 5 ( ii) (E(G)) = C4 and n = 5 (iii) (E(G)) = K4 e where e is an edge and n = 6 (iv) (E(G)) = K4 and n = 7 (v) Proof. LetGbeagraphofordern 3withe(G) :=::; n1. Ife(G) then G is Hamiltonian by Theorem 2.10. So assume n2 ::; e(G) ::; n1. If J(G) :=::; 1 (or equivalently K1,n_2 G), then clearly G is not Hamiltonian. This is exceptional case (i). So assume J(G) 2. If n = 3, then J(G) 2 implies that G = C3 which is Hamiltonian. If n = 4, then J(G) 2 implies that G E {K4 K4 e, C4}, all of which are Hamiltonian. If n 8, then by Theorem 5.4, G is Hamiltonian So, assume 5 ::; n ::; 7. Now n2 :=::; e(G) :=::; n1 and 5 :=::; n :=::; 7 imply G is not complete. So let u and v be a pair of nonadjacent vertices in G. Let H = G{ u, v }. Next, we break the proof into cases based on the number of edges in G incident to u or v. Case 1. Suppose that the number of edges in G incident to u or v is n1. Since the total number of edges in G is at most n1, His complete. If n = 5, then having four edges incident to u or v in G forces J(G) :=::; 1, which contradicts the assumption that J (G) 2. Ifn = 6, then there are four edges from {u,v} to H in G. Moreover, J(G) 2 implies that :=::; 3 and Therefore G is, up to isomorphism, one of the graphs in Figure 5.1. If G is graph (a) in Figure 5.1, then uwyvzxu forms a Hamiltonian cycle. If G is graph (b) in Figure 5.1, then uwyzvxu forms a Hamiltonian cycle. If G is graph (c) in Figure 5.1, then clearly G is not Hamiltonian since N(u) = 49
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y X (a) y X (b) y X (c) Figure 5.1. The nonisomorphic possibilities for G when the number of edges incident to u or v is 5 and n = 6. N(v) = {w,x} and the largest cycle containing vertices u and vis C4 Further, (E(G)) = K4 e, where e is an edge, and we get exceptional case (iv). y (a) y (b) y (c) w X Figure 5.2. The nonisomorphic possibilities for G when the number of edges incident to u or v is 6 and n = 7. If n = 7, then there are five edges from { u, v} to H in G. Moreover, 5( G) 2 implies that ::; 4. Therefore G is, up to isomorphism, one of the graphs in Figure 5.2. If G is graph (a) or (b) in Figure 5.2, then uwavzyxu forms a Hamiltonian cycle. If G is graph (c) in Figure 5.2, then uwvazyxu forms a Hamiltonian cycle. This addresses all the graphs where the number of missing edges incident to { u, v} is n 1. Case 2. Suppose that the number of edges in G incident to u or v is exactly n2. Since the total number of edges in G is at most n1, H = 50
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G{ u, v} is missing at most one edge. If n = 5, then there are two edges from { u, v} to H in G. Moreover, 6 (G) 2 implies :S 2. Therefore G is, up to isomorphism, one of the graphs in Figure 5.3. The two nonisomorphic possibilities for G when G has three edges incident u v u v u v y0w X X X (a) (b) (c) u v u v u v y? w yz w X X X (d) (e) (f) Figure 5.3. The nonisomorphic possibilities for G when the number of edges incident to u or v is 3 and n = 5. to u or v and e(fl) = 0 are depicted in the first column in Figure 5.3. As stated previously, the subgraph H can be missing at most one edge. The graphs in the second and third column depict all possible nonisomorphic placements of that additional edge. If G is graph (a) or (c) in Figure 5.3, then uxvywu forms a Hamiltonian cycle. If G is graph (b) in Figure 5.3, then clearly G is not Hamiltonian since x is a cut vertex in G. Further, (E(G)) = C4 and we get exceptional case (iii). If G is graph (d), (e), or (f) in Figure 5.3, then G is not Hamiltonian since N(u)UN(v) = { w, x }. Further, K3 G and we get exceptional case (ii). If n = 6, then there are three edges from { u, v} to H in G. Moreover, 6 (G) 2 implies :S 3. Therefore G is, up to isomorphism, one of the graphs in Figure 51
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5.4. The two nonisomorphic possibilities for G when G has four edges incident u v u v zLF,w z
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u v u v u v u v u v u v W W W afVw "117 X Z X Z X X Z X Z y y y y y y (a) (b) (c) (d) (e) (f) u v u v u v u v u v af:w af:w afJw Z X Z X Z X Z X Z X y y y y y (g) (h) (i) (j) (k) u v u v u v u v W W X Z X X y y y y (l) (m) (n) (o) u v u v u v u v u v u v af?':w Z X af6:w Z X Z X Z X Z X aj?":w Z /X y y y y y y (p) (q) (r) (s) (t) (u) u v u v u v u v af\\w Z X aAw Z X a0w Z X aAw Z X y y y y (v) (w) (x) (y) Figure 5.5. The nonisomorphic possibilities for G when the number of edges incident to u or v is 5 and n = 7. 53
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The five nonisomorphic possibilities for G when G has 5 edges incident to u or v are depicted in the first column in Figure 5.5. Since the total number of edges in G is at most 6 and the number of edges in fi incident to u or v is exactly 5, H is missing at most one edge. The graphs in the second and subsequent columns depict all possible nonisomorphic placements of that additional edge. If G is graph (n) in Figure 5.5, then clearly G = J4 V K3 is not Hamiltonian since the largest cycle we can form is c6 obtained by alternately traversing between the vertices of K3 and J4 until we use all the vertices in K3 Further, (E(G)) = K4 and we get exceptional case ( v). If G is graph (a), (b), (c), (d), or (j) in Figure 5.5, then uwyzvaxu forms a Hamiltonian cycle. If G is graph (e) or (f) in Figure 5.5, then uwvazyxu forms a Hamiltonian cycle. If G is graph (g), (h), (i), (k), (p), (q), (r), or (t) in Figure 5.5, then uxvazywu forms a Hamiltonian cycle. If G is graph (l), (m), or (o) in Figure 5.5, then uwvyzaxu forms a Hamiltonian cycle. If G is graph (s) or (u) in Figure 5.5, then uwzyavxu forms a Hamiltonian cycle. If G is graph (v), (w), (x), or (y) in Figure 5.5, then uwzxavyu forms a Hamiltonian cycle. This addresses all the graphs where the number of missing edges incident to u or v is at least n2. Therefore, we can assume for all nonadjacent vertices u and v in G that the number of edges in G incident to u or v is at most n 3. Case 3. Suppose that the number of edges in G incident to every nonadjacent pair of of vertices u, v is at most n 3. There are at most n4 edges missing from { u, v} into H. Therefore, there are at least 2(n2)(n4) = n edges from { u, v} into H. So, da(u) + da(v) 2: n and by Ore's Theorem (Theorem 2.6), G is Hamiltonian. D 54
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Now we can proceed with determining X(G, 'Cj) when e(G) (j1) for all j 3. Theorem 5.7 If e(G) (j1), then r n21 111 r n11 111 if (E(G)) = 2K2 and j = 3, if (E(G)) = P3 and j = 3, Kl,j2 G and j 4, K3 G and j = 5, (E(G)) = C4 and j = 5, (E(G)) = K4 e where e is an edge and j = 6, or (E(G)) = K4 and j = 7, and I otherwise. Proof. If e(G) > (j1), then the result follows from Theorem 5.1. So assume e(G) = (j1). Further, the result follows when j 8 by Theorem 5.5. Therefore, we only need to determine X(G, 'Cj) when j = 3, 4, 5, 6 or 7. If j = 3, then (E(G)) E {2K2, P3}. If (E(G)) = 2K2, then x(G, ,c3) = I j::::n by Theorem 4.1. If (E(G)) = P3 then X(G, ,C3 ) = I j::::il by Theorem 4.1. If j = 4, then (E(G)) E {3K2 P3 + K 2 K1 3 P4 K3}. If (E(G)) = 3K2 then P3 C. G and, by Theorem 4.2, we get X( G, ,C4) = max (I I I l) = I (E(G)) E {P3 + K 2 K1 3 P4 }, then by Theorem 4.2, X(G, 'C4) = I j::::i 1 if n j, and X(G, ,C4 ) = I I 1 if n < j since I j::::il I I 1 if and only if n j. But, when n < j, we know that I I 1 = 1 = I j::::i l Therefore, X( G, 'C4) = I j::::i l If (E(G)) = K3 then by Theorem 5.3, X(G, ,C4 ) = I j::::i 1 ifn j, and X(G, ,C4 ) = I I 1 if n < j since I j::::i 1 I I 1 if and only if n j. But, when n < j, we know that II1 = 1 = lj::::il Therefore, X(G,,C4 ) = lj::::il Assume 5 :::; j :::; 7. Let A be a color class in a 'Crcoloring of G. Suppose that IAI j + 1. Let B A with lEI = j + 1. If e((B)) (j2), 55
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then we get Cj (B) by Theorem 2.10 and reach a contradiction. Therefore, e( (B)) (1!1 ) (j 1). But since G is missing j 1 edges, we must have e((B)) = (1!1)(j1). If fl((B)) 1, then e((B)) < (j + 1)/2. Now, j1 = e((B)) (j + 1)/2, which implies j 3. But this contradicts the assumption that j 5. Therefore, fl((B)) 2, which implies J((B)) j2. Choose v E V((B)) so that d(B)(v) j2. Consider (B)v. Now, e((B)v) = e((B))d(v) (1!1)(j1)(j2) = G) (j3) and by Theorem 2.10, we have Cj (B) and reach a contradiction. Therefore, IAI j. Let a be the number of color classes of size j in a ,Crcoloring of G and A be a color class of size j. By Theorem 2.10, (A) must be missing at least j2 edges. Since G is missing at most j1 edges, we must have a(j2) j1, or simply a 1. Therefore, all ,Crcolorings of G can contain at most one color class of size j, which implies that X(G,,Cj) 1;::::11 + 1 = fj::::il By Theorem 5.6, we know which graphs missing at most j1 edges are nonHamiltonian. If G contains one of these nonHamiltonian graphs as an induced subgraph, then we can color G so that there is a color class of size j and get D The previous theorem produces an upper bound for determining how many edges need to removed from a complete graph to obtain a graph with X(G, 'Cj) = I j::::n. Determining X(G, 'Cj) when G is missing j edges can be accomplished, we believe, using the above methods. However, extrapolating on the complexity of the proofs for removing j1 edges versus the complexity of the proof for removing only j 2 edges leads us to believe that the proof for removing j edges will 56
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be extremely long and complicated. We have already determined the minimum number of edges to be removed to obtain a conditional chromatic number of I j.::::n in Theorem 5. 7. Removing j edges from a complete graph will probably not allow the conditional chromatic number to decrease again. Thus, there is little to be gained by proving the above type of result for graphs missing j edges. In fact, further research in the area of determining X(G, Cj) should focus on determining general upper and lower bounds on the number of edges for a graph to attain a particular conditional chromatic number. 57
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6 Determining x(G, 'Pj) We discovered in Theorem 5. 7 that it is necessary to remove j 2 edges from a complete graph to decrease the ,Crchromatic number from I to I j=i l and that it is necessary to remove j 1 edges to attain a value of I ;=il for the ,Crchromatic number when j = 3. We would like to determine the minimum number of edges to remove from a complete graph to attain a value of I j=i l for the ,Prchromatic number and determine the minimum number of edges to remove from a complete graph to attain a value of I ;=il for the ,Prchromatic number. While attempting to find these numbers, we will obtain information about the ,Prchromatic number for a large set of graphs. We begin by determining which graphs of large size have a Hamiltonian path, then determine the minimum numher of edges which need to be removed from a complete graph for X(G, ,pj) to decrease from I to I j=i l, and then determine the chromatic number for all graphs missing 2j5 or fewer edges. Finally, we will determine a lower bound on X(G, ,pj) in terms of the size of G. 6.1 Determining which graphs of large size have a Hamiltonian path To determine the ,Prchromatic number for graphs whose complements have size at most 2j 5, we need to determine which graphs of order n missing at most 2n5 edges have a Hamiltonian path. The following is an easy and useful result. 58
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Lemma 6.1 Let G be a graph of order n and S V (G). If G has a Hamiltonian path, then the number of components in GS is at most lSI+ 1. Proof. Let S V(G) with 0 ::; lSI ::; n, and let H be a Hamiltonian path in G. If we remove S from G, then H breaks apart into at most lSI+ 1 subpaths. Therefore, GS contains at most lSI+ 1 components. D The following sets will be used repeatedly in this section. Let :F = {K2 V (K2 + I3), K1 V (Kn3 + I2), Ir V Kr2, where r 4 and n 4}, g = {GIG is a spanning subgraph of a graph in :F}, and 1l = {K2 V (K2 + I3), K1 V (Kn3 +h), I4 V K2, (I4 V K2) e, Is V K3, (Is V K 3 ) e, h V K4, where e is an edge and n 4}. Lemma 6.2 If a graph G E Q, then G does not have a Hamiltonian path. Proof. Let G = F V (K2 +h) where F = K2 LetS= V(F). Then the number of components in GSis 4 and 4 > 3 = lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = F V (Kn_3 +h) where F = K1 LetS= V(F). Then the number of components in GSis 3 and 3 > 2 = lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = Ir V F where F = Kr_ 2 and r 4. Let S = V(F). Then the number of components in GSis rand r > (r2) + 1 = lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = F, where F is a spanning subgraph of a graph in :F. Since each graph in :F does not contain a Hamiltonian path, F cannot contain a Hamiltonian path. D 59
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Now Q contains more graphs than just those graphs of order n missing at most 2n5 edges. Since we are interested in determining which graphs missing at most 2n5 edges have no Hamiltonian path, we will determine the largest subset of graphs of order n in Q that are missing at most 2n5 edges. Lemma 6.3 Let G be a graph of order n 1 with e(G) :::; 2n5. Then G E Q if and only if G E 1i. Proof. Let G be a graph of order n 1 with e(G) :::; 2n5. Assume that G E Q. If G is a spanning subgraph of K2 V (K2 + 13), then since K2 V (K2 + 13 ) is missing 9 = 2(7)5 edges, we must have that G = K2 V (K2 +h) E 1i. If G is a spanning subgraph of K1 V (Kn_3 + 12), then since K1 V (Kn_3 + 12 ) is missing 2n5 edges, we must have that G = K1 V (Kn_3 + 12 ) E 1i. So assume that G is a spanning subgraph of 1r V Kr2, where r 4. Now (2n5) :::=; e(G) :::=; e(1r V Kr2) = (;), or (;) :::; 2n5 = 2(2r2) 5. Simplifying, we get r2 9r + 18 :::; 0 or (r6)(r3) :::; 0. This implies that 3 :::; r :::; 6. We have already assumed that r 4. If G is a spanning subgraph of 14 V K2 then since 14 V K2 is missing 6 = 2(6)6 edges, G can be missing up to one more edge. Thus, G = 14 V K2 or G = (14 V K2 ) e where e is an edge. Both of these graphs are in 1i. If G is a spanning subgraph of h V K3 then since h V K3 is missing 10 = 2(8) 6 edges, G can be missing up to one more edge. Thus, G = h V K3 or G = (h V K3 ) e where e is an edge. Both of these graphs are in 1i. If G is a spanning subgraph of h V K4 then since h V K4 is missing 15 = 2(10)5 edges, we must have that G = h V K4 E 1i. Therefore, if Q 1i. Clearly, 1i Q. D By Lemma 6.2, 1i consists of graphs of order n missing at most 2n5 edges with no Hamiltonian path. The next step is to prove that a connected graph 60
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missing at most 2n5 edges is either in 1t or has a Hamiltonian path. To prove this, we need the following theorem which appears in Chartrand and Lesniak [17]. Theorem 6.1 Let G be a connected graph of order 3 or more that is not Hamil tonian. If for all distinct nonadjacent vertices u and v, d( u) + d( v) m, where m is a positive integer, then P m+l G. The following theorem completely characterizes which graphs of order n miss ing at most 2n5 edges have a Hamiltonian path. Theorem 6.2 If G is a connected graph of order n 1 and e( G) :::; 2n5, then either G E 1t or G has a Hamiltonian path. Proof. Let G be a connected graph of order n with e( G) :::; 2n 5. If n :::; 2, then the theorem is vacuously true. If n = 3, then G E {P3 K3}, both of which have a Hamiltonian path. If n = 4, then since e( G) :::; 3 and G is connected, G E {KI,3, P4, K1V(K1+K2), C4, K4e, K4}. lfG = K1,3, then G = K1V(K1+I2) E 1t. Otherwise, G has a Hamiltonian path. So assume n 5. If G = Kn, then obviously G has a Hamiltonian path. So assume G is not complete. Let u and v be a pair of nonadjacent vertices in G and H = G{ u, v }. Since G is connected, each of u and v has a neighbor. If N ( u) = N ( v) = { w}, then the number of edges missing between u and H is n3 and the number of edges missing between v and His n3. Since we assumed that uv tj. E(G), we have a total of 2n5 missing edges incident to u or v. This implies that G = K1 V (Kn_3 +h) E 1t. Therefore, we can assume that there exist a, b E V(H) such that a E N(u), b E N(v) and a =I b. We break the proof into cases based on the number of edges in G incident to u or v. 61
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Case 1. Suppose that the number of edges in G incident to u or v is n + 1 or more. Then His a graph on n2 vertices missing at most (2n5)(n+1) = (n2)4 edges. By Theorem 2.9, His Hamiltonian connected, and hence there is a Hamiltonian (a,b)path in H. So, ua together with the Hamiltonian (a,b)path in H together with bv is a Hamiltonian path in G. Therefore, we can assume that every pair of nonadjacent vertices is incident to no more than n edges in G. Case 2. Suppose the number of edges in G incident to u or v is exactly n. Then H is missing at most (2n5) n = (n2) 3 edges. By Theorem 2.9, the subgraph H has a Hamiltonian cycle C. Let C = u1u2 ... un_2u1 If u (or v) is adjacent to consecutive vertices on C, then we could construct a ( n1 ) cycle using a detour through u (or v) and could then attach v (or u) to this cycle and produce a Hamiltonian path in G. So assume neither u nor v is adjacent to consecutive vertices on C. Thus, d(u) :::; n22 and d(v) :::; n;2 ; for otherwise, by the pigeonhole principle, u (or v) would be adjacent to consecutive vertices on C. Next, we determine all possible values for the degrees of u and v in G. There are 2(n2) possible edges from { u, v} to H and since n1 of these edges are missing, there are exactly n 3 edges from { u, v} to H. Now d( u) + d( v) = n 3 and the fact that each of u and v has degree at most n22 implies that either n is even, d(u) = n24 and d(v) = n22 (Case 2A); or n is odd and d(u) = d(v) = n;3 (Case 2B). Case 2A. Assume n is even, neither u nor v is adjacent to a pair of consecutive vertices on C and d( u) = n24 and d( v) = n22 For v to be adjacent to half of the vertices m H and not have two of its neighbors adjacent on C, the vertex v must be adjacent to alternating vertices 62
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on C, say the vertices on C with odd subscripts. Since d(u) = n24 we can say without loss of generality that the vertex u is not adjacent to u1 Now we break this case into subcases based on whether or not N(u) N(v). Case 2A 1. IN ( u) n N ( v) I = n24 First we will show that if there is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If u2u4 E E(G), then we can construct a Hamiltonian path in Gas folthen we can construct a Hamiltonian path in Gas follows: uu3 u 4 ... Un_2 u 2 u 1v. If u2u2 k E E(G) for some 2 < k < (n2)/2, then we can construct a Hamiltonian path in G as follows (see Figure 6.1): u4u5 ... u 2ku2 u 1vu3uu2k+l ... Un_2 If Un_4un_2 E E(G), then we can construct a Hamiltonian path in G as follows: Figure 6.1. The graph G when n is even and u2u2k E E(G) for some 2 < k < n22. 63
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we can construct a Hamiltonian path in G as follows: If u 2ku2k+2 E E( G) for some 1 < k < then we can construct a Hamiltonian E(G) where 1 < k < l < and l > k + 1, then we can construct a Hamiltonian path in G as follows (see Figure 6.2): Figure 6.2. The graph G when n is even and u2ku21 E E(G), 1 < k < l ::; n22 and l > k + 1. Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus { u2 u4 ... Un_2 u, v} is an independent set of size nt2 Therefore, G is a spanning subgraph of I n+2 V K n2 = Ir V Kr_2 where r = 2 2 (n + 2)/2. Now n 2: 5 and even implies that r 2: 4. Therefore, G E Q and by Lemma 6.3, G E 1l. 64
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Case 2A2. IN(u) n N(v)l < n;4 Since vis adjacent to all vertices in H with odd subscripts, the vertex u cannot be adjacent to all vertices in H with odd subscripts excluding u1 or else IN(u) n N(v)l = n;4 Therefore, u must be ad jacent to u2k for some 1 :::; k :::; n22 If UUn2 E E(G), then UUn2Un3 ... u1v is a Hamiltonian path in G. If uu2 k E E(G) for some 1 :::; k < n22 then uu2ku2k1 ... ulun2Un3 ... u2k+lv is a Hamiltonian path in G. Case 2B. Assume n is odd, neither u nor v is adjacent to a pair of consecutive vertices on C, and d( u) = d( v) = n;3 Recall that n 5 and there exist a, b E V(H) such that a E N(u), bE N(v) and a =I b. If n = 5, then H = K3 and G is K3 with two pendant vertices, each joined to a different vertex in K3 This graph contains a Hamiltonian path. So, assume Figure 6.3. The graph G when n is odd and u2u2i E E(G) for some 2 :S i :S n;3 and n > 7. 65
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Since d( u) = n;3 and there are n2 vertices on C, the vertex u is not adjacent to two consecutive vertices on C, say uu1 tJ_ E(G) and uu2 tJ_ E(G). For u to have degree n23 and not have two adjacent neighbors on C, the vertex u must be adjacent to the remaining vertices on C with odd (or even) subscripts, say odd. Thus, we can assume that N(u) = { u 3 u 5 ... Un_2}. Next we break this case into subcases based on whether or not N(u) = N(v). Case 2Bl. N(u) = N(v). First we will show that if there is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If u2u2i E E(G) for some 2 i (n3)/2, then we can construct a if n > 7 (see Figure 6.3). If u4u21 E E(G) for some 2 < l n;3 then we can construct a Hamiltonian path in G as follows: u1 U2U3VU5 ... U2t1 uun2 ... u2l+l u21U4. If u2ku21 E E( G) for some 2 < k < l n;3 then we can construct a Hamiltonian path in G as follows (see Figure 6.4): Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus, e(fl) Recall that e(fl) n 5. Therefore, n 5 e(fl) Ct), which implies that n 11. Recall that the number of missing edges incident to u or v is n. If n = 7, then N(u) = N(v) = { u 3 u5}. We have assumed that u 2 u 4 tJ_ E(G). If u1u 4 E E(G), then uu5vu3 u 4 u1u 2 is a Hamiltonian path in G. If u1u 4 tJ_ E(G), 66
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Figure 6.4. The graph G when n is odd and u 2ku2 1 E E(G) for some 2 < k < l :S n;3 then we have accounted for all nine missing edges in G. Therefore, G = (u3 u5 ) V ( (u1, u2) + u + v + u4) = K2 V (K2 + !3) E 1l. If n = 9, then N(u) = N(v) = {u3,u5,u7}. We have already assumed that {u2u 4,u2u6,u4 u6 } E(G). This leaves at most one additional edge in fl. If u1u 4 E E(G), then u6u 7uu5vu3u 4 u1u2 is a Hamiltonian path in G. If u1u 4 tj. E(G), then we have accounted for all thirteen missing edges in G. Thus, u6u1 E E(G) If n = 11, then N(u) = N(v) = {u3,u5,u7,u9}. Since we have assumed that there are no edges between any pair of vertices in H with even subscripts, the graph induced by { u2 u4 u6 u8 } is a copy of !4 and we have accounted for all Hamiltonian path in G. 67
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Case 2B2. N(u) #N(v). Recall that N(u) = { u3 u5 ... Un_2}. Now either vu1 E E(G) or vu2 E E(G), otherwise, by the pigeonhole principle, v would be adjacent to consecutive vertices on C. If vu1 E E(G), then vu1u2 ... Un_2u is a Hamiltonian path in G. If vu2 E E(G), then vu2u1un_2 ... u4u3u is a Hamiltonian path in G. Therefore, we can assume that every pair of nonadjacent vertices is incident to no more than n 1 edges in G. Case 3. Suppose that the number of edges m G incident to u or v is exactly n1. There are 2(n 2) possible edges from { u, v} to H, and since n2 of these edges are missing, we get d(u) + d(v) = n2. The graph H is missing at most (2n 5) (n 1) = n 4 edges. Therefore, e(H) (n4) = edges. By Theorem 2.8, either His Hamiltonian, H = K1 V (K1 + Kn4) for n 5, or H = K2 V !3. Case 3A. Assume H is Hamiltonian. Again, let C = u1 u2 ... Un_2u1 be a Hamil tonian cycle in H. If u (or v) is adjacent to consecutive vertices on C, then we could construct a ( n 1 )cycle using a detour through u (or v) and could then attach v (or u) to this cycle and produce a Hamiltonian path in G. So assume neither u nor vis adjacent to consecutive vertices on C. Now n must be even, for otherwise either u or v would be adjacent to at least n2l vertices in H, say u, and by the pigeonhole principle, u would be adjacent to consecutive vertices on C. So assume that n is even, neither u nor v is adjacent to a pair of consecutive vertices on C, and d(u) = d(v) = n22 Since u is adjacent to exactly half of the vertices on C and u is not adjacent to a pair of consecutive vertices on C, the vertex u is adjacent to either all the vertices on C with odd subscripts or all the vertices on C with even subscripts. The same is true for v. By symmetry, there are two 68
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(Case 3Al); and N(u) = N(v) = { u1 u 3 us, ... Un3 } (Case 3A2). case, vu1 u 2 ... Un_2 u is a Hamiltonian path in G. Case 3A2. N(u) = N(v) = { u1 u 3 us, ... Un_3 }. First we will show that if there is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If Un4Un2 E E(G), then Un4Un2Un3uul ... UnsV is a Hamiltonian path in G. If u 2 kun_ 2 E E(G) for some 1 :::; k < (n4)/2, then is a Hamiltonian path in G. If u 2 ku 2k+2 E E(G) for some 1 :::; k < (n4)/2, u2ku2z E E(G) for some 1:::; k < l < n;2 and l > k + 1, then is a Hamiltonian path in G (see Figure 6.5). Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus { u2 u 4 ... un_ 2 u, v} is an independent set of size n!2 Therefore, G is a spanning subgraph of I n+2 V 2 K n2 = Ir V Kr_2 where r = (n + 2)/2. Now n 5 and even implies that r 4. 2 By Lemma 6.3, G E 1. Case 3B. Assume that H = K1 V (K1 + Kn_4 ) and n 5. Let w be the vertex of degree 1 in H, let x be the neighbor of w in H, and let U = { u1 u2 ... Un_4 } be the set of remaining vertices in H. Since G is connected, there are two cases to consider: d(u) 2 and d(v) 2 (Case 3Bl); and d(u) = 1 or d(v) = 1, without loss of generality, say d(u) = 1 (Case 3B2). 69
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Figure 6.5. The graph G when n is even and u2ku21 E E(G) for some 1 :S k < l < n22 and l > k + 1. Case 3Bl. Assume that d(u) 2 and d(v) 2. Suppose that wE N(u). If there exists ui E U such that uiv E E(G), without loss of generality, say Un_4v E E(G), then uwxu1 ... un_4v is a Hamiltonian path in G. If vis not adjacent to any vertex in U, then N(v) = {w,x} and uwvxu1 ... un_4 forms a Hamiltonian path in G. So we can assume that w tj. N(u). By symmetry, we can assume that w tj. N(v). Suppose that x E N(u). If u and v were each adjacent to only one ui E U, say u1 then since d(v) 2, we get N(u) = N(v) = { u1 x }. Since d(u) = 2, d(v) = 2, and d(u) + d(v) = n2, we must have n = 6. Thus, G = ((u + v + w + u2 ) V {u1,x})u1w = (!4 V K2)e E 1i. So we can assume that there exist distinct vertices ui and uk such that uk E N ( u) and ui E N ( v). Without loss of generality, say u1 E N(u) and Un_4 E N(v). Now wxuu1 Un_4v is a Hamiltonian path in G. So we can assume that x tj. N(u). Further, by symmetry, we can assume that x tj. N(v). Therefore, the only remaining case to consider is when N(u) U and 70
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N(v) U. Suppose that N(u) U and N(v) U. Since d(u) 1, the vertex u is adjacent to a vertex in U, say u1 Since d(v) 2, vis adjacent to a vertex ui in U where 1 < i n2. Without loss of generality, say uu2 E E(G). If vu1 E E(G), then uu1vu2 u3 ... Un_4xw forms a Hamiltonian path in G. If vu1 E(G), then v must be adjacent to another vertex in U, without loss of generality, say u3 Now uu1 u 2vu3 ... Un4Xw forms a Hamiltonian path in G. Case 3B2. Assume that d(u) = 1. Since d(u)+d(v) = n2, we get d(v) = n3, which implies that v is adjacent to all but one of the vertices in H. Therefore, either {x, w} N(v), x N(v), or w N(v). Suppose that N(u) = {w}. If {x,w} N(v), then uwvxu1 ... un_ 4 forms a Hamiltonian path in G. If x N(v) or w N(v), then v is adjacent to a vertex in U, say Un_4 Now uwxu1 ... Un_4 v forms a Hamiltonian path in G. Suppose that N ( u) = { x}. If v has no neighbor in U, then since v is adjacent to all but one vertex in H, N(v) = {x,w} and U = {u1}. Thus, G = {x} V ((v,w) + u + u 1 ) = K 1 V (K2 + !2 ) E 1t. So assume that v has a neighbor in U, say u 1 If wE N(v), then uxwvu1 ... Un4 forms a Hamiltonian path in G. If w N(v), then x E N(v) and G = {x }V( (u1, u2, ... Un4, v)+u+w) = K1 V(Kn3+12) E 1t. The last case to consider is when N(u) U. Without loss of generality, suppose that N ( u) = { u1}. If w E N ( v), then uu1 ... Un_4xwv forms a Hamiltonian path in G. Ifw N(v), then N(v) = {ul,u2 ... ,un4,x} and uu1 ... un4VWX forms a Hamiltonian path in G. Case 3C. Suppose that H = K2 V !3 Let U = { u1 u2 } be the set of vertices of degree 4 in H and V = { u3 u4 u5 } be the set of vertices of degree 2 in H. In this 71
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case there are five edges between { u, v} and the vertices of H. Without loss of generality, either d(u) = 1 and d(v) = 4, or d(u) = 2 and d(v) = 3. Case 3Cl. Suppose that d(u) = 1 and d(v) = 4. Then by the symmetry of H, we can assume that either vus tj_ E(G) or vu1 tj_ E(G). If vus E(G), then u1usu2u 3vu4u1 forms C6 and we can attach u to this cycle to form P7 G. If vu1 E(G), then u1u4vu3 u2usu1 forms C6 and we can attach u to this cycle to form P7 Case 3C2. Suppose that d( u) = 2 and d( v) = 3. Then either v is adjacent to at least two of the vertices in V or v is adjacent to at exactly one vertex in V. If v is adjacent to at least two of the vertices in V, say u 4 and us, then UIUsVU4U2U3Ul forms c6 and we can addu to form p7 If vis adjacent to exactly one vertex in V, say us, then N(v) = {u1,u2,us}. Now, if u is adjacent to at least one of u 3 or u 4 say u 3 (a similar construction works for u 4 ) then uu3 u1u4u2usv is a Hamiltonian path in G. On the other hand, if u is not adjacent to at least one of u3 or u 4 then without loss of generality, either N(u) = {u1,u2 } or N(u) = {u1,us}. If N(u) = {u1,u2 }, then G = (u1, u2) V ( (v, us)+ u + u3 + u4) = K2 V (K2 +h) E 1i. If N(u) = { u1, us}, then uusvu1u4u2u 3 is a Hamiltonian path in G. Case 4. Suppose that the number of edges in G incident to u or v is at most n2. If G is Hamiltonian, then obviously G has a Hamiltonian path. So assume G is not Hamiltonian. There are at most n 3 edges missing from { u, v} into H. Therefore, there are at least 2(n2)(n3) = n1 edges from { u, v} into H. So, d(u) + d(v) 2: n1. By Theorem 6.1, Pn D 72
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The next theorem builds on the previous theorem by stating which graphs of order n missing at most 2n5 edges have a Hamiltonian path. Theorem 6.3 Let G be a graph of order n 1 with e(G) 2n5. Then G has a Hamiltonian path if and only if 6(G) > 0 and G tj.1t. Proof. Let G be a graph of order n 1 with e(G) 2n5. Assume that 6(G) > 0 and G tJ. 1t. To apply Theorem 6.2, we must show that G is connected. Suppose that G is not connected. Let A be a component of G with IAI = a and B = GA. Since 6(G) > 0, we have 2 n2. Now, e(G) a(na)= ana2 The minimum for this quadratic function occurs when a = 2 or a = n 2 and we get e( G) an a2 2n 4. Thus, the size of G must be at least 2n 4, which contradicts the assumption that e( G) 2n 5. Therefore, G is connected. By Theorem 6.2, G has a Hamiltonian path. Assume that G has a Hamiltonian path. Now 6(G) > 0 or else G would not have a Hamiltonian path. By Lemma 6.2 and Lemma 6.3, G tJ. 1t. D Now that we know which graphs of large size have a Hamiltonian path, we will address the problem of determining X(G, ,pj) when e(G) 2j5. 6.2 Graphs missing complete subgraphs or a star As we have seen in the previous chapter, graphs of large size missing a star or complete subgraphs are the graphs which need to handled as special cases when de termining X(G, ,Cj) The same will prove to be true when determining X(G, ,pj) for graphs of large size. First, let's prove a result for a graph whose complement contains a star as a subgraph. 73
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Theorem 6.4 Let G be a graph of order n. If j 2, e(G) (2j5), and K1,j_1 G, then X(G, ,pj) = I :;=il Proof. If j = 2, then the theorem is vacuously true. So assume j 3. Let X be a set of vertices which induce K1,j_1 Since K1,j_1 G and e(G) :=::; 2j5, there is exactly one vertex v0 of degree at least j 1 in G. Therefore, v0 E X. Let A be a color class in a 'Prcoloring of G. Suppose v0 tJ. A. If IAI j, then (A) would be missing at most 2j5(j1) = j 4 edges, and by Theorem 2.9, we know that (A) has a Hamiltonian path. Therefore, for (A) to be Prfree, we must have IAI :=::; j1. Suppose v0 E A, and consider Av0 If lAvoi j, then by the same argument, we get Pj (Av0). Therefore, lAv01 :=::; j1, which implies IAI :=::; j. Thus, there can be at most one color class of size j (one containing v0), and X( G, ,pj) I j= + 1 = I j=i l To show the inequality in the other direction, form one color class B of size j by using the vertices in X together with one vertex from GX and form as many other color classes as possible of size j 1 with the remaining vertices in GX. Now (B) is Prfree since v0 is not adjacent to any other vertex in (B). Therefore, X(G,'Pj) :=::; 1:;=11 +1 = l:;=il D The previous theorem along with most of the following theorems will be used in the next section. The following theorems are more general than we need, but we wish to present them in their entire generality. Their proofs are straightforward and use the same proof methodology. Therefore, to quickly reach the main results of this chapter, the remaining theorems in this section will be presented here without proof. Their proofs can be found in Appendix A. 74
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Theorem 6.5 Let G be a graph of order n 1. If j 2 is even, m 0, and (E(G)) = then X(G, Pj) =max (I j=_71 I y l) Theorem 6.6 Let G be a graph of order n. If j 3 is odd and (E( G)) = Kill. e, 2 where e is an edge, then X ( G, ,pj) = I j:::: i l Theorem 6. 7 Let G be a graph of order n. If j 4 and (E( G)) = Ij_3 V K2 then X(G,.Pj) = lj::::il Theorem 6.8 Let G be a graph of order n 1. If j 4 is even, m 0, and (E(G)) =mKii!, thenX(G,.Pj) lj:1l). Theorem 6.9 Let G be a graph of order n. If j 2 zs even and (E( G)) Kii!e, where e is an edge, then X(G,.Pj) = l;::::il Theorem 6.10 Let G be a graph of order n. If j 6 is even and (E(G)) K 1 V + K1), then X(G, Pj) = I j::::i l Theorem 6.11 Let G be a graph of order n. If j > 6 is even and (E(G)) + K2 then X(G, .Pj) = I j::::i l We will use most of these theorems to determine X ( G, ,pj) when e (G) is large. 6.3 Determining x(G, .Pj) when e(G) is large First of all, we can remove j2 edges and show that X(G, .Pj) does not decrease by applying a well known result about Hamiltonian paths. As an aside, if j 3, then Theorem 6.12 is an immediate result of Theorem 6.13. Thus we did not need to prove Theorem 6.12 in full generality. However, it was easier to prove Theorem 6.12 this way. 75
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Theorem 6.12 Let G be a graph of order n. If j 2 and e(G) (j2), then X( G, .Pj) = I Proof. Let A be a color class in a Prcoloring of G. Suppose that IAI j. Let B A with IBI = j. Now (B) is missing at most j2 edges. By Theorem 2.9, we get Pj (B), a contradiction. Therefore, IAI < j, which implies that X(G,.Pj) Equality follows from Theorem 2.1. D We can see that this theorem is best possible since, if we remove the edges of a copy of K1,jl from a complete graph of order n j, then by Corollary 4.1, we can obtain a graph whose Prchromatic number is I j=i l What is the minimum number of edges we need to remove from a complete graph to obtain X(G, .Pj) = I j=il? One way to obtain X(G, .Pj) = I j=il is to remove the edges of 2K1,jl from a complete graph, i.e., remove 2j2 edges. With this in mind, we will attempt to close in on graphs missing 2j2 edges by examining graphs missing up to 2j5 edges. Theorem 6.13 Let G be a graph of order n 1 and e be an edge. If j 2 and e(G) (2j5), then I j=i l if G = Kn5 V (K2 +h) and j = 7, Kl,j1 G, G = Kn(jl) V (Kj3 +h) and j 4, G = Kn4 V I4 and j = 6, X(G, .Pj) = G = (Kn4 V I4)e and j = 6, G = Kn5 V Is and j = 8, G = (Kn5 VIs) e and j = 8, or G = Kn_6 V h and j = 10, and I otherwise. 76
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Proof. Let j 2 and G be a graph of order n with e(G) (2j5). If j = 2, then the theorem is vacuously true. So assume j 3. Suppose that G = Kn_ 5 V (K2 + I 3 ) and j = 7. Then (E(G)) = K 5 e, where e is an edge, and by Theorem 6.6, we get X(G,,P7) = l;.::::il Suppose that KI,jI Then by Theorem 6.4, we get X(G, ,pj) = I j.::::i l Suppose that G = Kn(jI) V (Kj3 +h) and j 4. Then (E(G)) = Ij3 V K2, and by Theorem 6.7, we get X(G, 'Pj) = I j.::::i l Suppose that G = Kn4 V I4 and j = 6, or G = Kn_ 5 V Is and j = 8, or G = Kn_ 6 V h and j = 10. Then (E(G)) = Kill, and by Theorem 6.5, we get 2 X(G,,Pj) = l;.::::il ifn j, and X(G,,Pj) = 171 ifn < j since l;.::::il 171 if and only if n j. But, when n < j, we know that I y 1 = 1 = I j.::::i 1 Therefore, X(G, ,pj) = I j.::::i l Suppose that G = (Kn4 V I4) e and j = 6 or that G = (Kn_ 5 VIs) e and j = 8, where e is an edge. Then G = Kn_ 5 V (KI, 3 + I1 ) and j = 6; or G = Kn6 V (I4 V I2) and j = 6; or G = Kn6 V (KI,4 +II) and j = 8; or G = Kn7 V (Is V I2) and j = 8. If G = Kn5 V (KI,3 +II) and j = 6 or if G = Kn6 V (KI,4 +II) and j = 8, then (E(G)) = KI V + KI) By Theorem 6.10, we get X(G, ,pj) = I j.::::i l If G = Kn6 V (I4 V I2) and j = 6 or if G = Kn7 V (Is V h) and j = 8, then (E(G)) = + K2 By Theorem 6.11, we get X(G, ,pi)= I j.::::i l So assume that G and j do not meet one of the conditions specifically listed in the statement of this theorem. In this case we will show that X( G, ,pi) = I Let A be a color class in a ,Prcoloring of G. If we can show that IAI < j, then X(G,,Pj) and equality follows from Theorem 2.1. Suppose that IAI j. 77
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Let B such that IBI = j. We want to use Theorem 6.3 to show that (B) has a Hamiltonian path. Since K1,j_1 C1 G, J( (B)) > 0. If (B) = K2 V (K2 + !3), then j = 7. Now 9 = 2(7) 5 e(G) ::; 2j5 imply that G = Kn_5 V (K2 + !3), a contradiction. e( (B)) and If (B) = K1 V (Kj_3 + !2), then j 4. Now 2j5 = e( (B)), which implies that G = Kn(j1 ) V (Kj_3 + !2), a contradiction. If (B) = !4 V K2 then j = 6 and 2j 5 = 7 e((B)) + 1. If (B) = (!4 V K2)e, then j = 6 and 2j5 = 7 = e((B)). Now either G = Kn4 V !4 or G = (Kn_4 V !4 ) e, and we reach a contradiction. If (B) = Is V K3 then j = 8 and 2j 5 = 11 e( (B)) + 1. If (B) = (Is V K3 ) e, then j = 8 and 2j 5 = 11 = e( (B)). Now either G = Kn_5 VIs or G = (Kn_5 VIs) e, and we reach a contradiction. If (B) = h V K4 then j = 10. Now 2(10) 5 = 15 = e( (B)), which implies that G = Kn_4 V h, a contradiction. Therefore (B) 1t and by Theorem 6.3, (B) has a Hamiltonian path, a contradiction. Thus IAI < j. The previous theorem has instances when X(G, ,pj) = D r n11 and instances 111 when X(G,'Pj) = We can see that fj::::il = in many cases as follows: if n = p(j 1) + q, where p 0, j 2, 0 ::; q < j 1 and q =J 1, then r n11 = fp(j1_)+(q1)1 = r ,B1 111 I 11 111 To continue with the approach of using a Hamiltonian cycle on n 2 vertices and attaching two vertices to this cycle to create a Hamiltonian path requires us to know which graphs of a fixed size are Hamiltonian. This is labor intensive and, even if we obtain this knowledge, will not yield a general result for conditional 78
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coloring. Therefore, we abandon this approach to find X(G, Pj) for all graphs of large size with the knowledge that in order to produce a graph whose Pr chromatic number is I j::::il, we need to remove somewhere between 2j 4 and 2j2 edges. 6.4 Determining bounds on x(G, .Pj) given the number of edges in a graph In this section, we find an upper bound on the size of a graph given the constraint that the Prchromatic number is at most J. To find the upper bound, we first need to determine the maximum size of a Prfree graph of order n. The first upper bound on the size of a Prfree graph was discovered in 1959, by P. Erdos and T. Gallai [24]. Their theorem states if G is a Prfree graph, then e( G) ( y )n. In 1975, R. Faudree and R. Schelp [25] improved on this bound for the maximum size of a Prfree graph. In fact, they found the least upper bound and determined which graphs met that bound. The theorem is restated here for completeness and reference. Theorem 6.14 (Faudree and Schelp, [25}) IfG is a graph of ordern = q(j1)+r (0 q, 0 r < j1) and G contains no Pj, then e(G) q(i;1 ) + (;) with equality if and only if G = qKj_1 + Kr or (when j is even, q > 0, and r = j /2 or (j 2) /2), G = lKi1 + ( K i?VI 0 l q 1. We would like to know how many edges we need to remove from a complete graph of order j + i ( i 0, j 2) to guarantee that a graph missing at least this number of edges is Prfree. To accomplish this, we tailor Faudree and Schelp's result to our needs with the following corollary. 79
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Corollary 6.1 If G is a Prfree graph of order n j 2, then e(G) (nj + 1)(j1). Proof. Let mn be the minimum number of edges in the complement of a Prfree graph of order n = q(j1) + r, where q 0 and 0:::; r:::; j2. We will show by induction that for n j, we have mn ( n j + 1) (j 1), which establishes the statement of the corollary. Now n j implies that q 1. By Theorem 2.9 (iii), mj j1. By Theorem 6.14, mn (;) q e ; 1) (;) (q(j21)+r) qe;1) (;) (q(j1) + r)(q(j1) + r1)q(j1)(j2)r(r1) 2 q2(j1)2 + rq(j1) + q(j1)(r1)q(j1)(j2) 2 q(j1)[q(j1) + 2r1j + 2] 2 q(j1)[(q1)(j1) + 2r] 2 When r < j2, we have n + 1 = q(j1) + (r + 1). Thus, q(j1)[(q1)(j1) + 2(r + 1)] q(j1)[(q1)(j1) + 2r] 2 2 q(j1) > j 1. When r =j2, we haven+ 1 = (q+ 1)(j 1), so that (q + 1)(j1)q(j1) q(j1)[(q1)(j1) + 2(j2)] 2 2 q2(j1)2 + q(j1)2q2(j1)2 + q(j1)22q(j1)(j2) 2 80
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q(j1)2 q(j1)(j2) q(j1) > j 1. Now by the induction hypothesis, (mn+l mn) +mn j 1 + (nj + 1)(j 1) = (n + 1j + 1)(j1). D We get the following immediate result for the ,Prchromatic number from Theorem 6.14. Theorem 6.15 Let j 2. If G is a graph of order n = k(j 1) + r (0 ::::; k, 0::::; r < j1) and e(G) > k(1;1 ) + (;), then X(G, 2. Proof. Let j 2 and let G be a graph of order n = k(j 1) + r (0 ::::; k, 0 ::::; r < j1) with e(G) > k(1;1 ) + (;). By Theorem 6.14, we get Pj G. Therefore, X(G, ,pi) 2. D Next we use Corollary 6.1 to derive an upper bound on the size of G given Theorem 6.16 LetG be a graph ofordern 1, j 2 andk 0. IfX(G,,Pj)::::; then e(G)::::; k(j1). Proof. Let G be a graph of order n satisfying X(G, ,pi) ::::; with the maximum number of edges. Now G must contain all edges between any two color classes. For, if not, we could add an edge between two color classes to form a new graph H with X(H, ,pj) ::::; J using the same coloring we used for G. But this contradicts the maximality of G. Therefore, we can assume all missing edges 81
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reside within the color classes. Furthermore, by maximality, any color class of size at most j1 is not missing any edges. Let aj+i be the number of color classes of size j + i for i = 0, ... nj and bi be the number of color classes of size i for i = 1, ... j 1. To minimize the number of missing edges in our graph G, we need to minimize the number of missing edges in all color classes of size at least j. By Corollary 6.1, a color class of size j + i must be missing at least (i + 1)(j1) edges for all i = 0, ... nj. Therefore, we want to find a lower bound for the function (j 1)aj + 2(j 1)aj+l + ... + (nj + 1)(j1)aj+nj = (j 1) + 1)aj+i Let's leave this for a moment and derive another inequality. j1 nj n = L ibi + L:U + i)aj+i i=1 i=O j1 j1 nj nj L:U1)biL[(j1)i]bi + L:(i + 1)aj+i + L:U1)aj+i i=1 i=1 i=O i=O j1 nj j1 nj u1)(L: bi + L: aj+i)L:[U1)iJbi + L:(i + 1)aj+i i=1 i=O i=1 i=O l kj j1 nj (j 1) ; = 1 [ (j 1) i] bi + ( i + 1) aj+i j1 nj < nkL[(j1)i]bi + L:(i + 1)aj+i i=1 i=O Now, rearranging the above equation, we get nj j1 L:(i + 1)aj+i k + L[(j1)i]bi k i=O i=1 since L,{;:{[(j1) i]bi 0. Using the above inequality, we get (j 1) (i+ 1)aj+i (j 1)k. Therefore, the number of missing edges in all color classes is at least k(j1), which implies that e(G) ::; (;) k(j1). D 82
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It is immediately seen from the following example that this bound is attainable. Let G be a graph of order n with G = kK1,j_1 where k 0 and j 2. Since n = kj = k(j1) + k, we have that j1 divides nk and by Theorem 4.3, we get X( G, ,pj) = I l = J. We have found an upper bound on the size of G given a bound for X(G, .Pj) for j 2. The next theorem gives a lower bound on the size of G given a particular Prchromatic number. Theorem 6.17 Let G be a graph of order n and j 2. If X(G, Pj) > I j::::i l then e(G) 2). Proof. We will prove the contrapositive. Let G be a graph with e( G) < Suppose 6(G) nj+l. Now, e(G) = L:d(v) L:6(G) 1) 2). This contradicts the assumption that e(G) < 2). Therefore, there exists a vertex v such that d( v) ::; n j. In other words, the vertex v is nonadjacent to at least j 1 other vertices in G. Color G as follows: Form a color class of size j by using v and j 1 nonneighbors of v. With the remaining n j vertices, create as many color classes of size j 1 as possible. With this coloring, we get X(G,.Pj)::; 1;::::11 + 1 = l}::::il D It is still an open question whether or not the lower bound is attainable. In other words, can we find a graph edges with X(G, Pj) = I It is also open as to what the lower bounds are for other values of X(G, .Pj) Finally, given X(G, .Pj), we would like to find an upper bound on the size of G that is tight when j1 does not divide nk. 83
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A Appendix All these theorems determine the Prchromatic number for a graph when its complement contains a very specific set of edges. The first three theorems and the final two theorems are used in the proof of Theorem 6.13. Theorem 6.5 Let G be a graph of order n 1. If j 2 zs even, m 0, and (E(G)) = then X(G, .Pj) =max (I l, I I l). Proof. If j = 2, then (E(G)) = mK2 which implies n 2m, or nm I% l We will show there is no color class of size at least 3. If A were a color class of size at least 3, then the graph induced by any three vertices in A would be missing at most one edge and P3 (A), a contradiction. Also, a color class of size 2 must consist of the endpoints of an edge in G. Therefore, X(G, .P2 ) nm. Since n we get nm l%l Thus, X(G, .P2 ) =max (nm, l%1). lfj 4, then by Theorems 5.3 and 2.2, we get X(G, .Pj) X(G, .Cj) =max (I l, I I l). To show the inequality in the other direction, we produce a minimum Pr . colormg. Observe that G = Vi=l Si where Si = IJ..B for z = 1, 2, ... m 2 and Si = /1 fori= m + 1, m + 2, ... n( J..) m. LetS= V(G)U:1 V(Si) Now, n = m + s. We consider two cases: j;2m::::; s and 0::::; s < j;2m. If ::::; s, then n mj and I l I I l For each i = 1, ... m, form a color class of size j by using all the vertices from Si together with from S not yet assigned to a color class. Since ::::; s, S contains a sufficient number of vertices to form these m color classes. Now, all the subgraphs induced by these 84
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color classes are isomorphic to I iH. V K which by Lemma 6.2, is Prfree. We 2 2 partition the remaining sym vertices into as many color classes of size j1 as 'bl Th h (G P) l(i)m+sml_ lnm1 poss1 e. us, we ave X , j :S m + I j1 1 j1 1 j1 If 0 < s < j2m then n < mJ and 1.!!1 > lnm1 Let r = 2 I J I J1 I J and T = S U ( 1 V ( Si)) For each i = 1, ... m r, form a color class of size j by using all the vertices from Si together with y vertices from T not yet assigned to a color class. Note that m r = m ( )ms l = l mj(if )m+s J = l ( i]m+s J = l7 J and there are sufficient elements in T to form l7 J color classes since ITI = s+r (i) = (i) 2: (i) = (y) (7) 2: (1;2 ) l7 J. Again, all the subgraphs induced by these color classes are isomorphic to I iH. V K which by Lemma 6.2, is Prfree. 2 2 If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Prfree, we will show that IBI < j. Now, lEI ITI ly j (j; 2 ) s + r (j; 2 ) (mr) e; 2 ) s + rjm (j; 2 ) < s+ +l)jme;2) J. If B = 0, then there were no remaining vertices after forming m r color classes of size j. Therefore, j divides n and X(G, ,pj) :S l7 J = I 71 If B #0, then j does not divide nand we get X(G, ,pj) :S l7 J + 1 = I 7l In either case, we have produced a ,Prcoloring with I 71 colors. D 85
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Theorem 6.6 Let G be a graph of order n. If j 3 is odd and (E( G)) = K w_e, 2 where e is an edge, then X(G, .Pj) = I j=i l Proof. Observe that (E(G)) = Kw_ e if and only if G = R V (S + T), 2 where R = K (ill.)' S = Ii::::..l and T = K2 Let A be a color class in a Pr n 2 2 coloring of G. Suppose that IAI j + 1. Let B A such that lEI = j + 1. Now, B contains up to 1vertices from S U T and at least vertices from R. For all combinations of vertices from these sets, we get (C) = I ill. V 2 K i::::..l (B). Let U = { ui, u2 ... uw_} be the set of vertices of degree in 2 2 (C) and W = {WI, w2 ... w i::::..l} be the set of vertices of degree j in (C). Now, 2 uiwiu2w2 UJ..:::...!WJ..:::...!Ui! forms Pj a contradiction. Therefore, IAI::; j. 2 2 2 Now, each color class of size j must contain strictly more than i.:pvertices from SUT. Suppose by way of contradiction that a color class A of size j contains at most i.:pvertices from S U T. Since j is odd, either j + 3 or j + 1 is divisible by 4. If j + 3 is divisible by 4, then A contains at most i.:pvertices from S U T and at least 3j,;3 vertices from R. For all combinations of vertices from these sets, we get (D)= Iw_ V K3i3 Let xi,x2 ... ,xw_ be the vertices of degree 4 4 4 37,;3 in (D) and YI, y2 ... Y3i3 be the vertices of degree j1 in (D). Now, 4 xiyix2y2 ... xw_yw_yi]_yi+n ... Y3i3 forms Pj (D), a contradiction. There4 4 4 4 4 fore, j + 1 must be divisible by 4, A contains at most i.:pvertices from S U T, and A must contain at least 3j;I vertices from R. For all combinations of vertices from these sets, we get (D) = Ii! V K3i1 (A). Let XI, x2 ... X.i...!. be the vertices 4 4 4 of degree 37;I in (D) and YI, y2 ... Y3i1 be the vertices of degree j1 in (D). 4 Now, XIYIX2Y2 ... X.i...!.Y.i...!.YillYill ... Y3i1 forms Pj (D), a contradiction. 4 4 4 4 4 Therefore, each color class of size j must contain strictly more than j!3 vertices 86
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from the S U T, which is strictly more than half the vertices of S U T. Therefore, there can be at most one color class of size j and X(G, .Pj) f;.=11 + 1 = I j.=i l Next, we will show the inequality in the other direction. Color G as follows: form one color class A of size j by using all the vertices in S together with y vertices from Rand both vertices from T. Now, (A) = V (I?+ K2 ) and since (A) contains vertices of degree y with exactly y common neighbors, (A) is Prfree. With the remaining vertices, form as many color classes of size j 1 as possible. With this coloring, we get X(G,.Pj):::::; f;.=n + 1 = fj.=il So, X(G,.Pj) = fj.=il D Theorem 6. 7 Let G be a graph of order n. If j 4 and (E( G)) = Ij_3 V K2 thenX(G,.Pj) = fj.=il Proof. Let j 4 and G be a graph of order n such that (E( G)) = Ij_3 V K2 Then G = R V (S + T) where R = Kn(j1), S = Kj_3 T = h. Let {r1,r2 ... ,rnj+1 } be the vertices of R, {s1,s2 ... ,sj_3 } be the vertices of S, and { t1 t2 } be the vertices of T. Let A be a color class in a Prcoloring of G. Suppose that IAI j + 1. Let B A such that lEI = j + 1. If t1 tJ_ B and t2 tJ_ B, then (B) is complete, which implies that Pj (B), a contradic tion. If t1 E B and t2 tJ_ B, then B = {t1 r1 ... rp, s1 s2 ... Sj1p} for some 3 :::::; p :::::; j 1. Now, t1r1r2r3 ... rps1s2 ... Sj1p forms Pj (B), a contradic tion. Therefore, we must have { t1 t2 } B. Since B can contain at most j 3 vertices from S, the subgraph (B) must contain at least two vertices r1 and r2 from R. So B = { t1 t2 r1 ... rt, s1 s2 ... Sj1p} for some 2 :::::; p :::::; j 1. Now, t1r1t2r2r3 rps1s2 Sj1p forms Pj (B), which again is a contradic87
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tion. Therefore, we must have IAI :::; j. Suppose IAI = j. By Theorem 2.9, if (A) is missing fewer than j 1 edges, then (A) has a Hamiltonian path. Since G is missing at most 2j 5 edges, there can be at most one color class of size j. Thus, X(G,.Pj) l;::::tl + 1 = l;::::il Next we produce a minimum Pr coloring. Form one color class A of size j with A = { t1 t2 r1 s1 s2 ... Sj_3}. Now (A) = K1 V (Kj_3 +h), which by Lemma 6.2, is Prfree. We partition the remaining nj vertices into as many color classes of size j1 as possible. Thus, we have X(G,.Pj):::; l;::::tl +1 = l;::::il D In earlier theorems we have derived a value of I j:=_71 for the Prchromatic number for graphs of order n whose complements consist of m pairwise vertex disjoint copies of a complete graph of a particular order. The following theorem derives a new value for the Prchromatic number, namely I l, for graphs whose complements consist of m pairwise vertex disjoint copies of a complete graph of a new order. Theorem 6.8 Let G be a graph of order n 1. If j 4 is even, m 0, and (E(G)) = then X(G, .Pj) =max (I l, I j:ll) Proof. Observe that G = Vi=l Si where Si = Iii for = 1, 2, ... m 2 and Si = /1 fori= m+1,m+2, ... ,n(i) m. Let A be a color class in a Pr coloring of G. Suppose that IAI > j + 1. Let B such that IBI = j + 2. Then either (B)= Iq; where q1 = (j +4)/2, and 1:::; Qi:::; j/2 for all2:::; i:::; k (1), or (B) = Iq; where 1 :::; q1 :::; (j + 2) /2, 1 :::; q2 :::; (j + 2) /2, and 1 :::; Qi :::; j /2 for all3:::; i:::; k (2). If (1), then form cl B such that ICll = j by removing two vertices from Iq1 If ( 2), then form C2 B such that I C2l = j by removing one 88
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vertex from each of Iq1 and Iq2 Now, (C1 ) = Iq1 2 V Iq;), and maxi{qi}::; Also, (C2 ) = Iq1 1 V Iq2 1 V Iq;), and maxi{qi} ::; By Lemma 5.1, (C1 ) and (C2 ) are Hamiltonian. This contradicts the assumption that (A) is Prfree. So IAI :s;j+1 andX(G,,Pj) 2: Let A be a color class of size j + 1. We will show that I i. (A). Suppose by 2 way of contradiction that Ii. C1 (A). Then (A) = Iq; where q1 ::; (j + 2)/2, 2 and 1 ::; Qi ::; j /2 for all 2 ::; i ::; k. Form B A such that lEI = j by removing one vertex from Iq1 Now, (B)= Iq1 1 V Iq;), and maxi{qi}::; By Lemma 5.1, (B) is Hamiltonian. This contradicts the assumption that (A) is Prfree. So Ii. (A). 2 Let C be a color class of size j. By Lemma 5.1, either (C) is Hamiltonian or I ill (C). Since (C) is not Hamiltonian, we must have that I ill (C). 2 2 Let a be the number of color classes of size j + 1 and b be the number of color classes of size j in a ,Prcoloring of G. Since each color class of size j + 1 must contain all the vertices from Si for some 1 ::; i ::; m, and since color classes are vertex disjoint, we must have a ::; m. We have just shown that each color class of size j must contain i independent vertices, which is more than half the vertices from some Si (1 ::; i ::; m) since jt2 > (Jt4 ) when j 2: 2. Thus we cannot use the vertices of Si ( 1 ::; i ::; m) to form two different color classes of size j. Since a color class of size at least j requires more than half the vertices from some Si (1 ::; i ::; m) and color classes are vertex disjoint, we get a+ b ::; m. This inequality and a ::; m imply that 2a + b ::; 2m. Now, X(G ,p.) > rna(j + 1) bj1 a b = rn2ab1 > rn2m1 'J"1 ++ "1 "1 JJJ89
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Hence, X(G, Pj) (I I l). To show the inequality in the other direction, we produce a minimum Pr ( +4) coloring. Recall that G = v:_:n.z..p m Si where Si =Jill fori= 1, 2, ... m and 2 Si = 1 1 fori= m + 1, m + 2, ... n( J.) m. LetS= V(G)V(Si) Now, n = (it4 ) m + s. We consider two cases: j;2 m:::; sand 0:::; s < j;2 m. If ym:::; s, then n m(j + 1) and I l I For each i = 1, ... m, form a color class of size j + 1 by using all the vertices from Si together with y vertices from S not yet assigned to a color class. Since ym:::; s, the setS contains enough vertices to form these m color classes. Now, all the subgraphs induced by these color classes are isomorphic to C = I ill V K and since C contains 2 2 vertices of degree y with exactly y common neighbors, C is Prfree. We partition the remaining sym vertices into as many color classes of size j1 as 'bl Th h (G P) is(J.?)ml_l(i{=i)m+s2ml_fn2ml poss1 e. us, we ave X , j :::; m + I j _1 1 j _1 1 j 1 If 0 < s < 1=1m then n < m(J' + 1) and r ,!!..1 > r n:2 ml Let r = I ( )msl 2 I J+1 I J1 I J+l and T = S U V(Si) ). For each i = 1, ... mr, form a color class of size j + 1 by using all the vertices from Si together with y vertices from T not yet assigned to a color class using the vertices from S first. Note that m r = m l = l )m+s J = l ( J = and there are sufficient elements in T to form color classes since ITI = s + r ( J.) = s + (j+4) > s + (J.?)ms (j+4) = (j2) (,!!..) > (j2) l,!!J. Again 1 1+1 2 1+1 2 2 1+1 2 1+1 all the subgraphs induced by these color classes are isomorphic to C = I ill V K 2 2 and we have already shown that (C) is Prfree. If there are vertices remaining in T, form one more color class B containing 90
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those vertices. To show (B) is Prfree, we will first show that lEI ::; j. Now, lEI = ITIlj: 1 j (j; 2 ) s + r (j; 4)(mr) (j; 2 ) s + r (j + 1) m ( j ; 2 ) < s + ( ( s + 1) (j + 1) m e ; 2 ) j + 1. If I B I = j, then (B) = I .i.i V I .i.::::..i since we formed color classes by using vertices 4 4 from S first. By Lemma 6.2, I .i.i VI .i.::::..i is Prfree. 4 4 If B = 0, then there were no remaining vertices after forming m r color classes of size j + 1. Therefore, j + 1 divides nand X(G, ,pj) :=::; = I If B 10, then j + 1 does not divide n and we get X( G, ,pj) :=::; + 1 = I In either case, we have produced a ,Prcoloring with I colors. D The following theorem is needed for the proof of Theorem 6.10. Theorem 6.9 Let G be a graph of order n. If j 2 is even and (E( G)) = K .i.i e, 2 where e is an edge, then X(G, ,pj) = I j=i l Proof. Observe that (E(G)) = K.i.i e if and only if G = R V (S + T), where 2 R = K (.i.i), S = I i, and T = K2 Let A be a color class in a ,Prcoloring n 2 2 of G. Suppose that IAI > j. Let B A such that lEI = j + 1. Now, B contains up to jt4 vertices from S U T and at least j;2 vertices from R. For all combinations of vertices from these sets, we get (C) = + K2 ) V K y (B). Let u1 u2 ... Ui be the vertices of degree y in (C), v1 v2 be the vertices of 2 91
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degree m (C), and WI, w2 be the vertices of degree j in (C). Now, 2 Now each color class of size j must contain strictly more than i vertices from S U T. Suppose by way of contradiction that a color class A of size j contains at most j!4 vertices from S U T. Since j is even, either j or j + 2 is divisible by 4. If j is divisible by 4, then A contains at most i vertices from S U T and therefore, at least 3t;4 vertices from R. For all combinations of vertices from these sets, we get (D) = Iii V K3i4 (A). Let xi, x2 Xii be the vertices 4 4 4 of degree 37 ; 4 in (D) and y1 y2 ... Y3i4 be the vertices of degree j 1 in (D). 4 Now, XIYIX2Y2 ... XiiYiiYiH_Yi+12 ... Y3i4 forms Pj (D), a contradiction. So 4 4 4 4 4 j + 2 must be divisible by 4, A contains at most t. vertices from S U T, and A must contain at least 3t;2 vertices from R. For all combinations of vertices from these sets, we get (D) =JiB V K3i2 (A). Let xi, x2 XtB be the vertices 4 4 4 of degree 37 ; 2 in (D) and y1 y2 ... Y3i2 be the vertices of degree j 1 in (D). 4 Now, XIYIX2Y2 ... XtBYiBYiH_Yi+lO ... Y3i2 forms Pj (D), a contradiction. 4 4 4 4 4 Therefore, each color class of size j must contain strictly more than i vertices from SUT, which is strictly more than half the vertices in SUT. Therefore, there can be at most one color class of size j and X( G, ,pj) I ;.::::{1 + 1 = I j.::::i l Next, we will show the inequality in the other direction. If j = 2, then (E( G)) = P3 Form one color class of size 2 using the two nonadjacent vertices in G and put the remaining n2 vertices into n2 color classes. Therefore, X(G, ,pj) :::; n1. If j 4, then color G as follows: form one color class A of size j by using all the vertices in S together with vertices from R and both vertices from T. Now, (A)= Ky. V K2 ) and (A) is a spanning subgraph of 92
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IJ..B V By Lemma 6.2, (A) is Prfree. With the remaining vertices, form as 2 2 many color classes of size j 1 as possible. We get X(G, ,pj) I ;::::tl + 1 = I j::::i l So,X(G,,Pj)=lj::::il D Theorem 6.10 Let G be a graph of order n. If j > 6 is even and (E(G)) K1 V + KI), then X(G, 'Pj) = I j::::i l Proof. Assume j 6 is even and G is a graph of order n such that (E( G)) = K1 V + K1). Then G = R V (S + T) where R = S = and T = I1 Let {r1,r2 ... be the vertices of R, {s1,s2 ... be the vertices of degree 1 in S, s be the vertex of degree in S, and t be the vertex in T. Now H = V (K2 + G and by Theorems 6.9 and 2.4, X(G, ,pj) X(H, ,pj) = I j::::i l To show the inequality in the other direction, color G as follows: form one color class A of size j with A= {s1 s2 ... Si, s, r1 r2 ... r.i.::::..i, t}. Now (A) = 2 2 K.i.::::..i V (K1 i + I1), which is a spanning subgraph of IJ..B V and by Lemma 2 '2 2 2 6.2, is Prfree. With the remaining vertices form as many color classes of size j 1 as possible. With this coloring, we get X(G,,Pj) l}::::il Theorem 6.11 Let G be a graph of order n. If j 6 is even and (E(G)) +K2 thenX(G,,Pj) = l}::::il Proof. Assume j 6 is even and G is a graph of order n such that (E( G)) = KJ..B + K2 Then G = R V (S V T) where R = Kn(J..B+2), S = IJ..B and T = I2 2 2 2 Let A be a color class in a ,Prcoloring of G. Suppose that IAI j + 1. Let B with lEI = j + 1. Now B contains up to 1+ 2 vertices from S U T and 93
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at least j;4 vertices from R. For all combinations of vertices from these sets, we get C = K:i.::::..i V (Jill V h) (B). Let x1 x2 ... X:i.::::..i be the vertices of degree j 2 2 2 in C, y1 y2 ... y ill be the vertices of degree in C, and z1 z2 be the vertices of 2 IAI '5:_j. To form a color class of size j, we must use all the vertices in S. For if A contains at most vertices from S, then for v E A, we have d(A)(v) and, by Dirac's Theorem (Theorem 2. 7), (A) is Hamiltonian, a contradiction. Thus, there can be at most one color class of size j. Therefore, we get X ( G, ,pi) r;=n + 1 = r;=il To show the inequality in the other direction, color G as follows: form one color class of size j by using all i vertices from S, one vertex from T, and vertices from R. Now (A) =Jill V which by Lemma 6.2, is Prfree. With 2 2 the remaining vertices form as many color classes of size j 1 as possible. With this coloring, we get X(G, Pj) '5:_ I j=i l D 94
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B References 1. J. Akiyama, H. Era, S. Grevacio and M. Watanabe, Path chromatic numbers of graphs, J. Graph Theory 13(5): 569575 (1989). 2. M. 0. Albertson, R. Jamison, S. T. Hedetniemi and S. C. Locke, The sub chromatic number of a graph, Discr. Math. 74:3349 (1989). 3. J. A. Andrews and M.S. Jacobson, On a generalization of chromatic number, Congr. Numer. 47:3348 (1985). 4. J. A. Andrews and M.S. Jacobson, On a generalization of chromatic number and two kinds of Ramsey numbers, Ars Combin. 23:97102 (1987). 5. K. I. Appel and W. Haken, The solution of the fourcolormap problem, Freeman, California, 1977. 6. P. Baldi, On a generalized family of colorings, Graphs and Combinatorics 6:95110 (1990). 7. L. W. Beineke and A. T. White, Selected Topics in Graph Theory, Academic Press, London, 1978. 8. L. D. Bodin and A. J. Friedman, Scheduling of committees for the New York State Assembly, Technical report, State University of New York, 1971. 9. J. A. Bondy, Variations on the Hamiltonian theme, Can. Math. B. 15:5762 (1972). 10. J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, Elsevier North Holland, Inc., New York, 1976. 95
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CONDITIONAL COLORING by Mark R. Dillon B.S., Purdue University, 1982, 1985 M.S., Purdue University, 1984 A thesis submitted to the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Do.:;tor of Philosophy Applied Mathematics 1998
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This thesis for the Doctor of Philosophy degree by Mark R. Dillon has been approved by David Fisher ren Tom Altman Date
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Dillon, Mark R. (Ph.D., Applied Mathematics) ConditionaL Coloring Thesis directed by Associate Professor Kathryn L. Fraughnaugh ABSTRACT The conditional chromatic number X( G, P) of a graph G with respect to a graphical property P is the minimum number of colors needed to color the vertices of G such that each color class induces a sub graph of G with property P. When P is the property that a graph contains no sub graph isomorphic to a graph F, we write X( G, F). The conditional chromatic number of a graph has been studied by various authors since 1968. We focus on two conditional chromatic numbers, specifically X( G, Ci) and X( G, ,pi), where Ci is a cycle oflength j for some fixed j 3 and Pj is a path of length j 1 for some fixed j 2. We find X( G, Cj) for graphs missing at most j1 edges and X( G, ,pi) for graphs missing at most 2j5 edges. To accomplish this, we characterize all Hamiltonian graphs of order n with at least ( n 1) edges and all graphs with no Hamiltonian paths with at least (;) (2n5) edges. We determine both conditional chromatic numbers for all graphs with acyclic complements. We also determine a lower bound on X(G, Pj) in terms of the size of G. Finally, we show the problem of determining if X(G, P3 ) :::; k, for some k 0, is NPcomplete. This abstract accurately represents the content of the candidate's thesis. I recom mend its publication. Ill
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DEDICATION To my wife, Wendy Tweten Dillon, whose love, understanding, patience, and encouragement made it possible to complete this dissertation.
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ACKNOWLEDGEMENTS I would like to express my deep gratitude to my advisor, Professor Kathryn Fraughnaugh for her tireless guidance, patience, inspiration and encouragement during the course of my research and throughout the preparation of this disser tation. My sincere thanks to Professor Richard Lundgren and Professor David Fisher who were always there to answer questions. I would also like to thank my mother and father, Pat and Russell Dillon who always gave me the support I needed and my brothers and sister, Steve, Dave, Scott, Laura and Paul Dillon for their constant encouragement. I also thank Mr. John Drubert, my sixth grade teacher, who was the first person to interest me in the field of mathematics and my close friends Doug Mckissack, Arthur Shulman and Kevin Williams for their constant encouragement. Finally, I received, and gratefully acknowledge financial assistance from Hughes Aircraft Company without whose support, this thesis would never have been com pleted.
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Contents 1 Introduction to coloring and its applications 1.1 Overview of Thesis Results 1.2 Definitions and Notation 1.3 Conditional Coloring .. 2 Basic Results for x(G, Pj) and x(G, Cj) 3 NPcomplete 4 Determining the conditional chromatic number for graphs with acyclic complements 4.1 Determining x(G, Cj) when G is acyclic 4.2 Determining x(G, Pj) when G is acyclic 4.3 The difference between the Cjand Pjchromatic numbers of a 1 4 5 9 16 22 29 29 35 graph. . . . . . . . . . . . . . . . . 37 5 Determining x(G, Cj) 39 5.1 Preliminaries . . 39 Vl
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5.2 Determining x(G, .Cj) when e(G) (j 1) 46 6 Determining x(G, .Pj) 58 6.1 Determining which graphs of large size have a Hamiltonian path 58 6.2 Graphs missing complete subgraphs or a star 6.3 Determining x(G, .Pj) when e(G) is large .. 6.4 Determining bounds on x(G, .Pj) given the number of edges in a 73 75 graph. . . . . . . . . . . . . . . . . 79 A Appendix 84 B References 95 Vll
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1 Introduction to coloring and its applications Let G be a graph. A vertex coloring of G is an assignment of colors to its vertices so that no two adjacent vertices receive the same color. The chromatic number X( G) is the minimum number of colors needed to color G. An equivalent definition of the chromatic number is the minimum integer k such that there is a partition of the vertices into k sets so that the subgraph induced by each set is an independent set. Vertex coloring has a wide variety of applications. One such application is the following. In the United States Government, there are congressional committees with members of Congress serving on multiple committees. When assigning meet ing times for these committees, one must not schedule simultaneous meetings for two committees that have a common member. A schedule solution is found by de termining the minimum number of time slots required for the committees to meet. We can model this problem by constructing a graph G whose vertices represent committees. We draw an edge between two vertices if the committees represented by these vertices have a common member. Determining the minimum number of time slots required for the committees to meet is equivalent to determining X( G). For further details, see Roberts [47] or Bodin and Friedman [8]. Another application is the channel assignment problem. Radio stations in a region are to be assigned transmitting frequencies so that radio stations which are geographically close, say 50 miles, receive different frequency assignments. The problem of assigning frequencies is a graph coloring problem. Let each vertex 1
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represent a radio station and draw an edge between two vertices if the radio stations represented by those vertices are within 50 miles. The number of frequencies required so that the radio stations do not interfere with each other is the vertex chromatic number of this graph. For more information regarding the channel assignment problem, see Cozzens and Roberts [20], Hale [28], Opsut and Roberts [43], or Pennotti [46]. The last application of vertex coloring we will discuss is the classic map coloring problem. This is a much studied problem and is discussed in most graph theory textbooks. For example, see Bondy and Murty [10], Harary [29], Chartrand and Lesniak [17), or Roberts [47). Given a map with various countries, we would like to color the countries in such a way that we use the fewest number of colors, and if two countries share a common border, they receive a different color. This problem can be translated into a graph coloring problem by building a graph with each vertex representing a country and drawing an edge between two vertices if the countries represented by those vertices have a common border. To determine the minimum number of colors required to color the map, we find X(G). This graph has the property that it is planar. The Four Color Theorem states that all planar graphs can be colored using at most four colors. Now, suppose we relax the condition that there be no edges in each color class and allow certain configurations of edges. For example, in the committee scheduling application, we could allow each committee to have a time conflict with at most one other committee with a common member. In this case, each color class can contain isolated vertices and edges. We will call this problem the relaxed committee scheduling problem. 2
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This is an example of a generalization for graph coloring called conditional graph coloring. We define the conditional chromatic number X( G, P) of G with respect to a graph theoretical property P to be the minimum integer k such that there is a partition of the vertices into k sets so that the subgraph induced by each set has the property P. Another application for conditional coloring is the circuit manufacturing prob lem. A designer draws an electrical circuit to be manufactured. Several circuit boards sandwiched one on top of the other may be required to build the entire circuit since, for this circuit to work properly, each circuit board must be built without intersecting edges. Now, the manufacturer would like to determine the minimum number of required circuit boards. The problem can be modeled as a conditional graph coloring problem. Let each vertex in our graph represent a junc tion in the circuit. Draw an edge between two vertices in the graph if the junctions represented by those vertices are connected in the circuit design. Determining the minimum number of circuit boards required is equivalent to determining X( G, P), where P is the property that each color class be drawn with no intersecting edges, i.e., is planar. This particular conditional chromatic number is also known as the vertex thickness of a graph. For more information regarding vertex thickness, see Beineke and White [7] or Cimikowski [18]. For other electrical circuit problems, see Hutchinson [37] or Garey and Johnson [26]. We will study two types of conditional chromatic numbers. The first permits no paths of order j in a color class, and the second permits no cycles of order j in a color class. The relaxed committee scheduling problem is an example of conditional coloring with the property that no color class contains P3 Studying 3
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these two numbers may give insight into and/or bounds on other conditional chro matic numbers or the original chromatic number. It is often the case that the understanding of a generalization leads to a better understanding of the original concept. 1.1 Overview of Thesis Results This thesis will present some new results about the conditional chromatic number X(G, Cj) with respect to the property of having no cycles of a fixed length j and some new results about the conditional chromatic number X(G, Pj) with respect to the property of having no paths of a fixed length j 1. The first chapter will provide some basic definitions from graph theory and review the existing literature regarding X( G, CJ and X( G, ,pi). The second chapter will present and prove some basic results regarding X ( G, Cj) and X(G, Pj). Some of these results will be needed in later chapters. The third chapter will show that the problem of determining if a graph can be P3colored using k colors is NPcomplete. The fourth chapter will answer the question: given any graph with acyclic complement, what is X(G, Cj) and X(G, Pj)? A construction which illustrates that X( G, ,pi) X( G, Cj) 2:: a for any positive integer a is also provided. The fifth chapter will determine X( G, Cj) for all graphs of large size. Dargen and Fraughnaugh [21] characterized X(G, Cj) when a graph is missing at most j 2 edges. In this chapter, we extend this result to graphs missing at most j 1 edges. To accomplish this, we characterize all Hamiltonian graphs of order n with at least ( n 1) edges. 4
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The sixth chapter will determine X( G, Pj) for all graphs of large size and determine an upper bound on the size of G given a bound for X(G, Pj) To determine X( G, Pj) for all graphs of large size, we will characterize all graphs of order n with no Hamiltonian paths having at least (2n5) edges. 1.2 Definitions and Notation A graph G consists of a finite nonempty set V = V( G) of vertices and a collection E = E(G) of distinct pairs of vertices, called edges. Throughout this paper, let G be a graph. The size or number Of edges of G is denoted by e = e( G), and the order or number of vertices of G is denoted by n = n(G). A graph is simple if there is at most one edge between any distinct pair of vertices and there are no loops. For this paper, we will assume that all graphs are simple. If u and v are vertices of G, we write the edge joining u and vas uv and call u and v neighbors. The open neighborhood N ( u) of a vertex u is the set of neighbors of u, and the closed neighborhood N [ u] of u is N ( u) U { u}. The degree dG ( v) of a vertex is defined to be the number of edges in G incident with that vertex. When it clear to which graph we are referring, we may write d(v). The minimum degree o( G) is the minimum degree among all vertices of G while the maximum degree Ll(G) is the maximum degree among all vertices of G. If o(G) = Ll(G) = r, then all vertices have the same degree and G is regular of degree r or rregular. A 3regular graph is a cubic graph. If a vertex is not incident to any edge, then this vertex is an isolated vertex. A subgraph H of G is a graph having all of its vertices and edges in G, and we write H G. For any set S of vertices in G, the induced subgraph (S) is the 5
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maximal subgraph of G with vertex set S. If Sis a nonempty set of edges in G, then (S) is the subgraph whose edge set is S and whose vertex set is the set of ends of edges in S. A graph is complete if every pair of vertices is joined by an edge. The complete graph on n vertices is I
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the only vertex at level 0. Further, all adjacent vertices differ by exactly one level, and each vertex at level i + 1 is adjacent to exactly one vertex at level i. The maximum level is the height hu(T) of the tree. The complement G of G also has V(G) as its vertex set, but two vertices are adjacent in G if and only if they are not adjacent in G. A clique of G is a complete subgraph of G. The clique number w( G) of G is the maximum order among all cliques of G. An independent set of vertices of G is a set V such that xy E for all x, y E /. An independent set of edges of G has no two of its edges incident, and such a set is a matching. Next, we will discuss some operations defined on graphs. Let G1 and G2 be two graphs with disjoint vertex sets. The union G1 + G 2 of G1 and G2 has V(Gt + G2) = V(Gt) U V(G2 ) and E(G1 + G 2 ) = E(G1 ) U E(G2). In general, mG is the pairwise vertex disjoint union of m copies of G. The join G1 V G2 of two graphs G1 and G2 has V(G1 V G2 ) = V(G1 ) U V(G2 ) and E(G1 V G2 ) = E(Gt) U E(Gz) U {uv: u E V(G1 ) and v E V(G2)}. The cartesian product G1 X G 2 of two graphs G1 and G 2 has V(G1 X G2 ) = {(u,v) : u E V(Gt) and v E V(G2)}, and (u1,v1 ) and (u2,v2 ) are adjacent whenever either UtUz E E(Gt) and v1 = v2, or Ut = u2 and v1vz E E(G2). The strong product G1 o G2 of two graphs G1 and G2 has V ( G1 o G2 ) = { ( u, v) : u E V(Gt) and v E V(G2)}, and (ut, v 1 ) and (u2 v2 ) are adjacent whenever either u1u2 E E(Gt) and Vt = Vz, or Ut = Uz and VtVz E E(G2), or UtUz E E(Gt) and v1v2 E E(G2). The conjunction G1 1\ G 2 of two graphs G1 and Gz has V(Gt 1\ Gz) = {(u, v) : u E V(G1 ) and v E V(Gz)}, and ( Ut, Vt) and ( u2, v2) are adjacent whenever UtUz E E(Gt) and VtV2 E E(G2). 7
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A vertex coloring is an assignment of labels to the vertices of G so that any two adjacent vertices receive different labels. We think of each distinct label as a color and call each set of vertices a fixed color a color class. The vertex chromatic number X( G) of G is the minimum k for which G has a kcoloring, i.e., one using k colors. A graph is kcolorable if we can color it using k colors. An edge coloring is an assignment of labels (or colors) to the edges of G so that any two incident edges receive a different label. The edge chromatic number of a graph is defined to be the minimum number of colors needed to edge color G. If we use the term "chromatic number" or "coloring" in this paper, we mean vertex chromatic number or vertex coloring. A graph is said to be embedded in a surface S when it is drawn on S so that no two edges intersect. A graph is planar if it can be embedded in the plane; a plane graph has already been embedded in the plane. We will refer to the regions defined by a plane graph as its faces. The Four Color Theorem states that every planar graph is 4colorable. For a proof of this result, see Appel and Haken [5). A bipartite graph G is a graph whose vertex set V can be partitioned into two subsets Vi and "\12 such that every edge of G joins Vi and V2 The bipartition number b( G) of a graph G is given by b( G) = max{ e( B) : B G and B is bipartite}. For other basic graph theory definitions and terminology, the reader is referred to Harary [29). 8
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1.3 Conditional Coloring An equivalent definition of vertex coloring is a partition of the vertex set so that the subgraph induced by each set of this partition is an independent set. Stated differently, each color class contains no path on two vertices. This led several authors [15], [16], [35] to a more general concept of vertex colorings. Let P be any graph theoretic property. For example, P could be the property that a graph contains a clique of a certain order, that a graph does not contain a cycle of order 6, that a graph does not contain an induced cycle of order 6, or the maximum degree of a graph is 5. We define a Pcoloring of a graph to be an assignment of colors to its vertices so that the sub graph induced by each color class satisfies the property P. The P conditional chromatic number X( G, P) of G (or briefly Pchromatic number) is the minimum k for which G has a Pcoloring with k colors. When P is the property that a graph consists entirely of isolated vertices, the Pchromatic number is the usual chromatic number. When P is the property that a graph contains no subgraph (not necessarily induced) isomorphic to a graph F, we write X( G, F) for the Pchromatic number and refer to a Pcoloring as a .F coloring and the Pchromatic number as the Fchromatic number. If X(G, F) s; k, then we say G is .F kcolorable. When P is the property that a graph contains no induced subgraph isomorphic to a graph F, we write X(G, .F!) for the Pchromatic number and refer to a Pcoloring as a F!coloring and the Pchromatic number as the .F!chromatic number. The concept of partitioning the vertex set of a graph so that the subgraph induced by the vertices in each set in the partition has the property P seems to 9
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have been independently discovered by several authors around 1968 (see [15], [16], [35]) and again by several authors around 1985 (see [3], [4], [12], [14], [23], [30], [42]). Mathematicians have studied various conditional chromatic numbers since 1968. In 1968, Chartrand, Kronk, and Wall [16] studied a conditional coloring number called the arboricity of a graph where color classes are acyclic. Hedetniemi [35] proved that the arboricity of a planar graph is at most three. In 1968, Chartrand, Geller and Hedetniemi [15] proved several results about X(G, Pi) For example, if the diameter of G is d, then X(G, Pi) s; dj + 3 for j 2. Further, for every j 2, the authors constructed a planar graph G such that X( G, Pi) = 4. This proved that there is no stronger result than the Four Color Theorem for X( G, ,pi) for j 2. Also, in 1968, Sachs and Schauble (48] proved that given j 2 and /{ k 1, there exists a graph G with X( G, ,/{i) = k and a ,/{i Kcoloring of G containing at least k color classes isomorphic to /{il In 1969, Kramer and Kramer [39] discussed X(G, P), where Pis the property that the subgraph induced by each color class has minimum degree j for some fixed j 0. In 1970, Lick and White [41] published a paper with results for the same conditional chromatic number. In 1975, Cook [19] studied X( G, K1,i) for j 1. In 1977, Harary and Kainen [34] discussed X(G, ,1{3 ) for planar graphs. Also, LesniakFoster and Straight [40] published results for X(G, P), where P is the property that each color class induces a complete graph or a graph with no edges. Sampathkumar, Prabha, Neeralagi, and Venkatachalam (50] published results for X( G, ,pi) for j 2. In 1978, Beineke and White (7] discussed the thickness of a graph. The concept 10
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of thickness is equivalent to X( G, P), where P is the property that each color class is planar. The authors determined the thickness of complete and complete bipartite graphs. In 1985, Harary [30] published a paper providing an overview of conditional coloring. This paper discussed the conditional vertex and edge chromatic numbers for several different properties and has become the standard for conditional coloring terminology. In 1985, Andrews and Jacobson [3] studied X(G, 6.t), where the maximum vertex degree in each color class is at inost t. The authors related X( G, 6.t) to X (G) by proving that X ( G, 6.t) X (G)/ ( t + 1). Furthermore, they showed X( G, 6.t) n2 j(tn + n2 2e), where n and e are the order and size of G. Harary and Fraughnaugh [31) also published results for X( G, 6.t) in 1985. Also in 1985, Mynhardt and Broere [42] discussed.conditional coloring with re spect to the property that each color class is a disjoint union of complete subgraphs. They also studied the problem of finding a graph subject to certain restrictions for which the conditional chromatic number is arbitrarily large. Mynhardt and Broere [12] also studied conditional coloring with respect to the property that each color class has no induced subgraph isomorphic to a graph F. In 1986, Harary and Fraughnaugh [32) generalized the concept of bipartition number to that of a conditional bipartition number. Specifically, given a property P, a graph is conditionally bipartite with respect to P if V (G) is the disjoint union of sets X and Y where the induced subgraphs (X) and (Y) both have property P. The conditional bipartition number b(G, P) is max{e(B) : B G and B is conditionally bipartite with respect to P}. The authors studied b( G, P) for several 11
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minimum and maximum degree properties. Also in 1986, Domke, Laskar, Hedetniemi, and Peters [23] discussed X(G, P), where Pis the property that the subgraph induced by each color class is a complete rpartite graph for any rand X(G,Q), where Q is the property that each color class is a disjoint union of complete subgraphs. In 1987, Andrews and Jacobson [4] published a paper with results for X(G, .6.t) Also in 1987, Brown and Corneil [14] studied conditional coloring with respect to the property that each color class has no induced subgraph isomorphic to a set of graphs. In 1989, Akiyama, Era, Gervacio, and Watanabe [1] discussed the kpath chro matic number. The kpath chromatic number X(G, Pk) of G is the smallest number c of distinct colors with which V( G) can be colored such that each connected com ponent of (Vi) is a path of order at most k, 1 i c. The authors proved that X(G, Pk) f !f' l for rregular graphs. Since the 2path chromatic number is equivalent to our ,P3chromatic number, we immediately get that X(G, ,P3 ) 2 for cubic graphs. Also in 1989, Albertson, Jamison, Hedetniemi, and Locke [2] discussed conditional coloring with respect to the property that each color class is a disjoint union of complete subgraphs. In 1990, Baldi [6] published results for X( G, ...,pj) for j 2. In 1991, Harary and Hsu [33] provided theorems relating the conditional chro matic number of the Cartesian product, join, strong product, and conjunction of two graphs to the conditional chromatic number of the original pair of graphs. In 1992, Borodin [11] discussed the kcyclic chromatic number. A coloring of the vertices of a planar graph is kcyclic if whenever two vertices lie in the 12
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boundary of the same face of size at most k, their colors are different. There are also conditional coloring papers published in 1992 where a color class is not permitted to contain some induced subgraph. For example, Brown and Corneil [13], discussed uniquely H! kcolorable graphs where His any graph. A graph G is uniquely kcolorable if G is kcolorable and there is only one kcoloring (up to a permutation of colors). In fact, Brown and Corneil conjectured that for all graphs of order at least two and for all nonnegative integers k, there exist uniquely H! kcolorable graphs. So far, they have shown this result whenever G is 2connected or G is 2connected. In 1992, Johns and Saba [38] considered X(G,Pj). They proved that given integers l 1 and j 2, there exists a graph G such that X(G, Pj) = l, namely K(jt)l Also, there exists a graph such that X( G, ,pi)X( G, Pj+t) = l, namely J{j(j+t)l In 1993, Dargen and Fraughnaugh [21] determined X(G, CJ for graphs missing up to j 2 edges and Sampathkumar [49] discussed X(G, P), where P is the property that the subgraph induced by each color class has independence j for some fixed j 0. Also in 1993, Hutchinson [37] applied thickness results to testing printed circuit boards for electrical shorts. In 1995, Cimikowski [18] presented some heuristics for the graph thickness problem, i.e., decomposing a graph into the minimum number of planar subgraphs. The heuristics are based on some algorithms for finding a maximal planar subgraph of a nonplanar graph. He also proved that T(G)::; + 3/2J, where T(G) is the thickness of G and e is the size of G. For a survey of different properties studied, who studied them, and when they 13
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were studied, see Table 1.1. Table 1.1. Conditional Coloring Properties studied Color Class Property Year Reference disconnected or trivial 1968 Hedetniemi [35] 1970 Hedetniemi [36] acyclic 1968 Chartrand, Kronk, Wall [16] 1968 Hedetniemi [35] has no /{3 1977 Harary, Kainen [34] has no /{j for some fixed j 1968 Sachs, Schauble [48] has no Pj for some fixed j 1968 Chartrand, Geller, Hedetniemi [15] 1977 Sampathkumar, Prabha, Neeralagi, Venkatachalam [50] 1990 Baldi [6] 1992 Johns, Saba [38] complete or a graph without edges 1977 LesniakFoster, Straight [40] has maximum degree j for some 1985 Harary, Fraughnaugh [31] fixed j 1985 Andrews, Jacobson [3] 1986 Harary, Fraughnaugh [32] 1987 Andrews, Jacobson [4] complete rpartite graph for any r 1986 Domke, Laskar, Hedetniemi, Peters [23] has no induced subgraph 1985 Broere, Mynhardt [12] isomorphic to graph F 1985 Mynhardt, Broere [42] 1987 Brown, Corneil [14] has no induced subgraph lSO1987 Brown, Corneil [14] morphic to any graph F in a set of graphs F disjoint union of complete 1985 Mynhardt, Broere [42] sub graphs 1986 Domke, Laskar, Hedetniemi, Peters [23] 1989 Albertson, Jamison, Hedetniemi, Locke [2] contains no K1,j 1975 Cook [19] minimum degree j for fixed j 1969 Kramer, Kramer [39] 1970 Lick, White [41] disjoint union of paths of order at 1989 Akiyama, Era, Gervacio, Watanabe most k [1] j independent for fixed j 1993 Sampathkumar [49] 14
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Table 1.1. Conditional Coloring Properties studied (cont.) Color Class Property Year Reference has no Cj for fixed j 1993 Dargen, Fraughnaugh [21] planar 1978 Beineke, White [7] 1993 Hutchinson [37] 1995 Cirnikowski [18] kcyclic 1992 Borodin [11] contains no induced subgraph iso1992 Brown, Corneil [13] rnorphic to H survey of properties 1985 Harary [30] In this thesis, we concentrate primarily on two conditional chromatic numbers, X(G, Ci) where Ci is a cycle of order j 3 and X(G, Pi) where Pi is a path of order j 2. 15
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2 Basic Results for x(G, Pj) and x(G, Cj) This chapter provides the necessary background for the topics presented in Chapters 3, 4, 5, and 6. We start by proving some basic relationships for X( G, .Ci) and X(G, Pi) The following straightforward result has been known since 1968 [15]. Theorem 2.1 Let G be a graph of order n. If j 21 then X( G, ,pi) :::; r l Proof. Let n = a(j1)+r, where 0:::; r < j1. Color Gas follows: create a color classes of size j 1 and one color class of size r if r > 0. Hence X( G, ,pi) :::; r i::1 l 0 Corollary 2.1 If j 21 then X(Kn, Pi) = Proof. Since a complete subgraph of order at least j contains Pi, each color class of Kn can contain at most j1 vertices. Therefore, X(Kn, Pi) and equality follows from Theorem 2.1. D The next theorem provides the most basic relationship between X(G, Ci) and X( G, ,pi). Theorem 2.2 If j 3 and G is a graph1 then X(G, Ci) :::; X(G, Pj) Proof. Let C be any minimum Picoloring of G. Since a color class that contains no Pi contains no Ci, the coloring C is also a Cicoloring of G. So, X ( G, Ci) :::; D Now that we have an upper bound for X( G, ,pi) and a relationship between X(G, Ci) and X(G, Pj), we can find an upper bound for X(G, Ci) 16
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Corollary 2.2 If j 3 and G is a graph of order n, then X(G, Cj) :=; I Further, X(Kn,'Ci) = Proof. By Theorem 2.2 and Theorem 2.1, we get X(G, Cj) ::::; X(G, Pj) :=; I Let G = Kn. A subgraph induced by j or more vertices of Kn contains Cj, and therefore cannot be a color class of Kn. Hence, X(Kn, Cj) I 0 The next theorem relates two different conditional coloring numbers. Theorem 2.3 If j k 2, then X(G, Pi)::; X(G, .Pk) Proof. Let C be any minimum .Pkcoloring of G. Since a color class that con tains no Pk contains no Pj for k :=; j, the coloring C is also a Picoloring of G. Therefore, X(G, Pi) ::::; X(G, .Pk) 0 One might ask if the above type of theorem is true for X( G, Cj ). In fact, there is no relationship in general in either direction. For example, X( C4 .C4 ) = 2 and X(C4, .C3) = 1, which implies that X(G, .C4 ) 1:. X(G, .C3 ) for at least one graph. Further, X(C3, C4) = 1 and X(C3, C3 ) = 2, which implies that X(G, C4 ) 'l X( G, .C3 ) for at least one graph. The next theorem provides a relationship between the conditional chromatic numbers of a graph and its subgraphs. Theorem 2.4 Let G be a graph with H G. If j > 2, then X(H, Pj) < X(G, .Pi) If j 3, then X(H, Cj) ::::; X(G, .Ci) Proof. Let C be any minimum Cj(Pj)coloring of G. Since H G, the coloring C is also a Cj(Pj)coloring of H. So, X(H, .Ci) :=; X(G, Cj) and 0 17
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Now that we have results for subgraphs for X(G, Cj) and X(G, Pj), we can derive an elementary lower bound for X(G, ;Pj) and X(G, Cj) Theorem 2.5 If G is a graph and j 2:: 3, then X( G, Pi) 2:: X( G, Ci) 2:: r l Proof. First of all, Kw(G) By Corollary 2.2, X(Kw(G), Ci) = By Theorem 2.4, we get X(G, Cj) 2:: X(Kw(G), Ci) = By Theorem 2.2, we have X(G, Pi) 2:: X(G, Ci) o This bound is attained when G = Kn, and therefore the bound is tight. In order to derive some upper bounds on the order of a color class of some graph G, we need some well known results regarding Hamiltonian graphs, Hamiltonian connected graphs, and graphs which have a Hamiltonian path. First, we state Ore's Theorem, Dirac's Theorem, and another theorem which appears as Corollary 4.6 in Bondy and Murty [10]. Theorem 2.6 {Ore [44]) If G is a graph of order n 2:: 3 such that for all distinct nonadjacent vertices u and v, d( u) + d( v) 2:: n, then G is Hamiltonian. Theorem 2. 7 (Dirac {22}) If G is a graph of order n 2:: 3 and each vertex has degree at least then G is Hamiltonian. Theorem 2.8 {Ore {45}, Bondy {9]} If G is a graph of order n 2:: 3 and e( G) then G is Hamiltonian. Moreover, the only nonHamiltonian graphs with n vertices and n2;n edges are K1 V (K1 + Knz) and Kz V I3. The following theorem is also well known. Parts ( i) and (iii) appear as an exercise in West [51] and part (ii) is simply a restatement of the first sentence in Theorem 2.8. 18
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Theorem 2.9 Let G be a graph of order n. If n 2 and e( G) (;) ( n 4), then G is Hamiltonian coTJ,nected. ( i) Ifn 3 and e(G) (n3), then G is Hamiltonian. (ii) Ifn 2 and e(G) (n2), then G has a Hamiltonian path. (iii) Proof. ( i). (by induction on n). If n ::; 3, then ( i) is vacuously true. If n = 4, then G = K4 and a complete graph is Hamiltonian connected. If n = 5, then G E {K5 K5e} where e is an edge. These two graphs are Hamiltonian connected. If n = 6, then G E {K6 K6 e, K6 E(2K2), K6 E(P3)} = g where e is an edge. Each G E g is Hamiltonian connected. So assume ( i) holds for appropriate graphs on n 1 vertices. Let G be a graph on n 7 vertices and e( G) (;) ( n 4). We will show G is Hamiltonian connected. Let { u, v} V (G) with u =f v. If d(u)::; n 2, then e(Gu) e(G)(n2) (n 4)(n 2) = (n;l) (n5) and, by the induction hypothesis, Gu is Hamiltonian connected. Now e(G) (n4) implies that o(G) 3 for we would need to remove at least n3 edges from a complete graph in order for o(G) = 2. Since d(u) 3, we can choose z E N ( u) { v} and add uz to a Hamiltonian ( z, v )path in G u to form a Hamiltonian ( u,v )path in G. If u is adjacent to every other vertex in G (i.e., d(u) = n 1), then e(Gu) = e(G)(n1) (;)(n4)(n1) = (n;l)(n4) and, by Theorem 2.8, Gu is Hamiltonian. Break an edge involving v (say vw) on the Hamiltonian cycle in Gu and add the edge wu to obtain a Hamiltonian (u,v)path in G. In either case, G has a Hamiltonian ( u, v )path. Therefore, G is Hamiltonian connected. (ii). For a proof of Ore's Theorem, see Roberts (47]. 19
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(iii). If G is complete, then G is Hamiltonian. So, assume G is not complete. Since e(G) (;) (n2), we get n 3. Let uv be any missing edge of G. Consider G + uv. Now, e(G + uv) = e(G) + 1 (;)(n3). By (ii), the graph G + e has a Hamiltonian cycle. Therefore, G has a Hamiltonian path. 0 We can see that all three statements in the previous theorem are exact since K2 V (K1 + Kn3 ) has (;) (n 3) edges and is not Hamiltonian connected, K1 V(K1 +Kn2 ) and K2 VIs have(;) (n2) edges and are not Hamiltonian (in fact, Theorem 2.8 points out that these are the only such graphs), and K1 + I
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P3colored using k colors? In the next chapter, we show that this problem is NPcomplete. 21
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3 NPcomplete A polynomial time algorithm is an algorithm whose worstcase running time is O(nk), where the input to the algorithm is of cardinality nand k is some.constant. A problem II is defined to be a binary relation on a set I of problem instances and a setS of problem solutions. For example, consider the problem SHORTEST PATH of finding a shortest path between two given vertices in G. An instance for SHORTESTPATH is a triple consisting of a graph and two vertices. A solution is a sequence of vertices in the graph G with perhaps the empty sequence denoting that no path exists. The problem SHORTESTPATH itself is a relation that associates each instance of a graph and two vertices with a shortest path in the graph that connects the two vertices. For simplicity, the theory of NPcompleteness restricts itself to decision problems, those having a yes/no solution. In this case, we can view an abstract decision problem as a function that maps the instance set I to the solution set {0, 1}. For example, a decision problem related to SHORTEST PATH is as follows: Given a graph G, two vertices {u,v} V(G), and a positive integer k, does there exist a path of length at most k '? There are decision problems which can be solved in polynomial time and those which require superpolynomial time. A polynomial time solvable problem is one which can be solved using a deterministic polynomial time algorithm. The complexity class NP is the class of problems where given a "yes" solution, we can verify this solution in polynomial time. A basic idea in the theory of NP completeness is that of a polynomial transformation. Let II1 and II2 denote two 22
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decision problems. We say that there is a polynomial transformation from Ih to II2 written II1 ex II2 if the following two conditions hold: (a) There exists a function F transforming any instance I of II1 to an instance F(I) of II2 such that the answer to I with respect to II1 is "yes" if and only if the answer to F(I) is "yes" with respect to II2 (b) There exists an polynomial time algorithm to compute F(I). A decision problem II is NPcomplete if II E N P and for every problem II' E N P, II' ex II. The kCOLORING problem is stated as follows: given a graph G = (V, E) and integer 3 k lVI, is G kcolorable? We know from Garey and Johnson [27] that the kCOLORING problem is NPcomplete for k 3. The goal of this chapter is to determine how difficult it is to solve the problem of conditionally coloring a graph. Observe that the .P2 kCOLORING problem is the usual kCOLORING problem. Does relaxing the condition for each color class being an independent set to each color class containing no Pj for j 3 change the computational complexity of the colorability problem? We will show the answer is "no" when j = 3. The .P3 kCOLORING problem is stated as follows: given a graph G = (V, E) and integer 3 k flfll (Note: By Theorem 2.1, a graph can always be .P3colored with at most colors), does there exist a kpartition of the vertex set so that each subgraph induced by a set in the partition does not contain P3 as a subgraph (not necessarily induced)? Theorem 3.1 The .P3 kCOLORING problem II is NPcomplete 3. Proof. Let G = (V, E) be a graph of order n, and C a .P3 kcoloring of G. The following is a .polynomial time algorithm which can be used to check that C is a 23
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valid P3coloring of G. Examine all triples of distinct vertices in each color class and determine whether the graph induced by each triple contains P3 Determining if a triple contains P3 is 0(1). Since each set in the partition can contain at most n vertices, there are O(n3 ) triples. This algorithm is of order n3 and this shows that II E NP. Next, we will show that a known NPcomplete problem can be transformed to II. Construct a graph F(G) from G = (V, E) as follows: for each v E V, let F( G) contain two vertices vi and v2 We say that VI and v2 are associated with the vertex v. In F(G), join VI and v2 to the vertices associated with each neighbor of v and join VI to v2 Observe that this transformation is polynomial. See Figure 3.1 for the transformation of G = C4 WewillshowthatX(G):::; kimpliesthatX(F(G),P3):::; k. AssumeX(G):::; k. X W D u v Figure 3.1. The construction of F( C4). Let H = F( G) and C be a kcoloring of G. Color H as follows: For each vertex v E V, assign C ( v) to the two vertices associated with v in H. This coloring uses at most k colors to color H. To see that this coloring is a valid of H, 24
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suppose, by way of contradiction, that a color class of H contains P3 = uvw. By construction, abE E(H) if and only if either a and bare associated with the same vertex in G or there exists cd E E( G) such that a is associated with c and b is associated with d. Therefore, if two adjacent vertices in H are assigned the same color, then they must be associated with the same vertex in G since G has a valid kcoloring. Thus, uv E E(H) implies that there exists c E V(G) such that u and v are associated with c. Similarly, vw E E(H) and v being a vertex associated with c imply that v and w are associated with c. This is a contradiction since, by construction, there are exactly two vertices in H associated with a vertex in G. Therefore, X(H, P3 ) k. Next, we will show that X(H, P3 ) k implies that X( G) k. Let d: V(H) 7 {1, 2, ... k} be a P3 kcoloring of H. We color the vertices of G using Algorithm 3.1 presented in Figure 3.1. Let c: V(G) 7 {1,2, ... ,k} be the function that assigns colors to the vertices of G based on Algorithm 3.1. Note 1: Observe that Algorithm 3.1 assigns a color to z E V(G) from one of the colors assigned to the two vertices in H associated with z. Note 2: For every v E V( G) with associated vertices VI and v2 there is at most one w E N H( VI, v2 ) colored c( v). Suppose by way of contradiction that there exist x,y E V(H) such that {x,y} NH(v17v2 ) and d(x) = d(y) = c(v). By Note 1, we know that either VI or v2 is colored c(v), say VI But XVIY forms P3 in H, which is a contradiction. Each time through the loop (Lines 722), a vertex in G is assigned a color in either Line 9 or Line 17. Since the number of vertices in a graph is finite, all the vertices of G are assigned a color. We will show that c is a valid kcoloring of G. 25
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Input: Graphs G and H, and a P3coloring d: V(H) + {1, 2, ... k} of H. Output: A kcoloring c : V( G) + {1, 2, ... k} of G. 1. For every z E V(G) with associated vertices z1 and z2 in H. 2. If d(z!) = d(z 2), then let c(z) = d(z!). 3. If d(z1 ) tt d(N(z1)) and z is uncolored, then let c(z) = d(z!). 4. If d(z 2 ) tt d(N(z 2)) and z is uncolored, then let c(z) = d(z2). 5. If G has uncolored vertices, then 6. Let previous_color = 1 and z be an uncolored vertex in G. 7. Repeat steps 822 until all vertices of G are colored. 8. If previous_color =J. d(z1), then 9. let c(z) = d(z1), 10. let previous_color = d(z1), and w be the unique vertex in V(G) with which the vertex in NH(z1 ) with color d(z!) is associated. 11. If w is uncolored, then 12. let z = w. 13. else if there are uncolored vertices in G, then 14. let previous_color = 1 and z be an uncolored vertex in G. 16. else (previous_color = d(z1)) 17. let c(z) = d(z 2), 18. let previous_color = d(z2), and w be the unique vertex in V(G) with which the vertex in NH(z 2 ) with color d(z 2 ) is associated. 19 If w is uncolored, then 20. let z = w. 21. else if there are uncolored vertices in G, then 22. let previous_color = 1 and z be an uncolored vertex in G. 23. Output c and stop. Figure 3.1. Algorithm 3.1 26
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Let v E V(G) with c(v) =a, u E N(v), vi and v2 be the vertices in H associated with v, and ui and u2 be the vertices in H associated with u. We will show that c(u) =1a. Assume v was assigned color a in Line 2 of Algorithm 3.1. Now, u cannot be assigned color a or else the subgraph induced by the color class in H containing VI, v2 and one of the vertices associated with u would contain P3 Thus, all vertices assigned a color in Line 2 of Algorithm 3.1 receive a valid color. Assume v was assigned color a in Line 3 of Algorithm 3.1, that is, d(vi) =a and the neighbors of VI (which include Ui and u2 ) are not colored a. Since u is assigned its color from one of the colors assigned to UI or u2 (Note 1), we have c( u) =1a. Thus, all vertices colored in Line 3 receive a valid color. A similar argument proves that every vertex colored in Line 4 receives a valid color. We have just shown that if u or v were assigned a color in Lines 2, 3, or 4, then c( u) =1c( v). So assume u and v were assigned colors in Lines 9 or 17 of Algorithm 3.1. Since only one vertex is colored at at time, let's assume that u was already colored when v is colored. Suppose that c( u) = c( v) = a. Assume u was assigned color a m Line 9 of Algorithm 3.1. In the loop (Lines 722) with z = u, in Line 9 we have c(u) = d( ui) = a and we let previous_color = a (Line 10). Since we assumed that c( v) = a, either vi or v2 is colored a (Note 1). Further, since there is at most one neighbor of ui colored a (Note 2), we must have z = v (Line 12), i.e., v is the next vertex to be colored in the loop (Lines 722). When we go through the loop to color v, either it gets colored in Line 9 or 17. If v is colored in Line 9, then since previous_color = a, we know that the color of c( vi) =f. a (Line 8) and c( v) =1a 27
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(Line 9). So v must be colored in Line 17 and we know that d( vt) = a (Line 16) and c(v) = d(v2 ) =a (Line 17). Now, d(v2 ) =f. a or else v would have been colored in Line 2. Thus c( v) =f. a, which contradicts the assumption that c( v) = a. By applying the previous argument replacing x1 with x2 and starting with z = x in Line 17, we again reach a contradiction. Thus, for every adjacent u, v E V(G), we have c(u) =f. c(v), which implies that Algorithm 3.1 produces a kcoloring of G. We have transformed a known NP complete problem to our problem. Therefore, the ..,p3 kCOLORING problem is NPcomplete. D This construction for H does not help to prove that the ..,pi kcolorability problem is NPcomplete for j 2: 4. We have tried other constructions for H for the ..,pi kcolorability problem with no success. Suggested areas for future research include determining whether the ,pi kcolorability problem, the ,Cj kcolorability problem, and the ,Kj kcolorability problem are NPcomplete for j 2: 3. Now, we turn our attention to finding X( G, ,pi) and X( G, ,Cj) for graphs with acyclic complements. 28
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4 Determining the conditional chromatic number for graphs with acyclic complements In this chapter we will show that X( G, Cj) and X( G, ,pj) may have different values for many graphs. First, we will examine the values of X( G, Cj) for graphs whose complements are acyclic and then perform a similar analysis for X( G, ,pi). Finally, the main theorem of this chapter shows that the difference between X( G, Ci) and X( G, Pj) can be made arbitrarily large in a particular family of graphs. 4.1 Determining x(G, Cj) when G is acyclic To determine X( G, Cj) for j 2: 3, it is necessary to determine how large any color class can be in a minimum Cjcoloring of G. If j = 3, then we will see the largest color class must be of size 4 or less and if j 2: 4, then the largest color class must be of size j or less. Therefore, we determine X( G, C3 ) and X( G, Cj) for j 2: 4 in two separate theorems. We first address the size of the largest possible color class for X(G, C3 ) with the following lemma. Lemma 4.1 If G is a graph of order n 2: 5 and G is acyclic, then G contains C3 as a subgraph. Proof. Let G be a graph of order n 2: 5 with acyclic complement. Let H be a sub graph of G with five vertices. We may assume fi is a tree (otherwise we remove edges from H until ii is a tree). If ii has a vertex v such that dn(v) 2: 3, then since ii contains no cycles, N n( v) is an independent set in ii of size at least 29
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3. If every vertex in fi has degree at most 2, then fi = P5 which clearly contains an independent set of size 3. In either case, the vertices from the independent set in fi forms C3 in H. Since H G, the graph G contains C3 o The following theorem evaluates X( G, C3 ) when G is acyclic. Theorem 4.1 Let G be a graph with acyclic complement. Let m be the number of edges in a maximum matching of G. If m 0, then X(G, 'C3) = r l Proof. Let A be a color class in a C3coloring of G. By Lemma 4.1, IAI :::; 4. Let a be the number of color classes of size 4 and b the number of color classes of size 3. The only C3free graphs of order 4 whose complements are acyclic are P4 and C4 Each of these graphs is missing two independent edges. Therefore, the subgraph induced by each color class of size 4 must be missing two independent edges. The only C3free graphs of order 3 whose complements are acyclic are P3 and /{2 + /{1 Each of these graphs is missing an edge. Therefore, the subgraph induced by each color class of size 3 must be missing an edge. Since color classes are disjoint, we can form a matching M1 in G with a pair of edges from each color class of size 4 and one from each color class of size 3. Thus I M11 = 2a + b, and since m is the size of a maximum matching, we get 2a+b= IMII :::;m. Thus, X(G,C3) rn4;3bl +a+b= rn22abl To construct a coloring, let M be a maximum matching in G. We form as many color classes of size 4 as possible by using the endpoints of two edges in M in each, then based on the parity of m, zero (if m is even) or one (if m is odd) color class of size 3 using the endpoints of one edge in Mandan additional vertex, and finally as many color classes of size 2 as possible. With this construction, the 30
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graph induced by a color class of size 4 is either C4 or P4 Further, the graph induced by a color class of size 3 is either P3 or 1<2+1<1 Each of these graphs is C3free and are valid color classes. If m is even, there are ; color classes of size 4, and the rest are of size at most 2. With this coloring, we get X(G,C3)::; +; = If m is odd andn > 2m, there are m;1 color classes of size 4 and one color class of size 3. The rest are of size at most 2. With this coloring, we get X( G, C3 ) ::; r n4(r)3l + m;1 + 1 = r l If m is odd and n = 2m, there are m;1 color classes of size 4 and one color class of size 2. With this coloring, X( G, C3 ) ::; m;1 + 1 = mt1 = r; l = r l 0 Next we will complete the study of graphs whose complements are acyclic by determining X(G, Cj) for j 2:: 4. Again, we need to determine how large any color class can be in a minimum Cjcoloring of a graph. Recall that Theorem 2.10 guarantees cycles of all orders if a graph is missing at most j3 edges. The following theorem shows that we still have long cycles for some graphs missing up to n1 edges. Lemma 4.2 Let G be a graph of order n 2:: 5 with acyclic complement. Then Cn1 G. Furthermore, if I<1,n2 !l G, then G is Hamiltonian. Proof. We may assume G = T where T is a tree and K1,n_2 T only if I<1,nz G. (Otherwise, remove edges from G until we get a graph whose complement is a tree. Since n 2:: 5, this can be done without creating K1,n_2 in the complement.) Choose a root vertex v0 ofT so that the height of the tree hv0(T) is maximum. 31
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Since n 5, the diameter of the tree T is at least 2 and therefore, hvo (T) 2. Further, dr(v0 ) = 1 and there is exactly one vertex v1 at Ievell. Let TE be the subgraph of G induced by the vertices on even levels ofT, and To be the subgraph of G induced by the vertices on odd levels of T. Since TE and To are complete graphs, every induced subgraph of TE or To is Hamiltonian connected. Further, all vertices on level i are adjacent to all vertices on levels i2, i3, ... 1, 0 in G. In each of the following cases, we will show Cn1 G and either Kl,n2 T or Cn G. Case 1. hv0 (T) 4. Let Vi be a vertex on the tree at level i for i = 2, 3, 4. Now, a Hamiltonian ( v0,v4)path from TEv 2 together with v 4 v 1 together with a Hamiltonian (v 1,v3 )path from To together with v3v0 forms Cnl Further, a Hamiltonian (vo,v4)path from TE together with v4v1 together with a Hamiltonian (v1,v3)path from To together with v3v0 forms Cn Case 2. hv0(T) = 3. Let w 1,w2, ... be the vertices on level2 and ... the vertices on level 3. Case 2A. The number of vertices in level 2 IS 1. Since n 5, there are at least two vertices x1 and x2 on level 3. Then Cnl is formed in G by using a Hamiltonian (x1,x2)path in To together with x1v0x2. In this case, we have K1,n2 T induced by the vertices on levels 1, 2, and 3. Case 2B. The number of vertices in level 2 is at least 2. Since hv0 (T) = 3, there is at least one vertex x1 on level 3. Assume without loss of generality that WIXl E E(T). If level3 has only one vertex, then a Hamiltonian ( v0,w2)path from TE together with w2x1vo forms Cn1 G. Further, K1,n2 Tis induced by the vertices on 32
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levels 0, 1, and 2. Assume level 3 has at least two vertices x 1 x2 (and w1 x1 E E(T)). If x2w1 E E(T), then form Cn_ 1 G by using a Hamiltonian (v0,w2)path in TE together with w2x2 together with a Hamiltonian (x2,xi)path in Tov1 together with x1vo. A Hamiltonian cycle is formed in G by using a Hamiltonian (v0,w2)path in TE together with w2x2 together with a Hamiltonian (x2,x 1 )path in To together with Otherwise, without loss of generality assume x2w2 E E(T), and we can form Cn_ 1 G by using a Hamiltonian (v0,w1 )path in TE together with w1 x2 together with a Hamiltonian (x2,x 1 )path in To v1 together with x1v0 A Hamiltonian cycle is formed in G by using a Hamiltonian ( v0,w1 )path in TE together with w1 x2 together with a Hamiltonian (x2,x 1 )path in To together with x1v0 Case 3. hv0 (T) = 2. Then G = I<1,n1 and TE = I
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Suppose IAI = j. We will show K 1 ,j_ 2 (A). Suppose j = 4. Every graph of order 4 with acyclic complement contains either C4 or K1 ,2 Since A is a color class, we must have K 1 2 (A). If j 5, then by Lemma 4.2, we have K 1 ,j_ 2 (A). Let a be the number of color classes of size j. Since color classes are vertex disjoint, we get a::; m. Therefore, X(G,.Cj) lj=_;i1 +a= lj=.71 and X( G, Cj) max (I j=.71 I j 1). To show the inequality in the other direction, we produce a minimum Cj coloring. Consider {Ub u2, ... 'Um}, where each ui is a K1,j2 in G and the Ui's are pairwise vertex disjoint. LetS= V(G)V(Ui) Now, n = (j 1)m + s. We consider two cases: m ::; s and 0 ::; s < m. If m ::; s, then n mj, which implies that I j=.71 r j 1 For i = 1, ... 'm, let V(Ui) together with a vertex from S form a color class of size j (the subgraph induced by each such color class is Cjfree since it contains a vertex of degree at most one). We partition the remaining sm vertices into as many color classes of size j 1 as possible. Clearly each such subgraph is Cjfree. Thus, we have X(G C) < m + rsml = = rnml J 31 I 31 31 If 0 ::; s < m, then n < mj' which implies that r j l I j=.71 Let r = I mjs 1 and U = {U1, U2, ... Umr }. Further, let T = S u V(Ui)). For i = 1, ... m r, let V ( Ui) together with a vertex from T form a color class of size j. Note that mr = mf mjs 1 = l mir+s j = l]' j, and there are sufficient elements in T to form l]' J color classes since ITI = s + r(j 1) = s + r m_;s l (j 1) s + mjs (j 1) = j. If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Cjfree, we will show that IBI < j. 34
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Now, IBI ITIlyJ s+r(j1)(mr) rj(ms) ms < ( + 1 )j ( ms) J J If B = 0, then there were no remaining vertices after forming mr color classes of size j. Therefore, j divides n and X( G, ,Cj) l I J = I I l If B # 0, then j does not divide nand we get X(G, ,Cj)::; li J + 1 = I I l In either case, we have produced a ,Cjcoloring with I I l colors. 0 Now that we have determined X( G, ,Cj) for graphs with acyclic complements, the next section gives X( G, ,pi) for this same class of graphs. 4.2 Determining x(G, Pj) when G is acyclic The following lemma determines the longest path in a graph whose complement is acyclic. This lemma will be used in the main theorem of this section which determines X( G, ,pi) when G is acyclic. Lemma 4.3 If G is a graph of order n 2 with acyclic complement, then Pn1 G. Furthermore, if G # K1,n_1 then G has a Hamiltonian path. Proof. If n = 2, the result is trivial. If n = 3, then G is missing at most two edges and P2 Further, if G =J. K1 2 then G = P3 or G = K3 and both contain 35
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where e is an edge. All graphs contain P4 except K4 E(K1 3), which contains P3. So, assume n 5. By Lemma 4.2, the graph G contains Cn1 as a subgraph and therefore Pn1 G. If G =f. K1,nb then 8( G) 1 and the vertex in G not on Cn1 can be added to Cn1 to form Pn G. 0 The following theorem determines X( G, Pj) for graphs whose complements are acyclic. Theorem 4.3 Let G be a graph of order n with acyclic complement. Let m 0 be the maximum number of pairwise vertex disjoint copies ofK1,j1 in G. If j 2J then X( G, Pj) = I j=.71 Proof. Let A be a color class in a Pjcoloring of G. Notice that (A) is acyclic since G is acyclic. If IAI > j, then by Lemma 4.3, we get Pj (A), which contradicts the assumption that A is a color class. Therefore, IAI ::; j. Suppose IAI = j. Now, (A) is acyclic and isPjfree. Therefore, by Lemma4.3, we get K1,jl = (A). Let a be the number of color classes of size j. Since color classes are vertex disjoint, we get a ::; m. Therefore, X( G, Pj) I +a = r > r n_m1 I 31 I 31 To show the inequality in the other direction, we produce a minimum ..Pj coloring. Consider {U1 U2 Um}, where each Ui is a K1,j_1 in G and the Ui's are pairwise vertex disjoint. For each i = 1, ... m, form a color class of size j using V(Ui) (the subgraph induced by each such color class is Pjfree since it contains a vertex of degree zero). With the remaining vertices, form as many color classes of size j 1 as possible. With this coloring, we get X( G, Pj) ::; I 1 +m = I j=.71 Therefore, X( G, ..Pj) = r j=.71 o 36
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The following corollary will be used several times in upcoming chapters and is stated here for reference. Corollary 4.1 lfG is a graph with (E(G)) = K1,m and j 2:2, then ifO:::; m < j 1, if j 1 :::; m :::; n 1. Proof. If m < j1, then G contains no copy of K1,jl and by Theorem 4.3, X(G;.Pj) = If m 2: j 1, then G contains exactly one K1,j1 and by Theorem 4.3, X( G, ,pj) = I l 0 Now, we can state the main theorem of this chapter. 4.3 The difference between the Cjand Pjchromatic numbers of a graph Theorem 4.4 If a 2: 0, j 2: 4 and n 2: a(j 1 )j, then there exists a connected graph G of order n such that X(G, .Pi)X(G, .Ci) 2: a. Proof. This is a proof by construction. Let G be the disjoint union of Ina(j1)2 and a(j 1) pairwise vertex disjoint copies of K1,j_2 See Figure 4.1 for the construction of G when j = 5, a = 1 and n = 24. In this case, G consists of Is and four copies of K1 3 Further, X(G, .C5 ) = 5 and X(G, .P5 ) = 6. Observe that G contains a(j 1) pairwise vertex disjoint copies of K1,j_2 and no copy of KI,j_1 By Theorem 4.3, X( G, ,pi) = I l Since n 2: a(j 1 )j, Theorem 4.2 implies that X(G, .Ci) = Therefore, X(G, .Pi) X(G, .Ci) I l I l = I l I l + a = a. o Since we have shown these two conditional chromatic numbers may be different, we will address each separately in the following two chapters. 37
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Figure 4.1. The construction of G for j = 5, a= 1 and n = 24 Some possible directions for further research include answering the following questions. What happens if we remove the restriction that the complement is acyclic and examine all graphs missing n 1 edges or examine graphs whose complements are bipartite? Will the difference increase and, if so, by how much? Also, are there families of graphs where X( G, Pj)X( G, Cj) is 0( nk) for k ?: 1? The following questions also remain open. For the set of graphs of a fixed order n, how much can these two conditional chromatic numbers differ if we remove half of the edges from the complete graph? And, in general, given the size of G, what are upper and lower bounds for X( G, Pj)X( G, Cj )? Which graphs attain these bounds? 38
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5 Determining x(G, Next we address the problem: given a graph of order n withe edges, what are the bounds on the conditional chromatic number? In 1995, Dargen and Fraughnaugh published the following theorem. Theorem 5.1 (Dargen and FraughnaughJ {21}) If e(G) (j2)) then if G = K1,j2 or G = K3 and j = 5 otherwise. We can deduce from this theorem that we need to remove at least j 2 edges from a complete graph to obtain a graph whose Cjconditional chromatic number is f l A natural question one may ask is how many edges do we need to remove so that X( G, Cj) = r j=i l? We will show that this number can be obtained by removing only two edges from a complete graph when j = 3. In this chapter, we will determine which graphs are Hamiltonian given that the size of their complement is at most n 1. Using this information, the final theorem of this chapter will expand the knowledge of X( G, Cj) to the class of graphs missing exactly j 1 edges. 5.1 Preliminaries To prove the final theorem, we will first address some special cases. We will soon see that the special cases for the final theorem are graphs missing a star of a particular order and graphs missing pairwise vertex disjoint complete subgraphs. First, we will address graphs missing a star of a particular order. 39
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Theorem 5.2 Let G be a graph of order n. If j 2: 3 and e( G) 2: (;) (2j 6) and K1,j_2 G, then X( G, Cj) = f l Proof. Let X be a set of Vertices which induce K1,j_ 2 G. Since K1,j2 G and e( G) ::::; 2j 6, there is exactly one vertex va of degree at least j 2 in G. Therefore, va E X. Let A be a color class in a Cjcoloring of G. Suppose va rJ. A. If IAI 2: j, then (A) would be missing at most j 4 edges and by Theorem 2.10, we get Cj (A). Therefore, for A to be a color class, we must have IAI ::::; j1. Suppose va E A, and consider Ava. If lAval 2: j, then by the same argument, we get Cj (Ava). Therefore, lAval ::::; j1, which implies IAI ::::; j. Thus, there can be at most one color class of size j (one containing va), andX(G,Cj) 2: fj={l + 1 = To show the inequality in the other direction, form one color class B of size j by using the vertices in X together with one vertex from GX and form as many other color classes as possible of size j 1 with the remaining vertices in GX. Now (B) is Cjfree since va is adjacent to at most one other vertex in (B). Therefore, X(G,Cj)::::; fj=fl + 1 = fj=;l. D Next, we would like to determine X(G, Cj) when G is missing a set of mutu ally disjoint complete graphs of a specific order. To accomplish this, we use the following two lemmas. Lemma 5.1 Let G be a graph of order n 2: 3 such that G = Vf=1 Si where Si = Ir; (ri 2: 1,p 2: 1) for each i. Then either G is Hamiltonian or maxi{ri} 2: f ntll Proof. Let v E V(G). Then v E Si for somei and d(v) = (n1)(ri1) = nri. Therefore, since each vertex in G is in some Si, we get b(G) = nmaxi{ri} If 40
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maxi{ri} < f n!1l, then b"( G) > n f nt1l = l J. Thus, b"( G) 2:: n/2 and by Dirac's Theorem (Theorem 2.7), G is Hamiltonian. 0 Lemma 5.2 Let G be a graph of order n 2:: 4 such that G = Vf=1 Si where Si = Ir, (1 ::::; ri ::::; r I l 'p 2:: 1) for each i. Then either G contains an ( n1) cycle or there exist r1 and r2 such that rl = r2 = r l Proof. Assume r1 ::::; r l and Ti < r l when i i= 1. Let v E sl and G' = Gv. Then G' = Irl1 v (Vf=2 Si) and maxi;:::2{ri, rl1} < r I l By Lemma 5.1, G' is Hamiltonian, and therefore G contains an ( n 1 )cycle. 0 The following theorem determines X(G, Cj) when G is missing a set of mu tually disjoint complete graphs of a specific order. Theorem 5.3 Let G be a graph of order n. If j 2:: 3 and (E(G)) = then Proof. Observe that (E( G)) = mK I i.:}!1 if and only if G = Vi=1 2 Si where si =/ft.:}! l fori= 1, 2, ... 'm and si = /1 fori= m + 1, m + 2, ... 'nr 2 l m. Assume j is odd. Then r l = Let A be a color class in a Cjcoloring of G. Suppose that IAI > j + 1. Let B A such that IBI = j + 2. Then (B) = Vf=1 lq; where 1 ::::; QI ::::; (j + 1)/2, 1 ::::; q2::::; (j + 1)/2, and 1 ::::; Qi::::; (j1)/2 for 3::::; i::::; k since IBI = j + 2 implies that (B) can contain at most two copies of I i!. By removing one vertex from each 2 of Ig1 and Ig2 form C B such that ICI = j. Now, (C)= lq1 1 V lq21 V (Vf=3 lq,), 41
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and maXi{qi} ::::; j;1 By Lemma 5.1, (C) is Hamiltonian. This contradicts the assumption that (A) is Cjfree. So IAI ::::; j + 1. If !AI = j + 1, then by Lemma 5.2, either 2Ji! <;; (A) or (A) contains a 2 jcycle. Because (A) cannot contain Cj, we must have 2Jj+t <;; (A). Now, if 2 IAI = j, then by Lemma 5.1, we have Ji! <;; (A). Let a be the number of color 2 classes of size j + 1 and b be the number of color classes of size j in a Cj coloring of G. Since color classes are vertex disjoint, we get 2a + b ::::; m. Now, X(G C) 2: +a+ b = I n_2abl > I n_ml J I 31 I 31 I 31 To show the inequality in the other direction, we split the proof into three cases. Recall that G = Vi=1 Si where Si = I ill for 1, = 1, 2, ... m and Si = J1 2 .c . 1 2 (j+1 ) 10r 1, m + m + ... n 2 m. Case 1. m is even. Color Gas follows: For 1 ::::; l ::::; m/2, form a color class of size j + 1 by using all the vertices from S2t_1 and S2t. The subgraph induced by each of these color classes is isomorphic to M = Ji! V Ji..!., which is bipartite. 2 2 Since bipartite graphs only contain even cycles, M is Cjfree. With the remaining vertices, form as many other color classes of size j 1 as possible. With this 1 t x(G C ) < rn(j+l)Tl + m I nml co ormg, we ge j 1 j1 2 = 1 j1 Case 2. m is odd and n 2: m () + Color Gas follows: For 1 ::::; l::::; m;1 form a color class of size j + 1 by using all the vertices from S21_1 and S2t. Again, since the subgraph induced by each of these color classes is bipartite, it is Cjfree. Observe that n 2: m ( +? guarantees that there are at least j vertices not yet assigned to color classes. Form one color class C of size j using all the vertices from Sm together with? vertices from V(Si) Now, (C)= ]j+! V Ki.:2 and 2 2 the largest cycle we can form is Cj_1 obtained by alternately traversing between 42
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the vertices of K 2.2 and I j+1 until we use all the vertices in K j1 Therefore, (G) 2 2 2 is Gjfree. With the remaining vertices, form as many other color classes of size j 1 as possible. Then X(G,.Ci):::; + m;1 + 1 = fj=71 Case 3. m is odd and n < m + j;1 This means that 0 :::; nm < We will show that nm is not divisible by j1. Suppose that nm = z(j1) for some nonnegative integer z. Then 0 < nm (j; 1 ) j1 <2 iff 0 < (" ) (j+l) j1 z J 1 + mm 2 <2 iff 0 < 2z(j 1) + 2m mj m < j 1 iff 0 < 2z(jI)m(j1) < j1 iff 0 < (2z m )(j 1) < j 1 iff 0 < 2zm < 1 iff m 2z. However, this contradicts the assumption that m is odd. Therefore, nm is not divisible by j 1. This implies that r = r j=71 Now back to the proof. Form a new graph F by taking G and adding back in all the missing edges in S1 Now F is missing m 1 pairwise vertex disjoint copies of K d b k. C 1 S X(F C) < rn(m1)1rnm+11_ rnm1 41 an we are ac In ase o, , 3 j1 j1 j1 and by Theorem 2.4, X(G, .Ci):::; X(F, .Ci) = r j:=.71 This completes the proof when j is odd. ( '+2) r '+11 '+2 m+nm Next, let's assume j is even. In this case, 2..f' = T and G = vi=1 2 si where Si =It fori= 1, 2, ... m and Si = I1 fori= m+l, m+2, ... m. 43
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Let A be a color class in a .Cjcoloring of G. Suppose that IAI > j. Let B A such that IBI = j + 1. Then (B) = Vf=1 lq; where 1 :S q1 :S (j + 2)/2, and 1 :S qi :S j /2 for 2 :S i :S k since IBI = j + 1 implies that (B) can contain at most one copy of I H2. By removing one vertex from lq1 form C B such that 2 ICI = j. Now, (C)= Ig,1 V (Vf=2 lq;), and maxi{qi} :S By Lemma 5.1, (C) is Hamiltonian. This contradicts the assumption that (A) is Cjfree. So IAI :S j and X(G, .Cj) I I l If IAI = j, then since (A) is Cjfree, by Lemma 5.1 we have (A). Let a 2 be the number of color classes of size j in a .Cjcoloring of G. Since color classes are vertex disjoint, we get a :S m. Therefore, X( G, Ci) I nj.!:"; l + m = I j=.7l Hence, X( G, .Ci) max (I I l I j=.71) To show the inequality in the other direction, we produce a minimum Cj coloring. Let U = V(G)Ui:1 V(Si) Now, n = + s. We consider two cases: i.:;}m :S s and 0 :S s < i;2 m. If i;2m :S s, then n mj and I j=.71 I I l For 1 :S i :S m, form a color class of size j by using all the vertices from si together with j;2 vertices from u not yet assigned to a color class. Since i.:;}m :S s, U contains a sufficient number of vertices to form these m color classes. Now, all the subgraphs induced by these color classes are isomorphic to C = I ill V K cl and the largest cycle we can form is 2 2 Cj2 obtained by alternately traversing between the vertices of K ;2 and I i.:E until 2 2 we use all the vertices inK cl Therefore, Cis Cjfree. We partition the remaining 2 s i;2m vertices into as many color classes of size j 1 as possible. Thus, we h X(G C) < + rs(l?)mlrmi(l?)m+smlr(i)m+smlfnml ave . 3 m i1 i1 i1 1 j1 44
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If 0 :::; s < i1m, then n < mj and I I l 2: I j=:7l Let r = I (i?)ms l and T = U u V ( Si)). For 1 :::; i :::; m r, form a color class of size j by using all the vertices from Si together with i;2 vertices from T not yet assigned to a color class. Again these subgraphs are Cjfree since their longest cycle is Cj_2 Notethatmr=ml(i?Jmsl = lmi(if)m+sj = l()m+sj = liJ,and there are sufficient elements in T to form li J color classes since ITI = s+r = s + ( Jms l ( 2: s + ( Jms ( = ( (I) 2: ( l I j. Again, all the subgraphs induced by these color classes are isomorphic to C = I ill V K cl, 2 2 and we have already shown that Cis Cjfree. If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Cjfree, we will show that IBI < j. Now, IBI = ITIly j (j; 2 ) j+2 (j2) s + r(2)(mr) 2 s + rjm (j; 2 ) < s+ +l)jm(i;2) J If B = 0, then were no remaining vertices after forming mr color classes of size j. Therefore, j divides n and X( G, Cj) :::; l I j = I I l If B =f. 0, then j does not divide nand we get X(G,Cj):::; lJJ + 1 = IIl In either case, we have produced a Cicoloring with r j l colors. 0 45
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5.2 Determining x(G, ,Cj) when e(G) > (j1) To determine the ,Cjchromatic number when j 8 for graphs given the size of the complement is at most j 1, we first determine which graphs of order n 8 are Hamiltonian given the size of their complement is at most n 1. Theorem 5.4 Let G be a graph of order n 8 with e( G) :s; n 1. Then either Kt,n2 G or G is Hamiltonian. Proof. Assume K1,n_2 CJ:. G. If G = Kn, then G is obviously Hamiltonian. So assume G is not complete and let u and v be a pair of nonadjacent vertices. Since K1,n_2 CJ:. G, we must have d(u) 2 and d(v) 2. Let H = G{u, v }. We break the proof into cases based on the number of edges in G incident to u or v. Case 1. Suppose that the number of edges in G incident to u or v is n 1. Since e( G) = n 1, H is complete. If IN(u) n N(v)l = 0, then let x, y E N(u) and a, bE N(v) be distinct vertices. These vertices exist since d(u) 2 and d(v) 2. Now xuybva together with a Hamiltonian ( a,x )path from H{y, b} forms Cn So assume IN( u) n N( v) I 1. If IN(u) n N(v)l = 1, then let w, x, and y be distinct vertices in H such that w E N(u) n N(v), x E N(u) andy E N(v). Then xuwvy together with an (y ,x )path from H { w} forms Cn. So assume IN( u) n N( v) I 2. Since there are n2 edges missing from { u, v} into H out of a possible 2( n 2), there must be n 2 edges from { u, v} into H. Since n 8, there are at least six edges from { u, v} into H. Therefore, we can assume d(u) or d(v) is at least 3. Without loss of generality, assume d(u) 3. Let w,x,y E N(u), and w,x E N(v). Then xvwuy together with a Hamiltonian 46
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(y,x)path from H{w} forms Cn Therefore, we can assume that every such pair of vertices can be incident to no more than n2 edges in G. Case 2. Suppose that the number of edges in G incident to u or v is exactly n2. Since there are n 3 missing edges from { u, v} into H, If is missing at most one edge, and there are 2(n2)(n3) = n1 edges from {u,v} into H. Suppose IN( u) n N( v) I = 0. Then in order to haven 1 edges from { u, v} into Hand IN(u) n N(v)l = 0, the subgraph H needs to have at least n1 vertices. But H only contains n2 vertices, and we reach a contradiction. Therefore, we can assume IN(u) n N(v)l 1. Let wE N(u) n N(v). Consider J = G{ u, v, w }. Then J is a graph of order n 3 missing at most one edge. Since n 8, we have e(J) (n;3 ) 1 (n;3 ) (n7) and J is Hamiltonian connected by Theorem 2.9. If N( u) :f. N( v ), then since d( u) 2, d( v) 2,and n 8, there exist distinct vertices x,y E V(J) such that x E N(u) andy E N(v). Now xuwvy together with a Hamiltonian (x,y)path in J forms Cn. So we can assume that N ( u) = N ( v). Recall that there are n 1 edges from { u, v} into H. Therefore, d( u) = d( v) = n;t. The integer n must be odd in order for n;l to be an integer. Therefore, n 2: 9 and d( u) = d( v) 4. This implies that there exist distinct vertices x, y E V(J) such that x, y E N(u) n N(v ), x :f. w, and y :f. w. Then xuwvy together with a Hamiltonian (x,y)path in J forms Cn. Thus, we may assume that for every nonadjacent pair of vertices u and v, the number of edges in G incident to u or v is at most n3. 47
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Case 3. Suppose that the number of edges in G incident to every nonadjacent pair of of vertices u, v is at most n 3. There are at most n 4 edges missing from { u, v} into H. Therefore there are at least 2(n2)(n4) = n edges from {u,v} into H. So, d(u) + d(v) since u and v are arbitrary nonadjacent vertices, by Ore's Theorem (Theorem 2.6) G is Hamiltonian. 0 Now, armed with Theorems 4.1, 4.2, 5.2, 5.3, and 5.4, we can proceed with determining X(G, .. ci) when e(G) (;) (j1). We begin by determining X(G, .. ci) when j 8. Theorem 5.5 If j 8 and e(G) (;)(j1), then { r ;=: 1 if Kl,j2 a X(G, .,Ci) = r l otherwise. Proof. Suppose K1,j_2 G. We have j 1 ::; 2j 6 since j 8. Therefore, e(G) = (;)(j 1) (2j6), and by Theorem 5.2, we have X(G, .. ci) = rn11 j1 Assume K1,j2 CZ:. G. Let A be a color class in a ,Cicoloring of G. We will show that IAI < j, which implies that X(G, .. cj) r and equality follows from Corollary 2.2. Suppose that IAI j. Let B A such that IBI = j. Now Kt,j2 CZ:. G implies that K1,j2 CZ:. (B). By Theorem 5.4, (B) is Hamiltonian, a contradiction. 0 We wish to extend the above result for all j 3. To accomplish this, we need to determine which graphs of order n are Hamiltonian given the size of their complement is at most n 1. 48
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Theorem 5.6 Let G be a graph of order n 2: 3 with e(G) :::; n1. Then G is Hamiltonian except in the following cases: K1,n2 G ( i) K3 G and n = 5 ( ii) (E(G)) = C4 and n = 5 (iii) (E(G)) = K4 e where e is an edge and n = 6 (iv) (E(G)) = K4 and n = 7 (v) Proof. Let G be a graph of order n 2 3 with e(G):::; n1. If e(G) 2 (;) (n3), then G is Hamiltonian by Theorem 2.10. So assume n 2 :::; e(G) :::; n1. If o(G) :::; 1 (or equivalently K1,n_2 G), then clearly G is not Hamiltonian. This is exceptional case (i). So assume o(G) 2 2. If n = 3, then o(G) 2 2 implies that G = C3 which is Hamiltonian. If n = 4, then o( G) > 2 implies that G E {K4 K4e, C4 }, all of which are Hamiltonian. If n 2 8, then by Theorem 5.4, G is Hamiltonian So, assume 5 :::; n :::; 7. Now n2 :::; e(G) :::; n1 and 5 :::; n :::; 7 imply G is not complete. So let u and v be a pair of nonadjacent vertices in G. Let H = G{ u, v }. Next, we break the proof into cases based on the number of edges in G incident to u or v. Case 1. Suppose that the number of edges in G incident to u or v is n 1. Since the total number of edges in G is at most n 1, H is complete. If n = 5, then having four edges incident to u or v in G forces o( G) :::; 1, which contradicts the assumption that o( G) 2 2. If n = 6, then there are four edges from { u, v} to H in G. Moreover, o( G) 2 2 implies that 6.( G) :::; 3 and Therefore G is, up to isomorphism, one of the graphs in Figure 5.1. If G is graph (a) in Figure 5.1, then uwyvzxu forms a Hamiltonian cycle. If G is graph (b) in Figure 5.1, then uwyzvxu forms a Hamiltonian cycle. If G is graph (c) in Figure 5.1, then clearly G is not Hamiltonian since N(u) = 49
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y X (a) y X (b) y X (c) Figure 5.1. The nonisomorphic possibilities for G when the number of edges incident to u or v is 5 and n = 6. N(v) = {w,x} and the largest cycle containing vertices u and vis C4 Further, (E(G)) = K4 e, where e is an edge, and we get exceptional case (iv). y (a) y (b) y (c) w X Figure 5.2. The nonisomorphic possibilities for(; when the number of edges incident to u or v is 6 and n = 7. If n = 7, then there are five edges from { u, v} to H in G. Moreover, o( G) 2 implies that .6.( G) ::; 4. Therefore G is, up to isomorphism, one of the graphs in Figure 5.2. If G is graph (a) or (b) in Figure 5.2, then uwavzyxu forms a Hamiltonian cycle. If G is graph (c) in Figure 5.2, then uwvazyxu forms a Hamiltonian cycle. This addresses all the graphs where the number of missing edges incident to { u, v} is n 1. Case 2. Suppose that the number of edges in G incident to u or v is exactly n 2. Since the total number of edges in G is at most n 1, H = 50
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G { u, v} is missing at most one edge. If n = 5, then there are two edges from { u, v} to H in G. Moreover, o (G) ;?: 2 implies ::::; 2. Therefore G is, up to isomorphism, one of the graphs in Figure 5.3. The two nonisomorphic possibilities for G when G has three edges incident u v u v u v Y0w X X X (a) (b) (c) u v u v u v y? w y e W X X X (d) (e) (f) Figure 5.3. The nonisomorphic possibilities for G when the number of edges incident to u or v is 3 and n = 5. to u or v and e(fl) = 0 are depicted in the first column in Figure 5.3. As stated previously, the subgraph H can be missing at most one edge. The graphs in the second and third column depict all possible nonisomorphic placements of that additional edge. If G is graph (a) or (c) in Figure 5.3, then uxvywu forms a Hamiltonian cycle. If G is graph (b) in Figure 5.3, then clearly G is not Hamiltonian since x is a cut vertex in G. Further, (E(G)) = C4 and we get exceptional case (iii). If G is graph (d), (e), or (f) in Figure 5.3, then G is not Hamiltonian since N(u)UN(v) = { w, x }. Further, I<3 G and we get exceptional case (ii). If n = 6, then there are three edges from { u, v} to H in G. Moreover, o( G) 2 G) ::::; 3. Therefore G is, up to isomorphism, one of the graphs in Figure 51
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5.4. The two nonisomorphic possibilities for G when G has four edges incident u v u v u v z/v. W z
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u v u v u v u v u v uv W "47 W W afV X Z X Z X X Z X Z y y y y y y (a) (b) (c) (d) (e) (f) u v u v u v u v uv a!f:w aff.w a!pw alflw Z X Z X Z X Z X Z X y y y y y (g) (h) (i) (j) (k) u v u v u v u v w afL:w W X Z X X y y y y (1) (m) (n) (o) u v u v u v u v u v u v af/:w Z X af6:w Z X Z X a{7:w Z X Z X a0w Z /X y y y y y y (p) (q) (r) (s) (t) (u) u v u v u v u v af\\w Z X aAw Z X a0w Z X ar\w Z X y y y y (v) (w) (x) (y) Figure 5.5. The nonisornorphic possibilities for G when the number of edges incident to u or vis 5 and n = 7. 53
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The five nonisomorphic possibilities for G when G has 5 edges incident to u or v are depicted in the first column in Figure 5.5. Since the total number of edges in G is at most 6 and the number of edges in fl incident to u or v is exactly 5, H is missing at most one edge. The graphs in the second and subsequent columns depict all possible nonisomorphic placements of that additional edge. If G is graph (n) in Figure 5.5, then clearly G = 14 V K3 is not Hamiltonian since the largest cycle we can form is c6 obtained by alternately traversing between the vertices of K3 and /4 until we use all the vertices in K3 Further, (E(G)) = K4 and we get exceptional case ( v ). If G is graph (a), (b), (c), (d), or (j) in Figure 5.5, then uwyzvaxu forms a Hamiltonian cycle. If G is graph (e) or (f) in Figure 5.5, then uwvazyxu forms a Hamiltonian cycle. If G is graph (g), (h), (i), (k), (p), (q), (r), or (t) in Figure 5.5, then uxvazywu forms a Hamiltonian cycle. If G is graph (1), (m), or (o) in Figure 5.5, then uwvyzaxu forms a Hamiltonian cycle. If G is graph (s) or (u) in Figure 5.5, then uwzyavxu forms a Hamiltonian cycle. If G is graph (v), (w), (x), or (y) in Figure 5.5, then uwzxavyu forms a Hamiltonian cycle. This addresses all the graphs where the number of missing edges incident to u or v is at least n2. Therefore, we can assume for all nonadjacent vertices u and v in G that the number of edges in G incident to u or v is at most n 3. Case 3. Suppose that the number of edges in G incident to every nonadjacent pair of of vertices u, v is at most n 3. There are at most n4 edges missing from { u, v} into H. Therefore, there are at least 2(n2)(n4) = n edges from {u,v} into H. So, da(u) + da(v) n and by Ore's Theorem (Theorem 2.6), G is Hamiltonian. D 54
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Now we can proceed with determining X(G, ,Cj) when e(G) 2: (;) (j1) for all j 2: 3. Theorem 5.7 If e(G);?:: (;)(j1), then I j:;l if (E(G)) = 2/{2 and j = 3, 131 if (E(G)) = P3 and j = 3, I<1,j2 G and j 2: 4, /{3 G and j = 5, (E(G)) = C4 and j = 5, (E(G)) = K4 e where e is an edge and j = 6, or (E(G)) = I<4 and j = 7, and r n l otherwise. li1 Proof. If e( G) > (;) (j 1), then the result follows from Theorem 5.1. So assume e(G) = (;) (j1). Further, the result follows when j;?:: 8 by Theorem 5.5. Therefore, we only need to determine X(G, ,Cj) when j = 3, 4, 5, 6 or 7. If j = 3, then (E(G)) E {2K2, P3} If (E(G)) = 2/{2, then x(G, ,C3) = I j:;l by Theorem 4.1. If (E(G)) = P3 then X(G, ,C3 ) = I l by Theorem 4.1. If j = 4, then (E(G)) E {3K2,P3 + K2,I
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then we get Cj (B) by Theorem 2.10 and reach a contradiction. Therefore; e((B)) :::; (j1). But since G is missing j1 edges, we must have e((B)) = (j 1). If :::; 1, then e((B)) :::; (j + 1)/2. Now, j1 = e((B)) :::; (j + 1)/2, which implies j :::; 3. But this contradicts the assumption that j 5. Therefore, !l((B)) 2, which implies cS((B)):::; j2. Choose v E V((B)) so that d(B)(v):::; j2. Consider (B)v. Now, e((B)v) = e((B))d(v) = (j3) and by Theorem 2.10, we have Cj (B) and reach a contradiction. Therefore, JAJ :::; j. Let a be the number of color classes of size j in a Cjcoloring of G and A be a color class of size j. By Theorem 2.10, (A) must be missing at least j2 edges. Since G is missing at most j1 edges, we must have a(j2) :::; j1, or simply a :::; 1. Therefore, all Cjcolorings of G can contain at most one color class of size j, which implies that X( G, Cj) I j=fl + 1 = I l By Theorem 5.6, we know which graphs missing at most j1 edges are non Hamiltonian. If G contains one of these nonHamiltonian graphs as an induced subgraph, then we can color G so that there is a color class of size j and get D The previous theorem produces an upper bound for determining how many edges need to removed from a complete graph to obtain a graph with X(G, Cj) = I i=i l Determining X( G, Cj) when G is missing j edges can be accomplished, we believe, using the above methods. However, extrapolating on the complexity of the proofs for removing j 1 edges versus the complexity of the proof for removing only j2 edges leads us to believe that the proof for removing j edges will be extremely 56
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long and complicated. We have already determined the minimum number of edges to be removed to obtain a conditional chromatic number of r j::::n in Theorem 5.7. Removing j edges from a complete graph will probably not allow the conditional chromatic number to decrease again. Thus, there is little to be gained by proving the above type of result for graphs missing j edges. In fact, further research in the area of determining X( G, Cj) should focus on determining general upper and lower bounds on the number of edges for a graph to attain a particular conditional chromatic number. 57
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6 Determining x(G, Pj) We discovered in Theorem 5. 7 that it is necessary to remove j 2 edges from a complete graph to decrease the .Cichromatic number from f l to f i=i l and that it is necessary to remove j 1 edges to attain a value of r ;=; l for the .Cjchromatic number when j = 3. We would like to determine the minimum number of edges to remove from a complete graph to attain a value of r i=i l for the .Pichromatic number and determine the minimum number of edges to remove from a complete graph to attain a value of f j=; l for the .Pichromatic number. While attempting to find these numbers, we will obtain information about the .Pichromatic number for a large set of graphs. We begin by determining which graphs of large size have a Hamiltonian path, then determine the minimum number of edges which need to be removed from a complete graph for X( G, ,pi) to decrease from r to r i=i l' and then determine the chromatic number for all graphs missing 2j 5 or fewer edges. Finally, we will determine a lower bound on X( G, ,pi) in terms of the size of G. 6.1 Determining which graphs of large size have a Hamiltonian path To determine the .Pichromatic number for graphs whose complements have size at most 2j 5, we need to determine which graphs of order n missing at most 2n5 edges have a Hamiltonian path. The following is an easy and useful result. 58
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Lemma 6.1 Let G be a graph of order nand S V(G). If G has a Hamiltonian path, then the number of components in GS is at most lSI+ 1. Proof. Let S V(G) with 0 :::; lSI :::; n, and let H be a Hamiltonian path in G. If we removeS from G, then H breaks apart into at most lSI+ 1 subpaths. Therefore, GS contains at most lSI+ 1 components. 0 The following sets will be used repeatedly in this section. Let :F = {K2 V (K2 + I3), K1 V (Kn3 + I2), Ir V Kr2, where r 4 and n 4}, Q = { G I G is a spanning sub graph of a graph in :F}, and 1i = {K2 V (K2 +h), K1 V (Kn3 + Iz), I4 V K2, (!4 V Kz) e, Is V K3, (/5 V K3)e, Is V K4 where e is an edge and n 4}. Lemma 6.2 If a graph G E Q, then G does not have a Hamiltonian path. Proof. Let G = F V (K2 + I3 ) where F = K2 Let S = V(F). Then the number of components in GSis 4 and 4 > 3 = lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = F V (Kn_3 + 12 ) where F = K1 LetS= V(F). Then the number of components in GSis 3 and 3 > 2 = lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = Ir V F where F = Krz and r 4. Let S = V(F). Then the number of components in GSis rand r > (r2) + 1 =lSI+ 1. By Lemma 6.1, G does not have a Hamiltonian path. Let G = F, where F is a spanning subgraph of a graph in :F. Since each graph in :F does not contain a Hamiltonian path, F cannot contain a Hamiltonian path. 0 59
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Now g contains more graphs than just those graphs of order n missing at most 2n5 edges. Since we are interested in determining which graphs missing at most 2n 5 edges have no Hamiltonian path, we will determine the largest subset of graphs of order n in g that are missing at most 2n 5 edges. Lemma 6.3 Let G be a graph of order n 1 with e(G) :::; 2n5. Then G E g if and only if G E 1i. Proof. Let G be a graph of order n 1 with e(G) :::; 2n5. Assume that G E Q. If G is a spanning subgraph of K 2 V (K2 + /3), then since K 2 V (K2 + /3 ) is missing 9 = 2(7)5 edges, we must have that G = K2 V (1<2 +fa) E 1i. If G is a spanning subgraph of Kt V(Kna+/2), then since Kt V(Kn3 +lz) is missing 2n5 edges, we must have that G = K1 V(Kna+l2 ) E 1i. So assume that G is a spanning subgraph of Ir V Kr2, where r 4. Now (;) (2n5) :::; e( G) :::; e( Ir V Kr2) = (;) (;), or (;) :::; 2n 5 = 2(2r 2) 5. Simplifying, we get r2 9r + 18 :::; 0 or (r6)(r3) :::; 0. This implies that 3 :::; r :::; 6. We have already assumed that r 4. If G is a spanning subgraph of /4 V K2 then since /4 V K2 is missing 6 = 2(6)6 edges, G can be missing up to one more edge. Thus, G = /4 V K2 or G = (/4 V K2 ) e where e is an edge. Both of these graphs are in 1i. If G is a spanning subgraph of /5 V K3 then since /5 V K3 is missing 10 = 2(8) 6 edges, G can be missing up to one more edge. Thus, G = /5 V K3 or G = (!5 V K3)e where e is an edge. Both of these graphs are in 1i. If G is a spanning subgraph of Is V K4 then since Is V K4 is missing 15 = 2(10) 5 edges, we must have that G = Is V K4 E 1i. Therefore, if g 1i. Clearly, 1i g. D By Lemma 6.2, 1i consists of graphs of order n missing at most 2n 5 edges with no Hamiltonian path. The next step is to prove that a connected graph 60
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missing at most 2n5 edges is either in 1i or has a Hamiltonian path. To prove this, we need the following theorem which appears in Chartrand and Lesniak [17]. Theorem 6.1 Let G be a connected graph of order 3 or more that is not Hamilto nian. If for all distinct nonadjacent vertices u and v, d( u) + d( v) 2 m, where m is a positive integer, then P m+t G. The following theorem completely characterizes which graphs of order n missing at most 2n 5 edges have a Hamiltonian path. Theorem 6.2 If G is a connected graph of order n 2 1 and e( G) :S 2n5, then either G E 1i or G has a Hamiltonian path. Proof. Let G be a connected graph of order n with e( G) :S 2n 5. If n :S 2, then the theorem is vacuously true. If n = 3, then G E {P3 K3}, both of which have a Hamiltonian path. If n = 4, then since e( G) :S 3 and G is connected, G E {KI,3, P4, K1V(K1+K2), C4, K4e, K4}. lfG = K 1,3, then G = K1 V(K1 +I2) E H. Otherwise, G has a Hamiltonian path. So assume n 2 5. If G = Kn, then obviously G has a Hamiltonian path. So assume G is not complete. Let u and v be a pair of nonadjacent vertices in G and H = G{ u, v }. Since G is connected, each of u and v has a neighbor. If N ( u) = N ( v) = { w}, then the number of edges missing between u and H is n 3 and the number of edges missing between v and H is n3. Since we assumed that uv E( G), we have a total of 2n5 missing edges incident to u or v. This implies that G = K1 V (I
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Case 1. Suppose that the number of edges in G incident to u or v is n + 1 or more. Then H is a graph on n2 vertices missing at most (2n5) ( n + 1) = (n2)4 edges. By Theorem 2.9, His Hamiltonian connected, and hence there is a Hamiltonian (a,b)path in H. So, ua together with the Hamiltonian (a,b)path in H together with bv is a Hamiltonian path in G. Therefore, we can assume that every pair of nonadjacent vertices is incident to no more than n edges in G. Case 2. Suppose the number of edges in G incident to u or v is exactly n. Then His missing at most (2n5)n = (n2)3 edges. By Theorem 2.9, the sub graph H has a Hamiltonian cycle C. Let C = u1 u2 un_2u1 If u (or v) is adjacent to consecutive vertices on C, then we could construct a ( n 1 )cycle using a detour through u (or v) and could then attach v (or u) to this cycle and produce a Hamiltonian path in G. So assume neither u nor v is adjacent to consecutive vertices on C. Thus, d( u) :=; and d( v) :::; for otherwise, by the pigeonhole principle, u (or v) would be adjacent to consecutive vertices on C. Next, we determine all possible values for the degrees of u and v in G. There are 2(n2) possible edges from { u, v} to H and since n1 of these edges are missing, there are exactly n 3 edges from { u, v} to H. Now d( u) + d( v) = n 3 and the fact that each of u and v has degree at most implies that either n is even, d(u) = and d(v) = (Case 2A); or n is odd and d(u) = d(v) = (Case 2B). Case 2A. Assume n is even, neither u nor v is adjacent to a pair of consecutive vertices on C and d( u) = and d( v) = For v to be adjacent to half of the vertices in Hand not have two of its neighbors adjacent on C, the vertex v must be adjacent to alternating vertices on C, say the 62
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vertices on C with odd subscripts. Since d( u) = n;4 we can say without loss of generality that the vertex u is not adjacent to u1 Now we break this case into subcases based on whether or not N(u) N(v). Case 2Al. IN( u) n N( v) I = n;4 First we will show that ifthere is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If u 2 u4 E E( G), then we can construct a Hamiltonian path in G as follows: u4u2u1vuau if n = 6 and u4u2u1vuauu5 ... Un2 if n > 6. If u2un2 E E(G), then we can construct a Hamiltonian path in G as follows: uu3u4 ... Un2u2u1v. If u2u2 k E E(G) for some 2 < k < (n2)/2, then we can construct a HamiltoFigure 6.1. The graph G when n is even and u2u2k E E(G) for some 2 < k < n22. Un_4un_2 E E(G), then we can construct a Hamiltonian path in G as follows: Un4Un2Unavulu2 ... Un5U. If U2kUn2 E E(G) for some 1 < k < (n 4)/2, then 63
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we can construct a Hamiltonian path in G as follows: If u 2ku2k+2 E E( G) for some 1 < k < n;4 then we can construct a Hamiltonian E(G) where 1 < k < l < n;2 and l > k + 1, then we can construct a Hamiltonian path in Gas follows (see Figure 6.2): Figure 6.2. The graph G when n is even and u2ku21 E E(G), 1 < k < l n;2 and l > k + 1. Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus { u2 u4 ... Un2 u, v} is an independent set of size n!2 Therefore, G is a spanning subgraph of I !!:. V K n2 = Ir V Kr2, where r = 2 2 (n + 2)/2. Now n 2: 5 and even implies that r 2: 4. Therefore, G E 9 and by Lemma 6.3, G E 11... 64
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Case 2A2. IN(u) n N(v)l < Since vis adjacent to all vertices in H with odd subscripts, the vertex u cannot be adjacent to all vertices in H with odd subscripts excluding u1 or else IN(u) n N(v)l = Therefore, u must be adjacent to u2k for some 1:::; k:::; If UUn2 E E(G), then UUn2Un3ulv is a Hamiltonian path in G. If uu2k E E( G) for some 1 :::; k < then uu2kU2kl ... u1 Un2Un3 ... U2k+l v is a Hamiltonian path in G. Case 2B. Assume n is odd, neither u nor v is adjacent to a pair of consecutive vertices on C, and d( u) = d( v) = Recall that n 2: 5 and there exist a, b E V(H) such that a E N(u), bE N(v) and a =f. b. If n = 5, then H = K3 and G is K3 with two pendant vertices, each joined to a different vertex in K3 This graph contains a Hamiltonian path. So, assume n 2:7. Figure 6.3. The graph G when n is odd and u 2u2i E E(G) for some 2 i n23 and n > 7. 65
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Since d( u) = n;3 and there are n2 vertices on C, the vertex u is not adjacent to two consecutive vertices on C, say uu1 fj. E( G) and uu2 fj. E( G). For u to have degree n;J and not have two adjacent neighbors on C, the vertex u must be adjacent to the remaining vertices on C with odd (or even) subscripts, say odd. Thus, we can assume that N(u) = { u 3 u5 Un2 }. Next we break this case into subcases based on whether or not N(u) = N(v). Case 2Bl. N(u) = N(v). First we will show that if there is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If u2u2 i E E(G) for some 2 ::; i ::; (n3)/2, then we can construct a Hamiltonian path in G as follows u1 u 2 u 4 uavu 5 u if n = 7 and if n > 7 (see Figure 6.3). If u4 u2 1 E E (G) for some 2 < l ::; n;3 then we can construct a Hamiltonian path in G as follows: u1 u2uavus ... u211 uun2 ... u21+1 u21U4. If u2ku21 E E( G) for some 2 < k < l ::; n;3 then we can construct a Hamiltonian path in G as follows (see Figure 6.4): Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus, e( H) Cf). Recall that e( H) ::; n 5. Therefore, n 5 e( H) Cf), which implies that n ::; 11. Recall that the number of missing edges incident to u or v is n. If n = 7, then N(u) = N(v) = {u3,u5}. We have assumed that u 2 u 4 fj. E(G). If u 1 u 4 E E(G), then uu5vu3 u 4 u 1 u 2 is a Hamiltonian path in G. If u1u4 fj. E(G), 66
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Figure 6.4. The graph G when n is odd and u 2 ku21 E E(G) for some 2 < k < l n23. then we have accounted for all nine missing edges in G. Therefore, G = (u3 u5 ) V ((u1,u2) + u + v + u4) = K2 V (K2 + /3) E 1l. If n = 9, then N(u) = N(v) = {u3,u5,u7 }. We have already assumed that {u2u4,u2u6,u4u6} E(G). This leaves at most one additional edge in fl. If u1u4 E E(G), then u6u1uu5vu3u4u1u2 is a Hamiltonian path in G. If u 1 u 4 E(G), then we have accounted for all thirteen missing edges in G. Thus, u6u1 E E( G) If n = 11, then N(u) = N(v) = {u3,u5,u7,u9}. Since we have assumed that there are no edges between any pair of vertices in H with even subscripts, the graph induced by { u2 u4 u6 u8 } is a copy of /4 and we have accounted for all Hamiltonian path in G. 67
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Case 2B2. N(u) =f. N(v). Recall that N(u) = {u3,u5 ... ,un2}. Now either vu1 E E(G) or vu2 E E(G), otherwise, by the pigeonhole principle, v would be adjacent to consecutive vertices on C. If vu1 E E(G), then vu1u2 Un2u is a Hamiltonian path in G. If vu2 E E(G), then vu2u1un2 ... u4u3u is a Hamiltonian path in G. Therefore, we can assume that every pair of nonadjacent vertices is incident to no more than n 1 edges in G. Case 3. Suppose that the number of edges in G incident to u or v is exactly n1. There are 2(n 2) possible edges from { u, v} to H, and since n 2 of these edges are missing, we get d( u) + d( v) = n 2. The graph H is missing at most (2n 5) (n 1) = n 4 edges. Therefore, e(H) ;::: (n;2)(n4) = edges. By Theorem 2.8, either His Hamiltonian, H = K1 V (K1 + Kn4) for n;::: 5, or H = K2 V /3. Case SA. Assume His Hamiltonian. Again, let C = u1 u2 un_2u1 be a Hamilto nian cycle in H. If u (or v) is adjacent to consecutive vertices on C, then we could construct a ( n 1 )cycle using a detour through u (or v) and could then attach v (or u) to this cycle and produce a Hamiltonian path in G. So assume neither u nor vis adjacent to consecutive vertices on C. Now n must be even, for otherwise either u or v would be adjacent to at least n;l vertices in H, say u, and by the pigeonhole principle, u would be adjacent to consecutive vertices on C. So assume that n is even, neither u nor v is adjacent to a pair of consecutive vertices on C, and d(u) = d(v) = n;2 Since u is adjacent to exactly half of the vertices on C and u is not. adjacent to a pair of consecutive vertices on C, the vertex u is adjacent to either all the vertices on C with odd subscripts or all the vertices on C with even subscripts. The same is true for v. By symmetry, there are two 68
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(Case 3Al); and N(u) = N(v) = {u1,u3,u5 ... ,una} (Case 3A2). Case 3Al. N(u) = {u2, u 4 ... Un2} and N(v) = {ur, ua, us, ... Una}. In this case, vu1 u 2 ... Un_2 u is a Hamiltonian path in G. Case 3A2. N(u) = N(v) = {ur,u3,u5 ,un3}. First we will show that ifthere is an edge in H between a pair of vertices with even indices, then we can construct a Hamiltonian path in G. If Un4 Un2 E E( G), then Un4Un2UnaUU1 ... UnsV is a Hamiltonian path in G. If u 2 kun_ 2 E E(G) for some 1 k < (n4)/2, then is a Hamiltonian path in G. If u 2 ku 2k+2 E E(G) for some 1 k < (n4)/2, u2ku 21 E E(G) for some 1 k < l < n;2 and l > k + 1, then is a Hamiltonian path in G (see Figure 6.5). Next, assume that there are no edges between any pair of vertices in H with even subscripts. Thus { u 2 u 4 ... un_ 2 u, v} is an independent set size n;2 Therefore, G is a spanning sub graph of V K n2 = Ir V Kr2, where r = (n + 2)/2. Now n 5 and even implies that r 4. 2 By Lemma 6.3, G E 1. Case 3B. Assume that H = K1 V (K1 + Kn4 ) and n 5. Let w be the vertex of degree 1 in H, let x be the neighbor of w in H, and let U = { ur, u 2 ... Un4} be the set of remaining vertices in H. Since G is connected, there are two cases to consider: d(u) 2 and d(v) 2 (Case 3Bl); and d(u) = 1 or d(v) = 1, without loss of generality, say d(u) = 1 (Case 3B2). 69
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Figure 6.5. The graph G when n is even and u2ku21 E E(G) for some 1 :::; k < l < n;2 and l > k + 1. Case 3B 1. Assume that d( u) 2 and d( v) 2. Suppose that w E N ( u). If there exists Ui E U such that UiV E E(G), without loss of generality, say Un4v E E(G), then uwxu1 Un_4 v is a Hamiltonian path in G. If vis not adjacent to any vertex in U, then N(v) = {w,x} and uwvxu1 un_4 forms a Hamiltonian path in G. So we can assume that w N(u). By symmetry, we can assume that w N(v). Suppose that x E N(u). If u and v were each adjacent to only one ui E U, say then since d(v) 2, we get N(u) = N(v) = Since d(u) = 2, d(v) = 2, and d(u) + d(v) = n2, we must haven= 6. Thus, G = ((u + v + w + u2) V { x}) u1 w = (/4 V K2 ) e E 1. So we can assume that there exist distinct vertices Ui and Uk such that uk E N(u) and Ui E N(v). Without loss of generality, say u1 E N ( u) and Un4 E N ( v). Now wxuu1 Un_4 v is a Hamiltonian path in G. So we can assume that x N(u). Further, by symmetry, we can assume that x N(v). Therefore, the only remaining case to consider is when N(u) U and 70
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N(v) U. Suppose that N(u) U and N(v) U. Since d(u) 1, the vertex u is adjacent to a vertex in U, say u1 Since d(v) 2, vis adjacent to a vertex Ui in U where 1 < i::::; n2. Without loss of generality, say uu 2 E E(G). If vu1 E E(G), then uu 1vu2 u3 Un_4xw forms a Hamiltonian path in G. If vu1 rf. E(G), then v must be adjacent to another vertex in U, without loss of generality, say UJ. Now uu 1 u2vu3 .. Un_4xw forms a Hamiltonian path in G. Case 3B2. Assume that d(u) = 1. Since d(u) + d(v) = n2, we get d(v) = n3, which implies that v is adjacent to all but one of the vertices in H. Therefore, either { x; w} N ( v), x rf. N ( v), or w rf. N ( v). Suppose that N(u) = {w }. If {x, w} N(v), then uwvxu1 ... Un4 forms a Hamiltonian path in G. If x rf. N(v) or w rf. N(v), then vis adjacent to a vertex in U, say un_ 4 Now uwxu1 ... Un_4 v forms a Hamiitonian path in G. Suppose that N(u) = {x }. If v has no neighbor in U, then since vis adjacent to all but one vertex in H, N(v) = {x,w} and U = {ul}. Thus, G = {x} V ((v,w) + u + ul) = K1 V (K2 + /2 ) E 1l. So assume that v has a neighbor in U, say u1 If w E N( v ), then ... Un_4 forms a Hamiltonian path in G. If w rf. N( v ), then x E N(v) and G = {x} V ((u1, u2, ... Un4, v) + u + w) = K1 V (Kn3 + l2) E 1l. The last case to consider is when N( u) U. Without loss of generality, suppose that N ( u) = { ul}. If w E N ( v), then uu 1 ... un_4xwv forms a Hamiltonian path in G. If w rf. N( v ), then N( v) = { u1, u2 ... Un4, x} and uu1 ... Un4vwx forms a Hamiltonian path in G. Case 3C. Suppose that H = K2 V 13 Let U = { u1 u2 } be the set of vertices of degree 4 in Hand V = { u3 u4 u5 } be the set of vertices of degree 2 in H. In this 71
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case there are five edges between { u, v} and the vertices of H. Without loss of generality, either d( u) = 1 and d( v) = 4, or d( u) = 2 and d( v) = 3. Case 3Cl. Suppose that d( u) = 1 and d( v) = 4. Then by the symmetry of H, we can assume that either vus
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The next theorem builds on the previous theorem by stating which graphs of order n missing at most 2n 5 edges have a Hamiltonian path. Theorem 6.3 Let G be a graph of order n 1 with e(G) ::::; 2n5. Then G has a Hamiltonian path if and only if o( G) > 0 and G rf. 1. Proof. Let G be a graph of order n 1 with e(G) 2n5. Assume that o(G) > 0 and G rf. 1. To apply Theorem 6.2, we must show that G is connected. Suppose that G is not connected. Let A be a component of G with IAI = a and B = GA. Since o(G) > 0, we have 2 n2. Now, e(G) a(na)= ana2 The minimum for this quadratic function occurs when a= 2 or a= n2 and we get e( G) an a2 2n 4. Thus, the size of G must be at least 2n 4, which contradicts the assumption that e( G) 2n 5. Therefore, G is connected. By Theorem 6.2, G has a Hamiltonian path. Assume that G has a Hamiltonian path. Now o (G) > 0 or else G would not have a Hamiltonian path. By Lemma 6.2 and Lemma 6.3, G rf. 1. D Now that we know which graphs oflarge size have a Hamiltonian path, we will address the problem of determining X(G, Pj) when e(G) ::::; 2j5. 6.2 Graphs missing complete subgraphs or a star As we have seen in the previous chapter, graphs of large size missing a star or com plete subgraphs are the graphs which need to handled as special cases when de termining X(G, Cj). The same will prove to be true when determining X(G, Pj) for graphs of large size. First, let's prove a result for a graph whose complement contains a star as a subgraph. 73
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Theorem 6.4 Let G be a graph of order n. If j 2: 2, e(G)'2: (2j5), and K1,j_1 G, then X( G, .pi) = r j::::: l Proof. If j = 2, then the theorem is vacuously true. So assume j 2: 3. Let X be a set of vertices which induce K1,j_1 G. Since K1,j_1 G and e( G) 2j 5, there is exactly one vertex v0 of degree at least j 1 in G. Therefore, v0 EX. Let A be a color class in a .Picoloring of G. Suppose vo rt A. If lA I 2: j, then (A) would be missing at most 2j 5(j 1) = j 4 edges, and by Theorem 2.9, we know that (A) has a Hamiltonian path. Therefore, for (A) to be Pjfree, we must have IAI j 1. Suppose vo E A, and consider Av0 If lAv01 2: j, then by the same argument, we get Pi (Avo). Therefore, lAvol j1, which implies IAI j. Thus, there can be at most one color class of size j (one containing vo), and X( G, .Pi) 2: r j::::{l + 1 = r j::::i l To show the inequality in the other direction, form one color class B of size j by using the vertices in X together with one vertex from GX and form as many other color classes as possible of size j1 with the remaining vertices in GX. Now (B) is Pjfree since v0 is not adjacent to any other vertex in (B). Therefore, x(G, .PJ r I=tl + 1 = r ;=n. 0 The previous theorem along with most of the following theorems will be used in the next section. The following theorems are more general than we need, but we wish to present them in their entire generality. Their proofs are straightforward and use the same proof methodology. Therefore, to quickly reach the main results of this chapter, the remaining theorems in this section will be presented here without proof. Their proofs can be found in Appendix A. 74
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Theorem 6.5 Let G be a graph of order n 1. If j 2 is even, m 0, and (E(G)) = mKi, then X(G, Pj) =max (f 'r I l). Theorem 6.6 Let G be a graph of ordern. If j 3 is odd and (E(G)) = KHa e, 2 where e is an edge, then X( G, Pj) = r l Theorem 6.7 Let G be a graph of order n. If j 4 and (E(G)) = Ij_3 V K2 then X( G, Pj) = r i=il Theorem 6.8 Let G be a graph of order n 1. If j 4 is even, m 0, and (E(G)) = mKi, then X(G, Pj) =max (f l, f Theorem 6.9 Let G be a graph of order n. If j 2 even and (E(G)) e, where e is an edge, then X(G, Pj) = r i=il Theorem 6.10 Let G be a graph of order n. If j 6 is even and (E(G)) K1 V (Kt + K1), then X(G, Pj) = r i=il Theorem 6.11 Let G be a graph of order n. If j 6 zs even and (E(G)) Ki + K2 then X(G, Pj) = r l We will use most of these theorems to determine X( G, Pj) when e( G) is large. 6.3 Determining x(G, Pj) when e(G) is large First of all, we can remove j2 edges and show that X( G, Pj) does not decrease by applying a well known result about Hamiltonian paths. As an aside, if j 3, then Theorem 6.12 is an immediate result of Theorem 6.13. Thus we did not need to prove Theorem 6.12 in full generality. However, it was easier to prove Theorem 6.12 this way. 75
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Theorem 6.12 Let G be a graph of order n. If j 2:: 2 and e(G) 2:: (;) (j2), then X(G, P3) = I Proof. Let A be a color class in a Pjcoloring of G. Suppose that IAI 2:: j. Let B A with IBI = j. Now (B) is missing at most j2 edges. By Theorem 2.9, we get Pj (B), a contradiction. Therefore, IAI < j, which implies that X( G, Pj) 2:: I Equality follows from Theorem 2.1. D We can see that this theorem is best possible since, if we remove the edges of a copy of K1,j_1 from a complete graph of order n 2:: j, then by Corollary 4.1, we can obtain a graph whose P3chromatic number is I j=; l What is the minimum number of edges we need to remove from a complete graph to obtain X( G, Pj) = I j=i l? One way to obtain X( G, Pj) = I j=i l is to remove the edges of 2K1,jl from a complete graph, i.e., remove 2j 2 edges. With this in mind, we will attempt to close in on graphs missing 2j 2 edges by examining graphs missing up to 2j 5 edges. Theorem 6.13 Let G be a graph of order n 2:: 1 and e be an edge. If j 2:: 2 and e(G) 2:: (2j5), then r I 31 if G = I
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Proof. Let j 2:: 2 and G be a graph of order n with e(G) 2:: (;) (2j5). If j = 2, then the theorem is vacuously true. So assume j 2:: 3. Suppose that G = Kns V (K2 + I3 ) and j = 7. Then (E(G)) = Kse, where e is an edge, and by Theorem 6.6, we get X( G, .P7) = r i=: l Suppose that Kt,jt G. Then by Theorem 6.4, we get X( G, ,pi) = f i=: l Suppose that G = Kn(jl) V(Kia+I2 ) andj 2::4. Then (E(G)) = IjaV K2, and by Theorem 6.7, we get X(G, .Pi)= r i=il. Suppose that G = Kn4 V I 4 and j = 6, or G = Kn5 V Is and j = 8, or G = Kn6 V I 6 and j = 10. Then (E(G)) = Ki+2, and by Theorem 6.5, we get 2 X(G, .Pj) = r i=il if n 2:: j, and X(G, .Pj) = r I l if n < j since r i=il 2:: r I l if and only if n 2:: j. But, when n < j, we know that r I l = 1 = r i=il. Therefore, X(G, .Pi) = r j=: 1 Suppose that G = (I
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Let B such that IBI = j. We want to use Theorem 6.3 to show that (B) has a Hamiltonian path. Since K1,j_1 % G, J"( (B)) > 0. If (B) = K2 V (K2 + I3), then j = 7. Now 9 = 2(7) 5 e(G) 2j5 imply that G = I
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coloring. Therefore, we abandon this approach to find X( G, ...,pj) for all graphs of large size with the knowledge that in order to produce a graph whose ...,pj_ chromatic number is I j=il, we need to remove somewhere between 2j4 and 2j2 edges. 6.4 Determining bounds on x(G, Pj) given the number of edges in a graph In this section, we find an upper bound on the size of a graph given the constraint that the ...,p3chromatic number is at most li=; J. To find the upper bound, we first need to determine the maximum size of a P3free graph of order n. The first upper bound on the size of a P3free graph was discovered in 1959, by P. Erdos and T. Gallai [24]. Their theorem states if G is a P3free graph, then e( G) ( )n. In 1975, R. Faudree and R. Schelp [25] improved on this bound for the maximum size of a Pjfree graph. In fact, they found the least upper bound and determined which graphs met that bound. The theorem is restated here for completeness and reference. Theorem 6.14 (Faudree and Schelp, {25}) lfG is a graph of ordern = q(jl)+r (0 q, 0 r < j1) and G contains no Pj, then e(G) q(i;1 ) + (;) with equality if and only if G = qKjl + Kr or (when j is even, q > 0, and r = j /2 or (j2)/2), G = lK31 + ( Ki?V /f+(q1l)(il}+r) 0 :::; l:::; q1. We would like to know how many edges we need to remove from a complete graph of order j + i ( i 0, j 2) to guarantee that a graph missing at least this number of edges is Pjfree. To accomplish this, we tailor Faudree and Schelp's result to our needs with the following corollary. 79
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Corollary 6.1 If G is a Pifree graph of order n j 2, then e(G) (nj + 1)(j1). Proof. Let mn be the minimum number of edges in the complement of a Pjfree graph of order n = q(j1) + r, where q 0 and 0 r j2. We will show by induction that for n j, we have mn (nj + 1)(j1), which establishes the statement of the corollary. Now n j implies that q 1. By Theorem 2.9 {iii), mj j1. By Theorem 6.14, mn qe; 1)(;) ( q(j ;) + r) q e ; 1) (;) (q(j1) + r)(q(j1) + r1)q(j1)(j2)r(r1) 2 q2(j1)2 + rq(j1) + q(j1)(r1)q(j1)(j2) 2 q(j1)[q(j1) + 2r1j + 2] 2 q(j1)[(q1)(j1) + 2r] 2 When r < j2, we haven+ 1 = q(j1) + (r + 1). Thus, q(j1)[(q1)(j1) + 2(r + 1)] 2 q(j1) > j 1. q(j1)[(q1)(j1) + 2r] 2 When r = j2, we haven+ 1 = (q + 1)(j1), so that (q + 1)(j1)q(j1) q(j1)[(q1)(j1) + 2(j2)] 2 2 q2(j1)2 + q(j1)2q2(j1)2 + q(j1)2 2q(j1)(j2) 2 80
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q(j1?q(j 1)(j2) q(j1) > j 1. Now by the induction hypothesis, (mn+lmn) +mn 2: j 1 + (nj+ l)(j 1) = ( n + 1 j + 1) (j 1). D We get the following immediate result for the Pjchromatic number from Theorem 6.14. Theorem 6.15 Let j 2: 2. If G is a graph of order n = k(j1) + r (0 ::=::; k, 0 ::=::; r < j1) and e(G) > k(i;1 ) + (;), then X(G, Pj) 2: 2. Proof. Let j 2: 2 and let G be a graph of order n = k(j 1) + r (0 ::=::; k, 0 ::=::; r < j1) with e(G) > k(i;1 ) + (;). By Theorem 6.14, we get Pj G. Therefore, X(G, Pj) 2: 2. D Next we use Corollary 6.1 to derive an upper bound on the size of G given Theorem 6.16 Let G be a graph of ordern 2: 1, j 2: 2 and k 2: 0. IfX(G, P3) ::=::; l;=:J, then e(G) ::=::; k(j 1). Proof. Let G be a graph of order n satisfying X( G, ,pi) ::=::; J with the maximum number of edges. Now G must contain all edges between any two color classes. For, if not, we could add an edge between two color classes to form a new graph H with X(H, Pj) ::=::; J using the same coloring we used for G. But this contradicts the maximality of G. Therefore, we can assume all missing edges 81
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reside within the color classes. Furthermore, by maximality, any color class of size at most j 1 is not missing any edges. Let ai+i be the number of color classes of size j + i for i = 0, ... nj and bi be the number of color classes of size i for i = 1, ... j 1. To minimize the number of missing edges in our graph G, we need to minimize the number of missing edges in all color classes of size at least j. By Corollary 6.1, a color class of size j + i must be missing at least ( i + 1 )(j 1) edges for all i = 0, ... n j. Therefore, we want to find a lower bound for the function (j 1 )ai + 2(j 1)ai+I + ... + (nj + 1)(j 1)ai+nj = (j1) I:f;J(i + 1)ai+i Let's leave this for a moment and derive another inequality. j1 nj n = I: ibi + L:U + i)ai+i i=1 i=O j1 j1 nj nj L:U1)bi2:[U1)i]bi + 2:(i + 1)aj+i + L:U1)aj+i i=1 i=1 i=O i=O j1 nj j1 nj (j1)(2: bi +I: ai+i)2:[(j1)i]bi + 2:(i + 1)ai+i i=1 i=O i=1 i=O l kj j1 nj (j1) = 1 ?:[(j1)i]bi + ?:(i + 1)ai+i J t=1 t=O j1 nj < nk2:[(j 1)i]bi + 2:(i + 1)ai+i i=1 i=O Now, rearranging the above equation, we get nj j1 2:(i + 1)ai+i 2:: k + 2:[(j1)i]bi 2:: k i=O i=1 since I:t::i [(j 1) i]bi 2:: 0. Using the above inequality, we get (j 1) I:f:J ( i+ 1 )ai+i 2:: (j 1 )k. Therefore, the number of missing edges in all color classes is at least k(j 1 ), which implies that e( G) ::; k(j 1 ). 0 82
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It is immediately seen from the following example that this bound is attainable. Let G be a graph of order n with G = kK1,il, where k 0 and j 2. Since n = kj = k(j 1) + k, we have that j 1 divides n k and by Theorem 4.3, we get X(G, Pi) = I j:;.l = l;:; J. We have found an upper bound on the size of G given a bound for X( G, ,pi) for j 2. The next theorem gives a lower bound on the size of G given a particular Pichromatic munber. Theorem 6.17 Let G be a graph of order n and j 2. If X( G, ,pi) > I j:il, then e(G) (;) 2). Proof. We will prove the contrapositive. Let G be a graph with e(G) < (;) 2). Suppose8(G) nj+l. Now, e(G) = (;) 2). This contradicts the assumption that e(G) < 2). Therefore, there exists a vertex v such that d( v) ::; n j. In other words, the vertex v is nonadjacent to at least j 1 other vertices in G. Color G as follows: Form a color class of size j by using v and j1 nonneighbors of v. With the remaining n j vertices, create as many color classes of size j 1 as possible. With this coloring, we get X( G, ,pi) ::; I j=fl + 1 = I j:=i l D It is still an open question whether or not the lower bound is attainable. In other words, can we find a graph with 2) edges with X( G, ,pi) = J It is also open as to what the lower bounds are for other values of X(G, Pi). Finally, given X(G, Pi), we would like to find an upper bound on the size of G that is tight when j 1 does not divide n k. 83
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A Appendix All these theorems determine the Pichromatic number for a graph when its complement contains a very specific set of edges. The first three theorems and the final two theorems are used in the proof of Theorem 6.13. Theorem 6.5 Let G be a graph of order n 1. If j 2 even, m 0, and Proof. If j = 2, then (E( G)) = mK2 which implies n 2m, or n m 1 We will show there is no color class of size at least 3. If A were a color class of size at least 3, then the graph induced by any three vertices in A would be missing at most one edge and P3 (A), a contradiction. Also, a color class of size 2 must consist of the endpoints of an edge in G. Therefore, X(G, P2 ) nm. Since n we get nm l Thus, X(G, P2) =max (nm, Ifj 4, then by Theorems 5.3 and 2.2, we get X( G, ,pi) X( G, Cj) = max (I j=.71 I j 1). To show the inequality in the other direction, we produce a minimum ,pi_ ( +2) coloring. Observe that G = v:_:nl m Si where Si = I ill for i = 1, 2, ... m 2 and Si = /1 fori= m + l,m + 2, ... ,n(i) m. LetS= V(G)Ui:1 V(Si) Now, n = ( m + s. We consider two cases: ?m:::; s and 0 :::; s < ?m. If i;2m :::; s, then n mj and f j=.71 f }l For each i = 1, ... m, form a color class of size j by using all the vertices from Si together with ? vertices from S not yet assigned to a color class. Since :::; s, S contains a sufficient number of vertices to form these m color classes. Now, all the subgraphs induced by these 84
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color classes are isomorphic to liE. V Ki2, which by Lemma 6.2, is Pifree. We 2 2 partition the remaining svertices into as many color classes of size j1 as r ( j2 ) 1 r(i.!) 1 'bl Th h X(G P) < s22 m+sm rnml poss1 e. us, we ave , 3 m + j1 :j1 j1 If 0:::; s < then n < mj and l:tl Let r = r(i?]msl and T = S U V(Si)). For each i = 1, ... ,mr, form a color class of size j by using all the vertices from Si together with vertices from T not yet assigned to a color class. Note that m r = m r ( Jmsl = l mj( r )m+s J = l ( ]m+s J = l ]J and there are sufficient elements in T to form l ]J color classes since IT!= s+r (i) = s+f()msl s+()ms (i) = (9) (:T) 2 ( l ]J Again, all the subgraphs induced by these color classes are isomorphic to Jill V K2=1.., which by Lemma 6.2, is Pjfree. 2 2 If there are vertices remaining in T, form one more color class B containing those vertices. To show (B) is Pjfree, we will show that IBI < J. Now, IBI = !TIly J (j 2 2 ) s + r (j; 2 ) (mr) (j; 2 ) s+rjm(j;2 ) < +l)jm(i;2) J. If B = 0, then there were no remaining vertices after forming mr color classes of size j. Therefore, j divides n and X( G, ,pj) :::; l :tJ = r T l If B =10, then j does not divide n and we get X( G, ,pj) :::; l:t J + 1 = r T l In either case, we have produced a .Pjcoloring with r T l colors. D 85
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Theorem 6.6 Let G be a graph of ordern. If j 3 is odd and (E(G)) = Ki+a e, 2 where e is an edge, then X( G, .Pi) = r l Proof. Observe that (E(G)) = ]{i.:!e if and only if G = R V (S + T), 2 where R = ]{ (ill)' S = li.=! and T = K2 Let A be a color class in a ,pi_ n 2 2 coloring of G. Suppose that IAI j + 1. Let B A such that lEI = j + 1. Now, B contains up to i.::p vertices from S U T and at least j;l vertices from R. For all combinations of vertices from these sets, we get (C) = I ill V 2 Ki.=! Let U = {u1,u2 ... ,uill} be the set of vertices of degree j;l in 2 2 (C) and W = { w1 w2 . w i.=!} be the set of vertices of degree j in (C). Now, 2 u1w1u2w2 ... Ui.=!Wi.=!Ui!. forms Pi (C), a contradiction. Therefore, IAI :::; j. 2 2 2 Now, each color class of size j must contain strictly more than vertices from S U T. Suppose by way of contradiction that a color class A of size j contains at most 2.. vertices from S U T. Since j is odd, either j + 3 or j + 1 is divisible by 4. If j + 3 is divisible by 4, then A contains at most vertices from S U T and at least 3i_;3 vertices from R. For all combinations of vertices from these sets, we get (D) = fill V K3i3 (A). Let X1, x2 ... X ill be the vertices of degree 4 4 4 3i_;3 in (D) and y11 y 2 ... y3i3 be the vertices of degree j1 in (D). Now, 4 X1Y1X2Y2 ... XillYillYi.I.Yi..ll. ... ya1a forms Pi (D), a contradiction. There4 4 4 4 4 fore, j + 1 must be divisible by 4, A contains at vertices from SUT, and A must contain at least 3i_;1 vertices from R. For all combinations of vertices from these sets, we get (D) = Ji! V Ka11 (A). Let x1 x2 ... Xi! be the vertices 4 4 4 of degree 3i_;1 in (D) and y11 y2 .. ya11 be the vertices of degree j1 in (D). 4 Now, X1Y1X2Y2 ... Xi.:i!Yi+l YillYill ... Y3i1 forms Pi (D), a contradiction. 4 4 4 4 4 Therefore, each color class of size j must contain strictly more than vertices 86
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from the S U T, which is strictly more than half the vertices of S U T. Therefore, there can be at most one color class of size j and X( G, Pj) f j=fl + 1 = f j=i l Next, we will show the inequality in the other direction. Color G as follows: form one color class A of size j by using all the vertices in S together with j;3 vertices from Rand both vertices from T. Now, (A) =I< V (I?+ 1<2) and since (A) contains vertices of degree j;3 with exactly i:{common neighbors, (A) is Pjfree. With the remaining vertices, form as many color classes of size j 1 as possible. With this coloring, we get X( G, Pj) :'S f j=fl + 1 = f i=:l. So, X(G, Pj) = r j=i 1 o Theorem 6.7 Let G be a graph of order n. If j 4 and (E(G)) = Ij3 V K21 then X(G, Pj) = f j=i l Proof. Let j 2 4 and G be a graph of order n such that (E(G)) = lj3 V 1<2. Then G = R V (S + T) where R = Kn(jI)l S = Kj3, T = /2. Let and { t1 t2 } be the vertices of T. Let A be a color class in a Pjcoloring of G. Suppose that IAI j + 1. Let B A such that IBI = j + 1. If t1 tj. B and t 2 tj. B, then (B) is complete, which implies that Pj (B), a contradic tion. Ift1 E Band t 2 tj. B, then B = {t1,r1 ... ,rp,sl,s2, ... ,sj1p} for some 3 :'S p :'S j 1. Now, t1r1r2r3 .. rps1s2 Sjlp forms Pj (B), a contradic tion. Therefore, we must have {t1 t 2 } B. Since B can contain at most j3 vertices from S, the subgraph (B) must contain at least two vertices r 1 and r2 from R. SoB= {t1,t2,r1 ... ,rt,s1 ,s2, ... ,sjlp} for some 2 :'S p :'S j 1. Now, t1 r 1t2 r 2 r3 ... rps1s2 Sjlp forms Pj (B), which again is a contradic87
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tion. Therefore, we must have \A\ j. Suppose \A\ = j. By Theorem 2.9, if (A) is missing fewer than j 1 edges, then (A) has a Hamiltonian path. Since G is missing at most 2j 5 edges, there can be at most one color class of size j. Thus, X( G, ,pi) 2:: I j.={l + 1 = I j:i l Next we produce a minimum ,pi_ coloring. Form one color class A of size j with A = {t1, t2 r1 s1 s2 ... Sj_3}. Now (A) = K1 V (Kja + /2), which by Lemma 6.2, is Pjfree. We partition the remaining nj vertices into as many color classes of size j 1 as possible. Thus, we have X(G, Pj) I j.=fl + 1 = r i=il 0 In earlier theorems we have derived a value of I l for the Pichromatic number for graphs of order n whose complements consist of m pairwise vertex disjoint copies of a complete graph of a particular order. The following theorem derives a new value for the Pjchromatic number, namely I n;_:;,n l, for graphs whose complements consist of m pairwise vertex disjoint copies of a complete graph of a new order. Theorem 6.8 Let G be a graph of order n 2:: 1. If j 2:: 4 is even, m 2:: 0, and (E(G)) = mK, then X(G, Pj) =max (I n;_:;,n l I m+n(l:)m Proof. Observe that G = Vi=t Si where Si = li.i for z = 1, 2, ... m and 2 Si = 11 fori= m + 1,m + 2, ... ,nm. Let A be a color class in a Picoloring of G. Suppose that \A\> j + 1. Let B such that \B\ = j + 2. Then either (B)= Vf=1 Iq; where q1 = (j +4)/2, and 1 qi j/2 for all2 i k (1), or (B)= Vf=1 lq; where 1 q1 (j + 2)/2, 1 q2 (j + 2)/2, and 1 qi j/2 for all3 i k (2). If (1), then form C1 B such that \C1 \ = j by removing two vertices from lq1 If (2), then form C2 B such that \C2 \ = j by removing one 88
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vertex from each of lq1 and lq2 Now, (C1) = lq12 V (Vf=2 lq,), and maxi{qi} ::S f. Also, (C2) = lq11 V lq2t V (Vf=3 lq,), and maxi{qi} ::S f. By Lemma 5.1, (Ct) and (C2 ) are Hamiltonian. This contradicts the assumption that (A) is Pifree. So IAI :::; j + 1 and X(G, Pj) 2: r Let A be a color class of size j + 1. We will show that li.i (A). Suppose by 2 way of contradiction that li.i (A). Then (A) = Vf=1 lq, where q1 ::S (j + 2)/2, 2 and 1 ::S q i ::S j/2 for all 2 ::S i ::S k. Form B such that IBI = j by removing one vertex from lq1 Now, (B) = lq1 _1 V (Vf:2 Iq,), and maxi{ qi} ::S f. By Lemma 5.1, (B) is Hamiltonian. This contradicts the assumption that (A) is Pifree. So li.i (A). 2 Let C be a color class of size j. By Lemma 5.1, either (C) is Hamiltonian or IJ..:E (C). Since (C) is not Hamiltonian, we must have that IJ..:E (C). 2 2 Let a be the number of color classes of size j + 1 and b be the number of color classes of size j in a Picoloring of G. Since each color class of size j + 1 must contain all the vertices from Si for some 1 ::S i ::S m, and since color classes are vertex disjoint, we must have a ::S m. We have just shown that each color class of size j must contain t independent vertices, which is more than half the vertices from some Si (1 ::S i ::S m) since i.:p> ( i) when j 2: 2. Thus we cannot use the vertices of si (1 :::; i :::; m) to form two different color classes of size j. Since a color class of size at least j requires more than half the vertices from some Si (1 :::; i :::; m) and color classes are vertex disjoint, we get a+ b :::; m. This inequality and a ::S m imply that 2a + b ::S 2m Now, ( G .P ) > rn a(j + 1) bjl b = rn 2a bl > rn 2ml X J 1 +a+ 1 1 JJJ89
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Hence, X( G, ,pi) max (I r l). To show the inequality in the other direction, we produce a minimum Pj(ill) coloring. Recall that G = v"::.:n2 m Si where Si = I ill for i = 1, 2, ... m and 2 Si = I 1 fori= m + l,m + 2, ... ,n(i) m. LetS= V(G)V(Si) Now, n = ( .i.:}i) m + s. We consider two cases: s and 0 s < i;2m. If s, then n m(j +I) and f l f For each i = 1, ... m, form a color class of size j + 1 by using all the vertices from si together with vertices from S not yet assigned to a color class. Since i;2m s, the setS contains enough vertices to form these m color classes. Now, all the subgraphs induced by these color classes are isomorphic to C = I ill V K and since C contains f 2 2 vertices of degree with exactly i;2 common neighbors, C is Pifree. We partition the remaining svertices into as many color classes of size j 1 as r 1 r(.i!) + 2 1 'bl Th h (G P) s2m 2m sm rn2ml poss1 e. us, we ave X i m+ i1 = i1 = _;1 If 0 s < i;2m, then n < m(j +I) and r n;!;n l Let r = and T = S U V(Si)). For each i = 1, ... mr, form a color class of size j + 1 by using all the vertices from si together with j;2 vertices from T not yet assigned to a color class using the vertices from S first. Note that m 1' = m r ( = l )m+s J = l ( J = J and there are sufficient elements in T to form color classes since ITI = s + 1' = s + r ( )msl (ill) > s + ( )ms (ill) = (ci) (.,.!!) > (cl) l.,.!!.J. Again J+l 2 J+l 2 2 J+l 2 J+1 all the subgraphs induced by these color classes are isomorphic to C = I ill V K 2 2 and we have already shown that (C) is Pjfree. If there are vertices remaining in T, form one more color class B containing 90
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those vertices. To show (B) is Pjfree, we will first show that IBI :::; j. Now, IBI = ITIlj: 1J (j 2 2 ) s + r (j; 4)(mr) (j; 2 ) s + r(j + 1)m (j; 2 ) < s+ +1) (j+l)m(i;2 ) j+l. If IBI = j, then (B) = ltH. V Ii=i since we formed color classes by using vertices 4 4 from S first. By Lemma 6.2, ltH. V /i4 is Prfree. 4 4 If B = 0, then there were no remaining vertices after forming mr color classes of size j + 1. Therefore, j + 1 divides n and X( G, ,pj) :::; J = r If B =f. 0, then j + 1 does not divide n and we get X( G, ,pi) :S j + 1 = r l In either case, we have produced a PjCOloring with r colors. 0 The following theorem is needed for the proof of Theorem 6.10. Theorem 6.9 Let G be a graph of ordern. lfj 2 2 is even and (E(G)) = KtH. e, 2 where e is an edge, then X( G, ,pi) = r l Proof. Observe that (E(G)) = KtH.e if and only if G = R V (S + T), where 2 R = K (ill)' S = h, and T = K2 Let A be a color class in a Pjcoloring n 2 2 of G. Suppose that IAI > j. Let B A such that IBI = j + 1. Now, B contains up to vertices from S U T and at least vertices from R. For all combinations of vertices from these sets, we get (C) = (It + K2) V K i;_2 (B). Let ub u 2 ... Ui. be the vertices of degree in (C), vb v 2 be the vertices of 2 91
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degree in (C), and w1 w2 .. Wi2 be the vertices of degree j in (C). Now, 2 u1 w 1u2w2 ... Ui2Wi2V1 v 2 forms Pj a contradiction. Therefore, IAI :S j. 2 2 Now each color class of size j must contain strictly more than j!4 vertices from S U T. Suppose by way of contradiction that a color class A of size j contains at most vertices from S U T. Since j is even, either j or j + 2 is divisible by 4. If j is divisible by 4, then A contains at most j!4 vertices from S U T and therefore, at least 3j;;4 vertices from R. For all combinations of vertices from these sets, we get (D) = hH V Kai4 (A). Let x2 .. Xi.! be the vertices 4 4 4 of degree 3j;;4 in (D) and Y2, .. yai4 be the vertices of degree j 1 in (D). 4 Now, X1Y1X2Y2 . ... yai4 forms Pj (D), a contradiction. So 4 4 4 4 4 j + 2 must be divisible by 4, A contains at most lvertices from S U T, and A must contain at least 3j;;2 vertices from R. For all combinations of vertices from these sets, we get (D) = /ill V Kai2 (A). Let x2 X ill be the vertices 4 4 4 of degree 3j;;2 in (D) and y1 y2 ... yai2 be the vertices of degree j 1 in (D). 4 Now, X1Y1X2Y2 .. Xi..:H_Yi+2Yi+eYi+lo ... y3i2 forms Pj a contradiction. 4 4 4 4 4 Therefore, each color class of size j must contain strictly more than j!4 vertices from S U T, which is strictly more than half the vertices inS U T. Therefore, there can be at most one color class of size j and X( G, ,pj) I i=tl + 1 = I;=: l Next, we will show the inequality in the other direction. If j = 2, then (E(G)) = P3. Form one color class of size 2 using the two nonadjacent vertices in G and put the remaining n2 vertices into n2 color classes. Therefore, X( G, ,pi) :S n 1. If j 4, then color G as follows: form one color class A of size j by using all the vertices in S together with j;4 vertices from R and both vertices from T. Now, (A) = I
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By Lemma 6.2, (A) is Pifree. With the remaining vertices, form as many color classes of size j 1 as possible. We get X( G, ,pi) :::; I j.=fl + 1 = I l So, Theorem 6.10 Let G be a graph of order n. If j > 6 ts even and (E(G)) K1 V (Kf + K1)} then X(G, .Pi)= I l D Proof. Assume j 2 6 is even and G is a graph of order n such that (E(G)) = K1 V + K 1 ). Then G = R V (S + T) where R = Kn(f+ 2), S = K 1,t and T = /1. Let r2, ... r n(f+2 )} be the vertices of R, { s 1 s2, ... be the vertices of degree 1 in S, s be the vertex of degree in S, and t be the vertex in T. Now H = Kn(f+ 2 ) V (K2 +If) G and by Theorems 6.9 and 2.4, x(G, .Pj) 2 x(H, .Pj) = 1 ;=n. To show the inequality in the other direction, color G as follows: form one color class A ofsizej with A= ... ,s,t.,s,r1,r2 ... ,rj41t}. Now (A)= 2 2 Ki4 V (K1 i. + /1), which is a spanning subgraph of Jill V K;2 and by Lemma 2 '2 2 2 6.2, is Pifree. With the remaining vertices form as many color classes of size j 1 as possible. With this coloring, we get X( G, ,pi) :::; I l Theorem 6.11 Let G be a graph of order n. If j 2 6 is even and (E(G)) Kl + K2) then X(G, .Pi)= I l Proof. Assume j 2 6 is even and G is a graph of order n such that (E(G)) = Ki + K2. Then G = R V (S V T) where R = Kn(i+2), S = Ii and T = /2. Let A be a color class in a .Picoloring of G. Suppose that IAI 2 j + 1. Let B with IBI = j + 1. Now B contains up to + 2 vertices from S U T and 93
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at least ? vertices from R. For all combinations of vertices from these sets, we get C = Ki=! V (Jill V 12 ) (B). Let x1 x2 Xi=! be the vertices of degree j 2 2 2 inC, yr, y2, ... Yill be the vertices of degree in C, and z 1 z2 be the vertices of 2 IAI :::;j. To form a color class of size j, we must use all the vertices in S. For if A contains at most vertices from S, then for v E A, we have d(A)(v) :;::: and, by Dirac's Theorem (Theorem 2.7), (A) is Hamiltonian, a contradiction. Thus, there can be at most one color class of size j. Therefore, we get X( G, ,pi) :;::: I i=f l + 1 = I 'J=i l To show the inequality in the other direction, color G as follows: form one color class of size j by using all i vertices from S, one vertex from T, and j;4 vertices from R. Now (A) = li+2 V Ki..2, which by Lemma 6.2, is Pjfree. With 2 2 the remaining vertices form as many color classes of size j 1 as possible. With this coloring, we get X( G, ,pi) :::; I 'J=i l 0 94
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