ON THE PERFECT ORDERABILITY OF UNIONS OF TWO GR4PHS

SUBMITTEDTO THE FACULTYOF GRADUATESTUDIES AND RESEARCH

IN PARTIALFULFILLMENT OF THE REQUIREMENTS

FOR THE DEGREEOF

MASTEROF SCIENCE

IN COMPUTERSCIENCE FACULTYOF ARTS AND SCIENCE LAKEREADUNIVERSITY

BY Xiaodan Tu Thunder Bay, OntKio August 1996

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Table of Contents

List of Tables

List of Figures

Chapter 1 INTRODUCTION 1 1.1 Problem ...... 1 1.2 Applications ...... 2 1.3 Overview of the Thesis ...... 3

Chapter 2 GRAPH COLORING TECHNIQUES AND PERFECTLY ORDERABLE GWHS 4 2.1 NP-Completeness ...... 4 2.2 Maximal Independent Set Approach ...... 5 2.3 Sequentid Approach ...... 7 2.4 Perfectly Orderable Graph ...... 9 2.5 DefinitionandTheorem ...... 9 2.6 Recognizing Perfectly Orderable Graphs is NP-Complete ...... 11 2.7 Orientations and Perfect Orderability ...... 12 2.8 Some Known Subclasses of Perfectly Orderable Graphs ...... 13 2.S. 1 Comparability Graphs ...... 13 2.8.2 Triangulated Graphs and Cotriangulated Graphs ...... 14 2.8.3 fi-Indifference Graphs ...... 16 2.8.4 Pd-Comparability Graphs ...... 16 2.8.5 P4-Simplicid Graphs ...... 17 2.5.6 RaspailGraphs ...... 17

2.8.7 Welsh-Powell Perfect. Graphs and Matula Perfect Graphs ... 18 2.8.8 Graphs with Dilworth Number at Most Three ...... 19 2.8.9 The Union of Two Threshold Graphs ...... 19 2.8.10 Intersection of Two Threshold Graphs ...... 22 2.S.11 D-Graphs ...... 23 2.8.12 BrittIeGraphs ...... 73

Chapter 3 INVESTIGATION OF THE PERFECT ORDERABIL- ITY OF THE UNION OF TWO GRAPHS 25 3.1 Motivation ...... 25 3.1.1 The List of Problems ...... 26 3.1.2 Explanation of the List ...... 26 3.2 The Perfect Orderability of the Union of Two Graphs in Three Cases 30 3.2.1 Some Minimally Non-perfectly Orderable Graphs ...... 30 3.2.2 Comparison with Known Classes of Perfectly Orderable Graphs 36 3.2.3 Conjectures ...... 40 3.3 Proof of the Theorerns ...... 41 3.4 Discussion of the Proof of Theorem 3.3 ...... 50 List of Tables

2.1 SixPossibleSubclassesofPerfectlyOrderableGraplis...... 16

3.1 The List of the Union of Two Graphs...... 27 List of Figures

2 ...... 11 22 An obstruction ...... 12 3.3 ...... 13

2.4 Transitive orientation ...... a...... 14 2.5 Semi-transitive orientation ...... 14 2.6 Simplicidorientation ...... 15 2.7 ...... 16 2.8 Ahouse...... 1'7 2.9 ...... 20 2-10 ...... 21 2.11 ...... 21 2.12 ...... 21

3.1 ...... ,...... 26 3.2 Cs...... 28 3.3 Cs...... 29 3.4 Cs...... 39 3.5 Possible Gi...... 31 3.6 ...... 32 3.8 Example 1...... 33 3.9 Example 2...... 34 3.10 Example 3 ...... 35 3.11 Examplel ...... 36 3-13 ...... 3s

3.24 ...... 48 3.25 Two possible Pq...... 49

3.27 A counter-exampleto the rule ...... 51 3.28 A counter-exampleto the question ...... 52 I would like to express my sincere thanks to my thesis supervisor, Dr. C.T. Hoàng, for his patience, constant encouragement, and guidance in this study. 1 will remain indebted to him for introducing me to scientific research. 1 thank the two anonymous reierees for pointing out several mistakes in the first version of this thesis. I would also like to thank others who have helped me in many different ways. The Depart ment of Cornputer Science gave me academic instructions and pro\-ided convenient facilities. My classrnates and friends, Jian Wang and Jun Wang, offered me help and advice. The Office of Graduate Studies and Research and my supervisor, Dr. Hoàng, provided financial assistance during the period of my study. My parents and sister gave me special encouragement and care al1 the time. Without these help, I would find it difficult to undertake and cornplete this study. To al1 of them 1 express rny gratitude. Abst ract

A gaph G is perfectly orderable if it adrnits an order < on its vertices suc11 that the sequential colonng algorithm delivers an optimum coloring on each (H,c) of (G,<). A graph is a threshold graph,ifit contains no Pd, 2K;. or C4 as induced subgraph. A theorem of Chvatal, Hoàng, Mahadev and de \Verra states that a graph is perfectly orderable if it can be wntten as the union of tiw threshold graphs. In this thesis, we investigate possible generalizations of the above t theorem. We conjecture that if G is the union of two graphs GI and G2 then C: is perfectly orderable whenever (i) Giand Gzare both P4-freeand 21&-free, or (ii) Gl is P4-free, 2K2-freeand Gz is P4-free,Cd-free. We show that the complement of the chordless cycle with at least five vertices cannot be a counter-example to Our conjecture and we prove, jointly with Hoàng, a specid case of (i): if G1 and GZ are two edge- disjoint gaphs that are P4-kee and 2K2-freethen the union of Gland G2 is perfectly orderable. Chapter 1

INTRODUCTION

In this chapter, a bnef introduction to the graph coloring problem and its appli- cation is given, followed by an overview of the thesis. s

1.1 Graph Coloring Problem

Graph coloring problem is to color the vertices of a gaph using a minimum number of colors, subject to the restriction that no two adjacent vertices get the sarne color.

Ever since it was originally formulated in the last century, finding an efficient way to optimally color a graph has attracted the interests of mathematicians. Although rnany significant results have been derived in this field, it is still left to be one of the most intractable problems in discrete mat hematics. We now give the definition of the gaph colohg problem in the term of graph t heory :

Definition 1.1 A graph G is said to be r-colomble if ifs vertices can be colored with r colors in such a way so that no two adjacent vertices are of the same color. The smallest number r for which the graph is r-colorable is called the chromatic nurnber x(G) of the gmph and finding this number is referred to as the coloring problem. Here, G is assumed to be a simple graph, that is. an iindirected graph with neither loops nor multiple edges.

1-2 Applications

The earliest application of gaph coloring can be traced baclc to the last century when it was origindly formulated. The cartographers asked for a minimum number of colors to color a political map in such a way that no two neighbouring countries get the same color. In this exarnple, different countries can be regarded as the set of l vertices V in G, and the two vertices of V are adjacent if and only if the two countries they represent are neighbours. They predicted that four colors always suffice. This is known as the Four Color Problem. mhich was solved by Appel and Haken ([l], [2]) in 1977- A contemporary application of the coloring problem is the scheduling and loading problem, as illustrated in the following example. A computer program stores the values of its variables in memory. For arithmetic computâtions, the values must be entered in "registers". Registers are expensive, so we want to use them efficiently. If two variables are not used at the some tirne, we cm allocate them to the sarne register. For each vaxiable, we compute the first and last time it is used. A variable is active during the interval between these times. We define a graph with the variables as vertices, in which two vertices are adjacent if they are active at a common the. The number of registers needed is the optimum colorhg of the corresponding gaph. Such a graph is called an interual graph:

. Definition 1.2 Givea any fumily of intervals, we can define a graph whose vertices are the intervals, mith vertices adjacent when the intervals intersect. A graph fomed in this way is an , and the fanily of intervals is an intenta1 representa- tion of the graph. Intervol graphs belong to the class of triangulated graphs, the optimum coloring of which can be obtained in potynomial time ([g]). We will discuss triangulated graphs in more detail later.

1.3 Overview of the Thesis

This thesis is concerned with the perfect orderability of the union of two grapiis.

The concept of perfectly orderable graph \vas developed with the motivation to salve the graph colonng problem. After a brief introduction to the graph coloring problem and its application, in chapter 2, we shall give a brief survey of two graph coloring techniques: maximal independent set and sequential approaches. We shall introduce "perfectly ordered graphsn which are the ordered graphs lor which the sequential algorithm delivers an optimum coloring. Chvttal, Hoàng, Mahadev and de Werra proved that a graph is perfectly orderable if it is the union of two threshold graphs. In chapter 3, we shdl invest igate possible generalizations of this t heorem. In particular, we conjecture that if G is the union of two graphs G1 and G2 then G is perfectly orderable whenever (i) Giand G2axe both Pd-freeand 2h;-free, or (ii) G1is P4-free, 216-free and G2is P4-free, Cd-free.We show that the complement of the chordless cycle with at least five vertices cannot be a wunter-example to our conjecture and

we prove, jointly with Hoàng, a special case of (i): if Gi and G2 are two edge- disjoint graphs that are P4-free and 2K2-free then the union of G1and G2is perfectly orderable. - - Chapter 2

GRAPH COLORING TECHNIQUES AND PERFECTLY ORDERABLE

GRAPHS 6

In this chapter, a further discussion of the graph coloring is given. Despite the NP- complete property of this problem, we introduce two basic approaches to color a graph: maximal independent set and sequential, followed by a more detailed addressing on the concept and study of perfectly orderable graphs.

The colonng problem was proved to belong to the class of NP-complete problems ([7]),and worse than that, a polynomial approximation algorithm with a constant error ratio cannot exist unless P = NP ([16]). As a result of its complexity, dthough the number of papers on the coloring problem exceeds that on any other graph problem, no formula has been found for the chromatic number of an arbitrary graph ond we must thus be satisfied with bound estimates. Here, we are going to introduce only two simple bounds: i)lower bound : u(G) 5 x(G) (2-1)

w(G) is the number of vertices in the largest of G. Since al1 vertices of any clique of G must have different colors, this lower bound is obvious.

ii)upper bound : x(G) 5 A(G) + 1 (2.2)

A(G) denotes the maximum degree of a vertes of G. This inequality follows fsom the observation that if A(G) + 1 colors are amilable, then at each vertex u of the graph G at least one of the colors can be used, since at most A(G) colon are used to color the neighbours of v. ~rooks([la]) proved further that there are only two classes of graphs for which the upper-bound holds with equality: odd cycles and complete graphs. Despite the great difficulty, many efforts have ben given to tackie the coIoring problem. There are basicly two approaches: independent set and sequential.

2.2 Maximal Independent Set Approach

A k-colorhg of G is a partition of the set of vertices of G into k independent sets K, V,, ..., such that KnCJ =0 fori #j,i,j = 1,2 ,..., k, and Ui,,Kk = V. Such a partition is cdled a k-coloring partition of V. Thus, the colonng problem is equivalent to finding a minimum k in a k-coloring partition of G. Let 6,i;, ..., be a k-coloring partition of a k-colorable graph G. Then we can constnict a Z-coloring partition I/', I/;, ..., 6'of G such that is a maximal (in the sense of set-inclusion) independent set of G, and 2 < k in the following way: First set

1/i = h,and for each vertex 3 in U fi... (J &. we put z in h' if and only if s has no neighbour in V;. Then we set y' = - {zlz E \/;) for 2 5 i 5 k. It follows t.liat there exists a maximal independent subset LI of the vertices of G such that

(Gv-u is the subgraph of G induced by V - U.) There is a finite nurnber of maximal independent sets W in G. minimizing over al1 such subsets, me obtain

Equation 2.4 is the basis for the following algorithm (Maximal Independent Set approach, or MIS) for cornputing the chromatic number:

Procedure MIS(G, k); Input: a non-empty graph G. output: chromatic number k of G. begin {procedure} if G= 0 then k :=O else begin k := IV(G)I; for each maximal independent set S of G do begin MIS (G - S, Z); if (1 + 1) < k then k := I + 1; end; end; end; Iprocedure} It is easy to see that procedure MIS can be modified to produce a k-coloring of a k-chromatic graph. The complexity of this method is O(mn2.445") (see [19]: as usual n and m denote the number of vertices and edges respectively).

2.3 Sequential Approach

Coloring function is oiten used when discussing the sequentid approach for col- onng a graph. I

Definition 2.1 A function f detemines a k-coloring of G. if

f : V - {l,2, ..., k} (2.5) with f (i) # f (j)for al1 (i,j) E E. A function that defines a k-coloring is called the k-coloring function.

A sequential approach can be stated as the following greedy algorithm: Algorithrn (Greedy coloring) The greedy coloring with respect to a vertex order-

to Y; the smallest-indexed color not dready used on its lower-indexed neighbours. From the above sequential approach, it is easy to determine an upper bound

us(G; VI, vz ,. . ., v,) for the number of colors x,(G) used by the sequential dgont hm applied to G and the ordering of its vertices vl, vz, ..., v,. Every vertex v; can be colored by color i, therefore f(vi) 5 i. On the other hand, at Ieast one of the first deg(vi) + l colors can be assigned to vi. Hence, for every i = 1.2, .... n, and thus

xS(G) 5 uS(G;VI, ~2: -.-, un) = max min{i, deg(vi) 1} (2.7) tsn + This inequality is obtained without any assumption about the ordering of vertices of G. It is obvious that in a sequential approach? the key point is to order the vertices of G, because it esclusively decides the hehaviour of the greedy algorithm. Welsh and Powell ([6]) gave the first version of sequential method by ordering the vertices according to nonincreasing degree. deg(vl) 2 dey(u2) 1 ... 2 dey(u,). Siicli an ordering is cded the largest-first ordering, or LF. A closer inspection of the algorithm and the proof of the inequality(3.7) reveals that for a given ordering v17v*,..-, v, of the vertices of a graph G, instead of f(a;) 1 + deg(ui) we have in fact f(ci) 5 I + degi(vi), where degi(vi) denotes the degree of vertex vi in the subgaph of G induced by VI,vzl-.-, vi. Therefore, the algorithm never requires more than max{l + degi(vi) : 1 5 i 5 n) colors; hence

The following procedure fmds a vertex ordering which minimizes u:(G;VI, v~~ ..., v,):

1. v, is a minimum degree vertex of G.

2. For i = n - 1, n - 2, ..., 2,1, vi is a minimum degree vertex in the subgraph of G induced by V - {v,, .. ., vi+l) -

Such an ordering is cailed smallest-Iast SL. Both LF and SL algorithms may improve the upper bound (2.2) substantially. But neither of them can guarantee an optimum coloring of an arbitrary graph G. It is natural for one to ask such a question: for which kind of ordered graphs does the greedy algorithm produce on optimum coloring? (An ordered graph is a graph with a gîven total order < on its vertices.) This is what motivated V-Chvital to propose the concept of perfectly orderable gaph.

In this section, we introduce the concept of perfectly orderable grapli, and the well-known theorem developed by Chvital that reveals the equivalent nature of per- 1

1

I fectly orderability and admissive order. Knowing that to recognize perfectly orderable

1 graphs is NP-complete, we int roduce some su bclasses of perfectly orderable grap hs. many of which con be recognized in polynomial time.

2.5 Definition and Theorem

Definition 2.2 (Chvital 131) An obstruction in an ordered graph (G,<) is a set of four vertices a, b,c,d with edges ab, bc, cd (and no other edges) and a < b, d < c. A hear order on the set of vertices of a graph will be called i) admissible if it creates no obstruction and ii) perfect if, for each induced ordered subgraph H, the greedy algorithm produces an optimum colorhg of H. A graph wiIl be called perfectly orderable if it admit5 a perfeet order.

Chviital also revealed the nature between perfect order and admissible order by proving the following theorem, which becomes the criterion for identifying perfectly orderable graphs:

Theorem 2.1 (Chvcital [w A linear order of the set of vertices of a graph is perfect if and only if it is admissible. Proof: Since the class of graphs having obstruction-iree orderings is Iiereditary (the inherited ordering for an induced subgraph is obstruction-free), it suffices to show that an obstruction-free ordering L gives a greedy coloring of G that is optimum. If the greedy coloring us'és k colors for the ordering Ltthen optimality can be established by showing that G has a k-clique (a clique on k vertices). To show that such a clique does exist, we introduce the following lemma:

Lernrna 2.1 Suppose that G has a clique Q and a stable set S disjoint /rom Q, crnd suppose thal for each vertez w E Q there is a vertex p(w) E S such fhat p(w) and w are adjacent. If L ik an obstruction-free ordering of G such that p(w) < lu for al1 w E Q, then some ~(w)E S is adjacent to al1 of Q.

Proof: By induction on IQI. The lemma holds trivially for IQ1 = 1, so we may assume IQI > 1. For each w E Q, the graph Q - w satisfies the hypotheses using the clique Q - w and the stable set {p(u): u E Q - w}. %y the induction hypothesis there is a vertex w' E Q - w such that p(w*) is adjacent to al1 Q -W. We may assume that p(w') is not adjacent to w for every PU E Q (for otherwise, p(w*) is adjacent to ail of Q and we are done). This assigns a unique w' to every w, since p(w') is nonadjacent only to w among Q . So setting cr(w ) = wu defines a bijection O on the vertices of Q . Let v be the least vertex of Q in L. Let b, c E Q be the vertices such that bi = v and c* = b. Let a = p(b) and d = p(v). Because p(w') is not adjacent to w and p(w) < w, we have that d is not adjacent to 6, a is not adjacent to c, and d < v and a < 6. Since a,d belong to the stable set S, a is not adjacent to d either (see the

Figure 2.1 bellow. An oriented edge of the form x -t y means x < y and xy E E.) Because d = p(b*), the only vertex of Q nonadjacent to d is b, which implies that c is adjacent to d. Since d = p(v) < v < c in L, the edge dc is onented as d + c, which implies that a, b, c and d induce an obstruction, a contradiction. D Now we continue to prove the Theorem 2.1. Figure 3.1

Let f : V(G)+ {1,2. .. .: k} he the coloring generated by the greedy algori thm with this ordering. Let i be the smallest integer such that G has a clique consisting of

vertices wio,...: tuk such that f (wj)= j. Since f uses k on some vertex, such a clique exists. If i = O, then G has a k-clique. Suppose i > O. For each wj there is a vertex p(w,) such that p(wj) < wj in L knd f(p(wj)) = i; otherwise the greedy colonng would have used a lower color on wj. Since the vertices in S = {p(~i+~), ..., p(wc)}all have color i, S is a stable set. Hence, the conditions of the Lemma 2.1 are satisfied, and there is a vertex of S that can be added to the clique and called wi, which contradicts the minimality of i. Perfectly orderable graphs generalize many well-known classes of gaphs, such as comparability graphs, triangulated graphs and their complements. We shall discuss this fact later.

2.6 Recognizing Perfect ly Orderable Graphs is NP-Complete

When Chvital introduced the notion of perfectly orderable graphs, he also posed the question: how difficult is i t to recognize perfectly orderable graphs? Middendorf and Pfeiffer answered t his question by proving the following t heorem:

Theorem 2.2 (Middendorf, Pfeifer [17]) To deczde whether a g~uphadmits a perfecf

order 2s NP-complete. EI - - They proved the theorem by gïving a reduction of 3SAT to the probiem of deciding whether a graph admits a perfect order.

2.7 Orientations and Perfect Orderability

When studying perfect ly orderable graphs. sometimes it is convenient to rvork with orientations instead of orders.

Definition 2.3 An orientation U of a gruph G is a directed graph obtained /rom G 6 y asszgning a direction to each edge of G. To- an ordered graph (Gt <): there corresponds an orientation D(G,<) of G such that ab E D(G,<)if and only if ab E E(G)and a < 6. So a graph is perfectly orderable if and only if it admits an acyclic orientation that does not contain an induced subgraph isomorphic to the Figure 2.2:

Figure 2.2: An obstmction.

Equivalently: a graph is perfectly orderable if and only if it admits an acyclic orientation in which each induced path of length three is one of the three types in Figure 2.3: Figure 2.3 6

Although the class of perfectly orderable graphs has very nice properties in the

sense of op t imizat ion, there is no known polynomial-time algori thm to recognize it. However, many subclasses of perfectly orderable graphs with special characteristics have been studied, many of which can be recognized in polynomial time. Next, we are going to introduce some of the known classes of perfectly orderable graphs.

2.8 Some Known Subclasses of Perfectly Orderable Graphs

2.8.1 Comparability Graphs

Definition 2.4 A simple graph G is a if it hm a transitive orien-

tation, which is an acyclic orientation such that ifxy,-yz E E(G),x + y and y -, z,

then xz E E(G) and x + z.

From Definition 2.4, a transitive orientation implies that each induced P3 in a comparahility graph is of the types in Figure 2.4, and it is clear there is no obstruction. So comparabi li ty graphs are perfectly orderable.

13 Figure 2.4: Transitive orientation.

The following theorern of Ghouila-Houri ([g]) is the key to a polynomial-tirne algori thm to recognize comparabili ty graphs.

Theorem 2.3 A graph is a comparability graph if and only if it adnrils an orientotiori that contains no induced subgraph isornorphic to the gtxpli in Figure 2.5 (a scrni- transitive on-entation). O

I Note that in the above theorem, the orientation may contain directed cycles-

Figure 2.5: Semi-transitive orientation.

2.8.2 Triangulated Graphs and Cotriangulated Graphs

Definition 2.5 A trïanguiated gruph G is a graph such that euery cycle of length 2 4 in G has a chord, that is, an edge joining two non-consecutive vertices of the cycle. It is also called chordal, rigid-circuit, monotone transitive, and perfect elimination grap h.

A triangulated graph G bas the property that every induced subgraph contains a vertex whose neighbourhood induces a clique (a simplicial vertex). It follows that it admits an order <, such that vi is simplicial in the subgraph induced by H = {xlx < ui)U{vi}. Such an order is called a szmpliciul ordel: In a simplicial order, each P3 is of the types shown in Figure 2.6. So it is obvious that a simplicial order is a perfect order. Testing whether a graph is triangulated is polynomially equivalent to testing whether a graphcontains a sirnplicial vertex which can be obviously solved in polynomial time.

Figure 2.6: Simplicial orïentat ion.

Definition 2.6 A cotriaagulatéd graph is the cornplernent of a triangulated graph.

In a triangulated graph G, the simplicial order implies that each Pq is of the Type 1 or 2 in Figure 2.3. In the complement G of G, define an order

By abcd, we denote the P4 with vertices a, 6, c, d, edges ab, bc, cd (and no other edges). If abcd is a P4 of G, then its complernent in G is bdac. Suppose abcd is a P4 of Type 1 in G (see Figure 2.7 (a)). Without ioss of generality, we may assume that b

5 P4-indifference Polynomid CL a. J 11 6 1 1 1 ,/ 1 Pa-com~arabilitv1 Polvnomial II

Table 2.1: Six Possible Subclasses of Perfectly Orderable Graplis.

2.8.3 Pd-IndifferenceGraphs

Definition 2.7 A graph G is P4-indifFerence if there is an acyclic orientation of G in which every P4 is of Type 2.

2 -8-4 P4-ComparabilityGrap hs

Definition 2.8 A graph G is P4-compôrability if it admits an acyclic orientation in which each P4 is of Type 3.

In a transitive orientation, every Pq is of Type 3. Thus the class of P4-comparabiiity graphs contaiiis al1 cornparabili ty graphs. Figure 2.8: A house.

2.8.5 P4-S implicial Graphs

Definition 2.9 A graph G is P4-simplicial if there is an acyclic orientation of G in ~ul~iclleuery P4 of G is of Type 1 or 2.

In a simplicial order, every -Pdis of Type 1 or 2. Tlius the class of Pd-simplicial gaphs contains al1 triangulated gaphs.

2.8.6 Raspail Graphs

Definition 2.10 A graph is Raspail i/ it admits an acyclic orientation in which every P4 is of Type 1.

Hertz and de Werra ([Il]) gave a characterization of Raspail graphs by forbidding induced subgraphs, among which is the house, as illustrated in Figure 2.8.

It is easy to see that in a house abcde, the two P4s abcd and cdea cannot be both of Type 1. We shall refer to this property of the house later in the thesis. Hoàng and Reed ([15]) also developed polynomial-time algorithms to recognize the above four classes of perfectly orderable graphs. Hoàng ([l3]) also showed that recognizing "generalized CR" gaphs is NP-complete. It is not known whether "one-in-one-out" graphs can be recognized in polynomial t ime. 2.8.7 Welsh-Powell Perfect Graphs and Matula Perfect Graphs

We consider graphs G with linear orders < on the vertices. Welsh and Powell choose < in such a way that,

dc(x) 2 d&) whenever x < y

with dc(x) standing for the degree of x in G; while Matula chooses < in such a way that.

dH(x) 2 dH(g) whenevei x < y (2.10)

and H is the subgraph of G induced by al1 z with z 5 y.

Definition 2.11 G is called Welsh-Powell perfect if the linear order < satisfying (2.9) is perfect (the greedy algorithm produces an optimum coforing for each induced subgraph of G), and Matula ~erfectif the linear order < satisfying (2.10) is perfect.

Chvital, Hoàng, Mahadev and de Werra ([5])proved theorems to characterize Welsh-Powell perfect and Matula perfect graphs by forbidding certain induced sub- graphs. Again a house is such a forbidden subgraph for both Welsh-Powell and Mat- da perfect graphs. They dso showed that these two classes of perfectly orderable graphs can be recognized in polynomial time, and presented an algori t hm t O prove the following theorem:

Theorem 2.4 Given any graph G that is Welsh-powelZ perfect or Matula perfect, one can find in time O(m+ n) a minimum ~010nngand a largest clique in G. Given any graph G whose complement is Welsh-Powell perfect or Matula perfect, one can find in time O(m+ n) a minimum clique cover and a largest stable set in G. 2.8.8 Graphs with Dilworth Number at Most Three

Definition 2.12 Let N(x) stand for the set of al1 the neighbours of a vertez x in G; we say that a vertex y dominates x if N(x) C N(y) U{y). z and y are comparable if z dorninates y or y dominates z, and incomparable if neither of them dominotes the other. The Dilworth number of G is the Zargest number of pairtoise incomparable vertices in G.

It is obvious that the class of graphs with Dilworth number at most three is recognizable in polynomial time. In [5], it is proved that these graphs are perfectly orderable.

2.8.9 The Union of Two Threshold Graphs

Threshold graphs were introduced to study the stable set polytope of a graph.

Definition 2.13 Let G = (V,E) be a graph on n vertices. Then G is a if there ezists a linear inequality

with a,ai E 8 and n = IV[such that the follotoing holds: S c V is a stable set of G if and ody if (2.11) is satisfied by the characteristic vector x, = (xi,zz, .-.,x,) of S where for al1 i

Theorem 2.5 (Chvatal, Hammer [4]) A necessary and suficient condition for G to be threshold is that G does not contain 216, Pq or C4 as an induced subgraph (Figure 2.9.) Figure 2.9

Cliv&al, Hoàng, Mahadev and De \Verra proved the following theorem:

Theorern 2.6 If G is the union of two threshold graphs then G is perfectly orderable.

We are going to give a new proof of the Tlieorem 2.6. But before that, let us first give some definitions about P4: Let abcd denote the Pq with vertices a, 6,c, d and edges ab, bc, cd, the edges ab and cd are called wings of P4. Proof: Let G be the union of two threshold graphs Gl and G2.Consider an' P4

w0v1v2v3 in G (if it exists). It is clear that the two wings vovl and v2v3 cannot be in the same Gi(i = 1,2), otherwise there will be either a 21i2 or a P4 in Gi.

VP4 E G, we now impose a partial orientation on the edges of G:

1. direct + vo if and only if v~vlE Gi and ~1~2E Gi,

2. direct v2 + v3 if and only if vlu2 E Gi and V22)3 E Gi.

It is obvious that such a partial orientation does not create an obstruction in G.

Claim 2.1 Such an orientation is unique on each edge.

Proof: Suppose there is an edge cd that receives two directions by the above orientation. Let us assume cd E GI. We must hâve a P4 ahcd with ab E G2, bc E G1 and another P4 cde f with de E G1,e f E G2-Since bd @ Gland ce 4 Gi,{bl c, d, e} form either a Pq or C4 in Gi, a contradiction (see Figure 2.10.) a (In al1 of the following figures in this thesis, we use solid lines to represent Gi and dashed lines to represent G2.)

Figure 2.10

- - - Claim 2.2 There are- no two directed edges ab. bc (Figure 2.11) such that ab E Gi (respectively G2)and bc E G2 (respectively Gi).

Figure 2.11

@ - -t - Proof: Suppose there are directed edges ab, bc with ab E G1and bc E G2-

Figure 2.12

Since a + b in G1,there must be a P4 deab with de E G2,and eu, ab E G1 (c is not identical to either d or e); similady, since b + c in G2,there must be a Pq gf b~

SI with gf E Gi and fb. bc E G2 (g may be identical to e, but / cannot be identical to

either d or e). Since G2 is a threshold graph, to have both edges ed and bc belonging to G2,there must be ce E G2and cd E G2. Now consider the subgraph consisting of the three edges dc, cb and' bf in G2.We have db 4. E(G)and cf 4 E(G). There id1 be either an induced Pq (if df 4 E(G2)) or Cq (if df E E(G2))in G2,a contradiction (see Figure 2.12.) fi

Claim 2.3 The partial orientation creates no directed cycle in C:. - Proof: We note that for any directed edge ab E G;, a strictly dominates b in G, (adominates b, but b does not dominate a). Therefore, there will be no direct.ed cycle in the same Gi. Furthermore, Claim 2.2 guarantees that.there is no directed cycle consisting of edges in both G1-and G2- a

Now WC can easily extend this acyclic partial orientation to a linear order < on G by the following procedure:

2. Find a vertex v with indegree (the number of directed edges pointing to c) O in H, assign to it the number i. If no such vertex v exists, order the remaining vertices in H randomly and then STOP.

3. H + H - v, i + i + 1, if H is not empty, goto step 2.

Such an order < is obstruction-free and thus is perfect. Therefore, G is perfectly orderable.

2.8.10 Intersection of Two Threshold Graphs

Definition 2.14 A graph G is the intersection of two threshold graphs if there are two threshold graphs Gi and G2 such that e E E(G) e E E(Gi) and e E E(G& Hammer and Mahadev ~rovedA [IO] that intersections of two threshold graphs are perfectly orderable.

2.8.il D-Graphs

Definition 2.15 A vertex x is a d-verter if for any edge yz with {y, z) n N(x) = y and z are comparable. A graph G is a D-graph if each of its induced subgraphs contains n d-verter.

Hoàng ([El)showed that the class of D-graphs contins dl graphs with the Dil- worth number at most three and a11 ~otrian~ulatedgraphs. He olso proved that every D-graph is perfectly orderable, and used this result to prove the the following t heorem:

Theorem 2.7 Let G1 be a threshold graph and let G2 be a graph contazning no zn- duced Pq and no induced C4. Then the union of Gi and G2 is perfectly orderable. 0

A polynomial algorithm with the complexity of O(nm) has been developed to recognize D-graphs ([12]).

2.8.12 Brittle Graphs

Definition 2.16 Let abcd be a P4 of a graph G. The vertices a,d are called endpoints of the P4 and the vertices 6, c are cafled rnidpoints of the P4. G is called bnttle if each induced subgraph H of G contains a vertex which is not an endpoint or a rnidpoint of aP4inH.

Chvatal introduced brittle graphs and pointed out that they are perfectly order- able. For more information on bnttle graphs, see ([Ml). The class of brittle graphs

23 cont ains dl Pd-simplicid graphs, al1 P4-indifference graphs, al1 t riôngulated graphs and their complements, dl Raspail graphs, al1 Welsh-Powell perfect graphs. al1 Matula perfect graphs and al1 graphs with Dilworth number at most three. Chapter 3

INVESTIGATION OF THE PERFECT ORDERABILITY OF THE UNION OF TWO GRAPHS

With the motivation to generalize Theorem 2.6 (that the union of two threshold graphs is perfectly orderable), we study the perfect orderability of the unions of the two graphs, in which some or al1 of the three induced subgraphs P4, C4 and 21\; are forbidden. For three of the au 28 possible csses of the unions, we pose the conjecture that they form a new subdass of perfectly orderable graphs. Two theorems are proved to support our conjecture.

3.1 Motivation

In the last chapter, we introduced a subclass of perfectly orderable class, the union of the two threshold graphs, and also mentioned that by allowing a 21\; in one of - the graphs, Hoàng used the property of D-graph to prove the Theorem 2.7, which generalizes the Theorem 2.6. We know that a threshold graph is a graph that contains no Pq,21' or Cq as induced subgraph. Given two graphs Gi and G2,Theorem 2.7 implies that the restriction t hat bot h Giand G2be threshold in Theorem 2.6 is st ronger tlian necessary for the union G = GiUG2 to be perfectly orderable. If we somehow weaken the restriction on Giand (or) GÎ,will the union G still be ~erfectlyorderable? Wit h this question in mind, we are going to list dl the possible. . unions of Giand G2, in "hich we forbid 1-3 of these three structures (Pd,2K2 and $) as induced subgraphs, and study in eacli. case the perfect orderability of G.

3.1.1 The List of Problems

Weuse Table 3.1 to list Our problems:

3.1.2 Explanation of the List

Xlthough there are 28 different cases of two graphs Gi and G2in the list, many of them can be easily eliminated by finding a counter-example showing that the union G is not necessarily perfectly orderable.

Figure 3.1

The graph in Figure 3.1 is not perfectly orderable. (If we direct a + h, then two edges gc and y f are forced to be directed from g 4 c and g + f, resulting in an 1 contains no 1 contains no 1 perfectly orderable? 1

s P4, C4 p4 NO case I 9 p4, cq (74 NO case 3 10 P4, C4 2 NO case 3 11 P4, c4 P4 ,€4 NO Cs I 12 p4r c4 P4: 2 I& ? 13 p4, c4 C4,2 IC2 NO case 6 14 P4,2 IC2 p4 NO case 1 15 P4,2K2 c4 NO case 2 16 p4i 21(2 2rc2 NO case 3 17 P4,2K2 P4,2 IC2 ? 18 P4,2h; c4,21c2 NO case 6 19 Cd, 2IG p4 NO case 1 20 C4,2K2 c4 NO case 2

21 c4,2K2 2 K2 NO - - case 3

22 C4,-2~~ c 2 -- NO case 6 23 p4 p4 NO case 1 24 p4 c4 NO case 2 25 p4 2 IC2 NO case 3 26 c4 c4 NO case 2 27 c4 2Ii; NO case 3 28 2& 2 K2 NO case 3

Table 3.1: The List of the Union of Two Graphs. impossible orientation on edge de without obstruction.) In this Figure 3.1, we see that Gi is a threshold grapli, and G2 is P4-free, so it is a counter-example to case 1 in the list. Since restrictions on Giare weaker in cases

S, 14, 19 and 23 than in case 1, we must Say "NOn to al1 these cases. - -

Figure 3.2: Cg.

Cs is a well-known simple graph that is not perfectly orderable. If any of these unions in the list contains a Csas induced subgraph, then they are also not perfectly orderable. in Figure 3.3, CS is written as the union of G1and G2 such that G1is threshold and G2 is 2&-free and Cd-&. So it is a counter-example to cases 2, 3 and 6 in the list. It follows that the answers to cases 9, 10, 13, 15, 16, 18, 20-22, 24-28 are "NO". In Figure 3.4, Csis written as the union of Giand G2such that G1and G2 are

both Pd-free and C4-free. So it is a counter-example to cases 11 in the list. Up to this point, we have given a definitive answer "YESn or "NOn to 25 out of

- 28 cases in the list, yet there are still3 cases labeled with "?" that we cannot so easily answer.

Let us list again these three cases for clarity: Figure 3.3: Cs.

Figure 3.4: Cs.

1. Case 5: Gt is threshold, Gz is P4-free and 2IG-free;

2. Case 12: Giis P4-freeand Cd-ire, G2 is P4-freeand 2K2-free;

3. Case 17: Giand Gz are both Pa-free and 2&-free.

It is obvious that the union G in case 5 is a subdass of the unions in both cases 12 and 17. 3.2 The Perfect Orderability of the Union of Two Graphs in Three Cases

In this section, we shall-investigate in more detail the perfect orderabiiity of the union of two graphs in the cases 5, 12, 17. First, we verify that no known minimally non-perfectly orderable graphs can be written as the union of two graphs with the

properties in these three cases. Second, we verify that no known class of perfectl?

orderable graphs can contain such unions. Third. we pose the conjecture that such three kinds of unions are new classes of perfectly orderable graphs, and with an estra

restriction that Gi and G2 are edge disjoint, we prove the union in case 17. hence case 5, is perfectly orderable.

3.2.1 Some Minimally Non-perfectly Orderable Graphs

In this subsection, we introduce some known minimally non-perfectly orderabie gaphs and show that they cannot be the unions of two graphs with the properties in case 5, 12 and 27.

0 Odd Hole

Definition 3.1 An odd hole is a chordless cycle &th odd length at least 5, namely

c2k+1 (k 2 2)-

Csis the sirnplest odd hole, which we used as a counter-example to many cases in the list, as explained in the tast subsection. To show that the odd hole is not a counter-example to the three unanswered cases, we prove the following observation:

Observation 3.1 Let G1 be a graph conlaining no P4,no 2& as indxced subgraphs, and G2 be a graph containing no P4 as an induced subgraph, then the union G = GiU G2 contains neither Cr.(k 2 5) nor fi (k 2 8). Proof: Suppose that G contains a Ci, (k 3 5) or a Pk (k2 8)- Since Gi contains no P4 and 2K2 as induced subgraphs, it is easy to see that Gt can be of only two forms, P2 and P3 (Figure 3.3) in Ck (k 2 5) or Pk (k 2 8) (not considering the isolated vertices of Gl ).

Figure 3.5: Possible Gl.

But in either case, Ck- G1 or Pk - Gl contains a P4 and is a subgraph of G2.a contradiction.

RecalI that G is the complement of a graph G.

Definition 3.2 An anti-hole is the cornplenzent of the cycle Ck.

Theorem 3.1 Anti-hole (k 2 5) is not perfectly orderable.

ProoE B y contradiction. Suppose t here is an admissible acyclic orientation on - Ck (k 2 5).

Let the vertices of an anti-hole (k 2 5) be denoted as va, VI, ..., vk-1, such that vivi+l is not an edge (subscript is taken modulo k).

- Proof: Note that both vivi+2vi-ivi+l and ~)i+2~ivi+~~i+lare P~s in Ck. By assurnption, there is no obstruction in CL, So if Di + Vi+zt there is a forcing vi-1 4 Figure 3.6

v;+i in the P4 zfi~~i+p~i-lt'i+i;if ui+2 4 Oii there is a forcing r;+3 + 1.,+, in the P,

Zt;+~ZiiVi+~üi+l(See Figure 3.6.) CI CVe continue the proof of the theorem. Without loss of generali ty. let os assume

VOV~is directed from vo to q.

Observation 3.2 implies that if k is even, will have a directed cycle i:~z:~c+..q-~CO: if k is odd, will have a directed cycle VOU~V~-..VI;-~V~V~...V~~~V~ (Figure 3.7). a con- tradiction,

k is even A k is odd

Figure 3.7: An anti-hole.

To show that an anti-hole is also not a counter-example to cases 5, 12. and 11. ive shall prove the following theorem:

Theorem 3.2 Let G1 be a graph containing no Pd, no 21i2 as induced subgraphs, and G2be a graph containing no P4 as induced subgraph, then the union G = G, U G2 The proof of Theorem 3.2 will be presented in section 3.3 of this chapter.

a Other Examples

Here, we give some more specific examples of rninimally non-perfectly orderable graphs: il

Figure 3.8: Example 1.

In Figure 3.8, without loss of generality, we may assume that 6 -t 7. There is

a chah reaction forcing 8 -, 4, 3 j 9, 10 -, 1 and 2 + 3, whereupon the vertices 2,3,4 and 8 constitute an obstruction.

l It is clear that the graph G in Figure 3.8 contains a Pa, namely 678439(10)1. and using the Observation 3.1, we know that G cannot be the union of two graphs in case 5, 12 and 17. ii ) The graph G in Figure 3.9 is the union of a q 1234567 and and a Pq 3895. Without loss of generality, we may assume 8 -, 9. Then there is a chain reaction

forcing 5 + 7, 4 -t 6, 3 + 5, 2 -t 4, 1 + 3 and 5 + 1 (by 8 -, 9), whereupon t here is a directed cycle 1351 in G. Figure 3.9: Esample 2.

To show that G cannot be the union of GI and G2in cases 5, 12 and 17. we sliall prove the following two observations:

Observation 3.3 G cannot le-the union of lwo graphs Ci and G2 in case 17.

Proof: By contradiction. Suppose G is the union of G1and Ga in case 17: that is Giand G2are both P4-free and 2K2-free. We notice that there is a P6, namely 895746, in G. Since Gl and Gp are both P4-free and 2h;-free, they can be of only two forms, Pz or P3, in Pb (not considering the isolated vertices). In either case, G2 (or G1)= G - G1 (or Ga)contains a P.4, a contradiction. O

Observation 3.4 G cannot be the union of Gi and Gz in case 12.

Proof: By contradiction. Suppose G1and G2are in case 12: that is Gl is P4-free and C4-free, G2is P4-free and 21G-free. Consider the Ps 895746 induced in G. We must have two outside edges 89 and 46 belonging to Gt only and the middle edge 57 belonging to G2only. (We Say that an edge e belongs to Gionly if c E E(Gi)- E(Gj)with i # j.) In the C4 8953, since 89 is in G1and G1contains no P4 and C4, we can see that either i) 35, 38 E GZonly, and 59 E Glonly, or ii) 35,95 E G2only, and 35 E G1only. Without loss of generality. we may assume the first case occurs. Now, if 25 E Gl,then there is a Pq 259s in Cl; if 25 E GZ,then there is a P4 2538 in G2,a contradiction. O

Since the classes of unions described in cases 12 and 17 contain al1 unions in case 5, the above two observations are sufficient to show G cannot be the union of trvo graphs in cases 5. iii)

G Figure 3.10: Example 3.

The graph G in Figure 3.10 is a minimally non-perfectly orderable graph (the forcing on the edges of G creates a directed cycle 1471).

Suppose G = G, UG2 as in âny of cases 5, 12, or 17. If both edges 34 and 45 belong to Gl, then consider the PS 23456 of G. Since G1contains no Pd, the edges 23 and 56 must belong to G2and they fom a 2K2 in G2,a contradiction. Othenvise, at least one of the edges 34 or 45 must be from Gz, so by symrnetry assume 34 is in G2.Similarly, at least one of the edges 67, or 78 is in Gz.In each case, the set (3, 4, 6, 7) (or (3, 4, 7, 8)) induces a 2h; or P4 in G2,a contradiction. 3.2.2 Cornparison wit h Known Classes of Perfectly Order- able Graphs

In this subsection, we will show that the union of two graphs in cases .5, 1'2 and 17 does not belong to any known clâss of perfectly ordeiable graphs introduced in Chapter 1.

Since the union in case 5 has more strict restrictions on G1, it is sufficient to show that such a union does not belong to an' class of perfectly orderable graphs. Nest.

we are going to give some esamples to illustrate Our conclusion-

Figure 3.11: Example 1.

Ln figure 3.11, the graph G is the union in case 5 (Giconsists of the solid lines, Gz consists of the dashed lines). i) G is not a comparability graph. It cannot be directed without constructing a semi-transitive orientation in Figure

2.5. (If we direct b -t c, then d -t e. Comparability orientations irnply f -t e, a -t e and u -r b, then we have a directed P3 a + b, b 4 c that is not a comparability orientation.) ii) G is not a triangulated graph. It contains a C4. iii) G is not a CO-triangulatedgraph. The two edges e f and bc form a 2K2 in G, so there is a C4 becf in Gowhich is not a triangulated graph. iv) G is not a P4-comparability graph.

Suppose that G is a P4-comparability graph. Without loss of generality: we may

assume b -, c. This implies d + E and E + a. But then one of the P4%feab or fedc is not of Type 3, a contradiction. v) G is not a Pd-indifference graph. In Figure 3.11, if P4 eabc is of Type 2, P4 bcde cannot be of Type 2, otherwise there will be a directed cycle bcdeab. vi) G is not a Raspail graph. - G contains a a house which is a forbidden subgraph in Raspail graphs. vii) G is not a Welsh-Powell perfect nor a Matula . A house is also a forbidden structure in both Welsh-Powell and Matula perfect graphs. viii) G is not a union of two threshold graphs. This is obvious.

Example 2

The graph G in Figure 3.12 (a) is the union of two threshold graphs G1and G2, with G1consisting of the edges ab, ac, bc and Gz consisting of the remaining edges.

Claim 3.1 G is not an intersection of two threshold graphs.

Proof: Suppose G is the intersection of two threshold graphs G1,G2.

Observation 3.5 Let et,ez be two edges that indvce a 21& in G. Then sorne edge ei (i = 1,2) must have one of its endpoints being adjacent to the two endpoints of ej Figure 3.12

(j# i) in Gi and the other endpoint being adjacent to the truo endpoints orej in G2

(Figure 3.1 3). 0

s (In Figures 3.13 and 3.12 (b), the thick lines denote edges in both G1and G2,the solid lines denote edges in G1,and the dashed lines denote edges in G2.)

Figure 3.13

Consider the 2K2formed by {at,c', a, c}. By Observation 3.5, we may assume that an', ac' E E(G1)and ca',cct E E(G2).Now consider the 21G formed by {a, b,at,C}, Observation 3.5 irnpties that bu', bi E E(G2. But the set {at,$, b, c} induces a 21(2 in Gi,a contradiction. O

We conclude that there exists a union in cases 5, 12, and 17 that does not beiong - to the class of intersections of two threshold graphs. Figure 3.14: A non-D-graph.

The graph G in Figure 3.14 is a union of GI (in solid line) and Gz (in dashed line). It is easy to verify that G1is threshold, and G2is P&ee and 21Y&ee. The graph G i) has Dilworth number greater than 3. Let S = {5,6,3,1). It is easy to verify that the vertices in S are pairwise incom- parable, so the Dilworth number of G is at least 4. ii) is not a D-graph.

Recall that a d-vertez w is such a vertex that for every edge ab with {a,b) N(w)= 0, a and L are comparable. There is no d-vertex in G (for vertices 1, 2, 3 and 4, we have that 5 and 6 are incomparable; for vertices 3' and 4', we have that 5' and 6' are incomparable; for vertices 5,6,7 and 8 (respectively 5', 6', 7', and 8'), we have that 3 (3') and 1 are incomparable). Figure 3.15

The grapli G shown in Figure 3.15 is a union of two threshold graplis. The reader may easily check that each vertex is the endpoint of some P4 and the midpoint of some other Pd.Thus G is not bnttle (and therefore not P4-simplicial). The above four examples show that no known class of perfectly orderable graphs

contains al1 unions of two graphs in case 5, or 12, or 17.

3.2.3 Conjectures

Since we are unable to find a non perfectly orderable graph that ran be written as the union of two graphs in case 5, or 12, or 17, we would like to propose the following conjecture:

Conjecture 1 If G is the union of two graphs G1 and G2 satisfying any of the following conditions:

1. Gi is threshold, G2 is P&ee and 2IG-free;

2 Gi is Pd-free and C4-free, Gz is Pd-fieeand 21i2-free; 3. G1 and G2 are both Pd-freeand 2&-free.

then G is perfectly orderable.

If Conjecture 1 is true, then the three classes of gaphs described by it would fom new classes of perfectly orderable graphs (by the results in subsection 3.2.2). Although we have not found a way to prove this conjecture, we do prove. joint 1- with Hoàng, the following theorem by adding the extra constraint that Giand G2 are edge-disjoint for case 3.

Theorem 3.3 ifGl and G2 are both P4-free and SI<*-free, and Giand G2 are edge : disjoint, then G = Gi U G2is perfectly orderable.

s The proof of this theorem is given in the next section.

3.3 Proof of the Theorems

Theorem 3.4 Let Gl be a graph containing no P4, no 2& as induced subgraphs, and GÎ be a graph containing no fi as induced subgraph, then G = Gl U G2 contains no c; (k 2 5).

ProoE By contradiction. Suppose G contains (k 2 5). Number the vertices

of a as vo, vl ... vk-1, such that vivi+l is not an edge (subscript is taken modulo k). Since it is easy to verify that the Theoran is true for k = 5, we may assume that k > 5. Before making any further xgument, we are going to introduce the following observation, which is frequently used in the proof (whenever we say an edge e zs in Gl(I = 1,2) only, we mean e E E(Gl)- E(Gt),t # 1):

Observation 3.6 In any induced C4 of (k 2 j), namely abcd, if ab is in G1 oniy, and bc is in G2 only, then ad is in Glonly, and cd is in G2 only (Figure 3.16). 0 Figure 3.16

Proof: Suppose r?;vi+z and vi+l vi+3 are in the same GL.we need to consider only four cases:

1. The- are both in G1 only;

2. They are both in G2oniy;

3. One is in both G1 and G2,the other one is in Gr (1 = 1,2) only;

4. They are both in Giand G2.

Since G1contains no Pq and ZIG, it is obvious that case 1 and case 4 are not possible. We are now going to show that case 2 and case 3 are not possible either.

case 2: Suppose vivi+z and vi+lvi+s are in G2ody. Since vi+lvi+3uivi+z is a Pq of G, V;V~+S must be in G1 only. Consider the Cq v~-~v~+zv~v~+J:using Observation

3.6, we have vi-lt.i+l belonging to G2 OIJYand vi+svi-l belonging to G1ody. NOW vi-lvi+l cannot belong to either Gior G2 (if Vi-*Vi+l is in Gi then ~i+l~i-1~i+3~iis

a P4 in G1; if Ui-ivi+l is in G2then Vi+lVi-1Vi+2vi is a P4 in G2),a contradiction (see Figure 3.17.)

case 3: Suppose vivi+z is in both G1and G2,~i+l~i+3 is in Gl (1 = 1,2) only- Figure 3.11

Since G1contains no Pq and 2&, vi+lci+3 cannot be in G1 (otherwise, the set

{vi,vi+l, ~i+*,v~+~) induces a P4 .or 21iP2 in Cl),and must be in G2only. and ri~i+3is in G1 only.

4- J Figure 3.18

Consider the C4z?ivi+jvi+,Vi,: Observation 3.6 implies that vi+dvi+l is in G2only, and ViVi+4 is in Glonlg. For j = i + 4, i + 5, ..., k - 1, O, 1, ..., i - 3, by considering the

C4 Z>ivjVi+li)j+l, we see that every edge vivj+l is in Gionly and every edge vi+i vj+l is in G2only. Consider the P4 viVi-~Vi+lVi-l: with ~i+lv;-.Z belonging to G2only and ~i-2~;belonging to G, only, it is obvious that ~;+~vi-lis in Ga only. So al1 the edges going out from vi (except for vi~i+2)are in G1only, and al1 the edges going out from vi+l are in G2only (see Figure 3.18.)

Figure 3.19

Consider the P4 t.i+~Vi+l Vi+~Ui+z: wi t h both edges vi+l vi+3 and vi+, vi+4 belonging to G2 only: it is clear that v+zz';+~ is in G1only- For j = i + 4, .-., - 1,0,1,..., i - 2: consider the C4 vi+1~jvi+2~j+1:where vi+lvj and vi+ivj+i are in Gz only and ui+2t1j is . . in G1 only, we have that every edge vi+zvj+l is in G1only. For j = 2, z - 1, ..., O. k -

1, ..., i + 4 , by considering the Cq Vi+~vjVi+~Vj-l,where vi+3vj7 vjvi+* and ui+zvj-l are in Gl,we see that vi+3vj-l is in G1.NOW consider the P4 ~i+~~i+~~i+~~i+4:we see that dl three edges belong to G,,a contradiction. This completes the proof of Claim 3.2 (see Figure 3.19.)

Observation 3.7 In any induced C4 abcd of E, if ab is in Gi dy, and ad and bc are in Gî, then ad and bc must belong to Cl.

Proof: It is clear that if ad and bc are both in GZonly? we either have a WC2 ab and cd in G1,or a P4 adcb in Gz,a contradiction. Without loss of generality, assume ad belongs to GI and bc does not. Then either adcb is a Pq in G2or badc is a P4 in Figure 3.20

Now, we continue the proof of the theorem. By Claim 3.2 may assume. without loss of generality, that vovz is in Gl only

(if VOV~is in Gî then vlvs is in Gi only, by renumbering the vertices of c,we could arrive at the same conclusion). Again, Claim 3.2 implies that ulv~-land vlv3 are in

G2only. Consider the C4 vov2vi-lv3 in G. By Observation 3.7, one of the following

two cases must occur: (i) either vov3 belongs to G1only or v2vk-l belongs to G1only,

or (ii) both VOV~and vzvk-1 belong to G1and G2.Suppose that case (ii) occurs. Then

we have v~vk-1belonging to G1 only; for otherwise vov3vk-lv2 is a P4 in G2.But now

v3vpt-lu2 is a P4 in G2. Thus we know that case (i) must occur. Without Ioss of generality, we may assume that vov3 is in G1only (see Figure 3.20).

For i = 3,4, .-.,k - 3, consider the C4 voUjv1Vj+l: urith vovj belonging to G1 only

and vlvj belonging to G2 only, by Observation 3.6 we have vovj+l belonging to G1

only and vlvj+l belonging to GÎ only (see Figure 3-21.)

Consider the P4 VJV~V~V;>:with two edges v3vl and vlv4 belonging to G2 only, we

. must have v22)4 belonging to G1only. For j = 4,5, ..., k - 2, by considering the Cq

~1vj~2vj+l,Urith both edges vjvl and vivj+l belonging to Ga only, and v2vj bebnging

to G1 only, we have V2vj+l belonging to G1onIy (see Figure 3.22.) I Figure 3.21

I Consider the C4 VOV~V~-~Z~~:with II~VO. 710~2~ and v2vk-l belonging to ÇI we liave

1 v3vk-1 belonging to G1.For j = - 2; k - 3, ..., 5, considering the C4vzvjvjvj-1: with

three edges v3vj-1, vj-1v2 and ~2t~jbelonging to Gl,we rnust have ~3vjbelonging to Gl as well. Now, look at the Pq v3v5v2v4, al1 three edges of P4 are in Gl, a contradiction to the definition of Cl(se Figure 3.23.) 0 Before we start the proof of Theorem 3.3, we restate it:

Theorem 3.3 If Gl and G2 are both P4-free and 2K2-free, and G1 and G2 are edge disjoint, then G = GiU G2 is perfectly orderable. Proof: Let Gi, G2 and G be as specified in the statement of the Theorem.

Claim 3.3 For euery edge ab E E(Gi) (respectioely ab E E(G2)),the two vertices a, b must be comparable in G2 (respectively G1).

Proof: Suppose ab E E(Gi)and a, b are not comparable in G2.It follows that there exist two edges ad E E(G2),bc E E(G2),such that ac 6 E(G2)and bd 4 E(G2)- Then there will be either a 21i2 or Pq in G2,which is a contradiction (see Figure 3.24.)

O Figure 3-22

Now define a partial orientation on G (we say that a strictly dominates b if a dominates b but 6 does not dominate a).

1. Vab E E(Gi),direct a -, b if and only if a strictly dominates b in G2 ;

2. Vcd E E(GÎ),direct c -t d if and only if c strictly dominates d in G1.

Claim 3.4 G has no obstruction under the partial orientation.

Proof: Since Gl and Ga are PI-& and 21i2-free,the Pq abcd in G, if there is any, rnust be of the following two forms (see Figure 3.25):

By definition of the partial orientation, we have c + d in 1) and b + a in 2). In either case, there is no obstruction. O

Claim 3.5 The partial orientation creates no cycle in G. Figure 3.24

ProoE Clearly, there are no cycles in the same Gi(1 = 1,2) since the domination relation is transitive. Now we are going to show there is no mized cycle, the cycle made up of directed edges in CIand G2. - Observation 3.8 Ifs; E E(GI)(respectiaely, ab E E(G2))and bc E E(G2)(respec-

-C tively, bc E f (GGi)), then there must be & E E(G2)(respectively, E E(Gi)).

-C e Proof: ab E E(Gi) implies a strictly dominates b in G2.Since bc E E(G2),we have ac E E(G2),and by Claim 3.3, a and c are comparable in Gt. Since ab E E(G1) and cb $! E(Gi),a strictly dominates c and there is an orientation from a to c ( see Figure 3.26.) O Figure 3.25: Two possible Pd-

Observation 3.8 implies that there is no mixed cycle on three vertices. Suppose there is a cycle in G, then it must be a mixed cycle. Let C be a cycle - vovl ...vl-1 with smallest length 2. Select vi (i E {O, 1, ...? 1 - l}), SUC~that vi-lvi E

E(G1), and E E(G2). By the Observation 3.8, we have ~i-zi+~E E(G2). Then there is a new cycle vovl...vi-lvi+l---vi_1of length 2 - 1, contradicting with the assumption that C is the smdest cycle. 0 Based on this acyclic partial orientation, we can easily construct a linear order < on G by the following procedure:

2. Find a vertex v in H with indegree O and assign the order i to v; if no such a v exists, order the remaining vertices in H randomly and then exit;

3. i + i+ l,H + H -v, if H # 0 goto step 2. Such an order < is sure ta be obstruction-free, therefore. C is perfectly orderalle.

3.4 Discussion of the Proof of Theorem 3.3

In the proof of Theorem 3.3, we first impose a partial orientation on the edges of G according to a certain rule, which guarantees the orientation to he unique on each edge, obstruction-free and acyclic, later construct a perfect linear order < on the vertices of G based on this partial orientation, thus get the conclusion that C: is perfectly orderable. The key point is how to find such an effective rule for orienting edges. We found it because we noticed the fact that in the two possible P4s in Figure 3.25, if we direct the edges in G.l (respectively G2)accordiig to the strict domination in G2 (respectively Gi),it guarantees there is at least one wing of P4 going out from the joint to the tip, preventing an obstruction. With the conditions provided in Theorem 3.3, we are able to show that such a rule is good in the since it can produce a perfect order on G. Here, the constraint that Gt and & are edge-disjoint is important, otherwise such a dewiD not work, even for the union of two threshold graphs. In the graph of Figure 3.27, a union of two threshold graphs (one in solid lines and one in dashed lines), this strict domination rule causes a directed cycle. Realizing the effect of such a constra.int on the rule, we naturally ask this ques- tion: can a union G of two threshold gaphs G1 and G2 be decomposed into two edge-disjoint graphs and G;, such that cl and are P4-free and 2K2-free? Un- fortunately, a counter-example is found in Figure 3.28 to give a "NOnanswer to the question above. The graph in Figure 3.28 is a union of two threshold graphs with a common edge in both G1 and G2.Since both and G; are P4-free and PI(,-free, if we put edge ef in G;, al1 the edges in G2except ab must be put in G;similarly, if e'f' is in G;, Figure 3.27: A counter-example to the rule. al1 the edges in Giexcept ab rnust be in G;. If ab is not in G;, there will be a Pq ac'bd in G;; if ab is not in G;, there will be a P4 dacb in G;, a contradiction. Figure 3.28: A counter-example to the question. Bibliography

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