Structure of Graph Homomorphisms

Roman Bac& bISc., Comenius' University, Czechoslovakia, 1988

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@) Roman Batik 1997 SIMON FRASER UNIVERSITY August, 1997

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In this thesis we study finite graphs and graph homomorphisms from both, a theoretical and a practical view point. A homomorphism between two graphs G and H is a function from the vertex set of G to the vertex set of H, which maps adjacent vertices of G to adjacent vertices of H.

in the first part of this thesis we study homomorphisms, which are equitable. Suppose we have a fixed graph H. An equitable H-coloring of a graph G is a homomorphism from G to H such that the preimages of vertices of H have almost the sarne size (they differ by at most one). We consider the complexity of the following problem: IXSTANCE: A graph G. QUESTION: Does G admit an equitable H-coloring? We givc a complete charactcrization of the complexity of the equitable H-coloring problem. In particular, we show that the problem is polynomial if II is a disjoint union of complete bipartite graphs, and it is NP-complete otherwise.

To get a better insight into a combinatorial problem, one often studies relaxations of the problem. The second part of this thesis deats with relaxations of graph homomorphisms. In particular, we define a fractional homomorphism and a pseudo- homomorphism as natural relaxations of graph homomorphism. We show that our pseudo-homomorphism is equivalent to a semidefinite relaxation, defined by Feige and Lov&z. We also show that there is a simple forbidden subgraph characterization for our fract ional homomorphism (the forbidden subgraphs are cliques). As a byproduct , we obtain a simpler proof of the NP-hardness of the fractional chromatic number, a result which was first proved by Grotschei, Lovkz and Schrijver using the ellipsoid method. We also briefly discuss how to appty these results to the directed case.

In the last part of this thesis we consider equivalence classes of graphs under the following equivalencer two .graphs G and H are equivalent, if there exist homomorphisms from G to H and from H to G. We study the multiplicative structure of these equivalence classes, and give a necessary and sufficient condition for the existence of a finite factorization of a class into irreducible elements. We also relate this problem to some graph theoretic conjectures concerning graph product. Acknowledgment s

The author is extremely grateful for the continuing support and guidance of his corn- mittee and in particular his senior supervisor, Professor Pavol Hell. The author also wishes to thank Professor Tibor Katrifiik for pointing out the connection with projective lattices in Chapter 4. Contents

... Abstract ...... iii Acknowledgments ...... v 1 Introduction ...... 1 1.1 Basic Definitions and Notations ...... 1 1.2 Background ...... 8 Equitable Graph Homomorphism ...... 15 2.1 Introduction ...... 15 2.3 Overview ...... 16 3.3 PreliminaryResults ...... 17 3.4 Polynomiality ...... 19 2.5 NP-hardness ...... 22 3.6 Conclusion ...... 32 Relaxations of Graph Homomorphism ...... 37 3.1 Introduction ...... 37 3.1.1 Notations ...... 35 3.2 Overview ...... 40 3.3 Relaxations of Homomorphism ...... 40 3.4 Fractional Homomorphisms ...... 41 3.4.1 Consequences ...... 44 3.5 Pseudo-homomorphisms ...... 46 3.6 Examples ...... 51 3.6.1 Graph Coloring ...... 52 3.6.2 Digraph Coloring ...... 56

Chapter 1

Introduction

1.1 Basic Definit ions and Notations

Let 2,N, No, Q and R be the sets of al1 integers, positive integers, nonnegative integers, rational numbers and real numbers, respectively. For a set S, we denote the cardinality of S by 151. For two sets S and TlS C T denotes the strict, or proper inclusion, i.e., S C T and S # T. The set of al1 subsets of a given set S is denoted by 2'.

A graph is an ordered pair (VlE), where V is a (possibly empty) finite set (IV1 < m) and E is a (possibly empty) collection of nonempty subsets of V of cardinality at most two. If G = (V, E) is a graph, then elements of V = V(G)are called vertices of G and the elements of E = E(G) are called edges of G. The sets from E of cardinality one are also called loops. A graph is called simple if it does not have any loop. Two vertices g, g' E V are called adjacent if {g, g') E E and we will wri te g - 9'.

A bijective mapping a : V(G)-, V(H) such that for al1 g,g' E V(G),g - g' if and only if a(g) - a(gf)is called an isomorphism. Since we are interested only in classes of isomorphic graphs, we Say that two graphs G and H are equal, denote G = H, if there exists an isomorphism from G to H.

Example 1 A graph G with k E N vertices V(G) = (1,. . ., k) and k - 1 edges CHAPTER 1. INTRODUCTlON 2

E(G) = {{1,2), {2,3),. . . , {k - 1, k)) is cded the path of length k - 1 and it is denoted by Pk,see Figure 1.1 (B).

Example 2 A graph G with k E N vertices V(G) = (1,.. . , k) and k edges E(G) = {{1,2), {2,3),. .. , {k - 1, k), {k,1)) is called the cycle of length A. and it is denoted by Ck, see Figure 1.1 (A), (C).

Example 3 A graph G with k E N verticcs V(G)= (1,.. . , k) and O edges E(G) = 0 is called the independent set of size k and it is denoted by Ik.

Example 4 A graph G with k E N vertices V(G) = {l,.. . , k) and (i) edges E(G)= {S C V(G) 1 !SI = 2) is called the clique, or the cornplete graph of sire k and it is denoted by ICk, see Figure 1.1 (C).

Example 5 A graph G with s + t E N uertices V(G) = (1,. .. ,s + t) where s, t f N and st edges E(G)= {{i,j) E V(G) 1 i 5 s and s < j 5 s + t) is called the cornplete bipartite graph of size s + t and it is denoted by Ka,t,see Figure 1.1 (LI).

Example 6 A graph G with 1 uertez V(G)= (1) and 1 edge E(G)= {{l)) is called the loop, and it is denoted by 1.

Example 7 A graph G with O vertices V(G)= 0 and O edges E(G) = 0 is called the empty graph, and tt is denoted by O. For a technical reason we also dejine the complete graph I\ro = 0.

Given two graphs G and H, we Say that G is a subgraph of H, if V(G) C V(N) and E(G) C E(H). If at least one of these inclusions is proper, than G is a proper subgraph of H and if E(G) = I"(') 17 E(HJ,then G is called an inderced subgraph of H. For a subset S E V(G) we denote by GIS the induced subgraph of G with the vertex set V(G1S) = S. If h is a vertex of H, we also write H - h for the subgraph HI(V(H) - {h)). Graph G is called a spanning subgraph of H if G is a subgraph of H and V(G) = V(H). A homomorphism is a mapping a : V(G)+ V(H)frorn the vertex set V(G)of a graph G to the vertex set V(H)of a graph H, which maps adjacent vertices of G to adjacent vertices of H, i.e., g - gt in G irnplies a(g) - a(gf) in H. We say that a graph G is homomorphie to a graph H, if such a homomorphism exists, and in this case we will write G -r H, or G 5 H. If there is no homomorphism from G to H, we write G ft H.

A homomorphism G = G is called an endomorphism.

For every graph G, there is a unique homomorphisrn eG to the graph 1 which maps al1 vertices of G to the unique vertex of 1. Also, for every graph G, there is a unique homomorphism ac hmO to G which is just the empty mapping.

If G H is a homornorphism and W is a subgraph of G, then a restriction of a to CV, denote al W, is a homomorphisrn W "'w H defined by (alW)(w) = a(w)for every w E V(W)2 V(G).

A homomorphism W G is called injective, if w # wf E V(W)implies a(w) # a(wt) E V(G), Le., if cr is one to one. An injective homomorphism is also called an monornorphism. A homornorphism G 5 H is called surjective, if h E V(H)implies that there is a vertex g E V(G)such that a(g) = h, i.e., if a is onto vertices. A surjective homomorphism is also called an epimorphism (sometirnes epimorphism is referred to a mapping which is ont0 both vertices and edges, cf. [27], but we will require from epirnorphism to be ont0 vertices only).

A homomorphism which is both injective and surjective, is called a bimorphism.

One can observe that there is a bimorphism from G to H if and only if G is isomorphic to a spanning subgraph of H. Note that every isomorphism is an injective and surjective homomorphism and hence bimorphism. But not every bimorphism is an isomorphism, since one cm define a bimorphism from C4 to K4,but these two graphs are not isomorphic.

An isomorphisrn from G = G is called an automorphism. The automorphism which maps every vertex of G to itself is denoted by lc, and it is also called the identity automorphism. Also, one can observe that for an endomorphism G G the following three conditions are equivalent (since we only consider finite graphs):

a is a monomorphism,

a a is an epimorphism,

a a is an automorphism. P If G H and H -, W are two homomorphisms, then their composition is the mapping Pa, where (/3a)(g)= p(u(g)) for every g E V(G). One can easily observe that G P= W is a homomorphism. Since the operation composition is a composition of mappings, it is associative. Moreover, plc = = IH/?. Also, one can observe that if pa is surjective, then necessarily P is surjective and if Ba is injective, then necessarily a is injective. Moreover, a is injective if and only if a/3 = ay implies B = 7 and a is surjective if and only if Ba = ya implies /3 = 7. CHAPTER 1. INTROD IJCTION 5

Note that if CV is a subgraph of G, then the natural inclusion z which maps vertices of G to the corresponding vertices of H is a monomorphism and ûz = albV.

Graphs G and H are called equivalent, denoted G - H, if G -, H and H -r G. One can observe that the relation - is an equivalence and if G1 - Gz,Hl - Hz and Gi-, Hl then also Gz -+ Hz. Equivalent classes of graphs are also called color families by Welzl [Ml.

P II G 5 H and H -, G are two homornorphisms such that Ba = lc, then we Say that G is a retract of H and we cal1 p a retraction and a a coretraction, i.e, a retraction is a homomorphism which has a right inverse and a coretraction is a homomorphisrn which has a left inverse. Therefore every coretraction is injective and every retraction is surjective. Let G H be a homomorphism, a homomorphism a' is called a weak coretraction, or a weak coretraction belonging to a, if ad is an automorphism on H. One can easily observe that a is a retraction if and only if there is a weak coretraction a' belonging to a, since given a weak coretraction a' belonging to a,d(aat)-' is a coretraction belonging to a. A composition of two retractions is also a retraction, since if aa' = 1 and Bb' = I then (B'at)(a@)= 1-

For a homomorphism G = fl and subsets S C_ V(G):T C V(H)we denote a(S)= {a(s)E V(H)( s E S} and a-'(T) = {t E V(G)1 a(t)E 2'). For a one el- ernerit set S = {s} we will write a(s)instead of ~({s))and a-'(s) instead of a-'({s)).

The product of two graphs G and H,denoted G x H, is a graph with V(G x H) = V(G)x V(H),in which {(s,u),(t, w)} E E(G x H) if and only if {s,t} E QG) and {u, w) E E(H). The homomorphisms G x H 3 G and G x H 3 H defined by

pi(s, u) = s and p2(s, U) = u are called natural projections. If al is a homomorphism

from G to HI and a2 is a homornorphism from G to Hz,then the product of al and a2 is the homomorphism G Hl x Hz mapping each vertex g of G to the vertex (al(g),az(g)) of Hlx Hz.One cm easily check that the product of two homomorphisms is indeed a hornomorphism. On the other hand, every homomorphism CHAPTER 1. 1NTRODUCTION

Figure 1.2: & V3,= C6

G % Hl x H2 is a natural product of homomorphisms al = plu and cr2 = pzu. In particular, G -+ Hl and G + Hz is equivalent tu G + Hl x Hz.

Note that this product is not the familiar cartesian product of two graphs but it is the categoricai product.

The coproduct or sum of two graphs G and H, denoted G + H,is a disjoint union of G and Hl i.e., the vertex set V(G + H) = (V(G) x (O)) U (V(H) x (1)) and for {(s,i), (t, j)) E E(G + II) if and only if i = j and {s, t)E E(G) U E(H). The homomorphisms G I$ G + H and H % G + H defined by il(s) = (s,O) and i2(t) = (t,1) are called natural inclusions. If al is a homomorphisrn from Gl to H and cr2 is a homomorphism from G2 to H, then the sum of al and a2 is the homomorphism Qi+Q2 Gr + G2 -' H mapping vertices of Gr by al and vertices of G2 by a2. One can easily check that the sum of two homomorphisms is indeed a homomorphisni. On the other band, every homomorphism Gl+ G2 -% N is a natural sum of homomorphisms = ail = u(Gi and a2 = ai2 = alG2 (the restrictions of u to G1 and G2). En

particular, Gi4 H and G2 -+ H is equivalent to G1+ Gz -, H.

For two graphs G and H with two distinguished vertices g E V(G) and h E V(H) we define the wedge (or the amalgarn) GVgWhH to be the graph obtained from G+H by identifying the vertices g and h (see Figure 1.2). The hornornorphisms G % G Vgc H and H 3 G V,,I, H defined by il(s) = (s, 0) and iz(t)= (t, 1) are called naturd indu- sions. If al is a homomorphism from Cl to H and a? is a homornorphism from G2 to

H such that for gl E V(GI)and 92 E V(G2), al(gl)= a2(g2),then the wedge of al and a* is the homomorphism Glv,,, G2 Q1''g1,g202- H mapping vertices of Glby al and vertices of G2by a*. One can easily check that the wedge of two homomorphisrns is indeed a homomorphism. On the 0th- hand, every homomorphism Gi V,,, Gz H is a naturai wedge of homomorphisms al =

A core is a graph G, such that every endomorphism from G to G is an automorphism.

A graph G is called bipartite if G + II2. Therefore if G is bipartite, then its vertex set can be partitioned into two (possibly empty) independent sets A and B, i.e., V(G)= A u B (disjoint union) such t hat no two vertices of A are adjacent as well as no two vertices of B are adjacent. If V(Ii3 = {1,2} and a is a homomorphism from G to h;, then A = &(l) and B = ~~(2).We will Say that A and B form a bipartilion of G. One can observe that if G is a connected bipartite graph, then it has exactly one such bipartition up to switching the roles of A and B (1 and 3). If this is the case, we will write G = {A, B}. Recall that if every vertex of the set A is connected to every vertex of the set B, G is called a complete bipartite graph and it is denoted by I'l.il,lsl,since it is determined by the sizes of the sets A and B. We allow A or B to be empty and hence O and In are also complete bipartite graphs.

A graph G is called disconnected if G = H + W where H # O and W # 0, and the subgraphs H and W are called components of G. A graph which is not disconnected is called connected. Since we only consider finite graphs, every graph is a sum of its connected components and this sum is unique up to permutation of summands. CHAPTER 1. INTROD UCTIOiV

1.2 Background

A homomorphism of a graph G into a graph H is also called an H-coloring of G and it is a generalization of a k-coloring of G, which can be thought of as a homomorphism of G into the complete graph IG. We will refer to both H-coloring and k-coloring as graph coloring.

Graph coloring originated with the four color problem (1852) of Francis Guthrie:

1s every planar graph 4-colorable? (cf. [36]).

The positive answer to this question was given by Appel, Haken and Koch [l] after more than a hundred years of development of graph theory.

Graph coloring deals with the fundamental problern of partitioning a set of objects into classes according to some rules. Therefore time tabling, sequencing, constrained scheduling problems or resource allocation problems can often be forrndated as graph coloring problems (cf. (441). For these applications, one wants to determine whether a given input graph is H-colorable.

It is well known that the 2-coloring problem is poiynomial tirne solvable, since one can greedily color vertices of a given graph by two colors (cf. [Ml). For a connected graph, one can easily observe that this coloring is unique up to a permutation of tl colors. Hence if G is connected and G 2 I', G 4 are two 2-colorings of G, then there is an automorphism Ic2 % 1\2 such that 1C, = a$. On the other hand k-coloring is known to be NP-complete for al1 1; > 2 [17].

The computational complexity of the H-coloring problem was completely characterized by Hel1 and Neetcil [30]. In particular, they proved that the H-coloring problem is polynomial time solvable when H is a bipartite graph and it is NP-complete otherwise. CHAPTER 1. I!VTRODUCTION

We would like to mention at this point that the H-coloring problem has proven to be much more difficult in the directed case. In [21] it is proved that there are oriented trees H and also some oriented cycles for which the H-coloring problem is KP-complete. On the other hand, for dl oriented paths and some oriented cycies H, the H-coloring problem is polynomiai time solvable (see [34, 431). According to the latest results of Hell, Zhu, and N&etfil[32] and independently of Feder and Vardi [15], for digraphs H with a property calied bounded treewidth duality (for the definition see [32]), the H-coloring is polynomial time solvable. It is conjectured that the H- coloring problern is not polynorniai otherwise 1'291. This conjecture is a counterpart of the châracterization in the undirected case, since bipartite graphs are the only graphs which have bounded treewidth duality in the undirected case [29].

Another class of problems, which are considered from the cornplexity point of view, is to determine whether a given input graph has some property. In particular, there is a polynomial time algorithm to determine whether a given graph is bipartite (this is the same as 3-coloring the graph). It is also well known that finding connected cornponents of a given graph can be donc in polynomial time [44] and therefore determining whether a given graph is connected is aiso polynomial time solvabIe. On the other hanci, determining whether a given graph is a core was shown to be NP- hard by Hell and Neietiil [31]. In Chapter 2 we witl study the equitable H-coloring prob1e.m and give a cornplete complexity classification of this problem. The equitable H-coloring problem is a natural generalization of the equitable coloring problem studied in [40, 45, 20, 231. Vie will prove that the equitable H-coloring problem is polynomial time solvable whenever H is a disjoint union of cornplete bipartite graphs and is NP-hard otherwise.

With each loopless graph G, we associate two numbers and w which are called the chromatic number and the clique number of the graph, respectively. The definitions are as follows: x = x(G) = min{n E No 1 G + K,) and w =w(G) =max{n E No 1 h;,+ G). Computing these numbers is known to be NP-hard 1171. If ,8 is a homomorphism H 5 t ben G "t H implies G 3 h7&) and hence x(G) C X( H). Therefore x is the rame for two equivalent graphs. Similarly, if 7. is a homomorphism K,(q G, then G % H irnplies % H and hence w(G) 5 w(H). Therefore w is the same for two equivalent graphs. The ~urnbersx and w also have the following integer linear program formulations:

In these formulations we take v to range over the vertex set of G and I to range over al1 maximal independent sets in G. If we replace the integrality conditions z, E {O, 11, yr E {O, 1) by x, >_ O, >_ O respectively, we obtain linear prograrns with optima wf(G),the fractional clique number of G, and %j(G),the jractional chromatic number of G. One can observe that these linear programs are duals of each other and hence wf (C) = xj(G) by the duality theorem of linear programming 11 11. Feasible solutions {x,) and {w*) of these linear programs are called fractional clique and fractional coloring respect ively. Therefore we have the inequalities:

4G) 5 wr(G) = xAG) C x(G)-

If G 5 H and {wr)is a fractional coloring of H, then one can define a fractional coloring {uJ)of G by w1 if J = a-'(1) O otherwise.

Therefore G -, H implies wf(G) = xf(G) 5 w,(H) = xf(H) and we conclude that wj = XI for equivalent graphs. To compute the fractional chromatic number is NP- hard. This result was first proved by Grôtschel, Lovisz and Schrijver in 1201 proving the fact that if one can optimize over a polytope in polynomial time then (using the CHAPTER 1. INTRODUCTION

ellipsoid method) one can optimize over its antiblocker (for the definition we refer interested readers to [20]). Since the antiblocker of the fractional clique polytope is the independent set polytope, this shows NP-hardness of fractional chromatic number. We will give an alternative algebraic/combinatorial proof of this result in Chapter 3.

Grotschel, Lovisz and Schrijver in 1201 also proved that one can compute in polynomial time a real number 6, which satisfies

The definition for 0 is as follows [42, 391: Let G be a graph with IV(G)I = n. An orthonormal corepresentation of G is a system {v~}~~~(~)of unit vectors in a vector space Rr, such that if {i, j) E E(G) then vivT = O. Every graph has an orthonormal corepresentation in Rn, for instance an orthonormal bais of Rn. Let u = {u~}~~~(~)range over al1 orthonormal corepresen- tations of G and el be the unit vector with first coordinate equal to 1, then 1 0(G) = min max ~w(G)(er~,)~'

One can observe that G = H implies that B(G) 5 B(H), since if {v~)~~~(~)is an orthonormal corepresentation of H,then {L~~(~)}~~~(~)is an orthonormal corepresentation of G. Tlierafore O is also the same for equivalent graptis and one can consider O(G) 5 8(H) to be a polynomially cornputable necessary condition for the existence of a homomorphism from G to H.

Another polynomially cornputable necessary condition called a hoaz was given by Feige and Lovk in [16]. They used a semidefinite relaxation of a quadratic program, which was obtained from a multi-prover interactive proof system for the graph homomorphism problem.

The mode1 of interactive proofs was introduced by Goldwasser, Micali and Rackoff [19] for cryptographic applications, and by Babai [4] as a game-theoretic extension of CHAPTER 1. INTRODUCTION

NP.

The model consists of a probabilistic polynomid-time verifier V and a prover P who tries to convince V that the input x is in a language L.

L E IP if there exists a verifier V that is always convinced when x E L, but if x # L then any prover P ha only a small probability of convincing V to the contrary.

The mode1 of multi-pmuer interactive proofs was introduced by Ben-Or, Gold- wasser, Kilian, and Wigderson [SI.

The model consists of a probabilistic polynomiai-time verifier V communicating with two provers who cannot communicate with each other during the protocol. The provers try to convince the verifier that the input x is in a language L.

L E lCl IP if there exists a verifier V that is always convinced when x E L, but if x 4 L then the provers have only a srnail probabiiity of convincing V to the contrary.

Shamir [48] proved that IP = PSPACE and Babai, Fortnow, and Lund [5] proved that MIP = NEXPTIME.

For the graph homomorphism problem, Feige and Lovkz in [16j made the following two prover interactive proof system.

Let us Say that the Verifier is trying to determine if a graph G is hornomorphic to a graph H. The Verifier chooses randornly and independently vertices s and t of G and sends them, respectively to the provers Pi and Pz. Then Pl replies with a vertex u [rom H as the claimed image of s, and P2 with a vertex w from H as the claimed image of t. The Verifier accepts just in the following situations: 1) Ifs = t then u = w 2) 11 s is adjacent to t in G, then u is adjacent to w in H. This system can be viewed as an optimization problem over probabilistic strate- gies of the provers and therefore it has a quadratic programming formulation, whose convex (positive semidefinite) relaxation can be solved in polynomial time using the ellipsoid method. In Chapter 3 we will define a pseudo-homomorphism in terms of û and a hom-product (to be defined) of graphs G and H,and we will show that our pseudo-homomorphism is equivalent to the notion of hoax of Feige and Lovisz.

The chromatic number is interesting also from the theoretical point of view. One of the most famous conjectures in graph theory is Hedetniemi's conjecture [26]

A graph W is called multiplicative if for every two graphs G and H,

G x H + W implies that G - GV or H -t W. Hedetniemi's conjecture is equivalent to the assertion that al1 complete graphs are multiplicative. One can easily prove that ICI and K2are multiplicative and the proof of the multiplicativity of K3 W~Saccomplished by El-Zahar and Sauer [14]. This conjecture \vas extensively studied by Burr, Erdos and Lovkz [9], He11 [18], Turzik [sl],Duffus, Sands and Woodrow [12], Wlzl [55] and others and they obtained some partial results. It is not even knoivn whether the function

p(n) = min{m 1 G f, Ii', and H 4, hi, but G x H -, Ii,.,,)

tends to infinity or not. Hedetniemi's conjecture is equivalent to the assertion that for al1 positive integers n, p(n) = n+l. Poljak and Rôdl[46] proved that p(n)either tends to infinity with n, or is bounded above by 16. Duffus and Sauer [13]observed that one can naturally define the product and the surn of the equivalence classes of graphs (color families) and that they form a Heyting algebra. With every class they associated a particular Boolean algebra such that the class contains multiplicative graphs if and only if the corresponding Boolean algebra has exactly two elements. They proved that if al1 Boolean algebras associated with classes which contain complete graphs CHAPTER 1. INTRODUCTiON 14 are finite, then p(n) tends to infinity with n. They also asked for conditions for the Boolean algebras to be finite. In Chapter 4 we will further follow this approach and show that the Boolean algebra associated wit h a particular color class is finite if and only if this class has a finite factorization into meet irreducible elements. We will also give a necessary and suficient condition for an element of a Heyting algebra to have a finite factorization into meet irreducible elements. Chapter 2

Equitable Graph Homomorphism

Introduction

In this chapter we consider the equitable coloring problem and a natural generalization of it, which we cal1 the equitable H-coloring problem for simple graphs. Equitable colorings were studied in [40,45, 10,231, but not from the viewpoint of complexity. Here we give a complete complexity characterization of the equitable H-coloring problem.

A graph G is said to be equitably colored with k colors, if the vertices of G are partitioned into k classes VI,.. . , Vk such that each L$ is an independent set and the classes I.$, i = 1, . . . ,k have almost the same size, i.e., 1 Il/, 1 - 16 II < 1 for al1 i and j. Recall that the graph G is hornomorphic to the graph H (G -, H), if there is a mapping a : V(G)-+ V(H)such that (s, t) E E(G) implies (a(s),a(t)) E E(H). Such mapping a is a homomorphism from G to H. CVe Say that a homomorphism cr is equitable, if the sets a-'(h) = {g E V(G)1 a(g) = h) have almost the same size for al1 h E V(H),i.e., I(a-l(h)l - la-L(h')ll 5 1 for al1 pairs h, h' E V(H). If there is an equitable homomorphism cr from G to H, we write G - H or G -CI H. An equitable homomorphism G - H is also called an equitable H-colon'ng of G. Given a graph H, by the equitable H-coloring problem we mean the following decision problem: INSTANCE: A graph G QUESTION: 1s G -u H? CHAPTER 2. EQUITABLE GRAPH HOh.IOMORPH1SbI

In this chapter we will prove that the equitable H-coloring problem is polynomial time solvable when H is a disjoint union of complete bipartite graphs, and it is NP- hard otherwise. A similar result holds if we restrict the instances of this problem to be connected graphs. In particular, the restricted version of the equitable H-coloring problem is polynomial time solvable whenever H is either not connected (for essen- tiany trivial reasons), or is a complete bipartite graph, and it is NP-hard otherwise. The restriction of equitable H-coloring problem to connected instances will be called the connected equitable H -coloring problem.

2.2 Overview

In this chapter, we give a complete complexity characterization of the equitable H- coloring problem. In Section 2.3, we introduce some definitions and make some easy observations about equitable homomorphisms which will be needed further. In Section 2.4, we observe that the connected H-coloring problem is trivial in the case when H is disconnected and it is polynomial time solvable if H is a complete bipartite graph. Consequently we prove that the equitable H-coloring problem is polynomial time solvable if H is a disjoint union of complete bipartite graphs. In Section 2.5, we will show that both problerns are XP-hard in a11 other cases. In particular we give a Turing reduction from the connected H-coloring problem to the connected equitable H-coloring problern which settles the case when H is not a bipartite graph. Then we give a polynomial reduction from the balanced complete bipartite subgraph problem to the connected equitable P4-coloring problem which settles the case of the smallest noncomplete bipartite graph. Then we will settle the case when H is a connected bipartite graph which is not a complete bipartite graph. This will be done by showing a Turing reduction from the connected equitable P4-coloring problem. To settle the last case, Le., the equitable H-coloring problem when H is not a disjoint union of complete bipartite graphs, we will show a polynomial reduction to H from the equitable Hf-coloring problem where H' is a connected component of H which is CHAPTER 2. EQUITABLE GRAPH HO~IOLMORPHISM 17 not a complete bipartite graph. In Section 2.6, we try to give some insight into ideas which are behind Our reductions.

2.3 Preliminary Results

In this section, we wiii make some simple observations about equitable homomorphisms. First of all, let us notice that if the number of vertices of G is n and the number of vertices of H is m, then n can be uniquely written as n = km + r, where O 5 r

Lemma 1 G 5 H is an equitable homomorphism ij and only if for al1 h E V(H):

Proof: If for some h E V(H)we had Iu-'(h)l 5 Lzj - 1 then

which is a contradiction. Similarly if for some h E V(H) we had la-'(h)l 3 [El + 1 t hen n n n n n'([-l+l)+(m-l)[-1 =m[-l+l?m-+l=n+l, m rn m m which is a contradiction. O

Corollary: G % H is an equitable homomorphism if and only if for r vertices h of H we have la-'(h)j = k + 1, and for n - r vertices h of H we have lu-' (h)! = k.

For a given graph H we Say that w : V(H)-+ N is a weight function on H, Le., if w maps vertices of H to positive integers. Let us fix the graph H and a weight function w on H; we write w(H) = ChEv(Hl~(h).We say that a is a w-weighted equitable homomorphism to H, written as G 3, H, if for al1 h E V(H) CHAPTER 2. EQ UITABLE GRAPH HOM Oh1 ORPHISM 15

We also write G --tu H if there is a w- weighted equitable homomorphism from G to Ji.

Let H be a graph. We write NH(h) for the neighborhood of the vertex h in H, i.e.,

NH(h) = {h' E V(H) ( h .Y h'}. We say that two vertices h,hl E V(H) are similar, denoted h G h', whenever NH(h)= NH(hf). This relation is an equivalence relation on the vertices of the graph H and partitions them into disjoint equivalence classes. Let H' be a graph, whose vertex set is equal to the set of equivalence classes of the vertices from H, and such that two equivalence classes c and c' are adjacent whenever there are two vertices h E c and h' E c' such that h .Y h' in H. One can easily check that H' is a well defined loopless graph and that there is a natural homomorphism sr from H to H' mapping a vertex h E V(H) to the equivalence class c with h E c.

Lemma 2 Let G and H be two graphs with (V(G)I = n and (V(H)I= m. Let w be a weight junction on Hr defined by w(c) = \cl for euery equivalence class c. Then G - H if and only if G -+, Hr. Prooj: Let a be an equitable homomorphism from G to H and let a be the natural homomorphism from H to Hr. Then for al1 h E V(H)

Therefore summing for al1 h E c, we obtain

On the other hand, let us assume that there is a homomorphism G P Hr. This implies, that we can (arbitrarily) partition vertices from ,û-'(c) into Ici disjoint sets A;, . . . ,Aicl such that 1-1 < IAi( < r$l for al1 i = 1,. .. , (cl. If we denote the vertices of the class c by h;, . . . ,hi,, then mapping al1 vertices of At to the vertex ht defines an equitable homomorphism from G to H. O

On the other hand, for a weight function w on H we can define a graph Hw with the vertex set V(H,) = UhEv(H){h)x (1,. . . ,w(h)} for which two vertices (h, i) and CHAPTER 2. EQUITABLE GRAPH HOMOICIORPHISICI

Figure 2.1: (Cs), for w = 2.

(h', j) are adjacent if and only if h and h' are adjacent in H. If w is a constant function, i.e., there is a constant k such that w(h) = k for al1 h E V(H),then we will simply write w = 6. (see Figure 2.1). Note that we have replaced each vertex h with w(h) copies of itself. One can observe that H = H, for w = 1 and there is a naturai homomorphism R from H, to H mapping the vertex (h,i) to the vertex h. The following lemma is now a straightforward consequence of these definitions.

Lemma 3 Let G and H be two graphs. Then (a) G' --+, H if and only if G u Hw, moreover if G 5 H,, theri G H where * is the natural homomorphism. (b) G --, H ij and only if there is a bimorphism jrom G to some H, such that [f J 5 w(h) 5 [$l for al1 h E V(H).

In this section, we will observe that the connected equitable H-coloring problem is trivially polynomial time solvable if H is not connected. Then we will show that if H is a complete bipartite graph, then the connected equitable H-coloring problem is polynomial time solvable. Finally we will prove that the equitable H-coloring problem CHAPTER 2. EQUITABLE GRAPH HOMOMORPHISM is in P whenever H is a disjoint union of complete bipartite graphs.

First, let us consider the connected equitable H-coloring problem and assume that H has at least two connected cornponents, Le., H = Hi + . .. + H,, q 2 2. Then any connected graph G must be mapped to one of Hi, Say HL. Therefore, if G 5 H, then lu-'(h)!= O for al1 h E V(H,)for 2 5 i 5 q. Since a is an equitable homomorphism,

where n = 1 V(G)[ and m = lV(H)I. We conclude that 151 = O and [ml 5 1. There- fore no two vertices can be mapped to the sarne vertex, and hence the homomorphism is injective. This implies that G is a subgraph of Hl and therefore also of H. Since H is fixed, it can be decided by brute force whether G is a subgraph of H in constant time (since G can contain at most IV(H)I vertices to be equitably homornorphic to H). FVe conclude that in the case that H is not connected, the connected equitable H-coloring problem is trivially polynomiai time solvable.

Next, let us consider the connected equitable H-coloring problem and assume that H is a cornplete bipartite graph I

vertex 2 or vice versa. Therefore G -c, I\',,t if and only if either

which cari be decided in polynomial time. Therefore we have just proved the following lemma. CHAPTER 2. EQU1T.LIBf.E GRAPH HOMOhlORPHISICl 21

Lemma 4 If H is a complete bipartite graph, then the connected equitable H-colo*ng problem is in P.

Now we can prove the main result of this section.

Theorem 5 1f H is a disjoint union of complete bipartite graphs then the equitable H-mloriag pmblem is in P.

Proof: We wiil modify the dynamic programming approach to the partition probiem from Garey and Johnson 1171 p. 90. Assume the connected components of H are I<,l,,I,.. . , and the connected components of G are Gi, . . . , Gk.One can observe that the graph H' consists of r disjoint copies of Kz and we wilI denote the two vertices

of the j-th copy by aj and ~. By Lemma 2, G -A H if and only if G --+, Hr, where w is defined by w(aj) = u, and w(P) = vj. Let a be a homomorphism from G to Hr and let si = la-'(aj)l denote the number of vertices mapped to aj by a and ti = la-'(@)[denote the number of vertices mapped to &' by a. According to our definition, cr is a w-equitable homornorphism if and only if for al1 q = 1,. .. ,r the following inequalities are satisfied:

and

By Lernma 2 we COlnclude that C w H if and only if there is a homomorphism û from G to H', such that these conditions are satisfied. For integers 1 5 nt <_ k, 0 5 il C U,[W]....,O 5 j, < ~~[ml,O 5 11 5 vlrHl,..., 0 6 1. 5 vrrlvw~~let t(m,jl, . . . ,j,, IL,. .. ,Ir) denote the truth value of the staternent: "there is a homomorphisrn a from G1 + . . . + G, to Hf, such that for al1 q = 1,. . . ,r,

sQ = j, and ta = 1,". One cm observe that G -t H if and only if there is an entry t(k, il,. . . ,j.,ll,. . ., 1,) which ha value T and for al1 q = 1,. . . ,r, j, > ~,l,~(~),ml and 1, > v, lm].We will show a recursive procedure that can be used to rompute t(k, jt, . . . ,j,, II,. .. ,Ir). First of dl, one can observe that t(1,jr,. .. ,j,, fi,. . .,Ir) = T CHAPTER 2. EQUITA BLE GRAPH HOMOhfORPHIS.hf 22

if and only if there is a q E (1,. .. ,r), such that j, = (AlIl 1, = (Blland j, = 1, = O for p # q, or j, = IBII, 1, = IAiI a.nd j,, = 1, = O for p # q, which encodes that G1 is mapped to the qth copy of h; and either the vectices of Al are rnapped to the vertex aq and the vertices of BI are mapped to the vertex bQ or the vertices of Al are mapped to the vertex bq and the vertices of Bl are mapped to the vertex a,. Having cornputed al1 t(m, jl, . . ., j,, 11,. . . ,Ir) with given m, t(m + 1,jl,. .. ,j,, 11, - .. , 1,) = T if and only if there is a q f (1,. .. ,r), such that either

which encodes that G,+l is mapped to the 9th copy of h;. The number of tuples (k, j,, . . . ,j,, il,.. . ,IV)is bounded by IV(G)12'f' and thereiore one can decide whether

G -A H in polynomial time. O

Both the equitable H-coloring problem and the connected equitable H-coloring problem are clearly in NP. In this section, we will focus our attention on finding reductions from known NP-hard problems. Since the connected equitable H-coloring problern is a restriction of the equitable H-coloring probiem, if we prove that for a given graph H the connected equitable H-coloring problem is NP-hard, then an immediate corollary is that the equitable H-coloring problem is also NP-hard. Similarly as in the equitable case, the version of the H-coloring problem restricted to the connected instances will be called the connected H-colon'ng problem.

Theorem 6 There is a Turing reduction jrom the connected H-coloring problem to the connected equitable H-coloring problern. CHAPTER 2. EQ UlTABLE GRAPH H0:IIOMORPHISM 23

Proof: Let us assume that the vertex set of H is V(H)= { 1,. . ., m). Let G be an instance of the connected H-coloring problem, that is, G is connected. Let us fix a vertex gr E V(G)and let IV(G)I = n. We will show that the following two conditions are equivalent : (a) G + H (b) G Vpt,h f.- H for at least one h E V(H)and at least one weight function w on H such that w(1) + . . . + w(m)= mn + 1 The implication (b) + (a) follows from the fact that G + GV,tIh Hu for al1 h E V(H) and al1 weight functions w on H. It remains to prove the ihplication (a) (b). Let a be a homomorphism irom G to H. If we denote xi = la-I(i)l for i = 1,. .. ,m, then xl+. . .+xm = n. Therefore mn = (m+l)n-n = ((m+l)-xl)+ ...+(( m+L)-x,). Without loss of generality we may assume that rr(gf) = 1. It follows, that for w such that w(1) = (rn + 1) - XI + 1 and w(h) = (rn + 1) - xh for h # 1 we have aVpt,ln GV,l,I H, - H where r is the natural homomorphism from Hw to H. (We added 1 to the first element, since gr and 1 are identified by the wedge.) Since G and H are connected, G v,~,~H, is also connected. To see that Our reduction is polynomial, one can observe that the graph G Vgi,hHw has (m + 1)n nodes which is 0(n)for the

fixed graph H. The nurnber of graphs G ~~f,~Hu is m (z), since m is the nurnber of ways we can choose h E V(H)and (r,)is the nurnber of solutions to the equation wl + . . . + w, = mn + 1 in positive integers. For a fixed graph H, this is O(nm-') and we conclude that our reduction is indeed polynomial. O

Hel1 and Neetfil (301 have shown that for nonbipartite H, H-coloring is NP- complete. For connected H, G -, H if and only if for al1 connected components C

of G', G' -+ H. Therefore there is a Turing reduction from the H-coloring problem to the connected H-coloring problem. This implies the following corollaries.

Corollary 7 If H is a connected nonbipartite graph, then the connected equitable H -coloring problem is NP-hard.

Corollary 8 IJ H is a connected nonbipartite graph, then the equitable H-colon'ng problem is NP-hard. CHAPTER 2. EQ UITABLE GRAPH HOA4OiLIORPHISiCI

The following problem, known as the balanced complete bipartite subgraph problem is NP-complete, see [l?]: INSTANCE: Bipartite graph G and a positive integer m 5 iIV(G)(. QUESTION: Does G have a subgraph isomorphic to Km,,?

We will actually need the following restriction of the balanced complete bipartite subgraph problem.

The restricted balanced cornpiete bipartite subgraph problem: INSTANCE: Bipartite graph G, its bipartition {A, B} and a positive integer m such that f rnax(lA1, IBI) c m 5 $lV(G)I. QUESTION: Does G have a subgraph isomorphic to Km,,?

Lemma 9 The restricted balanced complete bipartite subgraph problem is NP-cornplete.

Proof: Membership in NP is trivial. We will give a polynomial-time reduction frorn the balanced complete bipartite subgraph problem. Let G and m be an instance of the balanced cornplete bipartite subgraph problem. It is well known that ive can find a bipartition {A, B} of G in polynomial time. Let the number of vertices of G be n = IV(G)I. Let US construct a new graph G with bipartition {A', B') as follows. First, ive create 2n new vertices. Let A' contain a11 of the vertices of A and n of the new vertices and let B' contains al1 vertices of B and the remaining n new vertices. We connect a E A' with b E B' if and only if a is a new vertex, or b is a new vertex, or a - b in G. One can observe that G has a subgraph isomorphic to Km,, if and only if G' has a subgraph isomorphic to IL+,,,+,. Moreover iIV(Gt)1 = f 1 V(G)1 + n 2 m + n > n = '+y2 2 :+$ max{lA(,(BI) = $ max{n+IAl,n+IBI)= $ mm{lAtI, IB'I}.

Let us recàll that P4 denotes the path of length three on four vertices, Le., the graph obtained from 1\2,2= C4 by removing one edge (see Figure 1.1 (B)). Note that P4 is the smallest connected graph which is not a cornplete bipartite graph. The following proposition will serve as a basis for our inductive proof.

Proposition 10 The connected equifable P4-coloring problem is NP-complete. Proof: Recall that the vertex set of P4 is V(P4) = {1,2,3,4} and the edge set is E(P4) = {{1,2), {S, 3), (3,411. We will give a polynomial-time reduction from the restricted balanced complete bipartite subgraph problem. Let bipartite graph G, its bipartition {A, B}, and a positive integer rn such that f max{lAl, lBl} < m < f lV(G)l, be an instance of the restricted balanced complete bipartite subgraph problem. We construct a connected bipartite graph G' with a bipartition {A', B'} as follows. First, we create 4m - IV(G)I new vertices. Let A' contains al1 vertices of A and 2m - \Al of the new vertices and let Br contains al1 vertices of B and the remaining 2rn - IBI new vertices. LVe connect a E A' with b E B' if and only if a is a new vertex, or b is a new vertex, or a + b in G. The restrictive condition on m impiies that we have added a positive number of new vertices to both A and B and therefore our graph G' is connected. We now daim that G has a subgraph isomorphic to Km,, if and only

if G' --, P4. Let us assume that G has a subgraph isomorphic to A',,,. This irnplies that there are vertices al,. . ., a, E A and vertices 61,. . . ,6, E B such that for al1 i = 1,. . . ,rn and j = 1,. . . ,m, ai .- b, in G. Define a mapping A' u B' {1,2,3,4) so that the rn vertices a, are mapped to 1 and the remaining m vertices from A' are niapped to 3. Sirnilarly, the vertices bj E B' are mapped to 4 and the rest of the vertices from Br are rnapped to 2. One can easily observe that this rnapping is an equitable homomorphism from G' to P4. For the converse, let us assume that a is an equitable homomorphism from Gr to P4. It follows that ICI-'(i)(= rn for each i = 1,2,3,4. Without loss of generality we may assume that a(A') = {1,3) and a(B1) = {2,4). Let a-'(1) = {al,. .-,a,} E Ar and a-'(4) = {bl ,... ,b,) C B'. Since 1 cfs 4 in P.,, for al1 i = 1,. . . ,m and j = 1, ... ,m, we must have a; + bj in G'. Therefore none of al,. . . ,am,br, .. . ,b, can be a new vertex, as new vertices from A' or B' are connected to al1 of the vertices [rom the other set. This implies that al,. . .,a, E A and 61,. . . ,b, E B, and that a; - bj in G for al1 i = 1,. .. ,mand j = 1,. . . ,m. We conclude that G has a subgraph isomorphic to Km,,. O

Corollary 11 The equitable P4-coloring problem is NP-complete. CH.4PTER 2. EQ LrZTilBL E GRAPN HOM OMORPHIS:' 26

The following lemma will make our proofs a little bit easier.

Lemma 12 There is a Turing reduction from the connected equitable H-coloring problern to the restriction of the connected equitable H-coloring problem to connected instances G such that lV(H)I divides IV(G)(.

Proof: Let connected graph G be an instance of the connected equitable H-coloring problem, IV(G)(= n and IV(H)I = m. Let n = km f r, where O 5 r < m. If a is an equitable homomorphism from G to H, then for r vertices h of H: la-'(h)l = k + 1 and for the remaining rn - r vertices h of H: la-' (h)1 = k. For every subset S E V(G) of cardinality rn - r, define ws to be a weight function on G given by

2 ifgES -dg, = { 1 otherwise.

Al1 graphs G,, are connected and they have (k+ l)m vertices and obviously rn divides

(k + 1)m. We now daim that G --+ H if and only if for at least one S, G,, ?-, H. If G 5 H, then for rn - r vertices h of H we have Icr-'(h)\ = k. Let us choose one g E a"(h) for each such h to form the set S of cardinahty rn - r. One can observe that if we define S((g,i)) = a(g) from G,, to H, then for each h f V(H), I,F1(h)l = k + 1 and hence G,, -P H. On the other hand, let us assume that for some S, we have G,, H. Let G G, be the coretraction defined by ~l(~)= (g,l) for al1 g E V(G).Then G % H. Since there are (mlr)= O(nm)subsets of V(G)of cardinality r and m is fixed, our reduction is polynomial. O

For a technica! reason we also need the following four lemmaç. They will enable us to give a Turing reduction from the connected equitable P4-coloring problem to the connected equitable H-coloring problem in the case when H is connected bipartite graph which is not complete bipartite.

Lemma 13 The natural homomorphism R from fi, to H is a retraction. CHAPTER 2. EQ UTABLE GRAPH HOiLfOMORPHISM

Proof: To see this, one can define a coretraction x' from H to H, by mapping each vertex h E V(H)to any one of the vertices (h, l), (h,2),. . ., (h, w(h)) E V(H,). CJ

Lemma 14 Let n be a positiue integer, H be a graph on nt vertices and w be a weight Junction on H such that w(h) 2 mn for al1 h E V(h). Thea any homomorphisrn H, 5 H such that IC1(h)l 5 (rn + l)n for al1 h E V(H) is a retraction. More- ouer, there ezists a weak coretraction K' belonging to 4 such that n' is a coretraction belonging to the naturai homomorphism n from Hw to H.

Proof: Let us consider sets Bh = {(h, l), .. . ,(h,w(h))) C V(H,) and Ah = {#((h,l)), . . . , +((h,w(h)))} C V(H)for al1 h E V(H). We clairn that the sets Ah, h E V(H)have a system of distinct representatives, i.e., vertices al,.. . , a, E V(H) such that each ai E Ai. According to Hall's theorem [24],given a collection of m sets, there is a system of distinct representatives if and only if, for al1 1 5 s 5 m, the union

of any s of the sets has cardinality at least S. For a contradiction, suppose that this is not the case, i.e., that there are s 5 rn sets Ah,,. . ., Ah, such that

We may assume that s > 1, since al1 sets Ah are nonempty. This rneans that al1 vertices €rom U:=, Bh, are mapped to at most s - 1 vertices from H. The sets Bh, have cardinality at least rnn each and they are disjoint. Therefore at least vertices of H, are mapped to the same vertex, i.e., there is a vertex h E V(H)with srnn mn mn l#-l(h)l 2 ---mn+- >mn+-=mn+n, s-l s-1 ln which is a contradiction. We conclude that the sets Ah, h E V(H),have a system of distinct representatives ah, h E V(H).We may assume that ah = #((h, ih)) for some ih E {l,. .. ,w(h)}; we now define xf(h) = (h,ih). One can observe that r' is a homomorphism, since h - h' implies (h,ih)- (ht,ih+ Moreover (mf)(h) = r(h,ih) = h and hence T' is a coretraction belonging to r. Finally, since (dat)(h)= ah and since the ah, h E V(H)are distinct, we conclude that 47r' is an injective endomorphism CHAPTER 2. EQUITABLE GRAPH HOMO~fORPHISib1 28 and hence an automorphism. Hence r1is a weak coretraction belonging to 4 and we conclude that 4 is a retraction. O

b Lemma 15 Let w be a vreight function on H, H, 4 H a retraction, H, b+ H the natural homornorphism and for h E V(h), Bh = {(h,1), . . ., (h, w(h))} C V(Hw)- b1 Then /or euery corstraction H 4 & llelonging to 4 and for every coretraction H f; H, belonging to n: (a) rl(h)= (h,i) for every h E V(H)and some i E (1,. . ., w(h)), (6) NH,(~')= NH,((T'T)(~'))for euery hg E V(H,), (c) NH(~)Ç N~((dn'n#)(h)) for eveq h E V(H),and (d) NH(~)G N~(d(h*)) for every h E V(H)and eveq h' E B(,dqh).

Pro0j: (a): Let us assume that îrt(h) = (h',i) E V(H,) for some i E (1,. . . ,w(hl)). Since

KA' = idH,we have h = (nïrl)(h)= ?r(hl,i) = hl. We conclude that îrt(h)= (h,i). (b): If ho = (h,j) for some j E (1,. . ., w(h)), then by (a) at(h) = (h,i). There- fore (x1a)(h')= nl(h) = (h,i). From the definition of the graph H, it follows that NHw((h,i)) = NH,((h,j)) and we conclude that NHw(h')= NHw((nl~)(h')). (c): Since 4' is a homomorphism, u E NH(h)implies @(u)E NH(dl(h))= N,y((~l~)(d'(h) by (b).Therefore u = 4(d1(u))E NH($((ntaqb')(h))),since 4 is a homomorphism and ~$4= idH. (d): Let c#'(h) = (hl,i) for some i E (1,. . . ,w(hl)}. Then (x#)(h) = ~((h',i)) = hl and if h' E B(r+l)(h)= Bhl, then h' = (hl,j) for some j E (1,. . ., w(ht)). Let r' be a coretraction belonging to n defined by

(hl,j) = hg if h = h' = n(hD) nt(h)= ( (h,1) ot herwise. CHAPTER 2. EQ UlTABLE GRAPH HOiCIOMORPHISM 29

Lemma 16 Let C and H be two graphs, w a weight function on H such that w(h)5 V G+Hw LW]fit+ al1 h E V(H),Hu = H the natuml homomorphisrn, and G+ Hw 2 H an equitable honomorphism. Let t$ = PIH,, and for some coretraction x' belonging to n, &r' is an automorphism on H. Then there exists an equitable homomorphism G + Hw -2, H such that yjHw = 'IF.

Proof: Let us denote Bh = {(h,l), . .. ,(h,iu(h))} Ç V(Hw)for h E V(H)and a = (q5nf)-' be the inverse of the automorphism Qd,Le., &'a = id^. Thus 4 is a retraction and 4' = n'a is a coretraction belonging to 4. We will prove the lemma by induction on the number q = qp = ChEV(H)l{hœE &(hl 1 +(ha)# hl[. For q = O, al1 vertices from each Ba(h)are mapped to the vertex h by 4 and hence aq5 = x. Since a is an autornorphism, it follows that for 7 = ap we have G+& -Z* H and ylfl, = K. Assume that q > O. Let h' E be such that 4(h') # h. Since p is an equitable homomorphism,

Since h* E Ba(h)and h* 4 P-'(h), there is a vertex w E P-'(h) such that w 4 Note that w can be a vertex of G. Define a mapping from G+ H, to H by switching the images of w and ho in ,f3, i.e., for u E V(G + H,) define

Obviously, since @ is equitable, Pl is also equitable and qp, < qp. We will show that ,f3, is a homomorphism. Since B(,41)(hl= B(,,f,)(h) = Bo(h)and h' E B+), it follows from part (d) of Lemma 15 that NH(h)Ç NH(4(hœ)).If u, v E V(G+H,) and u - v then we distinguish the following four cases: Case 1: {u,v) n {w,h') = 0. In this case &(u) = B(u) - P(v) = Bt(v),since p is i homomorphism. Case 2: {u,v) n {w,h') = {zu). We may assume that w = v and let us recall thai CHAPTER 2. EQUITABLE GRAPH HOL\~OMORPHISM 30

@(w)= h, since w E P-'(h). Hence &(IL) = P(u) ,f?(v) = B(w) = h and therefore Pi(u) E NH(h) C NH(d(hm))= NH(Pl(w))= NH(Pi(v)). We conclude that P1(u) - Pl(v)- Case 3: {u,v) n {w,hg) = {h'}. We may assume that v = h* E V(Hw)and since u - v, also u E V(Hw).Therefore we have u - hg in Hu and hence u E NH,(hV) = NH,((rlr)(h*))by part (b) of Lemma 15. Since a+ is a homomorphism, this implies that (ad)(u) E NH((a+xtn)(h')) = NH(r(h')), since a = (q5r1)-'.From our assumption, h' E and hence n(h') = cr(h)- Hence (ad)(u)E NH(a(h)),and applying homomorphism a-' = +xr we conclude that r$(u) E NH(h). Therefore Ph)= B(u) = +(IL) - h = Pi(hf)= A(u). Case Li: {u,V) n {w, h*) = {w,hm). We will show that this case cannot occur, 0th- erwise u - v implies that w - h'. Therefore h = P(w) - P(hg)= +(hm)and hence $(hm)E NH(h) NH($(h*)),which is a contradiction, since H is bipartite and hence loopless. O

Now we are ready for the proof of the main theorem of this section.

Theorem 17 Let H be a connected bipartite graph which is not complete bipartite. Then the connected equitable H-coloring problem (and hence also the equitable H- coloring problem) is NP-hard.

Prooj: One can observe that a connected bipartite graph is a complete bipartite graph if and only if it does not contain an induced subgraph isomorphic to P4. Let the vertices of H be V(H)= (1,. . ., m} and we may assume that the first four vertices form an induced P4.We may also assume n > 4, since if m = 4 then H = Pd.We will give a Turing reduction from the connected equitable P4-coloring problem. Let a connected graph G be an instance of the connected equitable P4-coloring problem. By Lemma 12 we may assume that IV(G)I = n = 4b >_ 4. Let us define the weight function w on H by if j 5 4 w(j) = rnn { mn + k otherwise. CHAPTER 2. EQ UITABLE GRAPH HOMOikfORPHISM

LVe claim that G - P4 if and only if G + H, -, H.

Let H, f, H be the natural homomorphism. One implication of our claim is 7+r obvious, since if we assume that G 4 Hf,then G + Hw -+ H.

Note that G + Hw has n + 4nm + (m- 4)(nm + k) = (nm+ k)m vertices, and hence V C+Hw = nm + k. If we assume that G P4 and Hw H is the natural homomorpbisrn, then G+ Hw Ti-a H.

On the other hand, let us assume that G + H, -P H. PVe will show that the conditions of Lemma 16 are satisfied. Since G + Hw has n +4mn + (m- 4)(mn+ k) = m(mn + k) vertices, Lv(xnV(G+Hw = rnn + k. Since is an equitable hornomorphisrn, exactly rnn + k vertices of G + Hw must be mapped to the same vertex of H. Let $ = PlG and 4 = BIH,. Thus 14-l(h)l = mn+k = mn+? 5 mn+n for al1 h E V(H). From Lemma 14 it follows that there is a coretraction r' belonging to a such that 47r' V is an autornorphiçm. For the weight function w we have w(h) 5 mn + k = G+Hw Therefore Lemma 16 irnplies that there is an equitable hornomorphism G + H, H such that ylH, = x. Since a-l(h) = rnn + k for h 4 {1,2,3,4} and ~-'(h)= mn for h E { 1,2,3,4}, no vertex from the graph G is mapped to any of the vertices from (H)- {l,4. Lloreover, for h E {1,2,3,4) we have [(rlG)-l(h)( = k and we conclude that ylG is an equitable homomorphism from G to Pd.

To finish our reduction, let us fix a vertex g E V(G).For every vertex h' E V(Hw), let W. be the graph obtained from G+ Hw by connecting the vertex g with the vertex hm with an edge. One can easily observe that al1 Wh. are connected. We daim that G - H' if and only if for at least one of hm,Wh. -+ H. If G -3 H', then for any h E NH(~(g)),W(n,i) -+ H, if we map vertices of G by y and the vertices of Hw by the natural homomorphism. On the other hand, if L& Is* H and t denotes the obvious inclusion of G + Hw in Wh*,then G + H, 3 31 and hence G- Hf. a CHAPTER 2. EQUITABLE GRAPH HOMOMORPHISLV

Finally, we will prove the following theorem which concludes the comptete characterization of the complexity of the equitable H-coloring problem.

Theorem 18 If H is not a disjoint union of cornplete bipartite graphs, then the equitable H-coloring problem is NP-hard.

Proof: Let H have connected components Hl,. . . ,Hk. Without loss of generality we may assume that Hl is not a complete bipartite graph. We wiIl give a polynomial reduction from the equitable Hl-coloring problem to the equitable H-coloring problem. Let G be an instance of the equitable Hl-coloring problem. Let us denote n to be the number of vertices of G and m to be the number of vertices of Hl. We may also assume that m divides n by Lemma 12. Let w = 2 be the constant weight function. We will prove that G - Hiif and only if G + (Hz), + . . . + (Hk)ui- H. The implication + is obvious. Let us assume that G + (II2), + . . .+ (Hk),- Hl +Hz+ .. .+ Hk. Clearly no two graphs are mapped to the same graph. Therefore there is a permutation a on k elements such that the ith graph from the left hand side is mapped to the a(i)th graph from the right hand side. Let us decompose a into a p~oductof disjoint cycles. Let 1 be in the cycie (l,il,.. . ,i,), i-e., G - Hi,, (Hi,),--+ Hi,, . ., (Hir-l)w CT) Hi,, (Hi,)UI-+ Hl. By Lemma 3, from (Hi,), -.+ Hi,,,we get a bimor-

phism (Hi,), -t (H;,,,), and their composition gives us a bimorphism G -, (HI),. Q By Lemma 3 we conclude that for some ct G -, HLand since w is a constant, G fil for the natural homomorphism x. O

2.6 Conclusion

In this section we would like to clarify the ideas behind our NP-hardness proofs. Recall that a graph G is calred a core if it is not hornomorphic to any of its proper

subgraphs, or equivalently any homomorphism G 4 G is an isomorphism. Welzl in [54] proved that every graph has a retraction to uniquely defined graph which is a core. CHAPTER 2. EQ UITABLE GRAPH HOM OMORPHISICI 33

Hel1 and Ngetiil in [31] proved that determining whether a graph is a core is NP-hard.

We define a graph H to be an apparent core, if for every weight function w on H 4 and every two retractions H, -+ H and H, 5 H, there is an automorphism H % H on H such that 04 = 11,. Therefore there is an automorphisrn a such that the following diagram is commutative. 4 H, -H

We will show later that every core is an apparent core and we will also provide exarnples of apparent cores which are not cores. The motivation of the definition of apparent cores follows frorn the proof of the following proposition.

Proposition 19 If H is a connected apparent core mhich contains P4 as an induced subgraph, then the equitable H-coloring problem is NP-complete.

Proofr Let the vertices of H be V(H)= (1,. . ., m) and we rnay assume that the first four vertices form an induced Pd. We also may assume rn > 4, since if m = 4 then H = P.,. We will give a polynornial reduction frorn the connected equitable P4- cotoring problem. Let a connected graph G be an instance of the connected equitable P4-coloring problem. By Lemma 12 we may assume that IV(G)I = n = 4k 2 4. Let us define the weight function w on H by ifjS4 w(i) = nn+ k otherwise. We will prove that G + H, - H if and only if G - P4. Note that G + Hw has n + 4nm + (m- 4)(nm+ k) = (nm+ k)m vertices, and hence Iv(xnV C+Hw = nm + k. If we assume that G P4 and Hw 3 H is the natural hornomorphism, then

P On the other hand, let us assume that G + fi, -A H. Let q5 = PIH, be the restriction of 0 to H,. Since is an equitable hornomorphism, exactly nm+ k vertices CHAPTER 2. EQ UITABLE GRAPH HOM Oh1 ORPHISM 34 must be mapped to the same vertex. Therefore I+-'(h)l = mn+k = nm+ 5 mn+n. From Lernma 14 we conclude that q5 is a retraction. Since H is an apparent core, there is an automorphism CI such that the following diagram is commutative,

i.e., a# = r. Let us consider the homomorphism G + H, -% H. Since a is an automorphisrn, ap is also an equitable homomorphism. Moreover, a(P(H,) = ad = n and hence vertices from (h, 1),. . ., (h, w(h)) are mapped to the vertex h. There- fore r-'(h) = mn + k for h 4 {l,2,3,4) and R-'(h) = mn for h E {1,2,3,4). This impiies chat no vertex frorn the graph G is mapped to any of the vertices from V(H)- {1,2,3,4). Moreover, for h E (1,2,3,4) we have k = I(c@IG))-'(h)l and we concludc that a(P[G)is an equitable homomorphism from G to Pd. O

Thecefore the NP-hardness prouf for apparent cores is quite straightforward and it motivated their definition. We will give an alternative characterization of apparent cores but first we will prove the following simple lemma,

6 Lernma 20 Let w be a weight Junction on a graph H and let Hw + H be a retraction. Then iJ4 # r, there is a coretraction H H, which belongs to n but does not belong to 4.

Proof: Assume that 4 # K, i.e., there is a vertex (h', i) E V(H,) such that d((h1,i)) #

h'. Let us consider a mapping A' such that = { (hi if h = ht (h,1) otherwise.

One can observe that ?rl is a coretraction belonging to A but it does not belong to 4, since (&r)(ht)= d((h', 2)) # hl. a CHAPTER 2. EQ UITABLE GRAPH HOMOMORPHISM 35

Theorem 21 A graph H is an apparent con if and only iffor all distinct vertices h, ht E V(H),NH(~) If NH(~').

Proof: Let the vertex set of H be V(H)= { 1, .. . ,rn). First, let us assume that there are two distinct vertices h, ht € V(H)such that NH(~)C iVH(ht). Let w be a weight function on H defined by

2 ifj=h w(j)= 1 otherwise.

Define 4((j,1)) = j for j = 1,. . ., rn and #((h,2)) = h'. For every endomorphism H "t H such that CL+ = r we have a(h)= u($((h,1))) = ~((h,1)) = h = x((h,2))= û($((h,2)))= a(h') and hence u cannot be an automorphism.

On the other hand, let us assume that for al1 distinct vertices h, ht E V(H), !VH(h) fVK(ht).

First we will prove the claim for + = K. Let .$ be a retraction frorn H, to H and 4' : H -, Hw be a coretraction belonging to 4. Let us define û = 7rqS. Obviously fl 5 H is an endomorphism. To show that it is an automorphism, it is enough to prove that it is injective. Let r' be any cotetraction belonging to K. By Lemma 15, !\iH(h)C NH((4xt~q5')(h))for every h E V(H)and we conclude that h = (47rt7r4)(h) for every h E Cr(H) which means that dnflr$' = idK. Therefore #xt is a left inverse

to û = rdt and we conclude that a is an injective endornorphism and hence an automorphism. It remains to prove that ad = R. Since a is an automorphism, every left inverse has to also be a right inverse and hence a&rt = idH. This implies that a4 is a retraction and every coretraction x* belonging to r belongs to a$. We conclude that a4 = x by Lemma 20.

To finish the proof, let us assume that # and $ are two retractions from Hw to H. Let a and p be two automorphisrns on H such that 04 = n and P.S, = K. Then a# = @$and hence 11, = (@-'a)$for the automorphism /3-'a on K. U CHAPTER 2. EQ UITABLE GRAPH HOMOh,iORPHZSibf 36

We Say that an endornorphism is simple, if it differs from the identity automorphisrn on at most one vertex. If a graph H contains two vertices h, h' such that NH(h) C NH(hl),then one can define a simple endornorphisrn H 5 H by

h' ifj= h j otherwise.

This simple endornorphism is clearïy not an automorphism. Thereiore apparent cores are exactly those graphs which do not have any simple endomorphism which is not an autornorphisrn. This also implies that every core must be an apparent core. Note that every simple endomorphism H 5 H induces a retraction from H onto induced subgraph $(H) which has (V(H)I - 1 vertices. Therefore one can find an induced subgraph which is an apparent core in a polynomial time by applying simple retractions at most [V(H)Itimes, while there is a pair h, h' of distinct vertices such that N(h) C N(ht). Examples of apparent cores are cycles and complete graphs. Even cycles are examples of graphs which are apparent cores but not cores.

One can observe that our proof of Theorem 17 was inspired by the proofs of Proposition 19 and Theorem 21. Moreover, from our discussion above it follows that apparent cores, unlike cores (cf. [311) can be recognized in polynomial time. Moreover it also foIlows that for a given graph one can find a retract which is an apparent core in polynornial time and therefore we think that they may be of independent interest.

We would also like to mention that keeping the target graph H fixed, we have obtained polynomial time algorithms for the equitable H-coloring problem in the case H is a disjoint union of complete bipartite graphs and for the connected equitable H-coloring problem in the case when either H is disconnected or H is a disjoint union of complete bipartite graphs. In both of these cases the provided algorithms are exponential in (V(H)I and this may be a target for further improvement. Chapter 3

Relaxations of Graph Homomorphism

3.1 Introduction

Determining whether G + H for input graphs G and H is known to be an NP- complete problem (cf. [30]). This problem is a generalization of a large number of well-studied problems. For exarnple, determining if a graph G is k-colorable is equivalent to setting H to be a complete graph on k vertices and determining if G + H. Determining if a graph H ha a k-clique is equivalent to setting C to a complete graph

on k vertices and determining if G 4 H. Therefore it is desirable to identify families of instances for which a polynomial algorithm exists. Recent research in semidefinite optimization shows that it is a powerful tool towards obtaining approximation algorithms for NP hard problems (see [LS,351). Weiise semidefinite programming to study the graph homorriorphism problem. LVe define the notions pseudo-homomorphism and Jractional homomo~rphismwhich are, respectively, semidefinite and linear relaxations of a certain integer program corresponding to the graph homornorphism problem. In fact, the pseudo-homomorphism is defined in terms of the classical 6 number of Lovbz [42]. As semidefinite programs can be soIved in potynomial time, pseudo- homomorphism is a polynomial time notion. It turns out, as we will show in a later section, that our notion of pseudo-homomorphism coincides with another semidefinite CHAPTER 3. REL4,YATIONS OF GRAPH HOMOhIORPHISibf 35 relaxation defined by Feige and Lovisz in (161. Feige and Lovisz used this relaxation to show polynomial time solvability of various special cases of graph homomorphism. Moreover G is homornorphic to H implies G is pseudo-hornomorphic to H and this further implies that G is fractional homornorphic to H. It turns out that fractional homomorphism is a CO-NPcomplete problem. Therefore, if for some subset of graphs one can reverse any of these implications, then for this subset the corresponding problems are polynomial time solvable. One such example is the 2-coloring problem and another is the digraph homomorphism problem restricted to instances which can be soived using consistency checks. Although the general graph homornorphism does not admit a simple forbidden subgraph characterization, we show a surprisingly simple such characterization for the fractional homomorphism, which also yields several other interesting consequences. First, we obtain a much simpler proof of the NP hardness of fractional clique number, a result which was first proven in [20] using the ellipsoid algorithm. A longstanding conjecture in graph theory is the graph product conjecture [22]. Using our characterization of the fractional homomorphism, we show that the above conjecture is true, if we replace homomorphism by fractional homomorphisrn. The main results of this chapter appeared in [6] and they are a joint work with Sanjeev blahajan.

3.1.1 Notations

Al1 graphs in this chapter are finite, undirected and without loops, unless specified otherwise. We Say that two vertices are incident, if they are either adjacent or equal.

Let G and H be two graphs. The hom-product Go H is defined as the graph with

V(G O H)= V(G)x V(H),in which {(s,~),(t, w)) E E(G O H) if and only if (s # f) and ({s, t) E E(G) implies {u, w) E E(H)).

Recall, that the clique number, w(G),and the chromatic number, x(G),of a graph G, have the following integer linear program formulations. In t hese forrnulat ions we take v to range over the vertex set of G and I to range over al1 maximal independent CH-4PTER 3. R ELAXATlONS OF GRAPH HOikfOibfORPHISAf sets in G.

If we replace the integrality conditions x, E {O, 11, y1 f {O, 1) by x, 2 O, yl > O respectively, we obtain linear programs with optima wl(G), the fractional clique number of G, and xf(G), the fractional chromatic nurnber of G. One can observe that these linear programs are duals of each other and hence uI(G) = xf(G) by the duality theorem of linear programming (se [Il]). Therefore ive have the inequalities:

To compute any of these numbers is NP-hard. The NP-hardness of the fractional chromatic number was proved by Grotschel, Lovkz and Schrijver in [20]. They also proved that one can compute in polynomial tirne a real number 0 (see also [39]) which satisfies w(G) 2 8 1 (G).

One possible definition for O is as follows [50]: Let S be the set of al1 IV(G) x V(G)I positive semidefinite matrices, let I denote the unit matrix and J denote the matrix of al1 1's. The product of two matrices

A = (aij)and B = (bij) is the number A B = Cijaijbij. Then

Another semidefinite programming upper bound for w is the real number 01/2 of Schrijver [47], which can be expressed as:

Szegedy in [50] observed that is the same as the vector chrornatic number whose definition can be find in [3S]. CHAPTER 3. RELAX.4TIOiVS OF GRAPH HOMOMORPHISM

3.2 Overview

In Section 3.3, we define three relaxations of graph homomorphism and their rela- tions. In Section 3.4, we give a forbidden subgraph characterization of fractional homomorphism and its consequences which include an easier NP-hardness result for the fractional clique number problem and the fractional version of Hedetniemi's conjecture. In Section 3.5, we show that ounotion of pseud~homomorphisrnis equivalent to the notion of hoax of Feige and Lovhz [16] and give a necessary condition for the existence of a pseudo-homomorphism, We will also show that homomorphism, fractional homomorphism and pseudo-homomorphism are not equivalent in general. In Section 3.6, we wilf give some examples to demonstrate how our theory can be applied to give a polynomial time algorithm for some particular instances of graph homomorphism problem. Weconclude this chapter with Section 3.7 where we discuss other possibility how to define the pseudo-homomorphism.

3.3 Relaxations of Homomorphism

In this section we define three relaxations of graph homomorphism. Our motivation follows from the following theorem.

Theorem 22 For any two graphs G and H we have the following inequalities:

and u(GO H) = IV(G)I ij and only if there is a hornomorphzsm jrom G to H.

Proof: The first and the second inequalities follow from [SOI. The third inequaiity follows from Theorern 10 in [42]. From the definition of a fractional clique and from the duality of linear programming, it follows that w, = XI 5 X. The last inequality

follows from the fact that we can color the vertices of G O H by the vertices of the graph G, by assigning to a vertex (s, u) E V(G O H) the color s E V(G). From the definition of the hom-product it follows that no two adjacent vertices obtain the same

color. It remains to prove that G -+ H if and only if w(G O H)= IV(G)I. Let US CHAPTER 3. RELAXATIONS OF GMPN HOMOMORPHISAl 41 assume that ct is a homomorphism from G to H. One can easily observe that vertices

(s, f (s)),s E V(G)form a clique in G O H of size IV(G)I. On the other hand if we have a clique C in G O H of size (V(G)I,then the first coordinates have to span ail the vertices of G because (s, u) + (s,w), and hence C can be regarded as a function V(G)-, V(H). We define a homomorphism a from G to H as follows: a(s) = u if and only if (s,u) f C. To prove that a is a homomorphism, let s, t be vertices in G. Then (s,a(sf) and (t,a(t)) are €rom the clique C and hence they are adjacent. From the definition of the hom-product, it folows that s - t implies a(s) - a(t). We conclude that a is a homomorphism. 0

The previous theorem suggests the following relaxations of homomorphism: Let G' and H be two graphs. Definition 23 We say tkat G is fractionally homornorphic to H (denoted by G -+,H) ifar(GO H) = [V(G)I.

Definition 24 We say that G is pseudo-homomorphie to H (denoted by G 4, H)

if O(G O H) = (V(G)I.

Definition 25 CVe say dhat G is pseudolp-hornomorphic to H (denoted by G -+,p H)

$~i12(G0 Hl = JV(G)I-

Corollary 26 (G' + H)+ (G +,/2 H)+ (G+, H) + (G +j ,Y).

Later we will show that we cannot reverse the first and the last implications in general. This is not surprising because G +,, H, and G +,/z H are polynomiai, while both

G -t H and C +j H are NP hard. However, for certain families of graphs we can prove the reverse implications, and so obtain a polynornial algorithm for these families.

3.4 Fract ional Homomorphisms

In this section, we will give a forbidden subgraph characterization of fractionai homomorphism. In particular we will prove that G is fractionally homornorphic to H if and only if G does not contain a subgraph isornorphic to h&(H)+ll-This will enable us to prove a fractional version of the graph product conjecture [22]. As a consequence we will obtain an alternative proof of the NP-hardness of computing the fractional clique number of a gaph.

Theorem 27 G -q H if and only ifw(G) 5 wf(H).

-PraoJ: -Asmimefint Chat G ilfi. .Fora -contradiction let us also assume that there is a clique C with Lwf (H) + 11 vertices in G. We will show that wj(GO H) < IV(G)I by constructing a dual fractional coloring smaller than IV(G)I. Let w be a minimum fractional coloring of H. Its value is xj(H) = wf (H). Al1 independent sets in G O H are of the form I = ut==,si x Ai where sl, . . ., sr form a clique in G and for i # j, Vai E

Ai, aj E Aj : ai ++ aj. This is so, because two vertices (s, u),(t, w) from G O H are not connected by an edge whenever (s = t) A (u # w), or (s - t) A (u + w). Now we construct a fractional coloring w* of G O H as follows: wExT = WT for any maximal independent set T from H, w:~~~~)= 1 for u C, and wi = O otherwise. One can observe that this is a feasible fractional coloring and its objective value is at

On the other hand if we assume that w(G) 5 wr(H),we will show that q(GO H) = IV(G)J. Let x be a maximum fractional clique in H, and w a maximum fractional coloring of H dual to x. From complementary slackness [Il] it follows that

where J runs through the maximal independent sets in H. CVe construct a fractional '" E u E clique x* in G O H as follows: Let x~,~,)= q(W7where s V(G), V(H). Let us denote xs = xsESxsfor any subset S of vertices from V(G)and similarly

2;. = &u)ETxis,u) for any subset T of vertices from V(G O H). To prove that x'

is a fractional clique, we have to show that for every independent set I C G 0 Hl xi 5 1. For an independent set I = u:=, si x Ai,let us write B = UiU Ain A,, where i, j = 1,. . ., k and let Bi = Ai - B. Then I C @=,si x (BiU B),and therefore CHAPTER 3. RELAXATIONS OF GRAPH HOMOh1ORPHISXl

BY Our assumption k Iw,(H), therefore the following inequality holdç:

where J runs t hrough the maximal independent sets of H. One can observe t hat the set B u u~~(Jn Bi) is an independent set in H and therefore ~f=,XJ~B, + zs 5 1. Hence

It remains to show that the size of this fractional clique is 1 V(G)1.

This theorern shows that although the graph homomorphism problem is NP- complete, for any fixed non-bipartite graph H [30], the fractional graph homomorphism problem is polynomial, for every fixed graph H, since one can use brute force to check for a clique of size [wI(H)+ 11 in G. The theorem also shows that if H is part of the input, then fractional homomorphism is c+NP complete. Another interesting consequence of this theorem is that for a given k and a graph G, it is NP-hard to determine if G has fractional chromatic number at least k. This result was originally proved in [20] by the ellipsoid method. CH.4PTER 3. RELAXATIONS OF GRAPH iiOhfO~~fORPHlShf 44

3.4.1 Consequences

Theorem 27 has many interesting consequences and we devote this subsection to them.

Proposition 28 The following problem is NP-hard: Instance: A graph G and a number n Question: Is wf (G) at least n ?

Proof: We give a reduction from the clique problem. Let G and n be an instance of the clique problem. Since n - 1 = of by our theorem w(G) 2 n if and only if

G ++,l

A longstanding conjecture in graph theory States that whenever G f, IC, and W f, K,, then also G x H f,K,, cf. [22].

Proposition 29 G W and H fif CV imply G x H fif

ProoJ By Theorem 27 G j4 kV and H 74, W imply w(G) > wf(W) and w(H) > wl(W). One can easily observe that w(G x H) = min{w(G),w(H)) and therefore also w(G x II) > wf (W)which means G x H $+/ Ur. O

The fractional version of the product conjecture follows by taking W = A',.

We wodd like to mention some properties of homomorphism which are not shared by fractional homomorphism and vice versa.

The binary relation +J, like homomorphism, is not antisymmetric. Its quite bad behavior can be demonstrated by an example of two graphs with different chromatic number which are fractionally homomorphic to each other, for instance K2and Cs.

Unlike homomorphism, fractional homomorphism is not even transitive. As an exarnple let us choose an arbitrary graph G such that w(G) = 2 and wf (G) >- 3. Then

16 -.J G, G -+J h; but 16 ftl K2.An exarnple of such a graph G is the Mycielski CHAPTER 3. R ELAXATlONS OF GRAPH HOikfOMORPHiSiCI graph Mll from Figure 3.1.

On the other hand, fractional homomorphisrn is dichotornic, Le., for every two graphs G and H, either G +,H or H +f G. This follows €rom the fact that H j G implies that w(H)> wj(G) and therefore ive have w(G) 5 wl(G) < w(H) 5 wJ(N), which means that G +J H.

Welzl in 1541 proved that for every two graphs G and H such that G -+ H and H 4) G there exists a graph W such that G -+ W -+ H,W $, G and H f, W. This property is called density property of homomorphism. Similarly one can ask whether graphs are dense with respect to the fractional homomorphisrn, Le., whether for every two graphs G and H such that G +j H and H fif G there exists a graph W such that

G -+f W -if Hl bV f+ j G and H W. First of all, we have already observed that

H +j G implies G +j H and therefore we can omit G -1 H and G -+j 4j H parts of the question. The following proposition answers this question.

Proposition 30 11 H G then there ezists a graph CV such that LV fif G and H ftf CV (and hence G -+f CV +, H) if and only if w(H) - wf(G) > 1. rtloreover, i/ this condition is satisfied, we can choose CV to be a clique.

Proof: H +j C is equivalent to w(H) > wJ(G). Therefore we have the following inequali ties: 43 L w,(G) < w(H) 5 w,(H). Ci/' ft, G and H ftf CV if and only if w(W)> wj(G)and w(H)> wj(bV). Therefore there exists kV which satisfies these conditions if and only if the following inequalities are satisfied:

The necessity of our condition is now clear, since w(W) and w(H) are integers. The suficiency follows, since if the condition is satisfied, we can choose W = ï

Let V be an incidence matrix of the gaph G O H and C = Iv(&,l.V.Feige and Lovisz 1161 considered the foilowing optimization problem with convex const raints, which we denote by (*)clH:

s-t. Q is positive semidefinite Q is symmetric

The ellipsoid algorithm can be used to solve this optimization problem which is a semidefinite progam. This program is based on a two-prover interactive proof system. For more details we refer the interested reader to paper (161. A feasible solution to (*)G,Hfor which the objective function has value 1 is called a hoax (cf. [16]) and in their paper [16], Feige and Lovisz proved that if there is a homomorphisrn from G to H then (*)C,H has a hoax. They also gave the following necessary and sufficient conditions for (*)CIH to admit a hou.

Lernma 31 [16] The system (*)c,H has a hoaz if and only if there ezists a system of uectors v,,,s E V(G),uE V(H)satisfying the following conditions:

is independent of s, CHAPTER 3. RELAXATIONS OF GRAPH HOM OMORPHISICI

and T vsuvtW= O whenever V,u,t, = 0. (3.5)

The optimization problem obtained from (*)c,H by removing the last, nonnega- tivity condition, Vs, t, u, w : Q,,,t, > O will be denoted by (**)c,H and any feasible solution to (**)C,Hfor which the objective function has value 1 will be called a semi- hoax. One can similarly obtain the following necessary and sufficient conditions for

(**)G,~to admit a semi-hoax.

Lemma 32 The sgstem (**)c,H has a semi-houx if and only if there ezists a system 01 uectors vsu,s E V(G), u E V(H)satisfying the conditions (3.f),(3.Z), (3.3j, and (3.5).

We now prove that our notions of pse~do,/~-homomorphismand pseudo-homomorph coincide with the notions of ho^^ and semi-hoax respectively.

Theorem 33 The system (*)G,H has a houx if and only ij G +,12 H.

Proof: Let Q be a hoax of the system (*)C,H. By Lemma 31, there exists a systern of vectors v,, satisfying (1)-(5). One can observe that QsUvt,= v,,tlL = O whenever

(s, u) is not adjacent to (t,w) in G O H. Also, CHAPTER 3. RELAXATIONS OF GMPH HOhfOMORPHISrCI 4s

On the other hand let us assume that G -,/2 H and let 0112(GO H) = IV(G)(- Let B = AA= be the positive semidefinite matrix with nonnegative entries such that B I = 1, for which (s, u) + (t,w) irnplies bsu,tw = O and

We notice that B,,,, = O for u # w and therefore the u,,'s (the rows of matrix A) are orthogonal when s is fixed. If we denote a, = Cua,,, then from the condition B I = 1 we have

and therefore from B J = BI /2(GO H), it follows that

d 3 S J The last two inferences are a consequence of Cauchy's inequality. Because BI/?(G O H)= IV(G)1, it follows that bot h inequalities have to be equalities. That is,

and this happens if and oniy if aH a,'s have the same lengths. From B I = 1 it follows that for ail s E V(G) 1 IlasIl = lV(G)l-T. Also the inequality

must be an equality, Le., the angie between a, and al must be zero for each s, t E V(G). Therefore al1 a,'s are equal, Taking Q = (V(G)IB,we have proved the conditions (1) and (3). The conditions (4) and (5) follow from the definition of B. From the orthogonality of the as,'s for a fixed s, we have CWAPTER 3. RELAXATIONS OF GRAPH HOMOhIORPHIShI

implying (2). Therefore we can deduce from Lemma 31 that Q is a hoax of the system

(*)G.H- O One can similarly prove the following theorem.

Theorem 34 The system (+*)G,H has a semi-hoaz if and only if G +, H.

Although we do not have a forbidden subgaph characterization for pseudo-homomorpt as we have for fractional homomorphisrn, we can prove at least a necessary condition for existence of a pseudo-homomorphism which will be sufficient for our purposes. First, let us recall some definitions and results from [39].

If G and H are two graphs, then their stroag product G* H is defined as the graph with V(G * H) = V(G) x V(H), in which (s,u) is incident to (t,w) if and only if s is incident to t in G and u is incident to w in H. Let G denote the complement of G and 3(G)= O@). The following lemma is the monotonicity property from (39) p. 7.

Lemma 35 [39] Given Iwo graphs G and G', G C G' implies d(G) 1 i?(Gt).

The following lemma iç from (391 p. 22.

Lemma 36 L39] Giuen two graphs G and Cl, 9(G* G') = t9(G)d(Gf). Proposition 37

9(G)H) I(G)d(E).

Proof: Since G * H C G' o H,the inequality follows from the previous two lemmas. O

Corollary 38 If G +, H then S(G)B(H)2 (V(G)I.

In the last part of this section we prove that in general one cannot reverse implications in the Corollary 26. We need the following results (see [42, 391): for odd n CHAPTER 3. RELAXATIONS OF GRAPH WO~fO~1ORPHISAI

and for every n

where C, denotes the cycle on n vertices and Ii, denotes the complete graph on n vertices.

Proposition 39 For positive integers n,m such that n < m, Cz,+l frp h; and

C2n+1 +p C~m+î-

Proof: The proof follows from the fact that I~(C~,+~)is an increasing function of n 2 1, while IJ(C~,+~)is a decreasing and their product is 272 + 1. Also, we have OC)= . Therefore 9(C2,+1)9(C2,+1) < 9(C2,+1)d(Czn+1) = 2n + 1 and also 0(C2rn+l)b(%) = 29(C2m+l) c fid(c2m+i) = ~(C5)~(c2m+l)5 ~(c2m+i)fi(c2m+l) = 2m + 1. Corollary 38 implies the assertion of Our proposition. O

Since Cg +f I\;, and the above theorem shows that Cs +, K2, the pseudo- homomorphism is strictly stronger than the fractional homomorphism, i.e., the existence of the pseudo-hornomorphism implies the existence of the fractional homomorphism but not vice versa.

The next proposition shows that hornomorphism is strictly stronger than pseudolI2-

homomorphisrn. Let Ml 1 be Mycielski's graph from Figure 3.1.

Proof: The first assertion follows from the fact that hiIl is not three colorable. The odd cycle 1,2,3,4,5 requires at least three colors and the vertices 6,7,8,9,10 are forced to use al1 three colors. Hence the vertex 11 needs a fourth color. The second Figure 3.1: Mil. assertion follows from the fact that the following matrix Q is the hou. We will write its s,t blocks: 1/3 O O

O 116 116

for s t and 116 1/12 1/12 QQ= (l/lZ 1/12 1/121/6 l:;.), fors + t. One can check that al1 conditions for Q to be a hoax hold. a

3.6 Examples

In this section we will demonstrate how one can possibly reverse implications in the Corollary 26 for some families of graphs and hence have a polynomial algorithm for CHAPTER 3. RELAXATIONS OF GRAPH HOi%SOhfORPHIShI 53 these families. Unfortunately, for al1 our examples there is already a known polynomial algorithm by sorne other means. For a better readability, we will use the following notations.

For a graph H let us denote

(+H)= (G 1 G -, H), and for a family of graphs Cl let us denote

and if G f,H for al1 G E Clthen we will write C f,H.

We will use similar notations in the case of pseudo-hornomorphism and fractional homornorphism.

3.6.1 Graph Coloring

CVe will need a slight modification of Theorem S.1 in [16].

Proposition 41 Let C be a family of graphs such that C ++, H. Then the class (-+pH)is in P, and separates (C-t,) and (+H).

Proofr Obviously, (+ H) C (+, H). Assume, there is a graph G E (C-r) n (-rpH). Then there is a graph G' E C such that G' + G, and hence also G' +,G. We have also G +, H and from the transitivity of +, (see [16], for the transitivity of " has a hoax" relation), Gr -r, H and hence C +, G which is a contradiction. O

The ciass of al1 odd cycles will be denoted by Codd.

Proposition 42 For H bipartite, G + H if and only if G +,H. CHAPTER 3. RELAXATIONS OF GRAPH HO.MOMORPHISM 53

Proof: The graph H is bipartite if and only if it does not contain any odd cycle and this is if and only if H is 2-colorable. Therefore we have

By Proposition 39, CoddjSp I\r2 and hence from Proposition 41 it follows that the class

(+, IG) is in P, and separates (+ fc2) (dl bipartite graphs) from (Codd +) (g~aphs which contain an odd cycle, i-e. nonbipartite graphs). Therefore (tphr2) = (+&). But for bipartite graph H, one can easily observe that G + H if and only if G + h; and hence the theorem follows. O

Note that this example dso confirms the well known results that 2-colorability is polynomial. Similarly, one can use Proposition 39 to show that finding the odd girth of a graph (size of the smallest odd cycle) is polynomid. One can notice that 2-colorability has a duality property given by the equation (3.6). 1f (C+) = ('Hl

then we Say that H has C-duality property (c-f. [32]).

Proposition 43 If H satisfies the duality property (Cfi) = (+ H), and C +, H

/or some family oj graphs C, then (-t, H) = (+ H) and hence H-colorability is polynomial.

Prooj: This is because (C+) n (+,H) = 0 implies (4,H) = (CjS) n (+, H) = (+ H) n (+, H), and hence (+, H) E (+ H). The other direction is trivial. Therefore in this case, the semidefinite program (+)J gives us a polynomial algorithm for the H-coloring problem. O

We can slightly generalize this as follows. Let an almost complete graph be a graph obtained from a complete graph by removing one edge. Let us define a k-string as a graph Gkwith the following properties: a/ Gkis a union of almost complete graphs on k vertices Hi, i = 1,. . .,m and some edges between them CHAPTER 3. RELAXATIONS OF GRAPH HOMOMORPHZSM 54

b/ If (xi,Y;) is the missing edge from Hi then y; = zi+l for i = 1,. . . ,m- 1, and (xi, y,) is an edge in Gk. One can observe that if there is a k + 1-string as a subgraph of a gaph G, then G is not k-colorable, otherwise the vertices xi, and y; would have to have the same color and b/ would imply that XI,and y, have the same color but they are adjacent.

Let Skdenote the set of al1 k-strings, and . s= us,, k/l then the class of graphs with the following duality property is denoted by Cs,:

One can easily observe that (Sa+) = (Codd+). Therefore Cst contains the family of al1 bipartite graphs (see Example 1). Cs,contains also al1 graphs with w = x because

Example 9 1(, E Sn.

Lemma 44 Let v,, be vectors satisfying the conditions of Lemma 31 and O be a subset of these vectors which are mutually orthogonal. Then for v> = CuEOu we have

and the equality holds if and only if v> = 6.

which implies Ilvoll 5 1. We cm also see that if the equality holds then

and hence the angle between the vectors ü, and vk is O. But both of them have their length equal to 1 and so they are equal. On the other hand if they are equal it implies Ilv>ll = i. This finishes our proof. O , CHAPTER 3. RELAXATIONS OF GRAPH HOiCIOMORPHISM

Lernma 45 Let (*)G,H has a hoax, v,, be vectors satisfying the conditions of Lemma 31 and hn be a complete subgraph of G with n = IV(H)I uertices. Then Vu E

Proofr For a fixed vertex u E V(H),vectors v,,,s E 1cn are orthogonal. From the previous lemma it follows that II CSEKnvsUII 5 1. Therefore

Combining this with Lemma 2 yields our assertion. O

Now we can generalize Proposition 42 as follows.

Proposition 46 For G E Cst, G -, fi, if and only if G +,k.

Proof: Since G -, II, implies G -+,hn, it follows from Proposition 43 that it is

enough to prove Sn+1f+, fCn. To get a contradiction, assume G,+l -t, fi, for some Cnil E Sn+iand hence has a hoax. Let v, be vectors given by Lemma 31. The vectors us,, u E V(lin) are mutually orthogonal, since s cannot be mapped on distinct nodes, and hence

Since G,+l E Sn+i,it satisfies conditions a/ and b/ of the definition of n + 1-string and we will use notation from these conditions. The vectors us,, s E C'(Hi) - {xi} are also mutually orthogonal, since any two distinct nodes hmV(Hi) - {xi) are adjacent and hence they must be mapped on distinct nodes. Similarly the vectors us,, s E V(Hi)- {yi} are mutually orthogonal. On the other hand, IV(H;)I = n + 1 and hence from previous lemma follows

and similarly CHAPTER 3. RELAXATIONS OF GEWPH HOMO MOWHISM

From Lemma 44 it follows that CsEY(H,)-(r,)vSu = ü = CEV(K)-~O,~usY,and therefore fi - u,,. = ii - v,,, which implies = u,, . From condit ion b/ we have obtained

Vrlu = vymu. But vertices XI, and y, are adjacent and hence cannot be mapped to the same vertex u. Therefore u,,, and v,, have to be at the same tirne orthogonal, which is a contradiction, because their sum over al1 u is the unit vector 6. O

3.6.2 Digraph Coloring

A directed graph G is said to be homomorphie to a directed graph H, if there exists a function f : V(G)-+ V(H)called a homornorphisrn, such that whenever (s, t) is an edge in G,(f (s), f (t))is an edge in H. We can now similarly define the hom-product of two directed graphs G,H as the undirected graph GoH with V(GoH)= V(G)x V(H) and (s,u) -- (t, w) if and only if (s $ t)A ((s,t) E E(G) (u,w) E E(H))A ((t,s) E E(G)=+ (w,u) E E(H)).

Hell, NeSetiil, and Zhu found polynomial algorithms for classes of graphs which have trees-duality in [32, 34, 331, Le., C-duality where C is the family of al1 trees. They showed that the following labeling algorithm works for H-coloring (the problem whether G + H for a fixed graph if) for graphs H which have trees-duality. We will later refer to this algorithm as the consistency test (cf. [32]).

Instance: A digraph G; Question: 1s G -, If?

For any vertex s, the set LL+L(~)C Lk(s), k 2 O, and L&) = V(H). Therefore CHAPTER 3. RELAXATIOXS OF GRAPH HOMOMORPHISM there is some i such that Li(s) = Li+i(s). Output: G is H-colorable if for al1 s E V(G):Li(s) # 0.

We now prove that the class of graphs with trees-duality also admits a polynomial time solution by the theory described in this thesis.

Proposition 47 Let C be the class of al1 graphs tnhich have trees-duality. Then for every graph H E C and every graph G, G + H if and only ifG +r H.

Prool: For any H E C the H-coloring problem can be solved using the consistency test [32]. Therefore it is sufficient to show that if there is an s E V(G): Li(s) = 0, then G fij H.

Let us assume that G 4j H, and that a fractional clique x satisfies

First we will show by induction on k 2 1 that if u $ LL(s),then the corresponding weight x(,,,) = 0. If k = 1, and u $! L,(s), then there exists t f V(G)such that for al1 w E Lo(t) = V(H): (s - t) A (U + w). Therefore for al1 w E V(H): (s,u) rf! (t, w). Hence (s,u) together with vertices (t, w),w E V(H),form an independent set, and therefore

The equality (3.8) implies that CurGV(H)x(~,~) = 1 and hence x(,,,) = 0. Assume that u 4 L~(s)implies x(,,,) = O. For u Lk+I(~)we have t such that for al1 w E Lk(t) : (S - t) î\ (u + w), which means also that vertices (s,u), and (t, w),w E Lk(t) form an independent set. For w 9 Lk(t) we can use Our induction hypothesis that x(~,,) = 0, and therefore we can apply the equations from the basic case. If L;(s) = 0, then for al1 u E V(H) : x(,,,) = O, which contradicts (3.8). CHAPTER 3. RELAXATIONS OF GRAPH HOhf Ob1ORPHISM

3.7 Conclusion In the first chapter, we have proved that G - H implies w(G) 5 w(H),O(G) 5 O(H), x,(G) 5 xI(H) and x(G) 5 x(H). Therefore, if O(G) 5 B(H), then we have the following inequalities: w(G) I W)I W) I q(H), and therefore .G+ H implies -$(a)5 -B(fl) and this further implies u(G) 5 wj(H), which is equivalent to G +j H. Hence an alternative definition of (polynomially decidable) pseudo-homomorphism could be that G is pseudo7-homornorphic to H if and only if B(G) 5 8(H). VVe do not know if this is equivalent to our definition of pseudo-homomorphism and hence to the hoax of Feige and Lovkz. Although, for G with the transitive automorphism group, B(G)B(G)= IV(G)((c.f. [42]). Hence it follows from our necessary condition (Corollary 38) that for graphs G with transitive automorphism group, G +, H implies B(G) 5 B(H). Our investigation was inspired by work of Feige and Lovbz [16] and this determined our choice. The alternative definition is in terms of forbidden subgraphs.

Finally, from the proof of Theorem 22 it follows that there is a bijective corre-

spondence between cliques of size IV(G)I in G O H and homomorphisms from G to H. Therefore one can identify these cliques and homomorphisms from G to H. Let us define O(0) = O. From our definition of Go H it follows that Go H is always loopleçs, even if we extend the definition to ail graphs. If one consider a pseudo-homornorphism to be any symmetric, positive semidefinite matrix A of dimension 1 V(GoH) 1 x (V(GoH)I such that A 1 = 1, for al1 {(s,u), (t, w)} 6 E(G o H): A(s,u),(tIw)= O and O(G o H) = Am J, then one may consider a category PFG of graphs and pseudo-homomorphismç, A' if one define a composition of pçeudohornomorphiçms G 3, H and H -+, W to be A" *p G W such Ai',lu),(t*w) = Co1&~(H) A(s.a),(tIb)Ai.,~)l(bl~) (compare wit h [16],Lemma 5.22). The category FG of al1 finite graphs and homomorphism is actually ernbedded in PFG. To see this, we need a faithful functor T which maps homomorphisms to pseudo-homomorphisms. Let G % H be a homomorphism. Then CHAPTER 3. RELAXATIONS OF GRAPH HOM OMORPHISM

define T(a)= A such that

CVe conclude that T is an embedding of FG to PFG and one can observe that this embedding is isomorphic to the embedding of FG to the category of graphs and hoaxes of Feige and Lovisz, which is implicit in t heir paper [16]. Chapter 4

Structure of Color Classes

4.1 Introduction

In this chapter we will consider al1 finite graphs and we will denote the set of al1 finite graphs by S.

A graph CV is called multiplicative if for every two graphs G and H,

G x H -, W implies that G + W or H 4 W (4-1) Hedetniemi [26] proposed the following conjecture.

Conjecture 48 (Hedetniemi [26) Al1 complete graphs are multiplicative.

One can easily prove that Irl, and K2are multiplicative and the proof of the multiplicativity of K3 was accomplished by El-Zahar and Sauer [14]. An alternative form of (4.1) is the following:

G f, W and H f,W imply that G x H f,W. f 4-21 If one further restricts graphs G f, W and H f,W of (4.2), then it is sometimes possible to conclude that G x H f,W. This approach was considered by Burr, Erd6s and LovAsz [9], He11 (281,Turzik [5 11, Duffus, Sands and Woodrow [12], Welzl[55] and others. Particularly, they proved the following results: CHAPTER 4. STRUCTURE OF COLOR CLASSES 61

Proposition 49 (91 If G f, IL, H f, Ii', and each vertez of G is contained in a complete subgraph O/ order n, then G x H f,K,.

Proposition 50 [12, 551 IJG and H are connected, h', -, G f,I<, and II, -+ H f, ICn, then G x Hf,&.

Note that one cm assume connectivity of G and H, since if there are G and H such that G f, Ir',, H f, Kn but G x H -+ K,, then for some connected component G' of G, G' f, A',, and sirnilady for some connected component Hf of H, H' f, linbut

G' x H' -t K,.

Proposition 51 [Si] If G f, lc,, H f, Kn and for every pair of edges e, e' of C there ezists an edge e" of G which intersects both e and e', then G x H f,K,.

For a graph G, let @(G)be the simplicial complex whose vertices are the vertices of G and whose simplexes are those subsets of V(G)which have a common neighbor. A hornornorphism a from G to H induces a simplicial mapping, since a graph hornomorphism takes a set of vertices with a common neighbor to a set of vertices with a comrnon neighbor. It was shown by Lovisz [41] that if the geometric realization I@(G)Iof @(G)is (n - 2)-connected then G f, Kn. A homological version of this result was given by Walker in [52]. A topological space X is called n-acyclic mod 2, if the j-th reduced hornologv group of X with coefficients in 2/2 vanishes for each j 5 n. Walker [52] proved that if G is a graph such that Q(C)is (n -2)-acyclic mod 2, then G 41 K,. It follows that if [@(Gx H)(is (n-2)-connected then G x H f, K,,(cf. [%SI).However, it is not the case that I@(Gx H)I = (@(G)x @(H)I (cf. [?SI). In fact, Walker [52] showed that I@(G)x @(H)1has the same homotopy type as I@(G* H)1, where G* H is the join of G and H, i.e., the graph obtained from G+H by connecting al1 vertices of G with al1 vertices of H (cf. Proposition 4.1. in [52]). O ther multiplicative graphs were studied by Hiiggkvist , Hell, Miller and Neumann- Lara ['22] and they proved:

Proposition 52 [22] Each cycle is rnultiplicatiue. CHilPTER 4. STRUCTURE OF COLOR CLASSES

It is not even known whether the following function

p(n) = min{m 1 G f,K,, and H f, but G x H -, Km for some G, H}

tends to infinity or not. Note that the Hedetniemi's conjecture is equivalent to the assertion that for al1 positive integers n, p(n) = n + 1. Poljak and Red1 [46]proved that p(n) either tends to infinity with n, or is bounded above by 16.

Duffus and Sauer [13] proposed another approach. Let us recall that a graph G is equivalent to a graph H, denoted by G * H, if G + H and H -, G. The relation

H is an equivalence, since -, is reflexive and transitive relation. In their paper [13], Duffus and Sauer studied the equivalence classes of gaphs determined by the relation

H. They observed that if (G) denotes the class of graphs equivalent to the graph G, then for any two equivalence classes of graphs (G) and (H) one can naturally define their meet (G) x (H) to be the class (G x H), their join (G) + (H) to be the class (G + H) and their relative pseudocomplement (H)((G)) to be the class (H(G)), the class containing the map graph H(G) defined in [22] (we will provide the defini- t ion later). They have observed, that the set = Ç/ u of these equivalence classes with just mentioned operations forrn a Heyting algebra with the zero element (0) and the unit element (1). Respectively, ive will use the symbols x and + for operations meet and join in distributive lattices, unless specified otherwise. Recall that Heyting algebra is a distributive iattice with the minimum element such that for every two elements a and b there exists an element x for which a x c 5 b is equivalent to c 5 x for every element c. It follows that if such an element x exists then it is unique and it is colled the relative pseudocomplement of a with respect to 6. We will denote the relative pseudocomplement of a with respect to b by b(a).

Duffus and Sauer [13] also observed that if we fix a class w E c, then the set B, = {w(g) 1 g E C) is a Boolean algebra with meet x A y = x x y, join x V y = w(w(x+ y)), complement xC = w(x),the zero element w and the unit element (1) - the class of al1 graphs with a loop. They also proved that if B(K,,)is finite for al1 n, then the function p(n) tends to infinity with n. Consequently they asked for CHAPTER 4. STRUCTURE OF COLOR CLASSES which graphs N', B(r

We Say t hat CV has a finite factorization, if there are multiplicativegraphs Wl,. .. , CL, such t hat CV - CVl x . . . x CV,. Frorn the definition it follo~vst hat if W is multiplicative and CV - W*then W' is also multiplicative. Therefore if W is multiplicative, then al1 graphs in (W) are multiplicative. We Say that a class w E is multiplicative, if graphs from w are multiplicative. LVe will prove that the boolean algebra B(w)is finite if and only if CV has a finite factorization.

One can observe that multiplicative classes of graphs are meet irreducible elements in the lattice c. Therefore B, is finite if and only if w is a rneet of finitely many meet irreducible elements of F. This suggests that the fioite factorizations of elements of are closely related to Hedetniemi's conjecture, and that understanding of the multiplicative structure of would give us some insight about the distribution of multiplicative elements. In the following sections we focus our attention on the study of the multiplicative structure of C, and as a result we give necessary and sufficient conditions for a graph W to have a finite factorization. These conditions are presented in terms of ideals, to be defined later. Using ideals to study factorization is a standard approach to this problem in other areas of mathematics, such as commutative algebra.

We will also show that even the existence of one graph Ck' which has a finite factorization and satisfies & + CV would imply that p(n) -, m. This is a generalization of the result of Duffus and Sauer in [13],who proved that if al1 Bp-,)are finite, then

p(n) -r W. By our results, their assumption is equivalent to the assumption that al1 fi, have finite factorizations.

In Section 4.3, we will make our statements from Section 4.1 more precise and the results of this section were our motivation for investigating finite factorizations. In Section 4.4, we formally define ideals and make simple observations about operations CHAPTER 4. STRUCTURE OF COLOR CLASSES 61 on them. We will aiso prove the counterpart of the prime avoidance theorem from commutative algebra. Section 4.5 contains several quite trivial observations about additive structure of ç. In Section 4.6, we study the multiplicative structure of ideais. We define prime ideais and develop some tools which are useiul in later sections. Many of Our proois were motivated by the corresponding results from commutative algebra. In Section 4.6.1, we endow the set of al1 prime ideals with Zariski topology and we investigate topological properties of some of its subset, which are closely related with the iactorization problem. In Section 4.6.2, we give necessary and suficient conditions for the existence of irredundant and finite factorizations. Finaily, in Section 4.7, we summarize our results and interpret some of them in graph theoreticai terms.

4.3 Preliminary Result s and Motivation

In this section we will prove rnost of our claims from Section 4.1. We will use results and constructions from [22], 1461 and [13]. Probably the most important construction is the following one frorn [22].

Definition 53 [22] The map graph oftwo graphs G and W, denoted W(G), is defined

as follows: the vertices of W(G) are al1 possible mappings q5 : V(G)-t V(W)and (4, +) is an edge of W(G) whenever {s, t} E E(G) implies {rj(s),$(t)} f E(W).

The following properties of the map graph were also observed in 1121.

Proposition 54 [22] (a) PV(G) has a loop if and only ifG + W. (b) G x W(G)+ W.

(4 G + W(W(G))-

(d) PV -t I.V(G). (e) G x H + W if and only if H + I.V(G). (f) W is multiplicative if and only ifG f, PV implies W(G) + W. (g) G -+ G' implies W(G') -, W(G). (h) tV -+ IV' implies W(G)+ Wf(G). CHAPTER 4. STRIïCTClRE OF COLOR CLASSES 65

Therefore W is multiplicative if and only if bV(G) *-, N' or CV(G) ++ 1 for dl graphs G. Wewill need also the following properties of the map graph.

(1) CV is multiplicative if and only if G f, CV implies W(G) + PV for al1 connected nonbipartite graphs G.

ProoJ- (i): This was proved by Duffus and Sauer [13]. (j):Since F.Vl x CV2 + W1, Lemrna 54 (h) implies that (Wl x W2)(G) + CVl(G).

Simiiarly (bVl x LV2)(G) -' W2(G)and hence (LVl x CV2)(G) -, Wl(G) x Wz(G). On the other hand, the mapping a €rom CVl(G)x W2(G) to (Wl x W2)(G) defined by a((+*~))(g) = (+(g),$(g)) is easily seen to be a homomorphism. (b):Since b& + W,+ W2,Lemrna 54 (h) implies that Wl(G) + (l.ti + CV2)(G). Sim-

ilarly Lt;(G) + (tVi + +V2)(G)and hence Wl(G) + W2(G) -+ (Wl + +V2)(G). On the ot her hand let us fix an arbi trary vertex $ from the graph ÇVl (G)+ liV2(G). Let us define a mapping a from the vertices of (W1 + CV2)(G) to the vertices of CVl(G) + LVz(G) as follows:

We will show that a is a homomarphism. Et is enough to prove that vertices 4 of (CVl + W2)(G) such that #(V(G)) V(W1) and #(V(G)) V(iV2) are isolated vertices, i.e., their neighborhoods are empty- For a contradiction let us assume that for some g,gt E V(G)we have 4(g) E V(CVl) but #(gr) E V(W2),and that 9 - 4' for some vertex 4' of (CVl + W2)(G).Since G is a connected nonbipartite graph, there are (not necessarily distinct) vertices g = gl,gz,. .. ,g2k+l = gt E V(G)such that

gi - gi+l for i = 1,. . . ,2k. Since 4 -Y #, it follows that $(g) = #(gi) h. 4'(g2) - 4(g3) - . . - - 4'(gZk) - $(g2kfl) = #(gt). This implies that #(g) and d(g') belong to CHAPTER 4. STRUCTURE OF COLOR CLASSES the same connected component of Wl + CV2, which is a contradiction. (1): One can observe that al1 bipartite graphs are multiplicative, since every bipartite graph is either empty, or it is equivalent to either Kl or I(z and dl these graphs are multiplicative. If W is multiplicative than by Lemma 54 (f) it follows that G fr W implies W(G)-+ W and hence this implication is true for al! connected nonbipartite graphs G. On the other hand, if W is not a multiplicative graph than W is not bipartite and there are graphs G,H such that G f, W,H W but G x H + W. Let el,... ,Gn be the connected components of G and Hl,. .. , Hm be the connected components of H i.e., G = Gl+ . . .+ G, and H = Hl + .. . + Hm.Then G $, W and H f, W imply that at least one connected component of G and H is not homomorphic to W. We may assume that Gl f, W and Hl f, W. But G x H -, W implies G1 x Hl -, W and hence by Proposition 54 (e), Hl -, W(Gl). Since Hl f,W7 also W(Gl) ft CV and therefore Glis a connected nonbipartite graph such that Glf+ W and W(Gl) f, W. 0

We will show later that the connectivity of a graph is a dual notion of the multiplicativity of a graph. In this sense, corollary of the conditions (k) and (1) is the following dual of the fact that the product of two connected graphs is a connected graph if and only if at lest one of them is not bipartite (see [53]).

Corollary 56 The sum of Iwo multiplicative graphs is multiplicative.

The next two constructions are from [46].

Definition 57 [.6] The arc graph of a graph G, denote 6G,is defined as foliows: the vertices of 6G are the ordered pairs (x,Y) such that x - y in G, and (x,Y) is adjacent to (u,v) wheneuer y = u.

Definition 59 [.6] For each positiue integer n, let Sn be the jollowing syrnmetric

graph. The uertices of Sn are the subsets O/ (1,. .. ,n) and two vertices A, B are adjacent whenever A is not a proper subset of B and B is not a proper subset of A. CHAPTER 4. STRUCTURE OF COLOR CLASSES

Lemma 60 [46] Sn ++ fC( ln/2l 1. Proposition 61 [46] 6G + & if and only if G + Sn.

We will prove a more general version of this proposition as Proposition 64. But even the results mentioned above enable us to prove the following.

Proposition 62 If Kn is rnultipZicative, then K is also multiplicative. ( ~nÏ2j)

Proof: Assume that hi, is multiplicative and let G x H -, fi( n ) . This is equiv-

LnDl- - alent to 6(G x H) -, hi, and hence 6G x 6H -+ K,. Since Kn is multiplicative by Our assumption, either 6G -+ Kn or 6H + hn.But this is equivalent to that either

G + Ii'( n or H + K and we conclude that li' is multiplicative. O 1n/2l ( &j 1 ( ~nÎ2,)

Note that il n 2 4, then n < and hence if there is an n 2 4 such that Ti, is multiplicative, then there are infinitely many n such that fi, is multiplicative.

The following construction generalizes the already ment ioned graph Sn from [46].

For each graph H, let SH be the graph such that the vertices of SH are al1 subsets of V(H)and two vertices A, B are adjacent in SH whenever the following two conditions are satisfied: (a) there exists a E A - B such that for dl b E B,a - b, (b) there exists b E B - A such that for al1 a E A, b - a.

One can observe that Sn is equal to Sri,,.

Proof: If G H then let us map A to a(A). This mapping is a homomorphism, since if A - B then there exists a E A - B such that for al1 b E B,a - b and hence &(a) E a(A)- a(B)and for dl a(b)E a(B),cr(a) - ~(6).

Now we can prove the following proposition, which is a generalization of Proposi- tion 61. CHAPTER 4. STRUCTURE OF COLOR CLASSES

Proposition 64 bG + H iJ und only if G + SH.

The folfowing proposition shows how to construct new multiplicative graphs from old ones.

Proposition 65 If W is multiplicative, then also Sw is multiplicative.

Proof: Assume that IV is multiplicative and let G x H + Sw. This is equivalent to 6(G x H) CV and hence 6G x 6H + W. Since W is multiplicative by our assumption, either dG -, N' or 6H -, W. But this is equivalent to that either G -+ Sw or H -+ Slvand we conclude that Sw is multiplicative. O

Proving these results, let us recall that a graph G is equivalent to a graph H,

denoted G tt H, if G -, H and H -t G. The relation H is an equivalence, since + is reflexive and transitive relation. Following [13] we are going to study the equivalence

classes of graphs determined by the relation tt. If (G) denotes the class of graphs equivalent to the graph G, then one can define the meet (G) x (H) to be the class (C x II), the join (G) + (H) to be (G + H) and the relative pseudocomplement of (CV) and (G) is (W(G))to be the class containing the map graph W(G). Cire have

already rnentioned in Section 4.1 that the set = Ç/ H of these equivalence classes with just mentioned operations form a Heyting algebra. We would like to point out

that they are well defined, since G ++G' and H tt H' imply that G x H H G' x H'

and G + H H G' + H'. From Lemma 54 (g), (h) it also follows that W H W' and

G t-, G' impfy that W(G) H Wt(G'). Moreover, if considered as a poset, (G) 5 (W) if and only if G -, W.

We have also mentioned that if we fix a class w E then the set B, = {w(g) 1 g E - c, Ç) is a Boolean algebra with meet x A y = x x y, join x V y = w(w(x + y)), complement CHAPTER 4. STRUCTURE OF COLOR CLASSES

xC = w(x), zero element w and unit (1) - the class of al1 gaphs sith a loop [13]. Dufius and Sauer [13] proved that if B(K,,)is finite for al1 n, then the function p(n) tends to infinity with n. Consequently they asked for which graphs CV, B(w) is finite. Now we can provide an answer to this question.

Proposition 66 The booiean algebra B(w) is finite if and only if W hm a finite factorkation.

Proof: Let us assume that B(w) is finite. By Stone's representation theorem (see [49]), every finite Booiean algebra is isomorphic to a Boolean algebra of al1 subsets of some finite set. Therefore the zero element is equal to the meet of al1 coatoms. Duffus and Sauer (131 have shown that al1 coatoms in B(w) are multiplicative classes (the same conclusion is implicit in Proposition 99) and hence CV has a finite factorization. On the other hand, let us assume that W - LVl x .. . x CVn. Then CC'(G) (CVl x . . . x CV,)(G) H Wl(G)x . . . x M.',(G). But since each CV; is mul- t i plicat ive, we have (LVi(G)) E {(CVi),(l)), which implies t hat there are at most 2" different classes of graphs in the set {CV(G)1 G E Ç} and therefore B(w)is finite.

We have also the following proposition.

Proposition 67 If G and H have a Jinile factorizations, then both G x H and G+ fi also have a jinite factorizations. Illoreouer, if W has a finite factorization then the map graph W(G)has a Jinite factorization, for each graph G.

Proof: Let G H Pl x . .. x P, and H H Q1 x .. . x Qm be finite factorizations of G and H, i.e., assume that al1 graphs Pl,.. . ,P,, QI,.. . , Q, are multiplicative. Then

G x H H Pl x . . . x P, x Ql x . .. x Q, and hence G x H has a finite factorization. Also G + H ++ (Pl x . . . x P,) + (Q1x . .. x Q,) ct nij(Pi+ Qj)(se Lemma 74), and by Corollary 56 al1 sums Pi + Qj,i = 1,. .. ,n,j = 1,. . . ,m are multiplicative; we conclude that also G + H has a finite factorization. For the second part let us assume that W o CVl x . .. x W, be a finite factorization of W. Then by Proposition 55, W(G)- Wl (G) x . . . x bVr(G) H nc+w,Wi, since W.'S are multiplicative; we CHAPTER 4. STRUCTURE OF COLOR CLASSES conclude that W(C)has a finite factorization.

By the previous proposition, classes of graphs which have a finite factorization form a Heyting subalgebra of the Heyting algebra c.

Now we are going to prove that it is enough to assume a finite factorization of one gaph which contains K4for the proof that p tends to infinity. We need the following proposition.

Proposition 68 If Jor every graph G there ezists e multiplicative graph W such that G + W,then the function p(n) tends to infinity with n.

Proof: Assume that for every graph G there exists a multiplicative graph W such that G + tV. Let us denote Xn = {W 1 K, + W, W multiplicative). The assump-

tion implies X, # 0 and therefore ~(n)= min{m 1 W -t Km for some W E Xn} is a well-defined, positive integer valued function. In other words, ~(n)is the minimum chromatic number of al1 multiplicative graphs containing K,. We will prove that 7(p(n)) > n. Let CVn E X,, be one of the multiplicative graph such that

V + in.Let G f, li,, H ft hn and G x H + I\;(,). Since IV'(,) E &,), G' x H -+ -, W,(,). By assumption W&) is multiplicative and hence G -r IV,(,) or H + CVp(,). On the other band WP(,) -r Ii,(,(,)), which implies that ~(~(n))> n. One can observe that p is a nondecreasing function and if it was bounded, Say by c, then for ail n we would have n < r(p(n))5 max{y(l), .. . ,y(c)}, which is a contradiction. Ci

Note that if there is a sequence of multiplicative graphs {Pi)zosuch that for al1 n there is Pi which contains li,, then the conditions of Proposition 68 are satisfied. Moreover, if K4 + W, and W has a finite factorization, then there exists a multiplicative graph P, such that W + P (P can be chosen to be any graph from the factorization of W). If we denote Po = P and no = 4, then obviously IC, -, Po.By induction let us define ni+i = (Ln:îl) and Pi+1 = Spi. Assuming Ii,, -, Pi it fdlows CHAPTER 4. STRUCTURE OF CQLOR CLASSES 71 that Kn,+,- SKn,-+ Sp,= P*+l.Since no = 4, it follows that n; < n;+l and dl Pi's are multiplicative. Therefore we have the following corollary.

Corollary 69 If there is a graph W which has a finite factorizatioa and K4 -r W, then p(n) -, W.

This corolla,ry is a motivation for our investigation of finite factorizations in the consequent sections.

4.4 Ideals

In this section, we will define ideals and operations on them. We will also make some simple observations which are useful in tater sections.

Lernma 70 For any graphs G and H,G x H -+ G + G + H.

Proof: A homomorphism a from G x H to G can be defined as ~(g,h) = g. A homomorphism B from G to G + H can be defined as p(g) = (g,O). O

Lernma 71 The fol10 wing conditions are equiualent: (i)G-Gx H (ii) G + H (iii) G + H ++ H.

Proof:

(i) + (ii): From G 4 G x H and G x H + G (according to the Lemma 70) we conclude G -+ H. (ii) + (iii): Let ai be a homomorphism of G to H and let crz be any homomorphism of H to H (for instance the identity). A homomorphism B from G + H to H can be defined as p((s, i)) = ai(s) for (s, i) E V(G + H). (iii) (ii): From G + H -+ H and G -, G + H (according to the Lemma 70) we conclude G + H. CHAPTER 4. STRUCTURE OF COLOR CLASSES 7'2

(ii) + (i): Let a be a homomorphism of G to H. A homomorphism /3 of G to G x H can be defined as P(s) = (s, a(s)) for s E V(G).

Therefore we have the following corollary. Corollary 72 For al1 graphs G, G ++ G x G - G + G. Let us recdI that a graph G is called a core if it is not hornomorphic to any of its proper subgraphs, or equivalently if any endomomorphism G + G is an automorphism.

Lemma 73 [.](cf. also [31]) Euery equivalence class of graphs has a uniquely defined graph which is a core.

Recall that we consider two graphs equal if and only if they are isomorphic. One can easily observe that + and x are commutative and associative operations and we have the following distributive laws.

Proposition 74 For G, H, F E Ç (i) G x (H + F) = (G x H) + (G x F) (ii)G+(Hx F)-(G+H)x(G+F)

Pro0f: (i): One can check that the mapping which maps the vertex (g, (s, i)) of G x (H+ F) to the vertex ((g,s), i) of (G x H) + (G x F) is an isomorphism. (ii): The mapping, which maps the vertex (g, O) to the vertex ((g, O), (g, 0)) and the vertex ((h,f), 1) to the vertex ((h,l), (f, 1)) is a homomorphism from G + (H x F) to (G + H) x (G + F). On the other hand the mapping which maps the vertex ((9, O), (s, j))to the vertex (g, O), the vertex ((hl l),(g, O)) to the vertex (g,0) and the vertex ((hl l),(f, 1)) to the vertex ((hl f), 1) is a homomorphismfrom (G+H) x(G+F) to G + (Hx FI. a

As in [13], we identify equivalent graphs, and study the multiplicative structure of the classes of equivalent graphs c. As we have mentioned above, is a Heyting CHAPTER 4. STRUCTURE OF COLOR CLASSES 73 algebra, and hence a distributive lattice with meet x and join +. Also we will use symbols O and 1 for the minimum and the maximum elements of any lattice and al1 lattices considered are assumed to have O and 1. We will not use any specific property of the Heyting algebra G and hence we will state al1 results for a general distributive lattice L with O and 1. To interpret these results for graphs, the reader only bas to think of L as the lattice c. The only exception to this rule is in the next section. Recall that we use symbols x, + and 5 for meet, join and the partial order relation respectively. We now formally define ideals.

A subset a of a lattice L is an ideal, if a satisfies the following properties: (1) A~amdG~LimplyAxG~aand (3) A, B E a implies il + B E a.

One can observe that the first condition is equivalent to (1') if A E aand B 5 A then B E a. Vie will use this condition and condition (1) interchangeably.

A subset f of a lattice 1C is a filter, if f satisfies the following properties: (f)A E f and C E L imply A + G E f and (2) A, B E f implies A x B E f.

Again the first condition is equivalent to (7)if A E a and A 5 B then B E a.

Let L be a distributive lattice. Note that the set {O) is an ideal and it is contained in every ideal of L. It is easy to see that the intersection of ideals in C is always an ideal and that C itself is an ideal. We say that an ideal is proper, if it is not equal to L. Let S be a nonvoid subset of .Cl and let (S) denotes the smallest ideal containing S (the intersection of al1 ideals containing S). The ideal (S) is called the ideal generated by S. For S = 0 we define (S) = {O) = ({O)). An ideal is called principal if it is CHAPTER 4. STRUCTURE OF COLOR CLASSES 74

generated by a one element set and jinitely generated if it is generated by finite set. In this case we will write (Al,. . ., A,) instead of ({Al,. .. , A,)). In particular, we have (1) = L.

Similarly, the set (1) is a filter and it is contained in every filter of L. It is easy to see that the intersection of filters in L is always a filter and L is a filter. We say that a filter is propet, if it is not equal to L. Let S be a nonvoid subset of L. The smallest filter containing S (the intersection of al1 filters containing S) is called the filter generated by S.

The following observations are quite trivial but we will use them later.

Proof: Let a = {G 1 G 5 H}. First we will show that a is an ideal. The condition (1) follows from the transitivity of 5, since G E a implies that G 5 H,and therefore if G' 5 G then necessarily also G' 5 H. The condition (2) follows, since G E a and G' E a imply that G 5 H and G' 5 fI so we conclude G + Gr 5 H and hence G + G' E a Since H 5 Hl it follows that H E a and therefore we have proved that ( H) C a. The condition (1) implies the other inclusion. O

Proposition 76 Euery finitely generated ideal of L is principal.

Proof: Let a = (Gi,. . . , G,) be a finitely generated ideal. From (2) it follows that the principal ideal (Gi + .. . + G,) 5 a On the other hand (Cl+ . .. + G,) contains al1 Gi,i = 1,. . . ,n, since G;< Cl+. . . + G, for al1 i = 1,. . . ,n and we conclude that a = (Gi,.. . , G,) C (Gi + . . . + Gn).0

The following proposition shows that the existence of a homomorphism can be t ranslated to inclusion between corresponding ideals in the lat t ice 8.

Proposition 77 Let G, H E L. Then G 5 H if and only if (G)C (H). CHAPTER 4. STRUCTURE OF COLOR CLASSES 75

Proof: G 5 H implies that {F 1 F 5 G} C (F 1 F < H} and by Proposition 75 it follows that (G) C (H). On the other hand, let us assume that (G) C (H). Since G E (G), this implies that G E (H) and consequently by Proposition 75, G -+ H. 0

Note that the union of two ideals is not an ideal in general. For two ideals a and b of L: we define a+b={GIG=A+B, AE~, BE^).

One can easily check that a + b is an ideal.

Proposition 78 If a and b are tuio ideals then a + b = (au b).

Pro0f: If G E a+b, then G = A+B forsome A E aand B E 6). Therefore G 5 A+B E (a~b) and hence G E (au b). On the other hand (au 6) is the smallest ideai containing both a and b and therefore (a U b) Ç a + b. 0

Proposition 79 (G)n (H)= (Gx if) and (G)+ (H) = (G + H).

Proof: Frorn the previous proposition we have the second equality. Since G x H 5 G, we have (G x H) 5 (G), and similarly (G x H) Ç (H), which implies (G x H) C (G)f~ (H). On the other hand if CV € (G)il (W),then CV 5 G and CV 5 H, which implies that bV 5 G x H, and hence CV E (G x H). a

One can observe that n and + are commutative and associative binary operations on the set of al1 ideals ILof C, which satisfy both distributive laws and hence form a distributive lattice with the minimum {O} = (0) and the maximum C = (1).

A system of sets C is called a chain if for al1 a, b E C, either a E b or b a. The intersection of the chain C is the set naEca and the union O/ the chain C is the set UaEca.

Since the intersection of ideals is always an ideal, the intersection of a chain of ideals is also an ideal. One can easily observe that the union of a chain of ideals is CHAPTER 4. STRUCTURE OF COLOR CLASSES 76 also an ideal. Similariy one can observe that the intersection and the union of a chain of filters are filters.

We will also need the following theorem, whose proof was inspired by the proof of Theorem 51 from [37].

Theorem 80 If o is an ideal and Bj is a finite system of ideals, then

a C U b implies a C b for some b E Bi. b€Bf

Prooj: By induction on n = ]BI[. The proposition is obviously true for n = 1. Assume that n 2 2. For a contradiction, let us assume that for al1 b E Br, a 6. For every c E B we may assume that

4Ub, b#c otherwise we use the induction hypothesis to conclude that a C b for some b E BI. Therefore for every c E Bj there exists Gc E a such that Gc @ Ubfcb. Since a is an ideal and n 5 2, for any two distinct c # c' E P, G = Gc+ Gc E a. We clairn that G $! Ubaa,b. For a contradiction assume that C E b for some b E BI. Since c and c' are distinct, either b # c or b # 2. We rnay assume that b # c. Since Gc < G, it follows that also GcE 6, which is a contradiction proving our claim. Therefore G E a and G 4 UbEB, b, which is a contradiction with our assurnption a E UbEBfb.

The previous theorem has a counterpart in commutative algebra where it is called prime avoidance theorem, since in the case of commutative rings at most two ideals of Br may not be primes.

4.5 Additive structure

To rarm up, let us first have a brief look at the additive structure of ideals in c. Loosely speaking, it turns out that (join) irreducible elements correspond to connected graphs and hence every graph is the sum of finitely many (join) irreducible CHAPTER 4. STRUCTLTRE OF COLOR CLASSES 77 elements. To be more precise, let us recall that the graph with empty vertex and edge sets is denoted by O. For every graph G, O + G = G + O = G, O -, G and O x G = O.

Recall that a graph W is connected, if there are no two nonzero graphs G and H such that kV = G + H.

Defimition 81 We Say that an ideal a is additive, ifa C 6 + c implies a 5 b or a Ç c.

Lemma 82 An ideal a ts not additive if and only if there are ideals b and c such that a= b+c andbgc andcg b.

Prooj: If there are ideals b and c such that a = b + c and b c and c 6, then it is sufficient to show that a b and a c in order to prove that a is no additive. Let us assume that this is not true and Say a C b. Then

which implies that a = a + c, and we conclude that c Ç a C b. On the other hand, let us assume that a is not additive and hence there are two ideals bandcsuch that asbfcbut aebandae c. Thereforea =dIb+anc. Ifaflb E anc, then a = anc which implies that a Ç c, a contradiction. We conclude that anb anc and similarly a n c a n 6. O

Proposition 83 A principal ideal (G) E 9 is additive if and only i/ the class G contains a connected core.

Prooj: By Lemma 73 we may assume that G is a core. If G is connected, then

(G) b + c implies that G E b + c and therefore G 4 H + H' for some H E b and H' E c. Since G is connected, this implies that G + H or G + H' and hence G € b or C E c. Therefore (G) 2 b or (G) c. Let us assume that G is not connected and G1,... ,G, are its connected components. Therefore (G) = (Gi)+ (G2 + . . . + G,). If (Gi)C (Gl + . . . + Gn) or equivalently

GI+ Gz + .. . + G,, then G H G1 + . .. + G,, which is a contradiction with our CHAPTER 4. STRUCTURE OF COLOR CLASSES 78 asçurnption that G is a core. Similarly (G2+ . . . + G,) C (Gi) wodd imply that G cr G1,which is again a contradiction. O

This proposition shows that connectivity is a dud of rnultiplicativity in a sense that join irreducible elements of have connected cores and meet irreducible elements of C have multiplicative cores.

Proposition 84 Every principal ideal in con be uniquely wrïtten as a Jinite sum of additive principal ideak and hence every elernent is a finite sum of join irreducible elements.

Proof: This follows frorn the fact that every graph has a unique core, whose connected components have to be again cores and therefore generate additive principal ideals. O

4.6 Multiplicative structure

In this section, we will define prime ideals and give necessary tools to deal with them. We assume al1 ideals to be ideals of a distributive lattice C with O and 1.

For any two ideals a,b we define

Lemma 85 Let a, 6, c, ù, s. be ideals. Then (a) (a : 6) is an ideal, (6) (a: 6) = (1) if and only if b Ca, (4 (a : (1)) = a, (4 a G (a: b), (e) b 2 (a : (a : b)), (f)b n (a: b) = an b, CHAPTER 4. STRUCTURE OF COLOR CLASSES

(9) ((a : 6) : C) = (a : (b n c)),

(h) (a: b) n t (a: (b : c)), (j) (a : (Ui 4) = ni(a: s), (k) (ni oi : a) = : a), (1) a C b and c C 3 imply (a : D) C (6 : c), (m) a= (a: b)n (a: (a: b)). Pro0f: (a) LetusassumethatG~(a:b)andH~G.IfB~bthenHxB=(HxG)xB= H x (G x B) 5 G x B E a and hence H E (a : 6). To prove the second condition, let us assume that G, H E (a : 6). Let B E b. Then G x B and H x B are in a and therefore also their sum G x B + G x H is in a. But (G x B) + (Hx B) = (G+ H) x B and t herefore G + H E (a : 6).

(1) Let G E (a : a) and C E c Ç 3. Then G x C E G C b and therefore G E (b : c). (b) If (a: 6) = (l),then 1 E (a : 6) and therefore for B E 6, B = 1 x B E a. On the other hand if b C a, B E b and G E (1), then G x B 5 B E b C a. (d) Let A E aand BE6. Then A x B 5 AE aimplies that AE (a: b). (c) From (d) it follows t hat a C (a : (1)). Let G E (1) and A E a. Then A x G 5 A E a which implies A E (a : (1)). (e) Let B E b and G E (a: 6). Then B x G E a and therefore B E (a: (a: b)). (f) Let G' E bn(a: b), then GE band GE (a: b). ThereforeG = G x GEa. (g) Let G E ((a : 6) : c) and H E b n c. Then G x H E (a : b) since H E c. Also H E b and hence G x H = (Gx H) x H E a. We conclude that G E (a : (bnc)). On the other hand let us assume that G E (a : (bn c)) and H E c. Then if F E b, then F x H E b n c and hence G x (F x H) E a which implies that G x H E (a : b) and we conclude that G E ((a : b) : c). (h) Let G E (a : b) n c. Then G = G x G such that G E (a : 6) and G E c. Let fi E (b : c). Then G x H = (G x G) x H = G x (G x H). Since G x H E b, G x (G x H)E a and hence G E (a : (6 : c)). (j) Let G E (a : (Uiq)) and H E q. Then G x H E a which implies that G E ai for al1 i. On the other hand let us assume that G E ni(o : q) and H E Ui 4.. Then there is an i such that H E a; which implies that G x H E a and we conclude that CHAPTER 4. STRUCTURE OF COLOR CLASSES 80

G € (a : (Ui a;)). (k) Let G f (nia; : a). Then for H E a, G x H E (7; ai C ai. Therefore G E (ai :a) for al1 i, which implies that G E ni(% : a). On the other hand assume that G E ni(% : a). Then G E (ai : a) for al1 i and therefore for H E a, G x H E ai for al1 i, which implies that G x H E ni q. Therefore G E (ni ai : a). (rn) From (d) it folIows that a Ç (a : 6) n (a : (a : 6)). The other inclusion follows hm(f). a

This Lemrna particulady implies that the set of al1 ideals Zc is a Heyting algebra wit h relative pseudocomplement a(b) = (a : 6).

An ideal p is called prime if it is proper and for any two ideals a and 6, anb C p implies a p or b C p. An equivalent condition is that for every G, H € L, G x H E p implies G f p or H f p, which means that the complement L - p is a filter.

Loaking at the cornplements, one can easily observe that both the intersection and the union of a chain of prime ideals is a prime ideal.

By induction it follows that if p is a prime ideal and Aj is a finite system of ideals,

n a c p implies a C p for some a E A,. aEA The proof of the foilowing lemma is essentially the proof of Theorem 1 from [37].

Lemma 86 Let a be an ideal, f be a filter, anf = 0 and b be an ideal containing a, which is maximal with respect to the ezclusion off. Then b is prime, Le, the complement oj 6, (i.e. C - 6) is a filter.

ProoJ: Given G x H E b we must show that either G or H belongs to 6. Suppose the contrary. Then the ideal (G) + b is strictly larger than b and therefore intersects f. Thus there exists an element Si E f such that Si = Al + G', G' E (G) and Al E 6. Similarly we have S2 E f such that S2 = A2 + Hf, Hf E (H) and A2 E b. Hence CHAPTER 4. STRUCTURE OF COLOR CLASSES

SI xS2=(Al+C)x(A2+Hf)=(A~ XA~)+(A~XH')+(G~~A~)+(G'XH')E~, since Gr x Hf 5 G x H E b and the other three terms obviously belong to b. Since f is a filter, Six Sz E fn6, which is a contradiction. O

We will also need the following lemma. It is a dudizat ion of the previous Lemma, and hence its proof can be obtained from it by interchanging the roles of + with x, 5 with 2,and ideal with filter.

Lemma 87 Let a be an ideal, f be a filter, anf = 0 and r be a filter containing f, which is mazimal with respect to the ezclusion of a. Then the complement L - e is an ideal.

An ideal rn is called maximal if it is proper, and if there is no proper ideal a such that m c a c (1).

Given any ideal a disjoint from a filter f, by Zorn's lemma we can expand a to an ideal p maximal with respect to disjointness from f, and by the previous lemma it follows that p is prime. In particular if we take a = {O} and f = {l),then the resulting ideal is a maximal ideal. Hence we have the following proposition.

Proposition 88 Euery lattice with O # 1 has at ïeast one maximal ideal.

Proposition 89 If rn is mazimal ideal in L,tiien rn is prime.

Proof: Follows from Lemma 86 taking f = (1) and a = m. a

In our particular case there is exactly one maximai ideal in C, the ideal of classes - of equivalent graphs which contain loopless graphs, i.e. m = Ç - ((1)) (and we do not need Zorn's lemma in this case).

Proposition 90 The following conditions are equiualent: (i) p is prime, (ii) (p : a) E {p, (l)}for al1 ideals a, (iii) (p : a) E {p, (1)) for al1 principal ideals a. CHAPTER 4. SZXZTCTURE OF COLOR CLASSES 82

Proof: (i) a (ii) follows from Lemma 85 (m), (a) and (ii) (iii) is obvious. To prove the implication (iii) + (i) let us assume that p is not prime and hence there are G, H E L: which do not belong to p but G x H belongs to p. Then H E (p : (G)) and hence (p : (G)) # p, since H p and also (p : (G)) # (l),since G p.

In particular, this proposition and Lemrna 85 (b) implies that for prime ideal p,

(1) ifasp p otherwise.

The proof of the following lemma is essentially the proof of Theorem 10 from [37].

Lemma 91 Let a be an ideai which does not contain G. Then a is contained in a pr5me ideal which also does not contain G. Moreover, there ezists a prime ideal p satisfying these conditions which is minimal with respect to the inclusion.

ProoJ: Let f be the smallest filter which contains G, Le., f contains exactly those elements, which are greater or equal to G. One can observe that a and f are disjoint and by Zorn's lemma we can expand a to an ideal p which is maximal with respect to exclusion of f, and hence prime by Lemma 86. Let us embed p in a maximal chain P of prime ideals containing a (using Zorn's lemma). The intersection of the chain P is a prime ideal which is clearly minimal.

4.6.1 Prime Spectrum

In this section we study topological properties of prime ideals. It turns out, as we will see in the next section, that they are related to the factorization problem.

Let L be a distributive lattice with O and 1 and Spec(L) be the set of al1 prime ideals of f. For each subset E of L,let V(E)denote the set of al1 prime ideals which contain E. For G E L we write V(G)instead of V({G)).One can observe that (i) if a = (E) is the ideal generated by E, then V(E)= V(a), CHAPTER 4. STRUCTURE OF COLOR CLASSES

(ii) V(0) = Spec(&), V(l) = 0, (iii) V(4) = V(U;,I Ei), (iv) V(a) U V(b) = V(a n b), (u) V(b) C V(a) if and only if a C 6. PVe will prove only the condition (v). Assume th& a C 6. It follows that if p C b then p E a and hence V(b) C V(a). On the other hand let us assume that a b. Then there is G E a such that G 4 6. By Lemma 91 it follows that there is p E V(b) such that G 4 p and hence p cannot contain a as a subset. Therefore p 6 V(a) and we conclude that V(b) V(a).

Conditions (i) - (iv) imply that the sets V(E)satisfy the axioms for closed sets in a topological space. The resulting topology is called the Zariski topology. The topological space Spec(L) is calied the prime spectrum of L.

The open sets of the form Spec(L)- V(G), G E L are called basic open sets. They are denoted by D(G) and they form a basis for the open sets of the Zariski topology, since every closed set is of the form V(E) = nGEEV(G)- Moreover, one cm observe t hat (5) D(G) n D(H) = D(G x H), (z)D(G) = 0 if and only if G = 0, (lii)D(G) = Spec(L) if and only if G = 1, (Z)D(G) = D(H) if and only if G = H, (F)D(G) is compact, i.e., every open covering of D(G)ha a finite subcovering. In particular, (v) and (z)imply that SpecfL) is compact.

For an arbitrary subset A Ç Spec(C), the ideal J(A) = n PEA is called the ideal of A. One can easily observe that J(UiEIAi) = niErJ(Ai)- Lemma 92 For every E 5 C and A C Spec(L) we have E C J(A) if and only ij A V(E). CHAPTER 4. STRUCTURE OF COLOR CLASSES 84

Proof: Let us assume that E C J(A). It foliows by the condition (v) that V(J(A))E V(E]. But A C V(J(A)), since if p E A then J(A) = nqEAq E p and hence P E V(J(A)). Therefore A C V(J(A))C V(E}. On the other hand let us assume that A C V(E). This means that for al1 p E A, p _> E and hence also J(A) = bEAp C E. fi

Proposition 93 V(J(A))= A is fie dosure r>fA in Spec(C)for euery A Ç Spec(C), and J(V(E)) = (E) is the ideal generated by E in L for every E C L.

Prooj: The first equality follows, since

For the second equality let us denote a = (E), the ideal generated by E. The first equality and (i) imply that V(a) = V(E)= V(E) = V(J(V(E))) and hence by (v) we have (E) = a = J(V(E)). O

Let us recall that a topological space X is Hausdorf if for every two distinct points z, y E X there exist two disjoint open sets Cl&), Vy(x) X such that x E U&) and y E VY(x). In particular if X is Hausdorf, then al1 points x E X are dosed, since X - {x} = Uy+, V,(x) is open. The spectrum Spec(C) with the Zariski topology is not Hausdorf, since any two nonempty open sets have nonempty intersection by (T). Moreover, it contains nonclosed points, which is due to the fact that Spec(L) includes not just maximal ideals, but al1 prime ideals. Let us determine the closure of a point of Spec(L). If the point is the prime ideal p then its closure is by Proposition 93 V(J(p)) = V(p) and hence it consists of al1 prime ideals q with p q. In particular, a point of Spec(L) is closed if and only if it is a maximal ideal.

Let a be a proper ideal in C and let us denote the set of al1 minimal prime ideals containing a (with respect to the inclusion) by Ma, i.e., Me = (P E Spec(L) ( P 2 a and p 2 q 1 a implies p = q for al1 q E Spec(L)). Lemma 91 implies that Mo # 0 (take G = 1). Ma is a subset of SpecIL) and we endow Ma with the induced topology CHAPTER 4. STRUCTURE OF COLOR CLASSES 85 of Spec(Cj, Le., Y Ma is closed in Ma if and only if Y = V ri Ma for some cIosed subset V of Spec(C). One can observe that dl closed sets are of the form Va(E)= V(E)n Ma for some E C C and the sets Da(G) = D(G)n Ma, G E C form a bais for the open sets of the induced topology on Ma. It follows that al1 points of Ma are closed in Ma. We wil1 show that Ma is actually Hausdorf space, but first we will prove a criterion for a prime to belong to Ma. The proof of the following lemma was inspired by the proof of Theorem 2.1 from [35].

Lemrna 94 Let a be a proper ideal and p a subset ofC. Then theIollozuing conditions are epuivalent: (i P f lua, (ti) L - p is a Jilter that is maximal with respect to missing a, (iii) a C p, p is a prime ideal and for every G E p, there is an H p such that GxHE~.

Proof: (i) s (ii): If p E Ma then p is prime and hence L - p is a filter. By Zorn's lemma one cmexpand L - p to a filter f that is maximal with respect to missing a. If q is an ideal containing a that is maximal with respect to being disjoint from f then q is prime by Lemma 86. Note that q is disjoint from L - p which irnplies that q p and hence the minimality of p implies that q = p. Thus C - p = f. (ii) + (iii): Since f = L - p is a filter that is maximal with respect to missing a, a C p. Moreover, by Lemrna 57, p is an ideal and since its complement is a filter it is a prime ideal. Let G E p and let f = {G x H ( H E C - p). Then f is a filter that properly contains L - p. So there is some H E C - p such that G x H E a. (iii) 3 (i): Assume that a c q Ç p, where q is a prime ideal. If there exist some G E p - q, then there is an H 4 p such that G x H E a C q, which is a contradiction because q is a prime ideal. Therefore p = q. 0

Now we can prove the following theorem.

Theorem 95 The topological space Ma is Hausdorf. CHAPTER 4. STRUCTURE OF COLOR CLASSES

Proof: If a is a prime ideal, then Ma has only one point and Our theorem is true. Therefore we may assume that a is not a prime ideal. Let p and q be two distinct points of &. Since both of them are minimal and distinct, we have p g q and q p. Therefore there exist P,Q such that P E p, P $ q, Q E q and Q p. Moreover by Lemma 94 there exists G p such that P x G E a Since both Q and G do not belong to fi, also Q' = Q x G does not beloag to p but Q' belongs to q, since Q does. Hence p E D(Q'), q E D(P) and D(-P)nD(Q') = D(P x Q'). But D(P x Q')nl& = 0, since P x Q' = P x G x Q 5 P x G E a and hence every minimal prime ideal which contains a has to also contain P x Q'. We conclude that the open sets D,(P) = D(P)n Ma and Do(Q1)= D(Qf)n Ma are disjoint, p E &(QI), q E Da(P) and hence Ma is a Hausdorf space. O

Let us recall that a topological space is discrete if al1 of its subsets are clopen, Le., they are both closed and open. If X is a finite Hausdorf space, then al1 subsets of X are closed (and hence also open) since al1 points are closed, and therefore X is discrete. Note that a topological space is discrete if and only if all of its points are clopen. It turns out, as we will see in the next section, that discreetness of Ma is closely related to the factorization of a. Therefore Our aim is to identily al1 points of hla which are open (and hence clopen) in Ma.

Proposition 96 Va(J(A)) is the closure of A in bI, for every A 2 hk, J(Ma) = a, and J(Da(G)) = (a : (G))for every G E L.

Proof: From Proposition 93 it follows that Va(J(A))is the closure of A in Ma. To prove the equality J(Ma) = a, let us denote b = J(Ma) = npcM,p. Obviously a E b since a is a subset of every p from Ma # 0. Assume that G fZ a belongs to 6. By Lemma 86 there is a minimal prime ideal containing a which does not contain G, which is a contradiction, since b is intersection of al1 minimal prime ideals containing a. We conclude that a = 6. CH.4PTER 4. STRUCTURE OF COLOR CLASSES 87

The last equality follows, since

We Say that an ideal b is a-diuisorial, if b = (a : (a : 6)). We denote the set of dl prime ideals which are a-divisorial by Oa. We will prove later that Oa is a subset of n/r, and that Oa is the set of al1 clopen points of Ma.

Lemma 97 The ideal 6 is a-divisorial if and only if there exists an ideal c such that b = (a : c).

Prooj: If b is a-divisorial, then b = (a : (a : 6)) and therefore we can take c = (a : 6). On the other hand let us assume that b = (a : c) for some ideal c. Then Lemma S5 (e) implies that c (a : (a : c)) = (a : 6) and hence by Lemma 85 (1) we have (a : (a : 6)) C (a : c) = 6. The reverse inclusion follows from Lemma 55 (e). O

It follows from Lemma 85 (b) that L: = (1) = (a : a) and from Lernrna 85 (c)we have also a = (a : (1)). Therefore a and (1) are always a-divisorial. Lemma 85 (d) also implies that for every a-divisorial ideal 6, a E b C (1).

Lemma 98 If p is a prime ideal such that a C p, then a c (a : p) if and only ij p is a-diuisorial.

Proof: Let us assume that a c (a : p). Then we have (a : p) n (a : (a : p)) = a C p and since p is prime (a : p) p or (a : (a : p)) C p. But (a : p) C p would impiy that (a : p) = p n (a : p) = an p = a which is a contradiction with Our assumption a c (a : p). We conclude that (a : (a : p)) Ç p, which together with the (e) part of Lemma 85 imply that p is a-divisorial. On the other hand let us assume that p is a-divisorial. For a contradiction let a = (a : p). Then p = (a : (a : p)) = (a : a) = (l),which is a contradiction. O CHAPTER 4. STRUCTURE OF COLOR CLASSES 8s

An a-divisorial ided b is called a marimal a-dzuisorial ideal, if it is proper and if there is no proper a-divisorial ideal c such that b c c c (1).

Proposition 99 p E Oa ij and only ifp is a mazirnal a-divisorial ideal.

Prooj: Assume that p is prime a-divisorid ideal but it is not maximal a-divisorial ideal, i.e., p C b for some a-divisorial ideal b # (1). Lemma85 (1) implieç that (a : 6) 5 (a : p). Since p is a-divisorid, a C p and Lemma 85 (f) impIies that b n (a : b) = anb Ç a E p. But p is prime and hence (a : b) Ç p, since b p is impossible due to Our assumption b3p. Theinclusions(a: b) C(a:p)and(a:b)Epirnply(a: b) Gpn(a:p)=pna=a and hence by Lemma 85 (b) and (l), (1) = (a : a) C (a : (a : b)) = b, which is a contradiction. On the other hand assume that p is maximal a-divisorid ideal and suppose b and c are ideab such that b n c Ç p but b p and also c p. It can be assumed that p C b and p C c, ot herwise one can replace b with p + b and c with p + c. Tben p c b C (p : c) C (1). If wc prove that (p : c) is a-divisorial, then these inequations would imply that (p : c) is a-divisorial ideal properly containing p and therefore it is equai to (1). This would imply that c Ç p, which is a contradiction with our assurnptions. Therefore, to finish the proof that p is prime, it is enough to show that (p : c) is a-divisorial. But this follows from the equalities

(p: c) = ((a: (a: p)) : c) =(a: ((a: p) n c)),

and Lemma 97. O

Lemma 100 Oa C Ma.

Proof: Assume that p E Oa. If p is not minimal, then tkre exists a prime ideal q such that a C q c p c (1). Lemma 85 (1) and Lemma 98 impIy that a C (a : p) C_ (a : q) and hence by Lemma 9s it follows that q is a-divisorial. But this is a contradiction, since by Lemma 99 q is also maximal a-divisorial ideal. O

For open subsets of Ma we have the following lemma. CHAPTER 4. STRUCTURE OF COLOR CLASSES 89

Lemrna 101 If a subset A Ma is open in Ma then J(A) is a-diuisorial ideal. On the other hand, if b is a-divisorial ideal, then there ezists a subset A Ç Ma which is open in Ma and b = J(A).

Prooj: Let us assume that the subset A C Ma is open in Ma. Therefore A = UiElDa(G;) and hence by the Proposition 96 J(A)= J(U 4(G))= n J(&(G))= nt. : (ci))= (a : (U(Gi))). iEI iEI iEI iEI Therefore by Lemma 97, J(A) is a-divisorid ideal. On the other hand if b is a-divisorid ideal, then by Lemma 97 there exists an ideal c such that b = (a : c). Therefore

Now we are ready to identify al1 open (and hence clopen) points of Ma.

Theorem 102 Let p be a point of Ma. Then the foilowing conditions are equivalent: (i) (p) is clopen in Ma, (2;) P E Oa, (iii) (p) = Da(G) for sorne G E L, Le., {p) is a basic open set in Ma.

Proof: (i) =r (ii): Let us assume that (p) is clopen and hence open in Ma. Frorn Lemma 101 it follows that J({p)) = p is a-divisoriai ideal and hence p E Oa. (ii) (iii): Let us assume that p E O=. Since p is a-divisorial, p = (a : (a : p)). Moreover, by Lemrna 99 p is maximal a-divisorid ideal and by Lemrna 98 a c (a : p). Therefore there exists G E (a : p) such that G $ a. Then a c a + (G)C (a : p) and hence p = (a : (a : p)) Ç (a : (a + (G))) c (1). Since p is maximal, p = (a : (a -t (G)))= (a : a) n (a : (G)) = (1) n (a : (G)) = (a : (G)) = J(Da(G)) by Proposition 96. We claim that Da(G) = {p), since if q E Do(G), then p = J(Da(G)) C q and hence q = p. (iii) + (i): Assume that {p) is a basic open set in Ma. Since Ma is Hausdorf, {p) is also closed in Ma and we conclude that {p) is clopen in Ma. O

It follows from this theorem that hla is discrete if and only if Ma = Oa, i.e., al1 points of Ma are clopen. CHAPTER 4. STRUCTURE OF COLOR CLASSES

4.6.2 Factorization

In this section we define and give necessary and sufficient conditions for an existence of factorizations in distributive lattice with O and 1 and consequently in Heyting algebras.

Let a be a proper ideal. The representation

of a as an intersection of prime ideals is called a factorization. If A is finite, than it is called a finite factoriration. The factorization is called inedundant if

# n p for each q E A. PZ4

The existence of a factorization follows from Proposition 93, since we have a = J(V(a)) = Q,Ev(a) p. Unless a is an intersection of maximal ideals, t here are ide& p, q in V(a) such that p C q and hence this factorization is obviously redundant. Anothet factorization follows from the Proposition 96, since we have a = J(f%) = nPEMap and we will cal1 this factorization the natural jactorization.

Lemma 103 If a proper ideal a has a finite jactorization, then if has an irredundant (finite) factorization.

Proof: If a has a finite factorization, then one can omit the redundant factors from finite factorization to obtain an irredundant one. 0

We will show that an irredundant factorization is unique (if it exists), but we need first the following lemma.

LeKrma 104 Ifa = npEAp is an ivedundant lacton'zation, then A Ma.

Proof.. Let us assume that q E A, but q # Ma. Then by Zorn's Lemma there is q' E Ma such that q' C p. Since q n np+qp = a C- q' and q P q', it follows that CHAPTER 4. STRUCTURE OF COLOR CLASSES 9 1

Q,+q~ 2 q' and therefore npfqp C q' c q, which is a contradiction with our assumption that the factorization is irredundant. O

Now we wiil prove the uniqueness of irredundant factorization.

Theorem 105 The factorization a = bEAp is irredundant if and ody if A = (la.In particukar, the irredundant factoriration iç unique.

Proof: Let us assume that the factorization a = p is irredundant. To prove that O=C A, let us assume that q E Oa. Then q = (a : (G)) for some G E L by Proposition 102 and hence

for al1 p such that G 4 p. If G f p for al1 p E A, then q = (l), which is a contradiction. Since by Lemma 104, A C Ma we conclude that q = p E A. To prove the reverse inclusion A Oat let us assume that q E A. Since the factorization is irredundant, Q,#q p g q and hence there is G E bf p such that G 4 q. Therefore

This implies that q is a-divisorid and hence q E Oa. To prove the other implication we need to show that if a = npEoapi then this factorization is irredundant. Let q € Oa. Then by Lemma 98

and therefore this factorization is irredundant. O

An obvious corollary is the foilowing.

Corollary 106 The natural factorization of a is irredundant if and only if Ma = Oa.

Therefore an irredundant factorization of a exists (and is unique) if and only if J(Oa) = a and we have the following theorem. CHAPTER 4. STRUCTURE OF COLOR CLASSES 92

Theorem 107 A proper ideal a has an irredundant factorkation if and only if the set 0, is dense in AILI.

Proofi If a has the irredundant factorization then J(Oa) = a and hence by Proposi- tion 96 the closure of Oa in Ma is equal to Va(J(Oa)) = Va(a) = Ma. On the other hand if the set Oa is dense in Ma, then its closure in Mo is Ma and hence by Propo- sition .96 h& = Va(J(Oa))- This implies that a = J(Ma) = J(Va(J(Oo)))= J(Oa) by

Proposition 93. [7

We have the following equivalent conditions for the naturai factorization to be irredundant.

Theorem 108 The following conditions are equivalent: (i) the natural factoriration of a is irredundant, (ii) for each q E bIa, flpfqp SZ q1 (iii) Ma = Oa, i.e., hla is a discrete space.

Proof: (i) ++ (ii), since a = nPEAp = Op+qp if and only if bpqp Ç q. (i) ++ (iii) is Corollary 106.

For Finite factorizations we need the following lemma.

Lemma 109 If a has a finite factorizalion then Ma = Op.

Proof: If a has a finite factorization then it has an irredundant factorization and hence Oa is finite by Theorem 105. We have already proved that Oa Ma. To prove the other inclusion, assume that q E Ma. TThen p = a C q and since Oa is finite, there is p E Oa such that p Ç q. From the minimality of q it follows that q = p E Oa. O

Now we can prove the following equivalent conditions for the existence of a finite factorization. CHAPTER 4. STRC:CTURE OF COLOR CLASSES

Theorem 110 Let a be a proper ideal. Then the folIotuing conditions are equiualent: (i) a hm a finite factoriration, (4 for each s E s e UpZqP, (iii) the natural factorization of a is finite, Le., Ma is finite.

Proof: We will prove that (i) M (iii) and (ii) a (iii). fi) (iii): If a has a-finite factorization then it has an irredundant factorization and hence Oa is finite by Theorem 105. Moreover, Ma= Oa by Lemma 109 and hence Ma is finite. (iii) + (i): This implication is obvious. (iii) (ii): This implication follows from Theorem 80. (i) (2): .4ssurne that for each q E Ma, q g UpWp. For a contradiction, let us assume that Ma is infinite. Since Va(l) = 0 and Va(G) = Ada for every G E a, the set

is a subset of L which is disjoint from a and 1 E f. We daim that f is a filter. This claim follows, since if G E f and G 5 H then Va(H) 5 Va(G), which implies that H E f. If G,H E f then Va(G x H) = Va(G) u Va(H), which implies that G x H E f. By Zorn's lemma we can enlarge f to a maximal filter c which is disjoint from a and by Lemma 94, c = L - q for some minimal prime ideal q E Ma. Since q 55 UpZqp, t here is C E q such that G $ p for each p # q from Ma. Since G E q, C 6 c = L - q. On the other hand Va(G) = Iq} which implies G E f C c, which is the desired contradiction. O

The following proposition follows from definitions.

Proposition 111 If 'H is a Heyting algebra, then for W,G E 'H, ((W): (G))= (W(W

This rneans that if a and EI are principal ideals in a Heyting algebra, then also (a : 6) is a principal ideal. Together with Proposition 102 (iii) this irnplies that if a is principal, then al1 a-divisorial prime ideals are principal, since {p} = Da(@ implies that p = J({p}) = J(Da(G)) = (a : (G)) by Proposition 96. This means that if a CHAPTER 4. STWCTURE OF COLOR CLASSES 9-4

is principal, then al1 ideals from 0. are principal. The following proposition is aiso irnmediate from the definition.

Proposition 112 A principal ideal (P) C is prime if and only if the class of equivalen t graphs P is multiplicative.

Now we can give the promissed necessary and sufficient conditions for finite factorization of principal ideals in a Heyting algebra (and hence graphs up to the relation 4

Theorem 113 Let 'H be a Heyting algebra, 1 # W E 'H and let a = (W). Then the following conditions are equiualent: (i) a has a finite factorization, (ii) Ma = O,, (iii) al1 minimal prime ideals containing W are principaL

Proof: (i) * (ii): This implication follows from Lernma 109. (ii) * (iii): Let us assume that Ma = 0.. Since al1 ideals from Oa are principal also al1 minimal ideals containing W, i.e., ide& from Ilfa are principal.

(iii) + (2): Let us assume that al1 minimal ideals containing I.V (i.e. ideals from hla) are principal. We will show that the condition (ii) of Theorem 110 is satisfied. We may assume that IMQ[ = co, otherwise the natural factorization is a finite factorization. For a contradiction assume that for sorne q e Ma, q C_ Up2q~Since al1 minimal prime ideals containing W are principal, q = (G) for some G E %. Therefore G E q Upfqpand hence G E p for some p. # q. It follows that q = (G) C p, which is a contradiction with the minimality of p.

4.7 Conclusion

We will surnmarize what we have proved and interpret the results for graphs. Let us notice that there are Heyting algebras for which not every element is a meet of finitely CHAPTER 4. STRUCTURE OF COLOR CLASSES 95 many rneet irreducible elements. An example is any infinite Boolean algebra. Let us consider the following properties of a distributive iat tice L: Pl. Every element is a join of finitely many join irreducible elements. P2. Every element is a meet of finitely many meet irreducible elements. P3. The meet of any two join irreducible elements is join irreducible. P4. The join of any two meet irreducible elements is meet irreducible. It follows from Proposition 84, [53] and Coroilary 56 that the Heyting algebra is countable distributive Iattice and satisfies Pl, P3 and P4. In [7] it is shown that Pl and P3 imply P4 and that Pl, P2, and P3 are both necessary and sufficient conditions for a countable L to be projective in the variety of dl distributive lattices (for definitions we refer interested readers to 171). Therefore every graph has a finite factorization if and only if is projective in the variety of distributive lattices. We do not know whether this is indeed the case. We do not even know whether K4 has a finite factorization. Another condition satisfied by is: P5. if G E is a join irreducible elernent, then (CVT + fV2)(G) = CVl(G)+ tV2(G) for al1 LV,, \V2 E G. This condition foIlows from Proposition 55. The conditions Pl and P2 are not true in any infinite Boolean algebra. The conditions P3 and P4 are not true in any Boolean algebra and the condition P5 is true in every BooIean algebra. We do not know whether every Heyting algebra which satisfies four conditions Pl, P3, P4, and P5 has to also satisfy the condition P2.

To interpret the last theorem in the graph theoretical terms, let us fix a graph W. For a set S of graphs which contains the graph W we Say that S Ss IV-minimal, if it satisfies the following four conditions

1. (1st ideal condition) G E S and H -t G imply H E S, 2. (2nd ideal condition) G E S and H E S impiy G + H t~ S, 3. (primality condition) G x H E S irnplies G E S or H E S, 4. (W-rninirnality condition) G f S if and only if W(G)+! S.

For a set S of graphs we Say that S is principal, if it satisfies the condition CHAPTER 4. STRUCTURE OF COLOR CLASSES

5. (principdity condition) there exists G E S such that H + G for dl H E S.

It follows that W has a finite factorization if and only if every W-minimal set of graphs is principal. Chapter 5

Conclusions and Furt her Research

In this thesis we have studied graph homomorphisms, equitable graph hornomor- phisms, and the corresponding H-coloring problems from both theoretical and practical view points.

In Chapter 2, we have completely characterized the complexity of the equitable H-coloring problem and its restricted version, the connected equitable H-coloring problem. In particular, we have proved that both problems are polynomial if H is a disjoint union of cornpiete bipartite graphs. Since the complexity of Our algorithm is exponential in the number of connected components of H, one possible area for further research is to improve our aigorithm or to find a better one.

In Chapter 3, we have proposed relaxations of graph homomorphism. In particular we defined pseudo-homomorphism and fractional homomorphism as, respectively, semidefinite and linear relaxations of a certain integer program corresponding to the graph homomorphism problem. We have shown a simple forbidden subgraph characterization for the existence of fractional homomorphism but we could prove only a necessary condition for the existence of pseudo-homomorphism. It would be nice to have a similar characterization for the pseudo-homomorphism. Another possible area for further research is to give a nontrivial example of a class of gaphs for which either fractional homomorphism or pseudo-homomorphism would give a polynomial CHAPTER 5. CONCL LISIONS AND FURTHER RESEARCH 98 algorithm, since al1 of our examples are already known to be polynomial time solvable by other means. It seerns that using semidefinite progamming is overkill for these particular examples.

In Chapter 4, we have studied the multiplicative structure of equivalence classes of graphs. In particular, we have considered the problem of finite factorization. We gave necessary and sufEicient conditions for the existence of a finite factorization. However, we are not able to use these conditions even for simple graphs like K4. Curently it is known that KI, Kz, K3 and al1 cycles are multiplicative and that if G + H and H f, G then G x H is not multiplicative (cf. [22]). Therefore it is easy to construct a gaph which is not multiplicative, e.g., K3 x hil1,where Mll is the Mycielski gaph €rom Figure 3.1. We have shown that if Kq is multiplicative then there are infinitely many cornpiete gaphs which are multiplicative. We have also shown that if K4 has a finite factorization then ~(n)goes to infinity with n. (Whether p(n) goes to infinity with n is an open problem.) It is possible that K4 does not even have a finite factorization. Another possible problem for further research is to answer the factorization question for KA. Bibliography

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APPLIED A IMGE. lnc