Structure of Graph Homomorphisms

Roman Batik

MSc., Comenius' University, Czechoslovakia, 1988

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Name:

Degree:

APPROVAL .

Roman Bat ik

Doctor of Philosophy

Title of thesis: St ruc tu re of Graph I-lomomorphisms

Examining Committee: Dr. Fred Popowich

Chair

Dr. Pavol Hell

Senior Supervisor

Art Liestman

Supervisor

Dr . , Xrvind G u p t a

Examiner

Dr. l f ichael Fellows

External Examiner

August 1 1 , 1 9 9 7 Date Approved:

. . 11

Abstract

In this thesis we study finite graphs and graph homomorphisms from both, a theo-

retical 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 11-coloring of a graph G is a ho-

momorphism from G to H such that the preimages of vertices of H have almost the

same size (they differ by at most one). We consider the complexity of the following

problem:

INSTANCE: A graph G.

QUESTION: Does G admit an equitable 11-coloring?

We give a complete characterization of the complexity of the equitable If-coloring

problem. In particular, we show that the problem is polynomial if I1 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 relax-

ations of the problem. The second part of this thesis deals 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

LovAsz. We also show that there is a simple forbidden subgraph characterization for

our fractional 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 Grotschel, Lovrisz and Schrijver using the ellipsoid

method. We also briefly discuss how to apply these results to the directed case.

In the last part of this thesis we consider equivalence classes of graphs under the

following equivalence: two graphs G and H are equivalent, if there exist homomor-

phisms from G to N 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.

Acknowledgments

The author is extremely grateful for the continuing support and guidance of his com-

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 pro-

jective lattices in Chapter 4.

Contents

... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Abstract 111

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Introduction 1

1.1 Basic Definitions and Notations . . . . . . . . . . . . . . . . . 1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Background 8

. . . . . . . . . . . . . . . . . . . . 2 Equitable Graph Homomorphism 15

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Overview 16

. . . . . . . . . . . . . . . . . . . . . . . . 2.3 Preliminary Results 17

. . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Polynomiality 19

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 NP-hardness 22

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Conclusion 32

. . . . . . . . . . . . . . . . . . 3 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 I-Ionlomorphisms 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

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Conclusion

. . . . . . . . . . . . . . . . . . . . . . . . 4 Structure of Color Classes

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Overview

. . . . . . . . . . . . . . . 4.3 Preliminary Results and Motivation

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Ideals

. . . . . . . . . . . . . . . . . . . . . . . . 4.5 Additive structure

. . . . . . . . . . . . . . . . . . . . . 4.6 Multiplicative structure

. . . . . . . . . . . . . . . . . . . . 4.6.1 Prime Spectrum

. . . . . . . . . . . . . . . . . . . . . . 4.6.2 Factorization

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Conclusion

. . . . . . . . . . . . . . . . . . . . 5 Conclusions and Further Research

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography

vii

Chapter 1

Introduction

1.1 Basic Definitions and Notations

Let 2, N, No, Q and R be the sets of all integers, positive integers, nonnegative

integers, rational numbers and real numbers, respectively. For a set S, we denote the

cardinality of S by ISI. For two sets S and T, S c T denotes the strict, or proper in-

clusion, i.e., S T and S f T. The set of all subsets of a given set S is denoted by 2'.

A graph is an ordered pair (V, E), where V is a (possibly empty) finite set (IVI < CQ) 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 ver-

tices 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 i f it does not have any

loop. Two vertices g, g' E V are called adjacent if {g, g'} E E and we will write g - g'.

A bijective mapping cu : V(G) -+ V ( H ) such that for all g, g' E V(G), g g' i f and

only if a ( g ) a (g l ) 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 = •’1, if

there exists an isornorphism from G to H.

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

CHAPTER 1. LVTRODUCTION *> -

E(G) = {{1,2), {2,3), . . . , {k - 1, k ) ) is called 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) = { I , . . . , k ) and k edges E(G) =

{{1,2), {2,3) , . . . , {k - 1, k), {k, 1)) is called the cycle of length k and it is denoted

by Ck, see Figure 1.1 (A), (C).

Example 3 A graph G with k E N vertices V(G) = (1, . . . , k ) and 0 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) = (1, . . . , k) and (i) edges E(G) =

{S 2 V(G) I IS1 = 2) is called the clique, or the complete graph of size k and it is

denoted by Kk, see Figure 1.1 (C).

Example 5 A graph G with s + t E N vertices V(G) = (1, . . . , s + t ) where s , t E N

and st edges E(G) = {{i, j) C_ V(G) I i < s and s < j 5 s + t ) is called the complete

bipartite graph of size s + t and it is denoted b y K,,t, see Figure 1.1 (D).

Example 6 A graph G with 1 vertex V(G) = (1) and 1 edge E(G) = ((1)) is called

the loop, and it is denoted by 1.

Example 7 A graph G with 0 vertices V(G) = 0 and 0 edges E(G) = 0 is called

the empty graph, and it is denoted by 0. For a technical reason we also define the

complete graph KO = 0 .

Given two graphs G and H , we say that G is a subgraph of H , if V(G) C V ( H )

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) = 2"(') n E ( N ) , then G is called an induced subgraph of

11. For a subset S C 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

H J ( V ( H ) - { h ) ) . Graph G is called a spanning subgraph of H if G is a subgraph of

N and V(G) = V(H) .

Figure 1.1: (A): Cs, (B): P4, (C): K3 = C3, (D): 1<2,3

A homomorphism is a mapping a : V(G) -, V ( H ) from 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 N g1 in G implies a (g ) a (g t ) in H . We say that a

graph G is homomorphic to a graph H, if such a homomorphism exists, and in this

case we will write G --+ H, or G -% H . If there is no homomorphism from G to H,

we write G f t H .

A homomorphism G 3 G is called an endomorphism.

For every graph G, there is a unique homomorphism tG to the graph 1 which maps

all vertices of G to the unique vertex of 1. Also, for every graph G, there is a unique

homomorphism Oc from 0 to G which is just the empty mapping.

If G -% H is a homomorphism and W is a subgraph of G, then a restriction of a

to W, denote alW, is a homomorphism CV 3 H defined by (a lW)(w) = a ( w ) for

every w E V(W) C V(G).

A homomorphism W -% G' is called injective, if w # w1 E V(W) implies a ( w ) # a(wl) E V(G), i.e., if cr is one to one. An injective homomorphism is also called an

monomorphism.

CHAPTER 1. INTROD UCTION

A homomorphism G -% H is called surjective, if h E V ( W ) 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 (sometimes epimorphism is referred to

a mapping which is onto both vertices and edges, cf. [27], but we will require from

epimorphism to be onto 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 iso-

morphic 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 can define a bimorphism from C4 to K4, but these two

graphs are not isomorphic.

An isomorphism from G -% G is called an automorphism. The autornorphism

which maps every vertex of G to itself is denoted by lG, and it is also called the

identity automorphism. Also, one can observe that for an endomorphism G 5 G the

following three conditions are equivalent (since we only consider finite graphs):

a is a monomorphism,

a is an epimorphism,

a is an automorphism.

P If G -% H and H -+ CV are two homomorphisms, then their composition is the

mapping Pa, where (,f?a)(g) = P ( Q ( ~ ) ) for every g E V(G). One can easily observe P a that G ---+ W is a homomorphism. Since the operation composition is a composition

of mappings, it is associative. Moreover, plG = ,f? = lHP. Also, one can observe

that if pa is surjective, then necessarily p is surjective and if pa is injective, then

necessarily a is injective. Moreover, a is injective if and only if cup = a y implies

p = y and a is surjective if and only if pa = y a implies /3 = y.

CH'4PTER I . INTRODUCTION 5

Note that if W is a subgraph of G , then the natural inclusion z which maps vertices

of G to the corresponding vertices of H is a ~nonomorphism and az = alW.

Graphs G and H are called equivalent, denoted G ++ H , if G -+ H and H -t G.

One can observe that the relation ++ is an equivalence and if G1 +-+ G2, H I ++ & and GI t HI then also G2 -+ H2. Equivalent classes of graphs are also called color

families by Weld [54].

P If G -3 H and H t G are two homomorphisms such that Pa = lG, then we say

that G is a retract of H and we call p a retraction and a a coretraction, i.e, a retraction

is a homomorphism which has a right inverse and a coretraction is a homomorphism

which has a left inverse. Therefore every coretraction is injective and every retraction

is surjective. Let G -3 IT be a homomorphism, a homomorphism a' is called a weak

coretraction, or a weak coretraction belonging to cr, if aa' 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 , a'(aat)-' is a

coretraction belonging to cr. A composition of two retractions is also a retraction,

since if acr' = 1 and /?/?I = 1 then (/?'at)(cu/3) = 1.

For a homomorphism G -% W and subsets S C V ( G ' ) , 7' & V ( H ) we denote

a ( , $ ) = { a ( s ) E V ( H ) I s E S ) and a- ' (T) = { t E V ( G ) I a ( t ) E T ) . For a one el-

ement set S = { s ) we will write a ( s ) instead of a ( { 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 11) =

V ( G ) x V ( H ) , in which { ( s , u ) , ( t , w ) ) E E(G x H ) if and only if { s , t ) E E(G) and

{ u , w ) E E ( H ) . The homomorphisms G x H 3 G and G x H 2 H defined by

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

from G to H1 and a2 is a homomorphism from G to H z , then the product of a1

and a2 is the homomorphism G as2 H I x H2 mapping each vertex g of G to the

vertex (cr l (g) , ~ ~ ( 9 ) ) of I f 1 x 112. One can easily check that the product of two ho-

momorphisms is indeed a homomorphism. On the other hand, every homomorphism

CHAPTER 1. IIVTROD UCTIOIV

Figure 1.2: I{3 V3,a C6

G IH1 x Hz is a natural product of homomorphisms (1.1 = p l a and az = pza. In

particular, G -+ H1 and G -+ H2 is equivalent to G --t Ill x H 2 .

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

is the categorical product.

The coproduct or sum of two graphs G and H , denoted G + H , is a disjoint union

of G and H, i.e., the vertex set V(G + H) = (V(G) x (0)) U (V(H) x (1)) and for

{(s, i ) , ( t , j ) ) E E ( G + H) if and only if i = j and {s, t ) E E ( G ) U E(H). The homo-

morphisms G % G + I1 and H -% G + II defined by i l ( s ) = (s, 0) and iz( t) = ( t , 1)

are called natural inclusions. If o1 is a homomorphism from G1 to I1 and a2 is a

homomorphism from Gz to H , then the sum of ol and a 2 is the homomorphism a l + w G1 + Gz --) I1 mapping vertices of G1 by ol and vertices of G2 by ~ 2 . One can

easily check that the sum of two homomorphisms is indeed a homomorphism. On the

other hand, every homomorphism G1 + Gz 5 I1 is a natural sum of homomorphisms

ol = a i l = olGl and oz = a i z = alGz (the restrictions of a to GI and Gz). In

particular, G1 -, H and G2 -t H is equivalent to GI + G2 -+ 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 amalgam) GVgVh H to be the graph obtained from G + H by

identifying the vertices g and h (see Figure 1.2). The homomorphisms G 1: G Vg,h H

CHAPTER 1. INTRODUCTION 7

and H -3 G Vs,h H defined by i l ( s ) = (s, 0) and i2(t) = ( t , 1) are called natural inclu-

sions. If a1 is a homomorphism from GI to H and a 2 is a homon~orphism from G2 to

H such that for gl E V(G1) and g2 E V(G2), al (gl) = a2(g2), then the wedge of cul and a lV91,92a2

a 2 is the homomorphism G1 V,l,g2 Gz ---f H mapping vertices of G1 by al and

vertices of G2 by a2. One can easily check that the wedge of two homomorphisms is

indeed a homomorphism. On the other hand, every homomorphism GI V,,,,, G2 5 H

is a natural wedge of homomorphisms = ail = alGl and a 2 = a i 2 = alGz (the

restrictions of a to G1 and G2).

A core is a graph G, such that every endomorphism from G to G is an automor-

phism.

A graph G is called bipartite if G --+ K2. 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 that no two vertices of A are adjacent as well

as no two vertices of B are adjacent. If V(K2) = {1,2} and a is a homomorphism.

from G to K2, then A = aV1(l) and B = a-'(2). We will say that A and B form

a bipartition 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 2). If

this is the case, wc will write G = {A, B}. Recall that if every vertex of thc set A

is connected to every vertex of the set B, G is called a complete bipartite graph and

it is denoted by since it is determined by the sizes of the sets A and B. We

allow A or B to be empty and hence 0 and I, are also complete bipartite graphs.

A graph G is called disconnected if G = H + kt' where H # 0 and W # 0, and the

subgraphs 11 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 sun1 of its

connected components and this sum is unique up to permutation of summands.

CHAPTER 1. INTRODUCTION

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 Iilk. 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:

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

The positive answer to this question was given by Appel, Haken and Koch [I] after

more than a hundred years of development of gra.ph theory.

Graph coloring deals with the fundamental problem 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 formulated as graph

coloring problems (cf. [44]). 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 polynomial time solvable, since one

can greedily color vertices of a given graph by two colors (cf. [44]). For a connected

graph, one can easily observe that this coloring is unique up to a permutation of 4 II, colors. Hence if G is connected and G -+ K2, G -+ K2 are two 2-colorings of G, then

there is an automorphism K2 -5 K2 such that $ = a$. On the other hand k-coloring

is known to be NP-complete for all k > 2 [17].

The computational complexity of the H-coloring problem was completely charac-

terized by Hell and NeSetfil [30]. In particular, they proved that the H-coloring prob-

lem is polynomial time solvable when H is a bipartite graph and it is NP-complete

otherwise.

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

NP-complete. On the other hand, for all oriented paths and some oriented cycles H,

the H-coloring problem is polynomial time solvable (see [34, 431). According to the

latest results of Hell, Zhu, and Nesetiil [32] and independently of Feder and Vardi [15],

for digraphs H with a property called bounded treewidth duality (for the definition

see [32]), the If-coloring is polynomial time solvable. It is conjectured that the H-

coloring problem is not polynomial otherwise [29]. This conjecture is a counterpart of

the characterization 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 complexity 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 2-coloring the graph). It is also well known that finding connected

components of a given graph can be done in polynomial time [44] and therefore de-

termining whether a given graph is connected is also polynomial time solvable. On

the other hand, determining whether a given graph is a core was shown to be NP-

hard by Hell and NeSetiil [31]. In Chapter 2 we will study the equitable H-coloring

problem and give a complete complexity classification of this problem. The equi-

table II-coloring problem is a natural generalization of the equitable coloring problem

studied in [40, 45, 10, 231. We will prove that the equitable H-coloring problem is

polynomial time solvable whenever Ii is a disjoint union of complete bipartite graphs

and is NP-hard otherwise.

With each loopless graph G, we associate two numbers x and w which are called the

chi-omatic number and the clique number of the graph, respectively. The definitions

are as follows:

s = x(G) = min{n E No I G t I<,}

CHAPTER 1. INTRODUCTION

and

w = w(G) = max{n E No I I<, -+ G).

Computing these numbers is known t o be NP-hard [17]. If ,f3 is a homomorphism

H 3 then G J H implies G 3 and hence x(G) 5 x(H). Therefore x is the same for two equivalent graphs. Similarly, if y is a homomorphism I(,(c) A G,

then G H implies I<,(cl -% H and hence w(G) 5 w(H). Therefore w is the same

for two equivalent graphs.

The numbers x and w also have the following integer linear program formu1a.tions:

w (G) = max Cu xu x(G) = m i n C ~ Y I

CvE1 xu < 1 for each I XI,, y ~ > 1 for each v

xv E { O J ) Y I f {0,1)

In these formulations we take v to range over the vertex set of G and I to range over

all maximal independent sets in G. If we replace the integrality conditions xu E { O , l } ,

YT E { O , l ) by z, > 0, y ~ > 0 respectively, we obtain linear programs with optima

wj(G), the fractional clique number of G, and xj(G), the fractional chromatic number

of G. One can observe that these linear programs are duals of each other and hence

wj(G) = x (G) by the duality theorem of linear programming [I l l . Feasible solutions

{x,) and {wl) of these linear programs are called fractional clique and fractional

coloring respectively. Therefore we have the inequalities:

If G -% H and {wr} is a fractional coloring of H, then one can define a fractional

coloring {uJ) of G by

w1 if J = a-'(I) U J = {

0 otherwise.

Therefore G t H implies wj(G) = x (G) 5 wf ( H ) = xf (H) and we conclude that

w j = xf for equivalent graphs. To compute the fractional chromatic number is NP-

hard. This result was first proved by Grotschel, Lovisz and Schrijver in [20] proving

the fact that if one can optimize over a polytope in polynomial time then (using the

CHAPTER 1. INTROD CICTION 11

ellipsoid method) one can optimize over its antiblocker (for the definition we refer in-

terested 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, Lovcisz and Schrijver in 1201 also proved that one can compute in poly-

nomial time a real number 0, 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 sys-

tem {vg)gEv(c ) of unit vectors in a vector space RT, such that if {i, j } E E(G) then

v,vT % 3 = 0. Every graph has an orthonormal corepresentation in Rn, for instance an

orthonormal basis of Rn. Let u = { u ~ } ~ ~ ~ ( ~ ) range over all orthonormal corepresen-

tations of G and el be the unit vector with first coordinate equal to 1, then

1 O(G) = min max

~ E V ( G ) ( ~ T U ~ ) ~ '

One can observe that G -% I1 implies that U(G) 5 O ( H ) , since if { v ~ ) ~ ~ ~ ( ~ ) is a11

orthonormal corepresentation of N, then {v , (g)}gEV(G) is an orthonormal corepresen-

tation of G. Therefore 0 is also the same for equivalent graphs and one can consider

O(G) < U(H) to he a polynomially computable necessary condition for the existence

of a homomorphism from G to H.

Another polynomially computable necessary condition called a hoax was given by

Feige and Loviisz in [16]. They used a semidefinite relaxation of a quadratic program,

which was obtained from a multi-prover interactive proof system for the graph homo-

morphism problem.

The model 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. JNTROL) UCTIOIV

NP.

The model consists of a probabilistic polynomial-time verifier V and a prover P

who tries to convince V that the input x is in a language L.

L E I P if there exists a verifier V that is always convinced when x E L, but if

x $! L then any prover P has only a small probability of convincing V to the contrary.

The model of multi-prover interactive proofs was introduced by Ben-Or, Gold-

wasser, Kilian, and Wigderson [8].

The model consists of a probabilistic polynomial-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 M I P if there exists a verifier V that is always convinced when x E L, but

if x L then the provers have only a small probability 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 LovAsz in [16] made the follow-

ing two prover interactive proof system.

Let us say that the Verifier is trying to determine if a graph G is homomorphic to

a graph H. The Verifier chooses randomly and independently vertices s and t of G

and sends them, respectively to the provers Pl and P2. Then Pl replies with a vertex

u from 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) If s = t then u = w

2) If 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 sernidefinite) relaxation can be solved in polynomial time using the

ellipsoid method. In Chapter 3 we will define a pseudo-homomorphism in terms of

0 and a horn-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 -t W or H t W.

Hedetnierni's conjecture is equivalent to the assertion that all complete graphs are

multiplicative. One can easily prove that K 1 arid I<:! are multiplicative and the proof

of the multiplicativity of Ir13 was accomplished by El-Zahar and Sauer [14]. This

conjecture was extensively studied by Burr, Erdos and Lovisz [9], Hell [28], Turzik

[51], Duffus, Sands arid Woodrow [12], Welzl 1551 and others and they obtained some

partial results. It is not even known whether the function

p(n) = min{m I G f , I<, and H f , K, but G x H t I(,)

tends to infinity or not. I-Iedetniemi's conjecture is equivalent to the assertion that for

all positive integers n , p(n) = n+1. Poljak and Rod1 [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 sum 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 all Boolean algebras associated with classes which contain complete graphs

CNAPTER 1. INTRODUCTION

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 with 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 sufficient condition for an element of a Heyting algebra to have

a finite factorization into meet irreducible elements.

Chapter 2

Equitable Graph Homomorphism

2.1 Introduction

In this chapter we consider the equitable coloring problem and a natural generalization

of it , which we call the equitable H-coloring problem for simple graphs. Equitable col-

orings 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 Vl, . . . , Vk such that each V, is an independent set and the

classes V, , i = 1, . . . , k have almost the same size, i.e., 1 lV, 1 - I 1 1 5 1 for all i and

j . Recall that the graph G is homomorphic to the graph H ( G -+ I f ) , if there is

a mapping a : V(G) -+ V ( U ) such that (s, t ) E E ( G ) implies ( a ( s ) , cr(t)) E E(f1) .

Such mapping a is a homomorphism from G to H. We say that a homomorphism a

is equitable, if the sets a- '(h) = {g E V(G) I ~ ( g ) = h ) have almost the same size for

all h E V ( H ) , i.e., \la- '(h)l - la-'(hl)ll 5 1 for all pairs h , h' E V(I1). If there is an

equitable homomorphism cr from G to H, we write G --+ H or G % H. An equitable

homomorphism G 2.t H is also called an equitable H-coloring of G. Given a graph H ,

by the equitable H-coloring problem we mean the following decision problem:

INSTANCE: A graph G

QUESTION: Is G 2.t H?

CHAPTER 2. EQUITABLE GRAPH HO,LlO~VlORPHIS,2I

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 N-coloring

problem is polynomial time solvable whenever H is either not connected (for essen-

tially 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.

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

II is disconnected and it is polynomial time solvable if I1 is a complete bipartit,e

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 problems are NP-hard in all other cases. In particular we

give a Turing reduction from the connected 11-coloring problem to the connected

equitable 11-coloring problem 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 I1 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, i.e., the equitable H-coloring problem when N is not a disjoint

union of complete bipartite graphs, we will show a polynomial reduction to H from

the equitable 11'-coloring problem where H' is a connected component of 11 which is

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 will make some simple observations about equitable homomor-

phisms. First of all, let us notice that if the number of vertices of G is n and the

number of vertices of W is m, then n can be uniquely written as n = km + r , where

O < r < m a n d k = LEJ.

Lemma 1 G 3 H is an equitable homomorphism if and only if for all h E V ( H ) :

Proof: If for some h E V ( H ) we had Icr-'(h)\ 5 - 1 then

which is a contradiction. Similarly if for some h E V ( H ) we had la- '(h)l 2 121 + 1

then

which is a contradiction. 0

Corollary: G 3 N is an equitable homomorphism if and only if for r vertices h

of H we have la-'(h)l = k + 1, and for n - r vertices h of H we have (a - ' ( h ) l = k.

For a given graph H we say that w : V ( H ) --+ N is a weight function on N, i.e.,

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(H) w ( h ) . We say that cr is a w-weighted

equitable homomorphism to N, written as G 2, H , if for all h E V ( H )

CHAPTER 2. E Q UITABLE GRAPH _I-1OL110,L1 0RPH1SM 1 S

We also write G --t, H if there is a zu-weighted equitable homomorphism from G to H.

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 ) I h - h'). We say that two vertices h , h' E V(H) are similar,

denoted h h', whenever NH(h) = NH(hl). This relation is an equivalence relation

on the vertices of the graph H and partitions them into disjoint equivalence classes.

Let HT 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 N h' in H. One can easily check

that HT is a well defined loopless graph and that there is a natural homomorphism T

from H t o HT 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 IV(G)( = n and IV(H)I = m. Let w be

a weight function on HT defined by w(c) = Icl for every equivalence class c. Then

G -.., H if and only if G --tW HT.

Proof: Let o be an equitable homomorphism from G to H and let T be the natural

homomorphism from H t o H T . Then for all h E V(H)

Therefore summing for all h E c, we obtain

P On the other hand, let us assume that there is a homomorphism G --tw HT. This

implies, that we can (arbitrarily) partition vertices from P-'(c) into Icl disjoint sets

A;, . . . , A i l such that 121 5 /A:\ < for all i = 1,. . . , lcl. If we denote the vertices

of the class c by h i , . . . , hi,,, then mapping all vertices of A: to the vertex h: defines

an equitable homomorphism from G to H .

On the other hand, for a weight function w on H we can define a graph H , with

the vertex set V(H,) = UhtV(H){h} x { I , . . . , ~ ( h ) } for which two vertices (h, i ) and

Figure 2.1: (C5), 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 all h E V(H) , then we

will simply write w = k. (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 natural homomorphism T 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 g ~ a p h s . Then

(a) G -, H if and only if G ?-t II,, moreover if G 4 H,, then G Ga, H where 7 ~ . is

the natural homomorphism.

(b) G ?-t H if and only if there is a bimorphism from G to some H, such that

121 < w ( h ) 5 121 for all h E V ( H ) .

In this section, we will observe that the connected equitable Il-coloring problem is

trivially polynomial time solvable if H is not connected. Then we will show that if

IT is a complete bipartite graph, then the connected equitable H-coloring problem is

polynomial time solvable. Finally we will prove that the equitable N-coloring problem

CHAPTER 2. EQUITABLE GRAPH HOI~~OMORPIIISIZI

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 components, i.e., H = H1 + . . . + H,, q 2 2. Then any

connected graph G must be mapped to one of Hi, say HI . Therefore, if G -% H, then

la-l(h)l = 0 for all h E V(H;) for 2 i 5 q. Since a is an equitable homomorphism,

where n = IV(G)I and m = IV(H)I. We conclude that = 0 and 5 1. There-

fore no two vertices can be mapped to the same vertex, and hence the homomorphism

is injective. This implies that G is a subgraph of HI 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)\ vertices to be equitably homomorphic to

H). We conclude that in the case that H is not connected, the connected equitable

H-coloring problem is trivially polynomial time solvable.

Next, let us consider the connected equitable H-coloring problem and assume

that N is a complete bipartite graph I<,,t. Since = K2, our problem is to find w-

equitable homomorphism from G to K2 with weight function w(1) = s and 4 2 ) = t .

Let G' be a connected graph. One can decide in linear time whether G is bipartite

and find a 2-coloring if there is one. Therefore, if G is bipartite, then we can get the

unique bipartition { A , B ) of G in linear time. Hence if G -, K 2 , then all vertices of

A have to be mapped to the vertex 1 and all vertices of B have to be mapped to the

vertex 2 or vice versa. Therefore G -A if and only if either

which can be decided in polynomial time. Therefore we have just proved the following

lemma.

C H A P T E R 2. EQ IIITABLE GRAPH EIOM ORIORPHISM 2 1

Lemma 4 If H is a complete bipartite graph, then the connected equitable H-coloring

problem is in P.

Now we can prove the main result of this section.

Theorem 5 If N is a disjoint union of complete bipartite graphs then the equitable

H-coloring problem is in P.

Proof: We will modify the dynamic programming approach to the partition problem

from Garey and Johnson [17] p. 90. Assume the connected components of H are

I<ul,vl, . . . , Kur,Z/r and the connected components of G are G I , . . . , G k . One can observe

that the graph Eir consists of r disjoint copies of I<;! and we will denote the two vertices

of the j - th copy by aj and b3. By Lemma 2, G ?.t H if and only if G ?.t, H T , where

w is defined by w ( a j ) = uj and w ( P ) = v j . Let a be a homomorphism from G' to

If' and let s i = la-'(aj)l denote the number of vertices mapped to aj by a and

t i = la-'(@)( denote the number of vertices mapped to bi by a. According to our

definition, a is a w-equitable homomorphism if and only if for all q = 1,. . . , r the

following inequalities are satisfied:

and

By Lemma 2 we conclude that G ?.t 11 if and only if there is a homomorphism o

from G to I I r , such that these conditions are satisfied. For integers 1 < m < k, 0 5 jl 5 ul[&$)l ,..., 0 < j, z~,[Hl, o 5 1, < v l [ M l , . . . , 0 5 1, < v , [HI, let t ( m , j l , . . . , j,, 1 1 , . . . , I , ) denote the truth value of the statement: "there

is a homomorphism a from G I + . . . + G, to •’IT, such that for all q = 1 , . . . , r ,

sz = jq and t9, = I,". One can observe that G ?.t if if and only if there is an entry

. > u [U] t ( k , j l , . . . , j,, l l , . . . , I , ) which has value T and for all q = 1, . . . , r , 1, - , Iv(H)i

and I , 2 v, [HI. We will show a recursive procedure that can be used to compute

t ( k , j l , . . . , j,, 1 1 , . . . , I , ) . First of all, one can observe that t ( l , j l , . . . , j,, 11 , . . . ,1 , ) = T

C H A P T E R 2. EQUITABLE GRAPH IiOiZIOMORPIiISIli 22

if and only if there is a q E (1 , . . . , r) , such that j, = (A l l , I, = IB1 I and j, = 1, = 0

for p # q, or j, = lB1l, 1, = IAl( and j, = I, = 0 for p f q, which encodes that

GI is mapped to the qth copy of and either the vertices of AI are mapped to the

vertex a, and the vertices of B1 are napped to the vertex bq or the vertices of Al are

mapped to the vertex bq and the vertices of B1 are mapped to the vertex a,. Having

computed all t (m, jl, . . . , j,, E l , . . . ,1,) with given m , t (m + 1, jl, . . . , j,, 11,. . . ,1,) = T

if and only if there is a q E (1 , . . . , r), such that either

or

t ( m , j l 1 - ,.iq - IBrn+l\, - - , . iT , 11, + . ., lq - \ A ~ + I \ , . , j T ) = T ,

which encodes that Grn+l is mapped to the qth copy of K2. The number of tuples

( k , jl, . . . , j,, 11,. . . , I,) is bounded by IV(G)I2'+' and therefore one can decide whether

G -.-+ H in polynomial time. 0

Both the equitable H-coloring problem and the connected equitable H-coloring prob-

lem are clearly in NP. In this section, we will focus our attention on finding reductions

from known NP-hard problems. Since the connected equitable 11-coloring problem is

a restriction of the equitable H-coloring problem, 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 1I-coloring problem.

Theorem 6 There is a Turing reduction from the connected H-coloring problem to

the connected equitable ti-coloring problem.

CHAPTER 2. EQ CUTABLE GRAPH HORIOLVORPHISM 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 g' E V ( G ) and let IV(G)J = n. We will show that the following two conditions

are equivalent:

(a) G -+ H

(b) G Vg!,h ljrw v H for at least one h E V ( H ) and a t least one weight function tu on

H such that w(1) + . . . + w(m) = mn + 1

The implication (b) + (a) follows from the fact that G -+ GVglvh Hw for all h E V(H)

and all weight functions w on H. It remains to prove the implication (a) + (b). Let

a be a homomorphism from G to H. If we denote x, = la-'(;) 1 for i = 1, . . . , m, then

X I + ...+ x, = n. Thereforemn = ( m + l ) n - n = ( ( m + l ) - x l ) + ...+(( m+1)-x,).

Without loss of generality we may assume that a(gl) = 1. It follows, that for w

such that w(1) = (m + 1) - x1 + 1 and w ( h ) = (m + 1) - xh for h # 1 we have ff"g',1=

G V , I , ~ Hw H where T is the natural homomorphism from Hw to H. (We added

1 to the first element, since g' and 1 are identified by the wedge.) Since G and H

are connected, G V,ttl Hw is also connected. To see that our reduction is polynomial,

one can observe that the graph G V,,,h Hw has (m + l ) n nodes which is 0 ( n ) for the

fixed graph H. The number of graphs G V , I , ~ Hw is m ( z ) , since m is the number of

ways we can choose h E V(H) and (2) is the number of solutions to the equation

w1 + . . . + W, = mn + 1 in positive integers. For a fixed graph H , this is O(nm-')

and we conclude that our reduction is indeed polynomial.

Hell and Nesetiil [30] have shown that for nonbipartite 11, M-coloring is NP-

corriplete. For connected II, G --+ H if and only if for all connected components G'

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

Ii-coloring problem is NP-hard.

Corollary 8 If H is a connected nonbipartite graph, then the equitable H-colori~zg

problem is NP-hard.

C H A P l E R 2. E Q U I T A B L E G R A P H HOL!f OX1 ORPHISAf 24

The following problem, known as the balarzced cornplete bipartite subgraph problern

is NP-complete, see [IT]:

INSTANCE: Bipartite graph G and a positive integer m 5 fJV(G)I.

QUESTION: Does G have a subgraph isomorphic to I<,,,?

We will actually need the following restriction of the balanced complete bipartite

subgraph problem.

The restricted balanced complete bipartite subgraph problem:

INSTANCE: Bipartite graph G, its bipartition {A, B) and a positive integer m such

that f max{(AI, lBl) < m < fIV(G)I.

QUESTION: Does G have a subgraph isomorphic to K,,,?

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

Proof: Membership in N P is trivial. We will give a polynomial-time reduction from

the balanced complete bipartite suhgraph problem. Let G and m be an instance of

the balanced complete bipartite subgraph problem. It is well known that we 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, we create 2n new vertices. Let A' contain all of the vertices of A and n of the

new vertices and let B' contains all 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 Ir',,, if and only if G' has

a subgraph isomorphic to I<,+,,,+,. Moreover $IV(Gt)I = $IV(G)( + n 2 m $ n > n = q + > ; + f m a x { I ~ I , I B I ) = ! j m a x { n + I ~ l , n $ I ~ I } = f rnax(IA'1, IB'I).

Let us recall that P4 denotes the path of length three on four vertices, i.e., 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 complete bipartite graph. The

following proposition will serve as a basis for our inductive proof.

Proposition 10 The connected equitable 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), {2,3), (3,411. IVe will give a polynomial-time reduction from the re-

stricted balanced complete bipartite subgraph problem. Let bipartite graph G, its hi-

partition {A, B } , and a positive integer m such that : max{IAl, IBl} < m j :IV(G)l,

he 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 - JV(G)( new vertices. Let A' contains all vertices of A and 2m - IAJ

of the new vertices and let B' contains all vertices of B and the remaining 2m - IBI

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 f b in G. The restrictive condition on m implies that we have

added a positive number of new vertices to both A and B and therefore our graph G'

is connected. We now claim that G has a subgraph isomorphic to I<,,, if and only

if G' --+ P4. Let us assume that G has a subgraph isomorphic to I<,,,. This implies

that there are vertices al , . . . ,a, E A and vertices bl, . . . , b, E B such that for all

i = 1 , . . . , m and j = 1,. . . , m, a, N b, in G. Define a mapping A' u B' --+ {1,2,3,4}

so that the m vertices a, are mapped to 1 and the remaining m vertices from A' are

mapped to 3. Similarly, the vertices b, E B' are mapped to 4 and the rest of the

vertices from B' are mapped to 2. One can easily observe that this mapping is an

equitable homomorphism from G' to P4.

For the converse, let us assume that a is an equitable homomorphism from G' to P4.

It follows that la-'(i)l = m for each i = 1,2,3,4. Without loss of generality we may

assume that a(A1) = {1,3) and a(B1) = {2,4). Let a-'(1) = {a l , . . .,a,) C A' and

cr-'(4) = {bl, . . . , b,) C B'. Since 1 + 4 in P4, for a11 i = 1, . . . , m and j = 1 , . . . , m,

we must have a, + bJ in GI. Therefore none of a l , . . . ,a,, bl, . . . , bm can be a new

vertex, as new vertices from A' or B' are connected to all of the vertices from the

other set. This implies that a l , . . . , a m E A and bl,. . . , b, E B, and that a, b, in G

for all i = 1 , . . . , m and j = 1 , . . . , m. We conclude that G has a subgraph isomorphic

to C7

Corollary 11 The equitable P4-coloring problem is NP-complete.

CHAPTER 2. EQUITABLE GRAPH IiOiZf Oilf ORPHISlLf

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

Lemma 12 There is a Turing reduction from the connected equitable H-coloring

problem to the restriction of the connected equitable H-coloring problem to connected

instances G such that IV(H)I divides IV(G)I.

ProoJ Let connected graph G be an instance of the connected equitable H-coloring

problem, IV(G)( = n and IV(H)I = m. Let n = k m + r , where 0 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 m - r vertices h of H : la- ' (h) ( = k . For every subset S C V ( G )

of cardinality m - r , define w s to be a weight function on G given by

2 i f g ~ S w s ( d =

1 otherwise.

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

( k + 1 ) m . We now claim that G -.., H if and only if for a t least one S, G,, --+ H.

If G 5 H , then for m - r vertices h of H we have la - ' (h) ( = k. Let us choose

one g E a- ' (h ) for each such h to form the set S of cardinality m - r . One can

observe that if we define P((g , i)) = a ( g ) from G,, to H, then for each h E V ( H ) , P (P- ' (h )J = k + 1 and hence G,, --+ 11.

On the other hand, let us assume that for some S, we have G,, 3 H. Let G 5 G,,

be the coretraction defined by a ' ( g ) = ( g , l ) for all g E V ( G ) . Then G % H .

Since there are (zT) = O ( n m ) subsets of V ( G ) of cardinality r and m is fixed, our

reduction is polynomial.

For a technical reason we also need the following four lemmas. 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 n from H, to H is a retraction.

CEIL-1PTER 2. EQUITABLE GRAPH IIOXIOI~IORPHISLLI 2 7

Proof: To see this, one can define a coretraction d from H to H , by mapping each

vertex h E V ( H ) to any one of the vertices ( h , I ) , ( h , 2 ) , . . . , ( h , w ( h ) ) E V ( H w ) .

Lemma 14 Let n be a positive integer, H be a graph on m vertices and w be a weight

function on H such that w ( h ) 2 mn for all h E V ( h ) . Then any homomorphism

H , 3 H such that l4-'(h)l 5 ( m + 1)n for all h E V ( H ) is a retraction. More-

over, there exists a weak coretraction n1 belonging to $ such that n1 is a coretraction

belonging to the natural homomorphism n from H, to H .

Proof: Let us consider sets B h = { ( h , l ) , . . . , ( h , w ( h ) ) ) V (H, ) and Ah =

{ $ ( ( h , 1)), . . . , $ ( ( h , w ( h ) ) ) ) V ( H ) for all h E V ( H ) . We claim that the sets Ah,

h E V ( H ) have a system of distinct representatives, i.e., vertices a l , . . . , a, E V ( H )

such that each a; E A;. According to Hall's theorem [24], given a collection of nz sets,

there is a system of distinct representatives if and only if, for all 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 m sets Ah , , . . . , A h s such that

We may assume that s > 1, since all sets Ah are nonempty. This means that all

vertices from Uf=, Bh, are mapped to at most s - 1 vertices from H . The sets Bh,

have cardinality at least mn each and they are disjoint. Therefore at least 2 vertices

of If, are mapped to the same vertex, i.e., there is a vertex h E V ( H ) with

smn mn mn I P ( h ) I 2 - = nzn + - > m n + - = m n + n ,

s - 1 s - 1 m

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 , i h ) ) for some

i h E (1,. . . , w ( h ) } ; we now define n l ( h ) = ( h , i h ) . One can observe that n1 is a homo-

morphism, since h - h1 implies ( h , i h ) - (h l , ih l ) . Moreover ( n n l ) ( h ) = n ( h , i h ) = h

and hence n1 is a coretraction belonging to n . Finally, since ($.irl)(h) = ah and since

the uh, h E V ( H ) are distinct, we conclude that $nl is an injective endomorphism

CHAPTER 2. EQUITABLE GRAPH MO~ZIOMORPNZSiLI 2 8

and hence an automorphism. Hence 7r' is a weak coretraction belonging to $ and we

conclude that $ is a retraction.

4 Lemma 15 Let w be a weight function on H , H , -+ H a retraction, H, -5 H the

natural homomorphism and for h E V ( h ) , Bh = { ( h , 1 ) , . . . , ( h , ~ ( h ) ) ) C V(H,) . dJ1 Then for every coretraction H -+ H , belonging to 6 and for every coretraction H 1:

H, belonging to 7r:

(a) n l ( h ) = ( h , i ) for every h E V ( H ) and some i E (1,. . . , w ( h ) ) ,

(b) N H ~ ( ~ * ) = N H ~ ( ( T ' T ) ( ~ * ) ) for every h* E V ( & ) ,

(c) N H ( ~ ) C_ N~((qhr 'n$ ' ) (h ) ) for every h E V ( H ) , and

(d) N H ( h ) C N H ( $ ( h * ) ) for every h E V ( H ) and every h* E B( ,+I ) (~) .

Pro0 f:

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

7 r d = i d H , we have h = ( 7 r d ) ( h ) = 7r(h1, i ) = h'. We conclude that 7r1(h) = ( h , i ) . (b): If h* = ( h , j ) for some j E ( 1 , . . . , w ( h ) ) , then by ( a ) 7r1(h) = ( h , i ) . There-

fore (r17i)(h*) = d ( h ) = ( h , i ) . From the definition of the graph H, it follows that

NHw ( ( h , i ) ) = Nfrw ( ( h , j ) ) and we conclude that NH, ( h * ) = NH, ((.ir1x)(h*)).

(c): Since $' is a homomorphism, u E N H ( h ) implies $ ( u ) E N ~ j ( $ ' ( h ) ) = N ~ ( ( . n ; ' n ) ( $ ' ( h ) ) )

by (b). Therefore u = $($ ' (u) ) E NH(4((n17r$')(h))) , since $ is a homomorphism and

$4' = idEi.

(d): Let # ( h ) = ( h 1 , i ) for some i E ( 1 , . . . , w ( h l ) ) . Then ( r $ ' ) ( h ) = r ( ( h l , i ) ) = h'

and if h* E B ( n 4 ~ ) ( h ) = B h l , then h* = (h ' , j ) for some j E ( 1 , . . . , w ( h l ) ) . Let 7r' be a

coretraction belonging to 7r defined by

{ 1;:;; = h* if h = hi = i ( h * ) 7r1(h) =

otherwise.

By part (c) we have N H ( ~ ) C_ NH(($r l r$ ' ) (h ) ) = NH (($7r'7r)((h1, i ) ) ) = N f r ( ( $ d ) ( h l ) ) =

NH($((hl , . i )>) = N H ( $ ( ~ * ) ) .

CHAPTER 2. EQUITABLE GRAPH HO~LIOMORPHISXI 29

Lemma 16 Let G and H be two graphs, w a weight function on H such that w ( h ) 5 V G t H w P for all h E V ( H ) , H, 2, 11 the natural homomorphism, and G+ H, c H

an equitable homomorphism. Let 4 = / ? I & , and for some coretraction T' belonging

to T , d-ir' is an automorphism on H . Then there exists an equitable homomorphism

G + H, 4 EI such that ylHW = T .

Proof: Let us denote Bh = { ( h , 1 ) , . . . , ( h , w ( h ) ) ) C_ V(H,) for h E V ( H ) and

a = ( 4 ~ ' ) - I be the inverse of the automorphism 4n t , i.e., $ d a = idll . Thus 4 is a

retraction and 4' = ~ ' a is a coretraction belonging to 4 . We will prove the lemma by

induction on the number q = qp = Ch,V(H) I{h* E B,(h) / $ (h*) # h ) ] .

For q = 0, all vertices from each Bcr(h) are mapped to the vertex h by 4 and hence

a$ = n. Since a is an automorphism, it follows that for y = a,B we have G+ H, H

and yIHw = T .

Assume that q > 0. Let h* E Ba(h) be such that 4 ( h * ) # h. Since p is an equitable

homomorphism,

Since h* E Bu(h) and h* &I P-'(h), there is a vertex w E /3-'(h) such that w # Bn(h).

Note that w can be a vertex of G. Define a mapping P1 from G+ H, to H by switching

the images of w and h* in P, i.e., for u E V ( G + H,) define

if u = h*

if u = w

otherwise.

Obviously, since ,f? is equitable, P1 is also equitable and qp, < yo. We will show that

PI is a homomorphism. Since B(,4,)(h) = R(,,I,)(~) = Ba(h) and h* E Ba(h), it follows

from part (d) of Lemma 15 that N H ( h ) C N H ( 4 ( h * ) ) . 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 Pl (u ) = P(u) P(v) = Pl(v), since /? is a

homomorphism.

Case 2: { u , v} n { w , h* ) = { w ) . We may assume that w = v and let us recall that

CHAPTER 2. EQ UITABLE GRAPII IIOhl OXIORPZIISAI 3 0

p(w) = h , since to E P-'(h). Hence Pl(u) = P(u) - P ( v ) = P(w) = h and there-

fore /%(,) € N H ( ~ ) c N H ( $ ( ~ * ) ) = NH(,&(w)) = N H ( @ ~ ( v ) ) . We conclude that

A(.) P1(v).

Case 3: { u , v ) n {w, h*) = { h * ) . We may assume that v = h* E V(H,) and since

u v, also u E V(H,) . Therefore we have u - h* in H, and hence u E NH,(hS) =

NH,((7r1r)(h*)) by part (b) of Lemma 15. Since ogi is a homomorphism, this im-

plies that (crgi)(u) E NH((agint7r)(h*)) = N H ( n ( h * ) ) , since a = (gin1)-'. From our

assumption, h* E Ba(h) and hence ~ ( h * ) = a ( h ) . Hence ( a $ ) ( u ) E N H ( a ( h ) ) , and

applying homomorphism a-' = gin' we conclude that $ ( u ) E N H ( h ) . Therefore

P l ( 4 = P(u) = $ ( u ) h = Pl(h*) = P1(v). Case 4: { u , v) n { w , h*) = {w, h*) . We will show that this case cannot occur, oth-

erwise u v implies that w h*. Therefore h = P(w) - P(h*) = 4(h*) and hence

gi(h*) € N H ( h ) 2 N H ( 4 ( h * ) ) , which is a contradiction, since H is bipartite and hence

loopless.

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.

Proof: 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 I1 be V ( H ) = (1 , . . . , m ) and we may assume that the first four vertices

form an induced P4. We may also assume m > 4, since if m = 4 then H = P4. 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)( = n = 4k > 4. Let us define the weight

function w on H by i f j 5 4

4.i) = { + k otherwise.

CHAPTER 2. EQUITABLE GRAPII HOhfOMORPHIShl

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

Let H, -5 H be the natural homomorphism. One implication of our claim is

obvious, since if we assume that G 5 HI, then G + H, H .

Note that G -j- H, has n + 4nm + (m - 4)(nm + k) = (nm + k)m vertices, and

hence = n m + k.

If we assume that G P4 and Hw 5 H is the natural homomorphism, then Y+" G + H, H .

P On the other hand, let us assume that G + H, -A H . We will show that the

conditions of Lemma 16 are satisfied. Since G + N, has n + 4mn + (m - 4)(mn + k) = V G + H w m(mn + k) vertices, = mn + k. Since ,B is an equitable homomorphism,

exactly mn + k vertices of G + H, must be mapped to the same vertex of H. Let

$J = PIG and 4 = P(Hw. Thus I$-l(h)l = mn+k = mn+; 5 mn+n for all h E V(H).

From Lemma 14 it follows that there is a coretraction T' belonging to T such that $n' V G+Hw is an automorphism. For the weight function u, we have w(h) 5 mn + k = w.

Therefore Lemma 16 implies that there is an equitable homomorphism G + Hw 5 H

such that ylHw = n. Since r - l ( h ) = mn + k for h # {1,2,3,4) and n-'(h) = mn

for h E {1,2,3,4}, no vertex from 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 I(ylG')-'(h)l = k and we

conclude that y JG is an equitable homomorphism from G to P4.

To finish our reduction, let us fix a vertex g E V(G). For every vertex h* E V(lI,),

let Wh. be the graph obtained from G+ 11, by connecting the vertex g with the vertex

h* with an edge. One can easily observe that all Wh* are connected. We claim that

G --, HI if and only if for at least one of h*, Wh* -A H.

If G I]' , then for any h E NH(y(g)), I/V(h,l) u H, if we map vertices of G by y and 6

the vertices of H, by the natural homomorphism. On the other hand, if Wh* --t N

and z denotes the obvious inclusion of G + Hw in Mih*, then G + Hw 2 H and hence

G-, HI.

CHAPTER 2. EQ UITABLE GRAPH IIOMOMORPtllS,2f

Finally, we will prove the following theorem which concludes the complete char-

acterization of the complexity of the equitable H-coloring problem.

Theorem 18 If H is not a disjoint union of complete bipartite graphs, then the

equitable H-coloring problem is NP-hard.

Proof: Let H have connected components HI,. . . , Hk. Without loss of generality we

may assume that H1 is not a complete bipartite graph. We will give a polynomial re-

duction from the equitable HI-coloring problem to the equitable H-coloring problem.

Let G be an instance of the equitable HI-coloring problem. Let us denote n to be

the number of vertices of G and m to be the number of vertices of HI. We may also

assume that m divides n by Lemma 12. Let w = be the constant weight function.

We will prove that G -.A Hl if and only if G + (Hz), + . . . + (Hk), - H . The implica-

tion + is obvious. Let us assume that G + ( H 2 ) , + . . . + (Hk), -+ HI + H z + . . . + Hk.

Clearly no two graphs are mapped to the same graph. Therefore there is a permu-

tation a on k elements such that the i th 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 product of

disjoint cycles. Let 1 be in the cycle (1, i l , . . . , i,), i.e., G - Hi,, (Hi,), -+ Hi,, . . ., (Hir-,), Hir, (Hi?), -+ HI. By Lemma 3, from (Hi,), -.A Hi,,, we get a bimor-

phism (Hi,), -+ (Hi,+,), and their composition gives us a bimorphism G -+ (HI),.

By Lemma 3 we conclude that for some (I. G G, HI and since w is a constant,

G Z HI for the natural homomorphism n. 0

2.6 Conclusion

In this section we would like to clarify the ideas behind our NP-hardness proofs.

Recall that a graph G is called a core if it is not homomorphic to any of its proper

subgraphs, or equivalently any homomorphism G -+ 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 IIOAI OL~IORPHISAI 3 3

Hell and Negetfil 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, -+ H, there is an automorphism H 3 H

on H such that cud, = $. Therefore there is an automorphism cy such that the following

diagram is commutative. d,

Hw - H

We will show later that every core is an apparent core and we will also provide

examples of apparent cores which are not cores. The motivation of the definition of

apparent cores follows from the proof of the following proposition.

Proposition 19 If H is a connected apparent core which contains P4 as an induced

subgraph, then the equitable H-coloring problem is NP-complete.

Proof: Let the vertices of H be V(H) = (1 , . . . , m ) and we may assume that the

first four vertices form an induced P4. We also may assume m > 4, since if m = 4

then I j = P4. We will give a polynomial reduction frorn the connected equitable P4-

coloring problem. Let a connected graph G be an instance of the connected equitable

Pd-coloring problem. By Lemma 12 we may assume that IV(G')I = n = 4 k > 4. Let

us define the weight function w on H by

We will prove that G + If, -A H if and only if G P4. Note that G + 11, has

n + 4nrn + ( m - 4)(nm + k ) = (nm + k)m vertices, and hence = nm + 6 . IV(H)I

If we assume that G P4 and H, 5 I1 is the natural homomorphism, then

0 On the other hand, let us assume that G + H, -A H. Let d, = PIH, be the

restriction of /3 to H,. Since /3 is an equitable homomorphism, exactly nm+ k vertices

CHAPTER 2. EQUITABLE GRAPH HOMOMORPHIShI 34

must be mapped to the same vertex. Therefore \$-l(h)l = nm+k = nm+: 5 mn+n. From Lemma 14 we conclude that 4 is a retraction. Since H is an apparent core, there

is an automorphism cr such that the following diagram is commutative,

a0 i.e., cr4 = a. Let us consider the homomorphism G + Hw --+ H. Since cr is an au-

tomorphism, a/? is also an equitable homomorphism. Moreover, a(/?IHw) = cr4 = a

and hence vertices from (h , 1), . . . , (h, w(h)) are mapped to the vertex h. There-

fore a-'(h) = mn + k for h $ {1,2,3,4) and a-'(h) = mn for h € {1,2,3,4}.

This implies that no vertex from 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(cr(,BIG))-'(h)l and we

conclude that a(,B IG) is an equitable homomorphism from G to P4. 0

Therefore the NP-hardness proof 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.

dJ Lemma 20 Let w be a weight function on a graph H and let H, + H be a retraction.

Then if $ # a, there is a coretraction H d Hw which belongs to a but does not belong

to 0.

Proof: Assume that 4 # T , i.e., there is a vertex (h', i ) E V(Hw) such that 4((h1, i ) ) # h'. Let us consider a mapping a' such that

(h', i ) if h = h' 7rf(h) =

( h , 1) otherwise.

One can observe that a' is a coretraction belonging to a but it does not belong to 4, since (4a)(ht) = $((hf, i ) ) # h'.

C H A P T E R 2. EQUITABLE GRAPH NOAlOMORPHISnif 35

Theorem 21 A graph H is an apparent core if and only if for all distinct vertices

h , h' E V ( H ) , N H ( ~ ) N H ( ~ ' ) .

Proof: Let the vertex set of H be V ( H ) = (1,. . . , m ) .

First, let us assume that there are two distinct vertices h , h,' E V ( H ) such that

N H ( h ) N H ( h f ) . Let w be a weight function on H defined by

2 i f j = h w ( j > =

1 otherwise.

Define # ( ( j , 1)) = j for j = 1,. . . , m and $ ( ( h , 2)) = h'. For every endomorphism

H -5 H such that a+ = T we have a ( h ) = a ( d ( ( h , 1 ) ) ) = ~ ( ( h , 1 ) ) = h = ~ ( ( h , 2 ) ) =

a ( $ ( ( h , 2 ) ) ) = a ( h t ) and hence cu cannot be an automorphism.

On the other hand, let us assume that for all distinct vertices h , h' E V ( N ) ,

NH(h) g NH(hf) .

First we will prove the claim for II, = n. Let $ be a retraction from H, to E l and

#' : H t H , be a coretraction belonging to $. Let us define a = n#'. Obviously

H -% H is an endomorphism. 7'0 show that it is an automorphism, it is enough to

prove that it is injective. Let x' be any coretraction belonging to T . By Lernma 15,

Nr f (h ) E N H ( ( q ! m f ~ $ ' ) ( h ) ) for every h E V ( H ) and we conclude that h = ( $ T ' T $ ' ) ( ~ )

for every h E V ( H ) which means that $T'T& = i d H . Therefore 4 ~ ' is a left inverse

to a = T$' and we conclude that a is an injective endomorphism and hence an auto-

morphism. It remains to prove that a 4 = T . Since a is an automorphism, every left

inverse has to also be a right inverse and hence a $ ~ ' = idH. This implies that a4 is a

retraction and every coretraction T' belonging to n belongs to a$. We conclude that

a4 = T by Lemma 20.

To finish the proof, let us assume that $ and II, are two retractions from 11, to

fi. Let a and /? be two automorphisms on H such that a $ = T and /?$ = T . Then

a 4 = ,B$ and hence $ = (,B-'a)$ for the automorphism /?-'a on 11.

CHAPTER 2. EQUITABLE GRAPH MOMOMORPNISJI

We say that an endomorphism is simple, if it differs from the identity automor-

phism on a t most one vertex. If a graph H contains two vertices h, h' such that 4 N H ( h ) C N H ( h l ) , then one can define a simple endomorphism H -+ H by

h' if j = h Ki) =

j otherwise.

This simple endomorphism is clearly not an automorphism. Therefore apparent cores

are exactly those graphs which do not have any simple endomorphism which is not

an automorphism. This also implies that every core must be an apparent core. Note 4 that every simple endomorphism H -+ H induces a retraction from H onto induced

subgraph 4 ( H ) which has IV(H)l - 1 vertices. Therefore one can find an induced

subgraph which is an apparent core in a polynomial time by applying simple retrac-

tions at most (V(H)I times, while there is a pair h , h' of distinct vertices such that

N ( h ) C N ( h l ) . 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 (c.f. (311) can be recognized in polynomial time. Moreover

it also follows that for a given graph one can find a retract which is an apparent core

in polynomial 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

I1 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 IV(H)I and this may be a target for further improvement.

Chapter 3

Relaxat ions of Graph

Homomorphism

3.1 Introduction

Determining whether G t H for input graphs G and N is known to be an NP-

complete problem (cf. [30]). This problem is a generalization of a large number of

well-studied problems. For example, determining if a graph G is k-colorable is equiv-

alent to setting H to be a complete graph on k vertices and determining if G -+ H.

Determining if a graph H has a k-clique is equivalent to setting G to a complete graph

on k vertices and determining if G -+ 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 algo-

rithms for NP hard problems (see [18,38]). We use semidefinite programming to study

the graph homomorphism problem. We define the notions pseudo-homomorphism and

fractional homomorphism which are, respectively, semidefinite and linear relaxations

of a certain integer program corresponding to the graph homomorphism problem.

In fact, the pseudo-homomorphism is defined in terms of the classical 0 number of

Loviisz [42]. As semidefinite programs can be solved in polynomial time, pseudo-

honiomorphism 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. RELAS,ATIOLYS O F GRAPfi HORfOXIORPHISAtI 3s

relaxation defined by Feige and Loviisz in [16]. Feige and Lovcisz used this relaxation

to show polynomial time solvability of various special cases of graph homomorphism.

Moreover G is homomorphic to H implies G is pseudo-homomorphic to H and this

further implies that G is fractional homomorphic to H . It turns out that fractional

homomorphism is a co-NP complete problem. Therefore, if for some subset of graphs

one can reverse any of these implications, then for this subset the corresponding prob-

lems 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

solved using consistency checks.

Although the general graph homomorphism does not admit a simple forbidden sub-

graph 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 character-

ization of the fractional homomorphisn~, we show that the above conjecture is true,

if we replace homomorphism by fractional homomorphism. The main results of this

chapter appeared in [6] and they are a joint work with Sanjeev Mahajan.

3.1.1 Notations

All 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 iFf be two graphs. The horn-product G o H is defined as the graph with

V(G o H) = V ( G ) x V(H), in which { ( s , u ) , ( t , w)) E E(G o H) if and only if (s # t )

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 these formulations we

take v to range over the vertex set of G and I to range over all maximal independent

CHAPTER 3. RELAXATIONS O F GRAPH HOMOMORPHIShI

sets in G

If we replace the integrality conditions x, E { O , l ) , y l E { O , l ) by xu 2 0, y ~ > 0 respectively, we obtain linear programs with optima wj(G), the fractional clique

number of G, and x j(G), the fractional chromatic number of G. One can observe

that these linear programs are duals of each other and hence wj(G) = xj(G) by the

duality theorem of linear programming (see [ l l ] ) . Therefore we have the inequalities:

To compute any of these numbers is NP-hard. The NP-hardness of the fractional

chromatic number was proved by Grotschel, LovAsz and Schrijver in [20]. They also

proved that one can compute in polynomial time a real number 8 (see also [39]) which

satisfies

w(G) I 0 I Uj(G).

One possible definition for 8 is as follows [50]:

Let S be the set of all (V(G) x V(G)( positive semidefinite matrices, let I denote

the unit matrix and J denote the matrix of all 1's. The l product of two matrices

A = (aij) and B = (bi j ) is the number A B = aijbij. Then

8 = O(G) = max{B l JIB E S; B I = l ; V ( s , t ) @ E(G) : b,t = 0)

Another semidefinite programming upper bound for w is the real number OlJ2 of

Schrijver 2471, which can be expressed as:

OlJ2 = OIl2(G) = rnax{BoJIB E S; BoI = 1; V(s,t) $ E ( G ) : b,t = 0; Vs, t E V(G) : bst > 0).

Szegedy in [50j observed that OlJ2 is the same as the vector chromatic number

whose definition can be find in [38].

CHAPTER 3. RELAXATI0,VS OF GRAPH NOM01\10RPHISXI

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 ho-

momorphism and its consequences which include an easier NP-hardness result for the

fractional clique number problem and the fractional version of Hedetniemi's conjec-

ture. In Section 3.5, we show that our notion of pseudo-homomorphism is equivalent

to the notion of hoax of Feige and Lovisz [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 will 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. We conclude 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 w(G o H) = IV(G)I if and only if there is a homomorphism from G to H .

Proof: The first and the second inequalities follow from [50]. The third inequality

follows from Theorem 10 in [42]. From the definition of a fractional clique and from

the duality of linear programming, it follows that wf = xf < 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 horn-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 O F GRAPH IIOMOXIORPIIISI1I 4 1

assume that cr 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 IV(G)I, then the first coordinates have to span all

the vertices of G because (5, u) + (s, w), and hence C can be regarded as a function

V(G) -+ V(H). We define a homomorphism cr from G to H as follows: cr(s) = 21 if

and only if ( s 7 u ) E C. To prove that a is a homomorphism, let s, t be vertices in

G. Then (s, cr(s)) and ( t , cr(t)) are from the clique C and hence they are adjacent.

From the definition of the hom-product, it follows that s t implies ~ ( s ) cr(t). We

conclude that cr is a homomorphism.

The previous theorem suggests the following relaxations of homomorphism:

Let G and H be two graphs.

Definition 23 We say that G is fractionally homomorphic to H (denoted by G -+j H) if wr(G o H) = IV(G)I.

Definition 24 We say that G is pseudo-homomorphic to H (denoted by G -+, W )

if O ( G o H ) = IV(G)I.

Definition 25 We say that G is pseudoll2-homomorphic to H (denoted by G' -+,I:! H )

if Ql12(G 0 H) = IV(G)I.

Corollary 26 (G -+ H ) =+ (G -+,12 H) =+ (G --+, H) + (G --+f H ) .

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 --+,12 H are polynomial, while both

G -+ Ii and G -+f Ii are NP hard. However, for certain families of graphs we

can prove the reverse implications, and so obtain a polynomial algorithm for these

farnilies.

3.4 Fractional Homomorphisms

In this section, we will give a forbidden subgraph characterization of fractional homo-

morphism. In particular we will prove that G is fractionally homomorphic to H if and

CHAPTER 3. RELAXATIONS O F GRAPH MOrZIOXIORPIIISi~I 4 2

only if G does not contain a subgraph isomorphic to I<Lw,(q+ll. This will enable us

to prove a fractional version of the graph product conjecture [ E l . As a consequence

we will obtain an alternative proof of the NP-hardness of computing the fractional

clique number of a graph.

Theorem 27 G -+f H if and only i f w ( G ) 5 w j ( H ) .

Proof: Assume first that G -if H. For a contradiction let us also assume that there

is a clique C with Lwf ( H ) + 11 vertices in G. We will show that w j ( G o H ) < IV(G)I

by constructing a dual fractional coloring smaller than I V ( G ) [ . Let w be a minimum

fractional coloring of H . Its value is x j ( H ) = w j ( H ) . All independent sets in G o N are of the form I = u:=, s; x A; where sl, . . . , sk form a clique in G and for i f j , Va; E

A;, aj E Aj : a; + 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 ) /\ ( U + w). Now we construct a fractional coloring w* of G o H as follows: w;xT = WT for any

maximal independent set T from H , w:XV(H) = 1 for u $! C, and w; = 0 otherwise.

One can observe that this is a feasible fractional coloring and its objective value is at

most w f w ) + ( l V ( G ) l - l ~ j ( H ) + 11) < IV(G)I. On the other hand if we assume that w ( G ) 5 w f ( H ) , we will show that w f ( G o H ) =

IV(G)I. Let x be a maximum fractional clique in H, and w a maximum fractional

coloring of H dual to x . From complementary slackness [11] it follows that

where J runs through the maximal independent sets in H. We construct a fractional

clique x* in G o H as follows: Let x~,,,) = where s E V ( G ) , u E V ( H ) . w f ( w '

Let us denote x s = CsES x, for any subset S of vertices from V ( G ) and similarly

x; = ~ ( 3 , u ) E T x ~ 3 , 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 o H ,

z; 5 1. For an independent set I = ~ i ( = ~ si x Ai , let us write B = U i f j A; n A j , where . . 2 . 3 = 1 , . . . , k and let B, = Ai - B . Then I C ~ f ; = ~ s; x (B; U B ) , and therefore

CHAPTER 3. RELAXATIONS O F GIZA PI1 MOA101210RPHIShI

BY our assumption k 5 wj ( H ) , therefore the following inequality holds:

where J runs through the maximal independent sets of H. One can observe that the

set B U & ( J n B i ) is an independent set in H and therefore EL, X J ~ & + X B 5 1.

Hence

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

This theorem shows that although the graph homomorphism problem is NP-

complete, for any fixed non-bipartite graph ti [30], the fractional graph homomor-

phism problem is polynomial, for every fixed graph H, since one can use brute force to

check for a clique of size [ w f ( H ) + 11 in G. The theorem also shows that if H is part

of the input, then fractional homomorphism is co-NP complete. Another interesting

consequence of this theorem is that for a given k and a graph G , it is EP-hard to

determine if G has fractional chromatic number at least k. This result was originally

proved in [20] by the ellipsoid method.

CHAPTER 3. RELAXATIONS O F GRAPH HOMOMORPHISM

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 w j ( 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 = W ~ ( I < , - ~ ) , by our theorem w ( G ) 2 n if and only if

G f t j I<,-l, which is equivalent to w j ( G o I<,) < IV (G) ( . 0

A longstanding conjecture in graph theory states that whenever G f t I<, and

H f t I(,, then also G x H f t I<, cf. [22].

Proposition 29 G f t f W and H f t j W imply G x H f t f W

Proof: By Theorem 27 G f ; j W and H ++j W imply w ( G ) > w j ( W ) and

w ( H ) > w j ( W ) . One can easily observe that w ( G x H ) = min{w(G) , w ( H ) ) and

therefore also w ( G x H ) > w j ( W ) which means G x II f ; j W . 0

The fractional version of the product conjecture follows by taking W = I<,.

We would 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 K 2 and Cs.

Unlike homomorphism, fractional homomorphism is not even transitive. As an

example let us choose an arbitrary graph G such that w ( G ) = 2 and w f ( G ) 2 3. Then

K3 --+I G , G -+j K 2 but K3 f S I K2 . An example of such a graph G is the Mycielski

CHAPTER 3. REL.4SATIONS O F GRAPH IiOA10MORP11'IS1II

graph Mll from Figure 3.1.

On the other hand, fractional homomorphism is dichotomic, i.e., for every two

graphs G and H , either G -+I I f or H -+j G. This follows from the fact that H f t j G

implies that w ( H ) > w f ( G ) and therefore we have w ( G ) 5 w j ( G ) < w ( H ) 5 w r ( H ) ,

which means that G +j H.

Welzl in [54] proved that for every two graphs G and H such that G -+ H and

H f t G there exists a graph W such that G -+ W -+ H , W f t G and H f t W. This

property is called density property of homomorphism. Similarly one can ask whether

graphs are dense with respect to the fractional homomorphism, i.e., whether for every

two graphs G and H such that G + j H and H +if G there exists a graph W such that

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

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

Proposition 30 If H f t f G then there exists a graph W such that W f + j G and

H f t j W (and hence G -+f W 4 j H ) if and only if w ( H ) - w f ( G ) > 1 . Moreover,

if this condition is satisfied, we can choose W to be a clique.

Proof: H f t 5 G is equivalent to w(1-I) > w j ( G ) . Therefore we have the following

inequalities:

w ( G ) 5 w j ( G ) < w ( H ) < w f ( H ) .

W f t , G and H f t j W if and only if w ( W ) > w j ( G ) and w ( H ) > wj(bV) . Therefore

there exists W 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 ( N ) are integers. The

sufficiency follows, since if the condition is satisfied, we can choose T/V = Kw(N)- l .

CHAPTER 3. RELAXATIONS 0 F GRAPH HOhf 0MORPHISLV

In this section we will relate pseudo-homomorphism to the concept of a hoax intro-

duced by Feige and Lovisz [16].

V. Feige and Let V be an incidence matrix of the graph G o H and C =

Lovisz 1161 considered the following optimization problem with convex constraints,

which we denote by (*)G,H:

rmximize C ,,,, t , , Cs, , twQsu,tw

s.t.

Q is positive semidefinite

Q is symmetric

Qs, t : C,,, Qsu,tw = 1

Qs, t , U , w : Qs,,tw 2 0.

The ellipsoid algorithm can be used to solve this optimization problem which is a

semidefinite program. This program is based on a two-prover interactive proof system.

For more details we refer the interested reader t o paper [16]. A feasible solution to

(*)G,H for which the objective function has value 1 is called a hoax (c.f. [16]) and in

their paper [16], Feige and L o v k proved that if there is a homomorphism from G

to H then (*)G,N has a hoax. They also gave the following necessary and sufficient

conditions for (*)G,N to admit a hoax.

Lemma 31 [16] The system ( * ) G , ~ has a hoax if and only if there exists a system of

vectors v,,, s E V(G), u E V(H) satisfying the following conditions:

is independent of s,

CH14Pl'ER 3. R E L A X A T I O N S O F GRAPH NO,\.IOMORPIIISM 4 7

and T vsuvt, = 0 whenever KUjt , = 0 .

The optimization problem obtained from (*)G,H by removing the last, nonnega-

tivity condition; Vs, t, u, w : QaaVt, 2 0 will be denoted by (**)G,H and any feasible

solution to (**)G,H for 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,H to admit a semi-hoax.

Lemma 32 The system (**)G,H has a semi-hoax if and only if there exists a system

of vectors v su , s E V ( G ) , u E V ( H ) satisfying the conditions (3.1),(3.2),(3.3), and

(<3.5).

We now prove that our notions of pseudoll2-homomorphism and pseudo-homomorphism

coincide with the notions of hoax and semi-hoax respectively.

Theorem 33 The system (*)G,H has a hoax if and only if G -+,I;? H .

Proof: Let Q be a hoax of the system (*)G,H. By Lemma 31, there exists a systern

of vectors v,, satisfying (1 ) - (5 ) . One can observe that Qsu,t, = v S , v ~ = 0 whenever

( s , u) is not adjacent to ( t , w) in G o H. Also,

Moreover,

Taking B = &Q, we have proved that BL,2(G o H) 2 I V ( G ) / , and from Theorem

22 it follows that G --+,I2 Ii.

CHAPTER 3. RELAXATIONS OF GRAPH HO~IO~~_IORPIIISIZI 4 S

On the other hand let us assume that G -t,p H and let 81p(Go H ) = IV(G)I.

Let B = AAT be the positive semidefinite matrix with nonnegative entries such that

B 0 I = 1, for which (s, u) + (t, w) implies bsUttw = 0 and

We notice that B,,,,, = 0 for u $I w and therefore the aSu7s (the rows of matrix A )

are orthogonal when s is fixed. If we denote as = xu a,,, then from the condition

B 0 I = 1 we have

and therefore from B J = gIl2(G o H), it follows that

= (C llasll)2 = (C l-llasll)2 5 C l2 C llasl12 = IV(G)I* s s s s

The last two inferences are a consequence of Cauchy's inequality. Because OlI2(G o H ) =

IV(G)I, it follows that both inequalities have to be equalities. That is,

and this happens if and only if all as's have the same lengths. From B I = 1 it

follows that for all s E V ( G )

Also the inequality

must be an equality, i.e., the angle between as and a t must be zero for each s, t E V(G).

Therefore all as's are equal. Taking Q = IV(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 asu's for a fixed s , we have

CHAPTER 3. RELA AXTIOIL'S OF GRA PI1 IJOi\IIOAf ORPIIIS,\I 49

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

(*)G,H. 0

One can similarly prove the following theorem.

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

Although we do not have a forbidden subgraph characterization for pseudo- homomorphism,

as we have for fractional homomorphism, 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 strong 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 denote the complement of G

and d(G) = $(G). The following lemma is the monotonicity property from [39] p. 7.

Lemma 35 [39] Given two graphs G and GI, G C G' implies d (G) 2 d(Gf ) .

The following lemma is from [39] p. 22.

Lemma 36 [39] Given two graphs G and GI, d(G* GI) = 23(G)d(G1).

Proposition 37

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

Corollary 38 If G -+, H then d ( G ) Q ( H ) 2 IV(G)I.

In the last part of this section we prove that in general one cannot reverse impli-

cations in the Corollary 26. We need the following results (see [42, 391):

for odd n

CHAPTER 3. RELAXATIONS OF GRAPH HO~ZfOhIORPHISA.1

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, C2m+l f t , 1112 and

C2n+1 f t p C2rn+l.

Proof: The proof follows from the fact that 8(C2n+l) is an increasing function of

n 2 1, while d(CZntl) is a decreasing and their ~ r o d u c t is 2n + 1. Also, we have

d(Cs) = &. Therefore d(C2n+l)d(C2m+l) < d(C2n+l)d(C2n+l) = 2n + 1 and also

d(Czm+l)d(r(,) = 28(C2m+l) < &d(~2m+l ) = f i (Cg)d(C~m+~) L: 19 (C~rn+~)d(C~rn+l) =

2m + 1. Corollary 38 implies the assertion of our proposition.

Since Cs --if K2, and the above theorem shows that C5 f t , K2, the pseudo-

homomorphism is strictly stronger than the fractional homomorphism, i.e., the exis-

tence of the pseudo-homomorphism implies the existence of the fractional homomor-

phism but not vice versa.

The next proposition shows that homomorphism is strictly stronger than pseudoll2-

homomorphism. Let Mll be Mycielski's graph from Figure 3.1.

Proposition 40 hilIl f t but Mll --+p12 Ii'3.

Proof: The first assertion follows from the fact that Mll 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 all three colors. Hence the vertex 11 needs a fourth color. The second

CHAPTER 3. RELAXATIONS O F GRAPII N O ~ I ~ O ~ ~ O R P H I S L Z I

Figure 3.1: MI1.

assertion follows from the fact that the following matrix Q is the hoax. We will write

its s,t blocks: 113 0 0

for s N t , and

Q S y t = 'I!;G2 :;:;I, 1/12 1/12 116

for s + t .

One can check that all conditions for Q to be a hoax hold.

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 O F GRAPH HOiZfOMORPNISM 52

these families. Unfortunately, for all our examples there is already a known poly-

nomial algorithm by some other means. For a better readability, we will use the

following notations.

For a graph H let us denote

(+N) = {G ( G -, H ) ,

and for a family of graphs C, let us denote

(C-+) = {H I G -+ W for some G E C)

and if G f , H for all G E C , then we will write C f , H.

We will use similar notations in the case of pseudo-homomorphism and fractional

homomorphism.

3.6.1 Graph Coloring

We will need a slight modification of Theorem 8.1 in 1161.

Proposition 41 Let C be a family of graphs such that C j+, H . Then the class

( t , H ) is in P, and separates (C4,) and ( -+H) .

Proof: Obviously, (-+I$) C (-+,H). Assume, there is a graph G E (C4) n (--+,H).

Then there is a graph G' E C such that G' -t G, and hence also G' -+p G. We have

also G t, Ii and from the transitivity of t, (see [16], for the transitivity of "has a

hoax" relation), G' -+, I1 and hence C -+, G which is a contradiction.

The class of all odd cycles will be denoted by Codd.

Proposition 42 For N bipartite, G t H if and only if G -+, H .

CHAPTER 3. RELAXATIONS O F GRAPH NOiLIOAIORPHlShl 53

Proof: The graph W 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, Codd f+, K2 and hence from Proposition 41 it follows that the class

(-+,K2) is in P, and separates (-+I&) (all bipartite graphs) from (Codd--)) (graphs

which contain an odd cycle, i.e. nonbipartite graphs). Therefore (-+,Ii12) = (+li'2).

But for bipartite graph H, one can easily observe that G -+ H if and only if G -+ K2

and hence the theorem follows. 0

Note that this example also 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 polynomial.

One can notice that 2-colorability has a duality property given by the equation (3.6).

If

(Cf+) = (+HI

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

Proposition 43 If H satisfies the duality property (Cf t ) = (-+H), and C f+, Ii'

for some family of graphs C, then (-+, H ) = (-+ H ) and hence 11-colorability is

polynomial.

Proof: This is because ( C t ) n ( t , H ) = 0 implies (-+,N) = ( C f t ) n (-+,

11) = (-+ H) n (-+,If), and hence (t, H ) 5 (+ H). The other direction is

trivial. Therefore in this case, the semidefinite program (*)-,* gives us a polynomial

algorithm for the H-coloring problem.

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 Gk with the following properties:

a / Gk is a union of almost complete graphs on k vertices Hi, i = 1,. . . , m and

some edges between them

C H A P T E R 3. RELASATlONS OF GRAPH H O M O ~ ~ O R P N I S I M 53

b/ If (xi, y;) is the missing edge from H; then y; = x;+l for i = 1 , . . . , nz - 1, and

( X I , y,) is an edge in Gk. One can observe that if there is a k + 1-string as a subgraph of a graph G, then

G is not k-colorable, otherwise the vertices x;, and y; would have to have the same

color and b / would imply that x l , and y, have the same color but they are adjacent.

Let Sk denote the set of all k-strings, and

then the class of graphs with the following duality property is denoted by CSt

One can easily observe that (S3--+) = (Codd-+). Therefore C,t contains the family of

all bipartite graphs (see Example 1). CSt contains also all graphs with w = x because

€ s n .

Example 8 C2n+l E S3.

Example 9 I<, E S,.

Lemma 44 Let v,, be vectors satisfying the conditions of Lemma 31 and (3 be a

subset of these vectors which are mutually orthogonal. Then for 2% = CuEO u we have

and the equality holds if and only if v; = 6

and hence the angle between the vectors 6, and v s s 0. 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

11v>11 = 1. This finishes our proof. 0

CHAPTER 3. RELAXATIONS OF GRAPIJ HOMOA1ORPIiISi\l 55

Lemma 45 Let (*)G,H has a hoax, v,, be vectors satisfying the conditions of Lemma

31 and I{, be a complete subgraph of G with n = IV(H)I vertices. Then Vu E

V(H) : C,~K, vsu = 6.

Proof: For a fixed vertex u E V ( H ) , vectors v,,, s E Ir', are orthogonal. From the

previous lemma it follows that I( CsEK, v,,11 < 1. Therefore

Combining this with Lemma 2 yields our assertion. 0

Now we can generalize Proposition 42 as follows.

Proposition 46 For G E Cst, G t K, if and only i f G -+, I<,.

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

enough to prove Sn+l +, K,. To get a contradiction, assume G,+l t, Ii', for some

Gn+l E and hence (*)G,+l,Kn has a hoax. Let v,, be vectors given by Lemma

31. The vectors vsu,u E V(K,) are mutually orthogonal, since s cannot be mapped

on distinct nodes, and hence

Since Gn+l E Sn+l, it satisfies conditions a/ and b/ of the definition of n + 1-string

and we will use notation from these conditions. The vectors v,,, s E V(H;) - {z;}

are also mutually orthogonal, since any two distinct nodes from V(Hi) - {z;) are

adjacent and hence they must be mapped on distinct nodes. Similarly the vectors

v,,, s E V(H,) - {y;} are mutually orthogonal. On the other hand, IV(H,)I = n + 1

and hence from previous lemma follows

and similarly

1 = I / C vsu/I2. SEV(H,)-~Y,)

ClfAPTER 3. RELASr-1TlONS O F GRAPH HOLtfOrtIORPHIShf 56

From Lemma 44 it follows that CsEv(rr,)-(x,~ us, = 6 = ~ , t V ( H , ) - ( y , ) us,, and there-

fore fi - v,,, = fi - vy,,' which implies v,,, = v,,,. From condition b/ we have obtained

v,,, = u,,,. But vertices xl, and y, are adjacent and hence cannot be mapped to

the same vertex u. Therefore v,,, and v,,, have to be at the same time orthogonal,

which is a contradiction, because their sum over all u is the unit vector 6. 0

3.6.2 Digraph Coloring

A directed graph G is said to be homomorphic to a directed graph H , if there exists

a function f : V(G) -+ V(H) called a homomorphism, 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 Go H with V(GoH) = V(G) x V(H)

and ( s ,u ) N ( t , w) if and only if (s f. t ) A ( ( s , t ) E E(G) =+ (u, w) E E ( H ) ) A ( ( t , s ) E

E(G) + (w,u) E E(H)) .

Hell, NeSetfil, and Zhu found polynomial algorithms for classes of graphs which

have trees-duality in [32, 34, 331, i.e., C-duality where C is the family of all trees.

They showed that the following labeling algorithm works for H-coloring (the problem

whether G --+ H for a fixed graph H) for graphs H which have trees-duality. We will

later refer to this algorithm as the consistency test (cf. (321).

Instance: A digraph G;

Question: Is G -+ H?

Define labels Lk : V(G) -+ 2"(H), k 2 0 by induction as follows:

Lo(s) = V ( H ) for all s E V(G)

Lk+l(s) = {U E Lk(s) I Vt E V(G) 3w E Lk(t) :

((s, t ) E E ( G ) =+ (u, w) E E(H)) A (( t , 3) E E ( G ) =+ (w, 4 E E(H))h

For any vertex s , the set Lk+l (s) C Lk(s), k >_ 0, and Lo(s) = V(H). Therefore

CHAPTER 3. RELAXATlONS OF GRAPH I1012/101LfORPHISM

there is some i such that L ; ( s ) = Lit, ( s ) .

Output: G is N-colorable if for all s E V ( G ) : L i ( s ) f 8.

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 all graphs which have trees-duality. Then for

every graph H E C and every graph G , G t H if and only if G t j H.

Proof: 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 ) : L ; ( s ) = 8 , then G fir H.

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

First we will show by induction on k > 1 that if u $! L k ( s ) , then the corresponding

weight x(,,,) = 0.

If k = 1 , and u $! L l ( s ) , then there exists t E V ( G ) such that for all w E Lo( t ) =

V ( H ) : ( s t ) A ( u + w ) . Therefore for all w E V ( H ) : ( s , u ) # ((1, w ) . Hence ( s , u )

together with vertices ( t , w ) , w E V ( H ) , form an independent set, arid therefore

The equality (3.8) implies that CwEV(N) x ( ~ , ~ ) = 1 and hence x(,,,) = 0.

Assume that u $! L k ( s ) implies x(,,,) = 0. For u $! Lk t l ( s ) we have t such that

for all w E L k ( t ) : ( s t ) A ( U # w ) , which means also that vertices ( s , u ) , and

( t , w ) , w E L k ( t ) form an independent set. For w $! L k ( t ) we can use our induction

hypothesis that = 0 , and therefore we can apply the equations from the basic

case.

If L ; ( s ) = 8 , then for all u E V ( H ) : x(,,,) = 0, which contradicts (3.8).

C H A P T E R 3. R E L A S A T I O S S OF GRAPH HOltfOLtfORPI-fISI1f

3.7 Conclusion

In the first chapter, we have proved that G -+ H implies w(G) < w(H), O(G) < O(H), xJ(G) I x J ( H ) and x(G) 5 x(H). Therefore, if O(G) I O(N), then we have the

following inequalities:

w(G) I O(G) I O(H) 5 WJ(H),

and therefore G -+ H implies O(G) 5 O(H) and this further implies w(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 pseudo'-homomorphic to H if

and only if O(G) 5 O(H). We do not know if this is equivalent to our definition of

pseudo-homomorphisrn and hence to the hoax of Feige and Lovcisz. Although, for G

with the transitive automorphism group, O(G)d(G) = IV(G)I (c.f. [42]). Hence it

follows from our necessary condition (Corollary 38) that for graphs G with transitive

automorphism group, G -+, I-I implies O(G) 5 O(H). Our investigation was inspired

by work of Feige and L o v k [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 de-

fine O(0) = 0. From our definition of Go H it follows that Go H is always loopless, even

if we extend the definition to all graphs. If one consider a pseudo-homomorphism to

be any symmetric, positive semidefinite matrix A of dimension IV(Go1I)l x IV(GoH)I

such that A I = 1, for all {(s, u), ( t , w)) @ E ( G o 11): A(,,,),(t,,) = 0 and O(G o H) =

A J , then one may consider a category P F G of graphs and pseudo-homomorphisms, A A'

if one define a composition of pseudohomomorphisms G -+, H and H -+, W to be A" - 1 G -+, W such that A&,,),(,,,) - I V ( H ) ~ E a , b t v ( ~ ) A (s ,a ) , ( t . b ) A' (a ,u) , (b ,w) (compare with

[16], Lemma 5.22). The category FG of all finite graphs and homomorphism is ac-

tually embedded in P F G . To see this, we need a faithful functor T which maps

homomorphisms to pseudo-homomorphisms. Let G -5 H be a homomorphism. Then

CHAPTER 3. RELAXATIONS O F GRAPH IIO12-I OMORPHISM

define T ( a ) = A such that

if u = O ( S ) and w = a( t ) A(s,u) , ( t ,w) =

otherwise.

We 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 Loviisz, which is implicit in their paper [16].

Chapter 4

Structure of Color Classes

4.1 Introduction

In this chapter we will consider all finite graphs and we will denote the set of all finite

graphs by G .

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

G x H -, W implies that G -+ W or I1 -+ W

Hedetniemi [26] proposed the following conjecture.

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

One can easily prove that and are multiplicative and the proof of the multi-

plicativity of K3 was accomplished by El-Zahar and Sauer [14]. An alternative form

of (4.1) is the following:

G f t W and H f t W imply that G x H ft W. (4.2)

If one further restricts graphs G f t W and H ft W of (4.2), then it is sometimes

possible to conclude that G x II f + W. This approach was considered by Burr, Erdos

and LovAsz [9], Hell 1281, Turzik [51], Duffus, Sands and Woodrow [12], Welzl [55] and

others. Particularly, they proved the following results:

C H A P T E R 4. S T R U C T U R E OF COLOR CLASSES 6 1

Proposition 49 [9] If G ft Ii',, H ft I;, and each vertex of G is contained in a

complete subgraph of order n, then G x H ft Ii',.

Proposition 50 [12, 551 If G and H are connected, Ii', -+ G ft K , and Ii', + H ft K,, then G x H ++ Ii',.

Wote that one can assume connectivity of G and H, since if there are G and H such

that G f t I<,, H f t I<, but G x H + I<,, then for some connected component G'

of G, G' f t Ii', and similarly for some connected component H' of H, H' f-l Ii', but

G' x H1+ Ii',.

Proposition 51 [51] If G ft I<,, H ft I{, and for every pair of edges e , e' of G

there exists an edge e" of G which intersects both e and el, then G x H ft I<,.

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 homomorphism a from G to H induces a simplicial mapping, since a graph homo-

morphism takes a set of vertices with a common neighbor to a set of vertices with

a common neighbor. It was shown by Lovksz [41] that if the geometric realization

J@(G)I of @(G) is (n - 2)-connected then G f-1 Ii',. 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 homology group of X with coefficients in 212 vanishes for each

j n. Walker [52] proved that if G is a graph such that @(G) is (n - 2)-acyclic mod 2,

then G f t Ii',. It follows that if I@(G x H)I is (n-2)-connected then G x H f-l Ii', (cf.

[28]). However, it is not the case that I@(G x H)I = I@(G) x @ ( M ) I (cf. [28]). In fact,

Walker [52] showed that I@(G) x @ ( E l ) I has the same homotopy type as I@(G * H)I,

where G * H is the join of G and H, i.e., the graph obtained from G + N by connecting

all vertices of G with all vertices of II (cf. Proposition 4.1. in [52]).

Other multiplicative graphs were studied by Haggkvist, Hell, Miller and Neumann-

Lara [22] and they proved:

Proposition 52 [22] Each cycle is multiplicative.

CHAPTER 4. STRUCTURE OF COLOR CLASSES

It is not even known whether the following function

p(n) = min{m I G f t I<, and H ft I\', but G x H -+ I<, for some G, H )

tends to infinity or not. Note that the Hedetniemi's conjecture is equivalent to the

assertion that for all positive integers n, p(n) = n + 1. Poljak and Rod1 [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 tt H, if G --i H and H -+ G. The relation

t-t is an equivalence, since -t is reflexive and transitive relation. In their paper [13],

Duffus and Sauer studied the equivalence classes of graphs 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-

tion later). They have observed, that the set = G / t-t of these equivalence classes

with just mentioned operations form a Heyting algebra with the zero element (0) and

the unit element (1). Respectively, we will use the symbols x and + for operations

meet and join in distributive lattices, unless specified otherwise. Recall that IIeyting

algebra is a distributive lattice with the minimum element such that for every two

elements a and b there exists an element x for which a x c < 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 called the relative pseudocomplenzent of a with respect to b. 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 ( ~ ) I g E q) is a Boolean algebra with meet x A y = z 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 all graphs with a loop. They also proved that if B(K,) is finite for

all n , then the function p(n) tends to infinity with n. Consequently they asked for

CHAPTER 4. STRIiCTllRE OF COLOR CLASSES

which graphs W , B(rc.) is finite.

We say that W has afinite factorization, if there are multiplicativegraphs LVl, . . . , Wn, such that W tt Wl x . . . x W,. From the definition it follows that if W' is multiplica-

tive and W t, W' then W' is also multiplicative. Therefore if W is multiplicative,

then all graphs in (W) are multiplicative. We say that a class w E c is multiplicative,

if graphs from w are multiplicative. We will prove that the boolean algebra B(IY) is

finite if and only if W 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 meet of finitely many meet

irreducible elements of c. This suggests that the finite factorizations of elements

of are closely related to Hedetniemi's conjecture, and that understanding of the

multiplicative structure of c 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 W which has a finite factor-

ization and satisfies K4 --+ W would imply that p(n) -+ m. This is a generalization

of the result of Duffus and Sauer in [13], who proved that if all B(h-n) are finite, then

p(n) -+ cm. By our results, their assumption is equivalent to the assumption that all

K , have finite factorizations.

4.2 Overview

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 6 4

on them. We will also prove the counterpart of the prime avoidance theorem from

commutative algebra. Section 4.5 contains several quite trivial observations about

additive structure of c. In Section 4.6, we study the multiplicative structure of ideals.

We define prime ideals and develop some tools which are useful in later sections. Many

of our proofs were motivated by the corresponding results from commutative algebra.

In Section 4.6.1, we endow the set of all prime ideals with Zariski topology and we

investigate topological properties of some of its subset, which are closely related with

the factorization problem. In Section 4.6.2, we give necessary and sufficient conditions

for the existence of irredundant and finite factorizations. Finally, in Section 4.7, we

summarize our results and interpret some of them in graph theoretical terms.

4.3 Preliminary Results and Motivation

In this section we will prove most of our claims from Section 4.1. We will use results

and constructions from [22], [46] and [13].

Probably the most important construction is the following one from [22].

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

as follows: the vertices of W ( G ) are all possible mappings 4 : V(G) t V ( W ) and

( 4 , $11 is an edge of W ( G ) whenever { s , t ) E E(G) implies {$(s) ,$I( t ) ) E E(W) .

The following properties of the map graph were also observed in [22].

Proposition 54 [22]

(a ) W ( G ) has a loop if and only i f G --+ W .

(b) G x W ( G ) t W .

(c) G -+ W ( W ( G ) ) . ( d ) W 4 W ( G ) .

(e) G x N -+ W if and only if H -+ W(G) .

( f ) W is multiplicative if and only if G f , W implies W ( G ) t W .

(9) G -+ G' implies W(G1) + W ( G ) .

(h) W t W' implies W ( G ) t W1(G).

CHAPTER 4. STRUCTllRE OF COLOR CLASSES 65

Therefore kV is multiplicative if and only if W(G) tt It' or W(G) ++ 1 for all

graphs G. We will need also the following properties of the map graph.

Proposition 55 .

(2) bV(G1 + G2) ++ bV(G1) x W(G2),

(j) (Wi x W2)(G) +-+ Wi(G) x kV2(G), (k) i fG is a connected nonbipartite graph then (Wl + liV2)(G) +-+ Wl(G) + W2(G),

(I) W is multiplicative if and only if G f + W implies W(G) -4 W for all connected

nonbipartite graphs G.

Pro0 f:

(i): This was proved by Duffus and Sauer [13].

( j ) : Since kV1 x W2 -+ Wl, Lemma 54 (h) implies that (Wl x W2)(G) -+ Wl(G).

Similarly (Wl x W2)(G) t W2(G) and hence (Wl x W2)(G) -+ Wl (G) x W2(G). On

the other hand, the mapping a from Wl(G) x W2(G) to (Wl x W2)(G) defined by

~ ( ( 4 , $))(g) = ($(g), T+!I(g)) is easily seen to be a homomorphism.

(k): Since bV1 t Wl + W2, Lemma 54 (h ) implies that Wl(G) -t (Wl + kV2)(G). Sirn-

ilarly W2(G) t (Wl + W2)(G) and hence Wl(G) + W2(G) -+ (Wl + kV2)(G). On the

other hand let us fix an arbitrary vertex T+!I from the graph I/t; (G) + W2(G). Let us de-

fine a mapping a from the vertices of (Wl + W2)(G) to the vertices of T/Vl(G) + kV2(G)

as follows:

We will show that a is a homomorphism. It is enough to prove that vertices $ of

(Wl + W2)(G) such that $(V(G)) $7= V(Wl) and d(V(G)) $7= V(W2) are isolated ver-

tices, i.e., their neighborhoods are empty. For a contradiction let us assume that for

some g,gl E V ( G ) we have $(g) E V(Wl) but $(gl) E V(W2), and that $ - 4' for

some vertex 4' of (Wl + W2)(G). Since G is a connected nonbipartite graph, there

are (not necessarily distinct) vertices g = gl, g2,. . . , g2k+l = g1 E V(G) such that

g, - g;+l for i = 1 , . . . ,2k. Since $ - $I, it follows that $(g) = 4(gl) N 4'(g2)

$(93) N . - - - (f(g2k) - $(g2ktl) = 4(g1). This implies that 4(g) and 4(g1) belong to

CHAPTER 4. STRUCTURE OF COLOR CLASSES

the same connected component of I/ti + W 2 , which is a contradiction.

( I ) : One can observe that all bipartite graphs are multiplicative, since every bipartite

graph is either empty, or it is equivalent to either K1 or K 2 and all these graphs are

multiplicative. If W is multiplicative than by Lemma 54 ( f ) it follows that G f t W

implies W(G) -t W and hence this implication is true for all 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 t W , H ft W but G x H --+ W. Let G I , . . . , G, be the connected components of G and HI,. . . , H, be the connected

components of H i.e., G = GI + . . . + G, and H = H1 + . . . + H,. Then G f t W and

H f t W imply that a t least one connected component of G and H is not homomor-

phic to W. We may assume that G1 ft W and H1 f t W . Rut G x H -+ W implies

G1 x HI -+ W and hence by Proposition 54 (e), H1 --+ W(G1). Since H1 ft W , also

W(G1) f , W and therefore G1 is a connected nonbipartite graph such that G1 ft W

and W(G1) f t W .

We will show later that the connectivity of a graph is a dual notion of the mul-

tiplicativity 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 least one of them is not bipartite (see [53]).

Corollary 56 The sum of two multiplicative graphs is multiplicative.

The next two constructions are from [46].

Definition 57 [46] The arc graph of a graph G, denote SG, is defined as follows: the

vertices of SG are the ordered pairs (x, y ) such that x - y in G, and (x, y ) is adjacent

to (u,v) whenever y = u .

Lemma 58 [46] S(G x I!) = SG x 611.

Definition 59 [46] For each positive integer n, let S, be the following symmetric

graph. The vertices of S, are the subsets of (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. STRUCTIJRE O F COLOR CLASSES

Lemma 60 [46] Sn ++ 1. n/2l

Proposition 61 [46] SG -i K , if and only if G -i S,.

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 Iin is multiplicative, then Ir' is also multiplicative. ( LG2J )

Proof: Assume that K , is multiplicative and let G x H -i I ( ( [ . This is equiv- nl2J

alent to S(G x H) -+ Ir', and hence SG x SH 4 I<,. Since K, is multiplicative by

our assumption, either SG -+ Ii', or SH -+ I<,. But this is equivalent to that either

G--+ I<( n ) or H -+ I< and we conclude that ir' is multiplicative. Ln/2l ( ~n;2l ( L:~J )

Note that if n 2 4, then n < (,G~,) and hence if there is an n 2 4 such that IC,

is multiplicative, then there are infinitely many n such that K , is nxultiplicative.

The following construction generalizes the already mentioned graph S, from [46].

For each graph H , let SH be the graph such that the vertices of SH are all subsets

of V ( H ) and two vertices A, B are adjacent in SH whenever the following two condi-

tions are satisfied:

(a) there exists a E A - B such that for all b E B, a b,

(b) there exists b E B - A such that for all a E A, b a .

One can observe that S, is equal to SKn.

Lemma 63 G -+ H implies SG -+ SH.

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 all b E B, a N b and hence

a ( a ) E a ( A ) -a (B) and for all a ( b ) E a(B), a ( a ) -a(b) .

Now we can prove the following proposition, which is a generalization of Proposi-

tion 61.

CHAPTER 4. STRUCTURE OF COLOR CLASSES

Proposition 64 6G -+ H if and only if G -+ SH.

Proof: If 6G Z H then define /?(g) = a({g) x &(g)). If g - g', then (g, g') E

P(g) - P(gt) and (g, g') - (g' , g") E P(gt) which implies that P(g) - P(gl).

On the other hand if G -% SH then define a ( x , y) = h E @(x) - P ( y ) such that for all

h' E P ( y ) , h - h'. If (x, y) - (y, Z ) then a ( x , y) = h h' = Q ( Y , 4 E P ( y 1.

The following 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 W is multiplicative and let G x H -+ Sw. This is equivalent to

6(G x H) -+ W and hence SG x SH t W. Since W is multiplicative by our assump-

tion, either SG -+ W or 6 H -+ W. But this is equivalent to that either G -+ Sw or

H -+ Sw and we conclude that Sw is multiplicative. 0

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

denoted G t, H , if G + H and H -+ 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 (N) to be the class

(G x H), the join (G) + (H) to be (G + H) and the relative pseudocomplement of

(W) and (G) is (W(G)) to be the class containing the map graph W(G). We have

already mentioned in Section 4.1 that the set g = G/ t, 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 t, G' and H t, H' imply that G x H t, G' x H'

and G + N t, G' + HI. From Lemma 54 (g), (h) it also follows that W t, W' and

G H G' imply that W(G) t, W'(G1). Moreover, if considered as a poset, (G) (CV)

if and only if G t W .

We have also mentioned that if we fix a class w E c, then the set B, = {w(g) I g E

g) is a Boolean algebra with meet a: A y = s x y, join x V y = w(w(x + y ) ) , complement

CH'4PTER 4. STRIICl'IiRE OF COLOR CLASSES 69

xC = w(x), zero element w and unit (1) - the class of all graphs with a loop [13]. Duffus

and Sauer [13] proved that if B(K,) is finite for all 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 boolean algebra B(w) is finite if and only if W has a finite

factorization.

Proof: Let us assume that B p ) is finite. By Stone's representation theorem (see

[49]), every finite Boolean algebra is isomorphic to a Boolean algebra of all subsets

of some finite set. Therefore the zero element is equal to the meet of all coatoms.

Duffus and Sauer [13] have shown that all coatoms in B(w) are multiplicative classes

(the same conclusion is implicit in Proposition 99) and hence W has a finite fac-

torization. On the other hand, let us assume that i V H M/; x . . . x CV,. Then

W(G) H ( Wl x . . . x W,)(G) H 1% (G) x . . . x W,(G). But since each M/; is mul-

tiplicative, we have (Wi(G)) E {(W,), ( I ) ) , which implies that there are at most 2,

different classes of graphs in the set {W(G) I G E G } and therefore B(w) is finite.

We have also the following proposition.

Proposition 67 If G and H have a finite factorizations, then both G x H and G+ H

also have a finite factorizations. h4oreover, if W has a finite factorization then the

map graph W(G) has a finite factorization, for each graph G.

Proof: Let G * PI x . . . x P, and H H Q1 x . . . x Q m be finite factorizations of G

and H, i.e., assume that all graphs PI,. . . , P,, Q1,. . . , Q, are multiplicative. Then

G x H t, PI x . . . x P, x Q1 x . . . x Q, and hence G x H has a finite factorization.

Also G + H H (PI x . . . x P,) + (Q1 x . . . x Q,) H n i j ( P i + Qj) (see Lemma 74),

and by Corollary 56 all surris 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 H Wl x . . . x WT be a finite factorization of W. Then by Proposition

55, W(G) H Wl (G) x . . . x W,(G) H ncbw, W;, since Wi's are multiplicative; we

CHAPTER 4. STRllCTURE O F COLOR CLASSES

conclude that W(G) has a finite factorization. 0

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

graph which contains Ir14 for the proof that p tends to infinity. We need the following

proposition.

Proposition 68 If for every graph G there exists a 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 -+ W. Let us denote X, = { W I I(, -+ W, W multiplicative). The assump-

tion implies X, # 0 and therefore y(n) = min{m I W -+ K, for some W E X,)

is a well-defined, positive integer valued function. In other words, ?(n) is the mini-

mum chromatic number of all multiplicative graphs containing K,. We will prove

that y(p(n)) > n. Let W, E X, be one of the multiplicative graph such that

W, -+ KT(,). Let G ft I(,, H f t Ir', and G x H -+ Ir',(,). Since W,(n) E X p ( n ) ,

G x II -+ I{,(,) -+ W,(,). By assumption W,(,) is multiplicative and hence G -+ W,(,)

or H -+ W,(,). On the other hand W,(,) --+ KT(,(,)), which implies that y(p(n)) > n.

One can observe that p is a nondecreasing function and if it was bounded, say by c,

then for all n we would have n < y(p(n)) 5 max{y(l), . . . , y(c)), which is a contra-

diction. CI

Note that if there is a sequence of multiplicative graphs {P,)zo such that for all

n there is P, which contains I{,, then the conditions of Proposition 68 are satisfied.

Moreover, if -+ W, and W has a finite factorization, then there exists a multi-

plicative 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 I{,, 4 Po. By

induction let us define n ; + ~ = (Ln:,2J) and P,+, = Sp,. Assuming Ii,, -+ Pi it follows

CHAPTER 4. STRUCTURE: O F COLOR CLASSES 7 1

that I<,,+, t-t SK,, -+ SP, = Pl+l. Since no = 4, it follows that n, < n 1 + ~ and all P,'s

are multiplicative. Therefore we have the following corollary.

Corollary 69 If there is a graph W which has a finite factorization and K4 -+ I/V,

then p(n) --+ m.

This corollary 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 later sections.

Lemma 70 For any graphs G and 11, G x H -+ G -+ G + H .

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

homomorphism P from G to G + H can be defined as P(y) = (g, 0).

Lemma 71 The following conditions are equivalent:

( i ) G - G x H

(ii) G -+ H

(zii) G + H ++ H .

Proof:

( i ) =+ (ii): From G -+ G x H and G x I1 -+ G (according to the Lemma 70) we

conclude G -+ H.

(ii) =+ (iii): Let C Y ~ be a homomorphism of G to H and let az be any homomorphism

of H to •’I (for instance the identity). A homomorphism P from G + H to H can be

defined as P((s, i ) ) = CY, (s) for (s, i ) E V(G + H ) .

( i i i ) =+ (ii): From G + H -+ I1 and G --+ G + H (according to the Lemma 70) we

conclude G -+ H.

CHAPTER 4. STRUCTURE OF COLOR CLASSES 72

(ii) + (i): Let a be a homomorphism of G to H. A homomorphism ,L3 of G to G x H

can be defined as P(s) = (s, cr(s)) for s E V(G).

Therefore we have the following corollary.

Corollary 72 For all graphs G, G t, G x G ++ G + G.

Let us recall that a graph G is called a core if it is not homomorphic to any of

its proper subgraphs, or equivalently if any endomomorphism G 4 G is an automor-

phism.

Lemma 73 [54](cf. also [31]) Every equivalence class of graphs has a uniquely de-

fined 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 G (2) G x (I1 + F) = (G x H) + (G x F)

(ii) G + (H x F) H (G + H) x (G + F)

Proof:

(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, 0) to the vertex ((g, O), (g, 0)) and the

vertex ( (h , f ), 1) to the vertex ( ( h , I), ( 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

( ( g , 0), (s, j ) ) to the vertex (g, O ) , the vertex ( ( h , l) , (g, 0)) to the vertex (g, 0) and the

vertex ((h, I ) , (f, 1)) to the vertex ( ( h , f ) , 1) is a homomorphism from (G+H) x ( G + F )

to G + (H x F). 0

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

algebra, and hence a distributive lattice with meet x and join +. Also we will use

symbols 0 and 1 for the minimum and the maximum elements of any lattice and all

lattices considered are assumed to have 0 and 1. We will not use any specific property

of the Heyting algebra and hence we will state all results for a general distributive

lattice L with 0 and 1. To interpret these results for graphs, the reader only has 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 < for meet, join and the partial order relation

respectively.

We now formally define ideals.

A subset a of a lattice C is an ideal, if a satisfies the following properties:

(1) A E a and G E L imply A x G E a and

(2) A, B E a implies A + B E a.

One can observe that the first condition is equivalent to

(1') if A E a and B'< A then B E a.

We will use this condition and condition (1) interchangeably.

A subset f of a lattice C is a filter, if f satisfies the following properties:

(7) A E f and G 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

(it) if A E a and A 5 B then B E a.

Let L be a distributive lattice. Note that the set (0) 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 L 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 L, and let (S) denotes the smallest ideal containing S

(the intersection of all ideals containing S). The ideal (S) is called the ideal generated

by S. For S = 0 we define (S) = (0) = ((0)). An ideal is called principal if it is

CHAPTER 4. STRUCTURE OF COLOR CLASSES

generated by a one element set and finitely 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 proper, if it is not equal to L. Let S be a nonvoid subset of L. The

smallest filter containing S (the intersection of all filters containing S) is called the

filter generated by S .

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

Proposition 75 If H E L then (H) = {G E L I G 5 H ) .

Proof: Let a = {G I G < H) . First we will show that a is an ideal. The condition

(1) follows from the transitivity of <, since G E a implies that G < H , and therefore

if G' < G then necessarily also G' < H . The condition (2) follows, since G E a

and G' E a imply that G < H and G' < H so we conclude G + G' < H and hence

G + G' E a. Since H < H , it follows that H E a and therefore we have proved that

( H ) C a. The condition (1) implies the other inclusion. 0

Proposition 76 Every 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,) C a. On the other hand (GI + . . . + G,) contains

all G;, i = 1 , . . . , n, since G; < GI + . . . + G, for all i = 1,. . . , n and we conclude that

a = (GI , . . . ,G,) C (GI + . . . + G,).

The following proposition shows that the existence of a homomorphism can be

translated to inclusion between corresponding ideals in the lattice c. Proposition 77 Let G, H E L. Then G 5 H zf and only if (G) C ( H ) .

CHAPTER 4. STRUCTURE OF COLOR CLASSES 75

Proof: G 5 H implies that { F I F 5 G) 5 { F I F < 11) and by Proposition 75

it follows that (G) 5 (H) . On the other hand, let us assume that (G) C_ (N) . Since

G E (G), this implies that G E (H) and consequently by Proposition 75, G -+ H .

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 = { G [ G = A + B , A E a , BE^}.

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

Proposition 78 If a and b are two ideals then a + b = (a U b)

Proof:

If G E a+b , then G = A + B for some A E aand B E 6). Therefore G < A + B E ( aub )

and hence G E ( a u 6). On the other hand ( a ~ 6) is the smallest ideal containing both

a and b and therefore (a U b) C a + b. 0

Proposition 79 (G) n (H) = (G x H) and (G) + (H) = (G + H ) .

Proof: From the previous proposition we have the second equality. Since G x H < G,

we have (G x H) 5 (G), and similarly (G x H) C (H), which implies (G x H) C (G) n (11). On the other hand if W E (G) n ( H ) , then W 5 G and W 5 H, which

implies that W 5 G x H, and hence W E (G x H) .

One can observe that n and + are commutative and associative binary operations

on the set of all ideals ZL of L, which satisfy both distributive laws and hence form a

distributive lattice with the minimum {0} = (0) and the maximum C = (1).

A system of sets C is called a chain if for all a, b E C, either a C b or b C a. The in-

tersection of the chain C is the set naGc a and the union of the chain C is the set UaEC a.

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. STRliCTliRE OF COLOR CLASSES 76

also an ideal. Similarly 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 81 from [37 ] .

Theorem 80 I j a is an ideal and Bf is a finite system of ideals, then

a C U b i m p l i e s a b for s o m e b E B1. b a ,

Proof: By induction on n = IBf(. The proposition is obviously true for n = 1.

Assume that n 2 2. For a contradiction, let us assume that for all b E Bj, a 6. For

every c E 231 we may assume that

otherwise we use the induction hypothesis to conclude that a C_ b for some b E Of.

Therefore for every c E Bj there exists Gc E a such that Gc 6 Ubfc b. Since a is an

ideal and n 2 2, for any two distinct c f c' E P , G = Gc + Gcj E a. We claim that

G 9 Ubtof b. For a contradiction assume that G E b for some b E 231. Since c and

c' are distinct, either b j: c or b f c'. We may assume that b f c. Since Gc < G, it

follows that also Gc E 6, which is a contradiction proving our claim. Therefore G E a

and G' $ UbEof b, which is a contradiction with our assumption a C_ UbEof b.

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 Bj may not be primes.

4.5 Additive structure

TO warm up, let us first have a brief look at the additive structure of ideals in 8. Loosely speaking, it turns out that (join) irreducible elements correspond to con-

nected graphs and hence every graph is the sum of finitely many (join) irreducible

elements. To be more precise, let us recall that the graph with empty vertex and edge

sets is denot'ed by 0. For every graph G, 0 + G = G + 0 = G, 0 -+ G and 0 x G = 0.

Recall that a graph W is connected, if there are no two nonzero graphs G and H

such that bV = G + H .

Definition 81 W e say that an ideal a is additive, if a C b + c implies a C b or a G c.

Lemma 82 An ideal a is not additive if and only if there are ideals b and c such that

a = b + c a n d b g c a n d c g b .

Proof: If there are ideals b and c such that a = b + c and b g c and c 9 b, then it is

sufficient to show that a g 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 C a C 6.

On the other hand, let us assume that a is not additive and hence there are two ideals

band csuch that a c b+cbut a g b a n d a g c. Thereforea=anb+anc. I fanb C ant,

then a = a n c which implies that a C c, a contradiction. We conclude that a n b a n c

and similarly a n c g a n 6. 0

Proposition 83 A principal ideal (G) C_ is additive if and only if the class G

contains a connected core.

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

(G) 2 b + c implies that G E b + c and therefore G --+ N + H' for some N E b and

H' E c. Since G is connected, this implies that G -+ H or G -+ H' and hence G E b

or G E c. Therefore (G) b or (G) C_ c.

Let us assume that G is not connected and G I , . . . , G, are its connected components.

Therefore (G) = (GI) + (G2 + . . . + G,). If (GI) 2 (G2 + . . . + G,) or equivalently

GI -+ G2 + . . . + G,, then G et G2 + . . . + G,, which is a contradiction with our

CHAPTER 4. STRUCTURE OF COLOR CLASSES 7 8

assumption that G is a core. Similarly (G2 + . . . + G,) 5 (GI ) would imply that

G t+ GI, which is again a contradiction. 0

This proposition shows that connectivity is a dual of multiplicativity in a sense

that join irreducible elements of g have connected cores and meet irreducible elements

of have multiplicative cores.

Proposition 84 Every principal ideal in can be uniquely written as a finite sum

of additive principal ideals and hence every element is a finite sum of join irreducible

elements.

Proof: This follows from the fact that every graph has a unique core, whose con-

nected components have to be again cores and therefore generate additive principal

ideals. c7

4.6 Multiplicative structure

In this section, we will define prime ideals and give necessary tools to deal with them.

We assume all ideals to be ideals of a distributive lattice C with 0 and 1.

For any two ideals a, b we define

( a : b) = ( G E C I (G) n b C a).

Lemma 85 Let a, 6, c , 3, a, be ideals. Then

(a) (a : 6) is an ideal,

(b) (a : 6) = (1) if and only if b a,

( 4 (a : (1)) = a,

(d) a C (a : 6))

(e) b C (a : (a : 6)))

(f) b n ( a : 6) = a n b,

CHtlPTER 4. STRUCTURE OF COLOR CLASSES

(g) ( (a : b) : c) = (a : (b n c)) ,

(h) ( a : 6) n c C ( a : ( b : c)),

(3 (a : (u; 4)) = ni (a : a;),

(k) (0, a; : a) = ();(ai : a),

(I) a C b and c C a imply ( a : a) C (6 : c),

(m) a = ( a : b) n (a : (a : 6)).

Pro0 f:

(a) Let us assume that G E (a : 6) and H 5 G. If B E b then H x B = ( H x G) x B =

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 6. 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) + (H x B) = (G + H) x B and

therefore G + H E (a : 6).

(1) Let G E (a : a) and C E c 2 a. Then G x C E a C_ b and therefore G E (6 : c).

(b) If (a : 6) = (I), 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 ( I ) , then G x B < B E b C_ a.

(d) Let A E a and B E 6. Then A x B 5 A E a implies that A E (a : 6).

(c) From (d) it follows that a C (a : (1)). Let G E (1) and A E a. Then A x G' < 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 : 6)).

(f) Let G E b n (a : b), then G E b and G E (a : 6). Therefore G = G x G' E a.

(g) Let G E ( (a : 6) : c) and H E b n c. Then G x H E (a : 6) since H E c. Also H E b

and hence G x H = (G x H) x H E a. We conclude that G E (a : ( b n c ) ) . On the other

hand let us assume that G E (a : (b n c)) and H E c. Then if F E 6, 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 : 6) and we conclude that

G E ( ( a : 6) : c).

(h) Let G E (a : 6) n c. Then G = G x G such that G E (a : 6) and G E c. Let

H E (b : c). Then G x H = (G x G) x H = G x (G x N). Since G x H E 6,

G x (G x 11) E a and hence G E ( a : (6 : c)) .

(j) Let G E (a : (U, a;)) and H E 4. Then G x H E a which implies that G E a;

for all i. On the other hand let us assume that G E n i ( a : a;) and H E U, a;. 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

G E (a : (U; a;)).

(k) Let G E (n, a; : a). Then for H E a, G x H E n, a, C a;. Therefore G E (a, : a) for

all i , which implies that G E n,(a; : a). On the other hand assume that G E n,(a, : a).

Then G E (a, : a) for all i and therefore for H E a, G x H E a, for all i , which implies

that G x H E n, ai. Therefore G E (ni a; : a).

(m) From (d) it follows that a 2 (a : 6) n (a : (a : 6)). The other inclusion follows

from (f ) .

This Lemma particularly implies that the set of all ideals Zc is a Heyting algebra

with relative pseudocomplement a(b) = (a : 6).

An ideal p is called prime if it is proper and for any two ideals a and 6, a n b C p

implies a C p or b p . An equivalent condition is that for every G, H E L , G x H E p

implies G E p or H E p , which means that the complement L - p is a filter.

Looking at the complements, 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 Af is a finite system of ideals,

then

n a p implies a C p for some a E A,. a 0

The proof of the following lemma is essentially the proof of Theorem 1 from [37].

Lemma 86 Let a be an ideal, f be a filter, a n f = 0 and b be an ideal containing a, which

is maximal with respect to the exclusion off . Then b is prime, i.e, the complement of

6, (i.e. L - 6) is a filter.

Proof: 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 S1 E f such that S1 = A1 + GI, G' E (G) and A1 E 6.

Similarly we have S2 E f such that S2 = A2 + HI, HI E (H) and A2 E 6. Hence

CHAPTER 4. STRUCTURE OF COLOR CLASSES 8 1

SI x S2 = (A1 + GI) x (A2 + HI) =: (A1 x A2) + (A, x HI) + (GI x A,) + (G' x HI) E 6,

since G' x H' < G x H E b and the other three terms obviously belong to 6. Since f

is a filter, SI x S2 E f n 6, which is a contradiction. U

We will also need the following lemma. It is a dualization of the previous lemma,

and hence its proof can be obtained from it by interchanging the roles of + with X ,

< with 2, and ideal with filter.

Lemma 87 Let a be an ideal, f be a fdter, a n f = 0 and e be a filter containing f, which

is maximal with respect to the exclusion of a. Then the complement ,C - e is an ideal.

An ideal m 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 Zorn7s 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 = (0) and f = {I ) , then the resulting

ideal is a maximal ideal. Hence we have the following proposition.

Proposition 88 Every lattice with 0 # 1 has at least one maximal ideal.

Proposition 89 I f m is maximal ideal i n C, then m is prime.

Proof: F'ollows from Lemma 86 taking f = (1) and a = m. 0

In our particular case there is exactly one maximal ideal in c, the ideal of classes

of equivalent graphs which contain loopless graphs, i.e. m = c- {( I ) ) (and we do not

need Zorn7s lemma in this case).

Proposition 90 The following conditions are equivalent:

(i) p is prime,

( i i ) (p : a) E {p, (1)) for all ideals a,

(iii) ( p : a) E {p, (1)) for all principal ideals a.

CHAPTER 4. STRUCTURE OF COLOR CLASSES

Proof: (i) + (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)) f p , since H @ p and also ( p : (G)) f ( I ) , since G @ p. CJ

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

(1) i f a c p (p : a) =

p otherwise.

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

Lemma 91 Let a be an ideal which does not contain G. Then a is contained in a

prime ideal which also does not contain G. Moreover, there exists a prime ideal p

satisfying these conditions which is minimal with respect to the inclusion.

Proof: Let f be the smallest filter which contains G, i.e., 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. CI

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 0 and 1 and Spec(L) be the set of all prime

ideals of L. For each subset E of L, let V(E) denote the set of all prime ideals which

contain E. For G E ,C 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

(i i ) V(0) = Spec(L), V ( l ) = 0,

(i i i) n , , ~ ~ ( w = ~ ( u , , , E,) , (iv) V(a) U V(b) = V(a n b),

(v) V(b) C V(a) if and only if a s b. We will prove only the condition (v). Assume that a C 6. It follows that if p C b then

p C a and hence V(b) 2 V(a). On the other hand let us assume that a 6. Then there

is G E a such that G @ b. By Lemma 91 it follows that there is p E V(b) such that

G # p and hence p cannot contain a as a subset. Therefore p # V(a) and we conclude

that V(b) 9 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 called the prime spectrum of C.

The open sets of the form Spec(L) - V(G), G E ,C 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 ) = ncEE V(G). Moreover, one can observe

that

(;) D(G) n D(H) = D(G x I I ) ,

(z) D(G) = 0 if and only if G = 0,

(z) D(G) = Spec(L) if and only if G = 1,

(z) D(G) = D(H) if and only if G = H,

(c) D(G) is compact, i.e., every open covering of D(G) has a finite subcovering.

In particular, (a) and (z) imply that Spec(,C) is compact.

For an arbitrary subset A C Spec(L), the ideal

is called the ideal of A. One can easily observe that J(UiEI A;) = niEI J(A;).

Lemma 92 For every E C L and A Spec(L) we have E J ( A ) if and only if

A 5 V ( E ) .

CHAPTER 4. STRUCTURE OF COLOR CLASSES 84

Proof: Let us assume that E C J(A) . It follows by the condition (v) that V ( J ( A ) ) S

V ( E ) . 13ut A C V ( J ( A ) ) , since if p E A then J ( A ) = nqfAq C 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 all p E A, p > E and hence also J(A) = npEA p C-- E.

Proposition 93 V ( J ( A ) ) = 2 is the closure of A in Spec(L) for every A C_ Spec(C),

and J ( V ( E ) ) = ( E ) is the ideal generated b y E in L for every E C L.

Proof: 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 ) ) . CI

Let us recall that a topological space X is Hausdorf if for every two distinct points

x , y E X there exist two disjoint open sets U,(y) , V y ( x ) 5 X such that z E U x ( y ) and

y E Vy(x) . In particular if X is Hausdorf, then all points x E X are closed, since

X - ( x } = U y f x V y ( x ) is open. The spectrum Spec(L) with the Zariski topology is

not Hausdorf, since any two nonernpty 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 all 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 all prime ideals q with p C_ q. In ~ar t icular , a

point of Spec(L) is closed if and only if it is a maximal ideal.

Let a be a proper ideal in L and let us denote the set of all minimal prime ideals

containing a (with respect to the inclusion) by Ma, i.e., Ma = ( p E Spec(L) I P 2 a and p 2 q > a implies p = q for all q E Spec(L)). Lemma 91 implies that MU f 0 (take G = 1). Ma is a subset of Spec(C) and we endow Mu with the induced topology

CHAPTER 4. STRUCTURE OF COLOR CLASSES

of Spec(C), i.e., Y 5 hla is closed in Ma if and only if Y = V f l Ma for some

closed subset V of Spec(C). One can observe that all closed sets are of the form

Va(E) = V ( E ) n Ma for some E 5 C and the sets Da(G) = D(G) n Ma, G E ,C form

a basis for the open sets of the induced topology on Ma.

It follows that all points of Ma are closed in Ma. We will 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 1351.

Lemma 94 Let a be a proper ideal and p a subset of C . Then the following conditions

are equivalent:

( P E M ~ , (ii) C - p is a filter that is mazimal with respect to missing a,

(iii) a 5 p, p is a prime ideal and for every G E p, there is an H @ p such that

G x H E ~ .

Proof: (i) + (ii): If p E Ma then p is prime and hence C - p is a filter. By Zorn's

lemma one can expand C - 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 C - p which implies that q C p

and hence the minimality of p implies that q = p. Thus C - p = f.

(ii) + (iii): Since f = C - p is a filter that is maximal with respect to missing a, a C p.

Moreover, by Lemma 87, 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 L - p ) . Then f is a filter that properly

contairzs C - p. So there is some H E C - p such that G x H E a.

(iii) 3 (i): Assume that a c q C_ p, where q is a prime ideal. If there exist some

G E p - q, then there is an H @ 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 86

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 Ma. Since both of them are minimal and distinct, we have p 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 p , also Q' = Q x G does not belong to p but Q' belongs to q, since Q does. Hence

p E D(Q'), q E D ( P ) and D ( P ) n D(Q1) = D ( P x Q'). But D ( P x Q') n Ma = 8, 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 Da(P) = D ( P ) n Ma

and Da(Q1) = D(Qt) n Ma are disjoint, p E Da(Q1), q E Da(P) and hence Ma is a

Hausdorf space. 0

Let us recall that a topological space is discrete if all of its subsets are clopen,

i.e., they are both closed and open. If X is a finite Hausdorf space, then all subsets

of X are closed (and hence also open) since all 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 identify all points of

Ma which are open (and hence clopen) in Ma.

Proposition 96 Va(J(A)) is the closure of A in hfa for every A C_ Ma) 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) = nPEMap. Obviously a C b since a is a subset of every p from Ma # 0. Assume that G $ a belongs to b. 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 all minimal prime ideals containing

a. We conclude that a = b.

CHAPTER 4. STRUCTliRE OF COLOR CLASSES

The last equality follows, since

We say that an ideal b is a-divisorial, if b = (a : (a : 6)). We denote the set of all

prime ideals which are a-divisorial by Oa. We will prove later that Oa is a subset of

Ma and that Oa is the set of all clopen points of Ma.

Lemma 97 The ideal b is a-divisorial if and only if there exists an ideal c such that

b = ( a : c).

Proof: 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

85 (e) implies that c 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 85 (e). 0

It follows from Lemma 85 (b) that ,C = (1) = (a : a) and from Lemma 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 C b C (1).

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

a-divisorial.

Proof: Let us assume that a C (a : p). Then we have (a : p) f l (a : (a : p)) = a p

and since p is prime (a : p ) C p or (a : (a : p)) C_ p. But (a : p) C p would imply that

(a : p) = p n (a : p) = a n p = a which is a contradiction with our assumption a c (a : p).

We conclude that (a : (a : p)) C 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) = (1)) which is a contradiction. 0

CHAPTER 4. STRUCTURE OF COLOR CLASSES 88

An a-divisorial ideal b is called a ma.rima1 a-divisorial 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 if and only if p is a maximal a-divisorial ideal.

Proof: Assume that p is prime a-divisorial ideal but it is not maximal a-divisorial ideal,

i.e., p C b for some a-divisorial ideal b # (1). Lemma 85 (1) implies that (a : b) C_ (a : p).

Since p is a-divisorial, a 2 p and Lemma 85 ( f ) implies that b n (a : b) = a n b C_ a C_ p.

But p is prime and hence (a : 6) 2 p, since b C_ p is impossible due to our assumption

b > p. The inclusions (a : b) C (a : p) and (a : b) C p imply (a : 6) 5 p n (a : p) = p n a = a

and hence by Lemma 85 (b) and (I), (1) = (a : a) 2 (a : (a : 6)) = b, which is a

contradiction.

On the other hand assume that p is maximal a-divisorial ideal and suppose b and c are

ideals such that b n c C_ p but b p and also c p. It can be assumed that p C b and

p C_ c, otherwise one can replace b with p + b and c with p + c. Then p c b C_ (p : c) (1). If we 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 equal to (1). This would

imply that c C p, which is a contradiction with our assumptions. 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.

Lemma 100 Oa C Ma.

Proof: Assume that p E Oa. If p is not minimal, then there exists a prime ideal q such

that a C_ q c p c (1). Lemma 85 (1) and Lemma 98 imply that a c (a : p) (a : q) and

hence by Lemma 98 it follows that q is a-divisorial. But this is a contradiction, since

by Lemma 99 q is also maximal a-divisorial ideal. 0

For open subsets of Ma we have the following lemma.

CHAPTER 4. STRUCTURE OF COLOR CLASSES S9

Lemma 101 If a subset A 5 Ma is open in Ma then J ( A ) is a-divisorial ideal. On

the other hand, if b is a-divisorial ideal, then there exists a subset A c Ma which is

open in Ma and b = J(A).

Proof: Let us assume that the subset A C Ma is open in Ma. Therefore A =

UiE1 Da(Gi) and hence by the Proposition 96

J ( A ) = J(U Da(G)) = n J(Da(G,)) = n ( a : (Gi)) = (a : (U(Gi))) . i€I i€I i E 1 i E I

Therefore by Lemma 97, J ( A ) is a-divisorial 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 all open (and hence clopen) points of Ma.

Theorem 102 Let p be a point of Ma. Then the following conditions are equivalent:

( i) {p) is clopen in Ma,

(ii) P E Oa,

( 2 2 2 ) {p) = Da(G) for some G E &, i.e., {p} is a basic open set in Ma.

Proof: ( i) =+ (ii): Let us assume that {p) is clopen and hence open in Ma. From

Lemma 101 it follows that J ({p}) = p is a-divisorial ideal and hence p E Oa.

( 2 2 ) =+ (iii): Let us assume that p E Oa. Since p is a-divisorial, p = (a : (a : P)).

Moreover, by Lemma 99 p is maximal a-divisorial ideal and by Lemma 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 ) ) 2 (a : (a + (G))) c (1). Since p is maximal, p = (a : (a + (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 Da(G), then p = J (Da(G)) C q and hence q =

(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.

It follows frorn this theorem that Ma is discrete if and only if Ma = Oa, i.e., all

points of Ma are clopen.

4.6.2 Factorization

In this section we define and give necessary and sufficient conditions for an existence of

factorizations in distributive lattice with 0 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 factorization. The factorization is called irredundant if

a # n p for each q E A. P#'l

The existence of a factorization follows from Proposition 93, since we have a =

J(V(a)) = npEv(a) p. Unless a is an intersection of maximal ideals, there are ideals p, q

in V(a) such that p C q and hence this factorization is obviously redundant. Another

factorization follows from the Proposition 96, since we have a = J(Ma) = npEMa p and

we will call this factorization the natural factorization.

Lemma 103 If a proper ideal a has a finite factorization, then i t 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.

Lemma 104 Ifa = OPE, p is a n irredundant factorization, then A C_ Ma.

Proof: Let us assume that q E A, but q (t' Ma. Then by Zorn's Lemma there is

q' E Ma such that 4' c p. Since q n n p , e q ~ = a C_ q' and q 9 q', it follows that

CHAPTER 4. STRUCTURE OF COLOR CLASSES 91

npfqp C qt and therefore npfqp 2 qt c q, which is a contradiction with our assump-

tion that the factorization is irredundant.

Now we will prove the uniqueness of irredundant factorization.

Theorem 105 The factorization a = npEA p is irredundant if and only if A = Oa. In

particular, the irredundant factorization is unique.

Proof: Let us assume that the factorization a = npEA p is irredundant. To prove that

Oa C_ A, let us assume that q E Oa. Then q = (a : (G)) for some G E L by Proposition

102 and hence

= (a : (GI) = ( n : (GI) = n (P : (GI) = n 2 PEA PEA G@P

for all p such that G $ p. If G E p for all p E A, then q = ( I ) , which is a contradiction.

Since by Lemma 104, A C Ma we conclude that q = p E A. To prove the reverse

inclusion A C Oa, let us assume that q E A. Since the factorization is irredundant,

nPfq P q and hence there is G E npfq p such that G $ q. Therefore

This implies that q is a-divisorial and hence q E Oa.

To prove the other implication we need to show that if a = n p E o a p , then this factor-

ization is irredundant. Let q E Oa. Then by Lemma 98

and therefore this factorization is irredundant.

An obvious corollary is the following.

Corollary 106 The natural factorization o fa 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.

Theorem 107 A proper ideal a has an irredundant factorization if and only if the

set Oa is dense in Ma.

Proof: 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 Ma is Ma and hence by Propo-

sition 96 Ma = Va(J(Oa)). This implies that a = J(Ma) = J(Va(J(0a))) = J(Oa) by

Proposition 93. 0

We have the following equivalent conditions for the natural factorization to be

irredundant.

Theorem 108 The following conditions are equivalent.

(i) the natural factorization of a is irredundant,

( 2 2 ) for each q E Ma, npfq P $Z 4 ,

(iii) Ma = Oa, i.e., Ma is a discrete space.

Proof: ( i) * (ii), since a = n,,, P = n p t q p if and only if np+q P & 4 . ( i ) e (iii) is

Corollary 106. 0

For finite factorizations we need the following lemma.

Lemma 109 If a has a finite factorization then Ma = Oa.

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 C Ma. To prove

the other inclusion, assume that q E Ma. Then n p E o , p = a C q and since Oa is finite,

there is p E Oa such that p q. From the minimality of q it follows that q = p E Oa. 0

Now we can prove the following equivalent conditions for the existence of a finite

factorization.

CHAPTER 4. STRUCTURE OF COLOR CLASSES 9 3

Theorem 110 Let a be a proper ideal. Then the following conditions are equivalent:

( i ) a has a finite factorization,

( i i ) for each q t Ma, 4 9 Uppq P ,

( i i i ) the natural factorization of a is Jinite, i.e., Ma is finite.

Proof: We will prove that ( i ) e ( i i i ) and ( i i ) ( i i i ) .

(i) (iii): If a has a finite factorization then it has an irredundant factorization and

hence Oa is finite b y Theorem 105. Moreover, Ma = Oa by Lemma 109 and hence Ma

is finite.

( i i i ) =+ ( i ) : This implication is obvious.

( i i i ) + ( i i ) : This implication follows from Theorem 80.

( i i ) =+ ( i i i ) : Assume that for each q E Ma, q g U p f q p. For a contradiction, let us

assume that Ma is infinite. Since Va(l) = 0 and V a ( G ) = Ma for every G E a , the set

is a subset of C which is disjoint from a and 1 E f. We claim that f is a filter. This

claim follows, since if G E f and G < H then V a ( H ) < Va(G) , which implies that

H E f. If G , H € f then Va(G x H ) = V a ( G ) U V a ( H ) , which implies that G x H E f.

By Zorn's lemma we can enlarge f to a maximal filter e which is disjoint from a and

by Lemma 94, e = C - q for some minimal prime ideal q E Ma. Since q g U p f q p , there

is G E q such that G e p for each p f q from Ma. Since G E q , G e e = C - q. On the

other hand V a ( G ) = {q) which implies G E f C e, which is the desired contradiction. 0

The following proposition follows from definitions.

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

( W ( G ) ) .

This means that if a and b are principal ideals in a Heyting algebra, then also

( a : 6) is a principal ideal. Together with Proposition 102 ( i i i ) this implies that if a

is principal, then all a-divisorial prime ideals are principal, since {p) = Da(G) implies

that p = J ( { p ) ) = J ( D a ( G ) ) = ( a : ( G ) ) by Proposition 96. This means that if a

CHAPTER 4. STRUCTURE OF COLOR CLASSES 94

is principal, then all ideals from Oa are principal. The following proposition is also

immediate from the definition.

Proposition 112 A principal ideal ( P ) 2 is prime if and only if the class of

equivalent graphs P is multiplicative.

Now we can give the promissed necessary and sufficient conditions for finite factor-

ization of principal ideals in a Heyting algebra (and hence graphs up to the relation

4-

Theorem 113 Let 3-1 be a Heyting algebra, 1 # W E 3-1 and let a = ( W ) . Then the

following conditions are equivalent:

( i ) a has a finite factorization,

( i i ) Ma = Oa,

( i i i ) all minimal prime ideals containing W are principal.

Proof:

( i ) + ( i i ) : This implication follows from Lemma 109.

( i i ) j ( i i i ) : Let us assume that Ma = Oa. Since all ideals from Oa are principal also

all minimal ideals containing W , i.e., ideals from Ma are principal.

( i i i ) ( i ) : Let us assume that all minimal ideals containing W (i.e. ideals from

Ma) are principal. We will show that the condition ( i i ) of Theorem 110 is satisfied.

We may assume that (Mal = m, otherwise the natural factorization is a finite fac-

torization. For a contradiction assume that for some q E Ma, q 2 U p f q p . Since all

minimal prime ideals containing W are principal, q = ( G ) for some G E 'FI. Therefore

G E q 5 Upfq p and hence G E p for some p # q. It follows that q = ( G ) 2 p, which is

a contradiction with the minimality of p .

4.7 Conclusion

We will summarize 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

CM.4PTER 4. STRUCTURE OF COLOR CLASSES

many meet irreducible elements. An example is any infinite Boolean algebra. Let us

consider the following properties of a distributive lattice L:

P I . 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 Corollary 56 that the Heyting algebra c is

countable distributive lattice and satisfies P1, P 3 and P4. In [7] it is shown that

P1 and P 3 imply P4 and that P I , P2, and P 3 are both necessary and sufficient

conditions for a countable L to be projective in the variety of all 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 1{4 has

a finite factorization. Another condition satisfied by G is:

P5. If G E G is a join irreducible element, then (Wl + Wz)(G) = Wl(G) + Wz(G) for

all Wl, Wz E c. This condition follows from Proposition 55. The conditions P 1 and P 2 are not true in

any infinite Boolean algebra. The conditions P 3 and P4 are not true in any Boolean

algebra and the condition P 5 is true in every Boolean algebra. We do not know

whether every Heyting algebra which satisfies four conditions P I , P3 , P4, and P 5

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 is W-minimal, if it

satisfies the following four conditions

1. (1st ideal condition) G E S and H -+ G imply H E S,

2. (2nd ideal condition) G E S and H E S imply G + I i E S,

3. (primality condition) G x IJ E S implies G E S or H E S ,

4. ( W-minimality condition) G E 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. (principality condition) there exists G E S such that H -, G for all 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 Further Research

In this thesis we have studied graph homomorphisms, equitable graph homomor-

phisms, and the corresponding H-coloring problems from both theoretical and prac-

tical view points.

In Chapter 2, we have completely characterized the complexity of the equitable

N-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 complete 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 algorithm or to find a better one.

In Chapter 3, we have proposed relaxations of graph homomorphism. In particu-

lar 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 char-

acterization 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 graphs for which

either fractional homomorphism or pseudo-homomorphism would give a polynomial

CHAPTER 5. CONCLUSIONS AND FURTHER RESEARCH 9 S

algorithm, since all of our examples are already known to be polynomial time solvable

by other means. It seems that using semidefinite programming 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 sufficient 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 Kl, K2, Ic3 and all cycles are multiplicative and that if G f+ H and

H f , G then G x H is not multiplicative (cf. [22]). Therefore it is easy to construct

a graph which is not multiplicative, e.g., K3 x Mll, where Mll is the Mycielski graph

from Figure 3.1. We have shown that if is multiplicative then there are infinitely

many complete graphs which are multiplicative. We have also shown that if I(4

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 K4.

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