Hedetniemi's Conjecture Via Alternating Chromatic Number

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Hedetniemi’s Conjecture ViaAlternating Chromatic Number

Meysam Alishahi† and Hossein Hajiabolhassan∗

† Department of Mathematics

Shahrood University of Technology, Shahrood, Iran

meysam [email protected]∗ Department of Applied Mathematics and Computer Science

Technical University of Denmark

DK-2800 Lyngby, Denmark∗ Department of Mathematical Sciences

Shahid Beheshti University, G.C.

P.O. Box 19839-63113, Tehran, [email protected]

Abstract

In an earlier paper, the present authors (2013) [1] introduced the alternatingchromatic number for hypergraphs and used Tucker’s Lemma, an equivalentcombinatorial version of the Borsuk-Ulam Theorem, to show that the alternat-ing chromatic number is a lower bound for the chromatic number. In this paper,we determine the chromatic number of some families of graphs by specifyingtheir alternating chromatic number. We define matching-dense graph as a graphwhose vertex set is the set of all matchings of a specified size of a dense graphand two vertices are adjacent if the corresponding matchings are edge-disjoint.We determine the chromatic number of matching-dense graphs in terms of thegeneralized Turan number of matchings. Also, we consider Hedetniemi’s con-jecture which asserts that the chromatic number of the Categorical product oftwo graphs is equal to the minimum of their chromatic numbers. By topologicalmethods, it has earlier been shown that Hedetniemi’s conjecture holds for anytwo graphs of the family of Kneser graphs, Schrijver graphs, and the iteratedMycielskian of any such graphs. We extend these results to other graphs such asa large family of Kneser multigraphs, matching graphs, and permutation graphs.

Keywords: Chromatic Number, Hedetniemi’s Conjecture, General KneserHypergraph,Turan Number.Subject classification: 05C15

1 introduction

For a hypergraph H, the vertex set of the general Kneser graph KG(H) is the setof all hyperedges of H and two vertices are adjacent if the corresponding hyper-edges are disjoint. It is known that any graph can be represented by a generalKneser graph. In this paper, we can see that a graph has various representations,nevertheless we can determine the chromatic number of some graphs, by choosingan appropriate representation for them. In view of Tucker’s Lemma, an equivalentcombinatorial version of the Borsuk-Ulam Theorem, the present authors [1, 2] in-troduced the alternating chromatic number for hypergraphs which provides a tight

1The research of Hossein Hajiabolhassan is supported by ERC advanced grant GRACOL.

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lower bound for the chromatic number of hypergraphs. Next, they determined thechromatic number of Kneser multigraphs, multiple Kneser graphs and a family ofmatching graphs, see [1, 2]. A matching graph and a Kneser multigraph can berepresented by general Kneser graphs KG(H) and KG(G), respectively, such thatthe hyperedges of H are corresponding to all matchings of a specified size of a graphand the hyperedges of G are corresponding to all subgraphs isomorphic to some fixedprescribed simple graphs of a multigraph such that the multiplicity of each edge is atleast 2. Note that a Schrijver graph is a matching graph KG(H), where the hyper-edge set of H is corresponding to all matchings of a specified size of a cycle. Hence,by determining the chromatic number of matching graphs, we generalize Schrijver’sTheorem. A challenging and interesting problem in graph coloring is Hedetniemi’sconjecture which asserts that the chromatic number of the Categorical product oftwo graphs is the minimum of that of graphs. There are a few general results aboutthe chromatic number of the Categorical product of graphs whose chromatic num-ber is large enough. In view of topological bounds, it has earlier been shown thatHedetniemi’s conjecture holds for any two graphs for which the topological boundon the chromatic number is tight, see [8, 22, 25].

This paper is organized as follows. In Section 1, we set up notations and termi-nologies. In particular, we will be concerned with the definition of the alternatingchromatic number and the strong alternating chromatic number and we mentionsome results about them. In particular, we show that they provide tight lowerbounds not only for the chromatic number of graphs but also for some well-knowntopological parameters. Also, we introduce alternating Turan number which is ageneralization of the Turan number. Next, we introduce a lower bound for the chro-matic number in terms of the alternating Turan number. In fact, if one can showthat the alternating Turan number is the same as the Turan number for a family ofgraphs, then we can determine the chromatic number of some family of graphs. InSection 2, we determine the chromatic number of any matching-dense graph KG(H),where the hyperedges of H are corresponding to all matchings of a specified size ina dense graph. As a consequence, we determine the chromatic number of permu-tation graphs provided that the number of vertices is large enough. In Section 3,we generalize the definitions of alternating and strong alternating chromatic num-ber of graphs and in view of an appropriate representation for the Mycielskian ofa graph, we show that the generalized alternating chromatic number behave likethe chromatic number for the Mycielskian of a graph. Precisely, we show that thegeneralized alternating chromatic number of the Mycielskian of a graph G is at leastthe generalized alternating chromatic number of G plus one. Also, we show that thegeneralized strong alternating chromatic number satisfies Hedetniemi’s conjecture.By topological methods, it has earlier been shown that Hedetniemi’s conjecture holdsfor any two graphs of the family of Kneser graphs, Schrijver graphs, and the iteratedMycielskian of any such graphs. We extend this result to other graphs such as alarge family of Kneser multigraphs, matching graphs, and permutation graphs.

2

2 Notations and Terminologies

In this section, we setup some notations and terminologies. Hereafter, the symbol[n] stands for the set {1, 2, . . . , n}. A hypergraph H is an ordered pair (V (H), E(H)),where V (H) is a set of elements called vertices and E(H) is a set of nonempty subsetsof V (H) called hyperedges. Unless otherwise stated, we consider simple hypergraphs,i.e., E(H) is a family of distinct nonempty subsets of V (H). A vertex cover T of His a subset of its vertices such that each hyperedge of H is incident to some vertexof T . Also, a k-coloring of a hypergraph H is a mapping h : V (H) −→ [k] such thatfor any hyperedge e, we have |{h(v) : v ∈ e}| ≥ 2. In other words, no hyperedge ismonochromatic. The minimum k such that there exists a coloring h : V (H) −→ [k]is called the chromatic number of H and is denoted by χ(H). Note that if thehypergraph H has some hyperedge with cardinality 1, then there is no k-coloringfor any k. Therefore, we define the chromatic number of such a hypergraph to beinfinite. A hypergraph H is k-uniform, if all hyperedges of H have the same size k.By a graph G, we mean a 2-uniform hypergraph. Also, let o(G) denote the numberof odd components of G.

For two graphs G and H, a mapping f : V (G) −→ V (H) is called a homo-

morphism from G to H, if it preserves the adjacency, i.e., if xy ∈ E(G), thenf(x)f(y) ∈ E(H). For brevity, we use G −→ H to denote that there is a homomor-phism from G to H. If we have both G −→ H and H −→ G, then we say G andH are homomorphically equivalent and show this by G ←→ H. Note that χ(G) isthe minimum integer k for which there is a homomorphism from G to the completegraph Kk. An isomorphism between G and H is a bijection map f : V (G) −→ V (H)such that both f and f−1 are homomorphism. For brevity, we use G ∼= H to men-tion that there is an isomorphism between G and H. Also, if G ∼= H, then we sayG and H are isomorphic. For a subgraph H of G, the subgraph G \H is obtainedfrom G by deleting the edge set of H. Also, G−H is obtained from G by deletingthe vertices of H with their incident edges.

The general Kneser graph KG(H) has E(H) as vertex set and two vertices areadjacent if the corresponding hyperedges are disjoint. It is a well-known result thatfor any graph G, there is some hypergraph H such that KG(H) ∼= G.

A subset S ⊆ [n] is said to be s-stable if s ≤ |i−j| ≤ n−s for any distinct i, j ∈ S.Hereafter, for a subset A ⊆ [n], the symbols

(

Ak

)

and(

Ak

)

sstand for the set of all k-

subsets and all s-stable k-subsets of A, respectively. The Kneser graph KG(n, k) ands-stable Kneser graph KG(n, k)s−stab have all k-subsets and all s-stable k-subsets of[n] as their vertex sets, respectively. Also, in these two graphs, two vertices are

adjacent, when the corresponding sets are disjoint. If we set H1 =(

[n],([n]k

)

)

and

H2 =(

[n],([n]k

)

s

)

, then KG(H1) ∼= KG(n, k) and KG(H2) ∼= KG(n, k)s−stab. The

graph KG(n, k)2−stab = SG(n, k) is a well-known graph celled Schrijver graph. In1955, Kneser conjectured that χ(KG(n, k)) = n − 2k + 2. Lovasz [17], by usingthe Borsuk-Ulam theorem, proved this conjecture. Next, this was improved bySchrijver [20] who proved that the Schrijver graph SG(n, k) is a critical subgraph ofthe Kneser graph KG(n, k) with the same chromatic number.

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2.1 Alternating Chromatic Number

Suppose that H ⊆ 2[n] is a hypergraph, where n is a positive integer. Consider X ∈{−1, 0,+1}n \ {(0, 0, . . . , 0)}. Define X+ = {i ∈ [n] : xi = +1} and X− = {i ∈ [n] :xi = −1}. By abuse of notation, we set X = (X+,X−). Throughout this paper, Weuse interchangeably these two kinds of presentations of X, i.e., X = (x1, . . . , xn) orX = (X+,X−). The sequence x1, x2, . . . , xm ∈ {−1,+1} is said to be an alternating

sequence, if any two consecutive terms are different. For any X = (x1, x2, . . . , xn) ∈{−1, 0,+1}n \ {(0, 0, . . . , 0)}, the alternating number of X, alt(X), is the length ofa longest alternating subsequence of nonzero terms of (x1, x1, . . . , xn). Note thatwe consider just nonzero entries to determine the alternating number of X. Fora linear ordering (or a permutation) σ = (i1, i2, . . . , in) of [n] , define σ(j) = ij ,where 1 ≤ j ≤ n. Also, we sometimes use the usual notation for a linear orderingof [n], i.e., σ : i1 < i2 < · · · < in, and we use interchangeably these two kindsof presentations of any linear ordering. For a linear ordering σ = (i1, i2, . . . , in)of [n], set X+

σ = {σ(i) ∈ [n] : xi = +1}, X−σ = {σ(i) ∈ [n] : xi = −1}, and

Xσ = (X+σ ,X−

σ ).For any hypergraph H ⊆ 2[n] and σ ∈ Sn, define altσ(H) (resp. saltσ(H)) to

be the largest integer k such that there exists an X ∈ {−1, 0,+1}n \ {(0, 0, . . . , 0)}with alt(X) = k and that none (resp. at least one) of X+

σ and X−σ contains any

hyperedge of H. If for each X ∈ {−1, 0,+1}n \ {(0, 0, . . . , 0)}, either X+σ or X−

σ hassome hyperedge of F , then we define altσ(F) = 0. Note that altσ(H) ≤ saltσ(H)and equality can hold. Now, define alt(H) = min{altσ(H) : σ ∈ Sn} and salt(H) =min{saltσ(H) : σ ∈ Sn}. Now, we define the alternating chromatic number andstrong alternating chromatic number of a graph G, respectively, as follows

χalt(G) = maxF{|V (F)| − alt(F) : KG(F) ∼= G}

andχsalt(G) = max

F{|V (F)| + 1− salt(F) : KG(F) ∼= G} .

Remark 1. For any hypergraph H on n vertices, in view of the definition of

altσ(H) where σ is an ordering of the vertex set of H, throughout this paper, we

assume that V (H) was identified with the set [n]. We may represent V (H) with

different sets, nevertheless we can consider any representation as a relabeling of the

set [n].

It was proved in [1, 2] that both alternating chromatic number and strong alter-nating chromatic number provide tight lower bounds for the chromatic number ofgraphs.

Theorem A. [1] For any graph G, we have

χ(G) ≥ max {χalt(G), χslat(G)} .

In the sequel, we introduce another proof for the aforementioned theorem. Infact, we show that the alternating chromatic number and the strong alternating

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chromatic number provide tight lower bounds for some well-known topological pa-rameters. We refer the reader to [18, 21] for some basic definitions of algebraictopology such as two kinds of box complexes B(G) and B0(G). There are severalcharming topological lower bounds for the chromatic number of an arbitrary graphG as follows (see [18, 21])

χ(G) ≥ ind(B(G)) + 2 ≥ ind(B0(G)) + 1 ≥ coind(B0(G)) + 1 ≥ coind(B(G)) + 2.

The proof of the next lemma is based on ideas similar to those used in an in-teresting proof of Ziegler (see page 67 in [18]) for Gale’s Lemma and Proposition 8of [21].

Lemma 1. For any graph G, the following inequalities hold.

a) χ(G) ≥ coind(B0(G)) + 1 ≥ χalt(G),

b) χ(G) ≥ coind(B(G)) + 2 ≥ χsalt(G).

Proof. Let F ⊆ 2[n] and that KG(F) is isomorphic to G such that there existsa σ ∈ Sn with χalt(G) = n − altσ(KG(F)). Note that F may have some isolatedvertices and they seem like extra notations. Although, note that isolated vertices aresometimes useful to introduce an appropriate ordering to determine the alternatingchromatic number. Set m = χalt(G)− 1. Consider the following curve

γ = {(1, t, t2, . . . tm) ∈ Rm+1 : t ∈ R}.

Set W = {w1, w2, . . . , wn} where wi = γ(i), for i = 1, 2, . . . , n. Now, consider thesubset {v1, v2, . . . , vn} of points of Sm, where vi = (−1)i Wi

||Wi||, for any 1 ≤ i ≤ n.

Now, suppose that σ(i) ∈ [n] is identified with vi, for any 1 ≤ i ≤ n. It can bechecked that every hyperplane passing trough the origin intersects γ in no morethan m points. Moreover, if a hyperplane intersects the curve in exactly m points,then the hyperplane cannot be tangent to the curve; and consequently, at eachintersection point, the curve passes from one side of the hyperplane to the otherside.

Now, we show that for any x ∈ Sm, the open hemispheres H(x) or H(−x) con-tains some member of F . On the contrary, suppose that there is an x ∈ Sm suchthat neither H(x) nor H(−x) contains any member of F . Let h be the hyperplanepassing trough the origin which contains the boundary of H(x). We can move thishyperplane continuously to a position such that it still contains the origin and has ex-actly m points of W = {w1, w2, . . . , wn} while during this movement no points of Wcrosses from on side of h to the other side. Consequently, during the aforementionedmovement, no points of V = {v1, v2, . . . , vn} crosses from one side of h to the otherside. Therefore, without loss of generality, we may assume that h intersects W atexactly m points of W . Hence, at each of this intersections, γ passes from one side ofh to the other side. Assume that h+ and h− be two open half-spaces determined bythe hyperplane h. Now, consider X = (x1, x2, . . . , xn) ∈ {−1, 0,+1}n\{(0, 0, . . . , 0)}such that

xi =

0 if wi is on h

1 if wi is in h+ and i is odd1 if wi is in h− and i is even−1 otherwise.

.

5

Assume that xi1 , xi2 , . . . , xin−mare nonzero entries ofX, where i1 < i2 < · · · < in−m.

It is easy to check that any two consecutive terms of xij ’s have different sings. SinceX has n − m = altσ(F) + 1 nonzero entries, we have alt(X) = altσ(F) + 1; andtherefore, either X+

σ or X−σ has some member of F . Now, one see that it implies

that either H(x) or H(−x) has some member of F . For any vertex A of KG(F) andany x ∈ Sm, define DA(x) to be the smallest distance of a point in A ⊂ Sm fromthe set Sm \H(x). Note that DA(x) > 0 if and only if H(x) contains A. Define

D(x) =∑

A∈F

(DA(x) +DA(−x)) .

For any x ∈ Sm, either H(x) or H(−x) has some member of F ; and therefore,D(x) > 0. Thus, the map

f(x) =1

D(x)

(

∑

A∈F

DA(x)||(A, 1)|| +∑

A∈F

DA(−x)||(A, 2)||

)

from Sm to ||B0(KG(F))|| is a Z2-map. It implies coind(B0(G)) ≥ m.To prove the second part, assume that G ⊆ 2[n] and that KG(G) is isomorphic

to G such that there exists a σ ∈ Sn with χslat(G) = n+ 1− saltσ(G). Now, in theproof of part (a), we set m = χslat(G)− 2. With the same argument as in the proofof part (a), one can see that for any x ∈ Sm, both H(x) and H(−x) have some

member of G. Define D(x) =∑

A∈G

DA(x). For any x ∈ Sm, H(x) has some member

of G; and therefore, D(x) > 0. Thus, the map

f(x) =1

2D(x)

∑

A∈G

DA(x)||(A, 1)|| +1

2D(−x)

∑

A∈G

DA(−x)||(A, 2)||

from Sm to ||B(KG(G))|| is a Z2-map. It implies coind(B(G)) ≥ m. �

In [1, 2], the chromatic number of several families of graphs was determined, bycomputing their alternating chromatic number or their strong alternating chromaticnumber. Note that, by considering the natural ordering of positive integers, onecan see that χ(SG(n, k)) = χsalt(SG(n, k)) = n − 2k + 2. Now, we show thatχ(SG(n, 2)) = χalt(SG(n, 2)) = n−2 and χ(SG(2k+1, k)) = χalt(SG(2k+1, k)) = 3.Note that χalt(SG(n, 2)) ≤ χ(SG(n, 2)) and also it is simple to show χ(SG(n, 2)) ≤n − 2. Hence, it is sufficient to show n − 2 ≤ χalt(SG(n, 2)). To see this, firstassume that n = 2t. Consider a 2-uniform hypergraph H = (V (H), E(H)), whereV (H) = {1, 2, . . . , 4t} = {1, 2, . . . , 2t, a1, . . . , a2t} and E(H) =

([2t]2

)

2. Note that

a1, . . . , a2t are isolated vertices of H. Consider the ordering σ for V (H) as follows

σ : 1 < a1 < t+ 1 < a2 < 2 < a3 < t+ 2 < a4 < 3 < a5 < t+ 3 < a6 < 4 < · · ·< a2t−4 < t− 1 < a2t−3 < 2t− 1 < a2t−2 < t < a2t−1 < 2t < a2t

For any vector X, set l(X) to be the number of nonzero terms of X. Now, weshow that if X = (x1, . . . , x4t) ∈ {−1, 0,+1}4t \ {(0, 0, . . . , 0)} and alt(X) = 2t+ 3,then either X+

σ or X−σ contains some hyperedge of H. To see this, it is sufficient to

6

prove it for any X such that alt(X) = l(X) = 2t + 3. Moreover, one can supposethat for any even integer i, we have xi 6= 0. Since otherwise, one can check thatthere exists a Y = (y1, . . . , y4t) = (Y +, Y −) ∈ {−1, 0,+1}4t\{(0, 0, . . . , 0)} such thatalt(Y ) = l(Y ) = 2t+3, Y +

σ ∩{1, 2, . . . , 2t} ⊆ X+σ ∩{1, 2, . . . , 2t}, Y

−σ ∩{1, 2, . . . , 2t} ⊆

X−σ ∩ {1, 2, . . . , 2t}, and yj 6= 0 for any even integer j. Consequently, if we show

that either Y +σ or Y −

σ contains some hyperedge of H, then the same assertion holdsfor X+

σ or X−σ . Now, since alt(X) = l(X) = 2t + 3 and for any even integer i,

we have xi 6= 0, one can see that there are exactly 3 odd integers for which theircoordinates in X are nonzero. Also, one can see that for any two odd integers i andj, if |σ(i) − σ(j)| = 1 or |σ(i) − σ(j)| = 2t − 1, then either |i− j| = 4 or i ∈ {1, 3}.Now, we claim that there is no odd integer 1 ≤ i ≤ 4t− 5, such that xi = xi+4 = +1(resp. xi = xi+4 = −1). Otherwise, since alt(X) = l(X) = 2t + 3 and that for anyeven integer j, we have xj 6= 0, one can see that xi+2 = +1 (resp. xi+2 = −1). Inthis case, one can see that X+

σ (resp. X−σ ) contains some hyperedge of H. Also,

for a contradiction, suppose x1 = x4t−1 = +1 (resp. x1 = x4t−1 = −1). If weconsider the other nonzero coordinate xj 6= 0 (j is an odd integer), then, in view ofalt(X) = 2t+3, xj should have the same sign with exactly one of x1 or x4t−1 whichis impossible. Similarly, if x3 = x4t−3 = +1 (resp. x3 = x4t−3 = −1), then one canprove the assertion.

Also, for n = 2t+1, consider a 2-uniform hypergraph H = (V (H), E(H)), whereV (H) = {1, 2, . . . , 4t + 2} = {1, 2, . . . , 2t + 1, a1, . . . , a2t+1} and E(H) =

(

[2t+1]2

)

2.

Also, consider the ordering σ as follows

σ : t+ 1 < a1 < 1 < a2 < t+ 2 < a3 < 2 < a4 < t+ 3 < a5 < 3 < a6 < t+ 4 < · · ·< a2t−3 < t− 1 < a2t−2 < 2t < a2t−1 < t < a2t < 2t+ 1 < a2t+1

Similarly, one can show χ(SG(n, 2)) = χalt(SG(n, 2)) = n− 2.In [2], it was shown that χ(SG(5, 2)) = χalt(SG(5, 2)) = 3. Similarly, one can see

that χ(SG(2k + 1, k)) = χalt(SG(2k + 1, k)) = 3. To see this, consider a k-uniformhypergraph H = (V (H), E(H)), where V (H) = {1, 2, . . . , 4k + 2} = {1, 2, . . . , 2k +

1, a1, . . . , a2k+1} and E(H) =([2k+1]

k

)

2. Note that a1, . . . , a2k+1 are isolated vertices

of H. Consider the ordering σ for V (H) as follows

1 < a1 < 3 < a2 < · · · < 2k + 1 < ak+1 < 2 < ak+2 < 4 < ak+3 < · · · < 2k < a2k+1.

One can check that χ(SG(2k + 1, k)) = χalt(SG(2k + 1, k)) = 3.

2.2 Alternating Turan Number

A hypergraph labeling is an assignment of labels, from a set of symbols, to thehyperedges or vertices, or both, of a hypergraph. Formally, given a hypergraph H, avertex (resp. hyperedge) labeling is a function from the vertices (resp. hyperedges)of H to a set of labels. For a family F of unlabeled hypergraphs and an unlabeledhypergraph H, a subhypergraph of H is an F-subhypergraph, if it is a member ofF . Moreover, for a labeled hypergraph H and a family F of labeled hypergraphs,a subhypergraph of H is an F-subhypergraph, if there is an isomorphism f betweenthis subhypergraph and a member of F such that f preserves labels. Note that anydirected graph can be considered as a labeled graph. The general Kneser hypergraph

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KG(H,F) (resp. KGv(H,F)) has all F-subhypergraphs of H as vertex set and twovertices are adjacent if the corresponding F-subhypergraphs are hyperedge-disjoint(resp. vertex-disjoint). Hereafter, unless otherwise stated, we consider unlabeledhypergraph. A hypergraph H is said to be F-free, if it has no subhypergraphisomorphic to a member of F . For a hypergraph G, define ex(G,F), the generalizedTuran number of the family F in the hypergraph G, to be the maximum number ofhyperedges of an F-free spanning subhypergraph H of G. A subhypergraph of Gis called F-extremal if it has ex(G,F)-hyperedges and it is F-free. We denote thefamily of all F-extremal subhypergraphs of G with EX(G,F). It is usually a hardproblem to determine the exact value of ex(G,F). The concept of Turan number wasgeneralized in [2] in order to find the chromatic number of some families of graphsas follows. Let H be a hypergraph with E(H) = {e1, e2, . . . , em}. For any orderingσ = (ei1 , ei2 , . . . , eim), a 2-coloring of a subset of hyperedges of H (with 2 colors redand blue) is said to be alternating (respect to the ordering σ), if any two consecutivecolored edges (in the ordering σ) receive different colors. Note that we may assign nocolor to some hyperedges of H. In other words, in view of the ordering σ, we assigntwo colors red and blue alternatively to a subset of hyperedges of H. The maximumnumber of hyperedges of H that can be colored alternatively (respect to the orderingσ) by 2-colors such that each (resp. at least one of) color class has no member ofF , is denoted by exalt(H,F , σ) (resp. exsalt(H,F , σ)). Now, we are in a position todefine the the alternating Turan number and strong alternating Turan number of thefamily F in the hypergraph H, exalt(H,F) and exsalt(H,F), respectively, as follows

exalt(H,F) = min{

exalt(H,F , σ); σ ∈ SE(H)

}

andexsalt(H,F) = min

{

exsalt(H,F , σ); σ ∈ SE(H)

}

.

For a hypergraphH, let F be a member of EX(G,F) and σ be an arbitrary orderingof E(H). Now, if we color the edges of F alternatively with two colors with respect tothe ordering σ, one can see that any color class has no member of F ; and therefore,ex(H,F) ≤ exalt(H,F , σ). Also, it is clear that if we assign colors to more than2ex(H,F) edges, then a color class has at least more than ex(H,F) edges. It impliesexalt(H,F , σ) ≤ 2ex(H,F). Consequently,

ex(H,F) ≤ exalt(H,F) ≤ 2ex(H,F).

Next lemma was proved in [2].

Lemma A. [2] For any hypergraph H = (V (H), E(H)) and a family F of hyper-

graphs,

|E(H)| − exalt(H,F) ≤ χ(KG(H,F)) ≤ |E(H)| − ex(H,F),

|E(H)| + 1− exsalt(H,F) ≤ χ(KG(H,F)) ≤ |E(H)| − ex(H,F).

The previous lemma enables us to find the chromatic number of some families ofgraphs. Assume that F is a family of graphs and G is a given graph. If we find an

8

appropriate ordering σ of the edges of G and prove that we have either exalt(G,F) =ex(G,F) or exsalt(G,F) − 1 = ex(G,F), then we can conclude that

χ(KG(G,F)) = |E(G)| − ex(G,F).

In this regard, in [1, 2], the chromatic number of several families of graphs wascomputed by this observation.

For a given 2-coloring of a subset of hyperedges of H, a spanning subhypergraphof H whose hyperedge set contains all edges such that we have assigned color red(resp. blue) to them, is termed the red subhypergraph HR (resp. blue subhypergraph

HB). Furthermore, by abuse of notation, any edge of HR (resp. HB) is termed ared edge (resp. blue edge).

3 Matching Graphs

In this section, we investigate the chromatic number of graphs via their alternat-ing chromatic number. In [2], for a sparse graph G, the chromatic number of thematching-sparse graph KG(G, rK2) was studied. In contrast, in the sequel, we de-termine the chromatic number of the matching graph KG(G, rK2) provided that Gis a dense graph.

3.1 Matching-Dense Graphs

Let G be a graph with V (G) = {u1, . . . , un}. The graph G is termed (r, c)-locallyEulerian, if there are edge-disjoint nontrivial Eulerian subgraphs H1, . . . ,Hn of Gsuch that for any 1 ≤ i ≤ n, we have ui ∈ Hi and that for any u ∈ V (Hi), whereu 6= ui, we have degHi

(ui) ≥ (r − 1)degHi(u) + c.

Lemma 2. Let r ≥ 2 and s be nonnegative integers. Also, let G be a graph with n

vertices and δ(G) >(

r+22

)

+(r−2)s. If there exists an (r+s, c)-locally Eulerian graph

H such that G is a subgraph of H, |V (H)| = |V (G)|+s, and c ≥(

r−12

)

+(s+3)(r−1),then χ(KG(G, rK2)) = χalt(KG(G, rK2)) = |E(G)| − ex(G, rK2).

Proof. In view of Lemma A, it is sufficient to show that there exists an orderingσ of the edges of G such that exalt(G, rK2, σ) = ex(G, rK2).

Assume that V (G) = {u1, . . . , un} and V (H) = {u1, . . . , un+s}. In view ofdefinition of (r, c)-locally Eulerian graph, there are pairwise edge-disjoint nontrivialEulerian subgraphs H1, . . . ,Hn+s of H such that for any 1 ≤ i ≤ n + s, we haveui ∈ Hi and that for any u ∈ V (Hi), where u 6= ui, we have degHi

(ui) ≥ (r + s −

1)degHi(u) +

(

r−12

)

+ (s+ 3)(r − 1).To find the ordering σ, add a new vertex x and join it to all vertices of H by

edges with multiplicity two to obtain the graph H ′. Precisely, for any 1 ≤ i ≤ n+ s,join x and ui with two distinct edges fi and f ′

i . Now, if H′ has no odd vertices, then

set H = H ′; otherwise, add a new vertex z to H ′ and join it to all odd vertices ofH ′ to obtain the graph H. The graph H is an even graph; and therefore, it has anEulerian tour.

9

Also, note that the graph K = H − x is an even graph; and accordingly, any

connected component of K ′ = K \

(

n+s⋃

i=1

Hi

)

is also an Eulerian subgraph. Without

loss of generality, assume that K1, . . . ,Kl are the connected components of K ′.Construct an Eulerian tour for H as follows. At ith step, where 1 ≤ i ≤ n+ s, startfrom the vertex x and traverse the edge fi. Consider an arbitrary Eulerian tour ofHi starting at ui and traverse it. Next, if there exists a 1 ≤ j ≤ l such that ui ∈ Kj

and the edge set of Kj is still untraversed, then consider an Eulerian tour for Kj

starting at ui and traverse it. Next, traverse the edge f ′i and if i < n+ s, then start

(i+1)th step. Construct an ordering σ for the edge set of the graph G such that theordering of edges in E(G) are corresponding to their ordering in the aforementionedEulerian tour, i.e., if we traverse the edge ei ∈ E(G) before the edge ej ∈ E(G) inthe Eulerian tour, then in the ordering σ we have ei < ej.

Now, we claim that exalt(G, rK2, σ) = ex(G, rK2). Note that for any r − 1vertices {ui1 , . . . , uir−1

} ⊆ V (G), we have

ex(G, rK2) ≥r−1∑

j=1

degG(uij )−

(

r − 1

2

)

.

Consider an alternating 2-coloring of the edges of G with respect to the orderingσ of length ex(G, rK2) + 1, i.e., we assigned ex(G, rK2) + 1 blue and red colorsalternatively to the edges of G with respect to the ordering σ. For a contradiction,suppose that both red spanning subgraph GR and blue spanning subgraph GB arerK2-free subgraphs. In view of the Berge-Tutte Formula, there are two sets TR

and TB such that |V (GR)| − o(GR − TR) + |TR| ≤ 2r − 2 and |V (GB)| − o(GB −TB) + |TB | ≤ 2r − 2. In view of these inequalities, one can see that |TR| ≤ r − 1and |TB| ≤ r − 1. Moreover, the number of edges of GR not incident to some

vertex of TR (|E(GR − TR)|) is at most(2r−2|TR|−1

2

)

. To see this, assume thatOR

1 , OR2 , . . . , O

RtR

are all connected components ofGR−TR, where tR ≥ o(GR−TR) ≥|V (GR)|+ |TR| − 2r + 2. We have

|E(GR − TR)| ≤

tR∑

i=1

(

|V (ORi )|

2

)

≤

(

tR∑

i=1

|V (ORi )| − (tR − 1)

2

)

≤(2r−2|TR|−1

2

)

.

Similarly, we show |E(GB − TB)| ≤(

2r−2|TB|−12

)

. Also, in view of the ordering σ,any vertex u of G is incident to at most 1

2(degG(u)+ s+3) edges of GR (resp. GB).First, we show that if either |TR| ≤ r − 2 or |TB| ≤ r − 2, then the assertion holds.

10

If |TR| ≤ r − 2, then

|E(GR)| ≤

(

2r − 2|TR| − 1

2

)

+∑

u∈TR

1

2(degG(u) + s+ 3)

≤(2r−2|TR|−1

2

)

+ ex(G,rK2)2 − r−1−|TR|

2 δ(G) + 12

(

r−12

)

+ (s+3)|TR|2

<ex(G,rK2)

2 .

Since |E(GB)| ≤ |E(GR)| + 1, we have exalt(G, rK2, σ) = |E(GR)| + |E(GB)| ≤ex(G, rK2) which is impossible. Similarly, if |TB | ≤ r − 2, then the assertion holds.

If |TR| = |TB | = r − 1, in view of the Berge-Tutte Formula, one can concludethat all connected components of GR−TR and GB −TB are isolated vertices. Thismeans that TR and TB are vertex covers of GR and GB , respectively. Now, weshow that TR = TB. One the contrary, suppose that TR 6= TB . Without loss ofgenerality, let |E(GR)| ≥ |E(GB)| and choose a vertex ui ∈ TR \ TB .

Set LB = |E(HBi )|, i.e., the number of blue edges of Hi. Since TB is a vertex

cover of GB and all edges of Hi appear consecutively in the ordering σ (accordingto an Eulerian tour of Hi), one can check that at least 2LB edges of Hi are incidentto the vertices of TB . It implies that there is a vertex z ∈ V (Hi) \ {ui} with degree

at least 2LB

r−1 . However, by the assumption that for any u ∈ V (Hi) \ {ui},

degHi(ui) ≥ (r + s− 1)degHi

(u) +

(

r − 1

2

)

+ (s + 3)(r − 1),

we have

degHi(ui) ≥ (r + s− 1)

⌈

2LB

r − 1

⌉

+

(

r − 1

2

)

+ (s+ 3)(r − 1).

Thus, we have

LB ≤1

2

(

degHi(ui)−

(

r − 1

2

)

− (s+ 3)(r − 1)

)

.

Note that red color can be assigned to at most one edge between any two consec-utive edge of G in the ordering σ. Define lR to be the number of red edges incidentto ui in the subgraph Hi. One can see that lR ≤ LB + 1. If there exists a 1 ≤ j ≤ l

such that ui ∈ Kj and ur 6∈ Kj for any r < i, then set H ′i = Hi ∪ Kj ; otherwise,

define H ′i = Hi. Now, we show that the number of red edges incident to ui in the

graph H ′i is at most LB + 1 + 1

2degH′

i(ui) −

12degHi

(ui). If H ′i = Hi, then there is

nothing to prove. Otherwise, one can check that red color can be assigned to at most12degH′

i(ui)−

12degHi

(ui)+1 edges incident to ui in the subgraph H ′i \Hi. Moreover,

if lR = LB+1, then one can see that there are two red edges e and e′ in Hi such thate (resp. e′) appears before (resp. after) any edge of HB

i in the ordering σ. Conse-quently, in this case, red color can be assigned to at most 1

2degH′

i(ui)−

12degHi

(ui)

edges incident to ui in the subgraph H ′i \Hi. Thus, the number of red edges incident

11

to ui in the graph H ′i is at most LB + 1 + 1

2degH′

i(ui) −

12degHi

(ui). Moreover, in

view of the ordering of the Eulerian tour of H, one can conclude that red color canbe assigned to at most 1

2(degG(ui) + s + 1 − degH′

i(ui)) edges incident to ui in the

subgraph G \H ′i. Therefore, the number of red edges incident to ui in the graph G

is at most

LB + 1 + 12degH′

i(ui)−

12degHi

(ui) +12(degG(ui) + s+ 1− degH′

i(ui)) ≤

12(degG(ui)− (s+ 3)(r − 2)−

(

r−12

)

)

Consequently,

|E(GR)| ≤ 12(degG(ui)− (s + 3)(r − 2)−

(

r−12

)

) +∑

u∈TR\{ui}

1

2(degG(u) + s+ 3)

≤ −12

(

r−12

)

+∑

u∈TR

1

2degG(u)

≤ 12ex(G, rK2)

and it is a contradiction. Hence, TR = TB . Therefore, the number of blue and rededges of G, i.e., |E(GR)| + |E(GB)|, is at most the number of edges incident withvertices in TR or TB which is at most ex(G, rK2) and this is impossible. Accordingly,exalt(G, rK2, σ) = ex(G, rK2). �

Assume that G is a graph. A G-decomposition of a graph H is a set {G1, . . . , Gt} ofpairwise edge-disjoint subgraphs of H such that for each 1 ≤ i ≤ t, the graph Gi isisomorphic to G; and moreover, the edge sets of Gi’s partition the edge set of H. AG-decomposition of H is called a monogamous G-decomposition, if any distinct pairof vertices of H appear in at most one copy of G in the decomposition. Note that if agraph H has a decomposition into the complete graph Kt, then it is a monogamousKt-decomposition.

Theorem B. [16] Assume that m and n are positive even integers. The complete

bipartite graph Km,n has a monogamous C4-decomposition if and only if (m,n) =(2, 2) or 6 ≤ n ≤ m ≤ 2n − 2.

Lemma 3. Let r, t and t′ be positive integers, where 11 ≤ t ≤ t′ ≤ 2t − 2. If c is a

nonnegative integer and t ≥ 8r + 4c + 2, then the complete bipartite graph Kt,t′ is

(r, c)-locally Eulerian.

Proof. Assume that t = 2p+ q and t′ = 2p′ + q′, where 0 ≤ q ≤ 1 and 0 ≤ q′ ≤ 1.Extend the complete bipartite graph G = Kt,t′ to the complete bipartite graphH = KT,T ′ , where T = t + q and T ′ = t′ + q′. In view of Theorem B, consider amonogamous C4-decomposition of H. Call any C4 of this decomposition a block, ifit is entirely in G. Construct a bipartite graph with the vertex set (U, V ) such thatU consists of ⌊ t−3

8 ⌋ copies of each vertex of Kt,t′ and V consists of all blocks. Join avertex of U to a vertex of V , if the corresponding vertex of Kt,t′ is contained in the

12

corresponding block. One can check that the degree of each vertex in the part U isat least ⌈ t−3

2 ⌉ and the degree of any vertex in the part V is 4⌊ t−38 ⌋. In view of Hall’s

Theorem, one can see that this bipartite graph has a matching which saturates allvertices of U . For any vertex v ∈ Kt,t′ , consider ⌈

t−38 ⌉ blocks assigned to v through

the perfect matching and set Hv to be a subgraph of Kt,t′ formed by the union ofthese blocks. One can see that degHv

(v) = 2⌊ t−38 ⌋, while the degree of any other

vertices of Hv is 2. In view of Hv’s, one can see that Kt,t′ is an (r, c)-locally Euleriangraph. �

For a family of graphs F , we say a graph G has an F-factor if there are vertex-disjoint subgraphs H1,H2, . . . ,Ht of G such that each Hi is a member of F andt⋃

i=1

V (Hi) = V (G). Note that if a graph G has an F-factor, where each member of

F is an (r, c)-locally Eulerian graph, then G is also (r, c)-locally Eulerian. In viewof the aforementioned lemma, if a graph G has a Kt,t′ factor, then one can concludethat G is (r, c)-locally Eulerian. Now, we introduce some sufficient condition for agraph to have a Kt,t′ factor.

Graph expansion was studied extensively in the literature. Let 0 < ν ≤ τ < 1 andassume that G is a graph with n vertices. For S ⊆ V (G), the ν-robust neighborhood

of S, RNν,G(S), is the set of all vertices v ∈ V (G) such that |NG(v) ∩ S| ≥ νn. Agraph G is called robust (ν, τ)-expander, if for any S ⊆ V (G) with τn ≤ |S| ≤ (1−τ)nwe have |RNν,G(S)| ≥ |S|+νn. Throughout this section, we write 0 < a≪ b≪ c tomean that we can choose the constants a, b, and c from right to left. More precisely,there are two increasing functions f and g such that, given c, whenever we choosesome b ≤ g(c) and a ≤ f(b), for more about robust (ν, τ)-expander see [10]. A graphG with n vertices has bandwidth at most b if there exists a bijective assignmentl : V (G) −→ [n] such that for every edge uv ∈ E(G), we have |l(u)− l(v)| ≤ b.

Theorem C. [10] Let ν, τ , and η be real numbers, where 0 < ν ≤ τ ≪ η < 1, and∆ be a positive integer. There exist constants β > 0 and n0 such that the following

holds. Suppose that H is a bipartite graph on n ≥ n0 vertices with ∆(H) ≤ ∆and bandwidth at most βn. If G is a robust (ν, τ)-expander with n vertices and

δ(G) ≥ ηn, then G contains a copy of H.

Theorem 1. Let ν, τ and η be real numbers, where 0 < ν ≤ τ ≪ η < 1 . Suppose

that G is a robust (ν, τ)-expander graph on n vertices with δ(G) ≥ ηn. If n is

sufficiently large, then χ(KG(G, rK2)) = χalt(KG(G, rK2)) = |E(G)| − ex(G, rK2).

Proof. In view of Lemma 2, it is sufficient to show that the graph G is an (r, c)-locally Eulerian graph, where c =

(

r−12

)

+3(r−1). Set t = 2r2+14r−6 and let k andt′ be integers such that n = 2tk+t′ and 2t ≤ t′ ≤ 4t−1. Now, set H to be a bipartitegraph on n vertices with 1 + n−t′

2t connected components such that one componentis isomorphic to K

⌈ t′

2⌉,⌊ t′

2⌋and any other component is isomorphic to Kt,t. In view

of Theorem C, if n is sufficiently large, then H is a spanning subgraph of G. ByLemma 3, one can see that K

⌈ t′

2⌉,⌊ t′

2⌋and Kt,t are both (r, c)-locally Eulerian graph;

consequently, by Lemma 2, the theorem follows. �

13

Lemma B. [13] Assume that τ and η are positive constants, where τ ≪ η < 1. Let

G be a graph with n vertices and degree sequence d1 ≤ d2 ≤ · · · ≤ dn such that for

any i < n2 either di ≥ i + ηn or dn−i−⌊ηn⌋ ≥ n − i. If n is sufficiently large, then

δ(G) ≥ ηn and G is a robust (τ2, τ)-expander.

In view of the previous lemma and Theorem 1, we have the next corollary.

Corollary 1. Let γ > 0 be a real number. Assume that G is a connected graph

with n vertices and degree sequence d1 ≤ d2 ≤ · · · ≤ dn. Also, suppose that for each

i < n2 , we have either di ≥ i + γn or dn−i−⌊γn⌋ ≥ n − i. If n is sufficiently large,

then χ(KG(G, rK2)) = |E(G)| − ex(G, rK2).

For a graph property P, we say G(n, p) possesses P asymptotically almost sure, ora.a.s. for brevity, if the probability that G ∈ G(n, p) possesses the property P tendsto 1 as n tends to infinity. For constants 0 < ν ≪ τ ≪ p < 1, a.a.s. any graph G

in G(n, p) is a robust (ν, τ)-expander graph with minimum degree at least pn2 and

maximum degree at most 2np. This observation and Theorem 1 imply that a.a.s. forany graph G in G(n, p) we have χ(KG(G, rK2)) = |E(G)| − ex(G, rK2). Moreover,Huang, Lee, and Sudakov [9] proved a more general theorem. Here, we state it in aspecial case.

Theorem D. [9] For positive integers r,∆ and reals 0 < p ≤ 1 and γ > 0, thereexists a constant β > 0 such that a.a.s., any spanning subgraph G′ of any G ∈ G(n, p)with minimum degree δ(G′) ≥ p(12 + γ)n contains every n-vertex bipartite graph H

which has the maximum degree at most ∆ and bandwidth at most βn.

In view of the proof of Theorem 1 and the previous theorem, we have the followingresult.

Corollary 2. If 0 < p ≤ 1 and γ > 0, then a.a.s. for any spanning subgraph G′ of

any G ∈ G(n, p) with minimum degree at least p(12 +γ)n we have χ(KG(G′, rK2)) =|E(G′)| − ex(G′, rK2).

Assume that H is a graph with h vertices and χ(H) = l. Set cr(H) to be the sizeof the smallest color class over all proper l-colorings of H. The critical chromatic

number, χcr(H), is defined as (l − 1) hh−cr(H) . One can check that χ(H) − 1 <

χcr(H) ≤ χ(H) and equality holds in the upper bound if and only if in any l-coloring of H, all color classes have the same size. Assume that H has k connectedcomponents C1, C2, . . . , Ck. Define hcfc(H) to be the highest common factor ofintegers |C1|, |C2|, . . . , |Ck|. Let f be an l-coloring of H such that x1 ≤ x2 ≤ · · · ≤ xlare the size of coloring classes in f . Set D(f) = {xi+1 − xi| 1 ≤ i ≤ l − 1} and

D(H) =⋃

D(f) where the union ranges over all l-colorings f of H. Now, define

hcfχ(H) to be the highest common factor of the members of D(H). If D(H) = {0},then we define hcfχ(H) =∞. We say

H is in class 1 if

{

hcfχ(H) = 1 when χ(H) 6= 2,hcfχ(H) ≤ 2 and hcfc(H) = 1 when χ(H) = 2,

otherwise, H is in Class 2.

14

Theorem E. [12] For every graph H on h vertices, there are integers c and m0 such

that for all integers m ≥ m0, if G is a graph on n = mh vertices, then the following

holds. If

δ(G) ≥

{

(1− 1χcr(H))n+ c H is in Class 1,

(1− 1χ(H))n+ c H is in Class 2,

then G has an H-factor.

Theorem 2. For any integer r ≥ 2, there are constants α and β such that for any

graph G with n vertices, if δ(G) ≥ (12 − α)n + β and n is sufficiently large, then

χ(KG(G, rK2)) = χalt(KG(G, rK2)) = |E(G)| − ex(G, rK2).

Proof. Define t = 2r2+14r−6. Set H to be a bipartite graph with two connectedcomponents C1 and C2 isomorphic to Kt,t and Kt+1,t, respectively. One can checkthat H is in Class 1. Hence, by Theorem E, there are integers c1 and m1 such thatif |V (G′)| ≥ m1, |V (H)| divides |V (G′)|, and δ(G′) ≥ (1− 1

χcr(H))|V (G′)|+ c1, then

the graph G′ has an H-factor. Let T be an integer such that 4t + 1 ≤ T < 8t + 2and 4t + 1|n − T . It is known that if n is sufficiently large and δ(G) ≥ ηn, where0 < η < 1, then G contains a copy of the complete bipartite graph K⌈T

2⌉,⌊T

2⌋. Note

that 1χcr(H) =

12 +

18t+2 . Set α = 1

8t+2 and β = c1+8t−1. If δ(G) ≥ (12 −α)n+β and

n is sufficiently large, then G contains K⌈T2⌉,⌊T

2⌋ and also the graph G\K⌈T

2⌉,⌊T

2⌋ has

an H-factor. Hence, G can be decomposed into the complete bipartite graphs Kt,t,Kt+1,t, and K⌈T

2⌉,⌊T

2⌋. In view of Lemma 3, these graphs are (r, c)-locally Eulerian

graphs with c =(

r−12

)

+ 3(r − 1). Therefore, G is an (r, c)-locally Eulerian graph;and consequently, by Lemma 2, the assertion holds. �

3.2 Permutation Graphs

Assume that m,n, r are positive integers, where r ≤ m,n. For an r-subset A ⊆ [m]and an injective map f : A −→ [n], the ordered pair (A, f) is said to be an r-partial permutation [5]. Let Sr(m,n) denotes the set of all r-partial permutations.Two partial permutations (A, f) and (B, g) are said to be intersecting, if thereexists an x ∈ A ∩ B such that f(x) = g(x). Note that Sn(n, n) is the set of alln-permutations. The permutation graph Sr(m,n) has all r-partial permutations(A, σ) as its vertex set and two r-partial permutations are adjacent if and only ifthey are not intersecting. Note that Sr(m,n) ∼= Sr(n,m); and therefore, for thesimplicity, we assume that m ≥ n for all permutation graphs. One can see that thepermutation graph Sr(m,n) is isomorphic to KG(Km,n, rK2).

Next theorem gives a sufficient condition for a balanced bipartite graph to havea decomposition into complete bipartite subgraphs.

Theorem F. [26] For any integer q ≥ 2, there exists a positive integer m0 such that

for all m ≥ m0, the following holds. If G = (X,Y ) is a balanced bipartite graph on

2n = 2mq vertices, i.e., |X| = |Y | = mq, with

δ(G) ≥

{

n2 + q − 1 if m is evenn+3q2 − 2 if m is odd,

then G has a Kq,q-factor.

15

Now, we investigate the chromatic number of general Kneser graph KG(G, rK2)provided that G = (X,Y ) is a balanced (|X| = |Y |) dense bipartite graph. Inparticular, we determine the chromatic number of any permutation graph providedthat the number of its vertices is large enough. For more about permutation graphs,see [3, 6, 11].

Theorem 3. For any positive integer r, there exist constants q and m such that for

all n ≥ m, the following holds. If G is a graph on 2n vertices which has a bipartite

subgraph H = (U, V ) with |U | = |V | = n and δ(H) ≥ n2 + q, then χ(KG(G, rK2)) =

χalt(KG(G, rK2)) = |E(G)| − ex(G, rK2).

Proof. In view of Lemma 2, it is sufficient to show that the graph H is an (r, c)-locally Eulerian graph, where c =

(

r−12

)

+ 3(r − 1). Set t = 2r2 + 14r − 6. ByTheorem F, there are integers q1 and m1 such that if n ≥ m1 and t|n, then anybalanced bipartite graph H ′ with 2n vertices and δ(H ′) ≥ n

2 + q1 has a Kt,t-factor.Let t′ be an integer, where t ≤ t′ < 2t and t|n − t′. It is known that if n is

sufficiently large and δ(H) ≥ ηn, where 0 < η < 1, then H contains a copy of thecomplete bipartite graph Kt′,t′ . Define q = q1 + 2t− 1. Note that if n is sufficientlylarge, then H contains a copy of Kt′,′t and also, in view of Theorem F, H \ Kt′,t′

has a Kt,t-factor. This implies that H can be decomposed into complete bipartitegraphs Kt′,t′ and Kt,t. In view of Lemma 3, these graphs are (r, c)-locally Euleriangraphs, where c =

(

r−12

)

+ 3(r − 1). Therefore, H and G are (r, c)-locally Euleriangraphs; and consequently, by Lemma 2, the assertion holds. �

Corollary 3. Assume that m,n, r are positive integers, where m ≥ n ≥ r. If m is

large enough, then

χ(KG(Km,n, rK2)) = χalt(KG(Km,n, rK2)) = m(n− r + 1).

Proof. In view of Hall’s Theorem, any maximal rK2-free subgraph of Km,n has(r − 1)m edges. Hence, in view of Lemma A, we have χ(KG(Km,n, rK2)) ≤ m(n−r + 1). In view of Lemma 3, if m is sufficiently large, then χ(KG(Km,m, rK2)) =m(m− r+1). Now, we show that for any positive integer n < m, if m is sufficientlylarge, then χ(KG(Km,n, rK2)) = m(n− r+1). To see this, on the contrary, supposethat f : V (KG(Km,n, rK2)) −→ {1, 2, . . . , χ(KG(Km,n, rK2))} is a proper coloringof KG(Km,n, rK2), where χ(KG(Km,n, rK2)) < m(n − r + 1). Add m − n newvertices to the small part of Km,n and join them to all vertices in the other partto construct Km,m, and call the new edges e1, . . . , e(m−n)m. Extend the coloring f

to a proper coloring g for KG(Km,m, rK2) as follows. If a matching M is a subsetof Km,n, then set g(M) = f(M); otherwise, assume that i is the smallest positiveinteger such that ei ∈ M , in this case set g(M) = i + χ(KG(Km,n, rK2)). Thisprovides a proper coloring for KG(Km,m, rK2) with less than m(m − r + 1) colorswhich is a contradiction. �

The chromatic number of permutation graph KG(G, rK2) was studied in [2] andit was proved that if G is a connected (s, t)-regular bipartite graph with s ≥ t

and even s, then χ(KG(G, rK2)) = |E(G)| − ex(G, rK2). This result shows thatχ(KG(Km,n, rK2)) = Sr(m,n) = m(n − r + 1) provided that m is an even integer

16

and m ≥ n ≥ r. However, if m is a small odd value, then the chromatic number ofthe permutation graph Sr(m,n) is unknown. The aforementioned results motivateus to consider the following conjecture.

Conjecture 1. For any connected graph G and positive integer r, we have

χ(KG(G, rK2)) = |E(G)| − ex(G, rK2).

4 Hedetniemi’s Conjecture

In this section, first we study the alternating chromatic number of the Mycielskiconstruction of graphs. Then, we present some lower bounds for the chromaticnumber in terms of alternating chromatic number and strong alternating chromaticnumber of the Categorical product of graphs. In view of bounds, we determine thechromatic number of the Categorical product of some families of graphs.

In the sequel, we generalize the definitions of alternating and strong alternat-ing chromatic number of graphs. We can reformulate several well-known conceptsin graph colorings via graph homomorphism. One can see that if two graph arehomomorphically equivalent, then they have the same chromatic number, circularchromatic number, and fractional chromatic number. In this regard, we define thefollowing generalizations of the alternating chromatic number and the strong alter-nating chromatic number of a graph G, respectively, as follows

χhalt(G) = max {χalt(L); G←→ L} ,

χhsalt(G) = max {χsalt(L); G←→ L} .

Clearly, we have

χ(G) ≥ max{χhalt(G), χhsalt(G)} ≥ max{χalt(G), χsalt(G)}.

We do not know how much χhalt(G) can be far apart χalt(G); although, it seemsthat χhalt(G) = χalt(G). It is worth noting that we can sometimes easily determineχhalt(G) rather than χalt(G), see Lemma 4.

4.1 Mycielski Construction

For a given graph G with the vertex set V (G) = {u1, . . . , un}, the Mycielskian of Gis the graph M(G) with the vertex set V (M(G)) = {u1, . . . , un, v1, . . . , vn, w} andthe edge set E(M(G)) = E(G) ∪ {uivj : uiuj ∈ E(G)} ∪ {wvi : 1 ≤ i ≤ n}. In fact,the Mycielski graph M(G) contains the graph G itself as an isomorphic inducedsubgraph, together with n + 1 additional vertices. For any 1 ≤ i ≤ n, vi is calledthe twin of ui and they have the same neighborhood in G and also the vertex w istermed the root vertex. The vertex w and vi’s form a star graph. Some coloringproperties of the Mycielski graph M(G) have been studied in the literature. Forinstance, it is known that χ(M(G)) = χ(G) + 1 and χf (M(G)) = χf (G) + 1

χf (G) .

For any vector ~r = (r1, . . . , rn), the ~r-blow up graph G(~r) of G is obtained byreplacing each vertex ui of G by ri copies u

1i , . . . , u

rii , such that for any 1 ≤ i1 ≤ ri

17

and 1 ≤ j1 ≤ rj, ui1i is adjacent to u

j1j if ui is adjacent to uj in G. In other words, the

edge uiuj is replaced by the complete bipartite graphKri,rj . Note that for any vector~r with positive entries, two graphs G(~r) and G are homomorphically equivalent.

Lemma 4. For any graph G, we have χhalt(M(G)) ≥ χhalt(G) + 1.

Proof. Let F ⊆ 2[n] and that KG(F) is homomorphically equivalent to G suchthat there exists an ordering σ of [n] for which χhalt(G) = n−altσ(F). Assume thatF = {A1, . . . , Am}. Set ~r = (r1, . . . , r2m+1), where r1 = · · · = rm = 2t + 1, rm+1 =· · · = r2m =

(2t+1t+1

)

, r2m+1 = 1 and t = n − altσ(F). In what follows, we introducea Kneser representation for the ~r-blow up of M(KG(F)), i.e., M(KG(F))(~r). Notethat M(KG(F))(~r) and M(G) are homomorphically equivalent. Define

V ′ = {b1, b2, . . . , b2t+1, c1, c2, . . . , c(2t+1)(m−1)},

V ′′ = {a1,1, a1,2, . . . , a1,(2t+1), a2,1, a2,2, . . . , a2,(2t+1), . . . , am,1, am,2, . . . , am,2t+1},

where V ′, V ′′, and the set [n] are pairwise disjoint. Set V = [n] ∪ V ′ ∪ V ′′ andl =

(

2t+1t+1

)

. For any 1 ≤ i ≤ m and 1 ≤ j ≤ 2t + 1, define Ai,j = Ai ∪ {ai,j}.Moreover, for any 1 ≤ i ≤ m, consider distinct sets Bi,1, . . . , Bi,l such that for any1 ≤ k ≤ l, there exists a unique (t+1)-subset {bk1 , bk2 , . . . , bkt+1

} of {b1, b2, . . . , b2t+1}where Bi,k = Ai ∪ {bk1 , bk2 , . . . , bkt+1

}. Set

H = {Ai,j , Bi,k : 1 ≤ i ≤ m, 1 ≤ j ≤ 2t+ 1, 1 ≤ k ≤ l} ∪ {V ′′}.

Now, one can check that KG(H) provides a Kneser representation forM(KG(F))(~r).To see this, one can check that Aij ’s, Bij’s, and V ′′ are corresponding to the verticesof KG(F), their twins, and the root vertex, respectively. Note that H ⊆ 2V andc1, . . . , c(m−1)(2t+1) are the isolated vertices of H. As a benefit of using isolatedvertices, we present an ordering π to determine the alternating chromatic numberof M(KG(F))(~r). First, consider the ordering τ as follows

a1,1 < c1 < a2,1 < c2 < · · · < am−1,1 < cm−1 < am,1 < b1 <

a1,2 < cm < a2,2 < cm+1 < · · · < am−1,2 < c2m−2 < am,2 < b2 <...

a1,2t+1 < c2t(m−1)+1 < a2,2t+1 < · · · < c(2t+1)(m−1) < am,2t+1 < b2t+1

Construct the ordering π by concatenating the ordering σ after τ , i.e., π = τ ||σ.Note that the number of elements of π is (2t+1)m+(2t+1)+(2t+1)(m−1)+n =2m(2t+1)+n. Define p = 2m(2t+1)+n. Now, we claim that altπ(H) ≤ altσ(F)+2m(2t + 1) − 1. To see this, assume that X = (x1, x2, . . . , xp) ∈ {−1, 0,+1}p \{(0, 0, . . . , 0)} and alt(X) = altσ(F)+2m(2t+1). We show that X+

π or X−π contains

a hyperedge of H. If alt((x1, x2, . . . , x2m(2t+1))) = 2m(2t + 1), then X+π or X−

π

contains {a1,1, a1,2, . . . , am,2t+1}, i.e., the root vertex; and consequently, the assertionfollows. Hence, let alt((x1, x2, . . . , x2m(2t+1))) ≤ 2m(2t + 1) − 1; and consequently,alt(Y ) ≥ altσ(F) + 1, where Y = (x2m(2t+1)+1, x2m(2t+1)+2, . . . , xp). Hence, Y +

σ

or Y −σ contains a hyperedge of F . Without loss of generality, suppose that Y +

σ

contains Ai ∈ F . If ai,j = +1 for some 1 ≤ j ≤ 2t+1, then Ai,j ∈ H and Ai,j ⊆ X+π .

18

Moreover, if there exists a (t+ 1)-subset {bk1 , bk2 , . . . , bkt+1} such that bkj = +1 for

any 1 ≤ j ≤ t+1, then Ai ∪ {bk1 , bk2 , . . . , bkt+1} ∈ H and Ai ∪ {bk1 , bk2 , . . . , bkt+1

} ⊆X+

π . On the other hand, if for any 1 ≤ j ≤ 2t+ 1, ai,j 6= +1, and if for any (t+ 1)-subset {bk1 , bk2 , . . . , bkt+1

}, there exists some 1 ≤ j ≤ t + 1 such that bkj 6= +1,then one can conclude that alt((x1, x2, . . . , x2m(2t+1))) ≤ 2m(2t + 1) − (t + 1) =2m(2t + 1) − n + altσ(F) − 1. On the other hand, alt(X) = altσ(F) + 2m(2t + 1);hence, we should have alt(Y ) ≥ n+ 1 which is impossible. �

A graph G is said to be alternatively t-chromatic (resp. strongly alternatively t-

chromatic) if χhalt(G) = χ(G) = t (resp. χhsalt(G) = χ(G) = t). Note thatif graphs G and H are homomorphically equivalent, then M(G) and M(H) arehomomorphically equivalent as well. In other words, the previous theorem statesthat the graph M(G) is alternatively (t + 1)-chromatic graph provided that G isalternatively t-chromatic graph.

4.2 Chromatic Number of Categorical Product

There are several kinds of graph products in the literature. The Categorical ProductG×H of two graphs G and H is defined by V (G×H) = V (G)× V (H), where twovertices (u1, u2) and (v1, v2) are adjacent if u1v1 ∈ E(G) and u2v2 ∈ E(H). It is easyto check that by any coloring of G orH we can present a coloring of G×H; and there-fore, χ(G×H) ≤ min{χ(G), χ(H)}. In 1966, Hedetniemi [7] introduced his interest-ing conjecture, celled Hedetniemi’s conjecture, about the chromatic number of theCategorical product of two graphs, which states that χ(G×H) = min{χ(G), χ(H)}.This conjecture has been studied in the literature, see [14, 19, 23, 25, 29]. In viewof topological bounds for chromatic number, it was shown that Hedetniemi’s con-jecture holds for any two graphs for which the topological bound on the chromaticnumber is tight, see [8, 22, 25].

Theorem G. [8] If G and H are two graphs for which the topological bound on the

chromatic number is tight, then χ(G×H) = min{χ(G), χ(H)}.

Also, in [27], Hedetniemi’s conjecture was generalized to circular chromatic numberof graphs. Zhu [27] conjectured that χc(G × H) = min{χc(G), χc(H)}. For moreresults about this conjecture and circular chromatic number, one can refer to [15,24, 27, 28]. In this section, we present some lower bounds for the chromatic numberof the Categorical product of graphs.

Lemma 5. Let G and H be two graphs. If there exists a graph homomorphism

f : H −→ G, then χhalt(H) ≤ χhalt(G) and also, χhsalt(H) ≤ χhsalt(G).

Proof. We prove χhalt(H) ≤ χhalt(G) and similarly one can show χshalt(H) ≤χshalt(G). First, we assume that H is a subgraph of G and we prove a strongerassertion. In fact, we show that if H is a subgraph of G, then χalt(H) ≤ χalt(G). Tosee this, assume that F ⊆ 2[n] such that KG(F) is isomorphic to H. We show thatthere are Y and G ⊆ 2Y such that KG(G) is isomorphic to G and that n− alt(F) ≤|Y | − alt(G).

19

Without loss of generality, suppose that alt(F) = altI(F), where I is the identityordering, i.e., I : 1 < 2 < · · · < n. Let g : V (H) −→ F be an isomorphismbetween KG(F) and H. Also, we can assume that H is a spanning subgraph ofG. To see this, let v ∈ V (G) \ V (H). Now, add this vertex to H as an isolatedvertex to obtain H1. Set F1 = F ∪ {{1, 2, . . . , n + 1}} ⊆ 2[n+1]. One can seethat KG(F1) is isomorphic to H1. Also, altI(F1) ≤ altI(F) + 1; and therefore,n + 1 − alt(F1) ≥ n + 1 − altI(F1) ≥ n − altI(F) = n − alt(F). By repeating theprevious procedure, if it is necessary, we can find a spanning subgraph H of G and aKneser representation KG(H) for H with H ⊆ 2[n+l] and l = |V (G)| − |V (H)| suchthat n − alt(F) ≤ n + l − alt(H). Thus, it is enough to prove the lemma just forspanning subgraphs.

Now, we can assume that H is a spanning subgraph of G. Again, without loss ofgenerality, we can assume that there is an edge e = ab ∈ E(G) such that H+e = G.Assume that g(a) = A and g(b) = B. Since a and b are not adjacent, A ∩ B

is not an empty set. Assume that A ∩ B = {y1, y2, . . . , yt}. Consider 2t positiveintegers {y′1, y

′2, . . . , y

′t} ∪ {z1, z2, . . . , zt} disjoint from [n]. Let F0 = F , σ0 = I,

g0 = g, and Yi = [n]∪ {y′1, z1, y′2, z2, . . . , y

′i, zi}. Assume that Fi ⊆ 2Yi , σi ∈ SYi

, andgi : V (H) −→ Fi were defined when i < t. Set gi+1(a) = gi(a)∪ {y

′i+1} \ {yi+1} and

for u 6∈ {a, b}, if yi+1 ∈ gi(u), then gi+1(u) = gi(u) ∪ {y′i+1}. Let Fi+1 = {gi+1(v) :

v ∈ V (H)}. To obtain the ordering σi+1, replace yi+1 with yi+1 < zi+1 < y′i+1 inthe ordering σi, i.e., put zi+1 immediately after yi+1 and put y′i+1 after zi+1. Notethat Fi+1 ⊆ 2Yi+1 and σi+1 ∈ SYi+1

, where Yi+1 = [n]∪{y′1, z1, y′2, z2, . . . , y

′i+1, zi+1}.

Assume that Fi’s and σi’s were obtained for every i ∈ {0, 1, . . . , t}. One can see thataltσi+1

(Fi+1) ≤ 2 + altσi(Fi); and therefore, altσt(Ft) ≤ 2t + altσ0

(F0). Note thatKG(Ft) ∼= H + e = G. Set G = Ft and Yt = Y . Consequently,

|Y |−alt(G) ≥ n+2t−altσt(G) ≥ n+2t−(2t+altσ0(F0)) = n−altI(F) = n−alt(F)

Hence, if H is a subgraph of G, then we have χalt(H) ≤ χalt(G). Similarly, one canshow that χslat(H) ≤ χsalt(G).

Now, assume that for two given graphs G and H, there exists a graph homo-morphism f : H −→ G. Consider H and F ⊆ 2[n] such that H is homomorphicallyequivalent to H, KG(F) ∼= H and χhalt(H) = n − alt(F). Since H and H arehomomorphically equivalent and there exists a graph homomorphism f : H −→ G,the graph G ∪ H, i.e., the disjoint union of G and H, is homomorphically equiv-alent to the graph G and also, this graph has H as its subgraph. In view of theaforementioned discussion, there are Y and G ⊆ 2Y such that KG(G) ∼= G ∪ H and

χhalt(H) = n− alt(F) ≤ |Y | − alt(G) ≤ χhalt(G ∪ H) = χhalt(G).

Similarly, one can show χshalt(H) ≤ χshalt(G). �

In view of the proof of the aforementioned lemma, the next corollary follows.

Corollary 4. If H is a subgraph of G, then χalt(H) ≤ χalt(G) and χslat(H) ≤χsalt(G). In particular, for any graph G, we have min{χalt(G), χsalt(G)} ≥ ω(G),where ω(G) is the clique number of G.

20

In the next result, we show the accuracy of Hedetniemi’s conjecture for the strongalternating chromatic number of graphs.

Lemma 6. For any two graphs G and H, we have

a) χhsalt(G ×H) = min{χhsalt(G), χhsalt(H)},b) χhalt(G×H) ≥ max{min{χhalt(G), χhsalt(H)−1},min{χhsalt(G)−1, χhalt(H)}}.

Proof. First, we prove part (a). Assume that G ⊆ 2V and H ⊆ 2V′

such thatKG(G) and KG(H) are homomorphically equivalent to G and H, respectively. Also,assume that χhsalt(G) = 1 + |V | − salt(G), χhsalt(H) = 1 + |V ′| − salt(H), whereV = {1, 2, . . . , n} and V ′ = {n + 1, n + 2, . . . , n + m}. Without loss of gener-ality, we can assume that salt(H) = saltI(H) and salt(G) = saltI(G), where I

is the identity ordering. Let F = {A ∪ B : A ∈ G & B ∈ H} and V ′′ =V ∪ V ′. One can check that KG(F) ∼= G × H and also, χhsalt(G × H) ≥ 1 +|V ′′| − salt(F) ≥ min{χhsalt(G), χhsalt(H)}. To see this, it is enough to show thatsaltI(F) ≤ max{|V ′|+ salt(G), |V |+ salt(H)}. Define l = max{|V ′|+ salt(G), |V |+salt(H)}. Consider X ∈ {−1, 0, 1}n+m with altI(X) ≥ 1 + l.

Now, consider two vectors X(1),X(2) ∈ {−1, 0, 1}n+m such that the first n

coordinates of X(1) (resp. the last m coordinates of X(2)) are the same as X andthe last m coordinates of X(1) (resp. the first n coordinates of X(2)) are zero. If weshow that altI(X(1)) > salt(G) and altI(X(2)) > salt(H), then each of X+ and X−

has a hyperedge of F and it completes the proof. Suppose that altI(X(1)) ≤ salt(G)(resp. altI(X(2)) ≤ salt(H)). Therefore, altI(X) ≤ altI(X(1)) + altI(X(2)) ≤salt(G) + |V ′| ≤ l (resp. altI(X) ≤ altI(X(1)) + altI(X(2)) ≤ |V | + salt(H) ≤ l)which is a contradiction. Hence, χhsalt(G×H) ≥ min{χhsalt(G), χhsalt(H)}. On theother hand, there exists a graph homomorphisms from G × H to both G and H.Consequently, by Lemma 5, we have χhslat(G×H) ≤ min{χhslat(G), χhslat(H)}.

Now, we prove part (b). By symmetry, it suffices to prove χhalt(G × H) ≥min{χhalt(G), χhsalt(H) − 1}. Consider G ⊆ 2V , H ⊆ 2V

′

, σ ∈ SV , and γ ∈ SV ′

such that G←→ KG(G), H ←→ KG(H), χhalt(G) = |V | − altσ(G), and χhsalt(H) =|V ′|+1−altγ(H). Without loss of generality, we can assume that V = {1, 2, . . . , n},V ′ = {n+1, n+2, . . . , n+m}, σ : 1 < 2 < . . . < n, and γ : n+1 < n+2 < . . . < n+m.Define L = {A ∪B : A ∈ G & B ∈ H} ⊆ 2[n+m]. One can check that KG(L) ∼=KG(G) × KG(H); and therefore, KG(L) ←→ G ×H. Set I : 1 < 2 < . . . < m + n

and M = max{|V | + saltγ(H), |V′| + altσ(G)}. Now, we show that altI(L) ≤ M .

To see this, assume that we have an X ∈ {−1, 0,+1}m+n \ {(0, 0, . . . , 0)} such thatalt(X) ≥ M + 1. Let X(1),X(2) ∈ {−1, 0, 1}m+n be the same as in the proof ofprevious part. One can see that there exists an alternative subsequence of nonzeroterms in X(1) of length more than altσ(G). Therefore, either X(1)+I or X(1)−I hassome hyperedge of G. Now, we show that that both X(2)+I and X(2)−I have somehyperedges of H. On the contrary, suppose that this is not true. Therefore, wehave alt(X(2)) ≤ saltγ(H) and thus alt(X) ≤ |V | + alt(X(2)) ≤ M which is acontradiction.

Without loss of generality, we can assume that X(2)+I and X(2)−I contain A andB of H, respectively, and X(1)+I contains C of G. Now, in view of A∪C ⊆ X+

I andthat A ∪ C ∈ L, the assertion follows.

21

Hence, we have

χhalt(G×H) ≥ m+ n− altI(L)≥ m+ n−M

= min{n− altσ(G),m− saltγ(H)}= min{χhalt(G), χhsalt(H)− 1}

as desired. �

Lemma 7. Assume that there are G ⊆ 2V , H ⊆ 2V′

, σ ∈ SV , and γ ∈ SV ′ such that

χalt(KG(G)) = |V | − altσ(G) and χalt(KG(H)) = |V ′| − altγ(H). If

max{|V |+ altγ(H), |V′|+ altσ(G)} ≥ saltσ(G) + saltγ(H),

then χalt(KG(G)×KG(H)) ≥ min{χalt(KG(G)), χalt(KG(H))}.

Proof. Let G = KG(G) and H = KG(H). Without loss of generality, we canassume that V = {1, 2, . . . , n}, V ′ = {n + 1, n + 2, . . . , n +m}, σ : 1 < 2 < . . . < n,and γ : n + 1 < n + 2 < . . . < n + m. Define L = {A ∪B : A ∈ G & B ∈ H}.Note that KG(L) ∼= KG(G) × KG(H) ∼= G × H. Set I : 1 < 2 < . . . < m + n andM = max{|V |+ altγ(H), |V

′|+ altσ(G)}. In view of the assumption, we have

M = max{|V |+ altγ(H), |V′|+ altσ(G)} ≥ saltγ(H) + saltσ(G).

Now, we show that altI(L) ≤ M . To see this, assume that we have an X ∈{−1, 0,+1}m+n \ {(0, 0, . . . , 0)} such that alt(X) ≥ M + 1. Consider two vectorsX(1),X(2) ∈ {−1, 0, 1}m+n such that the first n coordinates of X(1) (resp. thelast m coordinates of X(2)) are the same as X and the last m coordinates of X(1)(resp. the first n coordinates of X(2)) are zero. One can see that there exists analternative subsequence of nonzero terms in X(1) (resp. X(2)) of length more thanaltσ(G) (resp. altγ(H)). Therefore, either X(1)+I or X(1)−I (resp. either X(2)+Ior X(2)−I ) has some hyperedge of G (resp. H). Now, we show that either bothX(1)+I and X(1)−I or both X(2)+I and X(2)−I have some hyperedges of G or H,respectively. On the contrary, suppose that this is not true. Therefore, we havealt(X(1)) ≤ saltσ(G) and alt(X(2)) ≤ saltγ(H). Note that these inequalities implythat alt(X) ≤ alt(X(1)) + alt(X(2)) ≤M which is a contradiction.

Without loss of generality, we can suppose that X(1)+I and X(1)−I contain A andB of G, respectively, and also X(2)+I contains C of H. Now, in view of A∪C ⊆ X+

I

and that A ∪ C ∈ L, the assertion follows.Hence, we have

χalt(G×H) ≥ m+ n− altI(L)≥ m+ n−M

= min{χalt(G), χalt(H)}

as desired. �

Remark 2. In view of the aforementioned lemma, one can determine the chro-

matic number of the Categorical product of some family of graphs. For instance,

22

one can consider matching-dense graphs with n vertices, where n is sufficiently

large. In view of Lemma 7, it is sufficient to introduce a good upper bound for

exsalt(G, rK2, σ), where σ is the same ordering presented in the proof of Lemma 2.We show exsalt(G, rK2, σ) ≤ ex(G, rK2) +

(

r−12

)

+ (s + 3)(r − 1) + 1. To see this,

consider an alternating 2-coloring of the edges of G with respect to the ordering σ of

length ex(G, rK2)+(

r−12

)

+(s+3)(r− 1)+ 2 and suppose that GR has no matching

of size r. By a similar argument as in the proof of Lemma 2, one can conclude that

|TR| = r − 1. Also,

|E(GR)| ≤∑

u∈TR

1

2(degG(u) + s+ 3) ≤

1

2

(

ex(G, rK2) +

(

r − 1

2

)

+ (s+ 3)(r − 1)

)

which is impossible.

In view of Lemmas 6 and 7, one can determine the chromatic number of the Cate-gorical product of some family of graphs. In particular, if both of them are stronglyalternatively t-chromatic graphs. Note that the following graphs are strongly alter-natively t-chromatic graphs.

1. Schrijver graphs and Kneser graphs

2. The Kneser multigraph KG(G,F): G is a multigraph such that all of its edgeshave even multiplicities and F is a family of its simple subgraphs, see [2].

3. The matching graph KG(G, rK2): G is a connected graph with n vertices, oddgirth at least g, and degree sequence degG(v1) ≥ degG(v2) ≥ · · · ≥ degG(vn).

Also, r ≤ max{g2 ,degG(vr−1)+1

4 }, {v1, . . . , vr−1} forms an independent set, anddegG(v1),degG(v2), . . . ,degG(vr−1) are even integers, see [2].

Note that, in [22], it has been shown that for any two graphs G and H we havecoind(B(G × H)) = min{coind(G), coind(H)}. Also, in [4], it was proved that forany positive integer r, we have coind(B(Mr(G))) ≥ coind(B(G)) + 1, where Mr(G)is the generalized Mycielskian of G. Consequently, in view of Lemma 1, one cansee that Hedetniemi’s conjecture holds for any two graphs of the family of stronglyalternatively t-chromatic graphs and the iterated generalized Mycielskian of anysuch graphs.

Also, the following graphs are alternatively t-chromatic graphs.

1. Kneser graphs and multiple Kneser graphs: In [1], multiple Kneser graphswere introduced as a generalization of Kneser graphs.

2. Kneser Multigraphs, see [2].

3. Matching-dense graphs and a family of matching-sparse graphs, see [2].

4. The permutation graph Sr(m,n): m is large enough.

5. The Schrijver graph SG(n, k): k = 2 or n = 2k + 1.

23

6. Any number of iterations of the Mycielski construction starting with any graphappearing on the list above.

Acknowledgement: The authors would like to express their deepest grati-tude to Professor Carsten Thomassen for his insightful comments. They also ap-preciate the detailed valuable comments of Dr. Saeed Shaebani. Moreover, theywould like to thank Skype for sponsoring their endless conversations in two coun-tries.

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