On ωψ-Perfect Graphs ∗

M. Gabriela Araujo-Pardogaraujo@math.unam.mx

Christian Rubio-Montielchristian@math.unam.mx

———————————————————Instituto de Matematicas

Universidad Nacional Autonoma de MexicoCiudad Universitaria, 04510, Mexico D.F.———————————————————

January 18, 2014

Abstract

In this paper we study a generalization for the perfection of graphs to other interestingparameters related with colorations. This generalization was introduced partially by Christenand Selkow in 1979 and Yegnanarayanan in 2001.Let a, b ∈ {ω, χ,Γ, α, ψ} where ω is the clique number, χ is the chromatic number, Γ is theGrundy number, α is the achromatic number and ψ is the pseudoachromatic number. A graphG is ab-perfect, if for every induced subgraph H, a(H) = b(H). In this work we characterize theωψ-perfect graphs.

Key words. Perfect graphs, Grundy, achromatic and pseudoachromatic numbers.

1 Introduction

Let G be a simple graph. A vertex coloring ς : V (G) → {1, . . . , k} is called complete if each pairof different colors appears in an edge. The pseudoachromatic number ψ(G) is the maximum k forwhich a complete coloring of G exists.

A vertex coloring of G is called proper if uv is an edge of G then ς(u) = ς(v). The achromaticnumber α(G) is the maximum k for which a proper and complete coloring of G exists.

A vertex coloring of G is called Grundy (whose colors, as usual, are positive integers), if it is aproper vertex coloring having the property that for every two colors i and j with i < j, every vertex

∗Research supported by CONACyT-Mexico under Projects 166306, 178395 and PAPIIT-Mexico under ProjectIN101912.

1

colored j has a neighbor colored i (consequently, every Grundy coloring is a complete coloring).The Grundy number Γ(G) is the maximum k for which a Grundy coloring of G exists.

Clearly,

ω(G) ≤ χ(G) ≤ Γ(G) ≤ α(G) ≤ ψ(G) (1)

where ω(G) denotes, as usual, the clique number of G and χ(G) denotes the chromatic number.

ω χ Γ α ψ

2 2 3 3 3

1

u1

2

u2

3

u3

1

u4

Figure 1: A coloring of P4 with 3 colors.

The Grundy number was introduced by Grundy in 1939 [14], the achromatic number by Harary,Hedetniemi and Prins in 1967 [16] and the pseudoachromatic number by Gupta in 1969 [15].

Interesting results on these invariants can be found in [1, 7, 8, 15, 16, 24, 25] and the referencestherein. More specifically, the authors of this paper and others have worked with the achromaticand pseudoachromatic index of complete graphs (see [1, 2, 3]).

Let a, b ∈ {ω, χ,Γ, α, ψ}. A graph G is called ab-perfect, if for every induced subgraph H of G,a(H) = b(H). This definition extends the usual notion of perfect graph introduced by Berge [5] in1961; with this notation a perfect graph is denoted by ωχ-perfect. Some important known resultsrelated to this are the following: Lovasz [17] proved in 1972 that G is ωχ-perfect if and only if Gc

is ωχ-perfect, where Gc denotes the complement of G (that is, if G is a graph of order n, thenV (Gc) = V (G) and E(Gc) = E(Kn)−E(G)). Chudnovsky, Robertson, Seymour and Thomas [10]proved in 2006 that G is ωχ-perfect if and only if G and Gc are C2k+1-free for all k ≥ 21; andSeinsche [20] proved in 1974 the following theorem, which we will use to prove our Theorem 3:

Theorem 1. (Seinsche, 1974) If a graph G is P4-free then G is ωχ-perfect.

The concept of the ab-perfect graphs was introduced by Christen and Selkow in 1979 [9], but theyonly consider ab-perfect graphs for a, b ∈ {ω, χ,Γα}. Yegnanarayanan in 2001 [24] also works inthis concept for a, b ∈ {ω, χ, α, ψ}, in this work he states the following theorem and corollaries:

Theorem 2. (Yegnanarayanan, 2001) For any finite graph G the following are equivalent:

⟨1⟩ G is ωψ-perfect,

⟨2⟩ G is χψ-perfect,

⟨3⟩ G is αψ-perfect,

1A graph G without an induced subgraph H is called H-free. A graph H1-free, H2-free, . . . , Hm-free is called(H1, H2, . . . , Hm)-free.

2

⟨4⟩ G is C4-free.

Corollary 1. (Yegnanarayanan, 2001) Every αψ-perfect graph is χα-perfect.

Corollary 2. (Yegnanarayanan, 2001) Every χα-perfect graph is ωχ-perfect.

Unfortunately, the Theorem 2 is false (a counterexample is P4 because it is C4-free but not ωψ-perfect, see Figure 1); i.e., ⟨4⟩ does not necessarily imply ⟨1⟩. Consequently the corollaries are notwell founded. However, while Corollary 1 is false (then again, the counterexample is P4, see Figure1), Corollary 2 is true (see Theorem 3). Note that if a graph G is ωψ-perfect then it is immediatelyωχ-perfect (see Equation 1). That is, G is perfect in the usual sense and it is well known that Gdoes not allow odd cycles (except triangles), or their complements (see [10]), and then the conditionC4-free is not sufficient. In Appendix I we exhibit the exact place where the proof of Theorem 2,given by Yegnanarayanan, is wrong. With the aim of finding the correctness of this theorem weobtain the following results:

Theorem 3. Let G be a graph and let a ∈ {Γ, α, ψ}. G is χa-perfect if and only if G is ωa-perfect.

Theorem 4. The connected graph G is (C4, P4)-free if and only if there exists a set of connected

graphs {G1, . . . , Gk} for some k ∈ N also P4-free and C4-free andm ∈ Z+, such that G = Km⊕k∪i=1Gi

and G−Km is disconnected (if k ≥ 2). 2

The graphs (C4, P4)-free are also called trivially perfect graphs by Golumbic (see [12] and [13]);comparability graphs of trees by Wolk (see [21] and [22]), or quasi-threshold graphs by Ma (see [18]and [23]).

In the following two theorems we prove some equivalences between graphs. Note that in Theorem5 we analyze some equivalence when G is a connected graph. In Theorem 6 we analize the same,but in this case G is any graph (not necessarily connected).

Theorem 5. For any connected graph G the following are equivalent:

⟨1⟩ G is ωψ-perfect,

⟨2⟩ G is χψ-perfect,

⟨3⟩ G is (C4, P4, P3 ∪K2, 3K2)-free,

⟨4⟩ G = Kn1 ⊕ (Kn2 ∪Kn3 ∪ n4K1) or G = Kn1 ⊕ (G′ ∪ n2K1) for n1 ∈ Z+ and n2, n3, n4 ∈ Nwhere G′ is a non complete connected graph and (C4, P4, P3 ∪K2, 3K2)-free.

Theorem 6. For any graph G the following are equivalent:

⟨1⟩ G is ωψ-perfect,

⟨2⟩ G is χψ-perfect,

⟨3⟩ G is (C4, P4, P3 ∪K2, 3K2)-free,

⟨4⟩ G = Kn1 ∪ Kn2 ∪ n3K1 or G = G′ ∪ n2K1 for n1 ∈ Z+ and n2, n3 ∈ N, where G′ is a noncomplete connected graph and (C4, P4, P3 ∪K2, 3K2)-free.

3

a)

G

v0

1v1

1v2

1v3

3

v4

1 v5

3

v62

v7

1

v8

2

v9

4v10

2v11

2v12

2b)

G

v0

1v1

2v2

2v3

2

v4

1 v5

2

v65

v7

3

v8

4

v9

5v10

4v11

3v12

5

Figure 2: Different coloration of G.

2 Preliminaries

In [16] Harary, Hedetniemi and Prins prove that for any graph G and for every integer a withχ(G) ≤ a ≤ α(G) there is a complete and proper coloring of G with a colors. In [9], Christen andSelkow prove that for any graph G and for every integer b with χ(G) ≤ b ≤ Γ(G) there is a Grundycoloring of G with b colors (see [8] pg. 349).In [25] Yegnanarayanan, Balakrishnan and Sampathkunar prove that if 2 ≤ a ≤ b ≤ c then thereexists a graph G with chromatic number a, achromatic number b, and pseudoachromatic number c.In [7] Chartrand, Okamoto, Tuza and Zhang prove that for integers a, b and c with 2 ≤ a ≤ b ≤ cthere exists a connected graphG with χ(G) = a, Γ(G) = b and α(G) = c, if and only if a = b = c = 2or b ≥ 3 (see [8] pg. 331 and 352).

Figure 2 shows a graph G with all of these parameters being different. In particular G has ω(G) = 2,χ(G) = 3, Γ(G) = 4, α(G) = 5 and ψ(G) = 6. Clearly, G does not contain C3 as an induced graph,because it is constructed identifying a K4,4 with a C7 cycle by an edge, then ω(G) = 2. Figure 2a)shows a proper vertex coloring of G with three colors (black, white and gray), then χ(G) ≤ 3, and,G contains a C7 then χ(G) = 3. Also to the left we show a Grundy vertex coloring of G with fourcolors (numbers), then Γ(G) ≥ 4. Figure 2b) shows a complete and proper vertex coloring of Gwith five colors and α(G) ≥ 5. The proof of Γ(G) ≤ 4, α(G) ≤ 5 and ψ(G) = 6 are the Claims 1, 2and 3 respectively, given in Appendix II.

In Figure 3 we show a directed transitive graph D where the vertices of D represent the classes ofab-perfect graphs and its label is ab respectively for a, b ∈ {ω, χ,Γ, α, ψ}. If two classes are equalthey define the same vertex (see Theorem 3). If a class of ab-perfect graphs is contained in a classof cd-perfect graphs then there is an arrow from vertex ab to vertex cd. To prove that D does notcontain another arrow we have the following counterexamples: P4, C4, C5 and P3 ∪K2.

In [9], Christen and Selkow give characterizations of ωa-perfect graphs when a is Γ or α.

Now we will establish some notation. For A ⊆ V (G) let < A > be the subgraph induced of G byA. Moreover, a graph H is an induced subgraph of a graph G, briefly denoted by H ≤ G, if thereis a set A ⊆ V (G), such that < A > is isomorphic to H. If we consider a set {v1, . . . , vn} ⊆ V thenwe define < v1, . . . , vn >:=< {v1, . . . , vn} >. The set of vertices adjacent to the vertex x is denotedby NG(x) (or only N(x)) and NG[x] = N(x)∪{x} (or only N [x]). The maximum degree of vertices

2For convenience, in this paper 0 ∈ N and ⊕ denotes the join of graphs.

4

Γψ

ωψ = χψ

ωα = χα

ωΓ = χΓ

ωχ

Γα

αψD

Figure 3: Relationship between ab-perfect graphs with a, b ∈ {ω, χ,Γ, α, ψ}.

αψ ↛ ωψ Γψ ↛ ωψ Γα↛ ωψ ωχ↛ ωψ

αψ ↛ ωα Γψ ↛ ωα Γα↛ ωα ωχ↛ ωα

αψ ↛ ωΓ Γψ ↛ ωΓ Γα↛ ωΓ ωχ↛ ωΓ

Table 1: P4 is a counterexample in these cases.

ωα↛ ωψ ωα↛ Γψ ωα↛ αψ ωχ↛ Γψ Γα↛ Γψ

ωΓ ↛ ωψ ωΓ ↛ Γψ ωΓ ↛ αψ ωχ↛ αψ Γα↛ αψ

Table 2: C4 is a counterexample in these cases.

αψ ↛ ωχ Γψ ↛ ωχ Γα↛ ωχ

Table 3: C5 is a counterexample in these cases.

αψ ↛ Γψ αψ ↛ Γα ωΓ ↛ Γα ωχ↛ Γα ωΓ ↛ ωα

Table 4: P3 ∪K2 is a counterexample in this cases.

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of G is denoted by ∆(G) and degG(v), or simply deg(v), denotes the degree of a vertex v in G.

3 Our results

Proof of Theorem 3.

Assume that G is χa-perfect. G is P4-free because χ(P4) = 2 and a(P4) = 3 (see Figure 1). ByTheorem 1, G is ωχ-perfect, then for all induced subgraphs H we have that ω(H) = χ(H) = a(H),i.e., G is ωa-perfect. The converse is automatically given from (1).

It is not difficult to note the following remarks:

Remark 1. If G is a connected graph and P4-free, then diam(G) ≤ 2.

Remark 2. If G is a H-free graph, then every induced subgraph of G is H-free.

The following lemma was proved in [22], we include the prove for the sake completeness.

Lemma 1. If G is a connected graph (P4, C4)-free of order n then ∆(G) = n− 1.

Proof. Suppose ∆(G) < n − 1 then, there exists a vertex x of maximum degree which has a non-neighbor z, since G is connected and P4-free, there exists a path between x and z which is ax − y − z for some y. Since deg(x) ≥ deg(y) and z is adjacent to y and not to x, there must bea vertex u adjacent to x and non to y. Then {u, x, y, z} induces a P4 or C4, a contradiction. So∆(G) = n− 1.

Proof of Theorem 4.

Assume that G does not contain an induced subgraph isomorphic to C4 and P4. By Lemma 1, there

exists w1 ∈ V (G), such that deg(w1) = n − 1. If G − {w1} is disconnected, then G = K1 ⊕k∪i=1Gi

where K1 = w1 and G1, . . . , Gk are the components of G−{w1}. As each component is an inducedsubgraph of G then, by Remark 2 it is C4-free and P4-free, hence, the implication is true. Now,if G − {w1} is connected and also, since it is an induced subgraph of G, yet again, by Remark2, it does not contain an induced subgraph isomorphic to C4 and P4; by Lemma 1 there existsw2 ∈ V (G − {w1}), such that deg(w2) = n − 2. Then again, if G − {w1, w2} is disconnected,

then G = K2 ⊕k∪i=1Gi where K2 =< w1, w2 > and G1, . . . , Gk are the components of G− {w1, w2}

and each component is an induced subgraph of G, applying the same argument than before theimplication is true. As G is a finite graph we can repeat this procedure obtaining a complete graphKm with V (Km) = {w1, . . . , wm}, and {G1, . . . , Gk} a set of connected graphs C4-free and P4-free,

which are the components of G− {w1, . . . , wm}, and G = Km ⊕k∪i=1Gi with G−Km disconnected

if k ≥ 2.

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To prove the converse we take a set X = {v1, v2, v3, v4} of vertices in V (G) and we analize threecases:Case 1) If X ⊆ V (Gi) for some i ∈ {1, . . . , k} then, by hypothesis, < X > is neither P4, nor C4.Case 2) If X is in two or more components of {G1, . . . , Gk} then again it can be neither P4, norC4 because < X > is disconnected.Case 3) If there exists vi in V (Km) for some i ∈ {1, 2, 3, 4} then we have a vertex of degree threein < X > and neither P4, nor C4 have a vertex of degree 3.

Now, we will prove a short, but also useful lemma:

Lemma 2. If H = Km1 ∪Km2 ∪m3K1 where m1 ∈ Z+ and m2,m3 ∈ N then ω(H) = ψ(H).

Proof. Note that ω(H) = max{m1,m2}. Suppose max{m1,m2} < ψ(H) and let ς : V (H) →{1, . . . , ψ(H)} a complete coloring of H. Note that each color class has at least two verticesin Km1 ∪ Km2 , in other case as deg(v) ≤ max{m1,m2} − 1 for any v ∈ V (H) then ς not becomplete, then ψ(H) ≤ m1+m2

2 ≤ max{m1,m2} which is a contradiction and we conclude thatψ(H) = max{m1,m2}.

Proof of Theorem 5.

The proof of ⟨1⟩ ⇒ ⟨2⟩ is immediate from (1).

To prove ⟨2⟩ ⇒ ⟨3⟩ note that, if H ∈ {C4, P4, P3 ∪K2, 3K2} then χ(H) = 2, and ψ(H) = 3, hencethe implication is true.

To prove ⟨3⟩ ⇒ ⟨4⟩ using the Theorem 4. We know that there exists k ∈ N and n1 ∈ Z+, such

that all the {G1, . . . , Gk} are P4-free, and C4-free, such that G = Kn1 ⊕k∪i=1Gi, and G − Kn1

is disconnected (if k ≥ 2). Without losing generality, if |E(Gi)| ≥ 2 for all i ∈ {1, 2, 3} then3K2 is an induced subgraph of G and this is not possible by hypothesis. Then we have thatG = Kn1 ⊕ (G1 ∪G2 ∪ n4K1) for n4 ∈ N. Now, if G1 is not a complete graph, it has P3 as inducedsubgraph as we note in the proof of Lemma 1 then necessarily G2

∼= K1 because in another casewe have P3 ∪K2 as induced subgraph of G. Using Remark 2 we have that G1 does not contain aninduced subgraph isomorphic to C4, P4, P3 ∪K2 and 3K2, hence the implication is true.

Finally, we will divide the proof of ⟨4⟩ ⇒ ⟨1⟩ into two cases and in both of them we will use induc-tion over |V (G)|:

Case 1) LetG = Kn1⊕(Kn2 ∪Kn3 ∪ n4K1), by induction, if |V (G)| = 1 thenG ∼= K1 is ωψ-perfect.Assume that, if |V (G)| ≤ m then G is ωψ-perfect and take a graph G = Kn1⊕(Kn2 ∪Kn3 ∪ n4K1),such that n1 + n2 + n3 + n4 = m + 1. If H is an induced connected subgraph of G, suchthat |V (H)| < |V (G)| then clearly H = Km1 ⊕ (Km2 ∪Km3 ∪m4K1) where mi ≤ ni for alli ∈ {1, . . . , 4} and m1 + m2 + m3 + m4 ≤ m, then by induction hypothesis H is ωψ-perfect, inparticular ω(H) = ψ(H).Now, if H is a disconnected induced subgraph of G, such that |V (H)| < |V (G)| then H =

7

Km1∪Km2∪m3K1 wherem1 ∈ Z+ andm2,m3 ∈ N, and using Lemma 2 we have that ω(H) = ψ(H).Now, we will prove that we also have that ω(G) = ψ(G): Let n = max{n2, n3, a} with a = 1 ifn4 > 0 and a = 0, if n4 = 0. If n = 0 then G = Kn1 , and it is ωψ-perfect. Suppose thatn = 0, clearly ω(G) = n1 + n. Suppose that ψ(G) > n1 + n and let ς : V (G) → {1, . . . , ψ(G)} acomplete coloring of G. First of all, we also suppose that n4 > 0; let u be a vertex of n4K1. Asthe neighbours of u are n1 then ς(u) meet at most n1 different chromatic class from u, then theremust exist another vertex v, such that ς(u) = ς(v). Clearly N(u) ⊆ N(v) then ψ(G) = ψ(G− u),but ω(G) = ω(G − u), and ω(G − u) < ψ(G − u), which is a contradiction because G − u is aninduced subgraph of G and by induction hypothesis it is ωψ-perfect. Then, if ψ(G) > n1 + nwe have that G = Kn1 ⊕ (Kn2 ∪Kn3) and n = max{n2, n3}. Let u ∈ V (Kn2 ∪ Kn3), such thatς(u) /∈ {ς(v) : v ∈ V (Kn1)}. As the neighbours of u are at most n1 +n− 1, then ς(u) meet at mostn1 + n − 1 different chromatic class from u, there must exists another vertex v in V (Kn2 ∪Kn3),such that ς(u) = ς(v), and ψ ≤ n1 +

n2+n32 < n1 +n, which is a contradiction. Then ω(G) = ψ(G),

hence G is ωψ-perfect.

Case 2) Let G = Kn1 ⊕ (G′ ∪ n2K1), if |V (G)| = 4, then G = K1 ⊕ P3 = K2 ⊕ (2K1), and by theprevious case G is ωψ-perfect. Now, assume that any graph G with this structure and order lessor equal than m is ωψ-perfect, and let G = Kn1 ⊕ (G′ ∪ n2K1), such that n1 + n′ + n2 = m + 1where n′ = |V (G′)|. If H is an induced connected subgraph of G, such that |V (H)| < |V (G)|, thenby Remark 2, H does not contain an induced subgraph isomorphic to C4, P4, P3 ∪K2 and 3K2.Hence, by ⟨3⟩ ⇒ ⟨4⟩ we have that H = Kn5 ⊕ (Kn6 ∪Kn7 ∪ n8K1), or H = Kn5 ⊕ (H ′ ∪ n6K1) forn5 ∈ Z+, and n6, n7, n8 ∈ N where H ′ is a connected graph F -free for all F ∈ {C4, P4, P3∪K2, 3K2}(*). In both cases H is ωψ-perfect; when H = Kn5 ⊕ (Kn6 ∪Kn7 ∪ n8K1) we use the previouscase and in the case that H = Kn5 ⊕ (H ′ ∪ n6K1) the argument is the induction hypothesis;therefore H is ωψ-perfect, in particular ω(H) = ψ(H) (**). Furthermore, if H is an induceddisconnected subgraph of G = Kn1 ⊕ (G′ ∪ n2K1), such that |V (H)| < |V (G)|, then H is aninduced disconected subgraph of G′ ∪ n2K1, and if H ≤ n5K1 consequently, ω(H) = ψ(H). Now,if H ≤ n5K1, then ψ(H) = ψ(H − n5K1), and ω(H) = ω(H − n5K1). Since H − n5K1 ≤ G′ andG′ is a proper induced connected subgraph of G, for (*) and (**) G′ is ωψ-perfect, in particularω(H − n5K1) = ψ(H − n5K1), then ω(H) = ψ(H).Finally, we will prove that ω(G) = ψ(G): Note that ω(G) = ω(G′) + n1 and ω(G′) ≥ 2. Supposethat ω(G′) + n1 < ψ(G). Let ς : V (G) → {1, . . . , ψ(G)} a complete coloring of G. If we supposethat n2 > 0 we arrive to a contradiction using exactly the same arguments than in the previouscase (for n4 = 0). Then we have that G = Kn1 ⊕ G′. Note that, also in the same part of thisproof, the contradiction is obtained by the hypothesis that there exist two vertices u and v, suchthat ς(u) = ς(v); we can use the same argumentation if we suppose that there exist two vertices{u, v} ∈ Kn1 , which ς(u) = ς(v) and once again arrive to a contradiction, hence ς(u) = ς(v) forall u, v ∈ V (Kn1). Now, since G′ is not a complete graph, ω(G′) ≥ 2 and by induction hypothesisψ(G) ≥ ω(G′)+n1 then there exist at least two colors i and j, such that they do not appear in anyvertex ofKn1 and there exists xy ∈ E(G′), such that ς(x) = i and ς(y) = j, then ψ(G′) = ψ(G)−n1;on the other hand ω(G′) = ω(G) − n1, which it is not possible for (*) and (**). Therefore G isωψ-perfect.

Proof of Theorem 6.

8

To prove Theorem 6 we only note that, if G is connected we apply Theorem 5, and in anothercase let {G1, . . . , Gn} be the connected components of G, then only at most two of them aredifferent from K1 because in another case we have 3K2 as induced subgraph of G, which is notpossible. If only one of them is a non-complete graph then the rest of them are also isolated verticesbecause in another case we have P3 ∪K2 as an induced subgraph of G. Then we have two differentcases, either G = Kn1 ∪ Kn2 ∪ n3K1, or G = G′ ∪ n2K1 where n1 ∈ N and n2, n3 ∈ Z+. Thearguments to prove this theorem are exactly the same than those used to prove Theorem 5, onlyfor ⟨4⟩ ⇒ ⟨1⟩ implication it is important to note that, if G = Kn1 ∪ Kn2 ∪ n3K1 then Lemma 2states that ω(G) = ψ(G), and it is only necessary to prove the equality for all induced subgraphsbut, this analysis is similar than those made in Lemma 2. To prove the ωψ-perfectness of G whenG = G′ ∪ n2K1 we also use similar arguments.

4 Appendix

I On Theorem 2

In this subsection we recover the proof of Yegnanarayanan’s Theorem. The implications ⟨1⟩ ⇒ ⟨2⟩,⟨2⟩ ⇒ ⟨3⟩ and ⟨3⟩ ⇒ ⟨4⟩ of Theorem 2 are true. The problem in the proof appears in the implication⟨4⟩ ⇒ ⟨1⟩. More specifically we will exhibit where the proof is incorrect. The analysis of theauthor is by contradiction: he takes a C4-free graph G with a complete coloration of size ψ(G),afterwards, he tries to find a complete graph with order ψ(G) starting with a complete graphKm−1 =< v1, . . . , vm−1 > (where all the vertices have different colors) and m − 1 ≤ ψ(G). Ifm− 1 = ψ(G) the proof is finished then he assumes that m− 1 < ψ(G) and defines, for each vi, aset Si, which elements are all the vertices in N(vi) with different colors than those used in Km−1.

He claims thatm−1∩i=1

Si = ∅, which is not necessarily true; to see this take a P4 like Figure 1 in the

introduction, this P4 has a complete coloring of size 3. Take in P4 the K2 = ({u2, u3}, {u2u3}) withboth vertices of different colors as in Yegnanarayanan’s proof. Note that here Su2 = {u1}, and

Su3 = {u4}, but3∩

i=2Sui = ∅. The rest of the proof is based in this false statement and obviously it

is not necessarily true.

II On Figure 2

Let G be the graph despited in Figure 2, constructed identifying a K4,4 with a C7 by an edge, then:

Claim 1. Γ(G) ≤ 4.

Proof. Suppose there exists ς : V (G) → {1, . . . , 5} a Grundy coloring of G. If v ∈ {v4, . . . , v8} thenς(v) = 5 because deg(v) = 2. If ς(v0) = 5 then ς : {v9, . . . , v12} → {1, . . . , 4} is a bijection, henceς(v) = 5 for all v ∈ {v0, v1, v2}. This is a contradiction because ς(v) = 4 for some v ∈ {v9, . . . , v12},and v has at most one vertex without color 5. Similarly, the vertices v1, v2, v10, v11, and v12 cannothave color 5.

9

If ς(v3) = 5 then its neighbors v4, v9, v10, v11 and v12 would have the numbers 1, 2, 3 and 4, but v4cannot have neither color 4, nor 3, because it has degree 2. Then the colors 3 and 4 are in v9, v10,v11 and v12. Suppose the vertex v12 has color 4, then its neighbors v0, v1 and v2 have the colors 1,2 and 3, once again we have a contradiction because the coloration is proper and the color 3 is insome vertex of {v9, v10, v11}. Similarly we conclude that neither v10, nor v11, have color 4. Then,it is necessary that ς(v9) = 4, but v8 cannot have color 3 because deg(v8) = 2, then color 3 appearsin one of the vertices v0, v1 or v2, and once again we have a contradiction to the proper colorationbecause color 3, is in one of the vertices v10, v11 and v12. Consequently, v3 cannot have color 5 andusing a similar analysis we conclude that v9 does not have color 5 either. Therefore Γ(G) ≤ 4.

Claim 2. α(G) ≤ 5.

Proof. Suppose there exists ς : V (G) → {1, . . . , 6} a proper complete coloring of G. First, as thevertices v4, . . . , v8 are five, then there exists a chromatic class, C1 in v0, . . . , v3, or in v9, . . . , v12because the coloring is proper, without losing generality C1 is contained in v0, . . . , v3 hence v3 ∈ C1

because the coloring is complete; N(vi) ⊆ N(v3) and |N(vi)| = 4 for i ∈ {0, 1, 2}. Clearly, v9 ∈ C2,v10 ∈ C3, v11 ∈ C4, v12 ∈ C5, v4 ∈ C6 where Ci are chromatic classes. As the coloring is properv0, v1, v2 ∈ C1 ∪ C6, then |C3| ≥ 3, |C4| ≥ 3 and |C5| ≥ 3 because |N(vi)| = 2 for i ∈ {5, 6, 7, 8}and we have a contradiction because we need at least six vertices, but there are only four. Thenα(G) ≤ 5.

Claim 3. ψ(G) = 6.

Proof. It is clear that there does not exist a complete coloring of G with seven or more colorsbecause, since G has thirteen vertices, there must be a chromatic class of size one, but ∆(G) = 5and there exits at least one color, which is not incident with this chromatic class, thus ψ(G) ≤ 6.Now we will show a complete coloring of G with six colors: Figure 2 (Right) shows a proper completecoloring ς of G with 5 colors; let ς ′ : V (G) → {black, dark gray, gray, light gray, white, red},such that ς ′(vi) = ς(vi) if i = {3, 12}, and ς ′(vi) = red if i = {3, 12} then ς ′ is a complete coloringof G. Therefore ψ(G) = 6.

5 Open problems

In conclusion we wish to present two problems that, in our opinion, might be of interest. The firstone is a generalization of the problem only for the chromatic, achromatic and pseudoachromaticparameters as we state in the introduction (see [7], [25] and [8]), and the second one can be seenas a problem in graph theory using the language of perfectness.

Problem 1. Let 0 < a ≤ b ≤ c ≤ d ≤ e. Does there exist a graph G, such that ω(G) = a, χ(G) = b,Γ(G) = c, α(G) = d, and ψ(G) = e?

Problem 2. Let G be a graph and let a, b be any two parameters of G. G is called ab-perfect, if forevery induced subgraph H of G a(H) = b(H). Is it possible to characterize the ab-perfect graphs?

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In Problem 2 we are considering every type of parameters, but to explain our motivation consider,for instance, the known relation κ ≤ κ′ ≤ δ between the connectivity κ, the edge-connectivityκ′ and the minimum degree δ and characterize ab-perfect graphs for a, b ∈ {κ, κ′, δ}. Another

interesting motivation is to consider the relation |M∗| ≤∣∣∣K∣∣∣ where M∗ is a maximum matching,

and K is a minimum covering. In [19], Rautenbach and Zverovich, consider the ab-perfect graphsfor a, b ∈ {γ, i, γs, is} when γ is the domination, i is the independent domination, γi is the strongdomination, and is is the independent strong domination numbers of a graph G respectively.

Acknowledgment

The authors want to thank the anonymous referees of [4] for their kind help and valuable suggestionswhich led to an improvement of this paper.

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