THE SET CHROMATIC NUMBER OF RANDOM GRAPHS

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1 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT Abstract. In this aer we study the set chromatic number of a random grah G(n, ) for a wide range of = (n). We show that the set chromatic number, as a function of, forms an intriguing zigzag shae. 1. Introduction A roer colouring of a grah is a labeling of its vertices with colours such that no two vertices sharing the same edge have the same colour. A colouring using at most k colours is called a roer k-colouring. The smallest number of colours needed to colour a grah G is called its chromatic number, and it is denoted by χ(g). In this aer we are concerned with another notion of colouring, first introduced by Chartrand, Okamoto, Rasmussen and Zhang [1]. For a given (not necessarily roer) k-colouring c : V [k] of the vertex set of G = (V, E), let C(v) = {c(u) : uv E} be the neighbourhood colour set of a vertex v. (In this aer, [k] := {1,,..., k}.) The colouring c is a set colouring if C(u) C(v) for every air of adjacent vertices in G. The minimum number of colours, k, required for such a colouring is the set chromatic number χ s (G) of G. One can show that log χ(g) + 1 χ s (G) χ(g). (1) Indeed, the uer bound is trivial, since any roer colouring c is also a set colouring: for any edge uv, N(u), the neighbourhood of u, contains c(v) whereas N(v) does not. On the other hand, suose that there is a set colouring using at most k colours. Since there are at most k ossible neighbourhood colour sets, one can assign a unique colour to each set obtaining a roer colouring using at most k colours. We get that χ(g) χs(g), or equivalently, χ s (G) log χ(g). With slightly more work, one can imrove this lower bound by 1 (see [8]), which is tight (see []). Let us recall a classic model of random grahs that we study in this aer. The binomial random grah G(n, ) is the random grah G with vertex set [n] in which 1991 Mathematics Subject Classification. 05C80, 05C15, 05C35. Key words and hrases. random grahs, set chromatic number. Project sonsored by the National Security Agency under Grant Number H The United States Government is authorized to reroduce and distribute rerints notwithstanding any coyright notation hereon. The research of the third author is suorted in art by NSERC and Ryerson University. 1

2 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT Figure 1. The function r = r() for (0, 1) and (0, 1/], resectively. every air {i, j} ( ) [n] aears indeendently as an edge in G with robability. Note that = (n) may (and usually does) tend to zero as n tends to infinity. All asymtotics throughout are as n (we emhasize that the notations o( ) and O( ) refer to functions of n, not necessarily ositive, whose growth is bounded). We also use the notations f g for f = o(g) and f g for g = o(f). We say that an event in a robability sace holds asymtotically almost surely (or a.a.s.) if the robability that it holds tends to 1 as n goes to infinity. Since we aim for results that hold a.a.s., we will always assume that n is large enough. We often write G(n, ) when we mean a grah drawn from the distribution G(n, ). For simlicity, we will write f(n) g(n) if f(n)/g(n) 1 as n (that is, when f(n) = (1 + o(1))g(n)). Finally, we use lg to denote logarithms with base and log to denote natural logarithms. Before we state the main result of this aer, we need a few definitions that we will kee using throughout the whole aer. For a given = (n) satisfying for some ε > 0, let 4 (log n)(log log n) log n s = s() = min and 1 ε { } [(1 ) l ] + [1 (1 ) l ] : l N, and let l 0 be a value of l that achieves the minimum (l 0 can be assigned arbitrarily if there are at least two such values). We will show in Section 3 that { } log(1/) log(1/) l 0,, () log(1 ) log(1 ) and that s(). (3)

3 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 3 Figure. The function s = s() for (0, 1) and (0, 1/], resectively. If is a constant, then r = r() is defined such that n s r lg n = 1, that is, r = r() = lg(1/s). (4) Observe that r tends to infinity as 1 and undergoes a zigzag behaviour as a function of (see Figure 1). The reason for such a behaviour is, of course, that the function s is not monotone (see Figure ). Furthermore, observe that for each = 1 (1/) 1/k, where k is a ositive integer, l 0 = k, s = 1/, and r =. Now we state the main result of the aer. Theorem 1.1. Suose that = (n) is such that (log n) (log log n) /n and 1 ε, for some ε (0, 1). Let G G(n, ). Then, the following holds a.a.s. (i) If is a constant, then χ s (G) r lg n. (ii) If = o(1) and n = n α+o(1) for some α (0, 1], then (iii) If n = n o(1), then (α + o(1)) lg n χ s (G) (1 + α + o(1)) lg n. (lg(n) lg log n lg log(n)) χ s (G) (1 + o(1)) lg n. Note that the result is asymtotically tight for dense grahs (that is, for n = n 1 o(1) ; see art (i) and art (ii) for α = 1). For sarser grahs (art (ii) for α (0, 1)) the ratio between the uer and the lower bound is a constant that gets large for α small. On the other hand, the trivial lower bound of lg χ(g) (see (1)) gives us the following: a.a.s. ( ) n χ s (G(n, )) lg χ(g(n, )) lg α lg n, log(n)

4 4 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT rovided that n as n, and = o(1); χ s (G(n, )) lg χ(g(n, )) = Ω(1) otherwise (see [6, 7]). So the lower bound we rove is by a multilicative factor of +o(1) larger than the trivial one, rovided that log(n)/ log log n. If n = log C+o(1) n for some C [, ), then our bound is by a factor of (C 1)/C + o(1) better than the trivial one. This seemingly small imrovement is imortant to obtain the asymtotic behaviour in the case α = 1, and in articular, to obtain the zig-zag for constant. The uer and the lower bounds are roved in Section 3 and Section 4, resectively. Let us also mention that, in fact, the two bounds roved below are slightly stronger. In articular, the uer bound holds for n (/ log )(log n)(log log n), the oint where the trivial bound of χ(g(n, )) becomes stronger.. Preliminaries We will use the following version of Chernoff s bound. Suose that X Bin(n, ) is a binomial random variable with exectation µ = n. If 0 < δ < 1, then ( ) P[X < (1 δ)µ] ex δ µ, and if δ > 0, ( ) P[X > (1 + δ)µ] ex δ µ. + δ These inequalities are well known and can be found, for examle, in [5]. We will also use Suen s inequality that was introduced in [9] and revised in [4]. For a finite set S, let I = {(x, y) : x, y S, x y}, and for any (x, y) I, let A x,y be some event with the corresonding indicator random variable I x,y. (In our alication, A x,y will be the event that vertices x and y have the same neighbourhood colour sets.) Let X = (x,y) I I x,y be the random variable counting how many such events occur. The associated deendency grah has I as its vertex set, and (x 1, y 1 ) (x, y ) if and only if {x 1, y 1 } {x, y }. Suen s inequality asserts that P(X = 0) ex ( µ + e δ), (5) where µ = = (x,y) I P(A x,y ), (x 1,y 1 ) (x,y ) δ = max (x 1,y 1 ) I E[I x1,y 1 I x,y ], (x,y ) (x 1,y 1 ) P(A x,y ).

5 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 5 We start by roving () and (3). Since 3. Uer bound [(1 ) l ] + [1 (1 ) l ] = [ (1 ) l 1/ ] + 1/, it follows that s 1/, and consequently () also holds. Now, let log(1/) l = = log(1/) log(1 ) log(1 ) + δ, where 0 δ < 1. Observe that s() [(1 ) l ] + [1 (1 ) l ] = [1 (1 )δ ] + 1 imlying the uer bound in (3). [1 (1 )] + 1 = + 1. We kee the definition of function r = r() for constant introduced above (see (4)). We extend it here for sarser grahs as follows: suose that tends to zero as n, and that n = n α+o(1) for some α [0, 1]. Then, we define r = r() such that n s r lg n = 1, that is, r = r() 1 + α, since it follows from (3) that s 1/. The uer bound in Theorem 1.1 follows immediately from the next lemma. Lemma 3.1. Suose that = (n) is such that (log n)(log log n) log n and 1 ε, for some fixed ε (0, 1). Let G G(n, ). Then, a.a.s. χ s (G) (r + o(1)) lg n. Before we move to the roof, let us note that the lower bound for is not necessary, and the result can be extended to sarser grahs. The reason it is introduced here is that for sarser grahs, the trivial uer bound of χ(g) is stronger; note that a.a.s. n χ s (G(n, )) χ(g(n, )) log(n), rovided that n as n, and = o(1); χ s (G) χ(g) = O(1) otherwise. Proof. The roof is straightforward. Let ω = ω(n) = o(log n) be any function tending to infinity with n (slowly enough). Before exosing the edges of the (random) grah G, we artition (arbitrarily) the vertex set into r lg n + ω sets, each consisting of l 0 imortant vertices, and one remaining set of vertices, these being not imortant. (For exressions such as r lg n + ω that clearly have to be an integer, we round u or down but do not secify which: the choice of which does not affect the argument.) Note that (r lg n + ω)l 0 = O(log n/) = O(n/ log log n) = o(n), and so there are enough vertices to erform this oeration. All vertices in a given set receive the same colour, and hence the total number of colours is equal to (r+o(1)) lg n.

6 6 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT For a given air of vertices, x, y, we need to estimate from above the robability (x, y) that they have the same neighbourhood colour sets. We do it by considering sets of imortant vertices that neither x nor y belong to. Let U be the set of (imortant) vertices of the same colour, and let l 0 = U. Then, either both x and y are not connected to any vertex from U, yielding the contribution [(1 ) l 0 ] to the robability (x, y), or both x and y are connected to at least one vertex from U, giving the contribution [1 (1 ) l 0 ]. Thus, ( ) r lg n+ω (x, y) [(1 ) l 0 ] + [1 (1 ) l 0 ] = s r lg n+ω. Hence, the exected number of airs of adjacent vertices that are not distinguished by their neighbourhood colour sets is at most ( n )s r lg n+ω n s r lg n sω = sω, where the last equality follows from the definition of r. Finally, by (3), we get that s() ( + 1)/ ((1 ε) + 1)/ < 1 and so s ω / tends to zero as n. Hence, the lemma follows by Markov s inequality. 4. Lower bound Before we move to the roof of the lower bound, we need the following technical lemma. Lemma 4.1. Let 1/ x, y 1 and β x, β y be any ositive real numbers. Then, there exist unique s and z such that 1/ s, z 1 and Moreover, (x + (1 x) ) βx (y + (1 y) ) βy = (z + (1 z) ) βx+βy = s βx+βy. (6) (x 3 + (1 x) 3 ) βx (y 3 + (1 y) 3 ) βy (x + (1 x) ) βx (y + (1 y) ) βy (z3 + (1 z) 3 ) βx+βy (z + (1 z) ) βx+βy = ( ) βx+β 3s 1 y. (7) s The lemma can be inductively alied to obtain the following corollary. Corollary 4.. Let 1/ x 1, x,..., x k 1. Then, there exist unique s and z such that 1/ s, z 1 and Moreover, k (x i + (1 x i ) ) = (z + (1 z) ) k = s k. i=1 k i=1 (x3 i + (1 x i ) 3 ) k i=1 (x i + (1 x i) ) (z3 + (1 z) 3 ) k (z + (1 z) ) k = ( ) k 3s 1. s

7 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 7 Proof of Lemma 4.1. Let 1/ x, y 1 and β x, β y be fixed ositive real numbers. First we show that there exist unique numbers z and s satisfying (6). Since f(t) := t +(1 t) is increasing on [1/, 1], we get (f(1/)) βx+βy (x + (1 x) ) βx (y + (1 y) ) βy (f(1)) βx+βy. Clearly, (f(t)) βx+βy is also increasing and continuous on t [1/, 1], and thus, there is a unique real number z such that 1/ z 1 and (x + (1 x) ) βx (y + (1 y) ) βy = (f(z)) βx+βy. To finish the roof of (6), set s = z + (1 z) and observe that 1/ s 1, since 1/ z 1. Now we move to the roof of (7). Let Since for every real number t, we get Furthermore, and so a = x + (1 x) and b = y + (1 y). t 3 + (1 t) 3 = 3(t + (1 t) ) 1, (8) x 3 + (1 x) 3 = 3a 1 and y 3 + (1 y) 3 = 3b 1. z + (1 z) = (x + (1 x) ) βx βx+βy (y + (1 y) ) βy βx+βy βx βy z 3 + (1 z) 3 3a βx+βy b βx+βy 1 =. In order to show the inequality in (7) it suffices to rove = a βx βx+βy b βy βx+βy, (z 3 + (1 z) 3 ) βx+βy (x 3 + (1 x) 3 ) βx (y 3 + (1 y) 3 ) βy, which is equivalent to βx βy 3a βx+βy b βx+βy 1 and subsequently to Set (3a) βx βx+βy (3b) βy βx+βy βx+βy ( ) βx ( 3a 1 3b 1 ) βy 1 (3a 1) βx βx+βy (3b 1) βy βx+βy. (9) = β x + β y β x and q = β x + β y β y. Now, showing (9) (and hence also (7)) is equivalent to showing (3a) 1 (3b) 1 q (3a 1) 1 (3b 1) 1 q + 1.

8 8 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT The latter inequality immediately from Hölder s inequality (see, for examle, [3]): indeed, let a 1 = (3a 1) 1, b1 = (3b 1) 1 q and a = b = 1. (Observe that a 1 and b 1 are well-defined since 3a 1 > 0 and 3b 1 > 0.) Then, since 1/ + 1/q = 1, > 1, and q > 1, Hölder s inequality yields (a 1 + a ) 1 (b q 1 + b q ) 1 q a1 b 1 + a b, as required. Finally note that the equality in (7) follows from (8) alied with s = z + (1 z). The roof of the lemma is finished. As we did in the revious section, we kee the definition of the function r = r() for constant (see (4)). We extend it here for sarser grahs as follows (in a different way than in the revious section): suose that tends to zero as n and that n = n α+o(1) for some α [0, 1]. This time, r = r() is defined such that (n) s r lg n = (log n)(log (n)), that is, r = r() = (lg(n) lg log n lg log(n)) (lg n)(lg s) (lg(n) lg log n lg log(n)) lg n α, since s 1/. Now we are ready to come back to the roof of the lower bound. The lower bound in Theorem 1.1 follows immediately from the following lemma,since the condition (log n) (log log n) /n is equivalent to the condition (log n) (log(n)) n. Lemma 4.3. Suose that = (n) is such that (log n) (log(n)) n and 1 ε, for some fixed ε (0, 1). Let G G(n, ). Then, a.a.s. χ s (G) r lg n, rovided that = o(1), and χ s (G) (1 + o(1))r lg n otherwise. Proof. First, let us note that, since the exected degree tends to infinity faster than (log n)(log (n)), it follows immediately from Chernoff s bound and the union bound that a.a.s. all vertices have degree at most, say, n. Hence, since we aim for a statement that holds a.a.s., we may assume that the maximum degree of G is at most n. In fact, the argument is slightly more delicate and will be exlained soon. Suose that we are given a colouring of the vertices. We artition all colours into imortant and unimortant ones: a colour is imortant if the number of vertices of that colour is at most log n/. First, let us show that unimortant colours can distinguish only a few edges. Formally, we claim the following: Claim: A.a.s. each set of log n/ vertices dominates all but at most log n/ vertices.

9 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 9 Proof of the claim. Note that the exected number of airs of disjoint sets of size log n/ with no edge between them is at most ( ) ( ) 4 log n/ ( n ne log n (1 ) ( log n/) ex 4 ) log n log n ( ( = o n 4 log n/ ex 4 )) log n = o(1). The claim follows from the first moment method. Hence, if is constant, then O(log n) unimortant colours can distinguish only O(log n) = o(n) vertices. All remaining vertices will have all unimortant colours resent in their neighbourhood colour sets; as a result, no edge in the grah induced by these vertices can be distinguished by unimortant colours. On the other hand, if = o(1), then at most ( + o(1)) lg(n) ( + o(1)) lg n unimortant colours can distinguish at most 6 log n/ = o(n) vertices, since n log n. As a consequence of the claim, we may concentrate on imortant colours from now. Suose that a colouring c : I [k] using k imortant colours is fixed; I V is the set of vertices coloured with imortant colours. Moreover, let us fix a set U V of O(log n/) = o(n) vertices that are (ossibly) distinguished by unimortant colours. Our goal is to estimate the robability q(c, U) that imortant colours distinguish endoints of edges in G[V \ (I U)], which is the grah induced by those vertices that are coloured with unimortant colours and that are adjacent to at least one vertex from each unimortant colour class. Since the number of configurations to investigate is at most ( n log n ) O(log n) ( n O( log n ) ) ( (n) O(log n/) = ex O it is enough to bound q(c, U) from above by, say, ( (log n) (log(n)) )), Q = Q() := ex( (log n)(log 3/ (n))/). (10) The result will then follow immediately by the union bound. The exected number of edges in G[V \(I U)] is ( ) (1 o(1))n n /3 n log n, and so it follows from Chernoff s bound that with robability at most ex( n log n) Q/ the number of edges is smaller than, say, n /4. On the other hand, if the maximum degree in G[V \ (I U)] is larger than n (for some configuration (c, U)), then we sto the whole argument and claim no lower bound for χ s (G). Recall that at the beginning of the roof, we showed that a.a.s. (G) n. Clearly, if this is the case, then (deterministically) the degree of each vertex in G[V \ (I U)] is at most n. Hence, we may condition on the event that the grah G[V \ (I U)] has the following two roerties: (i) the number of edges is at least n /4, and (ii) no vertex has degree more than n. It is imortant that no edge between V \ (I U) and I has been exosed yet.

10 10 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT Let us focus on constant first. Suose that the number of imortant colours is equal to 5 log log n k := r lg n + = r lg n O(log log n) r log n, log s where the error term follows from (3) and from our assumtion that 1 ε from which we get, as before, s ((1 ε) +1)/ < 1. Suose that for a given configuration (c, U), the robability (x, y) that two adjacent vertices x, y from V \ (I U) are not distinguished by imortant colours is t k for some t s. (Recall that s k is the lower bound for (x, y) which can be attained when all colour classes have size l 0.) We will use Suen s inequality to obtain an uer bound for q(c, U). Let I = {(x, y) : x, y V \ (I U), x y, xy E}. For any (x, y) I, let A x,y be the event (with the corresonding indicator random variable I x,y ) that the neighbourhood colour sets (restricted to imortant colours only) of x and y are equal. Let X = (x,y) I I x,y. We wish to estimate the robability that X = 0. Denote by κ i (for 1 i k) the number of vertices coloured by colour i. Suose first that t = s. This means k ( P(A x,y ) = [(1 ) κ i ] + [1 (1 ) κ i ] ) = s k. (11) In this case, i=1 µ = s k I s k n 4 = sr lg n log 5 n n 4 = log5 n log 4 n, 4 where the last equality follows from the definition of r. Observe that for (x, y) and (y, z) in I we get k ( P(A x,y A y,z ) = [(1 ) κ i ] 3 + [1 (1 ) κ i ] 3). i=1 Thus, since the number of airs (x, y) and (y, z) is bounded by I 4n, we have k ( I 4n [(1 ) κ i ] 3 + [1 (1 ) κ i ] 3) i=1 k = I s k i=1 4n ([(1 )κ i ] 3 + [1 (1 ) κ i ] 3 ) k i=1 ([(1 )κ i ] + [1 (1 ) κ i ] ) = µ 4n k i=1 ([(1 )κ i ] 3 + [1 (1 ) κ i ] 3 ) k i=1 ([(1 )κ i ] + [1 (1 ) κ i ] ). Now, using (11), we aly Corollary 4. with x i = (1 ) κ i if (1 ) κ i > 1/, and x i = 1 (1 ) κ i otherwise, to get ( ) k 3s 1 µ 4n. s

11 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 11 Now we will rove that µ. Using Taylor exansion at s = 1, one can show that for any s [1/, 1] we have ( 3s 1 ) s = 3s 1 = 1 3 s s 3/ 8 (1 s) 5 8 (1 s) (1 s) (1 s). Furthermore, since (3) together with 1 ε imlies we get ( 3s 1 s Finally, s ε, ) / s 1 3ε /3, and hence ( µ (4n)s k/ 1 3 ) k 3 ε ( ) = µ (4n) s (r lg n)/ log 5/ n = µ (4 log 5/ n)n Ω(ε) µ. n Ω(ε ) δ (n)s k = 4n 1 log 5 n = o(1). It follows from Suen s inequality (see (5)) that P(X = 0) ex ( µ + e δ) = ex( (1 + o(1))µ) Q/, and q(c, U) Q/ + P(X = 0) Q, as needed (see (10)). Suose now that t > s. In this case, µ is larger than before but, unfortunately, grows faster than µ (as t grows) and eventually becomes larger than µ. In order to avoid this undesired situation, we make the deendency grah sarser so that µ is still of order log 5 n. Let us note that if t u, where u is defined so that (u/s) k = n, then Suen s inequality can be avoided and we can simly use Chernoff s bound to obtain the desired uer bound for P(X = 0): indeed, it follows from the claim roved above that there exists a matching in G[V \ (I U)] consisting of at least n/3 edges; otherwise, the remaining n O(log n/) n/3 = n/3 o(n) log n/ vertices in V \ (I U) would form an indeendent set that could be slit into two sets of equal size and, clearly, no edge will be resent between them, contradicting the claim. Since the events are now indeendent, and the exected number of edge endoints not distinguished is µ tk n 3 = sk (t/s) k n sk n = log5 n log 4 n, the desired bound holds, since P(X = 0) ex( Ω(µ)) by Chernoff s bound. It remains to consider the case s < t < u. Our goal is to scale the degree in the deendency grah down by a multilicative factor of ξ = ξ(t) := (s/t) k (s/u) k = 1/(n). Let I ξ be a random subgrah of I: each (x, y) I is indeendently ut into I ξ with robability ξ. Since I n /4, E I ξ ξn /4 n/4, and thus a.a.s. I ξ ξn /5. Moreover, since the maximum degree in the deendency grah is at most n, by

12 1 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT Chernoff s bound together with a union bound over all vertices, it follows that a.a.s. the maximum degree in a random subgrah of it is at most max{4ξn, 10 log n} 40ξn log n. Therefore, the deterministic (non-constructive) conclusion is that there exists a subgrah of the deendency grah with at least ξn /5 airs and the maximum degree at most 40ξn log n. We restrict ourselves to this subgrah, stressing one more time that no edge between V \ (I U) and I is exosed yet. Now, µ = t k I ξ s k ( t s ) k ξn 5 = log5 n 5 log 4 n. Moreover, ( ) k ( 3t 1 3t 1 t k I ξ (40ξn log n) µ (80ξn log n) t t ( ) k 3t 1 = µ (80n log n) s k. t Let h(t) := (3t 1)/(t ). Note that h(1/) = 1, h(t) is increasing on the interval [1/, /3] attaining h(/3) = 9/8, and then is decreasing on the interval [/3, 1] going back to h(1) = 1. Hence, if s /3, then (h(t)s) k, as a function of t, is maximized for t = /3. Then, as a function of s, since s /3, r = / lg(1/s), and k r lg n, (h(/3)s) k is maximized for s = /3. We are back to the case t = s that we already checked. On the other hand, if s > /3, then, since t s, h(t) and therefore is maximized again for t = s. Hence, in both cases we have µ. Finally, δ (40ξn log n)t k = 80n(log n)s k = 80n 1 log 6 n = o(1). Hence, Suen s inequality can be alied as before, and the roof for constant is finished. ) k The case = o(1) can be verified exactly the same way. In fact, it is slightly easier since s = 1/ + O( ), r + o(1), and we do not have to worry about an increasing value of s (and therefore, neither about an increasing value of r). We oint out only the adjustments of the roof. Recall that the definition of r is extended to the case = o(1). The number of colours is now k = r lg n, and we have (n) s k = (log n)(log (n)). For the case t = s, we have µ = s k I s k n 4 = (log n)(log (n)) 4 (log n)(log 3/ (n)),

13 THE SET CHROMATIC NUMBER OF RANDOM GRAPHS 13 as desired for the union bound (see (10)). Since now s = 1/ + O( ), the argument for is much easier: s k ( 1 + O( )) k I 4n µ(4n) ( s + O( ) ) k = 4µns k ex(o( log n)) = 4µ(n) 1 (log n)(log (n)) ex(o( log n)) µ, since n (log n)(log (n)). (Note that for = Ω(1/ log n) but still = o(1), we have ex(o( log n)) = n o(1). However, this causes no roblem, as (n) 1 = n 1+o(1) and so µ. Otherwise, that is, if = o(1/ log n), we have ex(o( log n)) 1 and µ follows easily.) Finally, as before, and again using the same lower bound on n, we have δ (n)s k = 4(n) 1 (log n)(log (n)) = o(1), and Suen s inequality can be alied. Now let us consider the case t > s. The definition of u is not affected, and for t u we use Chernoff s bound since the events are indeendent. The only difference is that the new value of s k has to be used to get µ tk n 3 = sk (t/s) k n 3 sk n 3 = (log n)(log (n)) 3 (log n)(log 3/ (n)), as desired. For the case s < t < u, the definition of ξ remains the same and again, after adjusting the value of s k we get µ (log n)(log 3/ (n))/. The argument for µ is not affected. Finally, δ (40ξn log n)t k = 80n(log n)s k = 80(n) 1 (log 3 n)(log (n)) = o(1), rovided that n (log 3 n)(log (n)). For slightly sarser grahs, that is, when (log n)(log (n)) n = O((log 3 n)(log (n))), observe that we only have that δ = o(log n), but in fact we can show a stronger bound for : it follows that and thus (h(/3)s) k = (9/8) lg n/ lg(3/) n log 5 n = n lg(9/8)/ lg(3/) log 5 n, (80n log n)(h(t)s) k = O(n lg(9/8)/ lg(3/) log 9 n log log n) = n Ω(1). Hence, = µn Ω(1), and so e δ µn Ω(1) n o(1) = o(µ), as needed for Suen s inequality to be useful. The roof is finished. References [1] G. Chartrand, F. Okamoto, C.W. Rasmussen, and P. Zhang, The set chromatic number of a grah, Discuss. Math. Grah Theory, 9 (009), [] R. Gera, F. Okamoto, C. Rasmussen, and P. Zhang, Set colorings in erfect grahs, Math. Bohem. 136 (011)

14 14 ANDRZEJ DUDEK, DIETER MITSCHE, AND PAWE L PRA LAT [3] G. Hardy, J. Littlewood, and G. Pólya, Inequalities, Rerint of the 195 edition. Cambridge University Press, Cambridge, [4] S. Janson, New versions of Suen s correlation inequality, Random Structures & Algorithms 13 (1998), [5] S. Janson, T. Luczak, A. Ruciński, Random Grahs, Wiley, New York, 000. [6] T. Luczak, The chromatic number of random grahs, Combinatorica 11 (1991), no. 1, [7] C. McDiarmid, On the chromatic number of random grahs, Random Structures & Algorithms 1 (1990), no. 4, [8] J.-S. Sereni, Z. Yilma, A tight bound on the set chromatic number, Discuss. Math. Grah Theory, 33 (013), [9] S. Suen, A correlation inequality and a oisson limit theorem for nonoverlaing balanced subgrahs of a random grah, Random Structures & Algorithms 1 (1990), Deartment of Mathematics, Western Michigan University, Kalamazoo, MI, USA address: andrzej.dudek@wmich.edu Université de Nice Sohia-Antiolis, Laboratoire J-A Dieudonné, Parc Valrose, Nice cedex 0 address: dmitsche@unice.fr Deartment of Mathematics, Ryerson University, Toronto, ON, Canada address: ralat@ryerson.ca