Clique-chromatic number of dense random graphs
CClique-chromatic number of dense random graphs
Yu. Demidovich ∗ M. Zhukovskii † Abstract
The clique chromatic number of a graph is the minimum number of colors required to as-sign to its vertex set so that no inclusion maximal clique is monochromatic. McDiarmid, Mitscheand Pra (cid:32)l at proved that the clique chromatic number of the binomial random graph G (cid:0) n, (cid:1) is atmost (cid:0) + o (1) (cid:1) log n with high probability. Alon and Krivelevich showed that it is greater than log n with high probability. In this paper we show that the upper bound is asymptoticallytight. Keywords: random graphs, clique chromatic number
A clique coloring of a graph G = ( V, E ) is a coloring of V such that there is no inclusion maximalmonochromatic clique (i.e. every monochromatic clique has a vertex adjacent to all its vertices andcolored in another color). The clique chromatic number χ c ( G ) is the minimum number of colors in aclique coloring of G. The random graph G ( n, ) is a random uniformly distributed element of the set of all graphs on [ n ] (or, in other words, every pair of vertices is adjacent with probability p = independently). Wesay that a graph property Q holds with high probability (whp), if P (cid:0) G ( n, ) ∈ Q (cid:1) → as n → ∞ . Everywhere below we write log n for the logarithm to the base of and ln n for the natural logarithm.In 2016 [1], McDiarmid, Mitsche and Pra (cid:32)l at proved that whp χ c ( G ( n, )) (cid:54) (cid:0) + o (1) (cid:1) log n (i.e.,for every ε > whp χ c (cid:0) G (cid:0) n, (cid:1)(cid:1) (cid:54) (cid:0) + ε (cid:1) log n ) and asked about a lower bound. In 2017 [2], Alonand Krivelevich proved that log n is the right order of magnitude by showing that whp χ c (cid:0) G ( n, ) (cid:1) (cid:62) log n. They also suggested that the right constant in front of log n is . In the present work weprove that this is true. ∗ Moscow Institute of Physics and Technology (National Research University), Laboratory of Combinatorial andGeometric Structures, e-mail: [email protected] † Moscow Institute of Physics and Technology (National Research University), Laboratory of Combinatorial andGeometric Structures, e-mail: [email protected] a r X i v : . [ m a t h . C O ] D ec heorem 1 With high probability χ c ( G ( n, )) (cid:62) (cid:0) + o (1) (cid:1) log n. The proof of Theorem 1 is based on Lemma 2.3 from [2] that Alon and Krivelevich used to provetheir lower bound. We state its slightly modified version below. The proof given by Alon and Kriv-elevich works properly for this modification. However, for the convenience of readers we give its proofin Appendix.Denote by N ( v, U ) the set of non-neighbors of v in U (a host graph is always clear from thecontext). Let < α (cid:54) , k = (cid:100) (1 + α ) log n (cid:101) . Lemma 1
Fix a set Y ⊂ [ n ] of size | Y | (cid:62) n
12 +4 α log n . The probability that • N ( v, Y ) (cid:62) n +2 α for every v ∈ [ n ] \ Y, • there are no cliques K k ⊂ Y such that every vertex in [ n ] \ Y has a non-neighbor in K k is at most e − α | Y | log n (1+ o (1)) . In Section , we define a graph property that due to Lemma 1 holds for G (cid:0) n, (cid:1) whp. In Section , we prove Theorem 1 by showing that any graph with the described property has a clique chromaticnumber at least log n (1 + o (1)) . Put s := (cid:98) (1 − δ ) · log n (cid:99) , < δ < . Let G be a graph on the vertex set [ n ] . Denote by N ( v , . . . , v j ) the set of all non-neighbors of v , . . . , v j . Set N ( ∅ ) = [ n ] . Definition 1 G has Property C , if for any j ∈ [ s ] ∪ { } and vertices v , . . . , v j , | N ( v , . . . , v j ) | (cid:62) n j − √ n ln n,
2. any Y ⊆ N ( v , . . . , v j ) , s.t. • | Y | (cid:62) | N ( v ,...,v j ) | log n , • | N ( v, Y ) | (cid:62) n + δ for every v ∈ [ n ] \ Y, contains a maximal clique. Lemma 2
Whp G ( n, ) has Property C . roof of Lemma 2. Let α = δ , v , . . . , v j ∈ [ n ] . Denote by ξ the number of vertices in G ( n, ) nonadjacent to any of v , . . . , v j . Denote by B the event that ξ (cid:62) n j − √ n ln n. Since ξ has the binomialdistribution Bin (cid:0) n − j, j (cid:1) , by the Chernoff’s bound [3, Theorem 2.1], P (cid:0) B (cid:1) (cid:54) e − j − ln n (1+ o (1)) . Denote by A the event saying that the Property from Definition 1 holds for v , . . . , v j . Exposefirst all the edges having vertices among v , . . . , v j . Define G j := G | [ n ] \{ v ,...,v j } d = G ( n − j, ) . Choose Y ⊂ N ( v , . . . , v j ) ⊂ V ( G j ) of size at least ξ log n . Since the distribution of edges of G j does not dependon neighbors of v , . . . , v j , we may apply Lemma 1 to G j (a maximal clique in G j should be alsomaximal in G ) and the union bound (over Y ): P (cid:0) A ∩ B (cid:1) = P ( A ∩ B ) + P ( B ) (cid:54) P ( A | B ) + P ( B ) (cid:54)(cid:54) max N (cid:62) n j −√ n ln n (cid:88) y (cid:62) N log n (cid:18) Ny (cid:19) e − α y log n (1+ o (1)) + e − j − ln n (1+ o (1)) (cid:54)(cid:54) max N (cid:62) n j −√ n ln n (cid:88) y (cid:62) N log n e y (cid:16) ln log n +1 − α log n (cid:17) (1+ o (1)) + e − ln4 n (1+ o (1))2 (cid:54)(cid:54) max N (cid:62) n j −√ n ln n ne − α N (1+ o (1)) + e − ln4 n (1+ o (1))2 = e − ln4 n (1+ o (1))2 . By the union bound, the probability that G ( n, ) does not have Property C is at most s (cid:88) j =0 (cid:18) nj (cid:19) e − ln4 n (1+ o (1))2 (cid:54) sn s e − ln4 n (1+ o (1))2 → , n → ∞ . (cid:4) Let G be a graph on the vertex set [ n ] with Property C . Suppose there is a proper assignment of s = (cid:98) (1 − δ ) · log n (cid:99) , < δ < , colors to [ n ] . Due to Lemma 2, to prove Theorem 1 it is sufficient toget a contradiction with Property C .There exists a color class Y with at least n log n vertices. Find a vertex v ∈ [ n ] \ Y that has thesmallest number of non-neighbors in Y. Since the coloring is proper, N ( v , Y ) < n + δ by Property C .For j = 1 , . . . , s − repeat the following procedure.ut X = N ( v , . . . , v j ) , x = | X | . There exists a color class that has at least x log n vertices in X, choose a class that has the biggest number of vertices in X and denote its intersection with X by Y. Find a vertex v j +1 ∈ [ n ] \ Y that has the smallest number of non-neighbors in Y. Since the coloring of G is proper, N ( v j +1 , Y ) < n + δ by Property C .At the end of the process, we obtain the set N ( v , . . . , v s ) of size Ω( n + δ ) (by Property C ) coloredin s colors with every color class smaller than n + δ — contradiction. We have shown that the clique chromatic number χ c (cid:0) G (cid:0) n, (cid:1)(cid:1) is, whp, (cid:0) + o (1) (cid:1) log n. The sameproof can be applied to the random graph G ( n, p ) with constant probability < p < . So, whp χ c ( G ( n, p )) = log − p n (1 + o (1)) . Notice that our arguments also imply that all clique colorings in not too many colors are closeto the “greedy coloring”. Formally, for every δ > , ε > whp for every coloring Y , . . . , Y s ofthe vertex set of G ( n, p ) in at most log − p n colors ordered in a decreasing order from a largestcolor class to a smallest one for every j < (cid:0) − δ (cid:1) log − p n there exist vertices v , . . . , v j such that | ( Y ∪ . . . ∪ Y j ) \ ( N ( v ) ∪ . . . ∪ N ( v j )) | (cid:54) n + ε , where N ( v ) is the set of all neighbors of v in G ( n, p ) . This work is supported by the Ministry of Science and Higher Education of the Russian Federation inthe framework of MegaGrant no 075-15-2019-1926.
References [1] C. McDiarmid, D. Mitsche, P. Pra (cid:32)l at, “Clique coloring of binomial random graphs” , RandomStructures & Algorithms, (2016) 10.1002/rsa.20804.[2] N. Alon, M. Krivelevich, “Clique coloring of dense random graphs” , Journal of Graph Theory(2017), 1–6.[3] S. Janson, T. (cid:32)L uczak, A. Ruci´nski, “Random Graphs” , Wiley, New York, 2000. ppendix: proof of Lemma 1
Set | Y | = y (cid:62) n
12 +4 α log n . Let X ⊂ [ n ] \ Y be a set of size t := (cid:100) α log n (cid:101) . The probability that the number ofnon-edges between X and Y is at most (cid:0) − α (cid:1) yt is at most e − ( α ) yt (cid:54) e − α n y by the Chernoff’sbound [3, Theorem 2.1]. Thus the probability of existence of X with that many non-edges is at most (cid:18) n − yt (cid:19) e − α n y (cid:54) n t e − α n y (cid:54) e − y α n (1+ o (1)) . Let B be the set of all vertices in [ n ] \ Y that have less than (cid:0) − α (cid:1) y non-neighbors in Y. Asshown above, | B | (cid:62) t with probability at most e − y α n (1+ o (1)) . Expose all edges between Y and [ n ] \ Y. Assume that, for every v ∈ [ n ] \ Y, N ( v, Y ) (cid:62) n +2 α and b := | B | < t. Put m = n + α and choose for each v ∈ B a subset of Y of size m consisting of non-neighborsof v where all these subsets are pairwise disjoint. It is possible to choose them sequentially as eachvertex in [ n ] \ Y has at least n +2 α (cid:29) mt non-neighbors in Y. We denote them Z , . . . , Z b and put Y (cid:48) = Y \ b (cid:70) i =1 Z i . Let y (cid:48) = | Y (cid:48) | = y − bm = y (1 + o (1)) . For every vertex v ∈ [ n ] \ ( Y ∪ B ) , holds | N ( v, Y (cid:48) ) | (cid:62) (cid:18) − α (cid:19) y − mt = (cid:18) − α (cid:19) y (cid:48) (1 + o (1)) . Choose k − b disjoint sets Z b +1 , . . . , Z k in Y (cid:48) of size m uniformly at random. It is clear thatwhp every vertex of [ n ] \ ( Y ∪ B ) has at least (cid:0) − α (cid:1) m (1 + o (1)) non-neighbors in Z i for every i ∈ [ k ] \ [ b ] . Indeed, let v ∈ [ n ] \ ( Y ∪ B ) . If we choose sets Z b +1 , . . . , Z k sequentially then, for each i ∈ [ k ] \ [ b ] , the conditional distribution of the number of non-neighbors of v in Z i , given Z b +1 , . . . , Z i − , has hypergeometric distribution with expectation λ (cid:62) (cid:0) − α (cid:1) m (1 + o (1)) . By [3, Theorem 2.10], theprobability that this number is smaller than λ − m is at most e − √ m (1+ o (1)) ( − α ) . By the union bound,the probability of existence of a vertex v ∈ [ n ] \ ( Y ∪ B ) and i ∈ [ k ] \ [ b ] such that v has less than (cid:0) − α (cid:1) m (1 + o (1)) (for an appropriate choice of o (1) ) non-neighbors in Z i is at most nke − Ω( √ m ) → ,n → ∞ . So, for n large enough, we may choose disjoint subsets Z b +1 , . . . , Z k ⊂ Y (cid:48) of size m s.t. each v ∈ [ n ] \ ( Y ∪ B ) has at least (cid:0) − α (cid:1) m (1 + o (1)) in each Z i . Let Z b +1 , . . . , Z k be such sets.Denote by F the family of all subsets of size k of Y that contain one element in each set Z i , i ∈ [ k ] , and contain at least one non-neighbor of each vertex v ∈ [ n ] \ Y. By the definition of Z i , i ∈ [ b ] , anylement of F contains at least one non-neighbor of every v ∈ B. Then, by the union bound, − |F | m k (cid:54) n (cid:18)(cid:18)
12 + α (cid:19) + o (1) (cid:19) k − b (cid:54) n ( − α + ( α ) log ( α )) (1+ o (1)) (cid:54)(cid:54) n − α ( − ( α )
13 ln 2 ) (1+ o (1)) (cid:54) n − α (1+ o (1)) . Therefore |F | (cid:62) m k (1 − o (1)) . It remains to prove that, if we expose the edges inside Y, then with probability at least − e − n log n at least one set from F induces a clique.For each member K of F denote by I K the indicator random variable equal to iff K inducesa clique in G ( n, ) . Define I := (cid:80) K ∈F I K , µ := E I = |F | − ( k ) = (1 − o (1)) m k − ( k ) . Define ∆ := (cid:80)
K, K (cid:48) P ( I K = I K (cid:48) = 1) , where the summation is over all ordered pairs K, K (cid:48) of members of F thatsatisfy (cid:54) | K ∩ K (cid:48) | (cid:54) k − . Denote by ∆ i the contribution of pairs K, K (cid:48) ∈ F for which | K ∩ K (cid:48) | = i. Then ∆ µ = k − (cid:88) i =2 ∆ i µ (cid:54) k − (cid:88) i =2 |F | µ − ( k ) (cid:18) ki (cid:19) ( m − k − i − ( k ) + ( i ) (cid:54)(cid:54) k − (cid:88) i =2 m k − i i ) |F | (cid:18) ki (cid:19) (cid:54) (cid:32) k m + k − (cid:88) i =3 (cid:32) (cid:0) ki (cid:1) i ) m i (cid:33)(cid:33) (1 + o (1)) . If (cid:54) i < (cid:100) α (cid:101) , then (cid:0) ki (cid:1) i ) m i < (cid:32) k i m (cid:33) i (cid:54) (cid:32) k α m (cid:33) (cid:54) m − o (1) . If (cid:100) α (cid:101) (cid:54) i (cid:54) k − , then (cid:0) ki (cid:1) i ) m i < (cid:32) k i m (cid:33) i (cid:54) (cid:32) kn α n α (cid:33) α = (cid:18) n α − o (1) (cid:19) α = 1 n − o (1) . Hence, ∆ µ (cid:54) k m (1 + o (1)) . By the Janson’s inequality [3, Theorem 2.18], P ( I = 0) (cid:54) e − µ (cid:54) e − (1+ o (1)) m k = e − n α − o (1) ..