On the strong chromatic number of random graphs
aa r X i v : . [ m a t h . C O ] J un On the strong chromatic number of random graphs
Po-Shen Loh ∗ Benny Sudakov † Abstract
Let G be a graph with n vertices, and let k be an integer dividing n . G is said to be strongly k -colorable if for every partition of V ( G ) into disjoint sets V ∪ . . . ∪ V r , all of size exactly k , thereexists a proper vertex k -coloring of G with each color appearing exactly once in each V i . In thecase when k does not divide n , G is defined to be strongly k -colorable if the graph obtained byadding k (cid:6) nk (cid:7) − n isolated vertices is strongly k -colorable. The strong chromatic number of G isthe minimum k for which G is strongly k -colorable. In this paper, we study the behavior of thisparameter for the random graph G n,p . In the dense case when p ≫ n − / , we prove that the strongchromatic number is a.s. concentrated on one value ∆ + 1, where ∆ is the maximum degree of thegraph. We also obtain several weaker results for sparse random graphs. Let G be a graph, and let V , . . . , V r be disjoint subsets of its vertex set. An independent transversal with respect to { V i } ri =1 is an independent set in G which contains exactly one vertex from each V i .The problem of finding sufficient conditions for the existence of an independent transversal, in termsof the ratio between the part sizes and the maximum degree ∆ of the graph, dates back to 1975,when it was raised by Bollob´as, Erd˝os, and Szemer´edi [10]. Since then, much work has been done[1, 3, 5, 14, 15, 17, 18, 22, 26, 27], and this basic concept has also appeared in several other contexts,such as linear arboricity [4], vertex list coloring [23, 24, 8], and cooperative coloring [2, 19]. In thegeneral case, it was proved by Haxell [14] that an independent transversal exists as long as all partshave size at least 2∆. The sharpness of this bound was shown by Szab´o and Tardos [26], extendingearlier results of [18] and [27]. On the other hand, we proved in [19] that the upper bound can befurther reduced to (1 + o (1))∆ if no vertex has more than o (∆) neighbors in any single part. Such acondition arises naturally in certain applications, e.g., vertex list coloring.In the case when all of the V i are of the same size k , it is natural to ask when it is possible tofind not just one, but k disjoint independent transversals with respect to the { V i } . This is closelyrelated to the following notion of strong colorability. Given a graph G with n vertices and a positiveinteger k dividing n , we say that G is strongly k -colorable if for every partition of V ( G ) into disjointsets V ∪ . . . ∪ V r , all of size exactly k , there exists a proper vertex k -coloring of G with each colorappearing exactly once in each V i . Notice that G is strongly k -colorable iff the chromatic number of ∗ Department of Mathematics, Princeton University, Princeton, NJ 08544. E-mail: [email protected]. Re-search supported in part by a Fannie and John Hertz Foundation Fellowship, an NSF Graduate Research Fellowship,and a Princeton Centennial Fellowship. † Department of Mathematics, Princeton University, Princeton, NJ 08544, and Institute for Advanced Study, Prince-ton. E-mail: [email protected]. Research supported in part by NSF CAREER award DMS-0546523, NSFgrant DMS-0355497, USA-Israeli BSF grant, Alfred P. Sloan fellowship, and the State of New Jersey. G by adding a union of vertex disjoint k -cliques is k . If k does not divide n ,then we say that G is strongly k -colorable if the graph obtained by adding k (cid:6) nk (cid:7) − n isolated vertices isstrongly k -colorable. The strong chromatic number of G , denoted sχ ( G ), is the minimum k for which G is strongly k -colorable.The concept of strong chromatic number first appeared independently in work by Alon [4] andFellows [11]. It was also the crux of the longstanding “cycle plus triangles” problem popularized byErd˝os, which was to show that the strong chromatic number of the cycle on 3 n vertices is three. Thatproblem was solved by Fleischner and Stiebitz [12]. The strong chromatic number is known [11] tobe monotonic in the sense that strong k -colorability implies strong ( k + 1)-colorability. It is also easyto see that sχ ( G ) must always be strictly greater than the maximum degree ∆: simply take V to bethe neighborhood of a vertex of maximal degree, and partition the rest of the vertices arbitrarily. Theintriguing question of bounding the strong chromatic number in terms of the maximum degree has notyet been answered completely. Alon [5] showed that there exists a constant c such that sχ ≤ c ∆ forevery graph. Later, Haxell [15] improved the bound by showing that it is enough to use c = 3, and infact even c = 3 − ǫ for ǫ up to 1 / K ∆ , ∆ cannot be strongly (2∆ − K ∆ , ∆ into the sets V and V , respectively. Then these 2∆ vertices shouldget different colors. It is believed that this lower bound is tight and the strong chromatic number ofany graph with maximum degree ∆ should be at most 2∆.It is natural to wonder what is the asymptotic behavior of the strong chromatic number for therandom graph G n,p , relative to the maximum degree of the graph. As usual, G n,p is the probabilityspace of all labeled graphs on n vertices, where every edge appears randomly and independently withprobability p = p ( n ). We say that the random graph possesses a graph property P almost surely , ora.s. for brevity, if the probability that G n,p satisfies P tends to 1 as n tends to infinity. One of themost interesting phenomena discovered in the study of random graphs is that many natural graphinvariants are highly concentrated (see, e.g., [21] for the result on the clique number and [25, 20, 6] forthe concentration of the chromatic number). In this paper we show that the strong chromatic numberis another example of a tightly concentrated graph parameter. For dense random graphs, it turns outthat we can concentrate sχ ( G n,p ) on a single value, and for some smaller values of p we were onlyable to determine sχ ( G n,p ) asymptotically. In the statement of our first result, and in the rest of thispaper, the notation f ( n ) ≫ g ( n ) means that f /g → ∞ together with n . Also, all logarithms are inthe natural base e . Theorem 1.1
Let ∆ be the maximum degree of the random graph G n,p , where p < − θ for anyarbitrary constant θ > . (i) If p ≫ (cid:16) log nn (cid:17) / , then almost surely the strong chromatic number of G n,p equals ∆ + 1 . (ii) If p ≫ (cid:16) log nn (cid:17) / , then a.s. the strong chromatic number of G n,p is (1 + o (1))∆ . Unfortunately, our approach breaks down completely when p ≪ n − / . However, for this range of p , we have a different argument which shows how to find at least one independent transversal.2 heorem 1.2 Let ∆ be the maximum degree of the random graph G n,p . If p ≥ log nn , then almostsurely every collection of disjoint subsets V , . . . , V r of G n,p with all | V i | ≥ (1 + o (1))∆ has an inde-pendent transversal. This rest of this paper is organized as follows. In Section 2, we prove both parts of our firsttheorem concerning the strong chromatic number of relatively dense random graphs. We then shiftour attention to the sparser case, proving our second result about transversals in Section 3. The lastsection of our paper contains some concluding remarks. Throughout this exposition, we will make noattempt to optimize absolute constants, and will often omit floor and ceiling signs whenever they arenot crucial, for the sake of clarity of presentation.
In this section, we prove Theorem 1.1, which determines the value of the strong chromatic number ofa rather dense random graph. To this end, we first prove several lemmas that establish certain usefulproperties of random graphs. We will use these properties to find a partition of G n,p into independenttransversals. Lemma 2.1
Let θ > be an arbitrary fixed constant. If q log nn ≪ p < − θ then a.s. G n,p has thefollowing properties. (i) No pair of distinct vertices has more than (1 + o (1)) np common neighbors. (ii) The maximum degree is strictly between np and . np , and there is a unique vertex of maximumdegree. (iii) The gap between the maximum degree and the next largest degree is at least √ np log n . Proof.
For the first property, fix an arbitrary constant δ > u and v . Theircodegree X is binomially distributed with parameters n − p . Thus by the Chernoff bound (see,e.g., Appendix A in [7]), P (cid:2) X ≥ (1 + δ ) np (cid:3) ≤ e − Θ( δ np ) = o ( n − ). Taking a union bound over all O ( n ) choices for u and v , we find that the probability that the first property is not satisfied tends to0 as n → ∞ . The second and third claims are special cases of Corollary 3.13 and Theorem 3.15 in [9],respectively. (cid:3) Lemma 2.2
Let α > be an arbitrary fixed constant and let q log nn ≪ p ≤ . Then almost surely G n,p does not contain a set U of size αnp and
50 log n sets T i , | T i | ≤ (cid:6) p (cid:7) , such that all the sets aredisjoint and for every i all but at most αnp/ vertices in U have neighbors in T i . Proof.
Fix sets U and { T i } as specified above. If all but at most αnp/
50 vertices in U have neighborsin T i , we say for brevity that T i almost dominates U . For a given vertex v , the probability that it hasa neighbor in T i is 1 − (1 − p ) | T i | ≤ − (1 − p ) ⌈ /p ⌉ < / p ≤ /
5, since 1 − (1 − p ) ⌈ /p ⌉ ismaximal in that range when p → / [ T i almost dominates U ] ≤ (cid:18) αnpαnp − αnp/ (cid:19) (cid:18) (cid:19) αnp − αnp/ = (cid:18) αnpαnp/ (cid:19) (cid:18) (cid:19) αnp/ ≤ (cid:18) e (cid:16) (cid:17) (cid:19) αnp/ < − αnp/ . Since all sets T i are disjoint, the events that T i and T j almost dominate U are independent. Thisimplies that P [every T i almost dominates U ] ≤ (cid:16) − αnp/ (cid:17)
50 log n = 3 − αnp log n . Using that log n/p = o ( np ) and ⌈ /p ⌉ ≤ /p , we can bound the probability that there is a choice of { T i } and U which violates the assertion of the lemma by P ≤ (cid:18) nαnp (cid:19) (cid:20) p (cid:18) n /p (cid:19)(cid:21)
50 log n − αnp log n ≤ n αnp (cid:18) p (cid:19)
50 log n n
100 log np − αnp log n = e (1+ o (1)) αnp log n · − αnp log n = o (1) , so we are done. (cid:3) Lemma 2.3
Let α > be an arbitrary fixed constant and let q log nn ≪ p ≤ . Then almost surelyevery collection of at most (cid:6) p (cid:7) disjoint subsets of size αnp in G n,p has an independent transversal. Proof.
Fix a collection of disjoint subsets V , . . . , V r , r ≤ (cid:6) p (cid:7) , of G n,p , each of size αnp . A partialindependent transversal T is an independent set with at most one vertex in every V i , and we say thatit almost dominates some part if all but at most αnp/
50 vertices in that part have neighbors in T .For every V i , let { T ij } be a maximal collection of pairwise disjoint partial independent transversals,each of which almost dominates V i . Then, by Lemma 2.2, a.s. the total number of T ij must be at most r (50 log n ). Delete all the sets T ij from the graph, and let { V ′ i } be the remaining parts. Clearly, itsuffices to find an independent transversal among the { V ′ i } .Since log n/p = o ( np ) and each T ij is a partial transversal, each part loses a total of ≤ r (50 log n ) ≤ (cid:6) p (cid:7) log n = o ( np ) vertices from the deletions. We can now use the greedy algorithm to find anindependent transversal. Take v to be any remaining vertex in V ′ , and iterate as follows. Supposethat we already have constructed a partial independent transversal { v , . . . , v ℓ − } such that v i ∈ V ′ i forall i < ℓ . This partial independent transversal does not almost dominate V ℓ , or else it would contradictthe maximality of { T ℓj } above. So, there are at least αnp/
50 choices for v ℓ ∈ V ℓ that would extend thepartial independent transversal { v , . . . , v ℓ − } . Yet V ℓ lost only o ( np ) vertices in the deletion process,so there is still a positive number of choices for v ℓ ∈ V ′ ℓ as well. Proceeding in this way, we find acomplete independent transversal. (cid:3) Lemma 2.4
Let q log nn ≪ p ≤ . Then the following statement holds almost surely. For every choiceof s and t that satisfies np/ ≤ s ≤ np and
40 log n ≤ t ≤ s − (cid:6) p (cid:7) log n , G n,p does not contain acollection of disjoint subsets U, T , . . . , T t such that | U | = s , each of the | T i | ≤ (cid:6) p (cid:7) , and at least s − t vertices of U have neighbors in every T i . roof. Fix some ( s, t ) within the above range. As we saw in the proof of Lemma 2.2, for a givenvertex v the probability that it has a neighbor in T i is 1 − (1 − p ) | T i | ≤ − (1 − p ) ⌈ /p ⌉ < /
8, and bydisjointness these events are independent for all 1 ≤ i ≤ t . Therefore we can bound the the probabilitythat there is a collection of sets which satisfies the above condition by P ≤ (cid:18) ns (cid:19) (cid:20) p (cid:18) n /p (cid:19)(cid:21) t s (cid:18) (cid:19) ( s − t ) t ≤ n s s ! (cid:0) n /p (cid:1) t s (cid:18) (cid:19) ( s − t ) t ≤ n s +2 t/p (cid:18) (cid:19) ( s − t ) t . (1)Throughout this bound, we use (cid:6) p (cid:7) ≤ p . The first binomial coefficient and the quantity in the squarebrackets bound the number of ways to choose the sets U and { T i } . The 2 s bounds the number of waysto select a subset of size s − t from U , and the final factor bounds the probability that all vertices inthis subset have neighbors in every T i .The logarithm of (1) is quadratic in t with positive t -coefficient. Therefore, the right hand side of(1) is largest when t is minimum or maximum in its range 40 log n ≤ t ≤ s − (cid:6) p (cid:7) log n . Let us beginwith the small end, i.e., t = 40 log n . Then, since log n/p ≪ np and s ≥ np/
2, we have that n s +2 t/p (cid:18) (cid:19) ( s − t ) t ≤ e (1+ o (1)) s log n (cid:18) (cid:19) (40 − o (1)) s log n ≤ e (1+ o (1)) s log n e − (4 − o (1)) s log n = o (cid:0) n − (cid:1) . Similarly, if t = s − (cid:6) p (cid:7) log n , the bound is n s +2 t/p (cid:18) (cid:19) ( s − t ) t ≤ e s log n/p (cid:18) (cid:19) (40 − o (1)) s (cid:6) p (cid:7) log n ≤ e s log n/p e − (4 − o (1)) s (cid:6) p (cid:7) log n = o (cid:0) n − (cid:1) . Since the number of choices for t and s is at most n , we conclude that the probability that theassertion of the lemma is violated is o (1). (cid:3) We start by proving part (i) of Theorem 1.1. If ∆ is the maximum degree of G n,p , then the strongchromatic number must be at least ∆ + 1, as we already mentioned in the introduction. Suppose that G is a graph obtained from G n,p by adding (∆ + 1) ⌈ n ∆+1 ⌉ − n isolated vertices, and we have a partitionof V ( G ) into V ∪ . . . ∪ V r with every | V i | = ∆ + 1. By Lemma 2.1, ∆ ≥ np almost surely, so thisimplies that r ≤ (cid:6) p (cid:7) . Note that if 3 / ≤ p < − θ , then r ≤ Lemma 2.5
Let / ≤ p < − θ , where θ > is an arbitrary fixed constant, and let V ( G ) = V ∪ V be a partition of the vertices of G described above, with | V | = | V | = ∆ + 1 . Then a.s. V can beperfectly matched to V via non-edges of G . roof. Without loss of generality, we may assume that V contains at most n/ G n,p . Let B ⊂ V be those original vertices. The rest of V consists of isolated vertices, so any perfectmatching of B to V trivially extends to a full perfect matching between V and V . Therefore, byHall’s theorem, it suffices to verify that each subset A ⊂ B has at least | A | non-neighbors in V . If A = { v } is a single vertex, this is immediate because | V | > ∆ ≥ d ( v ). For larger A , the Hall conditiontranslates into checking that ∆ + 1 − | N ( A ) | ≥ | A | , where N ( A ) denotes the set of common neighborsof A in V . Since | A | ≥ N ( A ) is at most (1 + o (1)) np .So the Hall condition is satisfied for all A with 2 ≤ | A | ≤ θnp/ < ∆ − (1 + o (1)) np .Let c be a constant for which p − p c > / p in the range [3 / , − θ ). One can easily showusing a Chernoff bound that a.s. every set of c distinct vertices in G n,p has at most 2 np c commonneighbors. This implies that the Hall condition is also satisfied for all A of size at least c , since then∆ + 1 − | N ( A ) | > np − np c > n/ ≥ | B | ≥ | A | . Together with the previous paragraph, this completes the proof. (cid:3)
It remains to consider p < /
5, so we will assume that bound on p for the remainder of this section.We use the following strategy to produce a partition of ∪ V i into a disjoint union of independenttransversals.1. Find an independent transversal through the unique vertex of maximum degree ∆, and deletethis transversal from the graph.2. As long as there exists a vertex v which has at least 0 . np neighbors in some part V i , find anindependent transversal T through v , and delete T from the graph.3. As long as there exists a minimal partial independent transversal T such that all but at most np/
100 vertices in some part V i have neighbors in T , split T into two nonempty ( | T | ≥ T ∪ T . Note that by minimality of T , eachpart V i contains a subset U i of at least np/
100 vertices which have no neighbors in T . ByLemma 2.3, there is an independent transversal through { U i } , which can be used to extend T to a full independent transversal T ′ . Delete T ′ from the graph, and then perform the samecompletion/deletion procedure for T .4. Finally, we construct the rest of the independent transversals, building them simultaneously from V to V r using Hall’s matching theorem. Our deletions in Steps 1–3, together with the propertiesof G n,p which we established in the previous subsection, will ensure that this is possible.The following lemma, which we prove later, ensures that we will indeed find the independenttransversals claimed in Steps 1–2. Lemma 2.6
Let V ∪ . . . ∪ V r be the above partition of V ( G ) , and let x be any vertex in this graph. • If x is the unique vertex of maximum degree ∆ , then G contains an independent transversalthrough x . If x is not of maximum degree, then for all k ≤ (cid:6) p (cid:7) and for any collection of subsets V ′ i ⊂ V i , | V ′ i | = ∆ + 1 − k , one of which contains x , there exists an independent transversal through x withrespect to { V ′ i } . Let us bound the number of independent transversals we delete in the first 3 steps. Note that iftwo vertices have at least 0 . np neighbors in the same V i , since by Lemma 2.1 | V i | ≤ ∆ + 1 ≤ . np ,their codegree will be at least 0 . np ≥ . np , contradicting Lemma 2.1. Therefore, during thefirst two steps, we will delete at most r + 1 ≤ (cid:6) p (cid:7) + 1 transversals. Next, suppose that after deleting O (cid:0)(cid:6) p (cid:7) log n (cid:1) independent transversals from G , we have that for some set T all but at most np/ V i have neighbors in T . Since (cid:6) p (cid:7) log n ≪ np , this certainly implies that the numberof vertices in the original V i with no neighbors in T was bounded by np/
50. Together with Lemma 2.2,this ensures that for each fixed V i , 1 ≤ i ≤ r , we never repeat Step 3 more than 50 log n times. Sinceeach iteration deletes two independent transversals and r ≤ (cid:6) p (cid:7) , we conclude that by the time wereach Step 4, we have deleted at most 1 + (cid:6) p (cid:7) + 100 (cid:6) p (cid:7) log n < (cid:6) p (cid:7) log n independent transversalsfrom G .Let us now describe Step 4 in more detail. At this point, all parts V i have the same size | V i | = s =∆ + 1 − k , where k < (cid:6) p (cid:7) log n = o ( np ) is the total number of independent transversals deletedso far. We build the remaining s disjoint independent transversals simultaneously as follows. Start s partial independent transversals { T i } si =1 by arbitrarily putting one vertex of V into each T i . Nowsuppose we already have disjoint partial independent transversals { T i } si =1 through V , . . . , V ℓ . Createan auxiliary bipartite graph H whose right side is V ℓ +1 and left side has s vertices, identified with thetransversals { T i } . Join the i -th vertex on the left side with a vertex v ∈ V ℓ +1 if and only if v has noneighbors in T i . Then, a perfect matching in this graph will yield a simultaneous extension of each T i which covers V ℓ +1 .We ensure a perfect matching in H by verifying the Hall condition, i.e., we show that for every t ≤ s , every set of t vertices on the left side of H has neighborhood on the right side of size at least t . Observe that after Step 3, for every T i there are more than np/
100 vertices in V ℓ +1 which haveno neighbors in T i . Therefore every vertex on the left side of H has degree greater than np/
100 andhence the Hall condition is trivially satisfied for all t ≤ np/ np/ < t ≤ s − (cid:6) p (cid:7) log n , then by definition of H , there are t partial independent transversalsamong { T i } and a subset W of V ℓ +1 of size greater than s − t such that every vertex of W has neighborsin every one of these transversals (i.e., is not adjacent to them in H ). This contradicts Lemma 2.4,so the Hall condition also holds for these t . It remains to check the case when t > s − (cid:6) p (cid:7) log n .Note that given any vertex v in V ℓ +1 and any collection of disjoint partial independent transversals,the number of them in which v can have a neighbor is at most the degree of v . However, we deletedthe maximum degree vertex in Step 1, so by Lemma 2.1 d ( v ) ≤ ∆ − √ np log n . Since p ≫ (cid:16) log nn (cid:17) / ,this is less than ∆ − (cid:6) p (cid:7) log n ≤ s − (cid:6) p (cid:7) log n . Therefore, in the auxiliary graph H , any set of t > s − (cid:6) p (cid:7) log n vertices on the left side has neighborhood equal to the entire right side. HenceHall’s condition is satisfied for all t and we can extend our transversals. This completes the proof,since one can iterate this extension procedure to convert all T i into full independent transversals. (cid:3) Proof of Lemma 2.6.
First, consider the case when x is not the vertex of maximum degree ∆ andwe have a collection of subsets V ′ i ⊂ V i of size ∆ + 1 − k , where k ≤ (cid:6) p (cid:7) . Without loss of generality,7ssume that x ∈ V ′ , and recall that by Lemma 2.1, the maximum degree ∆ satisfies np < ∆ < . np .If the number of neighbors of x in every set V ′ i , i ≥
2, is at most 0 . np then delete them and denotethe resulting sets V ′′ i . Since each V ′′ i still has size at least ∆ + 1 − (cid:6) p (cid:7) − . np > . np , by Lemma2.3 there exists a partial independent transversal through V ′′ , . . . , V ′′ r , which together with x providesa full independent transversal containing x . Next, suppose that x has at least 0 . np neighbors insome part, say V ′ . Since the degree of x is less than ∆ < . np , it must then have less than 0 . np neighbors in every other V ′ i . Furthermore, since x is not of maximum degree and p ≫ (cid:16) log nn (cid:17) / ,Lemma 2.1 implies that (∆ + 1) − d ( x ) ≥ √ np log n ≫ ⌈ p ⌉ ≥ r + k . Therefore there are more than r vertices in V ′ not adjacent to x . Also by Lemma 2.1, the codegree of every pair of vertices is at most1 . np < . np , so in particular no two vertices can both have ≥ . np neighbors in any given V ′ i .By the pigeonhole principle, there must be a vertex y ∈ V ′ not adjacent to x with less than 0 . np neighbors in each of the other V ′ i . That means that every other part has less than 0 . np neighborsof x and 0 . np neighbors of y . Since | V ′ i | ≥ ∆ − (cid:6) p (cid:7) > . np , there are still at least 0 . np verticesleft in each V ′ i , i ≥
3, that are non-adjacent to both x and y . Thus we can apply Lemma 2.3 as aboveto complete { x, y } into an independent transversal.The case when x is the vertex of maximum degree has a similar proof but involves one more step.As in the previous paragraph, we may assume that x ∈ V and has at least 0 . np neighbors in V , orelse we are done. Let W be the set of vertices in V that are not adjacent to x . Since | V | = ∆ + 1, wehave W = ∅ . If there exists some y ∈ W that has < . np neighbors in each of the other V i , i ≥ { x, y } to a full independent transversal as above. Otherwise, by Lemma 2.1 thecodegree of every pair of vertices is at most 1 . np < . np and hence each y ∈ W is associatedwith a distinct part in which it has ≥ . np neighbors. Yet x has exactly | W | − V i , i ≥
3, so there must exist y ∈ W such that x has no neighbors in the part (withoutloss of generality it is V ) in which y has ≥ . np neighbors. Since x is the unique vertex of maximumdegree and p ≫ (cid:16) log nn (cid:17) / , Lemma 2.1 gives d ( y ) ≤ ∆ − √ np log n < ∆ − (cid:24) p (cid:25) ≤ ∆ − r. Therefore V contains a subset W of at least r + 1 vertices which are not adjacent to both x and y .Since for every i ≥ W can have more than 0 .
81 neighbors in V i (by anothercodegree argument), the pigeonhole principle ensures that there is a vertex z ∈ W such that z has at most 0 . np neighbors in each V i , i ≥
4. Also note that x has less than 0 . np neighborsin each such V i , and y has less than 0 . np . Therefore every V i , i ≥
4, has in total less than0 . np + 0 . np + 0 . np < (∆ + 1) − . np neighbors of any of { x, y, z } , so we can apply Lemma2.3 as before to complete { x, y, z } into an independent transversal. (cid:3) Proof of Theorem 1.1 (ii).
We may assume that p < n − / because the case p ≥ n − / is alreadya consequence of part (i) of this theorem. Fix an arbitrary ǫ >
0. Suppose that G is a graph obtainedfrom G n,p by adding (1 + ǫ )∆ (cid:6) n (1+ ǫ )∆ (cid:7) − n isolated vertices and V ( G ) is partitioned into V ∪ . . . ∪ V r with every | V i | = (1 + ǫ )∆. Since ∆ ≥ np a.s., we have that r ≤ (cid:6) p (cid:7) . We use the same Steps 1–4 toproduce a partition of ∪ V i into a disjoint union of independent transversals. Actually Steps 1–2 cannow be made into a single step, since there is no need here to treat the vertex of maximum degree8eparately. The codegree argument implies again that we perform Steps 1–2 at most r + 1 times.Moreover, the existence of the independent transversals claimed in these two steps follows easily fromLemma 2.3. Indeed, suppose that we have deleted O (cid:0)(cid:6) p (cid:7)(cid:1) independent transversals from G . Since p ≫ (cid:16) log nn (cid:17) / , we have 1 /p = o ( np ) and thus every part still has size at least (1 + ǫ/ x bean arbitrary remaining vertex. Since the degree of x is at most ∆, every part still contains at least ǫ ∆ / x . By Lemma 2.3, we can find an independent transversal through thesevertices which will extend { x } .There is no change in the analysis of Step 3 and the same argument as in the proof of part (i)shows that the total number of transversals deleted from G in Steps 1–3 is at most O (cid:0)(cid:6) p (cid:7) log n (cid:1) . Since p ≫ (cid:16) log nn (cid:17) / , this number is o ( np ), and therefore in the beginning of Step 4 each part V i still hassize s ≥ (1 + ǫ/ s disjoint independent transversalssimultaneously, extending them one vertex at time to cover each new part V ℓ +1 . So again we define anauxiliary bipartite graph H whose left part corresponds to the partial independent transversals { T i } on V , . . . , V ℓ , right part is V ℓ +1 , and the i -th vertex on the left is adjacent to v ∈ V ℓ +1 iff v has noneighbors in transversal T i . A perfect matching in H gives a simultaneous extension of each T i .Hence it is enough to verify the Hall condition for H , i.e., we must show that for all t ≤ s ,every set of t vertices on the left has at least t neighbors on the right. The proof that this holdsfor all t ≤ s − (cid:6) p (cid:7) log n is exactly the same as in part (i) and we omit it here. So suppose that t > s − (cid:6) p (cid:7) log n ≥ s − o ( np ) > (1 + ǫ/ v ∈ V ℓ +1 is at most ∆,it can have neighbors in at most ∆ < t transversals. Therefore there is at least one transversal in ourset of size t which has no neighbors of v , and hence every set of t > s − (cid:6) p (cid:7) log n vertices on the lefthas neighborhood equal to entire right side of H . This verifies the Hall condition and completes theproof. (cid:3) In this section, we prove our second theorem. We only need to consider here the range log nn ≪ p ≪ log / n √ n , since part (ii) of Theorem 1.1 implies Theorem 1.2 for larger values of p . Again, we begin byshowing that G n,p satisfies certain properties almost surely. Lemma 3.1 If log nn ≪ p ≪ log / n √ n , then a.s. G n,p has the following properties:1. No pair of distinct vertices has more than / n common neighbors.2. The maximum degree is strictly between np and . np . Proof.
The codegree X of a fixed pair of vertices is binomially distributed with parameters n − p . Therefore P h X ≥ / n i ≤ (cid:18) n −
23 log / n (cid:19) ( p ) / n ≤ (cid:18) enp / n (cid:19) / n ≪ ( e/ / n = o ( n − ) . O ( n ) pairs of vertices, we see that the first property holds a.s. Thesecond property is a special case of Corollary 3.13 in [9]. (cid:3) Lemma 3.2
Let C ≥ and let G be a graph obtained from the random graph G n,p by connecting everyvertex to at most n new neighbors. Then a.s. every subset S ⊂ V ( G ) of size | S | ≤ Cp − log n spans a subgraph with average degree less than C log n , i.e., contains < C | S | log n edges. Proof.
Since the edges which we add to the random graph can increase the number of edges inside S by at most | S | (8 log n ) / | S | log n , it suffices to show that in G n,p a.s. every subset S as abovespans less than eC | S | log n edges. The probability that this is not the case is at most Cp − log n X m =1 (cid:18) nm (cid:19)(cid:18) (cid:0) m (cid:1) eCm log n (cid:19) p eCm log n ≤ Cp − log n X m =1 n m (cid:18) em eC log n · p (cid:19) eCm log n ≤ Cp − log n X m =1 n m − eCm log n ≤ Cp − log n X m =1 (cid:0) n − eC log n (cid:1) m = o (1) , so we are done. (cid:3) Fix ǫ >
0, and suppose we have disjoint subsets V , . . . , V r of G n,p , with all | V i | = (1 + ǫ )∆. ByLemma 3.1, r < n/ ∆ < /p . If a vertex v has more than ∆log n neighbors in some V i , say that v is locally big with respect to V i . If it has more than ∆2 log n , call it almost locally big . For each i , let B i be the set of v that are almost locally big with respect to V i . We claim that | B i | < n . Indeed, if | B i | ≥ n , then Lemma 3.1 together with ∆ ≥ log n and the Jordan-Bonferroni inequality wouldimply that the union of neighborhoods in V i of vertices from B i is at least(4 log n ) ∆2 log n − (cid:18) n (cid:19) / n ≥
32 ∆ > | V i | , contradiction. Next, make each B i a clique by adding all the missing edges. However, ∆ will stillrefer to the maximum degree of the original graph. Since each vertex is almost locally big withrespect to less than 2 log n sets V i , this operation increases the degree of each vertex by less than2 log n · n = 8 log n ≪ ∆2 log n . Thus every vertex that is locally big after the additions was almostlocally big before. In particular, there is now an edge between every pair of vertices that are locallybig with respect to the same V i , and there are less than r (4 log n ) < p − log n locally big vertices intotal.Let I ⊂ [ r ] be the set of indices i such that V i contains more than ǫ ∆ locally big vertices, anddefine the notation V S to represent S i ∈ S V i . Note that | V I | < (1 + ǫ )∆ · (cid:16) ǫ (cid:17) − p − log n < ǫ − p − log n ǫ is sufficiently small). As long as there exist i I such that there are more than (240 ǫ − log n ) | V i | crossing edges between V i and V I , add i to I . Note that each such index which we add to V I increases the number of edges in this set by morethan (240 ǫ − log n ) | V i | . Therefore if in this process I doubles in size we obtain a set of size at most40 ǫ − p − log n with average degree more than 240 ǫ − log n , which contradicts Lemma 3.2. Thus atthe end of the process we have | I | ≤ ǫ − p − log n .Given I , for t ≥ I t +1 ⊂ I t as follows. By Lemma 3.2, V I t induces less than(120 ǫ − log n ) | V I t | edges. Thus, there are less than 2 (cid:0) ∆log ∆ (cid:1) − · (120 ǫ − log n ) | V I t | vertices in V I t with > ∆log ∆ neighbors in this set. To define I t +1 we consider the following process. Start with I t +1 to bethe set of all i ∈ I t for which V i has more than ǫ ∆ vertices that have > ∆log ∆ neighbors in V I t . Aslong as there exist i ∈ I t \ I t +1 such that there are more than (240 ǫ − log n ) | V i | edges between V i and V I t +1 , add i to I t +1 . As above, Lemma 3.2 ensures that this process must stop before I t +1 doubles insize. Therefore in the end we have | I t +1 | ≤ (cid:16) ǫ (cid:17) − · (cid:18) ∆log ∆ (cid:19) − · (120 ǫ − log n ) | V I t |≤ O (cid:18) log n log ∆∆ | V I t | (cid:19) ≤ O (cid:18) log n log ∆∆ | I t | (cid:19) ≪ n | I t | . Clearly, | I | ≤ r ≤ n . Therefore, when t ≥ n log log n , I t will be empty. Let σ be the smallest indexsuch that I σ = ∅ . We now recursively build partial independent transversals T σ , . . . , T , where T t isan independent transversal on V I t . Let us say that T t satisfies property P t if for every i I t , all thevertices in T t that are not locally big with respect to V i have together at most 300( σ − t ) ∆log n neighborsin V i . It is clear that T σ = ∅ satisfies P σ , so we can apply the following lemma inductively to construct T , an independent transversal on V I satisfying P . Lemma 3.3
Suppose t > , and T t is an independent transversal on V I t which satisfies P t . Then wecan extend T t to T t − , an independent transversal on V I t − which satisfies P t − . We postpone the proof of this lemma until Section 3.4. Suppose that we have T as describedabove. Let J be the set of all indices j I such that some v ∈ T is locally big with respect to V j .Then, as we did with I , as long as there exist ℓ I ∪ J such that more than (600 ǫ − log n ) | V ℓ | edgescross between V ℓ and V J , add ℓ to J . Since | T | = | I | and each vertex can be locally big with respectto at most (1 + o (1)) log n sets V i , we have that initially | J | ≤ (1 + o (1)) | I | log n ≤ ǫ − p − log n .Therefore as before, Lemma 3.2 ensures that this process stops before J doubles in size, so the finalset J has size at most 100 ǫ − p − log n .As before, we construct a sequence of nested index sets J ⊃ · · · ⊃ J τ = ∅ , where for t ≥
1, define J t +1 in terms of J t as follows. Let J t +1 ⊂ J t be the set of all j ∈ J t for which V j contains morethan ǫ ∆ vertices that have > ∆log ∆ neighbors in V J t . Next, as long as there exist j ∈ J t \ J t +1 suchthat more than (600 ǫ − log n ) | V j | edges cross between V j and V J t +1 , add j to J t +1 . Lemma 3.2 againensures that we stop before J t +1 doubles in size, and the same computation as we did for I t +1 shows11hat | J t +1 | ≪ n | J t | . Thus when t ≥ n log log n , J t is empty. Let τ be the smallest index for which J τ = ∅ .Next, delete all neighbors of T in V J and all vertices in V J that are locally big with respect to any V k with k I . Denote the resulting sets V ′ j , j ∈ J . We claim that each V ′ j still has size at least ǫ ∆.Indeed, at most one v ∈ T can be locally big with respect to V j , because T is an independent set andall vertices that are locally big with respect to the same part were connected by our construction. Thusdeleting neighbors of this v can decrease the size of V j by at most d ( v ) < ∆ + 8 log n = (1 + o (1))∆.As for the remaining vertices in T , which are not locally big with respect to V j , P ensures thattogether they have at most O (cid:0) σ ∆log n (cid:1) = o (∆) neighbors in V j , since σ ≤ n log log n . Also, by constructionof I , every part whose index is not in I has at most ǫ ∆ locally big vertices. Hence the size of V ′ j isat least | V j | − (1 + o (1))∆ − ǫ ∆ ≥ ǫ ∆, as claimed.Let us say that a set U t satisfies property Q t if for every k I ∪ J t , all the vertices in U t that arenot locally big with respect to V k have together at most 300( τ − t ) ∆log n neighbors in V k . We need thefollowing analogue of Lemma 3.3. Lemma 3.4
Suppose t > , and U t is an independent transversal on V ′ J t which satisfies Q t . Then wecan extend U t to U t − , an independent transversal on V ′ J t − which satisfies Q t − . We also postpone the proof of this lemma until Section 3.4. Starting with U τ = ∅ , we iterate thislemma until we obtain U , an independent transversal on V ′ J which satisfies Q . Since τ ≤ n log log n ,this property implies that each V k with k I ∪ J has O (cid:0) τ ∆log n (cid:1) = o (∆) vertices with neighbors in U . Finally, let K = [ r ] \ ( I ∪ J ). Delete all neighbors of T ∪ U and all locally big vertices from every V k with k ∈ K , and denote the resulting sets by V ′ k . All V ′ k will still have size at least (cid:0) ǫ (cid:1) ∆, butnow no vertex there has more than ∆log n neighbors in any single set V ′ k . Thus, the following result from[19] implies that for sufficiently large n , there is an independent transversal on V ′ K , which completes T ∪ U into an independent transversal through all parts. Theorem 3.5 (Loh, Sudakov [19]) For every ǫ > there exists γ > such that the following holds.If G is a graph with maximum degree at most ∆ whose vertex set is partitioned into r parts V , . . . V r of size | V i | ≥ (1 + ǫ )∆ , and no vertex has more than γ ∆ neighbors in any single part V i , then G hasan independent transversal. This completes the proof of Theorem 1.2, modulo two remaining lemmas. (cid:3)
We take a moment to record two results which we will need for the proofs of the remaining lemmas.The first is the symmetric version of the Lov´asz Local Lemma, which is typically used to show thatwith positive probability, no “bad” events happen.
Theorem 3.6 (Lov´asz Local Lemma [7]) Let E , . . . , E n be events. Suppose that there exist numbers p and d such that all P [ E i ] ≤ p , and each E j is mutually independent of all but at most d of the otherevents. If ep ( d + 1) ≤ , then P (cid:2)T E i (cid:3) > . Proposition 3.7 (Alon [4]) Let G be a multipartite graph with maximum degree ∆ , whose parts V , . . . , V r all have size at least e ∆ . Then G has an independent transversal. Proof.
Independently and uniformly select one vertex from each V i , which we may assume is of sizeexactly ⌈ e ∆ ⌉ . For each edge f of G , let the event A f be when both endpoints of f are selected. Thedependencies are bounded by 2 ⌈ e ∆ ⌉ ∆ −
2, and each P [ A f ] ≤ ⌈ e ∆ ⌉ − , so the Local Lemma impliesthis statement immediately. (cid:3) Since the proofs of Lemmas 3.3 and 3.4 are very similar, we only prove Lemma 3.3. We will simplyindicate the two places where the proofs differ.
Proof of Lemma 3.3.
Fix some t as in the statement of the lemma. To extend an independenttransversal T t on the set V I t , satisfying P t , to one on the larger set V I t − , satisfying P t − , we will usethe following key properties of our construction. (i) For every i ∈ I t − \ I t , the set V i contains at most ǫ ∆ vertices that have > ∆log ∆ neighbors in V I t − . (ii) Each set V i has size (1 + ǫ )∆. (iii) For every i I t − , there are at most ( β log n ) | V i | edges between V i and V I t − , where we definethe constant β to be 240 ǫ − .In the case of Lemma 3.4, property (ii) is that each set V ′ j has size at least ǫ ∆, and the constant β in property (iii) is β = 600 ǫ − .Let D = I t − \ I t . From every V i with i ∈ D , delete all vertices that have > ∆log ∆ neighborsin V I t − , and all neighbors of vertices in T t . Denote the resulting sets by V ∗ i . Note that now alldegrees in the subgraph on V ∗ D = S i ∈ D V ∗ i are at most ∆log ∆ . Furthermore, we claim that every | V ∗ i | ≥ ǫ ∆. To see this, recall that at most one vertex v ∈ T t can be locally big with respect to V i , because T t is independent and all vertices that are locally big with respect to the same part areconnected by our construction. Deleting neighbors of such v can decrease the size of V i by at most d ( v ) < ∆ + 8 log n = (1 + o (1))∆. The rest of the vertices in T t are not locally big with respect to V i ,so P t implies that they have less than O (cid:0) σ ∆log n (cid:1) = o (∆) neighbors in V i since σ ≤ n log log n . Finally, byproperty (i) above, in V i we will delete at most ǫ ∆ vertices that have > ∆log ∆ neighbors in V I t − , soproperty (ii) implies that | V ∗ i | ≥ (1 + ǫ )∆ − (1 + o (1))∆ − ǫ ∆ ≥ ǫ ∆, as claimed.In the case of Lemma 3.4, recall that by construction all V ′ j with j ∈ J contain no locallybig vertices with respect to any part (we deleted all of them). Thus, the partial transversal U t contains no locally big vertices with respect to V ′ j . Property Q t then implies that the total numberof neighbors that vertices of U t have in V ′ j is only O (cid:0) τ ∆log n (cid:1) = o (∆). Hence when we reduce V ′ j to V ∗ j by deleting all neighbors of U t , and all vertices that have > ∆log ∆ neighbors in V J t − , the total effectof U t is o (∆), not (1 + o (1))∆ as above. Combining this with properties (i) and (ii) , we see that | V ∗ j | ≥ | V ′ j | − o (∆) − ǫ ∆ ≥ ǫ ∆, so the claim is still true. This is the second and final place in which13he proofs of the two lemmas differ, and explains why Lemma 3.4 holds with part sizes of only ǫ ∆,while Lemma 3.3 requires part sizes of (1 + ǫ )∆.Returning to the proof of Lemma 3.3, randomly select a subset W i ⊂ V ∗ i for each i ∈ D byindependently choosing each remaining vertex of V ∗ i with probability log ∆∆ , and let W = S i ∈ D W i .Define the following families of bad events. For each i ∈ D , let A i be the event that | W i | < ǫ log ∆,and for each v ∈ V ∗ D , let B v be the event that v has more than 2 log ∆ neighbors in W . Also, for each j I t − , let C j be the event that the collection of vertices in W that are not locally big with respectto V j has neighborhood in V j of size > ∆log n . We use the Lov´asz Local Lemma to show that withpositive probability, none of these events happen.Let us begin by bounding the dependencies. Say that A i lives on V ∗ i , B v lives on the neighborhoodof v in V ∗ D , and C j lives on the neighborhood of V j in V ∗ D . Note that each of our events is completelydetermined by the outcomes of the vertices in the set that it lives on. Hence events living on disjointsets are independent. A routine calculation shows that for any given event, at most O (∆ ) otherevents can live on sets overlapping with its set; the worst case is that an event of C -type can live ona set that overlaps with the sets of ≤ (1 + ǫ )∆ other C -type events.It remains to show that each of P [ A i ], P [ B v ], and P [ C j ] are ≪ ∆ − . The size of W i is distributedbinomially with expectation ≥ ǫ log ∆, so by a Chernoff bound, P [ A i ] < e − Ω(log ∆) ≪ ∆ − . Similarly,for each v ∈ V ∗ D the expected value of the degree of v in W is at most ∆log ∆ · log ∆∆ = log ∆ so P [ B v ] < e − Ω(log ∆) ≪ ∆ − . For P [ C j ], we proceed more carefully. For each 0 ≤ k ≤
8, let Y k be theset of vertices in V ∗ D that have between ∆∆ ( k +1) / log n and ∆∆ k/ log n many neighbors in V j . By property (iii) , the number of edges between V I t − and V j is at most ( β log n ) | V j | ≤ β ∆ log n . Therefore, | Y k | ≤ β ∆ ( k +1) / log n . However, since ∆ ≥ np ≥ log n , the probability that at least 30∆ k/ vertices in Y k are selected to be in W is bounded by P ≤ (cid:18) β ∆ ( k +1) / log n k/ (cid:19) (cid:18) log ∆∆ (cid:19) k/ ≤ e · β ∆ / log n · log ∆∆ ! k/ ≤ (cid:18) eβ · log ∆∆ / (cid:19) k/ ≪ ∆ − . Therefore, with probability 1 − o (∆ − ), the collection of vertices in W that are not locally big withrespect to V j has neighborhood in V j of size less than P k =0 k/ k/ log n < ∆log n , and hence P [ C j ] ≪ ∆ − .By the Lov´asz Local Lemma, there exist subsets W i ⊂ V ∗ i for each i ∈ D such that none of the A i , B v , or C j hold. In particular, every | W i | is greater than 2 e times the maximum degree in thesubgraph induced by W , so Proposition 3.7 implies that there exists an independent transversal T ′ there. Letting T t − = T t ∪ T ′ , we obtain an independent transversal on V I t − . Since T ′ ⊂ W and no C j hold, we have that for every j I t − , the vertices in T t ∪ T ′ which are not locally big with respect to V j have together at most 300( σ − t ) ∆log n + 300 ∆log n = 300( σ − ( t − ∆log n neighbors in V j , i.e., T t ∪ T ′ satisfies P t − . (cid:3) Concluding remarks
A simple modification of our argument yields a slight improvement of Theorem 1.2, and shows thatthe theorem is in fact true for all p ≫ log α n , for any fixed α >
0. We decided not to prove thatresult here in such generality for the sake of clarity of presentation. Also, it is not very difficult, usingour approach, to prove a statement similar to Theorem 1.2 for the sparse case, when p ∼ cn for someconstant c . However, these extensions are not as interesting as the main problem that remains open,which is to study the behavior of the strong chromatic number of random graphs when p ≤ n − / . Weare certain that the strong chromatic number of the random graph G n,p is a.s. (1 + o (1))∆ for every p ≥ cn for some constant c . It would also be very interesting to determine all the values of the edgeprobability p for which almost surely sχ ( G n,p ) is precisely ∆ + 1. Acknowledgments.
The authors would especially like to thank Bruce Reed for interesting remarksand useful insights at the early stage of this project. The idea of studying the strong chromaticnumber of random graphs originated from a conversation the second author had with Bruce Reed,during which it was realized that the strong chromatic number of dense random graphs should be∆ + 1. We would also like to thank Michael Krivelevich for stimulating discussions.
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