Regular Induced Subgraphs of a Random Graph*

Regular Induced Subgraphs of a Random Graph* Michael Krivelevich, 1 Benny Sudaov, Nicholas Wormald 3 1 School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel; e-mail: rivelev@post.tau.ac.il Department of Mathematics, University of California, Los Angeles UCLA, Los Angeles, California 90095; e-mail: bsudaov@math.ucla.edu 3 Department of Combinatorics and Optimization, University of Waterloo, Waterloo, ON, Canada; e-mail: nwormald@uwaterloo.ca Received 14 August 008; accepted 1 August 009 Published online 13 April 010 in Wiley Online Library wileyonlinelibrary.com. DOI 10.100/rsa.034 ABSTRACT: An old problem of Erdős, Fajtlowicz, and Staton ass for the order of a largest induced regular subgraph that can be found in every graph on n vertices. Motivated by this problem, we consider the order of such a subgraph in a typical graph on n vertices, i.e., in a binomial random graph Gn,1/. We prove that with high probability a largest induced regular subgraph of Gn,1/ has about n /3 vertices. 010 Wiley Periodicals, Inc. Random Struct. Alg., 38, 35 50, 011 Keywords: regular induced subgraph; random graph; extremal graph theory; random regular graphs 1. INTRODUCTION A rather old and apparently quite difficult problem of Erdős, Fajtlowicz, and Staton see Refs. [4] or [3], page 85 ass for the order of a largest induced regular subgraph that can be found in every graph on n vertices. By the nown estimates for graph Ramsey numbers c.f., e.g., Ref. [5], every graph on n vertices contains a clique or an independent set of size at least c ln n, for some positive constant c > 0, providing a trivial lower bound of c ln n for the problem. Erdős, Fajtlowicz, and Staton conjectured that the quantity in question is ωlog n. So far this conjecture has not been settled. Some progress has been achieved in upper bounding this function of n: Bollobás in an unpublished argument showed as stated in Ref. [3] the existence of a graph on n vertices without an induced regular subgraph on Correspondence to: N. Wormald *Supported by USA-Israel BSF Grant; Israel Science Foundation; Pazy Memorial Award; NSF CAREER Award DMS-081005; Canada Research Chairs Program; NSERC. 010 Wiley Periodicals, Inc. 35

36 KRIVELEVICH, SUDAKOV, AND WORMALD at least n 1/+ɛ vertices, for any fixed ɛ>0 and sufficiently large n. A slight improvement has recently been obtained by Alon and the first two authors [1], who too the upper bound down to cn 1/ log 3/4 n. Given the simplicity of the problem s statement, its appealing character and apparent notorious difficulty, it is quite natural to try and analyze the behavior of this graph theoretic parameter for a typical graph on n vertices, i.e., a graph drawn from the probability space Gn,1/ of graphs. Recall that the ground set of the probability space Gn, p is composed of all graphs on n labeled vertices, where each unordered pair {i, j} appears as an edge in G, drawn from Gn, p, independently and with probability p. In the case p = 1/ all labeled graphs G on n vertices are equiprobable: Pr[G] = n. This is the subject of the present article. We say that a graph property P holds with high probability, or whp for brevity, if the probability of a random graph to have P tends to 1 as n tends to infinity. It was shown by Cheng and Fang [] that the random graph Gn,1/ whp contains no induced regular subgraph on cn/ log n vertices. We improve the upper bound, and give a nearly matching lower bound, as follows. Theorem 1.1. Let G be a random graph Gn,1/. Then with high probability every induced regular subgraph of G has at most n /3 vertices. On the other hand, for = on /3, with high probability G contains a set of vertices that span a 1/-regular graph. It is instructive to compare this result with the above-mentioned result of Ref. [1]. Alon, Krivelevich, and Sudaov also used a certain probability space of graphs to derive their upper bound of On 1/ log 3/4 n. Yet, their model of random graphs is much more heterogeneous in nature the expected degrees of vertices vary significantly there, see Ref. [1] for full details. As expected, the rather homogeneous model Gn,1/ produces a sizably weaer upper bound for the Erdős Fajtlowicz Staton problem. The difficult part of our proof is the lower bound, for which we use the second moment method. The main difficulty lies in getting an accurate bound on the variance of the number of d-regular graphs on vertices, where d = 1/. Our main tool for achieving this goal is an estimate on the number N, H of d-regular graphs on vertices which contain a given subgraph H, when H is not too large. Provided H has o vertices, and its degree sequence satisfies some conditions which are quite typical for random graphs, we obtain an asymptotic formula for N, H which is of independent interest see Theorem 5.1. In Section, we introduce some notation and technical tools utilized in our arguments, and then prove a rather straightforward upper bound of Theorem 1.1. A much more delicate lower bound is then proven in Section 3. The technical lemma used in this proof relies on the above-mentioned estimate of N, H. The proof of this estimate is relegated to Section 4. The final section of the article contains some concluding remars.. NOTATION, TOOLS, AND THE UPPER BOUND In this short section, we describe some notation and basic tools to be used later in our proofs. Then, we establish the upper bound part of Theorem 1.1. We will utilize the following standard asymptotic notation. For two functions f n, gn of a natural number n, we write f n = ogn, whenever lim n f n/gn = 0; f n = ωgn if gn = of n. Also, f n = Ogn if there exists a constant C > 0 Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 37 such that f n Cgn for all n; f n = gn if gn = Of n, and f n = gn if both f n = Ogn and f n = gn. We write f g if the ratio f /g tends to 1 when the underlying parameter tends to infinity. For a real x and positive integer a, define [x] a = xx 1 x a + 1. All logarithms in this article have the natural base. We will use the bound en/, valid for all positive n and. For a positive integer, let m be the largest even integer not exceeding 1/. Let Gd denote the number of labeled simple graphs on vertices with degree sequence d = d 1, d,..., d, where the degree of vertex i is d i. Also, we denote p = P[a random graph G, 0.5 is m-regular]. Clearly, p = Gd, with all di being set equal to m. We will repeatedly cite the following corollary of a result of McKay and the third author see Theorems and 3 of Ref. [6]. Theorem.1. Let d j = d j, 1 j be integers such that d j = λ 1 is an even integer where 1/3 <λ</3, and λ d j =O 1/+ɛ uniformly over j, for some sufficiently small fixed ɛ>0. Then Gd = f d λ λ 1 λ 1 λ 1 1 where fd = O1, and ifmax{ λ d j } = o, then f d e 1/4, uniformly over the choice of such a degree sequence d. It is a routine matter to chec that λ λ 1 λ 1 λ 1 d = O 1. j m One way to see this is as follows. We can show that the expression increases if the smallest d i is increased, unless they are all at least 1/ + O1. By a symmetrical argument, we may assume they have an upper bound of the same form. Then, for such a sequence, adding bounded numbers to all d i changes the expression by a bounded factor. Hence Gd is Op for every degree sequence d covered by Theorem.1. Also, using Stirling s formula it is straightforward to verify that p = 1 + o1 π/ and that p 1 /p =. To prove the upper bound in Theorem 1.1, we show that, for a given and r, the probability that a random graph on vertices is r-regular is Op. For future use we prove here a somewhat more general statement. We then use the above-mentioned estimate for p and apply the union bound over all possible values of r. Lemma.. For every degree sequence d = d 1,..., d, P[G, 0.5 has degree sequence d] =Op. Proof. Following the remar after Theorem.1, assume that Gd Cp for every degree sequence d covered by the theorem, where C > 0 is an absolute constant. Let d be a degree sequence of length for which Gd is maximal [which is obviously Random Structures and Algorithms DOI 10.100/rsa d j

38 KRIVELEVICH, SUDAKOV, AND WORMALD equivalent to choosing d to be a most probable degree sequence in G,1/]. Write P[G, 0.5 has degree sequence d] =a p. If all degrees in d satisfy d i / 1/+ɛ, then we can apply Theorem.1. Otherwise, there is d i, say, d, deviating from / by at least 1/+ɛ, for some fixed ɛ>0. To bound the probability that G,1/ has degree sequence d, we first expose the edges from vertex to the rest of the graph. By standard estimates on the tails of the binomial distribution, the probability that has the required degree is exp{ ɛ }. The edges exposed induce a new degree sequence on vertices 1,..., 1. The probability that the random graph G 1, 0.5 has this degree sequence is at most a 1 p 1 in our notation. We thus obtain: a p maxcp, a 1 p 1 Oexp{ ɛ }. Recalling that p 1 /p =, we obtain that a maxc, a 1 Oexp{ ɛ } for large enough. The result follows by induction starting from 0 for which the above factor Oexp{ ɛ } is < 1; the final bound is a maxc, a 0. To complete the proof of the upper bound of Theorem 1.1, note that by Lemma. the probability that a fixed set V 0 of vertices spans a regular subgraph in Gn,1/ is Op. Summing over all 0 = n /3 and all vertex subsets of size, we conclude that the probability that Gn,1/ contains an induced regular subgraph on at least 0 vertices is 0 n Op 0 en 1+o1 π/ n 1 + o1 en π 3/ 0 0 = o1. 3. A LOWER BOUND In this section, we give a proof of the lower bound in our main result, Theorem 1.1. To be more accurate, we give here most of the proof, deferring the proof of a ey technical lemma to the next section. The proof uses the so-called second moment method and proceeds by estimating carefully the first two moments of the random variable X = X, counting the number of 1/-regular induced subgraphs on vertices in Gn,1/. For convenience we assume throughout the proof that 1 mod 4. Since this estimate is used for proving the lower bound of Theorem 1.1, we can allow ourselves to choose in such a way without losing essentially anything in the lower bound. It is somewhat surprising to be able to apply successfully the second moment method to sets of such a large size, however two earlier instances of similar application can be found in Refs. [7] and [9]. So let X be the random variable counting the number of 1/-regular induced subgraphs on vertices in Gn, 0.5. We write X = A = X A, where X A is the indicator random variable for the event that a vertex subset A spans a 1/-regular subgraph. Then E[X] = n E[X A ]= p. A = Plugging in the estimate for p cited after the statement of Theorem.1, it is straightforward to verify that E[X] tends to infinity for = on /3 ; in fact, E[X] =ω1 in this regime. A corollary of Chebyshev s inequality is that P[X > 0] 1 Var[X]/E [X], and therefore to prove that whp Gn,1/ contains an induced regular subgraph on vertices, it is enough to establish that Var[X] =oe [X]. To estimate the variance of X we need to estimate the correlation between the following events: A spans a 1/-regular subgraph and B spans a 1/-regular subgraph, Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 39 where A, B are -element vertex subsets whose intersection is of size i. To this end, define p,i = max P[G, 0.5 is 1/-regular G[i] =H], VH =i where the maximum in the expression above is taen over all graphs H on i vertices, and G[i] stands for the subgraph of G,1/ spanned by the first i vertices. Since X = A = X A, we have: Var[X] =E[X ] E [X] = 1 Var[X A ]+ A = 1 E[X A ]+ A = E[X]+ i= n p l A = B = A B =i 1 i= i i= A = B = A B =i E[X A X B ] E[X A ]E[X B ] P[X A = 1]P[X B = 1 X A = 1] P[X A = 1]P[X B = 1] i p,i p. As a warm-up, we first show that a rather crude estimate for Eq. suffices to prove that Var[X] =oe [X] for = o n. We start with the following bound for p,i. Lemma 3.1. For i 1, i i p,i = O i ii p i. Also, p,i p Ce log i, for a sufficiently large constant C > 0. Proof. First, given H, expose the edges from H to the remaining i vertices denote the latter set by X. For every v H, we require dv, X = 1/ d H v. This happens with probability i i +i 1/ d H v i +i the middle binomial coefficient is the largest one. Hence the probability that all i vertices from VH have the required degree of 1/ ing is at most the i-th power of the right-hand side of the above expression. Now, conditioned on the edges from H to X, we as what is the probability that the subgraph spanned by X has the required degree sequence each v X should have exactly 1/ dv, H neighbors in X. Observe that by Lemma. the probability that G[X] has the required degree sequence is at most C 0 p i for some absolute constant C 0 > 0, providing the first claimed estimate for p,i. Random Structures and Algorithms DOI 10.100/rsa

40 KRIVELEVICH, SUDAKOV, AND WORMALD From Theorem.1, p t = 1 t t. t 1/ Therefore, the ratio p,i /p can be estimated as follows: p,i i C 0 p C 1 i i i 1 1 i ii p i i i i i 1 C 1 p i ii i 1 i 1 [ 1 3 = C i + 4 i + 1 1 1 i [ 1 C 1 1 i + 1 ] 1 i + 1 i 1 ] { 1 + 1 4 + + 1 }. C exp i + 1 In the third inequality above we used that i 1 i 1 1 1 + 1 i i. Observe that i 1 j= i+1 < dx = log. This completes the proof of the second part of the lemma. j i x i Now, we complete a proof of a weaer version of the lower bound of Theorem 1.1, by showing that whp Gn,1/ contains an induced 1/-regular subgraph on = o n vertices. Omitting the term p in the sum in Eq. and using E[X] = p, we obtain: 1 n i= i i p,i + 1 n i i p Denote Var[X] E [X] n gi = i i E[X] = 1 i= p,i p. p,i p + o1. 3 Let us first estimate the ratio of the binomial coefficients involved in the definition of gi. n i i e i n i i e i 1 i + 1 = i n + in + i 1 n + 1 e i i 3 i. i n + i in To analyze the asymptotic behavior of gi, we consider three cases. Case 1. i /. In this case, by Lemma 3.1 and the inequality log1 + x x for x 0 we have: p,i Ce log i = Ce log1+ i i Ce i i Ce i. p We thus get the following estimate for gi: n gi = i i p,i 3 i 3e Ce i C i. 4 p in in The above inequality is valid for all values of. When = o n it gives that gi = o1 i. Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 41 Case. / i. Recalling Lemma 3.1 again, we have p log,i/p Ce log Ce log log. Hence in this case i 3 i 6 / gi Ce log log log log Ce in n 1 / Ce log log = Ce log 4 + log log e. For future reference, it is important to note here that in the calculation above we used 6/n 1/. This inequality stays valid as long as n/6 /3. Case 3. i log. In this case it suffices to use the trivial estimate p,i 1. We also need that E[X] = p = ω1. Therefore, gi = n n i i p,i i i p ω1 n i ω1 n / log = eo ω1 ω1 e. In the above calculation we used the assumption log n = Olog. In the complementary case = n o1 the expression p behaves lie cn/ 3/ n /, while the numerator in the expression for gi is at most n / log = n o, and the estimate wors as well. Note that, as in Case, the inequality here remains valid even for as large as n /3. It thus follows that 1 i= gi is negligible, implying in turn that Var[X] =oe [X], and thus X is with high probability positive by the Chebyshev inequality. Now, we proceed to the proof of the real lower bound of Theorem 1.1, i.e., assume that satisfies = on /3. In this case estimating the variance of the random variable X, defined as the number of induced 1/-regular subgraphs on vertices, becomes much more delicate. We can no longer ignore the term p in the sum in Eq.. Instead, we show that for small values of i in this sum p,i is asymptotically equal to p. In words, this means that nowing the edges spanned by the first i vertices of a random graph G = G,1/ does not affect by much the probability of G being 1/-regular. We claim this formally for i = o in the following ey lemma. Lemma 3.. For i = o, p,i = 1 + o1p. The proof of this lemma is rather involved technically. We thus postpone it to the next section. We now show how to complete the proof assuming its correctness. We first repeat estimate : Var[X] E [X] 1 1 E[X] + o1 + i= t i= i i p,i p p n i i p,i p + p Random Structures and Algorithms DOI 10.100/rsa 1 n i i i=t p,i p, 5

4 KRIVELEVICH, SUDAKOV, AND WORMALD where t = t, n is chosen so that t = ω /n but t = o. Since = on /3 such a function is easily seen to exist. Due to our choice of t we can apply Lemma 3. to the first sum above. It thus follows that t i= i i p,i p p = t i= n i i op n i=0 o1 i i = o1. p As for the second sum in Eq. 5 we can utilize the same case analysis as done before for = o n. The only difference is in Case 1, that now covers all i from t till /. Therefore, for every i in this new interval we can use inequality 4 to conclude n i i p,i p This completes the proof of Theorem 1.1. 3e i 3e C C i = o1 i. in tn 4. PROOF OF KEY LEMMA The proof of Lemma 3. is overall along the lines of the proof of Lemma 3.1, though requiring a much more detailed examination of the probabilities involved. Throughout this section, let be odd and, for simplicity, denote 1/byd. Let D i be the set of integer vectors d = d 1,..., d i such that 0 d j i 1 for 1 j i, and d j d j is even. Given d D i, let Nd denote the number of graphs G on vertex set [] for which G[i] has no edges, and d G j = d d j for j [i], whilst d G j = d for i < j. Note that if d is the degree sequence of a graph H on vertex set [i], then Nd is the number of d-regular graphs G on vertex set [] for which G[i] =H. Of course, in this case Nd is nonzero only if d is even, and hence is congruent to 1 mod 4. Proposition 4.1. Assume i = o. Given d D i and a nonnegative vector s = s 1,..., s i with j even, put d = d s. Then, uniformly over such d and s with the additional properties that d D i and i 3/4, d dj + Nd Nd. d + 1 i dj Proof. We use a comparison type argument. Since it is quite complicated, we give the idea of the proof first. For any vector c = c 1,..., c j, write c for the vector d c 1,..., d c j. Let V 1 ={1,..., i} and V ={i + 1,..., }. For simplicity, suppose that s 1 = s = 1, and = 0 for j 3. We can compute Nd as the number of possible outcomes of two steps. The first step is to choose a bipartite graph B with bipartition V 1, V and degree sequence d in V 1. The second step is to add the remaining edges between vertices in V such that those vertices will have degree d. By comparison, to count Nd we choose in the first step B with degree sequence d in V 1, and then do the second step for each such B. The proof Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 43 hinges around the fact that there is a correspondence between the set of possible B and B such that the number of ways of performing the second step is roughly the same, at least for most of the corresponding pairs B, B. The correspondence is many-to-many. For a graph B we may add two edges, incident with vertices 1 and, to obtain a graph B. The number of ways this can be done, without creating multiple edges, is i d dj = d + 1 i dj. Conversely, each B comes from d d j + 1 different B. The ratio of these quantities gives the asymptotic ratio between Nd and Nd claimed in the theorem, ignoring the number of ways of performing the second step. Our actual argument gets more complicated because not only some bipartite graphs must be excluded, but also some sets of edges to be added to them. So, we will present equations relating to the above argument in a slightly different form to mae exclusion of various terms easier. Let B denote the set of bipartite graphs with bipartition V 1, V.ForB B, write D j B for the degree sequence of B on the vertices in V j in the natural order, so D 1 B = db 1,..., d B i and D B = d B i+1,..., d B.Ford = d 1,..., d i with d j d j even, let Bd denote {B B : D 1 B = d }. Let G D B denote the number of graphs with degree sequence D B. Clearly, Nd = G D B. 6 B Bd Suppose that we wish to add to B a set S of edgeoining V 1 and V, without creating any multiple edges, such that the degree of j V 1 1 j i in the graph induced by S is as given in the statement of the proposition. The family of all such sets S will be denoted by SB, s. Note that necessarily S 3/4 for S SB, s. The cardinality of SB, s is i d+1 i dj, because db j = d j = d d j, so as in the setch above j has d + 1 i d j spare vertices in V to which it may be joined. Hence, we can somewhat artificially rewrite Eq. 6 as Nd = 1 d + 1 i dj B Bd S SB,s G D B. 7 Also for B Bd define S B, s to be the family of sets S EB such that the degree of j V 1 in the graph induced by S is 1 j i. Since d B j = d d j and d j = d j, a similar argument gives Nd = 1 d dj + B Bd S S B,s G D B. 8 The rest of the proof consists of showing that the significant terms in the last two equations can be put into 1-1 correspondence such that corresponding terms are asymptotically equal. We first need to show that for a typical B Bd, the variance of the elements of D B as a sequence is small. Given d, define d = i 1 j V 1 d d j, and note that this is equal to i 1 j V d B j for every B Bd. Since d is determined uniquely by d, the value of d is the same for all B Bd. Random Structures and Algorithms DOI 10.100/rsa

44 KRIVELEVICH, SUDAKOV, AND WORMALD Lemma 4.. Let d D i, and select B uniformly at random from Bd. Then E d d B j i i. j V Proof. First observe that in B, the neighbors of any vertex t V 1 form a random subset of V of size dt, and these subsets are independent for different t. So for fixed j V, d B j is distributed as a sum of i independent 0-1 variables with mean t V 1 dt / i = d. It follows that the variance of d B j is < i. Hence E d d B j < i, and the lemma follows by linearity of expectation. Returning to the proof of the proposition, we will apply Theorem.1 to estimate G D B. This graph has i vertices, degree sequence {d d B j, j V }, and its degree sum is e := id eb, where eb = j V d B j = i d is the number of edges in the bipartite graph B. Consider λ from Theorem.1. We see that e λ = λd := i i 1 = d d i 1. 9 The product of binomials in 1 is in this case i 1. 10 d d B j j V For every B Bd, all components of the vector D B are at most V 1 =i. Thus We have x j := i 1 d d B j = d B j i/ = Oi = o. a a = exp x /a + Ox 3 /a 11 a/ + x a/ for x = o a, which may be established for instance by analyzing the ratio of the binomial coefficients. Hence j=i+1 i 1 = d d B j i 1 i exp i 1 x j i + oi. 1 Note that here and in the rest of the proof, the asymptotic relations hold uniformly over d D i. Since i = o we can choose a function ω of n such that ω and ω i = o. Define ˆB ω d to be the subset of Bd that contains those B for which j V d d B j ωi i. 13 Since j V d B j = eb is the same for all bipartite graphs B Bd, by definition of x j we have that j x j j d B j also does not depend on B. Similarly, the sum in 13 differs from j d B j by a constant independent of B. Therefore j x j for all B ˆB ω/ d Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 45 is smaller than the corresponding sum for B Bd \ ˆB ω d by an additive term of at least ωi i/. This implies that the product of binomials in Eq. 1 is larger, for all B ˆB ω/ d, than for any B Bd \ ˆB ω d. Also, from Lemma 4. and Marov s inequality, almost all members of Bd are in ˆB ω/ d. Moreover, since all degrees in degree sequence D B deviate from i/ by at most Oi = o, the function f D B from Theorem.1 is e 1/4 for all B Bd. Combining these observations, we conclude that the contribution to Eq. 6 from B / ˆB ω d is o Nd. Thus, the same observation holds for Eq. 7. That is, Nd 1 d + 1 i dj B Bωd ˆ S SB,s G D B. 14 We also note for later use, that by Eq. 13 and Cauchy s inequality, for all B ˆB ω d d d B j i ωi. 15 j V Fix B ˆB ω d. Consider S chosen uniformly at random from SB, s, and let r m S denote the number of edges of S incident with a vertex m V. Fixing m and using that S 3/4, we can bound the probability that r m S 5by {j 1,...,j 5 } V 1 5 t=1 t i d d j t s 5 j S 5 = = O 5/4. d i d i j V 1 Hence by Marov s inequality, with probability 1 O 1/4, S SB, s satisfies i max j V r j S 4. We would next lie to bound j V d d B j rj S. To do this, note that we may choose the edges in S incident with any given vertex sequentially, each time selecting a random neighbor from those vertices of V still eligible to be joined to. For each such edge joining to such a random vertex j V, by Eq. 13 there are, as a crude bound, at least i/3 vertices of V to choose from for sufficiently large. Amongst the eligible vertices, the average value of d d B j must be at most 3 ωi by Eq. 15. Hence, if X h denotes the value of d d B j for the h-th edge added, we have EX h 3 ωi. Thus E h X h 3 ωi S 3 ωi 3/4. Noting that h X h = j V d d B j r j S and using Marov s inequality, we deduce that almost all S SB, s more precisely all except the fraction 3/ ω = o1 of them, at most satisfy d d B j r j S ω i 3/4. ii j V Define ˆ SB, s to be the set of S SB, s satisfying both the properties i and ii. Then, since each S SB, s contributes equally to Eq. 14, Nd 1 d + 1 i dj Random Structures and Algorithms DOI 10.100/rsa B Bωd ˆ S ˆ SB,s G D B. 16

46 KRIVELEVICH, SUDAKOV, AND WORMALD Let B ˆB ω d and S SB, ˆ s. Then, using Eq. 13 together with i and ii, we get d d B j r j S ωi i + d d B j r j S + r j S j V j V j V ωi i + ω i 3/4 + 16 i ωi. Hence, for sufficiently large, those S appearing in the range of the summation in Eq. 16 satisfy B + S ˆB ω d, where B + S is the graph obtained by adding the edges in S to B and noting that d = d +s. Since, as we saw, the contribution to Eq. 6 from B / ˆB ω d is o Nd, we may also relax the constraint on B in the summation in Eq. 16, to become B ˆB ω d. Now redefining ω as ω, we obtain Nd 1 d + 1 i dj B,S W G D B, 17 where W denotes the set of all B, S such that B ˆB ω d, S SB, ˆ s and B + S ˆB ω d. Define S ˆ B, s, analogous to SB, ˆ s, to be the set of S S B, s with maximum degree in V at most five and also obeying property ii above, where B = B S. Since B has density close to 1/, B and its complement are more or less equivalent, and adding or deleting edges should have a similar effect. Indeed, the above argument applied to 8, with suitable small modification, gives Nd 1 d dj + B,S W G D B 18 where W denotes the set of all B, S such that B ˆB ω d, S SB ˆ, s and B S ˆB ω d. Observe that B, S W if and only if B + S, S W. So the summation in Eq. 18 is equal to G D B + S. B,S W Hence, comparing with Eq. 17, the proposition follows if we show that G D B G D B + S 19 uniformly for all B, S W. We may apply Eq. 1 to both sides of Eq. 19. Write gλ, n = λ λ 1 λ 1 λ. Notice that λd, as defined in Eq. 9, satisfies: j V λd = d d B j i i 1, Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 47 which is exactly λ for the degree sequence D B as defined in Theorem.1. The same applies to λd and the degree sequence D B + S. Using d d B j = id d B t = id id + j V t V 1 j V i d t = t=1 i 1 + i, it is easy to derive from Eq. 9 that both λd and λd are 1/ + Oi / = 1/ + o 1 for all d, d D i. For such λ the derivative of log gλ, i is i logλ/1 λ = O i / = o. Moreover, λd λd =O 5/4 because the values of d for B and B+S differ by O S /. Hence gλd, i gλd, i. It is now also easy to see that f D B, f D B+S from Theorem.1 satisfy: f D B f D B+S e 1/4, since all degrees in these two degree sequences deviate from i/byoi = o i. It only remains to consider the product of binomials in the two sides of Eq. 19 after the application of Theorem.1. Recalling the expression in Eq. 10, the ratio of these two products is i 1 / i 1. d d B j d d B j r j S Since B ˆB ω d, all r j S 4 and j r js 3/4, so using d = 1/ and d B j i, this expression is, up to a multiplicative factor of 1 + O 1 j r j S = 1 + o1, equal to i 1 d + db j rj S d i/ i/ db j rj S d d B j d i/ + i/ d B j j V = j V = j V 1 + O i/ db j rj S. By its definition, d = i 1 j V 1 d d j = i/ + Oi / = i/ + o1, and so using condition ii [the right-hand side of which is ω i 3/4 = o], and not forgetting rj S 3/4,weget r j S i/ d Bj j V j V d d B j r j S + j V d i/ r j S = o1. Hence, the expression above is asymptotic to one. This argument shows that Eq. 19 holds with the required uniformity. For a slightly simpler version of the formula in Proposition 4.1, put ˆd = d 1 i 1 = 1 i which is in some sense the average degree of vertices of side V 1 in the bipartite graph B and δ j = d j 1 i 1. Random Structures and Algorithms DOI 10.100/rsa

48 KRIVELEVICH, SUDAKOV, AND WORMALD Then the proposition gives Nd Nd [ˆd δ j + ] sj [ˆd + δ j ] sj. Recalling that δ j and are at most i = o and using log1 + x = x x / + Ox 3, we have [ˆd δ j + ] sj = ˆd exp δ j /ˆd + s j /ˆd + o1/, [ˆd + δ j ] sj = ˆd exp δ j /ˆd s j /ˆd + o1/. Thus, we may rewrite the assertion of Proposition 4.1 as { } i { i Nd Nd exp δj /ˆd + s j /ˆd } exp δ j /ˆd { }. 0 i exp δ j /ˆd To proceed, we extend this formula so that s is permitted to have negative entries. Corollary 4.3. Assume i = o. Given d D i and an integer vector s = s 1,..., s i with j even, put d = d s. Then, uniformly over such d and s with the additional properties that d D i and i 3/4, { Nd 1 i Nd exp } δj + s j. ˆd Proof. Define the vector s by turning the negative entries of s into zero; that is, the jth entry of s is if 0, and zero otherwise. If this causes the sum of entries to change parity, leave one of these entries as 1. It is easy to modify the following proof accordingly. Let s = s s. The jth entry of s is if < 0, and zero otherwise. We can now estimate the product Nd Nd s Nd s = Nd / Nd Nd Nd s Nd s using two applications of Eq. 0, noting that all entries of s = s s are nonnegative and that the δ j defined for degree sequence d equals δ j. Define d 0 to be the constant sequence of length i, all of whose entries are i 1/. If d i i 1/ is odd, adjust the first entry to i 1/ +1 to ensure that d 0 D i. We can use the following result to compare the number of graphs with an arbitrary degree sequence d on G[i] to the number with d 0. Recall, however, that Nd is defined even if d is not the degree sequence of any graph on [i]. Corollary 4.4. i If d D i then Nd Nd 0 1 + o1. ii If, in addition, i δ j = o, then Nd Nd 0. Proof. We treat the case that the first entry of d 0 was not adjusted for the parity reason above; in the other case, only trivial modifications are required, for which we omit the details. Random Structures and Algorithms DOI 10.100/rsa

REGULAR INDUCED SUBGRAPHS OF A RANDOM GRAPH 49 Part ii of the corollary follows immediately from Corollary 4.3 by putting = δ j for each j, since if i δ j = o then by Cauchy s inequality δ j =o i = o 3/4. Note also that ˆd /. For the first part, let d maximise Nd. If i δ j < say, the above argument shows that Nd Nd 0 1 + o1. So assume that i δ j >. Putting = δ j and applying Corollary 4.3 shows the result, provided δ j 3/4. If the latter condition fails, we can simply define = α j δ j for some 0 α j 1 such that j iust below 3/4 and is even. Since δ j and they both have the same sign, we can conclude that j δ j + sj j. By Cauchy s inequality, the sum of the squares of grows asymptotically faster than. Let s = s 1,..., s i and let d = d s. Then, by Corollary 4.3 we obtain Nd = ond, which contradicts the maximality assumption and proves the result. Define d 1 to be the constant sequence of length, all of whose entries are d = 1/. We can now determine the asymptotic value of Nd 0. Recall that is odd; we will now assume that 1 mod 4 to ensure that Nd 0 is not 0. If 3 mod 4, we could prove a similar result by subtracting 1 from the first entry of d 1. Corollary 4.5. For 1 mod 4 we have Nd 0 Gd 1 / i. Proof. Let H be one of the i graphs on vertex set [i] chosen at random, and let dh = {d 1,..., d i } be its degree sequence. Then d j is a binomially distributed random variable with expectation i 1/ and variance i 1/4. Hence for δ j = d j i 1/ wehave E[δ j ]=Var[d j]=i 1/4. Then E j [i] δ j i, and, by Marov s inequality, whp j [i] δ j = o. Thus, from Corollary 4.4ii it follows that for almost all graphs H, Nd H Nd 0. Part i of the same corollary shows that for all other graphs, Nd H 1+o1Nd 0. Since Nd H is the number of d-regular graphs G on vertex set [] for which G[i] =H, we have that Gd 1 = H Nd H = 1+o1Nd 0 i, and the corollary follows. Proof of Lemma 3.. From Corollary 4.4, p,i P[G, 0.5 is 1/-regular G[i] =H], where H is a chosen to be a graph with degree sequence d 0. Note that the number of random edges outside H to be exposed is i, and each of them appears independently and with probability 1/. Therefore, the above probability equals to Nd 0 / i.by Corollary 4.5, this is asymptotic to Gd 1 / = p. 5. CONCLUDING REMARKS Our technique for proving Proposition 4.1 is a rather complicated comparison argument somewhat related to the method of switchings used for graphs of similar densities in Ref. [8]. Random Structures and Algorithms DOI 10.100/rsa

50 KRIVELEVICH, SUDAKOV, AND WORMALD One might be tempted to try proving the result for S = i =, as setched in the first part of the proof, and then applying this repeatedly, as in the proof of Corollary 4.4, to go from one degree sequence to another. However, this seems to provide insufficient accuracy. Similarly, attempts to use switchings directly were not successful. Of independent interest is the following estimate for the probability that a regular graph with vertices and degree 1/ contains a given induced subgraph with degree sequence d 1,..., d i on its first i vertices. This gives an asymptotic formula provided the sum of the absolute values of δ j = d j i 1/ is a bounded multiple of 3/4, and otherwise gives less accurate bounds. Theorem 5.1. Assume i = o, with 1 mod 4. Let H be a graph on vertex set [i] with degree sequence d = d 1,..., d i. Then the probability that a random 1 1-regular graph G on vertex set [] has the induced subgraph G[i] equal to H is where δ j = d j i 1/. i exp i i δ j exp o 3/4 i δ i Proof. We may use the argument in the proof of Corollary 4.4 to jump from d to d 0, using at most 3/4 i δ i applications of Corollary 4.3, and then apply Corollary 4.5. ACKNOWLEDGMENTS We than the referees of the article for their careful reading and helpful remars. REFERENCES [1] N. Alon, M. Krivelevich, and B. Sudaov, Large nearly regular induced subgraphs, SIAM J Discrete Math 008, 135 1337. [] Q. Cheng and F. Fang, Kolmogorov random graphs only have trivial stable colorings, Inform Process Lett 81 00, 133 136. [3] F. R. K. Chung and R. L. Graham, Erdős on graphs: His legacy of unsolved problems, A. K. Peters, Ltd, Wellesley, MA, 1998. [4] P. Erdős, On some of my favourite problems in various branches of combinatorics, Fourth Czechoslovaian Symposium on Combinatorics, Graphs and Complexity Prachatice, 1990, Ann Discrete Math 51 199, 69 79. [5] R. L. Graham, B. L. Rothschild, and J. H. Spencer, Ramsey theory, nd edition, Wiley, New Yor, 1990. [6] B. D. McKay and N. C. Wormald, Asymptotic enumeration by degree sequence of graphs of high degree, Eur J Combinator 11 1990, 565 580. [7] S. Janson, The numbers of spanning trees, Hamilton cycles and perfect matchings in a random graph, Combin Probab Comput 3 1994, 97 16. [8] M. Krivelevich, B. Sudaov, V. Vu, and N. C. Wormald, Random regular graphs of high degree, Random Struct Algorithms 18 001, 346 363. [9] O. Riordan, Spanning subgraphs of random graphs, Combin Probab Comput 9 000, 15 148. Random Structures and Algorithms DOI 10.100/rsa