The number of K s,t -free graphs

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1 The number of K s,t -free graphs József Balogh an Wojciech Samotij August 17, 2010 Abstract Denote by f n (H the number of (labele H-free graphs on a fixe vertex set of size n. Erős conjecture that whenever H contains a cycle, f n (H = 2 (1+o(1 ex(n,h, yet it is still open for every bipartite graph, an even the orer of magnitue of log 2 f n (H was known only for C 4, C 6, an K 3,3. We show that for all s an t satisfying 2 s t, f n (K s,t = 2 O(n2 1/s, which is asymptotically sharp for those values of s an t for which the orer of magnitue of the Turán number ex(n, K s,t is known. Our methos allow us to prove, among other things, that there is a positive constant c such that almost all K 2,t -free graphs of orer n have at least 1/12 ex(n, K 2,t an at most (1 c ex(n, K 2,t eges. Moreover, our results have some interesting applications to the stuy of some Ramsey- an Turán-type problems. 1 Introuction Let H be an arbitrary graph. We say that a graph G is H-free if G oes not contain H as a (not necessarily inuce subgraph. Denote by f n (H the number of labele H-free graphs on a fixe vertex set of size n. Let ex(n, H enote the Turán number for H, i.e., the maximum size of an H-free graph on n vertices. Extening the classical theorem of Turán [35], Erős an Stone [13] prove that if H is not bipartite, then the orer of magnitue of ex(n, H epens only on the chromatic number of H, i.e., ( 1 n 2 ex(n, H = 1 χ(h o(n2. Since every subgraph of an H-free graph is also H-free, it follows that f n (H 2 ex(n,h. Erős, Frankl, an Röl [12] prove that this crue lower boun is in fact tight whenever χ(h 3, namely f n (H = 2 (1+o(1 ex(n,h. (1 Department of Mathematics, University of Illinois, 1409 W Green Street, Urbana, IL 61801, USA; an Department of Mathematics, University of California, San Diego, 9500 Gilman Drive, La Jolla, CA 92093, USA. aress: jobal@math.uiuc.eu. This material is base upon work supporte by NSF CAREER Grant DMS an DMS , UIUC Campus Research Boar Grants an 08086, an OTKA Grant K Department of Mathematics, University of Illinois, Urbana, IL, 61801, USA. Research supporte in part by the Trijtzinsky Fellowship an the James D. Hogan Memorial Scholarship Fun. aress: samotij2@illinois.eu. 1

2 For a brief survey an some relate results, also on the (ifferent problem of inuce containment, see, e.g., [1, 3, 4, 5, 27, 33]. The picture changes ramatically when one rops the χ(h 3 assumption. Erős aske [9] if (1 is still true if H is a bipartite graph containing a cycle, but his question remains unanswere for all such graphs an for most such H, not even the correct orer of magnitue of log 2 f n (H is known. The only results in this irection are ue to Kleitman an Winston [24], who prove that log 2 f n (C ex(n, C 4, Kleitman an Wilson [23], who showe that log 2 f n (C 6 = Θ(ex(n, C 6, an the authors, who showe [7] that log 2 f n (K 3, ex(n, K 3,3. It is worth mentioning that the 2 O(n5/4 boun on the number of C 8 -free graphs obtaine by Kleitman an Wilson [23] as well as the 2 O(n2 1/m boun on f n (K m,m obtaine by the authors [7] may turn out to be asymptotically tight once the orers of the Turán numbers ex(n, C 8 an ex(n, K m,m in the case m 4 are etermine. Here we prove the best possible result that one can expect for all complete bipartite graphs. Definition 1. The binary entropy function H : [0, 1] R is efine as H(x := x log 2 x (1 x log 2 (1 x. For every positive integer s with s 2, let ( C s := sup x 1+1/s H(x x (0,1 an observe that C s [sγ, (s+2γ], where γ = (log 2 e/e 0.531, which can be shown using elementary calculus. Theorem 2. For all s an t with 2 s t, the number of labele K s,t -free graphs on n vertices satisfies s(t 11/s log 2 f n (K s,t (1 + o(1 2s 1 C s n 2 1/s. Erős conjecture [11] that ex(n, K s,t = Θ(n 2 1/s for all s an t with 2 s t. If this conjecture is true, Theorem 2 woul be asymptotically sharp for all pairs (s, t. So far it has been resolve in the affirmative in the case when s 3 (see [8, 15, 16] or t > (s 1! (see [2, 28]. Therefore, Theorem 2 is sharp for most pairs (s, t. Fürei [15] prove that if t 2, then ex(n, K 2,t = 1 2 t 1 n 3/2 + O(n 4/3. Together with Theorem 2, it implies the following. Corollary 3. If t 2, then the number of K 2,t -free graphs of orer n satisfies ex(n, K 2,t log 2 f n (K 2,t ( o(1 ex(n, K 2,t. Let f n,m (H enote the number of H-free graphs on a fixe n-element vertex set, having exactly m eges. The methos use in the proof of Theorem 2 also give an upper boun on f n,m (K s,t. Theorem 4. For every s an t with 2 s t, let µ s,t = 1 s + s 1 s 2 (t 1(t s s. 2

3 There is an n 0 (epening on s an t such that for all n an m with n n 0 an m n 2 µs,t (log n 3t/s+2, the number f n,m (K s,t of labele K s,t -free graphs of orer n an size m satisfies ( 3tn 2 m f n,m (K s,t. Remaining interesting bipartite graphs, for which Erős conjecture is still wie open inclue C 2k for k 5, Q 3 the graph of the 3-imensional cube an the universal graphs U(k. Recall that for a positive integer k, the universal graph U(k is the bipartite graph with parts A := 2 [k] an B := [k], an ege set efine as follows: m s E ( U(k := { {a, b} : a A, b B an b a }. The remainer of this paper is organize as follows. Section 2 stuies various corollaries of Theorems 2 an 4. In Section 3 we introuce some notation an state a general counting lemma, which is one of the basic builing blocks in the proof of Theorem 2, given in Section 4. Theorem 4 is prove in Section 5. 2 Implications of the main results The main results of this paper, Theorems 2 an 4, have various interesting consequences, some of which we list below. Most of the statements in this section are straightforwar applications of the main results, an hence their proofs are omitte. 2.1 Balogh-Bollobás-Simonovits conjecture Let H be a fixe non-bipartite graph. For every positive constant ε, almost all H-free graphs on n vertices have between ( 1 ε ex(n, H an ( 1 +ε ex(n, H eges. It is not known whether 2 2 a similar concentration aroun one half still occurs when H is bipartite. Nevertheless, one woul expect that the number of eges in a typical H-free graph is at least boune away from the extremal values, 0 an ex(n, H. Balogh, Bollobás, an Simonovits [3] formalize this intuition by stating the following conjecture. Conjecture 5 ([3]. For every bipartite graph H that contains a cycle, there is a positive constant c H such that almost all H-free graphs on n vertices have at least c H ex(n, H an at most (1 c H ex(n, H eges. So far, Conjecture 5 has been prove only in the case H = C 4 [6, 14] an partially (only the lower boun for C 6 [14, 23] an K 3,3 [7]. In [3], the precise structure of almost all octaheron-free (K 2,2,2 -free graphs was characterize. The main obstacle to extening that result to other complete multipartite graphs was the lack of results establishing the lower boun in Conjecture 5 for complete bipartite graphs other than C 4. An immeiate corollary of Theorem 4 provies such a lower boun. Corollary 6. Let s an t be integers satisfying s {2, 3} an t s, or s > 3 an t > (!. There exists a positive constant c s,t such that almost all K s,t -free graphs of orer n have at least c s,t ex(n, K s,t eges. Moreover, if t 2, then we may choose c 2,t = 1/12. 3

4 Combining the methos evelope in the proof of Theorem 2 to obtain an upper boun on the number of one-vertex extensions of a K 2,t -free graph with the argument use in [6], one gets the following. Theorem 7. There exists a positive constant ε such that for every t with t 2, almost all K 2,t -free graphs of orer n have at most (1 ε ex(n, K 2,t eges. The proof of Theorem 7 in the case t > 2 is virtually ientical to the proof for the case t = 2 in [6]. The only non-trivial change is reformulating [6, Lemma 5] an reproving it using the coing algorithm evelope in the proof of Theorem Haxell-Kohayakawa- Luczak conjecture Given two graphs G an H, let us efine the generalize Turán number for H in G, ex(g, H := max{e(k : H K G}. A simple averaging argument implies that for every positive integer k, an arbitrary graph G has a k-partite subgraph with at least (1 1/k e(g eges. It follows that for every G an H, ex(g, H ( 1 1 e(g χ(h 1 ex(n, H ( n 2 e(g. It is natural to ask for which graphs G the above inequality becomes an equality. Haxell, Kohayakawa, an Luczak [21] conjecture that whenever p is large enough, so that the ranom graph G(n, p has many uniformly istribute copies of H, then asymptotically almost surely, ex(g(n, p, H = (1 1 + o(1 e(g(n, p. χ(h 1 Definition 8. Let H be a fixe graph. The 2-ensity of H, enote 2 (H, is efine by { } E(K 1 2 (H := max : K H, V (K 3. V (K 2 Conjecture 9 ([21]. Let H be a fixe balance graph an let G(n, p enote the Erős-Rényi ranom graph of orer n with ege probability p. If p(n n 1/2(H, then with probability tening to 1 as n, ex(g(n, p, H = ( 1 1 χ(h 1 + o(1 e(g(n, p. So far Conjecture 9 has been prove for all cycles [21, 22], K 4 [26], an K 5 [20]. Some partial results are also known for larger complete graphs. Recently, Conlon an Gowers [10] an, inepenently, Schacht [34] have announce that they have prove Conjecture 9 in its full generality an extene it to the setting of ranom uniform hypergraphs. A straightforwar application of Theorem 4 an the first moment metho gives the following relaxe version of Conjecture 9 when H is a complete bipartite graph. Corollary 10. Assume that 2 s t an let µ s,t be as in the statement of Theorem 4. If pn µs,t (log n 3t/s+2, then asymptotically almost surely ex(g(n, p, K s,t = o ( e(g(n, p. (2 4

5 Note that in orer to prove Conjecture 9, one has to show that (2 is still true if we only assume that pn s+t 2 st 1. Still, unless pn 1/s, an hence ex(n, K s,t = o ( E [ e(g(n, p ], the result prove in Corollary 10 is non-trivial. The proof for the case s = t given in [7] works for all s an t. Actually, Theorem 4 allows us to prove (2 in a stronger form. Namely, the little o in (2 can be replace with an explicit function of n an p. Corollary 11. Assume that 2 s t an let µ s,t be as in the statement of Theorem 4. There exists a constant C (epening only on s an t such that if p(n Cn 1 sµs,t (log n 3t+2s, then asymptotically almost surely ex(g(n, p, K s,t Cp 1/s n 2 1/s. (3 Since for arbitrary graphs G an H, one trivially has ex(g, H e(g/ ( n 2 ex(n, H, if Erős conjecture is true an ex(n, K s,t = Θ(n 2 1/s, then for some positive constant c, asymptotically almost surely ex(g(n, p, K s,t cpn 2 1/s. (4 Closing the gap between (3 an (4 remains an interesting problem. In the case of K 2,2 (an all even cycles, this is one in [25], where sharp estimates are obtaine for certain range of p. 2.3 Kohayakawa- Luczak-Röl conjecture Let G be a bipartite graph with parts V 1 an V 2. For two sets V 1 V 1 an V 2 V 2, we efine the ensity of the bipartite graph inuce by the pair (V 1, V 2, enote (V 1, V 2, to be the quantity e(v 1, V 2/( V 1 V 2, where e(v 1, V 2 is the number of eges of G between V 1 an V 2. We say that G is ε-regular if for all sets V 1 V 1 an V 2 V 2 that satisfy V 1 ε V 1 an V 2 ε V 2, the ensity (V 1, V 2 iffers from the ensity (V 1, V 2 of G by at most ε. Definition 12. For a graph H, let G(H, n, m be the family of graphs on the vertex set x V (H V x, where V x are pairwise isjoint sets of vertices of size n, whose ege set is {x,y} E(H E x,y, where E x,y V x V y an E x,y = m. Let G(H, n, m, ε enote the set of graphs in G(H, n, m in which each (V x V y, E x,y is an ε-regular graph. A graph G G(H, n, m, ε looks like H in which every vertex has been replace by an inepenent set of size n an every ege by a set of m eges which form an ε-regular bipartite graph. Kohayakawa, Luczak, an Röl [26] conjecture that whenever these bipartite graphs are ense enough, only a small fraction of graphs in G(H, n, m, ε oes not contain a copy of H. Conjecture 13. Let H be a fixe graph. For any positive β, there exist positive constants ε, C, an n 0 such that for all m an n satisfying m Cn 2 1/2(H an n n 0, we have ( n {G G(H, n, m, ε : H G} β m 2 E(H. m Conjecture 13 is known to be true when H is a tree, a cycle [17], or a complete graph on three [30], four [19], or five vertices [20]. Some partial results are also known for larger complete graphs [18]. A straightforwar application of Theorem 4 gives the following relaxe version of Conjecture 13 in the case when H is a complete bipartite graph. 5

6 Corollary 14. Let s an t be integers satisfying 2 s t, an let µ s,t be as in the statement of Theorem 4. For any positive β an ε, there exist positive constants C an n 0 such that for all n an m satisfying m Cn 2 µs,t (log n 3t/s+2 an n n 0, we have {G G(K s,t, n, m, ε : K s,t G} β m ( n 2 m E(Ks,t. (5 Note that in orer to prove Conjecture 13, one woul have to show that (5 is still true s+t 2 2 if we only assume that m Cn st Ranom Ramsey graphs A graph G is Ramsey with respect to H, G H, if every two-coloring of the eges of G results in a monochromatic subgraph isomorphic to H. Unsurprisingly, the smallest graphs that are Ramsey with respect to the four-cycle are saturate by C 4 s. Erős an Fauree aske (see [14] whether this is always the case, i.e., if there exists a graph G such that G C 4, but G oes not contain a K 2,3. Answering this question, Fürei [14] prove a much stronger result whenever m is large enough, there are K 2,3 -free graphs with m eges, whose largest C 4 -free subgraph has only m 1 c eges, where c 1/51+o(1. Clearly, all such graphs are Ramsey with respect to C 4. He also aske if similar results can be prove for other pairs of graphs. Using the ranom graph argument from [14] combine with Theorem 4, we can give an answer to this question. We woul also like to remark that the problem of Erős an Fauree mentione above was inepenently solve by Nešetřil an Röl [32]. Corollary 15. For all integers s an t with 2 s t, there exist an integer u with u > t an a positive constant c such that for all large enough m, there exists a K s,u -free graphs G with m eges, whose largest K s,t -free subgraph has only m 1 c eges. In particular, if s = t = 3, then one can take u = 4. 3 Notation an preliminaries For a graph G, we enote its vertex an ege sets by V (G an E(G, respectively. The number of eges in G is e(g. For a vertex v V (G, we enote the set of its neighbors by N G (v or simply N(v whenever G is clear from the context. The egree of v in G, enote G (v, is the size of its neighborhoo, i.e., G (v := N G (v. The minimum egree of G is δ(g. For a set A of vertices of G, by NG (A we will enote the set of common neighbors of all vertices in A, i.e., NG (A := v A N G(v, an refer to such sets as A -fol neighborhoos in G. When k is a nonnegative integer, the term k-set (or k-subset abbreviates the phrase k-element set (or k-element subset. For a hypergraph H, we enote its vertex an ege sets by V (H an E(H, respectively. We say that H is k-uniform if E(H consists only of k-subsets of V (H. The number of eges in H is e(h. For an arbitrary X V (H, the subhypergraph inuce on X is the hypergraph H[X] with V (H[X] = X an E(H[X] = {D E(H : D X}. Given a subset S V (H, H S abbreviates H[V (H S]. For a subset W V (H, the egree of W in H, enote eg H (W, is the number of eges of H that W is containe in, i.e., eg H (W := {D E(H : W D}. Given a vertex 6

7 w V (H, we enote the egree of w by eg H (w. For a positive integer l, the maximum l-egree of H is l (H := max{eg H (W : W V (H with W = l}. The maximum egree of H, enote (H, is its maximum 1-egree, 1 (H. Clearly, for every positive integer l, (H V (H l 1 l (H. (6 Given a hypergraph H on a linearly orere vertex set V, the max-egree orering of the vertices of H is the unique orering w 1,..., w V of V such that for each j, if we let W j = {w 1,..., w j }, then w j+1 is the vertex with the smallest label among all vertices in V W j minimizing eg H[V Wj ](w j+1. Finally, σ(h will enote the minimum size of a set of vertices that covers more than half of the eges of H, i.e., σ(h := min{ S : e(h S < e(h/2}. Since one vertex covers no more than (H eges of H, one clearly has σ(h > e(h 2 (H. (7 Throughout the paper, log will always enote the natural logarithm. One of the key ingreients in the proof of Theorem 2 is the following lemma, whose proof, a ouble counting argument in the spirit of Kövári, Sós, an Turán [29], can be foun in [7]. Lemma 16. Fix two integers s an t with 1 s t an a positive real ε such that ε(1 + ε t 1. Let G be an n-vertex graph with minimum egree at least, an A be any set of a vertices of G, where a (1 + ε(t 1 ( ( n s / s. Then the number of copies of Ks,t in G with the larger partite set completely containe in A, enote N s,t (A, satisfies N s,t (A β a t, where β = β(s, t, n,, ε = εt t! ( t / s ( n s t 1. 4 Proof of Theorem 2 Let G be a K s,t -free graph of orer n an let v be a vertex of minimum egree in G. Furthermore, let Ĝ = G {v}. Clearly, Ĝ is K s,t -free an δ(ĝ δ(g 1 = G(v 1. It easily follows that one can fin an orering v 1,..., v n of V (G, such that if we let G i := G[{v 1,..., v i }], then δ(g i Gi+1 (v i+1 1 for all i {1,..., n 1}. In other wors, every n-vertex K s,t -free graph can be obtaine from a single vertex by successively ajoining a vertex of egree + 1 to a graph with minimum egree at least, for some. The general iea of the proof is showing that the number of ways in which one 7

8 can obtain a K s,t -free graph of orer i + 1 from some i-vertex K s,t -free graph in the above process of ajoining vertices of minimum egree is 2 O(i1 1/s, an therefore the number of labele K s,t -free graphs of orer n satisfies n 1 f n (K s,t n! 2 O(i1 1/s = 2 O(n2 1/s. i=1 We start by introucing some notation. For a fixe n-vertex K s,t -free graph G, let f(g; K s,t enote the number of ways we can exten G to a K s,t -free graph of orer n + 1 by ajoining to G a new vertex of egree at most δ(g + 1. Then, we let f(n; K s,t := sup f(g; K s,t, G where the supremum is taken over all K s,t -free graphs with n vertices. The core of the proof is the escription an analysis of an algorithm that encoes the aforementione one-vertex extensions in an economical way, i.e., using only few bits. Precisely, we will achieve the following goal. Goal. Construct an algorithm A meeting the following specification: INPUT: An n-vertex K s,t -free graph G an a set N V (G of size at most δ(g + 1 such that the aition of a new vertex v with N(v = N yiels a K s,t -free graph of orer n + 1. OUTPUT: A bitstring of length at most (1 + o(1(t 1 1/s C s n 1 1/s that uniquely encoes N. By saying that A uniquely encoes N, we mean that there is another algorithm B, which given G an A(G, N, the coe of N in G prouce by A, outputs N. Although we will not explicitly construct such B, it will become clear that one can obtain such an algorithm by slightly moifying A. In particular, the existence of such coing an ecoing proceures implies that for a fixe K s,t -free graph G, the map A(G, is an injection of the set of all possible K s,t -free extensions of G by a single vertex of egree at most δ(g + 1 into a set of size 2 (1+o(1(t 11/s C s n 1 1/s. It then follows that for every n-vertex K s,t -free graph G, f(g; K s,t 2 (1+o(1(t 11/s C s n 1 1/s, an hence n 1 n 1 log 2 f n (K s,t log 2 (n! f(i; K s,t (1 + o(1(t 1 1/s C s i=1 s(t 11/s = (1 + o(1 2s 1 C s n 2 1/s. i=1 i 1 1/s In the remainer of the proof, we will escribe an analyze an algorithm that meets our requirements. To begin with, let us fix some vali input for A, i.e., an n-vertex K s,t -free graph G an a set N V (G with N δ(g + 1 such that making a new vertex v ajacent to all of N yiels a K s,t -free graph G of orer n + 1. Furthermore, let := N 1 an note that δ(g by our assumption. 8

9 Since V (G = n, we may clearly assume that there is an injective mapping of V (G into the set {0, 1} log 2 n or, in other wors, a istinct log 2 n -bit coe for each vertex of G. To simplify notation, from now on we will generally not istinguish vertices of G from their coes. For the most part, the output of our algorithm will be a sequence of vertex coes interwoun with numbers an short control sequences (strings like LOW DEGREE VERTEX, PREPROCESSING, etc. coming from a constant size set. Since all the numbers involve will come from the set {0,..., n 1}, in orer to avoi confusion, let us agree that by outputting a number we will mean outputting its unique coe of fixe length log 2 n, e.g., the binary representation of the number. Same convention applies to control sequences each of them will be assigne a unique coe of length log 2 n. Recall that + 1 = N an δ(g. If n 1 1/s / log 2 n, then A will simply output LOW DEGREE VERTEX, followe by + 1 an the list of all + 1 elements of N in an arbitrary orer. Clearly, the length of the output string is precisely ( + 3 log 2 n, which oes not excee (1 + o(1n 1 1/s. After having hanle the easy case, for the remainer of this section we will restrict our attention to the more interesting case > n 1 1/s / log 2 n. Since G, which we recall is the graph obtaine from G by ajoining v to the vertices in N, is K s,t -free, whenever a t-set D V (G is the larger partite set in a copy of K,t in G, N oes not contain D, i.e., N D t 1. Since n 1 1/s /(2 log n n 1 1/(, Lemma 16 implies that G contains many copies of K,t. Vaguely, this means that N cannot be an arbitrary (( + 1-subset of V (G, but is very restricte, an hence its entropy is much lower than log n Below, we try to make this intuition precise. For the sake of brevity, let us first introuce the following efinition. Definition 17. A t-set D V (G is angerous if N (D = s 1, i.e., D is the larger partite set in a copy of K,t in G. In other wors, a t-set D is angerous if an only if D N (U for some (s 1-set U V (G. The starting point in esigning of the algorithm are the following three simple observations an an estimate on the number of angerous sets. Observation 18. No angerous set is fully containe in N. Observation 19. Let U V (G be an arbitrary (-set of vertices. Then N N (U t 1. Observation 20. Let W V (G be an arbitrary s-set of vertices. Then N (W t 1 an hence N (W contains at most ( t 1 ifferent (s 1-subsets. Lemma 21. Fix some positive ε satisfying ε(1 + ε t 1 an let A be any set of a vertices in G with a (1 + ε(t 1 ( ( n /. There is a 0 such that for all with 0, the number D(A of angerous t-sets in A satisfies D(A α a t, where α = α(s, t, n,, ε = εt s!t! (t n. ((t 1 9

10 Proof. Since G is K s,t -free, every angerous t-set is the larger partite set of exactly one copy of K,t in G, an therefore by Lemma 16, D(A = N,t (A β(s 1, t, n,, ε a t, where β(s 1, t, n,, ε is efine in the statement of Lemma 16. It suffices to prove that β α. First let us observe that lim (1 s/(t = 1, an hence there is a 0 (epening only on s an t such that if 0, then s ( s (t (t. It follows that if 0, then ( t ( t 1 ( β = εt n / εt ( s t ( (s 1! t! s 1 s 1 t! (s 1! n εt t! (t = α. s(s 1!n ((t 1 t 1 Next, let us sketch the rough iea of how our algorithm works. Although this escription is not very formal or precise an misses out a lot of technical etails, we hope that it will make the unerstaning of the pseuocoe of A somewhat easier. At all times, A will maintain a list of alreay encoe elements of N (neighbors of v, enote by Q, an a set A containing the remaining neighbors the set N Q. We will refer to A as the set of eligible vertices an Q the set of alreay encoe vertices. Our goal will be to shrink the eligible set A as much as we can without growing Q too much at the same time. This will be achieve by moving to Q only very carefully chosen vertices from N. Since, as we will later see, encoing one element of Q requires approximately ( log 2 n bits, at all times we can encoe the entire set N using roughly Q log 2 n + log A 2 N Q bits. Once we are one shrinking A, this number will be small enough for our purposes. Before we procee with the explanation, let us efine a few parameters. Let ε = ε(n := 1/ log n, ω = ω(n := (log n 3, an b := t s t s+1. The target size of the eligible set A, i.e., the maximum number of elements we woul like A to have at the very en, is a 0, efine as ( ( n a 0 := (1 + ε(t 1 /. s 1 s 1 Note that a 0 is the lower boun on the carinality of a set A that surely contains a lot of angerous t-subsets (see Lemma 21. The algorithm will work in steps. During a single step, A will lose many elements, whereas Q (an the length of the coe generate by A will grow very little. Each step starts with preprocessing of the eligible set A, a proceure which makes sure that A is well-behave in terms of the sizes of inuce (s 1-fol neighborhoos. In step 3a, we simply remove 10

11 from A all (s 1-fol neighborhoos larger than ω A /, an encoe (an move to Q all neighbors of v (elements of N that those large neighborhoos contain (Observation 19 says that there are at most t 1 neighbors of v in each such neighborhoo. This will be of extremely high importance later, when we analyze the algorithm. When A no longer contains very large (s 1-fol neighborhoos, then we run the core part of A. In step 3c, we pick out a carefully chosen sequence of subsets Q t,..., Q s+1 N Q of size b each that we encoe an move to Q. At the same time, we construct a sequence H t,..., H s, where each H r is an r-uniform hypergraph on the vertex set A with E(H r { D A : {w t,..., w r+1 } D is angerous for some w t Q t,..., w r+1 Q r+1 }. The key property of each H r is that not only none of its eges is fully containe in N (see Observation 19, but also the fact that H r can be compute from only G an the sets Q t,..., Q r+1, without the full knowlege of N (this will allows us to ecoe A(G, N later. Our ultimate goal in the for loop 3c is to maximize the number of eges in H s. Since the eges of the r-uniform hypergraph H r are neighbors in the (r + 1-uniform hypergraph H r+1 of the vertices from Q r+1, i.e., D E(H r if D {w} E(H r+1 for some w Q r+1, we try to achieve this goal by maximizing e(h r in turn for all r {t 1,..., s}. In orer to o that, we try to a to Q r+1 vertices with highest egrees in H r+1. Since Q r+1 N, our choices are quite limite an it might happen that very few of the high-egree vertices in H r+1 belong to N. In this case, we will not be able to make e(h r very large, but we can use the extra information about N (the fact that N contains very few vertices that have high egree in H r+1 to shrink the eligible set we simply elete from A all the high-egree non-neighbors of v, which we keep liste in the set Y. Finally, the existence of very few vertices not belonging to N that cover most of the eges of H r+1 woul get us into trouble as only eleting them from A woul not shrink the eligible set well enough. We overcome this obstacle by keeping the maximum egree of each H r boune the auxiliary set X serves this purpose. By efinition, the eges of the s-uniform hypergraph H s will have the nice property that none of them is fully containe in N. In the for loop 3, we will exploit this fact to shrink the eligible set by working with H s with the use of methos evelope in [7]. The rough iea is the following. If many vertices of N have high egree in H s, an hence some (s 1b of them almost-cover many eges, i.e., many eges of H s contain s 1 of these (s 1b vertices from N, then we can remove all the uncovere vertices in these almost-covere eges from A. Otherwise, very few of the high-egree vertices in H s are members of N an we can significantly shrink A by eleting all the high-egree non-neighbors of v from A. More precisely, we will repeat the following b times. Using H s, we construct a sequence H,..., H 1, where each H r is an r-uniform hypergraph that can be compute from only G an Q an has the property that none of its eges is fully containe in N. In particular, each ege of H 1 has empty intersection with N an hence we can remove it from A. Similarly as before, either H 1 has many eges or in the process of computing the sequence H,..., H 1, we gain some extra information about N that we can use to shrink the eligible set. Either way, we will elete many elements from A. After this lengthy introuction, we present the algorithm in the high-egree case, i.e., when > n 1 1/s /(2 log n. Recall the efinition of the max-egree orering given in Section Output HIGH DEGREE VERTEX. 2. Set A := V (G an Q :=. 11

12 3. While A > a 0, o the following: (a If there exists an (-set U V (G with N (U A > ω A /, o the following: i. Let U = {u 1,..., u } an N (U N = {w 1,..., w k }. ii. Set A := A N (U an Q := Q {w 1,..., w k }. iii. Output PREPROCESSING : u 1,..., u, k, w 1,..., w k an go to step 3. (b Let H t := {D A : D = t an D is angerous}. (c For r = t 1,..., s, o the following: i. Set Q r+1 :=, X :=, an Y :=. ii. Let H r be an empty r-uniform hypergraph on A. iii. For i = 1,..., b, o the following: Let w i 1,..., w i A X Y be the max-egree orering of the vertices of H r+1[a X Y ] an let W i j = {w i 1,..., w i j} for every j. Let j i be the smallest j such that w i j N. H r := H r { D : {w i j i } D H r+1 [A X Y W i j i 1] }. Set Q r+1 := Q r+1 {w i j i } an Y := Y W i j i. Set X := X {w A : eg Hr (w > b t r s t A r 1 }. iv. Set Q := Q Q r+1. v. Suppose the vertices ae to Q r+1 were w 1,..., w b. Output w 1,..., w b. vi. If Y σ(h r+1 /2, then A := A Y, output SKIP an go to step 3. ( For i = 1,..., b, o the following: i. For r = s 1,..., 1, o the following: Let w r 1,..., w r A be the max-egree orering of the vertices of H r+1[a] an let W r j = {w r 1,..., w r j} for every j. Let j r be the smallest j such that w r j N. Set A := A W r j r an Q := Q {w r j r }. Set H r := { D A : {w r j r } D H r+1 }. ii. Let A := A { w : {w} E(H 1 }. iii. Output w j,..., w 1 j Let N := N Q. Clearly N A. The set N is one of the ( A N ifferent N -subsets of A. Output REMAINDER : A, N, followe by a log 2 ( A N -bit coe of N in A. For the remainer of this iscussion, let us fix G an N with = N 1 n 1 1/s /(2 log n an assume that we run A on the pair (G, N. Note that given G an the output A(G, N, one can reconstruct N. The key observation that reassures us that it is possible is noting that the final sets A an Q can be recompute step-by-step in the exact same way as they were compute by A as all the necessary information about N that is neee for it appears in A(G, N. Once we reconstruct A an Q, we can easily ecoe N = N Q using the last fragment of A(G, N starting with REMAINDER. The non-trivial part of the analysis is proving an O(n 1 1/s boun on the size of the output of A. Recall that our aim is to prove that the length of the output satisfies A(G, N (1 + o(1(t 1 1/s C s n 1 1/s. 12

13 We start by looking at the preprocessing stage. Let p enote the total number of times A preprocesses the eligible set A, i.e., the number of times an appropriate (-set U is foun in step 3a. Claim 22. The total number of preprocessing steps satisfies p log n ω + 1. Proof. Each time A preprocesses the eligible set, A loses more than ω A / elements. Hence, preprocessing the eligible set q times shrinks it by a factor of at most γ 1, where ( γ 1 ω q e q ω. Since A starts with A = n an after p 1 preprocessing steps A is still non-empty, it follows that (p 1ω/ log n. Let α be as in the efinition in Lemma 21. Moreover, for each r {t,..., s 1}, let ( ( r t t 1 B r := 4t an D r := 3(t s(t r. (8 s 1 The core of our analysis will be the following lemma. Lemma 23. Suppose that uring some iteration of the main while loop, step 3, A oes not preprocess the eligible set. Then uring that iteration, the eligible set A loses at least elements. B (log n D t s α A Let z be the total number of times A oes not preprocess the eligible set A in an iteration of the main while loop. The following corollary is an immeiate consequence of Lemma 23. Corollary 24. The total number of times A executes the main while loop without preprocessing A, z (log nd +1 B s t α Proof. By Lemma 23, uring each iteration of the main while loop, in which A oes not preprocess the eligible set, A loses at least B (log n D t s α A elements. Hence, as a result of q such iterations, the eligible set shrinks by a factor of at most γ, where ( q ( B γ 1 (log n B D t s α exp q (log n D t s α. Since A starts with A = n an after z 1 such iterations A is still non-empty, (z 1 B (log n D t s α log n. 1 The phrase A shrinks by a factor of at most γ means that the size of A rops from some a to at most γ a or, in other wors, that A loses at least (1 γ a elements. 13

14 Before we ive into the proof of Lemma 23, let us show how Corollary 24 implies that A outputs short coes. Lemma 25. For every input pair (G, N, the length of the output prouce by A oes not excee (1 + o(1(t 1 1/s C s n 1 1/s. (9 Proof. Note that by Observation 19, the k in the preprocessing step 3a never excees t 1. Hence the total length l 1 of the output prouce by A in step 3a satisfies ( log n l 1 p (1 + (s (t 1 log 2 n + 1 (s + t log ω 2 n, (10 where the boun on the total number p of preprocessing steps comes from Claim 22. Since ω = (log n 3 log n log 2 n, it follows that l 1 = o(. Each of the z executions of the main while loop with no preprocessing outputs either coes of at most (t 1b vertices or coes of at most (t sb vertices an the SKIP control sequence. Either way, this is never more than tb log 2 n bits. Therefore the total length l 2 of the output prouce by A in steps 3c an 3 satisfies ( (log n D +1 l 2 z tb log 2 n s t α tb log B 2 n, (11 where the secon inequality comes from Corollary 24. Recall that ε = 1/ log n, an we are in the high-egree, i.e., > n 1 1/s /(2 log n, case. Therefore, s t α 1 = s!t! (log n t n((t 1 s(t 1 2 s(t 1 s!t! (log n (s+1t s (12 an hence l 2 g(n b, where g(n is polylogarithmic in n. It follows that l 2 = o(. When A finally reaches step 4, then A a 0 an hence the total length l 3 of the output prouce by A in step 4 satisfies ( ( a0 a0 + Q l 3 3 log 2 n + log 2 4 log N 2 n + log 2 (13 N + Q ( a0 + Q = 4 log 2 n + log 2 N ( a0 + Q 5 log 2 n + log 2. Next, note that for n large enough, ( ( n n a 0 = (1 + ε(t 1 / (1 + ε(t 1 (14 s 1 s 1 ( s ( n (1 + 2ε(t 1. Since ex(n, K s,t = O(n 2 1/s, 2e(G n ex(n + 1, K s,t n = O(n 1 1/s.

15 Recall that Q gains at most t 1 elements in each of the p preprocessing steps, an at most (t 1b elements in each of the z non-preprocessing steps. From (10, (11, an (12 it follows that ( ( 1 ( n Q p(t 1 + z(t 1b = O = O. (15 (log n 2 (log n 2 Recall again that ε = 1/ log n. Inequalities (14 an (15 imply that ( n a 0 + Q (1 + 3ε(t 1. (16 Using (13, (16, an the following well-known estimate relating binomial coefficients with the binary entropy function (see, e.g., Lemma 9 in [31]: ( 1 n n + 1 2n H(k/n 2 n H(k/n, k we further estimate ( a0 3 log 2 n + log 2 N ( ( (1 + 3ε(t 1 n 5 log 2 n + log 2 ( 5 log 2 n + (1 + 3ε(t 1(n/ H s (1 + 3ε(t 1n (17. Substituting x := s / ( (1 + 3ε(t 1n in (17 yiels ( a0 3 log 2 n + log 2 5 log N 2 n + ( (1 + 3ε(t 1 1/s H(x x 1 1/s n1 1/s. (18 Recall that C s = sup x ( H(x/x 1 1/s. Since the total size of the output, l 1 + l 2 + l 3, is boune by the sum of the quantities in the right-han sies of (10, (11, an (18, we get (9. Before we are able to prove Lemma 23, we nee to make some preparations. For the sake of brevity, by i th iteration of any for loop, we will enote the iteration, where the loop variable takes the value i. The following claim explains why A constantly preprocesses the eligible set an maintains the oly efine set X. Claim 26. Assume that uring some iteration of the main while loop, step 3, at the time we reach step 3c, the eligible set A has a elements. Then throughout this iteration, for all r {s,..., t}, ( t 1 ( ω t s (H r b t r a r 1. s 1 Proof. First observe that at all times uring any iteration of the main while loop, for all r {s,..., t}, the eges of H r all come from the set { D A : {wt,..., w r+1 } D is angerous for some w t Q t,..., w r+1 Q r+1 }. Consier first the case r t s+1. Fix some w r,..., w t s+1 A, let W := {w r,..., w t s+1 }, an note that by our assumption, W = r + s t 1. The set W is containe in some 15

16 D H r only if there are w t Q t,..., w r+1 Q r+1, an an (s 1-set U V (G such that {w t,..., w r+1 } D N (U A (recall Definition 17. Let W := {w t,..., w t s+1 }. Then clearly W = s an N (U W or equivalently, N (W U. Hence by Observation 20, when W fixe, then there are at most ( t 1 such sets U. Moreover, since Qr b for all r, the number of such W that contain our fixe set W is at most b W W = b t r. Also, because A is preprocesse, for every (s 1-set U, N (U A ωa/. Putting all these inequalities together, eg Hr (W b t r ( t 1 s 1 ( ωa ( r W t 1 = s 1 b t r ( ωa t s. (19 Since the term in the right-han sie of (19 oes not epen on a particular choice of W, it follows that ( t 1 ( ω t s r+s t (H r b t r a t s, s 1 an hence by (6, ( t 1 ( ω t s (H r b t r a r 1. s 1 The case s r t s is much more elicate. First, consier how much the H r -egree of a vertex w A can increase in one particular, say i th, iteration of the for loop 3(ciii. Since all eges ae in the i th iteration contain wj i i, eg Hr (w increases by no more than the number of eges D H r+1 containing both w an wj i i. In orer for such an (r + 1-set D to be an ege of H r+1, there ought to be some w t Q t,..., w r+2 Q r+2, an an (s 1-set U V (G such that {w t,..., w r+2 } D N (U A. Note that r + 2 t s + 3 by our assumption on r an let W := {w t,..., w t s+3, wj i i, w}. Then clearly, W = s an N (U W or equivalently, N (W U. Hence by Observation 20, when W fixe, then there are at most ( t 1 such sets U. Moreover, since Qr b for all r, the number of such W that contain both wj i i an w is at most b W 2 = b s 2. Also, because A is preprocesse, for every (s 1-set U, N (U A ωa/. Putting all these inequalities together, we see that eg Hr (w cannot change by more than ( t 1 ( ωa ( D {w,w i b s 2 ji } t 1 ( ωa r 1 = b s 2 (20 s 1 s 1 ( t 1 = ω r 1 t s r+1 s 1 b t s r+2 bt r s t a r 1. Recall that b = t s t s+1 t s r+1 b t s r+2 = 1 an note that the right-han sie of (20 is o(b t r s t a r 1, as ( t s r+2 = 1 ( r 1 t s r+2 ( b 1 t s+1 1 t s+1 =, r 1 t s+1 is only polylogarithmic in n. Every time a vertex w A lans in the set X of vertices with high egree in H r, no more eges containing w get ae to H r, since A looks only at the eges of H r+1 [A X Y ]. Hence the egree of w cannot excee b t r s t a r 1, i.e., the quantity from the efinition of X, by more than the right-han sie of (20. It follows that > 0, t s+1 n1/3, an ω r 1( t 1 (H r (1 + o(1b t r s t a r 1 ( t 1 s 1 ( ω t s b t r a r 1. 16

17 The secon key step in proving Lemma 23 is the following simple estimate on the number of eges in H r. Recall that the constants B r an D r are efine in (8. Claim 27. Assume that uring some iteration of the main while loop, at the time we reach 3b, the eligible set A has a elements. Then for every r, with t 1 r s, either A outputs SKIP by the en of the r th iteration of the for loop 3c or e(h r B r (log n Dr bt r α a r. (21 Proof. By Lemma 21, (21 clearly hols when r = t. It suffices to prove that if (21 hols for r + 1 an A oes not output SKIP in the r th iteration of the for loop 3c, then (21 hols for r. Let x an y be the sizes of X an Y at the en of the r th iteration. There are two cases to consier. Case 1. x + y σ(h r+1. Since A i not output SKIP, y < σ(h r+1 /2, an hence by (7, x > σ(h r+1 2 e(h r+1 4 (H r+1 B r+1 (log n D r+1 t s 4 ( t 1 ω t s α a, where the last inequality follows from Claim 26 an the assumption that (21 hols for r +1. Recall that for every w X, eg Hr (w > b t r s t a r 1. Clearly, e(h r 1 r eg Hr (w > x r bt r s t a r 1 w X B r (log n Dr bt r α a r, since B r (4r 1( t 1 1 Br+1, D r = D r+1 + 3(t s, an ω = (log n 3. Case 2. x + y < σ(h r+1. In particular, for all i {1,..., b}, uring the r th iteration of the for loop 3c, e(h r+1 [A X Y W i j ] e(h r+1 /2. By the maximality of eg Hr+1 [A X Y W i j i 1 ] (w i j i, in each iteration of the for loop 3(ciii, H r acquires e eges, where e r e(h r+1 /2 A X Y Wj i i 1 e(h r+1 a B r+1 (log n D r+1 bt r 1 α a r. Unfortunately, some eges may get ae to H r more than once. How many times can we a to H r the same ege D? For each i {1,..., b}, let D i := D {wj i i }. The set D becomes an ege of H r precisely when D i H r+1 for some i. In particular, every such D i is fully containe in some angerous set, an hence there must be an (s 1-set U V (G, such that D D i N (U. Since D = r s, by Observation 20, there are at most ( t 1 such U. Also, because for each i, wj i i N, by Observation 19, for no (s 1-set U, N (U contains more than t 1 ifferent wj i i s. It follows that the maximum number of times D can be ae to H r is (t 1 ( t 1. Therefore, B r+1 1 e(h r (t 1 ( t 1 b (log n D bt r 1 α a r r+1 B r (log n Dr bt r α a r. 17

18 Lemma 28 an its immeiate consequence, Corollary 29, are the last missing ingreients neee in the proof of Lemma 23. Lemma 28. For every fixe i an r satisfying 1 i b an 0 r s 1, the following hols. Suppose that uring the i th iteration of the for loop 3, at the beginning of the r th iteration of the for loop 3(i, e(h r+1 [A] γa r+1 for some γ an a with 0 < γ 1 an a A. Then r e(h 1 + j q γa. (22 q=1 Proof. For a fixe i, we prove the claim by inuction on r. The inequality (22 hols trivially when r = 0. Suppose that r > 0 an (22 hols for r 1. Each of w r 1,..., w r j r 1 clearly belongs to no more than A r (r + 1-subsets of A, an hence e(h r+1 [A W r j r 1] e(h r+1 [A] (j r 1 A r γa r+1 (j r 1a r. (23 If j r γa, then (22 hols, so we may suppose that the reverse inequality is true, an therefore the rightmost term in (23 is positive. Since we have selecte wj r r to maximize its egree in H r+1 [A Wj r r 1], we have e(h r = eg Hr+1 [A Wjr 1 r ] (wj r r + 1 r A j r + 1 e(h r+1[a Wj r r 1] r + 1 a j r + 1 (γa j r + 1 a r γa j r + 1 a j r + 1 ar γa j r a r, a j r where the last inequality hols since γ 1, an hence γa j r a j r. By the inuctive assumption with γ = γa jr a j r, r 1 e(h 1 + j q γa j r a γa j r. a j r q=1 Corollary 29. Assume that at the beginning of the 1 st iteration of the for loop 3, A has a elements. If at the beginning of the i th iteration, e(h s [A] βa s for some positive β, then in that iteration A loses at least βa elements. Proof. During the i th iteration, we elete from A precisely e(h 1 + q=1 j q elements. Since certainly a A, an we have assume that e(h s βa s, the statement of Corollary 29 is just a irect application of Lemma 28. Finally, we are reay to give the proof of Lemma 23. Proof of Lemma 23. By Claim 27, either at the en of the s th iteration of the for loop 3c, e(h s B s (log n Ds bt s α A s, (24 18

19 or for some r with r s, A outputs SKIP at the en of the r th iteration. In the latter case, at the en of the r th iteration, Y σ(h r+1 /2. By Claims 26 an 27, Y σ(h r+1 /2 e(h r+1 2 (H r+1 B r+1 2 ( t 1 ω t s (log n D r+1 B r (log n Dr t s α a B (log n D t s α A, t s α a an since A outputs SKIP, the eligible set A loses exactly Y elements. Therefore we can assume that (24 is true an A executes the for loop 3. Similarly as in the proof of Claim 27, there are two cases to consier. Case 1. At the en of the b th iteration of the for loop 3, e(h s [A] e(h s /2. In particular, this is true in all the previous iterations. Hence, if a is the size of the eligible set A at the beginning of the step 3, by Corollary 29, as a result of a single iteration, A loses at least e(hs elements (apply Corollary 29 with β := e(hs. Since there are b iterations, 2a 2a s altogether A loses a elements, where (recall that t s = b t s+1 a b e(h s 2a B s 2(log n Ds bt s+1 α a B (log n D t s α A, where the secon inequality follows from (24. Case 2. At the en of the b th iteration of the for loop 3, e(h s [A] < e(h s /2. It means that in the step 3, A must have lost at least σ(h s elements an σ(h s e(h s 2 (H s B s (log n Ds ( t 1 t s where the secon inequality follows from (24 an Claim 26. α a B ω t s (log n D t s α A, Since in Lemma 25 we have alreay shown how Lemma 23 implies that A outputs short coes, the proof of Theorem 2 is now complete. 5 Proof of Theorem 4 As it was remarke at the beginning of the proof of Theorem 2, every n-vertex graph G can be constructe from an isolate vertex v 1 by successively connecting a vertex v i+1 to some i vertices in G[{v 1,..., v i }] in such a way that for all i {1,..., n 1}, i = δ(g[{v 1,..., v i+1 }] δ(g[{v 1,..., v i }] + 1. Moreover, if G is K s,t -free, so are all the intermeiate graphs G[{v 1,..., v i }]. Call the sequence ( i n 1 i=1 a egeneracy sequence of G an note that e(g = n 1 i=1 i. Let f(g;, K s,t be the number of ways one can ajoin to a K s,t -free graph G, with δ(g, a new vertex of egree + 1, so that the graph remains K s,t -free. If we let f(n;, K s,t := sup f(g;, K s,t, G 19

20 where the supremum is taken over all n-vertex K s,t -free graphs whose minimum egree is at least, then f n,m (K s,t n! n 1 f(i; i 1, K s,t, (25 ( i i=1 where the above sum is taken over all egeneracy sequences with sum m. If n 1 µs,t (log n 3t/s an n n 0, then we give a rather crue boun ( ( i n ( en ( f(i;, K s,t n n exp n 1 µ s,t (log n 3t/s+1. ( Suppose now that > n 1 µs,t (log n 3t/s an let α = α(s, t, n,, 1/(3t 3 be as in Lemma 21. Suppose we run the high-egree case in the algorithm A from the proof of Theorem 2 on some i-vertex K s,t -free graph G an a set N of size + 1, where G an N satisfy our usual assumptions. Note that Claim 22 an Corollary 24 are still true, since in their proofs we have not use any assumptions on. Reasoning along the lines of Lemma 25, we can see that the total length of the output prouce by A in the preprocessing step 3a is still o(. Moreover, recall that b = s t s t+1 an hence s t α 1 b = s!t!(3t 3 t i((t 1 s(t 1 t s t s+1 s!t!(3t 3t s!t!(3t 3 t (log n 3t s [s(t 1+ t s+1] 1, where the last inequality follows because µ s,t satisfies ( 1 (s 1(t 1 = (1 µ s,t s(t 1 +. t s + 1 n ((t 1 s(t 1+ 1 t s+1 By (11, the total length of the output prouce by A in steps 3c an 3 is at most ( ( t s+1 t 1 tb log 2 i + t 4t (log i 3(t s(t s+1+1 log s 1 2 i s t α 1 b. By (27, this is clearly o(, since b log 2 i = o( an (log i 3(t s(t s+1+1 log 2 i (log n 3t s [s(t 1+ t s+1] 1. By inequality (17, the total length of the output prouce by A in step 4 is at most ( ( t i 5 log 2 i + log 2 ( ( t n 5 log 2 n + log 2 5 log 2 n + log 2 ( etn Hence the total length of the output of A in the case > n 1 µs,t (log n 3t/s is ( etn log 2 + o(. s s (27. 20

21 Since with G fixe, A outputs a unique coe for every N, the above boun implies that if > n 1 µs,t (log n 3t/s, then the total number of vali ( + 1-sets N satisfies ( ( etn f(g;, K s,t exp log + o(. (28 Let I := {i: i > n 1 µs,t (log n 3t/s } an let m := i I ( i 1. Since the term in the right-han sie of (28 oes not epen on G, it is also an upper boun on f(i;, K s,t an hence for every egeneracy sequence ( i n 1 i=1 with sum m, ( n 1 f(i; i 1, K s,t exp n 2 µs,t (log n 3t/s+1 + ( etn ( i 1 log + o(m. ( i=1 i 1 s i I (29 The function [0, x x log x R is convex, an so ( i 1 log( i 1 I (m / I log(m / I m log(m /n. i I i I This yiels ( ( etn etn ( i 1 log m log(etn m s log(m /n = m 2 log. (30 ( i 1 s (m s Since (x log(y/x = log(y/x 1, m x m = n 1 + i I ( i 1 n + n 2 µs,t (log n 3t/s, an m n 2 µs,t (log n 3t/s+1, we get the estimate ( ( etn 2 etn 2 m log m log = O ( (m m log n = o(m, (m s m s which combine with (29 an (30 gives Since n 1 ( ( etn f(i; i 1, K s,t exp n 2 µs,t (log n 3t/s m log i=1 s m s + o(m. (31 m n 2 µs,t (log n 3t/s+1, e < 3, m ex(n, K s,t 1 2 (t 11/s n 2 1/s + O(n, an there are at most n! egeneracy sequences, combining (25 with (31 yiels ( 3tn 2 m f n,m (K s,t, whenever n is large enough. Acknowlegement We are inebte to the anonymous referee for pointing out an error in a previous proof of Theorem 4 an for their numerous valuable comments an suggestions that helpe to improve the presentation of the paper. 21 m s

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