Probabilistic Methods in Combinatorics Lecture 6

Transcription

1 Probabilistic Methods in Combinatorics Lecture 6 Linyuan Lu University of South Carolina Mathematical Sciences Center at Tsinghua University November 16, 2011 December 30, 2011

2 Balance graphs H has v vertices and e edges. 2 / 16

3 Balance graphs H has v vertices and e edges. ρ(h) = e/v. 2 / 16

4 Balance graphs H has v vertices and e edges. ρ(h) = e/v. H is called balanced of for any subgraph H, ρ(h ) ρ(h). 2 / 16

5 Balance graphs H has v vertices and e edges. ρ(h) = e/v. H is called balanced of for any subgraph H, ρ(h ) ρ(h). H is called strictly balanced of for any proper subgraph H, ρ(h ) < ρ(h). 2 / 16

6 Results Theorem: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph (not necessarily induced) of G. Then p = n v/e is the threshod function for A. 3 / 16

7 Results Theorem: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph (not necessarily induced) of G. Then p = n v/e is the threshod function for A. If H is not balanced then p = n v/e is the threshod function for A. 3 / 16

8 Results Theorem: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph (not necessarily induced) of G. Then p = n v/e is the threshod function for A. If H is not balanced then p = n v/e is the threshod function for A. Proof: Write X = S X S. Then E(X) = ( n v) p e. 3 / 16

9 Results Theorem: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph (not necessarily induced) of G. Then p = n v/e is the threshod function for A. If H is not balanced then p = n v/e is the threshod function for A. Proof: Write X = S X S. Then E(X) = ( n v) p e. If p n v/e, then E(X) = o(1); X = 0 almost surely. 3 / 16

10 Results Theorem: Let H be a balanced graph with v vertices and e edges. Let A(G) be the event that H is a subgraph (not necessarily induced) of G. Then p = n v/e is the threshod function for A. If H is not balanced then p = n v/e is the threshod function for A. Proof: Write X = S X S. Then E(X) = ( n v) p e. If p n v/e, then E(X) = o(1); X = 0 almost surely. If p n v/e, then E(X). We have v = O( n v i p e (ie/v) ) = o(e(x)). i=2 3 / 16

11 Two other results Theorem: Let H be a strictly balanced graph with v vertices and e edges and a automorphisms. Let X be the copies of H in G(n, p). Assume p n v/s. Then almost always X nv p e a. 4 / 16

12 Two other results Theorem: Let H be a strictly balanced graph with v vertices and e edges and a automorphisms. Let X be the copies of H in G(n, p). Assume p n v/s. Then almost always X nv p e a. Theorem: Let H be any fixed graph. For every subgraph H of H (including H itself) let X H denote the number of copies of H in G(n, p). Assume p is such that E(X H ) for every H. Then almost surely X H E(X H ). 4 / 16

13 Clique number of G(n, 1/2) ω(g): the clique number of G. 5 / 16

14 Clique number of G(n, 1/2) ω(g): the clique number of G. f(k) = ( n k) 2 ( k 2) : the expected number of k-cliques. 5 / 16

15 Clique number of G(n, 1/2) ω(g): the clique number of G. f(k) = ( n k) 2 ( k 2) : the expected number of k-cliques. Theorem: Let k = k(n) satisfying k 2 log 2 n and f(k). Then almost surely ω(g) k. 5 / 16

16 Clique number of G(n, 1/2) ω(g): the clique number of G. f(k) = ( n k) 2 ( k 2) : the expected number of k-cliques. Theorem: Let k = k(n) satisfying k 2 log 2 n and f(k). Then almost surely ω(g) k. Proof: For each k-set S, let X S be the indicator random variable that S is a clique and X = S =k X S. E(X) = ( ) n 2 2) (k = f(k). k 5 / 16

17 Continue = k i) where g(i) = i)( (k n k ( n k) k 1 i=2 ( k i )( ) n k 2 2) ( (i k 2). k i k 1 E( X ) = 2 (i 2). i=2 g(i), 6 / 16

18 Continue = k i) where g(i) = i)( (k n k ( n k) k 1 i=2 ( k i )( ) n k 2 2) ( (i k 2). k i k 1 E( X ) = 2 (i 2). Then i=2 g(i), g(i) max{g(2), g(k 1)} = o(n 1 ). Thus, = o(e(x)). 6 / 16

19 Remark For k 2 log 2 n, then f(k + 1) f(k) f(k + 1) f(k) = n k k k. = n 1+o(1). 7 / 16

20 Remark For k 2 log 2 n, then f(k + 1) f(k) f(k + 1) f(k) = n k k k. = n 1+o(1). Let k 0 be the value with f(k 0 ) 1 > f(k 0 + 1). For most of n, f(k) will jump from very large to ver small. With high probabilty, ω(g) = k 0. 7 / 16

21 Distinct sum A set x 1,..., x k of positive integers is said to have distinct sums if all sums x i, S {1,...,k} are distinct. i S 8 / 16

22 Distinct sum A set x 1,..., x k of positive integers is said to have distinct sums if all sums x i, S {1,...,k} are distinct. i S Let f(k) be the smallest k for which there is a set with distinct set. It is clear f(n) 1 + log 2 n. {x 1, x 2,...,x k } {1,...,n} 8 / 16

23 Distinct sum Erdős offered $300 for a proof or disproof that f(n) log 2 n + O(1). 9 / 16

24 Distinct sum Erdős offered $300 for a proof or disproof that f(n) log 2 n + O(1). An easy upper bound f(n) < log 2 n + log 2 log 2 n + O(1). 9 / 16

25 Distinct sum Erdős offered $300 for a proof or disproof that f(n) log 2 n + O(1). An easy upper bound f(n) < log 2 n + log 2 log 2 n + O(1). Theorem: f(n) < log 2 n log 2 log 2 n + O(1). 9 / 16

26 Lovász Local Lemma A 1, A 2,..., A n : n events in an arbitrary probability spaces. 10 / 16

27 Lovász Local Lemma A 1, A 2,..., A n : n events in an arbitrary probability spaces. A dependency digraph D = (V, E): if for each A i, A i is mutually independent to all the events {A j : A i A j E}. Lovász Local Lemma, general case: If there are real number x 1,..., x n such that 0 x i < 1 and Pr(A i ) x i (i,j) E (1 x j) for all 1 i n. Then Pr ( ) n n i=1āi (1 x i ) > 0. i=1 10 / 16

28 Symmetric Case Lovász Local Lemma, symmetric case: Let A 1, A 2,...,A n be events in an arbitrary probability space. Suppose that each event A i is mutually independent of a set of all the other event A j but at most d, and that Pr(A i ) p for all 1 i n. If ep(d + 1) < 1, then Pr( n i=1āi) > / 16

29 Property B Theorem: Let H = (V, E) be a hypergraph in which every edge has at least k elements, and suppose that each edge of H intersects at most d other edges. If e(d + 1) 2 k 1, then H has property B. 12 / 16

30 Property B Theorem: Let H = (V, E) be a hypergraph in which every edge has at least k elements, and suppose that each edge of H intersects at most d other edges. If e(d + 1) 2 k 1, then H has property B. Proof: Color each vertex in two colors randomly and independently. For each edge f E, let A f be the event that f is monochromatic. Then Pr(A f ) = 2 1 f 2 1 k. A f is independent to all event but at most d. Aplly LLL. 12 / 16

31 k-coloring of R Let c: R {1, 2,...,k} be a k-coloring of R. A set T R is multicolored if c(t) = {1, 2,..., k}. 13 / 16

32 k-coloring of R Let c: R {1, 2,...,k} be a k-coloring of R. A set T R is multicolored if c(t) = {1, 2,..., k}. Theorem: Let m and k be two positive intergers satisfying e(m(m 1) + 1)k(1 1 k )m 1. Then, for any set S of m real numbers there is a k-coloring so that each translantion x + S (for x R) is multicolored. 13 / 16

33 k-coloring of R Let c: R {1, 2,...,k} be a k-coloring of R. A set T R is multicolored if c(t) = {1, 2,..., k}. Theorem: Let m and k be two positive intergers satisfying e(m(m 1) + 1)k(1 1 k )m 1. Then, for any set S of m real numbers there is a k-coloring so that each translantion x + S (for x R) is multicolored. The condition is satisfied if m > (3 + o(1))k log k. 13 / 16

34 Proof First we use LLL to prove For any finite set X R, there is a k-coloring so that x + S (for all x X) is multi-colored. 14 / 16

35 Proof First we use LLL to prove For any finite set X R, there is a k-coloring so that x + S (for all x X) is multi-colored. Let Y = x X (x + S). Color numbers in Y in k-colors randomly and independently. Let A x be the event that x + S is not multi-colored. Pr(A x ) k(1 1 k )m / 16

36 Proof First we use LLL to prove For any finite set X R, there is a k-coloring so that x + S (for all x X) is multi-colored. Let Y = x X (x + S). Color numbers in Y in k-colors randomly and independently. Let A x be the event that x + S is not multi-colored. Pr(A x ) k(1 1 k )m 1. A x depends on A y if (x + S) (y + S). Equivalently, y x S S. There are at most m(m 1) such events. d m(m 1). 14 / 16

37 continue Apllying LLL, we get Pr( x X Ā x ) > 0. Then by Tikhonov s theorem, [k] R is compact. For any x R, let C x = {c [k] R : x + S is multi-colored}. 15 / 16

38 continue Apllying LLL, we get Pr( x X Ā x ) > 0. Then by Tikhonov s theorem, [k] R is compact. For any x R, let C x = {c [k] R : x + S is multi-colored}. Now C x is a closed set and x X C x for any finite X. Then x R C x. 15 / 16

39 Ramsey numbers Theorem (Spencer, 1975) R(k, k) (1 + o(1)) 2 e k2k/2. 16 / 16

40 Ramsey numbers Theorem (Spencer, 1975) R(k, k) (1 + o(1)) Theorem (Spencer, 1975) R(3, k) ck2 log k. 2 e k2k/2. 16 / 16

41 Ramsey numbers Theorem (Spencer, 1975) R(k, k) (1 + o(1)) Theorem (Spencer, 1975) R(3, k) ck2 log k. 2 e k2k/2. Best bounds for R(r, k) (for fixed r and k large), c ( ) (r+1)/2 k k r 1 < R(r, k) < (1 + o(1)) log k log r 2 k. 16 / 16