Rainbow matchings. Existence and counting. Guillem Perarnau. Universitat Politècnica de Catalunya Departament de Matemàtica Aplicada IV

Transcription

1 Rainbow matchings Existence and counting Guillem Perarnau Universitat Politècnica de Catalunya Departament de Matemàtica Aplicada IV 2nd September 2011 Budapest joint work with Oriol Serra

2 Outline 1 The problem 2 Counting with the Local Lemma 3 Our Approach 4 Random Models Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 2 / 19

3 Outline 1 The problem 2 Counting with the Local Lemma 3 Our Approach 4 Random Models Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 3 / 19

4 Rainbow matchings and Latin transversals Edge coloring. C : E(K n,n) N Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

5 Rainbow matchings and Latin transversals Edge coloring. Perfect matching: C : E(K n,n) N M = {e i indep} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

6 Rainbow matchings and Latin transversals Edge coloring. Perfect matching: C : E(K n,n) N M = {e i indep} Rainbow matching: no repeated colors in M. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

7 Rainbow matchings and Latin transversals Edge coloring. C : E(K n,n) N Integer square matrix A = {a ij } Perfect matching: M = {e i indep} Rainbow matching: no repeated colors in M C A Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

8 Rainbow matchings and Latin transversals Edge coloring. C : E(K n,n) N Integer square matrix A = {a ij } Perfect matching: M = {e i indep} Transversal T σ = {a iσ(i) } Rainbow matching: no repeated colors in M C A Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

9 Rainbow matchings and Latin transversals Edge coloring. C : E(K n,n) N Integer square matrix A = {a ij } Perfect matching: M = {e i indep} Transversal T σ = {a iσ(i) } Rainbow matching: no repeated colors in M. Latin Transversal: no repeated entries in T σ C A Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 4 / 19

10 Open problems on Latin squares - Existence Conjecture (Ryser, 1967) Every latin square of odd order admits a latin transversal. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 5 / 19

11 Open problems on Latin squares - Existence Conjecture (Ryser, 1967) Every latin square of odd order admits a latin transversal. Conjecture (Brualdi, 1975) Every latin square admits a partial latin transversal of size n 1. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 5 / 19

12 Open problems on Latin squares - Existence Conjecture (Ryser, 1967) Every latin square of odd order admits a latin transversal. Conjecture (Brualdi, 1975) Every latin square admits a partial latin transversal of size n 1. Theorem (Hatami and Shor, 2008) Every latin square admits a partial latin transversal of size n O(log 2 n). Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 5 / 19

13 Open problems on Latin squares - Existence Conjecture (Ryser, 1967) Every latin square of odd order admits a latin transversal. Conjecture (Brualdi, 1975) Every latin square admits a partial latin transversal of size n 1. Theorem (Hatami and Shor, 2008) Every latin square admits a partial latin transversal of size n O(log 2 n). Proposition (Erdős and Spencer, 1991) For every integer matrix, if no entry appears more than n times, then it 4e has a latin transversal. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 5 / 19

14 Open problems on Latin squares - Counting Conjecture (Vardi, 1991) Let z n be the number of latin transversals of the cyclic group of order n. Then there exists two constants 0 < c 1 < c 2 < 1 such that c n 1n! < z n < c n 2n! Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 6 / 19

15 Open problems on Latin squares - Counting Conjecture (Vardi, 1991) Let z n be the number of latin transversals of the cyclic group of order n. Then there exists two constants 0 < c 1 < c 2 < 1 such that c n 1n! < z n < c n 2n! Theorem (McKay, McLeod and Wanless, 2006 / Cavenagh and Wanless, 2010) Let z n be the number of latin transversals of the cyclic group of order n. Then a n < z n < b n nn! where a = and b = Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 6 / 19

16 Outline 1 The problem 2 Counting with the Local Lemma 3 Our Approach 4 Random Models Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 7 / 19

17 Poisson Paradigm A 1,..., A m bad events Pr(A i ) = p i, Pr m\ A i!? i=1 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 8 / 19

18 Poisson Paradigm A 1,..., A m bad events Pr(A i ) = p i, Pr m\ A i!? i=1 1 If A i are mutually independent Pr! m\ my A i = (1 p i ) e µ µ = i=1 i=1 mx i=1 p i expected number of bad events Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 8 / 19

19 Poisson Paradigm A 1,..., A m bad events Pr(A i ) = p i, Pr m\ A i!? i=1 1 If A i are mutually independent Pr! m\ my A i = (1 p i ) e µ µ = i=1 i=1 mx i=1 p i expected number of bad events 2 If µ < 1, by the union bound Pr! m\ mx A i 1 Pr(A i ) = 1 µ > 0 i=1 i=1 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 8 / 19

20 Poisson Paradigm A 1,..., A m bad events Pr(A i ) = p i, Pr m\ A i!? i=1 1 If A i are mutually independent Pr! m\ my A i = (1 p i ) e µ µ = i=1 i=1 mx i=1 p i expected number of bad events 2 If µ < 1, by the union bound Pr! m\ mx A i 1 Pr(A i ) = 1 µ > 0 i=1 i=1 Poisson paradigm: If the dependencies among A i are rare.! m\ Pr A i = (1 + o(1))e µ i=1 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 8 / 19

21 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

22 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

23 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events}, Pr(A i T j S A j) = Pr(A i ) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

24 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events}, Pr(A i T j S A j) = Pr(A i ) x 1,..., x m (0, 1) such that Y Pr(A i ) x i (1 x j ) A j N(A i ) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

25 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events}, Pr(A i T j S A j) = Pr(A i ) x 1,..., x m (0, 1) such that Y Pr(A i ) x i (1 x j ) A j N(A i ) Then, Pr! m\ A i > 0 EXISTENCE i=1 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

26 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events}, Pr(A i T j S A j) = Pr(A i ) x 1,..., x m (0, 1) such that Y Pr(A i ) x i (1 x j ) A j N(A i ) Then, Pr! m\ my A i > (1 x i ) COUNTING (lower bound) i=1 i=1 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

27 Lovász Local Lemma dependency graph H, V (H) = {A 1,..., A m} E(H) = {dependencies among events}, Pr(A i T j S A j) = Pr(A i ) x 1,..., x m (0, 1) such that Y Pr(A i ) x i (1 x j ) A j N(A i ) Then, Pr! m\ my A i > (1 x i ) COUNTING (lower bound) i=1 i=1 Lopsided version (Erdős and Spencer, 1991) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 9 / 19

28 Upper bound using local Lemma (Lu and Szekely, 2009) ε-near dependency graph H, V (H) = {A 1,..., A m} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 10 / 19

29 Upper bound using local Lemma (Lu and Szekely, 2009) ε-near dependency graph H, V (H) = {A 1,..., A m} Pr(A i A j ) = 0 if (i, j) E(H) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 10 / 19

30 Upper bound using local Lemma (Lu and Szekely, 2009) ε-near dependency graph H, V (H) = {A 1,..., A m} Pr(A i A j ) = 0 if (i, j) E(H) for any S [m] \ N(A i ) Pr(A i \ A j ) (1 ε) Pr(A i ) j S Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 10 / 19

31 Upper bound using local Lemma (Lu and Szekely, 2009) ε-near dependency graph H, V (H) = {A 1,..., A m} Pr(A i A j ) = 0 if (i, j) E(H) for any S [m] \ N(A i ) Pr(A i \ A j ) (1 ε) Pr(A i ) j S Then, Pr! m\ A i Y i i=1 (1 (1 ε) Pr(A i )) COUNTING (upper bound) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 10 / 19

32 Matchings of K n,n Let M be a collection of matchings of K n,n (or K n). Choose a perfect matching F u.a.r.. For any M M define, A M = {M F} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 11 / 19

33 Matchings of K n,n Let M be a collection of matchings of K n,n (or K n). Choose a perfect matching F u.a.r.. For any M M define, A M = {M F} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 11 / 19

34 Matchings of K n,n Let M be a collection of matchings of K n,n (or K n). Choose a perfect matching F u.a.r.. For any M M define, A M = {M F} V (H) = {A M } M M E(H) between A M and A N if M and N are in conflict (M N is not a matching) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 11 / 19

35 Matchings of K n,n Let M be a collection of matchings of K n,n (or K n). Choose a perfect matching F u.a.r.. For any M M define, A M = {M F} V (H) = {A M } M M E(H) between A M and A N if M and N are in conflict (M N is not a matching) Theorem (Lu and Szekely, 2009) Let M be a sparse set of matchings of K n,n (or K n). Then H is both a negative dependency graph and an ε-near dependency graph. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 11 / 19

36 Matchings of K n,n Let M be a collection of matchings of K n,n (or K n). Choose a perfect matching F u.a.r.. For any M M define, A M = {M F} V (H) = {A M } M M E(H) between A M and A N if M and N are in conflict (M N is not a matching) Theorem (Lu and Szekely, 2009) Let M be a sparse set of matchings of K n,n (or K n). Then H is both a negative dependency graph and an ε-near dependency graph. k-cycle free permutations Latin rectangles n r d-regular graphs (configuration model) Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 11 / 19

37 Outline 1 The problem 2 Counting with the Local Lemma 3 Our Approach 4 Random Models Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 12 / 19

38 Setting the bad events Set the following bad events Given an edge coloring Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

39 Setting the bad events Set the following bad events Given an edge coloring A e,f if e and f independent and same color Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

40 Setting the bad events Set the following bad events Given an edge coloring A e,f if e and f independent and same color In this matching A (11),(34) holds Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

41 Setting the bad events Set the following bad events Given an edge coloring A e,f if e and f independent and same color In this matching A (11),(34) holds If no event holds, we have a rainbow matching Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

42 Setting the bad events Set the following bad events Given an edge coloring A e,f if e and f independent and same color In this matching A (11),(34) holds If no event holds, we have a rainbow matching M = {A e,f : e, f E(K n,n), c(e) = c(f ), e and f independent} Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

43 Setting the bad events Set the following bad events Given an edge coloring A e,f if e and f independent and same color In this matching A (11),(34) holds If no event holds, we have a rainbow matching but... M = {A e,f : e, f E(K n,n), c(e) = c(f ), e and f independent} M is not sparse Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 13 / 19

44 Counting rainbow matchings Theorem Fix an edge coloring of K n,n such that no color appears more than n/k times. Let µ = M /n(n 1). If k 12 then there exist two constants γ 1 (k) < 1 < γ 2 (k), such that for a random matching M. e γ 2(k)µ Pr(M rainbow) e γ 1(k)µ Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 14 / 19

45 Counting rainbow matchings Theorem Fix an edge coloring of K n,n such that no color appears more than n/k times. Let µ = M /n(n 1). If k 12 then there exist two constants γ 1 (k) < 1 < γ 2 (k), such that for a random matching M. e γ 2(k)µ Pr(M rainbow) e γ 1(k)µ Corollary If z n is the number of latin transversals in a latin square of size n where each entry appear at most n/k times (k 12), then there exist 0 < c 1 < c 2 < 1 constants such that c n 1n! z n c n 2n! Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 14 / 19

46 Outline 1 The problem 2 Counting with the Local Lemma 3 Our Approach 4 Random Models Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 15 / 19

47 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

48 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

49 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares DIFFICULT! Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

50 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares DIFFICULT! Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares, Journal of Combinatorial Designs, 4, 1996, 6, McKay, B. D. and Wanless, I. M., Most Latin squares have many subsquares, Journal of Combinatorial Theory. Series A, 86, 1999, 2, Cavenagh, N. J. and Greenhill, C. and Wanless, I. M., The cycle structure of two rows in a random Latin square, Random Structures & Algorithms, 33, 2008, 3, Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

51 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares DIFFICULT! Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares, Journal of Combinatorial Designs, 4, 1996, 6, McKay, B. D. and Wanless, I. M., Most Latin squares have many subsquares, Journal of Combinatorial Theory. Series A, 86, 1999, 2, Cavenagh, N. J. and Greenhill, C. and Wanless, I. M., The cycle structure of two rows in a random Latin square, Random Structures & Algorithms, 33, 2008, 3, Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

52 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares DIFFICULT! Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares, Journal of Combinatorial Designs, 4, 1996, 6, McKay, B. D. and Wanless, I. M., Most Latin squares have many subsquares, Journal of Combinatorial Theory. Series A, 86, 1999, 2, Cavenagh, N. J. and Greenhill, C. and Wanless, I. M., The cycle structure of two rows in a random Latin square, Random Structures & Algorithms, 33, 2008, 3, Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

53 Ryser conjecture whp Ryser conjecture is difficult Is it true for almost all latin squares? First step: settle a model for random latin squares DIFFICULT! Jacobson, M. T. and Matthews, P., Generating uniformly distributed random Latin squares, Journal of Combinatorial Designs, 4, 1996, 6, McKay, B. D. and Wanless, I. M., Most Latin squares have many subsquares, Journal of Combinatorial Theory. Series A, 86, 1999, 2, Cavenagh, N. J. and Greenhill, C. and Wanless, I. M., The cycle structure of two rows in a random Latin square, Random Structures & Algorithms, 33, 2008, 3, Given integer matrix A Pr( latin transversal) Pr(A latin square) e n2 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 16 / 19

54 Random models of colorings Let s = nk the number of colors. Random Model 1: Uniform model Choose a color for each edge u.a.r. Prob space: [s] n2 Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 17 / 19

55 Random models of colorings Let s = nk the number of colors. Random Model 1: Uniform model Choose a color for each edge u.a.r. Prob space: [s] n2 Random Model 2: Regular model Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 17 / 19

56 Random models of colorings Let s = nk the number of colors. Random Model 1: Uniform model Choose a color for each edge u.a.r. Prob space: [s] n2 Random Model 2: Regular model Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 17 / 19

57 Random models of colorings Let s = nk the number of colors. Random Model 1: Uniform model Choose a color for each edge u.a.r. Prob space: [s] n2 Random Model 2: Regular model Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 17 / 19

58 Random models of colorings Let s = nk the number of colors. Random Model 1: Uniform model Choose a color for each edge u.a.r. Prob space: [s] n2 Random Model 2: Regular model Uniformly among s-edge-colorings, each color appearing n/k times. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 17 / 19

59 Results Proposition Let C be a random edge coloring of K n,n in the URM with s n colors. Then, Pr(M rainbow) = e c(k)µ, where µ n 2k and c(k) = 2k 1 (k 1) log k k 1. Let C be a random edge-coloring of K n,n in the RRM with s n colors. Then Pr(M rainbow) = e (c(k)+o(1))µ. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 18 / 19

60 Results Proposition Let C be a random edge coloring of K n,n in the URM with s n colors. Then, Pr(M rainbow) = e c(k)µ, where µ n 2k and c(k) = 2k 1 (k 1) log k k 1. Let C be a random edge-coloring of K n,n in the RRM with s n colors. Then Pr(M rainbow) = e (c(k)+o(1))µ. Theorem Every random edge coloring of K n,n with s n colors in the URM (RRM) has a rainbow matching with high probability. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 18 / 19

61 Results Proposition Let C be a random edge coloring of K n,n in the URM with s n colors. Then, Pr(M rainbow) = e c(k)µ, where µ n 2k and c(k) = 2k 1 (k 1) log k k 1. Let C be a random edge-coloring of K n,n in the RRM with s n colors. Then Pr(M rainbow) = e (c(k)+o(1))µ. Theorem Every random edge coloring of K n,n with s n colors in the URM (RRM) has a rainbow matching with high probability. In particular Pr( latin transversal) 1 n Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 18 / 19

62 Results Thanks for your attention. Guillem Perarnau MA4-UPC Rainbow matchings: Existence and counting 19 / 19