Ryser's conjecture and more Liana Yepremyan joint work with Peter Keevash, Alexey Pokrovskiy and Benny Sudakov University of Illinois at Chicago London School of Economics Oxford Discrete Mathematics and Probability Seminar, May 5, 2020 Conjecture (Ryser, Brualdi, Stein ) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. The Ryser-Brualdi-Stein conjecture A Latin square of order n is an n by n square with cells filled using n symbols so that every symbol appears once in each row and once in each column. A transversal of order k in a Latin square is a set of k cells from distinct rows and columns, containing distinct symbols. A transversal of order n is called full transversal. Liana Yepremyan Ryser's conjecture and more May 5, 2020 2 / 41 The Ryser-Brualdi-Stein conjecture A Latin square of order n is an n by n square with cells filled using n symbols so that every symbol appears once in each row and once in each column. A transversal of order k in a Latin square is a set of k cells from distinct rows and columns, containing distinct symbols. A transversal of order n is called full transversal. Conjecture (Ryser, Brualdi, Stein ) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. Liana Yepremyan Ryser's conjecture and more May 5, 2020 2 / 41 What size transversals we are guaranteed to have? 2n=3 + O(1) Koksma, '69 3n=4 + O(1) Drake, '77 p n − n Brouwer, De Vries, and Wieringa, '78 and Woolbright,'78 n − O(log2 n) Shor, '82, contained an error n − O(log2 n) Hatami, Shor, 2008 The Ryser-Brualdi-Stein conjecture Conjecture (Ryser, Brualdi, Stein ) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. Liana Yepremyan Ryser's conjecture and more May 5, 2020 3 / 41 2n=3 + O(1) Koksma, '69 3n=4 + O(1) Drake, '77 p n − n Brouwer, De Vries, and Wieringa, '78 and Woolbright,'78 n − O(log2 n) Shor, '82, contained an error n − O(log2 n) Hatami, Shor, 2008 The Ryser-Brualdi-Stein conjecture Conjecture (Ryser, Brualdi, Stein ) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. What size transversals we are guaranteed to have? Liana Yepremyan Ryser's conjecture and more May 5, 2020 3 / 41 The Ryser-Brualdi-Stein conjecture Conjecture (Ryser, Brualdi, Stein ) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. What size transversals we are guaranteed to have? 2n=3 + O(1) Koksma, '69 3n=4 + O(1) Drake, '77 p n − n Brouwer, De Vries, and Wieringa, '78 and Woolbright,'78 n − O(log2 n) Shor, '82, contained an error n − O(log2 n) Hatami, Shor, 2008 Liana Yepremyan Ryser's conjecture and more May 5, 2020 3 / 41 Theorem (Keevash, Pokrovskiy, Sudakov, Y.) log n Every n × n Latin square contains a transversal of order n − O( log log n ). Our result Conjecture (Ryser, Brualdi, Stein) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. Liana Yepremyan Ryser's conjecture and more May 5, 2020 4 / 41 Our result Conjecture (Ryser, Brualdi, Stein) Every n × n Latin square has a transversal of order n − 1. Moreover if n is odd it has a full transversal. Theorem (Keevash, Pokrovskiy, Sudakov, Y.) log n Every n × n Latin square contains a transversal of order n − O( log log n ). Liana Yepremyan Ryser's conjecture and more May 5, 2020 4 / 41 Rainbow matchings in Kn;n A transversal in a Latin square of order n with R rows, C columns and S symbols corresponds to a perfect rainbow matching in Kn;n with bipartition (R; C) and colours S such that color(ri cj ) = sij . Liana Yepremyan Ryser's conjecture and more May 5, 2020 5 / 41 Theorem (Keevash, Pokrovskiy, Sudakov, Y.) Every properly n-edge-coloured Kn;n has a rainbow matching of size log n n − O( log log n ). Rainbow matchings in Kn;n A matching in an edge-coloured graph is called rainbow if no two edges have the same colour. Conjecture (Ryser, Brualdi, Stein) Every properly edge-coloured Kn;n contains a rainbow matching of size n − 1. Moreover if n is odd it has a perfect rainbow matching. Liana Yepremyan Ryser's conjecture and more May 5, 2020 6 / 41 Rainbow matchings in Kn;n A matching in an edge-coloured graph is called rainbow if no two edges have the same colour. Conjecture (Ryser, Brualdi, Stein) Every properly edge-coloured Kn;n contains a rainbow matching of size n − 1. Moreover if n is odd it has a perfect rainbow matching. Theorem (Keevash, Pokrovskiy, Sudakov, Y.) Every properly n-edge-coloured Kn;n has a rainbow matching of size log n n − O( log log n ). Liana Yepremyan Ryser's conjecture and more May 5, 2020 6 / 41 V (H) = R [ C [ S where R are the rows of L, C the columns of L, and S the symbols of L. For i 2 R; j 2 C; s 2 S, fi; j; sg is a hyperedge of H if (i; j)-th entry of L has symbol s. H is 3-uniform, every pair of vertices lying in different parts is in exactly one edge, in other words, H is a 3-partite Steiner triple. A transversal of size k in L −! − a matching of size k in H. Matchings in hypergraphs Given an n × n Latin square L filled with symbols f1; 2;:::; ng, construct the following hypergraph H. Liana Yepremyan Ryser's conjecture and more May 5, 2020 7 / 41 H is 3-uniform, every pair of vertices lying in different parts is in exactly one edge, in other words, H is a 3-partite Steiner triple. A transversal of size k in L −! − a matching of size k in H. Matchings in hypergraphs Given an n × n Latin square L filled with symbols f1; 2;:::; ng, construct the following hypergraph H. V (H) = R [ C [ S where R are the rows of L, C the columns of L, and S the symbols of L. For i 2 R; j 2 C; s 2 S, fi; j; sg is a hyperedge of H if (i; j)-th entry of L has symbol s. Liana Yepremyan Ryser's conjecture and more May 5, 2020 7 / 41 A transversal of size k in L −! − a matching of size k in H. Matchings in hypergraphs Given an n × n Latin square L filled with symbols f1; 2;:::; ng, construct the following hypergraph H. V (H) = R [ C [ S where R are the rows of L, C the columns of L, and S the symbols of L. For i 2 R; j 2 C; s 2 S, fi; j; sg is a hyperedge of H if (i; j)-th entry of L has symbol s. H is 3-uniform, every pair of vertices lying in different parts is in exactly one edge, in other words, H is a 3-partite Steiner triple. Liana Yepremyan Ryser's conjecture and more May 5, 2020 7 / 41 Matchings in hypergraphs Given an n × n Latin square L filled with symbols f1; 2;:::; ng, construct the following hypergraph H. V (H) = R [ C [ S where R are the rows of L, C the columns of L, and S the symbols of L. For i 2 R; j 2 C; s 2 S, fi; j; sg is a hyperedge of H if (i; j)-th entry of L has symbol s. H is 3-uniform, every pair of vertices lying in different parts is in exactly one edge, in other words, H is a 3-partite Steiner triple. A transversal of size k in L −! − a matching of size k in H. Liana Yepremyan Ryser's conjecture and more May 5, 2020 7 / 41 2n=9 − O(1) Wang, '78 4n=15 − O(1) Lindner and Phelps, '78 n=3 − O(n2=3) Brouwer, '81 n=3 − O(n1=2 log3=2 n) Alon, Kim, and Spencer 1997 Theorem (Keevash, Pokrovskiy, Sudakov, Y.) Every Steiner triple system on n vertices has a matching of size at least n=3 − O(log n= log log n). Brouwer's conjecture Conjecture (Brouwer, 1981) Every Steiner triple system of order n contains a matching of size (n − 4)=3. Liana Yepremyan Ryser's conjecture and more May 5, 2020 8 / 41 Theorem (Keevash, Pokrovskiy, Sudakov, Y.) Every Steiner triple system on n vertices has a matching of size at least n=3 − O(log n= log log n). Brouwer's conjecture Conjecture (Brouwer, 1981) Every Steiner triple system of order n contains a matching of size (n − 4)=3. 2n=9 − O(1) Wang, '78 4n=15 − O(1) Lindner and Phelps, '78 n=3 − O(n2=3) Brouwer, '81 n=3 − O(n1=2 log3=2 n) Alon, Kim, and Spencer 1997 Liana Yepremyan Ryser's conjecture and more May 5, 2020 8 / 41 Brouwer's conjecture Conjecture (Brouwer, 1981) Every Steiner triple system of order n contains a matching of size (n − 4)=3. 2n=9 − O(1) Wang, '78 4n=15 − O(1) Lindner and Phelps, '78 n=3 − O(n2=3) Brouwer, '81 n=3 − O(n1=2 log3=2 n) Alon, Kim, and Spencer 1997 Theorem (Keevash, Pokrovskiy, Sudakov, Y.) Every Steiner triple system on n vertices has a matching of size at least n=3 − O(log n= log log n).