Transversals of Latin squares and covering radius of sets of permutations

We consider the symmetric group SnSn as a metric space with the Hamming metric. The covering radius cr(S) of a set of permutations S⊂SnS⊂Sn is the smallest rr such that SnSn is covered by the balls of radius rr centred at the elements of SS. For given nn and ss, let f(n,s)f(n,s) denote the cardinality of the smallest set SS of permutations with cr(S)⩽n−s.The value of f(n,2)f(n,2) is the subject of a conjecture by Kézdy and Snevily that implies two famous conjectures by Ryser and Brualdi on transversals in Latin squares. We show that f(n,2)⩽n+O(logn)f(n,2)⩽n+O(logn) for all nn and that f(n,2)⩽n+2f(n,2)⩽n+2 whenever n=3mn=3m for m>1m>1. We also construct, for each odd m⩾3m⩾3, a Latin square of order 3m3m with two rows that each contain 2m−22m−2 transversal-free entries. This gives an infinite family of Latin squares with odd order nn and at most n/3+O(1)n/3+O(1) disjoint transversals. The previous strongest upper bound for such a family was n/2+O(1)n/2+O(1).Finally, we show that f(5,3)=15f(5,3)=15 and record a proof by Blackburn that cr(AGL(1,q))=q−3 when qq is odd.