A Quantitative Approach to Perfect One-Factorizations of Complete Bipartite Graphs

Given a one-factorization F of the complete bipartite graph Kn,n, let pf(F) denote the number of Hamiltonian cycles obtained by taking pairwise unions of perfect matchings in F . Let pf(n) be the maximum of pf(F) over all one-factorizations F of Kn,n. In this work we prove that pf(n) > n2/4, for all n > 2. 1 Perfect one-factorizations A one-factorization of a graph G is a partition of its set of edges into perfect matchings (see Wallis [8]). The union of two distinct perfect matchings (or 1-factors) A and B in any one-factorization of a graph G induces a spanning subgraph GA,B which is the vertex disjoint union of even length cycles. When this graph GA,B is a Hamiltonian cycle, the pair A,B is called a perfect pair. A one-factorization F is perfect if the graph induced by every two distinct perfect matchings in F is a Hamiltonian cycle (see Seah [7]). The perfect one-factorization conjecture states that for any positive integer n, the complete graph K2n admits a perfect one-factorization. The conjecture seems to be first explicitly mentioned in [3] although it was mentioned informally in [5]. Additionally, in [6] determining the set of all n for which K2n has a perfect one-factorization appears as a problem. There are only two infinite families for which it is known that K2n has a perfect one-factorization: n ∈ {(p+ 1)/2, p}, for any odd prime p [1, 5]. A quantitative version of this conjecture was pursued by Wagner [9]. For even n, Wagner defined c(n) as the maximum over all one-factorizations F of Kn of the number of perfect pairs in F and for odd n, he defined c(n) = c(n + 1). Hence, for any even ∗This work was partially supported by program Basal-CMM (M.M.) and Núcleo Milenio Información y Coordinación en Redes ICM/FIC RC130003 (N.A. and M.M.). the electronic journal of combinatorics 22(1) (2015), #P1.72 1 n such that Kn has a perfect one-factorization, c(n) = ( n−1 2 ) . Wagner showed that c(nm) > 2c(n)c(m) whenever m and n are odd and coprime and c(n) > nφ(n)/2, where n is odd and φ is the Euler totient function. In this work we undertake a similar analysis for one-factorizations of complete bipartite graphs. For a positive integer n we define pf(n) to be the maximum over all one-factorizations F of Kn,n of the number of perfect pairs in F . As before, if the complete bipartite graph Kn,n has a perfect one-factorization, then pf(n) = ( n 2 ) . It is known that if Kn has a perfect one factorization then Kn−1,n−1 has a perfect one-factorization as well (see Wanless and Ihrig [12]). Then, from what we know for complete graphs, two infinite families of complete bipartite graphs are known to admit a perfect one-factorization: Kp,p and K2p−1,2p−1, for each odd prime p. In [2] it was proved that for each odd prime p, the complete bipartite graph Kp2,p2 also admits a perfect one-factorization. This evidence leads to the following version of the perfect one-factorization conjecture for complete bipartite graphs, first stated by Wanless in [10]. Conjecture 1 For every odd n > 3, pf(n) = ( n 2 ) . In contrast, for each even n it is known that pf(n) 6 n/4 and that this upper bound is achieved for each n = 2p, where p is an odd prime [11]. Our results are presented in the language of Latin rectangles since it makes the presentation easier. Given two positive integers m and n with m 6 n, a Latin rectangle L of size m × n is a matrix with m rows and n columns filled with symbols from an alphabet ΣL of size n, such that each row contains each symbol in ΣL once, and each column contains each symbol in ΣL at most once. When m = n a Latin rectangle is called a Latin square of order n. Two rows i and j of a Latin rectangle L of size m × n form a perfect pair if i 6= j and the permutation Li,j, which assigns to the symbol x in row i and column k the symbol y in row j and column k, is a cyclic permutation. We denote by pf(R) the number of perfect pairs in a Latin rectangle R and by pf(m,n) the maximum of pf(R) over all Latin rectangles R of size m× n. Then pf(m,n) 6 ( m 2 ) . When a Latin rectangle of size m × n achieves this upper bound, it is called a perfect or pan-Hamiltonian Latin rectangle ([4],[10]). It is known that there is a one-to-one correspondence between perfect Latin squares of order n and perfect one-factorizations of Kn,n ([12]). From this relation it is easy to derive that pf(n, n) is in fact the same as pf(n). Our main result is that pf(n) > n/4 for each n > 2. The proof is split into the cases n even and n odd. In Section 2 we explicitly construct, for each even n, a Latin square D of order n such that pf(D) = n/4. In Section 3 we prove that for each odd n, pf(n) > n/4.