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.


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 G A,B which is the vertex disjoint union of even length cycles. When this graph G A,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 K 2n 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 K 2n has a perfect one-factorization appears as a problem. There are only two infinite families for which it is known that K 2n 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 K n of the number of perfect pairs in F and for odd n, he defined c(n) = c(n + 1). Hence, for any even n such that K n 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 K n,n of the number of perfect pairs in F. As before, if the complete bipartite graph K n,n has a perfect one-factorization, then pf(n) = n 2 . It is known that if K n has a perfect one factorization then K n−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: K p,p and K 2p−1,2p−1 , for each odd prime p. In [2] it was proved that for each odd prime p, the complete bipartite graph K p 2 ,p 2 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].
In contrast, for each even n it is known that pf(n) n 2 /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 n, a Latin rectangle L of size m × n is a matrix with m rows and n columns filled with symbols from an alphabet Σ L of size n, such that each row contains each symbol in Σ L once, and each column contains each symbol in Σ L 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 = j and the permutation L i,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) 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 K n,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 2 /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 2 /4. In Section 3 we prove that for each odd n, pf(n) > n 2 /4.

Even order case
As we previously mentioned, in [11] it was proved that for each even n, pf(n) n 2 /4. In the next result we prove that equality holds. Proof: We only need to prove that for each even n, there is a Latin square of order n having n 2 /4 perfect pairs.
In the rest of this proof all calculations are carried out in Z n/2 . Let A, B and C be the following Latin squares of orders n/2 with Σ A = Σ B = Σ C = Z n/2 . For each i, j ∈ Z n/2 , • A(i, j) = j + i mod n/2.

Odd order case
From now on, for every Latin rectangle L we shall assume that each column is labelled by the corresponding symbol from its first row and that each row is labelled by the corresponding symbol from its first column. Thus the first row and the first column have the same label and we shall denote this label byL.
We also denote by L a the permutation L a,L , for each row a of L. With this notation, for every two rows a and a of L we have the following relations: In order to prove that pf(n) n 2 /4 for each odd n, we prove something slightly more general. Namely, we prove that for each pair of odd integers m and n with 3 m n we have that pf(m, n) m 2 /4.
When n is an odd prime the result is direct since we know that there is a Latin square N of order n such that pf(N ) = n 2 . Hence, for each m n the Latin rectangle N obtained by choosing m rows of N satisfies pf(N ) = m 2 . When n is an odd composite number we proceed in three steps. We first prove that pf(n) pf(p, n/p)(n/p) 2 , where p is the smallest prime divisor of n. Later, we prove that if pf(n) n 2 /4, then pf(m, n) m 2 /4, for each m n. Finally, by using the two first steps we prove that pf(n) n 2 /4, by induction on n.

Construction
Let n be an odd composite number. Given any Latin rectangle K of size p × (n/p), where p is the smallest prime divisor of n, we construct a Latin square T of order n such that pf(T) pf(K)(n/p) 2 . We first describe the construction of T in terms of K and give an example for n = 15. Later, in Theorem 2, we prove that pf(T) pf(K)(n/p) 2 .

Description
Let L be the Latin square of order n/p with Σ L = Z n/p andL = 0 ∈ Z n/p given by L(x, y) = x + y mod (n/p), where x, y ∈ Z n/p . Then, L a,a (x) = x + (a − a) mod (n/p) and L t a,a (x) = x + t(a − a) mod (n/p), for each x, a, a ∈ Z n/p and each t ∈ Z. Let P be a Latin square of order p such that pf(P ) = p 2 . Without loss of generality, we assume that Σ L = Σ K and that Σ P ⊆ Σ L . This is possible because p n/p. Let Σ T := Σ L × Σ P and let T be the Cartesian product of L and P given, for each (a, b), (c, d) ∈ Σ T , by T ((a, b), (c, d)) = (L(a, c), P (b, d)). For each a ∈ Σ L and each b ∈ Σ P , we define the permutation π a,b = K −1 b • L p a . Finally, we define T to be the n × n matrix obtained from T by changing in each row (a, b) of T the symbol (x,P ) to (π a,b (x),P ), for each x ∈ Σ L .
Example 1 Let K be the Latin rectangle of size 3 × 5 with Σ K = Z 5 andK = 0 ∈ Z 5 , defined by K(x, y) = x + y mod 5, for x ∈ {0, 1, 2} and y ∈ Z 5 . For each b ∈ {0, 1, 2} and each y ∈ Z 5 , we have that In Figure 2 we present the construction of the Latin square T of order 15 based on the two Latin squares P and L of sizes p = 3 and n/p = 5, respectively, where Σ L = Z 5 and L(x, y) = x + y mod 5, for each x, y ∈ Z 5 , and Σ P = Z 3 and P (x, y) = x + y mod 3. Additionally,P = 0 ∈ Z 3 andL = 0 ∈ Z 5 .
We now explicitly determine the permutation π a,b for each a ∈ Z 5 and each b ∈ Z 3 . By definition, π a,b is given by When we apply the modifications induced by the permutations π a,b to T we obtain the Latin square T given in Figure 3.
Roughly speaking the role of the modifications defined by the permutations π a,b is to glue cycles of some row permutations of T into cycles of length n.  In the next result we prove that this situation occurs for each permutation T (a,b),(a ,b ) , whenever K b,b is a cyclic permutation. It is not hard to see that the permutation T (a,b),(a ,b ) is given by Hence, symbols appearing in rows (a, b) and (a , b ) of T can be ordered (by permuting columns if necessary) as T ((a, b), ·) :(x, y) (L a,a (x), P b,b (y)) · · · (L p−1 a,a (x), P p−1 b,b (y)) · · · T ((a , b ), ·) :(L a,a (x), P b,b (y)) (L 2 a,a (x), P 2 b,b (y)) · · · (L p a,a (x), y) · · · As T is obtained by modifying only symbols of the form (x,P ) we know that rows (a, b) and (a , b ) in T are given by: T ((a, b), ·) :(π a,b (x),P ) (L a,a (x), P b,b (P )) · · · (L p−1 a,a (x), P p−1 b,b (P )) · · · T((a , b ), ·) :(L a,a (x), P b,b (P )) (L 2 a,a (x), P 2 b,b (P )) · · · (π a ,b (L p a,a (x))),P ) · · · From the definition of L a we get that It remains to show that T is a Latin square. It is clear that the Cartesian product T is a Latin square. By the definition of T we only need to check that the modification (x,P ) → (π a,b (x),P ) induces a permutation of Σ T in each row and each column. By definition, the permutation π a,b is an injective function. Hence, the first components of the symbols in each row of T form a permutation of Σ K . Therefore, each row of T is a permutation of Σ T . Let us assume that for some rows (a, b), (a , b ) and some column (c, d) of T we have that π a,b (L(a, c)) = π a ,b (L(a , c)) and P (b, d) =P = P (b , d). Since P is a Latin square we have that b = b which implies that L p a (L(a, c)) = L p a (L(a , c)). From its definition L −p a • L p a = L p a ,a and L(a , c) = L a,a (L(a, c)). Therefore, L(a, c) = L p a ,a • L a,a (L(a, c)) = L p−1 a ,a (L(a, c)). From this and the definition of L, the following equalities hold in Z n/p .
a,a (L(a, c)) = L(a, c) + (p − 1)(a − a) = L(a, c). Since p is the smallest prime divisor of n, we know that p − 1 does not divide n/p which implies that a = a . Hence each column of T is a permutation of Σ T . Therefore T is a Latin square. This finishes the proof.
From the previous result we immediately get the following consequence. Corollary 1 For each odd composite integer n we have that pf(n) pf(p, n/p) (n/p) 2 , where p is the smallest prime divisor of n.
Proof: Let K be a Latin rectangle of size p × (n/p) such that pf(K) = pf(p, n/p) and let T be the Latin square of order n defined above. Then, pf(n) pf(T) pf(K)(n/p) 2 = pf(p, n/p)(n/p) 2 .

A graph theoretical Lemma
When proving Theorem 3, we use induction on the size of Latin rectangles. The induction hypothesis will give us a Latin rectangle K of size p × (n/p) such that pf(K) p 2 /4. By plugging this into Theorem 2 we obtain that pf(n, n) n 2 /4. In order to get a similar result for Latin rectangles, we have to choose some rows of the Latin square so that the resulting Latin rectangle has the desired number of perfect pairs. This last step can be achieved by using Lemma 1 which proves that a Latin square of order n with more than n 2 /4 perfect pairs has m rows such that the subrectangle induced by these rows has at least m 2 /4 perfect pairs. In fact, the property is slightly more general since it corresponds to a density property of a subgraph of a dense graph. For a graph G = (V, E) we denote by e(G) the cardinality of the set of edges E and by v(G) the cardinality of the set of vertices V . For each vertex x of a graph G we denote by d G (x) the degree of x in G.

Proof:
The case 3 = m = k being obvious we can assume that k 4. We proceed by induction on k. By deleting edges we can assume that k 2 /4 < e(G) k 2 /4 + 1.
Therefore, the subgraph G − v 0 has k − 1 vertices and more than (k − 1) 2 /4 edges. By the induction hypothesis, for each m, 3 m k − 1 there is a subgraph H of G − v 0 such that v(H) = m and e(H) > m 2 /4. As H is also a subgraph of G, we get the conclusion. The result is tight in the sense that we cannot replace the strict inequality in the hypothesis by equality even if we relax the conclusion in the same manner. In fact, the complete bipartite graph K t,t has (2t) 2 /4 edges and 2t vertices but any subgraph of K t,t with 2t − 1 vertices has t(t − 1) edges which is less than (2t − 1) 2 /4.

Induction argument
Theorem 3 For each odd integer n and each integer m, with 2 m n, we have that pf(m, n) m 2 /4.

Proof:
The case m = 2 is direct since for each integer n 2 the Latin rectangle with two rows, each of length n, where the second one is a cyclic rotation of the first one, has one perfect pair. For 3 m n we proceed by induction on n. The basis case, n = 3, is obvious since there is a perfect Latin square of order 3. We already know that when n is an odd prime the result holds. So we assume that n is odd and composite. Let p be the smallest prime divisor of n. From Corollary 1 we know that pf(n, n) pf(p, n/p)(n/p) 2 . As n/p is odd, less than n and at least p, we can apply the induction hypothesis to get that pf(p, n/p) p 2 /4 which shows pf(n, n) n 2 /4.
Let N be a Latin square of order n such that pf(N ) n 2 /4. Let G be the graph on Σ N such that ab is an edge of G if and only if N a,b is a cyclic permutation. Then as n is odd we get that e(G) > n 2 /4. From Lemma 1 we know that for each odd m, with 3 m n, there is a subgraph H with e(H) > m 2 /4 and v(H) = m. Therefore, the Latin rectangle N obtained from N by choosing rows whose labels are in the set of vertices of H satisfies pf(N ) m 2 /4. This shows that pf(m, n) m 2 /4.

Conclusion
In this work we have considered a quantitative version of Conjecture 1 focusing on uniform lower bounds for the function pf(n)/n 2 : we have proved that pf(n)/n 2 1 4 , for each n 2. The proof of Theorem 3 suggests the following weakening of Conjecture 1.
Conjecture 2 For every two odd integers n and m such that n m 3 we have that pf(n)/n 2 pf(m)/m 2 .
If true, then our lower bound 1 4 for pf(n)/n 2 might be replaced by 1 3 when n is odd. From the proof of Theorem 2 we can prove that pf(n, n)/n 2 pf(p, n/p)/p 2 when p is a prime that divides n and n p 2 . Hence, for p a prime divisor of n such that p 2 n we have pf(n)/n 2 pf(p)/p 2 if the following is true.