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ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 17 (2019) 525-533 https://doi.org/10.26493/1855-3974.1610.03d
(Also available at http://amc-journal.eu)
On the Hamilton-Waterloo problem: the case of two cycles sizes of different parity*
Melissa S. Keranen
Michigan Technological University, Department of Mathematical Sciences, Houghton, MI 49931, USA
Adrian Pastine
Instituto de Matemática Aplicada San Luis (IMASL), Universidad Nacional de San Luis, CONICET, Ejército de los Andes 950 (D5700HHW), San Luis, Argentina
Received 19 February 2018, accepted 3 October 2019, published online 25 November 2019 Abstract
The Hamilton-Waterloo problem asks for a decomposition of the complete graph of order v into r copies of a 2-factor F and s copies of a 2-factor F2 such that r + s = \_v-r\. If F1 consists of m-cycles and F2 consists of n cycles, we say that a solution to (m, n)-HWP(v; r, s) exists. The goal is to find a decomposition for every possible pair (r, s). In this paper, we show that for odd x and y, there is a solution to (2kx, y)-HWP(vm; r, s) if gcd(x, y) > 3, m > 3, and both x and y divide v, except possibly when 1 G {r, s}.
Keywords: 2-factorizations, Hamilton-Waterloo problem, Oberwolfach problem, cycle decomposition, resolvable decompositions.
Math. Subj. Class.: 05C51, 05C70
1 Introduction
The Oberwolfach problem asks for a decomposition of the complete graph Kv into copies of a 2-factor F. To achieve this decomposition, v needs to be odd, because the vertices must have even degree. The problem with v even asks for a decomposition of Kv into ^jj2 copies of a 2-factor F, and one copy of a 1-factor. The uniform Oberwolfach problem (all cycles of the 2-factor have the same size) has been completely solved by
*The authors would like to thank Stefaan De Winter and Martín De Borbon for wonderful discussions about rings of polynomials that led to the original constructions used in previous versions of the paper, and the anonymous referees for helpful comments which greatly improved the readability of the final version.
This work was partially supported by the Universidad Nacional de San Luis, grant PROICO 03-0918. E-mail ¡addresses: msjukuri@mtu.edu (Melissa S. Keranen), adrian.pastine.tag@gmail.com (Adrian Pastine)
©® This work is licensed under https://creativecommons.org/licenses/by/4.0/
526	Ars Math. Contemp. 17 (2019) 493-514
Alspach and Haggkvist [1] and Alspach, Schellenberg, Stinson and Wagner [2]. The nonuniform Oberwolfach problem has been studied as well, and a survey of results up to 2006 can be found in [8]. Furthermore, one can refer to [6, 7, 9, 23, 24] for more recent results.
In [19] Liu first worked on the generalization of the Oberwolfach problem to equipartite graphs. He was seeking to decompose the complete equipartite graph K(m:n) with n partite sets of size m each into (n-21)m copies of a 2-factor F. For such a decomposition to exist (n - 1)m has to be even. In [14] Hoffman and Holliday worked on the equipartite generalization of the Oberwolfach problem when (n - 1)m is odd, decomposing into copies of a 2-factor F, and one copy of a 1-factor. The uniform Oberwolfach problem over equipartite graphs has since been completely solved by Liu [20] and Hoffman and Holliday [14]. For the non-uniform case, Bryant, Danziger and Pettersson [7] completely solved the case when the 2-factor is bipartite. In particular, Liu showed the following.
Theorem 1.1 ([20]). For m > 3 and u > 2, K(h:u) has a resolvable Cm-factorization if and only if hu is divisible by m, h(u — 1) is even, m is even if u = 2, and (h, u, m) G
{(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)}.
The Hamilton-Waterloo problem is a variation of the Oberwolfach problem, in which we consider two 2-factors, F1 and F2. It asks for a factorization of Kv when v is odd or Kv — I (I is a 1-factor) when v is even into r copies of F1 and s copies of F2 such that r + s = L^J, where F1 and F2 are two 2-regular graphs on v vertices. Most of the results for the Hamilton-Waterloo problem are uniform, meaning F1 consists of cycles of size m (Cm-factors), and F2 consists of cycles of size n (Cn-factors). If there is a decomposition of Kv into r Cm-factors and s Cn-factors we say that a solution to (m, n)-HWP(v; r, s) exists. The case where both m and n are odd positive integers and v is odd is almost completely solved by [11, 12]; and if m and n are both even, then the problem again is almost completely solved (see [5, 6]). However, if m and n are of differing parities, then we only have partial results. Most of the work has been done in the case where one of the cycle sizes is constant. The case of (m, n) = (3,4) is solved in [4, 13, 21, 25]. Other cases which have been studied include (m, n) = (3, v) [18], (m, n) = (3,3x) [3], and (m,n) = (4, n) [16,21] .
In this paper, we consider the case of m and n being of different parity. This case has gained attention recently, where it has been shown that the necessary conditions are sufficient for a solution to (m, n)-HWP(v; r, s) to exist whenever m | n, v > 6n > 36m, and s > 3 [10]. We provide a complementary result to this in our main theorem, which covers cases in which m \ n and solves a major portion of the problem.
Theorem 1.2. Let x, y, v, k and m be positive integers such that:
(i)	v, m > 3,
(ii)	x, y are odd,
(iii)	gcd(x,y) > 3,
(iv)	x and y divide v,
(v)	4k divides v.
Then there exists a solution to (2kx, y)-HWP(vm; r, s) for every pair r, s with r + s = |_(vm — 1)/2_|, r, s = 1.
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2 Preliminaries
The complete cyclic multipartite graph C(x:n) is the graph with n partite sets of size x, where two vertices (g, i) and (h, j) are neighbors if and only if i - j = ±1 (mod n), with subtraction being done modulo n. The directed complete cyclic multipartite graph ~C(x:n) is the graph with n parts of size x, with arcs of the form e(g, i), (h, i +1)) for every 0 < g, h < x - 1, 0 < i < n - 1.
One of the main tools in [17] is a Lemma that combines decompositions of C(x:k) to obtain decompositions of K(v:m). We present a version of the Lemma for uniform decompositions, as those are the focus of this manuscript.
Lemma 2.1 ([17]). Let m, x, y, and v be positive integers. Let si,..., sm-i be non-
2
negative integers. Suppose the following conditions are satisfied:
•	There exists a decomposition of Km into Cn-factors.
•	For every 1 < t < m-1 there exists a decomposition of C(v:n) into st Cxn-factors
and rt Cyn-factors.
Let
(m-1)	(m-1)
2 2
s = ^^ st and r = ^^ rt.
t=i t=i
Then there exists a decomposition of K(v:m) into s Cxn-factors and r Cyn-factors.
In order to decompose (x:n), x and n odd, into Cn-factors and Cxn-factors, the authors of [17] labeled the vertices by Zx x {0,..., n - 1}. They build a 2-factor F by providing n permutations of G. The ith permutation is used to connect vertices in column i - 1 to vertices in column i, in particular the n-th permutation is used to connect vertices in column n - 1 to vertices in column 0. It must be said that these permutations were used implicitly in [17], as no permutation language was used for this part of the construction.
Notice that in general, if the columns are labeled by an abelian group G, f is the ith permutation and g G G, in the 2-factor F, vertex (g, i - 1) is connected to vertex (f (g), i). Let F be the composition of all n permutations of the 2-factor F, such that (F(g), 0) is the vertex at which we finish if we start at vertex (g, 0) and move through F until we reach column 0 again. In the constructions in [17], G is abelian, and g - F(g) depends only on F and not on g. If this is the case, the length of the cycles of F is n times the order of the element g - F(g).
Lemma 2.2. Assume F is a 2-factor built with the permutation F, and g - F(g) depends only on F. If q is the order of g - F(g), then F is a qn-factor of ~C(xy4k:n).
As we will need to use the permutations of Zx, we will introduce them. For a G Zx, let fa be the permutation that adds a to every element of Zx, i.e. fa (g) = g + a. Let H(a, ft) be the 2-factor made with the following permutations:
•	fa from column i - 1 to column i if 1 < i < n - 3/2;
•	f-a from column i - 1 to column i if n - 3/2+1 < i < n - 3;
•	fa from column n - 3 to column n - 2;
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•	/-2a from column n - 2 to column n - 1 (this is a permutation because x is odd);
•	/p from column n — 1 to column 0.
In [17] the first n — 3 permutations were different, but the end result was the same. Notice that F(g) = g — a + 3. For every r e {0,1,2,..., x — 3, x — 2, x}, the authors of [17] gave permutations ^ of Zx that satisfied:
(a)	^>(a) = a for r elements of Zx;
(b)	gcd(a — ^>(a), x) = 1 for the remaining x — r elements of Zx.
Then, the decomposition of cC(x:n) was given by the 2-factors H(a, ^(a)), a e Zx.
In order for such a decomposition to work, for every a, 3 e Zx the permutations /a, /p needed to satisfy /a = /p if and only if a = 3, as otherwise some arcs would be repeated in the factor H(a, ^(a)) and the factor H(3, ^(3)).
Then, in [17], decompositions of it(x:n), cC(y:n), and (4:n) were combined using a graph product and permutations of Zx x Zy x Z4 to decompose G (4xy:n). Instead of doing
so, we will use group products to label the vertices of (4kxy:n), although we will make use of permutations of the group product.
In Section 3, we give permutations of Z2k x Z2k, and show that they satisfy the necessary conditions to be used for decompositions. In Section 4, we use multivariate bijections to give decompositions of (4fcxy:n) into 2kxk-factors and yk-factors. Finally, in Section 5, we use these decompositions to prove our main results.
3 The permutation fa(a, b) of Z2k x Z2k
Consider the group G = Z2k x Z2k, an element a = (a^ a2) and the function /a(a, b) = (—b + ai, a — b + a2).
Lemma 3.1. /a is a permutation of G.
Proof. As |G| is finite, it is enough to prove that /a is an injective function. Assume /a(a, b) = /a(c, d). Then
(—b + ai, a — b + a2) = (—d + ai, c — d + a2).
The equality —b + a1 = —d + a1 implies b = d. Using b = d, the equality a — b + a2 = c — d + a2 implies a = c. Therefore, /a is a permutation of G.	□
Lemma 3.2. /p(/2(a, b)) = (a, b) — a + 3.
Proof. We will prove this lemma by computing /p(/2 (a, b)).
/a(a, b) = (—b + ai, a — b + a2)
/2(a, b) = /(—b + ai, a — b + a2)
= (—a + b — a2 + ai, —b + ai — a + b — a2 + a2) = (—a + b — a2 + ai, ai — a)
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fg(fa(a, b)) = fg(-a + b - + ai, ai - a)
= (a — ai + pi, —a + b — a2 + ai — ai + a + ^2) = (a - ai + pi, b - a2 + —)
= (a, b) - a + p.	□
Letting p = a in Lemma 3.2 yields f(a, b) = (a, b).
Corollary 3.3. f(a, b) = (a, b).
As it was mentioned in Section 2, we need to show that if a = p, then fa(a, b) = fg (a, b) for every (a, b) G Z2k x Z2k; so that each arc is used exactly once. The statement of the following lemma is an equivalent claim.
Lemma 3.4. fa(a, b) = fg (a, b) for some (a, b) G Z2k x Z2k if and only if a = p. Proof. Assume fa(a, b) = fg(a, b). Then
(-b + ai, a - b + a2) = (-b + Pi, a - b + P2). Hence, ai = pi and a2 = p2. Therefore a = p.	□
4 Decomposing (C(4kxy:n) into (Cyn-factors and ctx2fcn-factors
Let G = Z2k x Z2k and label each column of " (4kxy:n) with the elements of the group
Gx X Zx X Zy .
Let R = G x Zx x Zy. For every A G R, let A = (a, p, 7), with a G G, p G Zx and 7 G Zy. For a G G, let fa be defined as in Section 3. For p G Zx let fg be the permutation of Zx defined by fg (a) = a + p. Similarly, for 7 G Zy let fY be the permutation of Zy defined by fY(a) = a + 7. Finally, for A = (a, p, 7) G R let fx be the permutation of R defined by fx (a, b,c) = (fa(a),fg(b),f7(c)).
Let ^ be a permutation of R, and for each A g G let H4kxy (A, y(A)) be the 2-factor formed with the following permutations:
1.	fx from column i to i + 1 if 1 < i < n - 3/2;
2.	f—i from column i to i +1 if n - 3/2 + 1 < i < n - 3;
3.	fx from column n - 2 to column n - 1;
4.	f (a, -2p, -27) from column n - 1 to column n;
5.	fv(a) from column n to column 1.
Notice that if you start in column 1 at vertex (a, b, c) the first time you reach column 1 again you reach vertex
(a, b, c) - (a, p, 7) + y>(a, p, 7) = (a, b, c) - A + <^(A).
Hence, we can apply Lemma 2.2 to obtain the length of the cycles in the 2-factor.
Let A G R. If A - <^(A) = (a,b, 0) with a G ±{(1,0), (0,1), (1,1)} and gcd(b,x) = 1, then by Lemma 2.2 the 2-factor H4k (A, y(A)) is a "t2kxn-factor. If A - y(A) = (0,0, c) with gcd(c, y) = 1, Lemma 2.2 implies that H4k (A, y(A)) is a "tyn-factor.
Therefore, to obtain a decomposition of "t(4kxy:n) into r "t2kxn-factors and s "tyn-factors, we need a permutation ^ satisfying
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(A)	A - <^(A) = (a, b, 0) with a G ±{(1,0), (0,1), (1,1)} and gcd(b,x) = 1 for r elements A g R;
(B)	A — ^(A) = (0,0, c) with gcd(c, y) = 1 for s = 4kxy — r elements A G R.
In order to obtain the permutation y>, consider the subgroup 2G of G of index 4, and let
K = {(0,0), (1, 0), (0,1), (1,1)}.
Notice that K is a set of representatives of the cosets of 2G in G. Let e G 2G, and let ^ be a permutation of G. If g, ^(g) G e + K, then either g = ^(g) or |g — ^(g)| = 2k because g — ^(g) G ±K. Hence, we can obtain ^ by providing 4k-1 permutations pe of K x Zx x Zy satisfying
(A') A — pe(A) = (a, b, 0) with a G ±{(1, 0), (0,1), (1,1)} and gcd(b,x) = 1 for re elements A g K x Zx x Zy;
(B') A—pe(A) = (0,0, c) withgcd(c, y) = 1 for se = 4xy —re elements A G KxZx xZy;
having r = J2ee2G r^, and having ^ act in each (e + K) x Zx x Zy as pe, i.e. if g = (e, c, d) + 0, 0), with ^ G K, y>(g) = (e, c, d) + pe(^, c, d). Notice that if a G K, a G ±{(1,0), (0,1), (1,1)} if and only if a = (0,0).
In [17], for every r G {0, 2,3,..., 4xy — 3,4xy — 2,4xy}, permutations ^ of Z4 x Zx x Zy were given satisfying:
(A'') A — ^(A) = (a, b, 0), with a = 0 and gcd(b, x) = 1 for r elements A G Z4 x Zx x Zy;
(B'') A — ^(A) = (0,0, c), with gcd(c, y) = 1 for the remaining 4xy — r elements A G
Z4 x Zx x Zy .
Let n: K ^ Z4 be a bijection such that n(0,0) = 0, and let ^: K x Zx x Zy ^ Z4 x Zx x Zy be the bijection that fixes the coordinates of Zx and Zy, and that behaves like n in the coordinate of K. Then if is a permutation of Z4 x Zx x Zy, satisfying Conditions (A'') and (B'') with re and se, pe =	is a permutation of K x Zx x Zy
satisfying Conditions (A') and (B') with re and se.
If we wanted either x = 1 or y = 1, we would need to change Conditions (A) and (B), but it is easy to see that the necessary permutations to decompose (C(4kxy:n) exist. Therefore we have the following.
Lemma 4.1. Let r G {1,4kxy — 1}, then there is a decomposition of ~C(4kxy:n) into r ~C2kxn-factors and s = 4kxy — r Cyn-factors.
5 Main results
The complete solution to the uniform case of the Oberwolfach problem will be vital to the proof of our main result.
Theorem 5.1 ([1, 2, 15, 22]). Kv can be decomposed into Cm-factors (and a 1-factor if v is even) if and only if v = 0 (mod m), (v, m) = (6, 3) and (v, m) = (12, 3).
We now apply the results from Section 4 to produce the following important result for the uniform equipartite version of the Hamilton-Waterloo problem where the two factor types consist of cycle sizes of distinct parities.
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Theorem 5.2. Let x, y, z, v, m, k be positive integers v,m,k > 3 satisfying the following:
(i)	v,m > 3,
(ii)	k > 2,
(iii)	x, y, z odd,
(iv)	z > 3,
(v)	gcd(x,y) = 1,
(vi)	vm = 0 (mod 4kxyz), v = 0 (mod 4kxy),
z v(m — 1) ■
(vii)	(4kxy ) is even,
(viii) ,m, z) £ {(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)},
then there is a decomposition of K(v:m) into r C2kxz-factors and s Cyz-factors, for any s,r =1.
Proof. Let v1 = v/4kxy. Consider K(vi:m). Item (vi) ensures that z divides v1m; and items (vii), (i), and (viii) give us v1(m - 1) is even, m = 2, and
m, z ) £ {(2, 3, 3), (6, 3, 3), (2, 6, 3), (6, 2, 6)}.
4k xy
Thus by Theorem 1.1 there is a decomposition of K(vi:m) into Cz-factors.
Replace each vertex in a £ K(vi:m) by 4kxy vertices (a, b), with 0 < b < 4kxy — 1, having an edge between (a1, b1) and (a2, b2) if and only if there was an edge between a1 and a2. This yields K(v:m). Even more, each Cz-factor becomes a copy of ^^C(4kxy:z). By Lemma 4.1, we have that each C(4kxy:z) can be decomposed into rp C2kxz-factors and sp Cyz-factors as long as rp, sp = 1. Choosing sp such that J2p sp = s and sp, rp = 1,
provides a decomposition of K(v:m) into r C2kxz-factors and s Cyz-factors by Lemma 2.1.
□
The next lemma, given in [17] shows how to find solutions to the Hamilton-Waterloo problems by combining solutions for the problem on complete graphs and solutions for the problem on equipartite graphs.
Lemma 5.3 ([17]). Let m and v be positive integers. Let F1 and F2 be two 2-factors on vm vertices. Suppose the following conditions are satisfied:
•	There exists a decomposition of K(v:m) into sa copies of F1 and ra copies of F2.
•	There exists a decomposition of mKv into sp copies of F1 and rp copies of F2.
Then there exists a decomposition of Kvm into s = sa + sp copies of F1 and r = ra + rp copies of F2.
We are now in a position to provide a proof of the main theorem.
Theorem 5.4. Let x, y, v, k and m be positive integers such that:
(i)	v, m > 3,
(ii)	x, y are odd,
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Ars Math. Contemp. 17 (2019) 493-514
(iii)	gcd(x,y) > 3,
(iv)	x and y divide v,
(v)	4k divides v.
Then there exists a solution to (2kx, y)-HWP(vm; r, s) for every pair r, s with r + s = |_(vm - 1)/2_|, r, s = 1.
Proof. Let r and s be positive integers with r + s = |_(vm - 1)/2J and r, s = 1. Write r = ra + rp and s = sa + sp, where ra,rp, sa, sp are positive integers that satisfy ra,sa = 1, ra + sa = v(m-1)/2, rp + sp = |(v - 1)/2j,andrp, sp G {0, |(v - 1)/2j}.
Start by decomposing Kvm into K(v:m) © mKv. Let z = gcd(x, y), xi = x/z, yi = y/z. By Theorem 5.2 there is a decomposition of K(v:m) into ra C2kxiz-factors and sa Cyiz-factors. This is a decomposition of K(v:m) into ra C2kx-factors and sa Cy-factors. By Theorem 5.1 there is a decomposition of mKv into rp C2kx-factors and sp Cy-factors. Lemma 5.3 shows that all of this together yields a decomposition of Kvm into r Cx-factors and s Cy-factors.	□
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