www.elsevier.com/locate/disc
Maximal sets of hamilton cycles in K
2p
− F
H.L. Fu
a, S.L. Logan
b, C.A. Rodger
baDepartment of Applied Mathematics, National Chiao Tung University, 1001 Ta Hseuh Road, Hsinchu, Taiwan bDepartment of Discrete and Statistical Sciences, Auburn University, 235 Allison, AL 36849, USA
Received 28 April 2003; received in revised form 9 May 2005; accepted 29 September 2006 Available online 6 June 2007
This paper is written in honour of Jennie Seberry on the occasion of her 60th birthday
Abstract
A set S of edge-disjoint hamilton cycles in a graph T is said to be maximal if the hamilton cycles in S form a subgraph of T such that T − E(S) has no hamilton cycle. The spectrum of a graph T is the set of integers m such that T contains a maximal set of m edge-disjoint hamilton cycles. This spectrum has previously been determined for all complete graphs, all complete bipartite graphs, and many complete multipartite graphs. One of the outstanding problems is to find the spectrum for the graphs formed by removing the edges of a 1-factor, F, from a complete graph, K2p.
In this paper we completely solve this problem, giving two substantially different proofs. One proof uses amalgamations, and is of interest in its own right because it is the first example of an amalgamation where vertices from different parts are amalgamated. The other is a neat direct proof.
© 2007 Elsevier B.V. All rights reserved. Keywords: Hamilton; Maximal; Amalgamations
1. Introduction
A hamilton cycle in a graph T is a spanning cycle of T. If S is a set of edge-disjoint hamilton cycles in T and if E(S)
is the set of edges occurring in the hamilton cycles in S, then S is said to be maximal if T − E(S) has no hamilton cycle.
In 1993, Hoffman et al.[5]showed that there exists a maximal set S of m edge-disjoint hamilton cycles in Knif
and only if m∈ {(n + 3)/4, (n + 3)/4 + 1, . . . , (n − 1)/2}. Using amalgamation techniques, Bryant et al.[2]
showed that there exists a maximal set S of m edge-disjoint hamilton cycles in the complete bipartite graph Kn,nif and
only if n/4 < mn/2. Later, Daven et al.[3]extended the use of amalgamation techniques by nearly showing that for
n3 and p 3, there exists a maximal set S of m hamilton cycles in the complete multipartite graph Knp (p parts of
size n) if and only if(n(p − 1))/4m(n(p − 1))/2, and m > (n(p − 1))/4 if n is odd and p ≡ 1(mod 4); the
case where the result is still in doubt is when n is odd and m((n + 1)(p − 1) − 2)/4.
In these results, if T=Knor T=Kn,n, then in every case the set S of m hamilton cycles is maximal because T−E(S)
is disconnected. However if T = Knp, then the construction in[3]usually results in T− E(S) being disconnected, but
in some cases it has edge-connectivity 1.
So we can put the results in[2,3,5]to prove the following result, as stated in[3].
E-mail address:rodgec1@auburn.edu(C. Rodger).
0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.09.059
Theorem 1.1. There exists a maximal set of m hamilton cycles in Knp(p parts of size n) if and only if
1. n(p − 1)/4mn(p − 1)/2 and 2. m > n(p− 1)/4 if
(a) n is odd and p≡ 1(mod 4), or (b) p= 2, n = 1
except possibly for the two undecided cases when n=2, and when n3 is odd, p is odd, and m((n+1)(p−1)−2)/4.
In this paper, we extend this result by removing the possible exception when n=2 (see Theorems 2.1 and 3.1). Clearly this complete multipartite graph with p parts of size 2 is simply formed from the complete graph on 2p vertices by removing the edges in a 1-factor. So, with T=K2p−F where F is a 1-factor of K2p, and with(p −1)/2mp −1,
we find a set S of m hamilton cycles in T which is maximal since T − E(S) is disconnected. To do so, we provide two substantially different proofs. One proof uses amalgamations, and is of interest in its own right because it is the first example of an amalgamation where vertices from different parts are amalgamated (see[1] for a survey of amalgamations, and see[6]for a related proof). The other is a neat direct proof.
Throughout the paper, loops will count two towards the degree of the incident vertex. If h is an edge-coloring of a graph G, then let hi(u, v)denote the number of edges colored i joining u and v in G. The subgraph of a graph G induced
by the edges colored i is known as the ith color class, and is denoted by Gi. An edge-coloring is said to be equitable
if|dGi(v)− dGj(v)|1 for all pairs of colors i, j and all vertices v ∈ V (G). If |hi(u, v)− hj(u, v)|1 for all pairs of colors i, j and all pairs of vertices u, v∈ V (G) then the edge-coloring h is said to be balanced. It has been shown that for all k1 and for any bipartite multigraph B there exists a k-edge-coloring of B (that is, an edge-coloring using
k colors altogether) that is both equitable and balanced[4]. Let(v, w) denote the number of edges joining vertices v and w in G, and let(v, C) denote the number of edges joining vertex v to vertices in some set C of vertices.
2. A proof using amalgamations
In this section we make use of the proof technique of amalgamations. The idea behind the method, informally
speaking, is as follows. An amalgamation of a graph T is the graph U defined by a homomorphism g: V (T ) → V (U).
Each vertex u in U can be considered to “contain” f (u)= |g−1(u)| vertices of T; f is called the amalgamation function
of (T , U ). In the following proof, we begin with a graph U together with an associated function f that could conceivably
be the amalgamation of (T = K2p− F, U). We then inductively prove that f is indeed this amalgamation function by
disentangling each vertex u into f (u) vertices one by one.
This proof is of particular interest since it is the first time amalgamations have been used where a vertex “contains”
some, but not all, of the vertices from several parts of the corresponding complete multipartite graph; see[3]for an
example where such an extension would be a great help. The more we know about disentangling, the simpler we can afford U to be, so the easier it is to construct U.
The following result is the focus of this paper.
Theorem 2.1. There exists a maximal set S of m hamilton cycles in T=K2p−F if and only if (p −1)/2mp −1, where F is a 1-factor of K2p.
Proof. It is shown in[3]that(p − 1)/2mp − 1 is a necessary condition for S to exist, so we now prove the sufficiency.
Clearly we can assume that p2. For 1y x p we first define the set of m-edge-colored graphs G(x, y) in which
each graph has vertex set V∪ W, where V = {v1, v2, . . . , vx−1, vp} and W = {w1, w2, . . . , wy−1, wp}; so x = |V | and
y= |W|. Each graph in G(x, y) has the associated function f defined on the vertices by
f (u)= ⎧ ⎨ ⎩ p− x + 1 if u = vp, p− y + 1 if u = wp, and 1 otherwise.
With this in mind, we define G with an m-edge-coloring to be a graph inG(x, y) if and only if it satisfies the following four defining properties.
(D1) The number of edges between vertices is
(vi, wj)= ⎧ ⎨ ⎩ 0 if 1i = j < y, f (vi)(f (wj)− 1) if i y when j = p, f (vi)f (wj) otherwise, and (u1, u2)f (u1)f (u2) if {u1, u2} ⊆ V or {u1, u2} ⊆ W.
(D2) For any u∈ V (G), the number of loops on u is at most (f (u)2 ). (D3) For any u∈ V (G), the degree of u in each color class is 2f (u). (D4) Each color class is connected.
Now observe that if there exists a graph G inG(p, p), then by the definition of f it must be the case that f (u) = 1 for all vertices u∈ V (G). Therefore, using the properties (D1–D4) it is easily seen by (D1–D2) that G is a loopless simple graph with no edges between vi and wi for 1i p, so G is a subgraph of K2p− F . Furthermore by (D1),
G contains all edges{vi, wj} where i = j, so the complement of G in K2p− F is disconnected. By (D3–D4) G has
an edge-coloring in which each of the m color classes is 2-regular and connected, so each color class Giis a hamilton
cycle. So proving thatG(p, p) is non-empty will prove Theorem 2.1.
To show thatG(p, p) contains a graph, we proceed by induction. We first show there exists a graph in G(1, 1). We then complete the proof by showing that for p and p − 1, if G(, ) contains a graph G, then we can use it to form a graph G inG(, + 1).
Constructing G ∈ G(1, 1): Let G be the multigraph with vertex set V ∪ W where V = {vp} and W = {wp}, with
associated amalgamation function satisfying f (vp)= f (wp)= p, defined as follows. As the definition proceeds we
also check that G∈ G(1, 1).
Join vpto wpwith p(p−1) edges. So clearly, G satisfies (D1). Next, notice that p(2m−p+1)/2=(2mp−p(p−1))/2
is an integer since p(p− 1) is even. Also p(2m − p + 1)/2(p(2(p − 1)/2 − (p − 1)))/2 since m(p − 1)/2. So if p is even then p(2m− p + 1)/2p(2(p/2) − (p − 1))/2 = p/2, which is non-negative since p is positive, and if p is odd then p(2m− p + 1)p(2((p − 1)/2) − (p − 1))/2 = 0; so p(2m − p + 1)/2 is non-negative. Therefore we can place p(2m− p + 1)/2 loops on each of vpand wp; so then each vertex has degree 2mp. To verify that (D2) is
satisfied, we should check that p(2m− p + 1)/2 (p2). Clearly p(2m− p + 1)/2 = (2mp − p(p − 1))/2p(p − 1) − p(p − 1)/2 = p(p − 1)/2, since we are given that mp − 1. Therefore (D2) is satisfied.
To define the m-edge-coloring h with colors in{1, . . . , m}, first arbitrarily name the edges joining vp to wp with
e1, . . . , ep(p−1). For 1i p(p − 1) color the edge ei with the color congruent toi/2(mod m). Since p(p − 1) is
even, it follows that:
(i) hi(vp, wp)is even for 1i m, and
(ii) |hi(vp, wp)− hj(vp, wp)|2 for i, j ∈ Zm.
So hi(vp, wp)is either p(p− 1)/m if this number happens to be an even integer, or is one of the two even integers
closest to (p(p− 1))/m otherwise. Next, since we are given that m(p − 1)/2, it follows that m(p − 1)/2, so 2pp(p − 1)/m. Therefore, being an integer, 2p is at least the smallest even number greater than or equal to
p(p− 1)/m, so is at least hi(vp, wp). This implies that 2p− hi(vp, wp)is non-negative and even, so for 1i m we
can color (2p− hi(vp, wp))/2 loops with color i on vpand also on wp. Then each of vpand wphas degree 2p in Gi,
so (D3) is satisfied. Alsoi(2p− hi(vp, wp))/2= (2mp − p(p − 1))/2 = p(2m − p + 1)/2, so each loop has been
colored.
Since p− 1m, p(p − 1)/2mp/21. Now observe that for 1i m, Gi is connected, since hi(vp, wp)2
(p(p− 1)/2m2. Therefore (D4) is satisfied. Because G satisfies (D1–D4), G ∈ G(1, 1).
Constructing G ∈ G(, + 1) from G ∈ G(, ): Let G ∈ G(, ). We can assume that there exists a vertex u ∈ V (G) such that f (u)2, for otherwise G ∈ G(p, p) and we are finished. Without loss of generality, we can
Let B be the bipartite multigraph with bipartition{C, Z} where C = {c1, . . . , cm} ∪ {d} and Z = (V (G) − {wp}) ∪ {l}
formed as follows:
(i) for each edge colored k joining z∈ Z to wpin G, let B contain an edge joining ckto z;
(ii) for each loop colored k on wpin G join ckto l in B with two edges; and
(iii) fori − 1 and for i = p join vi to d with f (vi)edges.
(The edges in B incident with vertex d “represent” edges{vi, wi} for i − 1, which are edges that, of course, are
not in K2p− F .)
By (D3), wphas degree 2f (wp)in each color class in G, so by (i) and (ii) it immediately follows that dB(ci)=2f (wp)
for 1i m. Also, dB(d)= ( − ) + (p − + 1) = p − + 1 = f (wp)by the definition of f, dB(l)2(f (w2p))by
(D2), and by (D1) dB(z)f (z)f (wp)for each z∈ V (Z), with equality holding if z ∈ {v1, . . . , v−1, vp}.
Give B an equitable and balanced f (wp)-edge-coloring with colors in{1, 2, . . . , f (wp)}. We can assume that the
edge{d, v} is colored 1 if < , and an edge {d, vp} is colored 1 if = .
At this point, it would be easy to use the edges colored 1 in B to determine which edges and loops to choose in
G in order to detach one of their ends from wp and then reattach to a new vertex winstead; in so doing (D1–D3)
would be satisfied by the new graph, G . Details of this are unnecessary, since unfortunately G may not satisfy (D4) if such an approach were to be used. To address the connectivity issue, note that the only way a color class G i can be disconnected in G is that there is exists a component H in Gi − wpthat is joined to wpwith exactly 2 edges (the
number of such edges is necessarily even, since each vertex in H has even degree in Gi), and both these edges are
selected to be detached from wpand joined to win G . To avoid selecting such pairs of “disconnecting edges”, we can
ensure that at most one edge from each such pair is chosen. This is accomplished by focusing on B , the subgraph of
B induced by the edges colored 1 and 2. Each vertex cihas degree 4 in B , so we then form B by splitting ci into two
vertices ci and c iin such a way that disconnecting pairs of edges in Gi correspond to adjacent edges in B (assuming
the corresponding pair of edges are even in B —they may receive colors other than 1 or 2 in B). Now an equitable 2-edge-coloring of B ensures that at most one of the edges in B corresponding to a disconnecting pair is colored 1. As we will see, it then turns out that the edges in B1 (that is, the edges colored 1 in B ) can be used to form G .
More formally, we consider the subgraph B of B induced by the edges colored 1 and 2 (these colors exist since
f (wp)2). B has the same bipartition of the vertices as does B. Now form a third bipartite graph B with V (B )=
V (B )∪{c i|i ∈ {1, . . . , m}} as follows. For 1i m, consider the four (since ciis incident with exactly dB(ci)/f (wp)=
2 edges of each color in B) edges incident with vertex ci in B . If ci is incident with two edges that correspond to
the unique pair of edges in G that join wp to the same component of Gi − {wp}, (clearly each edge is in at most
one such pair) or if there are at least two edges joining ci to l in B , then detach exactly two of them from ci and
join them to c i instead. Otherwise detach any pair of edges from ci in B and join them to c i instead to form B . So
dB (ci)= dB (c i)= 2.
Give B an equitable 2-edge-coloring with colors 1 and 2; again we can assume that the edge{d, v} is colored 1 if
< and an edge {d, vp} is colored 1 if = . Note that the pairing process and the balanced edge-coloring ensures
that B1 (l,{ci, ci }) B(l, ci) 2 , (1) soB 1(l,{ci, c
i}) is at most the number of loops colored i on wpin G.
We now construct a graph G from G as follows. For each edge in B1 joining a vertex ci or c ito a vertex z∈ Z − {l},
replace the corresponding edge in Gijoining z to wpwith an edge colored i joining z to a new vertex w. Also for each
edge in B1 joining a vertex cior c ito the vertex l, replace one loop on wpcolored i with an edge colored i joining wp
to w. (Recall each loop corresponds to two edges in B; but by (1) this step is possible.) So B1 corresponds to a subset of edges in G which are detached from wpand joined to winstead to result in a new graph G .
Now we must show G ∈ G(, + 1) by verifying that it satisfies the properties (D1–D4). Clearly we need only concern ourselves with the edges incident with the new vertex wand the modified vertex wp. Note that|V | = and
|W| = + 1 in G . Let f be the associated function of the graph G defined by f (w
)= 1, f (wp)= f (wp)− 1, and
For 1i − 1, we have that G (vi, w)= dB1 (vi)= dB(vi)/f (wp)= f (vi)= f (vi)f (w)since f (w)= 1.
Similarly, for 1i − 1, since G (vi, w)= f (vi)edges joining vi to wp in G are detached in forming G ,
G (vi, wp)=G(vi, wp)−f (vi)=f (vi)f (wp)−f (vi)=f (vi)(f (wp)−1)=f (vi)f (wp). In the same manner we see
that for 1i −1, G (wi, w)=dB1 (wi)dB(wi)/f (wp)f (wi)=f (wi)f (w), andG (w, wp)dB 1(l)=
dB(l)/f (wp)f (wp)(f (wp)− 1)/f (wp) = f (wp)− 1 = f (w)f (wp). Next, if < then G (v, w)= 0
since the edge incident with d in B1 is{d, v}, the only edge incident with vin B1 . This also means that when < we have:G (v, wp)= G (v, w)+ G (v, wp)= G(v, wp)= f (v)(f (wp)− 1) = f (v)f (wp);G (vp, w)=
dB
1(vp)= dB(vp)/f (wp)= f (vp)= f
(v
p)f (w); andG (vp, wp)= G(vp, wp)− f (vp)= f (vp)(f (wp)−
1)− f (vp)= f (vp)(f (wp)− 1). However, if = , then the edge incident with d in B1 is{d, vp}, in which case:
G (vp, w)= dB1 (vp)− 1 = f (vp)− 1 = (f (vp)− 1)f (w); and since = implies that f (vp)= f (wp), we
have thatG (vp, wp)= G(vp, wp)− (f (vp)− 1) = f (vp)(f (wp)− 1) − (f (vp)− 1) = f (wp)(f (vp)− 1) −
(f (vp)− 1) = (f (wp)− 1)(f (vp)− 1) = f (wp)(f (vp)− 1). These values of G (vp, w)andG (vp, wp)seem to
be reversed in the roles of V and W when compared to (D1), which is in fact the case because this is the one and only situation in which G has|W| > |V |. Finally, for < i − 1, since dB1 (vi)= dB(vi)/f (wp)= f (vi)edges joining vi
to wpin G are detached in forming G ,G (vi, w)= f (vi)= f (vi)f (w), andG (vi, wp)= G(vi, wp)− f (vi)=
f (vi)(f (wp)− 1) − f (vi)= f (vi)(f (wp)− 2) = f (vi)(f (wp)− 1). So (D1) is satisfied.
Clearly wis incident with no loops. By (D2), the number of loops on wp in G is (f (w2p))− , where is some
non-negative integer. Since dB(l)= (f (wp)(f (wp)− 1)) − 2, and therefore dB1 (l)(f (wp)− 1) − (2/f (wp)),
the number of loops on wpin G
= f (wp) 2 − − dB1 (l) f (wp)(f (wp)− 1) 2 − − (f (wp)− 1) − 2 f (wp) =(f (wp)− 1)(f (wp)− 2) 2 − − 2 f (wp) f (wp)(f (wp)− 1) 2 − − 2 2 = f (wp) 2 . Thus (D2) is satisfied.
Since ci is incident with two edges of each color in B and dB 1(ci)= dB1 (c i)= 1, these two edges of color 1 incident
with ci and c i correspond to the two edges of G 1incident with w. So clearly, wis incident with 2f (w)= 2 edges
in each color class. Likewise, since two edges of each color class are removed from wp, dG i(wp)= 2f (wp)− 2 =
2(f (wp)− 1) = 2f (wp). Thus (D3) is satisfied.
Notice that since each vertex in Gi has even degree, it follows that each edge-cut has even size. Therefore, since
exactly two edges colored i are detached from wpand reattached to w, the only way that G i could be disconnected
would be if there exists a component of Gi− wpthat is joined to wpin Giby exactly two edges, and those two edges
are the ones detached from wp and joined to w in forming G . Clearly the choice of edges incident with ci in B
prevents this. So by the construction of B we have ensured that G remains connected, thus verifying D4. Since (D1–D4) are satisfied, there exists a graph G ∈ G(, + 1). So the result now follows.
3. A direct construction
In this section we provide a direct construction for finding a maximal set S of m edge-disjoint hamilton cycles in the graph T = K2p− F . We ensure that each set is indeed maximal by showing that T − E(S) is a disconnected graph,
Let T = K2p− F with vertex set V ∪ W where V = {v0, v1, . . . , vp−1} and W = {w0, w1, . . . , wp−1}; and with
F= {{vj, wj}|0j p − 1}. We will define a maximal set S of m hamilton cycles by considering various cases in turn.
LetZs={0, 1, . . . s −1}. The hamilton cycles in S are built from hamilton cycles or hamilton paths in smaller graphs
in which the vertices are either elements ofZs,Zs∪ {∞} or Zs∪ {∞1,∞2} for some s. If H = (z0, . . . , zy)is such a
path or cycle then let Hs(i)= (z 0, . . . , z y)where if zj is inZs then z j= zj + i (mod s), and if zj = ∞, ∞1or∞2
then z j= zj.
Theorem 3.1. There exists a maximal set S of m hamilton cycles in T=K2p−F if and only if (p −1)/2mp −1,
where F is a 1-factor of K2p.
Proof. Again, we observe that it is shown in[3]that(p − 1)/2mp − 1 is a necessary condition for S to exist, so we now prove the sufficiency.
Case 1: p is odd. The classic hamilton cycle decomposition of Kp on the vertex setZp−1∪ {∞} is formed by
{Hp−1(i)|0i (p − 3)/2} where H is the hamilton cycle (z0, . . . , zp−1)defined by
zj= (−1)jj/2 + 1 (mod p − 1) for 0j (p − 3)/2,
zp−1−j = zj + (p − 1)/2 (mod p − 1) for 0j (p − 3)/2, and
z(p−1)/2= ∞.
Let H [i] be a copy of Hp−1(i)formed by renaming the vertex∞ with p − 1. Notice that since H contains the edges
{0, 1} and {0, 2}, it follows that for 0i (p − 3)/2
H [i] contains both the edges {i, i + 1} and {i, i + 2}. (2) Recall that T has bipartition V= {v0, . . . , vp−1} and W = {wo, . . . , wp−1}. For 0i (p − 3)/2, we can use H [i]
to define a hamilton cycleH[i] in T as follows. For each edge {k, l} ∈ E(H [i]), let H[i] contain the edges {vk, wl}
and{vl, wk}. If m = (p − 1)/2, the smallest possible value, then the set S = {H[0], . . . , H[(p − 3)/2]} forms the
desired maximal set of hamilton cycles since each edge in T joining a vertex in V to a vertex in W occurs in one of these hamilton cycles, thus T − E(S) is disconnected.
If m > (p− 1)/2 then for 0i m − ((p + 1)/2), H[i] is slightly altered and H[i + ((p − 1)/2)] is formed using two of the edges inH[i] as follows. Begin by letting H[i + ((p − 1)/2)] be the disconnected graph containing the edges{vk, vl} and {wk, wl} for each edge {k, l} ∈ E(H [i]). Next, by (2) H[i] contains the pair of edges {vi, wi+2} and
{vi+1, wi}, and H[i +((p −1)/2)] contains the pair of edges {vi, vi+1} and {wi, wi+2}; so we can now interchange the
pair of edges betweenH[i] and H[i + ((p − 1)/2)]. It is easy to check that since p is odd, this results in two hamilton cycles (seeFig. 1). Notice that the formation of each hamilton cycleH[i + ((p − 1)/2)] requires an alteration in the
original definition of each hamilton cycleH[i]. So the set S = {H[0], . . . , H[m − 1]} is a maximal set of m hamilton
cycles, since clearly all edges joining vertices in V to vertices in W occur in E(S).
Case 2: p is even. The classic hamilton path decomposition of Kpon the vertex setZpis formed by{Pp(i)|0i
(p− 2)/2} where P is the hamilton path (z0, . . . , zp−1)defined by
zj= (−1)jj/2 + 1 (mod p) for 0j p − 1. vi+1 wi+1 vi+2 wi+2 vi vi+1 vi vi+2 wi wi+1 wi wi+2
Notice that since P contains the edges{0, 1} and {0, 2}, it follows that for 0i (p − 2)/2
Pp(i)contains the edges{i, i + 1} and {i, i + 2}. (3)
The classic hamilton cycle decomposition of Kp−F on the vertex setZp−2∪{∞1,∞2} with F ={{j, ((p−2)/2)+
j}, {∞1,∞2}|j ∈ Z(p−2)/2} is formed by {Hp−2(i)|0i (p − 4)/2} where H is the hamilton cycle (z0, . . . , zp−1)
defined by
zj= (−1)j+1j/2(mod p − 2) for 0j (p − 4)/2,
zp−2−j= zj+ (p − 2)/2 for 0j (p − 4)/2,
zp−1= ∞2, and
z(p−2)/2= ∞1.
Notice that since H contains the edge{0, 1}, it follows that for 0i (p − 4)/2
Hp−2(i)contains the edge {i, i + 1} for 0i (p − 4)/2. (4)
For 0i (p − 4)/2, let H [i] be a copy of H
p−2(i)on the vertex setZp formed by renaming each vertex z ∈
Zp−2∪ {∞1,∞2} with f (z), where
f (j )= j for 0j (p − 4)/2,
f ((p− 2)/2 + j) = f (j) + p/2 for 0j (p − 4)/2, f (∞1)= p − 1, and
f (∞2)= (p − 2)/2.
It is crucial to notice that if{z1, z2} ∈ F , then f (z1)= f (z2)+ p/2 (mod p). Therefore this renaming produces a
hamilton decomposition of Kp− F where F = {{j, j + p/2}|j ∈ Zp/2}, in which by (4):
H [i] contains the edge {i, i + 1} for 0i < (p − 4)/2
and the edge{i, i + 2} for i = (p − 4)/2. (5) Let H [i] be another copy of Hp−2(i)on the vertex setZp, this time formed by renaming each vertex z∈ Zp−2∪
{∞1,∞2} with f (z), where f (j )= 2j for 0j p/4 − 1, f ((p− 4)/2 − j) = 3 + 2j for 0j p/4 − 2, f ((p− 2)/2 + j) = f (j) + (p/2) for 0j (p − 4)/2, f (∞1)= (p + 2)/2, and f (∞2)= 1.
Again, it is crucial to notice that if{z1, z2} ∈ F then f (z1)= f (z2)+ p/2 (mod p), so this renaming also produces a
hamilton decomposition of Kp− F , but this time it satisfies
H [i] contains the edge {2i, 2i + 2} for 0i p/4 − 2,
H [(p − 4)/2) − i] contains the edge {2i + 1, 2i + 3} for 0i p/4 − 2, and
Let H = {H [i] |0i (p − 4)/2}. We now match each H [i] with a hamilton cycle M(H [i]) in H in such a way that
one of the edges{i, i + 1} and {i, i + 2} occurs in H [i]
and the other occurs in a hamilton cycle in M(H [i]). (7) In view of (5) and (6) we can achieve (7) by matching H [i] with H [i/2] if i is even and matching H [i] with
H [(p − i − 3)/2] if i is odd.
We are now ready to define the hamilton cycles in T=K2p−F . For 0i (p −2)/2, define H[i] to be the hamilton
cycle containing the edges{vi+1, vi+((p+2)/2)} and {wi+1, wi+((p+2)/2)}, as well as the edges {vk, wl} and {vl, wk} for
each edge{k, l} ∈ E(Pp(i)). It is easy to check thatH[i] is a hamilton cycle. So if m = p/2, the smallest possible
value of m, then the set S= {H[0], . . . , H[(p − 2)/2]} forms the desired maximal set of hamilton cycles since each edge joining a vertex in V to a vertex in W occurs in one of these hamilton cycles, thus T − E(S) is disconnected.
If m > p/2 then for 0i m − ((p + 2)/2), H[i + (p/2)] is formed by using two of the edges in H[i] as follows. Begin by lettingH[i + (p/2)] contain the edges {vk, vl} for each edge {k, l} ∈ E(H [i]) and {wk, wl} for each edge
{k, l} ∈ E(M(H [i])). Now, by (3), H[i] contains the pair of edges {v
i, wi+2} and {vi+1, wi} and the pair of edges
{vi, wi+1} and {vi+2, wi}, whereas by (7) H[i + (p/2)] either contains the pair of edges {vi, vi+1} and {wi, wi+2} or
contains the pair of edges{vi, vi+2} and {wi, wi+1}. In either case we can interchange the pair of edges in H[i] with
the pair inH[i + (p/2)] so that both are still 2-regular. Then it is easy to check that both the resulting updated graphs H[i] and H[i + (p/2)] are hamilton cycles since p is even. So the set S = {H[0], . . . , H[m − 1]} is a maximal set of m hamilton cycles, since clearly all the edges joining vertices in V to vertices in W occur in E(S).
Acknowledgments
The authors wish to thank the referees for their thorough job in refereeing this paper, and for the changes that resulted from their suggestions.
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