Path And Cycle Decompositions
Brian Alspach
School of Mathematical and Physical Sciences
University of Newcastle
Callaghan, NSW 2308
Australia
[email protected]
November 26, 2014
1
General Decompositions
First I want to say a few words about my graph terminology. If I want to
allow loops, I use the adjective reflexive. If I want to allow multiple edges,
I use multigraph. Thus, a graph has no loops and no multiple edges.
I use valency rather than degree. If we say a graph is 4-valent (or tetravalent), it means it is regular of valency 4, for example.
A decomposition of a graph X is a partition of its edge set into subgraphs.
There are two typical situations. Either we want all the subgraphs to be
isomorphic to some fixed graph Y . We shall call this a Y -decomposition
of X, or a decomposition of X into subgraphs isomorphic to Y . The other
typical situation is that we are given a list L of subgraphs and we want a 1-1
correspondence between the parts of the decomposition and the members of
L.
There is an old saying that states “anything reasonable you can say
about decomposing complete graphs is probably true.” This suggests that
proving certain decompositions are not possible may be difficult. Let’s take
a look at one such result.
The next slide shows a picture of the well-known Petersen graph. The
complete graph of order 10 is regular of valency 9 and the Petersen graph P
has order 10 and is regular of valency 3. Hence, it is arithmetically possible
for K10 to be decomposed into three copies of P. We are going o prove that
this cannot happen. This very nice proof, due to Allen Schwenk, uses linear
algebra.
1
t
t
t
t
t
t
t
t
@
@t
t
Let A denote the adjacency matrix of P. The i, j entry of the matrix A2
gives the number of walks of length 2 between the vertex ui and uj because
X
A2i,j =
Ai,k Ak,j .
k
A2
Thus, the main diagonal of
contains 3 in each entry as every vertex has
valency 3. If ui uj is an edge of P, then there is a 0 in the i, j entry of A2
because P has girth 5. Finally, if ui uj is not an edge of P, then there is a 1
in the i, j entry of A2 because non-adjacent vertices are joined by a unique
path of length 2.
Hence, we have that A2 + A = 2I + J, where I denotes the identity
matrix and J denotes the matrix all of whose entries are 1. The matrix J
has rank 1 so that it has 0 as an eigenvalue with multiplicity 9. Let 1 denote
the vector with every entry equal to 1. Then J(1) = 101 so that 10 is an
eigenvalue with multiplicity 1. The identity matrix has 1 as an eigenvalue
with multiplicity 10.
Because every vector is an eigenvector for I, J and I have the same
set of eigenvectors. This means that 2I + J has 12 as an eigenvalue with
multiplicity 1 and 2 as an eigenvalue with multiplicity 9. In other words,
A2 + A has 12 as an eigenvalue with multiplicity 1, and 2 as an eigenvalue
with multiplicity 9. The vector 1 is the eigenvector corresponding to the
eigenvalue 12. Note that 1 also is an eigenvector of A corresponding to the
eigenvalue 3.
If λ is an eigenvalue of A, then λ2 + λ is an eigenvaule of A2 + A. Hence,
the equation λ2 + λ = 2 must hold for the remaining eigenvalues of A.
2
Thus, the eigenvalues of A are 3 with multiplicity 1, and either 1 or -2 with
multiplicities to be determined.
The trace of A is 0 so that the sum of all eigenvalues must be 0. This
forces 1 to have multiplicity 5 and -2 to have multiplicity 4. So we now know
that the Petersen graph has eigenvalue 3 with multiplicity 1, eigenvalue 1
with multiplicity 5, and eigenvalue -2 with multiplicity 4. We are going to
use this information.
Suppose that K10 has been decomposed into three copies of P. Let
A1 , A2 , A3 be the respective adjacency matrices of the three copies. Then
we have
A1 + A2 + A3 = J − I.
Note that the vector 1 is an eigenvector for all of the preceding matrices. Because A1 , A2 , A3 are all symmetric matrices, the dimension of the
eigenspace for the eigenvalue 1 is 5 and it is orthogonal to the 1-dimensional
subspace spanned by the vector 1. We conclude that the 5-dimensional
eigenspaces for A1 and A2 must have a non-trivial intersection. That is,
there is a non-zero vector x such that A1 (x) = A2 (x) = x.
Because x is orthogonal to 1, we have that J(x) = 0 which implies that
(J − I)(x) = −x. Therefore, (A1 + A2 + A3 )(x) = 2x + A3 (x) = −x which is
equivalent to A3 (x) = −3x. However, this implies that -3 is an eigenvalue of
A3 which is a contradiction. We conclude that K10 cannot be decomposed
into three Petersen graphs.
We are going to introduce a general technique loosely motivated by what
was just done. Let’s try to decompose K10 into some trivalent graph. We
know that P will not work, but there are many trivalent graphs of order 10.
Let X be some graph defined on the set of vertices {u0 , u1 , . . . , un−1 }.
Let f be a permutation defined on the vertex set. We define the graph f (X)
to have the same vertex set and for each edge ui uj in X, we have the edge
f (ui )f (uj ) in f (X). We think of the permutation f as acting on X.
Using this permutation action we know that X and f (X) are isomorphic
because we have inserted edges in f (X) precisely so that f is an isomorphism.
So if we can guarantee that X and f (X) are edge-disjoint, then we are on
our way towards a decomposition.
We return to the problem of finding a spanning trivalent graph decomposing K10 . It is clear that we need three copies of the graph as the valency
of K10 is 9. This suggests we use a permutation that has order 3. So let f
be the permutation
f = (u0 )(u1 u2 u3 )(u4 u5 u6 )(u7 u8 u9 ).
3
We now describe the orbits of the group hf i generated by f acting on
the unordered pairs of vertices. Because u0 is fixed by f , an unordered pair
such as u0 u1 lies in the orbit u0 u1 , u0 u2 , u0 u3 . An unordered pair like u1 u2
lies in the orbit u1 u2 , u2 u3 , u1 u3 . An unordered pair like u4 u8 lies in the
orbit u4 u8 , u5 u9 , u6 u7 .
It is not hard to see that every orbit has exactly three unordered pairs
in it. Thus, if we build a graph by choosing exactly one unordered pair from
each orbit to be an edge in the graph, then f and f 2 acting on this graph
give a decomposition of K10 into three subgraphs.
t
u4 P
tu1
PP
P
P
PPP tu2
u5 t
P
tu3
u6 t
A
A
A
u0 t A
A
A
A u7 t A A At
u8
Here are choices that
give a trivalent graph
decomposing K10 under hf i.
t
u9
2
Path Decompositions
Decomposing graphs into paths and cycles has a long history in graph theory.
Steiner triple systems were first studied in the early part of the nineteenth
century, and Hamilton decompositions near the end of the same century.
The interest in the general topic still is there because there are many wellknown conjectures and there also are interesting applications.
The first major path decomposition result is due to Walecki and appears
in volume 2 of Lucas’ classic series of books on recreational mathematics
from 1885–1895. The path decomposition arises out of a cycle decomposition
so that the next figure shows the cycle decomposition first.
4
t
t
t
HH
Ht
t
t
t
tH
It is clear that if we rotate
this cycle through 4 more
positions clockwise, we have
a decomposition of K11
into 5 Hamilton cycles.
t
t
HHt
It is easy to see that this same scheme works
for all complete graphs of odd order. We also can see
that there is group action going on.
The decomposition above is called the Walecki decomposition and let’s
state it as a theorem.
Theorem 1. The complete graph of odd order is Hamilton-decomposable.
Corollary 2. The complete graph of even order has a decomposition
into Hamilton paths.
Proof. Deleting the central vertex from the Walecki decomposition for
odd order complete graphs yields a decomposition of complete graphs of
even order into Hamilton paths.
Corollary 2 (also called the Walecki decomposition) gives us a decomposition of even order complete graphs into Hamilton paths. When n is even,
n(n−1)
n
we note that n − 1 divides
. This suggests the following natural
2 =
2
n
question: If ` divides 2 and ` < n, can Kn be decomposed into paths of
length `?
When n is odd, the Walecki decomposition of Kn into Hamilton cycles
provides a useful Euler tour of Kn . Start at the central vertex and traverse
the first Hamilton cycle until returning to the central vertex. Then move
into the next Hamilton cycle in clockwise order and continue in this way
until completing the tour. This Euler tour of Kn , n odd, has the very nice
property that any n − 3 (or fewer) successive edges form a path. Thus, if
` ≤ n − 3, we can chop the tour up into paths of length `.
When n is odd, it is easy to see that n − 1 does not divide n2 . Similarly,
when n is odd and n > 3, then n − 2 does not divide n2 . Hence, the
following result holds.
Theorem 3. If n is odd and ` < n divides n2 , then Kn has a decomposition into paths of length `.
The analogue for n even also holds but we shall not prove it here as the
proof is more intricate than the proof of Theorem 3.
Theorem 4. If n is even and ` < n divides n2 , then Kn has a decomposition into paths of length `.
5
In fact, the most general result possible actually holds and we now discuss
it.
Theorem 5. If `1 , `2 , . . . , `t are integers satisfying 1 ≤ `i ≤ n − 1,
1 ≤ i ≤ t, and `1 + `2 + · · · + `t = n2 , then the complete graph Kn admits
a decomposition into paths of lengths `1 , `2 , . . . , `t .
The proof of the preceding theorem took about thirty years to be completed. Tarsi made a large step by essentially proving it for odd n in 1983
and Bryant completed it in 2010.
The preceding result provides a decomposition for complete graphs.
They are rather special so what other kinds of path decompositions can
we find?
The minimum number of paths required to decompose a graph X is called
the path number of X and is denoted pn(X). This particular parameter has
not been studied a lot, but there are some results we can mention.
I attended a conference in Jamaica in 1969 at which Harary discussed
the path numbers of complete graphs. He pointed out that Kn , n even,
can be decomposed into n/2 Hamilton paths (the Walecki decomposition)
so that pn(Kn ) = n/2 for even n. He then said that pn(Kn ), n odd, must
be at least (n + 1)/2 by simply counting the number of edges in Kn .That
is, the maximum length of a path is n − 1 and (n − 1)(n − 1)/2 < n2 . His
talk was just before the lunch break and he asked if anyone could find a
decomposition of Kn , n odd, into (n + 1)/2 paths. Let’s see how easy this
is using the Walecki decomposition of Theorem 1.
t
t
t
H
Ht
t
t
t
t
H
It is clear that if we remove the
dashed edge from each cycle, we
have (n − 1)/2 Hamilton paths
and a single path of length
(n − 1)/2. This yields a
decomposition into (n + 1)/2
paths as required.
t
t
HHt
6
Theorem 6. If T is a tree of order at least 2, then pn(T ) is one-half
the number of vertices of odd valency in T .
Proof. There is more than one way to prove this and the direction
many people take is some kind of induction proof. However, we are going to
present a clever short proof. Note that a vertex of odd valency must be the
end of at least one path in a path decomposition. Hence, the path number
is at least one-half the number of vertices of odd valency. So it suffices to
show that there is a decompostion into the latter number of paths.
Let m be the number of vertices of odd valency. Adjoin a new vertex w
and insert an edge from w to each of the m vertices of odd valency. Every
vertex of the supergraph has even valency so that it possesses an Euler tour.
If we remove w and its incident edges from the tour, we obtain m/2 trails
that decompose T . However, a tree has no cycles so that the trails are, in
fact, paths and this completes the proof.
The following result is a corollary of a theorem Lovász proved when he
was quite young.
Theorem 7. If X is a graph of order n for which every vertex has odd
valency, then pn(X) = n/2.
If we try the same trick of adjoining a new vertex to obtain an eulerian
supergraph, we get a decomposition into trails which, of course, need not be
paths in this case because X may not be a tree. The proof, in fact, is rather
subtle.
We are going to look at one other path decomposition problem and this
deals with tournaments.
A tournament is an orientation of a complete graph, that is, a directed
graph in which there is precisely one arc between any pair of distinct vertices.
We use val+ (u) to denote the number of arcs going out from the vertex u,
and val− (u) to denote the number of arcs coming in to the vertex u.
The path number of a tournament A, denoted pn(A), is the fewest number
of directed paths into which the arc set of A can be decomposed. If A is a
tournament of order n, then val+ (u) + val− (u) = n − 1 for each vertex u of
A. It is easy to see that if val+ (u) > val− (u), then at least val+ (u) − val− (u)
directed paths must begin at u.
When n is even, then at each vertex u of a tournament of order n, val+ (u)
and val− (u) can never be equal. So we define the excess at vertex u to be
exc(u) = max{0, val+ (u) − val− (u)}. If A is a tournament, we define the
excess of A by
X
exc(A) =
exc(u).
u∈V (A)
7
Given a tournament A, then exc(A) is a lower bound for pn(A). This
lower bound need not be very good because a regular tournament has excess
0 as val+ (u) = val− (u) at each vertex. However, for even order tournaments,
we suspect something stronger holds.
Conjecture (Alspach).
pn(A) = exc(A).
If A is a tournament of even order, then
The preceding conjecture is closely related to a long unsolved conjecture
of Paul Kelly that regular tournaments can be decomposed into Hamilton
directed cycles. To see this, suppose the path number conjecture is true and
that A is a regular tournament of order 2m + 1. If we remove one vertex
from A along with its incident arcs, we obtain a tournament A0 of order 2m
such that exactly m of the vertices have excess 1.
Then A0 can be decomposed into m directed paths. Now A0 altogether
has m(m − 1) arcs and the maximum number of arcs in a directed path
is m − 1. So each directed path must be a spanning path. Then when
we reattach the deleted vertex, we obtain m Hamilton directed cycles that
decompose A. The next picture indicates what is happening.
We draw an arc from u to every vertex
with excess 1, and an arc from every
vertex of excess 0 to u.
+1
0
iP
P
PP
PP
8
PP
PPu
:t
Kelly’s Conjecture is hard to get hold of partly because working with
the family of regular tournaments is restrictive. That is, most operations
performed on a regualr tournament take you outside of the family. The path
number conjecture allows more freedom because it is over a much larger
family of tournaments. Something to think about is to try to extend the
path number conjecture to a conjecture covering all tournaments.
3
Cycle Double Cover Conjecture
If I were forced to pick one cycle decomposition problem as the most significant unsolved problem, I would choose the cycle double cover conjecture.
This is what we discuss next.
Cycle Double Cover Conjecture (Seymour and Szekeres). If X is
a 2-edge-connected graph, then there exists a family C of cycles in X such
that every edge of X belongs to exactly two members of C.
tu0
tv0
D
D
vt1
u4 t vt4
D
D
v3 t Dtv2
@
@t u2
u3 t
tu1
Above is the Petersen graph and here is a list of cycles covering every
edge exactly twice:
u0 v0 v3 v1 u1 u0 ; u1 v1 v4 v2 u2 u1 ; u2 v2 v0 v3 u3 u2 ;
u3 v3 v1 v4 u4 u3 ; u4 v4 v2 v0 u0 u4 ; u0 u1 u2 u3 u4 u0 .
We have mentioned the Petersen graph because it plays an interesting
role in this famous conjecture. The preceding example shows that the conjecture is valid for the Petersen graph. We now examine a much stronger
type of cycle decomposition. Suppose we are given a graph X equipped with
positive integer weights on its edges. We denote the weight of an edge e by
w(e).
A collection C of cycles in X is called a faithful w-cycle cover of X if
every edge e belongs to exactly w(e) elements of C. There are two obvious
9
conditions that must hold for a weight function w in order that it admits a
faithful w-cycle cover. They arise by examining edge cuts.
It is clear that a cycle must cross an edge cut an even number of times.
Hence, the sum of the weights on the edges in any edge cut must be even.
It is easy to see that if the sum of the weights on the edges incident with
each vertex is even, then the sum of the weights across every edge cut is
even. This follows because any edge not crossing the cut is counted twice.
If the sums of the edge weights at each vertex are even, then w is said to be
eulerian.
If we take an edge e of maximum weight w(e) across an arbitrary edge
cut, then the sum of the weights on the remaining edges of the edge cut
must be at least w(e) because each time we use the edge e in a cycle, there
must be another edge across the cut in order for us to return across the cut.
If every edge cut has the property that the sum of the weights of the edges
different from any edge e of maximum weight is not less than w(e), then the
weight w is said to be balanced.
If a weight function w on a graph X is both eulerian and balanced, then
we say that w is feasible. If X is 2-edge-connected and every edge is given
weight 2, then the weight function clearly is feasible and this special case is
precisely the cycle double cover conjecture.
Now consider the Petersen graph P. Give the spokes weight 2 and all
remaining edges weight 1. It is easy to check that this weight w is feasible.
This follows because the edges incident with each vertex have the sum of
the weights equal to 4, and there are are no edge cuts with a single edge.
We now wish to show that there is no faithful cycle cover of P for this
feasible weight. The sum of the edge weights is 20. Because P has no
Hamilton cycle, we must use at least 3 cycles. On the other hand, P has
girth 5 so that we may use at most 4 cycles.
If we try using 4 cycles, they all must be 5-cycles. But a 5-cycle uses at
most 2 spokes so that we cannot possibly use weight 10 on the spokes. If
we try using 3 cycles, then 2 of them must use 4 spokes each. This implies
they each have length at least 8 leaving at most length 4 for the third cycle.
This is impossible as the girth of P is 5. So there is no faithful cycle cover
for this weight.
So we have seen that not every feasible weight admits a faithful cycle
cover. The next result is the best known result in this area but first we
need a definition. Let X be a graph and let B1 , B2 , . . . , Bt be a partition of
V (X) such that X induces a connected subgraph hBi i for each part Bi , i =
1, 2, . . . , t. Define a reflexive multigraph X C with vertex set w1 , w2 , . . . , wt
such that the number of loops at wi is the number of edges contained in hBi i,
10
and the multiplicity of the edge joining wi and wj , i 6= j, is the number of
edges in X with one end in Bi and the other end in Bj . We call X C the
contraction of X with respect to the given partition.
The graph Y is a minor of X if Y is a subgraph of X C for some contraction of X.
Let C = u0 u1 · · · um−1 u0 be an m-cycle for m ≥ 3. If we take the
partition consisting of the singleton u0 , the singleton u1 , and the remaining
m − 2 vertices, then the contraction gives us a 3-cycle together with m − 3
loops at one of the vertices. Deleting the loops leaves us with a 3-cycle and
we see that the 3-cycle is a minor of any cycle.
Theorem 8. (Alspach, Goddyn and Zhang) If X is any graph without a
Petersen minor, then there is a faithful w-cycle cover for any feasible weight
w on X.
The proof of the preceding theorem is long and not easy. Thus, we shall
not present it here.
4
Oberwolfach Problem
Ringel gave us another problem that has attracted considerable attention
since it was first posed in1967. Ringel was attending a graph theory workshop at Oberwolfach and was undoubtedly motivated by the fact that people
are randomly placed around tables for meal service at the famous conference
center. He originally posed the problem for odd order complete graphs, but
because even order complete graphs minus a 1-factor behave very much like
odd order complete graphs with regard to cycle decompositions, we state
the problem for both families.
The Oberwolfach problem. Let F be a 2-factor whose components
are cycles of lengths `1 , `2 , . . . , `t . If the sum of the lengths is n, n odd,
the Oberwolfach problem OP(`1 , `2 , . . . , `t ) is to determine if Kn can be
decomposed into 2-factors isomorphic to F . When n is even, the problem is
the same except we want to decompose Kn − I, where Kn − I denotes the
complete graph with a 1-factor removed.
One of the first achievements on this problem was determining some parameters for which there are no solutions. For example, it is easy to see that
OP(3, 3) has no solution because if we remove a 2-factor composed of two 3cycles from K6 −I, the remaining graph is bipartite and certainly contains no
further 2-factors of the desired form. It turns out that OP(4, 5), OP(3, 3, 5)
11
and OP(3, 3, 3, 3) also do not have solutions. The proof for OP(3, 3, 5) is
rather ugly.
tP
t
PP
@P
PPt
t @
P
PP @
P
tPP
@t
It is conjectured that all other Oberwolfach problems have a solution. In
some sense there has not been a lot of progress on the Oberwolfach problem.
We’ll take a look at some of the positive results.
Some earlier work done before the Oberwolfach problem had been formulated provides solutions for a couple of special cases. For example, the
Walecki decomposition solves the special case for n odd and Hamilton cycles.
When n is even, if we perform a slightly modified construction, we obtain
a decomposition of Kn −I into Hamilton cycles. I usually call this a Walecki
decomposition as well so that Walecki gives us a solution of OP(n) for all
n. A typical example is shown next.
t
Q
t
aaQt
aa
t
at
X
XXX
X
t XX
t X
Xt
X
XXX
t
XXXt
a
aa
t aat
We rotate this clockwise through
five positions and are left with the
1-factor shown on the next page.
12
So this takes care of the
Oberwolfach problem for all
possible orders and
Hamilton decompositions.
t
t !t
!!
!
t !
t
t t
B t
t B
t
B !!
!
B
!
t ! Bt
In the early part of the 19th Century, Kirkman posed his famous “schoolgirl problem” which is equivalent to asking whether OP(3, 3, 3, 3, 3) has a
solution. A solution was found at the time and this led to the notion of
a Kirkman triple system. A Kirkman triple system is simply a solution of
OP(3, 3, . . . , 3), where there are an odd number of 3-cycles.
The existence of Kirkman triple systems turned out to be a very difficult problem and they were proved to always exist only in 1971 by Wilson
and Chaudhury. About the same time, people were considering the even
analogue of Kirkman triple systems. In doing so, it was discovered that
OP(3, 3, . . . , 3), where there are an even number of 3-cycles, has a solution
except for OP(3, 3) and OP(3, 3, 3, 3).
So we now have seen that 2-factors whose components are 3-cycles are
completely settled with respect to the Oberwolfach problem. There was one
other result that was known when the problem was first formulated. Let’s
take a look at it because it introduces a nice technique.
The wreath product of two graphs X and Y , denoted X o Y , is obtained
by replacing each vertex of X with a copy of Y , and joining every vertex
of one copy of Y to every vertex of another copy of Y if and only if the
corresponding vertices of X are adjacent. The next picture is an example of
C3 o P3 , where the double line between two copies indicates all edges between
the two copies are present.
13
t
t
t
t
t AAA
AA
AA
A
t
t
t
t
An important fact is that K2m − I is isomorphic to Km o K2 , where K2
denotes the complement of K2 . So as we go from Km to K2m − I, every
edge of Km blows up into K2,2 as shown next:
t
t
−→
t
t
t
t
We now introduce some simpler notation for the Oberwolfach problem.
We shall use exponent notation to indicate the number of cycles of a given
fixed length. For example, OP(32 , 5) denotes OP(3, 3, 5).
Theorem 9. The Oberwolfach problem OP(4a ) has a solution for all
a ≥ 1.
Proof. Let 2m = 4a to simplify notation a bit. We have seen that
K2m − I is isomorphic to Km o K2 . Now Km has a 1-factorization because
m is even. Each edge of a given 1-factor blows up to a 4-cycle in K2m − I
giving us a 2-factor in K2m − I composed of 4-cycles. Hence, the theorem
is proved.
At the time the Oberwolfach problem was posed, the preceding is essentially what was known in the direction of positive results: 3-cycles, 4-cycles
and Hamilton cycle. One approach that is almost always used on this kind of
problem is to examine small values of n and see how far you can get. This
usually is tedious but now and then something is discovered that allows
major progress towards a solution.
The computation has been carried out through n = 40. For the range
18 ≤ n ≤ 40, there had to be some ideas employed because the bruteforce method cannot handle values of n larger than 17. We have already
considered the main idea so let’s return to it.
14
The main idea that was used was group action. This allowed the search
space to be much smaller. We’ll illustrate with two examples for small values
of n.
Let’s suppose we are looking at OP(53 ). Because n = 15, we need seven
2-factors. So we use a permutation that is a product of two 7-cycles and a
fixed point. The next slide showns the initial 2-factor for the group action
and the permutation is
(u0 )(u1 u2 u3 u4 u5 u6 u7 )(v1 v2 v3 v4 v5 v6 v7 ).
tu1
v1 t
Z
Z
Z
Ztu2
v2 t
Z
Z
t
u0
tu3
v3 t
v4
v5
v6
t
J
t u4
J t J
tu5
J
J
t
Jt u6
v7 t
tu7
We now consider OP(3, 4, 5). The order of the graph is 12 so that we
are considering K12 − I which is regular of valency 10. This means that we
need five 2-factors. So we take a permutation of the form
f = (u0 )(v0 )(u1 u2 u3 u4 u5 )(v1 v2 v3 v4 v5 ).
It doesn’t matter that we have two fixed vertices because the edge joining
them is simply part of the 1-factor removed to form K12 − I. The edge
choices for the group hf i are shown in the next picture.
15
tv1
u1 t
u0
t
tv2
J
J
tv3
J
J
J
t
J t v4
v0
u2 t
u3 t
u4 t
tv5
u5 t
The first nice idea on this problem is due to Roland Häggkvist and what
I call the cannelloni stuffing lemma. It employs the wreath product idea we
considered earlier.
Take a cycle Cm of length m and consider the wreath product Cm o K2 .
This graph has even order 2m. Consider a 2-factor all of whose components
have even order, that is, it is composed of even length cycles. If we remove a
copy of the 2-factor from Cm o K2 , we shall find that what is left is a second
copy of the 2-factor. The next figure illustrates this.
t
t
t
t
t
t
t
t
@
@
@
@
@
@
@
@
@
@
@
@
@
@
@
@t
@t
@t
@t
@t
@t
@t
@t
t
@
@t
@
t
t
t
@
@t
t
t
t
@
t
t
t
@
@t
t
t
t
t
@
t
t
@
@t
t
@
t
@
t
@
@
@
@t
@t
@t
t
@
t
t
t
@
@
t
@t
Complementary 2-factors composed of cycles of lengths 4, 6 and 6
16
Thus, we can see that if Km has a decomposition into Hamilton cycles,
then K2m − I can be decomposed into 2-factors all of whose cycles have even
length. The Walecki decomposition gives us a Hamilton decomposition of
Km when m is odd. Even though Km does not have a Hamilton decomposition when m is even, there is a decomposition into Hamilton cycles and a
1-factor. We can perform a trick with the 1-factor left over and achieve the
following result.
Theorem 10. (Häggkvist). If `1 , `2 , . . . , `t are even integers greater
than 2, then OP(`1 , `2 , . . . , `t ) has a solution.
Theorem 10 was the first result that took a substantial bite out of the
Oberwolfach problem, but the cases covered by the result still essentially
have density zero over the entire problem. The next big step was a complete
solution for the case that all cycles have the same length. Of course, Theorem
10 takes care of the case for even length and the real progress was handling
odd length. The result was published in 1989.
Theorem 11. (Alspach, Schellenberg, Stinson and Wagner) The Oberwolfach problem OP(ma ) has a solution for all m ≥ 3 and all a ≥ 1 except
m = 3 and a = 2, 4.
The proof of Theorem 11 is long and somewhat intricate, but we shall
take a look at an outline of the proof. Because of the earlier work on
Kirkman triple systems and near-Kirkman triple systems, the answer for
m = 3 already was known. The interesting aspect of the proof is that it
does not work for m = 3. So we consider odd m ≥ 5.
First we break the vertex set into α sets of m vertices apiece, where
n = αm. On the next page is a figure depicting the case for α = 6. There
are then two cases depending on the parity of α, that is, whether α is even
or odd.
First we consider the case that α is odd. Because m is odd, each of
the complete subgraphs Km induced on the parts can be decomposed into
(m−1)/2 spanning cycles. This gives us (m−1)/2 2-factors composed of mcycles. The subgraph that is left over is Kα o Km which is just the complete
multipart graph Kα;m .
We now introduce a technique of projecting odd length directed cycles
onto multipartite graphs that are not bipartite (we cannot possibly get an
odd length cycle in a bipartite graph). Consider the wreath product Cs o Kt ,
where t ≥ s ≥ 3 and both s and t are odd. Coordinatize the vertices as ui,j ,
where 0 ≤ i ≤ s − 1 and 0 ≤ j ≤ t − 1. Let H = (uj1 , uj2 , . . . , ujt , uj1 ) be a
−
→
Hamilton directed cycle in Kt . The i-projection of H onto Cs o Kt is the
17
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
m
m
m
Breaking the vertex set into
sets of cardinality m.
m
m
t
t
m
t
cycle
ui,j1 ui+1,j2 · · · ui+s−1,js ui,js+1 ui+1,js+2 ui,js+3 · · · ui+1,jt ui,j1 .
Here is an example for s = 3 and t = 7.
t
t6
t
C@
C @
tC @t
t5
C
C t C t
t4
H
A HC H
A C H
t A C t HHt 3
A C
AC
t ACt t 2
t
t
t1
t
t
t0
t0
Q
Q
7
s t1
Q
6 tP
iP
PP PPt
?
t
5 * 2
t t
4
3
18
Looking back at the preceding example for s = 3 and t = 7, if we
interpret this as the 0-projection, then the 0-projection, the 1-projection and
the 2-projection together form a 2-factor composed of three 7-cycles. The
following lemma tells us that is always the case and the proof is immediate.
−
→
Lemma 12. If H is a Hamilton directed cycle of Kt , then the i-projections
of H onto Cs o Kt for i = 0, 1, . . . , s − 1 give a 2-factor of Cs o Kt composed
of s cycles of length t whenever 3 ≤ s ≤ t.
Lemma 12 certainly appears to be a useful tool for solving the problem
we are considering but there are two technical problems that are apparent.
First, edges of the form ui,j and uk,j never are used because of the projections. So we need to find a way to use up edges of this type between the
copies of Kt . Second, the length of the cycles we want must be at least as
big as the number of copies of Kt we are using.
−
→
If we take the complete directed graph Kt , we can write it as the arcdisjoint union of two circulant digraphs; one with connection set {±1, ±2}
and the other with connection set {±3, ±4, . . . , ±(t − 1)/2}. In the next
section we shall see that both of these circulant digraphs can be decomposed
into Hamilton directed cycles. So if we take the projections for the latter
circulant digraph onto Cs o Kt , we obtain a collection of 2-factors composed
of s cycles of length t. So we then consider the unused edges of Cs o Kt .
The unused edges form a subgraph of valency 10 arising from the projections of the directed edges coming from ±1 and ±2 in the connection set,
and from the edges of the form ui,j uk,j . Hence, if we can decompose this
subgraph into five 2-factors consisting of s cycles of length t, then we are
done.
We now give a verbal description of five special subgraphs built on a
t × s grid. Pictures are on the next page with s = 7 and t = 13. First of all,
the subscripts refer to the fact that the five graphs are built on the vertices
of columns 0 and 1 in the description of Cs o Kt . So Ai , for example, is
obtained by shifting A0 so that it sits between column i and column i + 1,
where column s is column 0.
Note that all edges corresponding to changes of 0, ±1, ±2 are used exactly
once as is required. Thus, if we provide a recipe for combining the matchings
so that we obtain five 2-factors composed of t-cycles, then we are done. Here
is such a description:
A0 ∪ D1 ∪ E2 ∪ D3 ∪ E4 ∪ · · · ∪ Ds−2 ∪ Es−1 is one 2-factor,
D0 ∪ E1 ∪ D2 ∪ E3 ∪ D4 ∪ · · · ∪ Es−2 ∪ As−1 is another 2-factor,
E0 ∪ A1 ∪ A2 ∪ · · · ∪ As−2 ∪ Ds−1 is another 2-factor,
19
t
t
t t
@
t @t
t t
@
t @t
s t t
@
t @t
@
t @t
@
t @t
@
t @t
@
t @t
@
t @t
@
t @t
A0
t t
@
t @t
t
t
t
t t
@
t @t
t
t t
t
t
t t
t
t
t
t
t
t
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t
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t
t
t
t
t
t
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t
t
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t
t
t
t
t t
@
t @t
B0
D0
t
A
tA
t
t
A
t At
A
tA t
A
t At
A
tA t
A A
tA At
A A
tA At
A A
tA At
A A
tA At
A A
tA At
A A
tA At
A
t At
E0
t
t
t t
A
tA t
A
t At
A
tA t
A
t At
t t
t t
t t
t t
t t
t t
t t
F0
B0 ∪ B1 ∪ B2 ∪ · · · ∪ Bs−1 is another one, and
F0 ∪ F1 ∪ F2 ∪ · · · ∪ Fs−1 is the fifth and final 2-factor.
The preceding results make the path clear when α is odd. It can be
proved that Kα , α odd, has a 2-factorization F1 , F2 , . . . , F(α−1)/2 so that the
components of each Fi are 3-cycles and 5-cycles except for α = 7 and α = 11.
Hence, Kα o Km can be written as the edge-disjoint union of Fi o Km as i
runs from 1 through (α − 1)/2. These wreath products are vertex-disjoint
unions of C3 o Km and C5 o Km . We’ve just seen how to decompose them
into 2-factors composed of m-cycles as long as m ≥ 5.
That takes care of all cases except α = 7 and α = 11 for which separate
arguments are required. It is not hard to handle them but we shall not do
so here.
We shall not discuss α even in much detail here but the basic idea is to
pair the α subsets in α/2 pairs. Then consider each as K2m − I. This can
be shown to have a decomposition into 2-factors made up of two m-cycles.
20
This uses all the edges inside the pairs and removes a 1-factor from Kαm .
The edges not used form the graph (Kα − I) o Km . An argument similar
to what was just used now works because we can prove, though we shall not
do so here, that Kα − I has a 2-factorization such that the components of
the 2-factors are made up of 3-cycles and 5-cycles except for α ∈ {4, 6, 12}.
The missing cases require separate arguments.
The most recent big idea on the Oberwolfach problem comes from Darryn
Bryant working with a couple of his students. Define a special graph Jr
which has vertices labelled u0 , u1 , u2 , . . . , ur+2 , ur+3 . It has edges between
any two vertices whose distance apart is 1, 2, 3 or 4 except
u0 u1 , u1 u2 , u2 u3 , u0 u3 , u1 u3 , ur ur+2 .
Suppose F is a 2-regular graph of order r in which we are interested.
One reason might be that we want to solve the corresponding Oberwolfach problem for F . It turns out that for almost all F we can find a 2factorization of Jr into four copies of F by placing one copy on the vertices u0 , u2 , u3 , . . . , ur−1 , ur+1 , another copy on u1 , u4 , u5 , . . . , ur , ur+2 , ur+3 ,
a third copy on u2 , u4 , u5 , . . . , ur , ur+1 , ur+3 , and the fourth and last copy
on the set of vertices u3 , u4 , . . . , ur , ur+1 , ur+2 .
Once we have the preceding, we then have a 2-factorization of circ(r; ±1,
± 2, ±3, ±4} by mapping ur+a to ua when mapping Jr to the circulant. If
you look at the missing edges in Jr everything fits to give the circulant.
Thus, to get solutions to OP(F ), we need to interpret the complete
graph as a circulant graph. With this viewpoint in mind, we require that
Kn or Kn − I may be written as an edge-disjoint union of graphs isomorphic
to circ(n; ±1, ±2, ±3, ±4}. This means we must consider isomorphisms of
circulant graphs.
If x is a unit in Zn and S is a connection set for a circulant of order
n, then the circulant graph with connection set xS is isomorphic. To see
this simply map the vertex ui to uxi for each i. Hence, if we can find
units x1 , x2 , . . . , xr in Zn such that {±1, ±2, . . . , ±(n − 1)/2} is partitioned
into x1 S 0 , x2 S 0 , . . . , xr S 0 , where S 0 = {±1, ±2, ±3, ±4}, then Kn has been
decomposed into subgraphs isomorphic to the special circulant.
This gives an approach whereby we can find infintely many values of n
for which the Oberwolfach problem has a solution for all 2-factors. Modifications of this approach have yielded further reults along this line.
21
5
Cayley Graphs On Abelian Groups
A circulant graph (defined earlier) is a special case of a Cayley graph on an
abelian group. The latter is defined by starting with an abelian group G
and a subset S ⊂ G satisfying i) 0 6∈ S, and ii) s ∈ S if and only if −s ∈ S.
We then label the vertices with the elements of G and join g and h with an
edge exactly when h − g ∈ S, that is, h = g + s for some s ∈ S. We use the
notation Cay(G; S) and call S the connection set.
At a workshop at Simon Fraser University in 1982, J.-C. Bermond gave
a one-hour invited talk on Hamilton-decomposable graphs. At the end of
the talk, I asked him if he had noticed that every example he gave in the
talk was, in fact, a Cayley graph on an abelian group. He said no and
thus was born what is now called a conjecture and which first appeared in
print in 1984. As a note, if the graph has odd valency, then a Hamilton
decomposition means we have edge-disjoint Hamilton cycles and a single
1-factor.
Conjecture (Alspach). Every connected Cayley graph on an abelian
group admits a Hamilton decomposition.
There have been three basic approaches to this conjecture. One approach has been to consider the valency of the graph, a second approach
has considered the order of the group, and the third approach has been to
consider special properties of the connection set.
It has been known for a long time that every connected Cayley graph
on an abelian group is hamiltonian. This appeared in Lovász’s first version
of his book entitled Combinatorial Problems And Exercises. That did not
prevent a small number of people submitting manuscripts over the years
containing the result. The preceding result means the conjecture is true for
the trivalent case.
The first new progress on the problem came in 1989 via the following
result.
Theorem 13. (Bermond, Favaron and Maheo) If X is a Cayley graph
on an abelian group and has valency 4, then X is Hamilton-decomposable.
There is a small problem with their proof in that they interpreted an
involution in the connection set as generating a multiple edge. This means
that a few graphs are omitted (such as the 4-dimensional cube), but the
omission is easily fixed.
Let’s now see how to use Theorem 12 to settle the conjecture for the 5valent case. This leads us to introduce a tool from which another unsolved
22
conjecture emerges. Let X = Cay(G; S) be a connected 5-valent Cayley
graph on the abelian group G. This implies |S| = 5 which, in turn, implies
that S contains an involution s.
Let S 0 = S − {s} and let X 0 be the 4-valent subgraph of X arising from
the connection set S 0 . If X 0 is connected, then it can be decomposed into
two Hamilton cycles by Theorem 12. The element s generates a 1-factor
and this gives us a Hamilton decomposition of X.
This leaves us with the interesting case that X 0 is disconnected. It is
not hard to see that X 0 must have two identical components joined by the
1-factor generated by s.
s
Because the subgraphs are identical, if we take a Hamilton cycle in a
component and its copy in the other component, we can perform a 4-cycle
switch and obtain a Hamilton cycle in X.
t
t
t
t
Hence, if we can find two vertex-disjoint edges with one in each spanning
cycle, then we can perform two 4-cycle switches to obtain two Hamilton
23
cycles in X leaving a 1-factor left over and complete the proof for valency 5.
In fact, something stronger is true. Let F1 , F2 be two edge-disjoint 2-factors
in a 4-valent regular graph Y . Note that Y has order at least 5.
Let uv be any edge of F1 .
t
u
t
v
Exactly 4 edges from F2 are incident with u and v, but F2 has strictly
greater than 4 edges. Thus, there is an edge of F2 that is vertex-disjoint
with uv. This is all we need for the 5-valent argument.
Let X be a regular graph of valency 2t, and let F1 , F2 , . . . , Ft be an
arbitrary 2-factorization of X. A t-matching that contains one edge from
each 2-factor is called an orthogonal matching.
Conjecture. (Alspach) Every 2-factorization of a regular graph of even
valency admits an orthogonal matching.
The valency argument comes to an end with valency 6.
Theorem 14. (Westlund, Liu and Kreher) If X is a 6-valent connected Cayley graph on an abelian group of odd order, then X is Hamiltondecomposable.
While they were able to prove most of the situations when G has even
order, they were unable to settle all the cases. This is as far as we have been
able to go by considering valency.
We now examine the approach via the order of the group. If the group
G has prime order p, then G is cyclic and any Cayley graph on G is a
circulant graph. If the connection set is {±s1 , ±s2 , . . . , ±st }, then all the
edges generated by si produce a Hamilton cycle. Hence, the graph has a
trivial Hamilton decomposition.
Theorem 15. If X is a connected Cayley graph on an abelian group
of order p2 or order pq, where p and q are primes, then X is Hamiltondecomposable.
The preceding results are all that is known by considering the order of
the group outside of some special results.
The last approach is based on the structure of the connection set for a
Cayley graph. We say that the connection set S is minimal if S generates
the group G (so that the graph is connected), but S − {±si } generates a
proper subgroup of G for every si ∈ S.
24
Theorem 16. (Jiuqiang Liu). Let X = Cay(G; S), where S is a minimal
connection set. Then X is Hamilton-decomposable if either
• G has odd order, or
• G has even order and 2s 6∈ S for every s ∈ S.
Corollary 17. (Aubert and Schneider) The cartesian product of an
arbitrary number of cycles is Hamilton-decomposable.
Proof. To prove Corollary 17, note that the cartesian product
C(1)2C(2)2 · · · 2C(t)
of cycles of the respective lengths `1 , `2 , . . . , `t is the Cayley graph on the
direct sum Z`1 ⊕ Z`2 ⊕ · · · ⊕ Z`t of cyclic groups of orders `1 , `2 , . . . , `t with
the standard basis and their inverses for the connection set. The connection
set clearly is minimal and the result follows.
Corollary 18. The n-dimensional cube Qn is Hamilton-decomposable
for all n.
Proof. When n is even, it is easy to see that Qn is isomorphic to the
cartesian product of n/2 4-cycles. Corollary 16 then tells us that there is a
Hamilton decomposition. So the interesting situation is for odd dimension.
The 1-dimensional cube Q1 is a single edge which is a trivial Hamilton
decomposition. The odd dimensional cube Q2m+1 consists of two vertexdisjoint copies of Q2m joined by a 1-factor. Each copy of Q2m has a decomposition into m Hamilton cycles. Let H1 , H2 , . . . , Hm be m Hamilton
cycles decomposing Q2m . If we can find an orthogonal matching for this set
of cycles, we can use the technique described earlier to obtain a Hamiltondecomposition of Q2m+1 .
Suppose we have chosen vertex-disjoint edges e1 , e2 , . . . , ek such that
ei ∈ Hi for i = 1, 2, . . . , k, where k < m. It suffices to show that we can
find an edge from Hk+1 that is vertex-disjoint from e1 , e2 , . . . , ek . The edges
e1 , e2 , . . . , ek have 2k vertices. Each of these vertices is incident with 2 edges
from Hk+1 so that altogether at most 4k edges of Hk+1 are incident with
vertices in the k-matching constructed so far. The number of edges in Hk+1
is exactly 4m and 4m > 4k because k < m. Thus, we may choose an edge
of Hk+1 which has no vertices in common with e1 , e2 , . . . , ek . Therefore, the
corollary follows.
Another application of Theorem 16 is to take an arbitrary connection
set and partition it into subsets such that each subset generates the group
25
and also is minimal. Each subgraph may then be decomposed into Hamilton cycles. This approach was the essential ingredient in proving the next
theorem, but first a definition is required.
Let q = pa , where p is a prime and q ≡ 1(mod 4). Define a graph as
follows. The vertices are the elements of the finite field with q elements and
two vertices are adjacent if and only if their difference is a quadratic residue
in the field. The resulting graph is known as the Paley graph of order q.
Theorem 19. (Alspach, Bryant and Dyer) The Paley graphs are Hamiltondecomposable.
26