In contrast with Kotzig's result that the line graph of a $3$-regular graph
$X$ is Hamilton decomposable if and only if $X$ is Hamiltonian, we show that
for each integer $k\geq 4$ there exists a simple non-Hamiltonian $k$-regular
graph whose line graph has a Hamilton decomposition. We also answer a question
of Jackson by showing that for each integer $k\geq 3$ there exists a simple
connected $k$-regular graph with no separating transitions whose line graph has
no Hamilton decomposition

Bryant, Darryn

Maenhaut, Barbara

Smith, Benjamin R.

English

arXiv

In contrast with Kotzig’s result that the line graph of a 3-regular graph X is Hamilton decomposable if and only if X is Hamiltonian, we show that for each integer k ≥ 4 there exists a simple non-Hamiltonian k-regular graph whose line graph has a Hamilton decomposition. We also answer a question of Jackson by showing that for each integer k ≥ 3 there exists a simple connected k-regular graph with no separating transitions whose line graph has no Hamilton decomposition

University of Queensland eSpace

AUSTRALASIAN JOURNAL OF COMBINATORICS
Volume 71(3) (2018), Pages 537–543
On Hamilton decompositions of line graphs
of non-Hamiltonian graphs and graphs
without separating transitions
Darryn Bryant Barbara Maenhaut Benjamin R. Smith
Department of Mathematics
University of Queensland
QLD 4072
Australia
db@maths.uq.edu.au bmm@maths.uq.edu.au bsmith.maths@gmail.com
Dedicated to the memory of Anne Penfold Street, 1932–2016
Abstract
In contrast with Kotzig’s result that the line graph of a 3-regular graph X
is Hamilton decomposable if and only if X is Hamiltonian, we show that
for each integer k ≥ 4 there exists a simple non-Hamiltonian k-regular
graph whose line graph has a Hamilton decomposition. We also answer
a question of Jackson by showing that for each integer k ≥ 3 there exists
a simple connected k-regular graph with no separating transitions whose
line graph has no Hamilton decomposition.
1 Introduction
In the 1960’s Kotzig [9] proved that the existence of a Hamilton cycle in a 3-regular
graph X is both necessary and suﬃcient for the existence of a Hamilton decompo-
sition of its line graph L(X). Hamilton decomposability of line graphs has subse-
quently been studied extensively, but the general question of classifying those graphs
whose line graphs have Hamilton decompositions remains open. This topic has been
considered from a number of diﬀerent perspectives. In particular, Hamilton decom-
posability of L(X) has been considered with imposed conditions on the connectivity
[4, 6, 7] or Hamiltonicity [2, 8, 10, 11, 12, 13] of X . Additional papers containing
results related to Hamilton decompositions of line graphs include [5, 14] and the
survey [1].
D. BRYANT ET AL. /AUSTRALAS. J. COMBIN. 71 (3) (2018), 537–543 538
In this paper we answer a question of Jackson [6] on Hamilton decomposability
of the line graphs of graphs with no separating transitions (a connectivity-related
condition deﬁned below), and we prove that the above-mentioned result of Kotzig
does not hold for k-regular graphs when k ≥ 4. If X is regular of degree 2k or 2k+1,
then a set of k pairwise edge-disjoint Hamilton cycles in X is called a Hamilton de-
composition, and a graph admitting a Hamilton decomposition is said to be Hamilton
decomposable.
In [6], Jackson calls a pair of half edges incident with a vertex u a transition at u,
and if t is a transition at u in a graph X , then he deﬁnes X t to be the graph obtained
from X by splitting u into two new vertices u1 and u2, joining the two half edges of
t to u1, and joining each other half edge at u to u2. A separating transition is then
deﬁned to be a transition t such that X t has more components than X. It is shown
in [6] that for k ≥ 3, the line graph of a connected k-regular graph X is (2k − 2)-
edge-connected if and only if X has no separating transitions (the result is actually
stated only for the case k is even, but the same argument works when k is odd). It
follows that if k ≥ 3 and X is any k-regular graph with a separating transition, then
L(X) has no Hamilton decomposition. We observe that the preceding statement is
not true without the requirement that X be regular. For example, any star with at
least three edges has both a separating transition and a Hamilton decomposable line
graph.
Having observed that absence of separating transitions in X is necessary for
Hamilton decomposability of L(X), Jackson asks (Problem 5.2 in [6]) whether it is
true that the line graph of a connected 2k-regular graph X has a Hamilton decompo-
sition if and only if X has no separating transitions. The same question could also be
asked for connected regular graphs of odd degree. The answer is no in the case of 3-
regular graphs because there are many connected non-Hamiltonian 3-regular graphs
that have no separating transitions, and Kotzig’s result tells us the line graphs of
these graphs have no Hamilton decomposition. In Section 2 we construct for each
integer k ≥ 3 a simple connected k-regular graph with no separating transitions
whose line graph has no Hamilton decomposition, thereby showing that the answer
to Jackson’s question is no for every degree greater than 3.
The authors [3] have recently shown that the existence of a Hamilton cycle in a
simple graph X is suﬃcient for Hamilton decomposability of L(X) when X is regular
of even degree, and that the existence of a Hamiltonian 3-factor in X is suﬃcient for
Hamilton decomposability of L(X) when X is regular of odd degree. Whether the
existence of a Hamilton cycle, rather than a Hamiltonian 3-factor, is suﬃcient for
Hamilton decomposability of L(X) when X is regular of odd degree remains an open
question. The results just mentioned partially extend Kotzig’s result to k-regular
graphs with k > 3, but only in the direction of suﬃciency. Going in the opposite
direction, we show in Section 3 that the existence of a Hamilton cycle in X is not
necessary for Hamilton decomposability of L(X) when X is regular of degree at
least 4.
The proofs of both of our main results involve construction of new graphs by
D. BRYANT ET AL. /AUSTRALAS. J. COMBIN. 71 (3) (2018), 537–543 539
deletion of an edge of a graph and insertion of the resulting graph into an edge of
another graph, and we now give the formal deﬁnition of this procedure. Let X and
X ′ be vertex-disjoint graphs (not necessarily simple), let u and v be adjacent vertices
in X, and let u′ and v′ be adjacent vertices in X ′. We deﬁne the insertion of X ′−u′v′
into an edge uv of X to be the graph obtained from X ∪ X ′ by replacing an edge
uv of X and an edge u′v′ of X ′ with an edge joining u to u′ and an edge joining
v to v′. In this deﬁnition the order in which the vertices of the edges uv and u′v′
are listed may change the resulting graph, but this will be of no consequence in our
constructions.
2 Separating transition-free graphs whose line graphs are
not Hamilton decomposable
Theorem 2.1 For each integer k ≥ 3, there exists a simple connected k-regular
graph with no separating transitions whose line graph has no Hamilton decomposition.
Proof For each integer k ≥ 3 and each even integer t ≥ 4, deﬁne Yk,t to be the
multigraph with vertices v1, v2, . . . , vt, and edge set given by joining vi to vi+1 with
two edges for i = 1, 3, . . . , t−1, and joining vi to vi+1 with k−2 edges for i = 2, 4, . . . , t.
Here, and throughout what follows, vt+1 is identiﬁed with v1. Let Xk,t be the graph
obtained from Yk,t by inserting a copy of Kk+1 − e into each edge of Yk,t. It is easy
to see that Xk,t is a simple k-regular graph that has no separating transitions. We
will show that L(Xk,t) has no Hamilton decomposition for t ≥ k, but ﬁrst we need
to introduce labels for various edges of Xk,t.
For i = 1, 3, . . . , t − 1, let X1i and X2i be the two copies of Kk+1 − e that are
inserted into the two edges joining vi to vi+1. For i = 1, 2, . . . , t, let e
1
i , e
2
i , . . . , e
k
i be
the k edges of Xk,t that are incident with vi. For i = 1, 3, . . . , t− 1 and for j = 1, 2,
let eji be the unique edge joining vi to X
j
i , and let e
j
i+1 be the unique edge joining vi+1
to Xji . For i = 1, 3, . . . , t−1 and for j = 1, 2, let f ji,1, f ji,2, . . . , f ji,k−1 be the k−1 edges
of Xji that are adjacent to e
j
i , and let f
j
i+1,1, f
j
i+1,2, . . . , f
j
i+1,k−1 be the k − 1 edges of
Xji that are adjacent to e
j
i+1. Finally, for i = 1, 2, . . . , t, let Ei be the set of 2(k − 2)
edges of L(Xk,t) having one endpoint in {e1i , e2i } and the other in {e3i , e4i , . . . , eki }.
For a contradiction, suppose t ≥ k andH is a Hamilton decomposition of L(Xk,t).
Note that H contains k− 1 Hamilton cycles. Since t > k− 1, in L(Xk,t) at least two
of the edges e11e
2
1, e
1
2e
2
2, . . . , e
1
t e
2
t are in the same Hamilton cycle of H. Let this cycle
be H ∈ H and let e1ae2a and e1be2b be distinct edges of H (so a, b ∈ {1, 2, . . . , t}).
Now, for i = 1, 3, . . . , t− 1 and for j = 1, 2, {eji , eji+1} is a vertex cut of L(Xk,t),
and it follows that for i = 1, 2, . . . , t and for j = 1, 2, each Hamilton cycle of H
contains exactly one of the k − 1 edges ejif ji,1, ejif ji,2, . . . , ejif ji,k−1 of L(Xk,t). But this
implies that H contains none of the edges of Ea and none of the edges of Eb. Since
Ea∪Eb is an edge cut of L(Xk,t), this is a contradiction, and we conclude that L(Xk,t)
has no Hamilton decomposition. 
D. BRYANT ET AL. /AUSTRALAS. J. COMBIN. 71 (3) (2018), 537–543 540
3 Non-Hamiltonian graphs whose line graphs are Hamilton
decomposable
Hamilton cycles in L(X) are related to certain Euler tours of X. If
v0, e1, v1, e2, v2, . . . , et, vt = v0
is any Euler tour of X (so the edge set of X is {e1, e2, . . . , et} and v0, v1, . . . , vt
are vertices of X), then (e1, e2, . . . , et) is a Hamilton cycle in L(X). However, not
every Hamilton cycle in L(X) corresponds to an Euler tour of X . For example,
(wz, wx, wy, yz, xy, xz) is a Hamilton cycle in the line graph of the complete graph
with vertex set {w, x, y, z}, but it clearly does not correspond to an Euler tour of
this complete graph. Indeed, there is no Euler tour of the complete graph of order 4.
We shall say that a Hamilton cycle in L(X) is Euler tour compatible if it cor-
responds to an Euler tour in X. In order to say more about the properties that
Hamilton cycles in L(X) must have in order that they be Euler tour compati-
ble, we make the following deﬁnitions. If u is a vertex in a graph X, then the
neighbourhood of u in X is denoted by NX(u). Suppose X is a simple graph and
NX(u) = {v, a1, a2, . . . , ak}. Then in L(X) the u-neighbourhood of the vertex uv is
{ua1, ua2, . . . , uak} and is denoted by NuL(X)(uv). Thus, NuL(X)(uv) ∩NvL(X)(uv) = ∅
and NL(X)(uv) = N
u
L(X)(uv) ∪NvL(X)(uv).
It is easy to see that a Hamilton cycle H in L(X) is Euler tour compatible if
and only if for each vertex uv in L(X), one neighbour in H of uv is from the u-
neighbourhood of uv and the other neighbour in H of uv is from the v-neighbourhood
of uv. If this propery holds for a vertex uv in a Hamilton cycle H of L(X), then
we say that H is Euler tour compatible at uv. Thus, a Hamilton cycle is Euler tour
compatible if and only if it is Euler tour compatible at each of its vertices. More
generally, a Hamilton decomposition of L(X) is Euler tour compatible at uv if each
of its Hamilton cycles is Euler tour compatible at uv, and is everywhere Euler tour
compatible if it is Euler tour compatible at every vertex of L(X).
A Hamilton decomposition of L(X) that is everywhere Euler tour compatible is
thus equivalent to a perfect set of Euler tours of X, where a set S of Euler tours
of X is perfect if each 2-path in X occurs in exactly one Euler tour in S. In [5],
Heinrich and Verrall construct perfect sets of Euler tours for each complete graph of
odd order, thus establishing the following theorem.
Theorem 3.1 [Heinrich and Verrall [5]] For each odd integer n ≥ 3, the line graph
of the complete graph of order n has a Hamilton decomposition that is everywhere
Euler tour compatible.
The complete graph of even order has no Euler tour. However, there is a natural
way to extend the above ideas by considering instead the multigraph Kn + I which
is obtained from the complete graph of even order n by duplicating each edge in
D. BRYANT ET AL. /AUSTRALAS. J. COMBIN. 71 (3) (2018), 537–543 541
a set I of edges that form a perfect matching. In [14], Verrall shows that Kn + I
has a perfect set of Euler tours for all even n ≥ 4, where the deﬁnition of perfect
set of Euler tours is suitably modiﬁed to accommodate the edges of multiplicity 2.
The modiﬁcation is exactly what is needed to ensure that perfect sets of Euler tours
of Kn + I correspond to Hamilton decompositions of L(Kn) that are Euler tour
compatible at each vertex of L(Kn) except those in I. Indeed, as stated in [14], the
modiﬁcation is made speciﬁcally to parallel Theorem 3.1, and it is easily veriﬁed that
the main result in [14] can be restated in our terminology as follows.
Theorem 3.2 [Verrall [14]] If n ≥ 4 is an even integer, K is a complete graph of
order n, and I is a perfect matching in K, then L(K) has a Hamilton decomposition
that is Euler tour compatible at each vertex of V (L(K)) \ I.
Lemma 3.3 Let X and X ′ be vertex-disjoint k-regular graphs, let uv be an edge in
X, let u′v′ be an edge in X ′, and let Y be the insertion of X ′ − u′v′ into the edge
uv of X. If H is a Hamilton decomposition of L(X) that is Euler tour compatible
at uv and H′ is a Hamilton decomposition of L(X ′) that is Euler tour compatible at
u′v′, then there exists a Hamilton decomposition H∗ of L(Y ) such that if H is Euler
tour compatible at a vertex xy = uv of L(X), then H∗ is also Euler tour compatible
at xy.
Proof Suppose H = {H1, H2, . . . , Hk−1} is a Hamilton decomposition of L(X) that
is Euler tour compatible at uv and suppose H′ = {H ′1, H ′2, . . . , H ′k−1} is a Hamilton
decomposition of L(X ′) that is Euler tour compatible at u′v′. For i = 1, 2, . . . , k− 1,
let the two neighbouring vertices of uv in Hi be uai and vbi, let the two neighbouring
vertices of u′v′ in H ′i be u
′a′i and v
′b′i, and let Ji be the graph obtained from the
union of Hi and H
′
i by replacing the vertices uv and u
′v′ with uu′ and vv′, replacing
the edge joining uai to uv with an edge joining uai to uu
′, replacing the edge joining
vbi to uv with an edge joining vbi to vv
′, replacing the edge joining u′a′i to u
′v′
with an edge joining u′a′i to uu
′, and replacing the edge joining v′b′i to u
′v′ with an
edge joining v′b′i to vv
′. It is easily seen that H∗ = {J1, J2, . . . , Jk−1} is the required
Hamilton decomposition of L(Y ). 
Theorem 3.4 For each integer k ≥ 4 there exists a simple non-Hamiltonian k-
regular graph whose line graph has a Hamilton decomposition.
Proof Let k ≥ 4, let v, u1, u2 and u3 be vertices in a complete graph X of order
k + 1. For each i ∈ {1, 2, 3}, let X ′i be a complete graph of order k + 1, let u′iv′i be
an edge in X ′i, and insert X
′
i − u′iv′i into the edge uiv of X. Let Y be the resulting
graph. We claim that Y is a non-Hamiltonian k-regular graph. To see that Y is
non-Hamiltonian, observe that for i ∈ {1, 2, 3}, {vv′i, uiu′i} is an edge cut, and so
any Hamilton cycle necessarily contains the three edges vv′1, vv
′
2 and vv
′
3, which is
impossible.
D. BRYANT ET AL. /AUSTRALAS. J. COMBIN. 71 (3) (2018), 537–543 542
We now use Theorems 3.1 and 3.2 and Lemma 3.3 to show that L(Y ) has a
Hamilton decomposition. Since k ≥ 4, L(X) has a Hamilton decomposition that is
Euler tour compatible at vu1, vu2 and vu3 by Theorem 3.1 (k even) or 3.2 (k odd).
Also by Theorem 3.1 (k even) or 3.2 (k odd), for each i ∈ {1, 2, 3}, L(X ′i) has a
Hamilton decomposition that is Euler tour compatible at u′iv
′
i. It thus follows by
Lemma 3.3 (applied three times) that L(Y ) has a Hamilton decomposition. 
Acknowledgements
The authors acknowledge the support of the Australian Research Council via grants
DP150100530, DP150100506, DP120100790, DP120103067 and DP130102987.
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(Received 17 Oct 2017; revised 25 Mar 2018)


On Hamilton decompositions of line graphs of non-hamiltonian graphs and graphs without separating transitions

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On Hamilton Decompositions of Line Graphs of Non-Hamiltonian Graphs and Graphs without Separating Transitions

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