On 2-factors with a bounded number of odd components
Jennifer Diemunsch1,2,4 , Michael Ferrara1,3,4 , Samantha Graffeo1 , Timothy Morris1,2,4
January 23, 2014
Abstract
A 2-factor in a graph is a spanning 2-regular subgraph, or equivalently a spanning collection
of disjoint cycles. In this paper we investigate the existence of 2-factors with a bounded number
of odd cycles in a graph. We extend results of Ryjáček, Saito, and Schelp (Closure, 2-factors,
and cycle coverings in claw-free graphs, J. Graph Theory, 32 (1999), no. 2, 109-117) and show
that the number of odd components of a 2-factor in a claw-free graph is stable under Ryjáček’s
closure operation. We also consider conditions that ensure the existence of a pair of disjoint
1-factors in a claw-free graph, as the union of such a pair is a 2-factor with no odd cycles.
Keywords: 2-factor, claw-free, closure, disjoint 1-factors
1
Introduction and Notation
Throughout this paper all graphs are simple and all cycles have an implicit clockwise orientation.
For some vertex v on a cycle C we will denote the first, second, and ith predecessor (respectively
successor) of v as v − , v −− , and v (−i) (respectively v + , v ++ , and v (+i) ) respectively. Given x and y
on C, C(x, y) is the set of vertices {x+ , x++ , . . . , y − } and C[x, y] is the set {x, x+ , . . . , y}. We also
let xCy (respectively xC − y) denote the path xx+ . . . y (respectively xx− . . . y). Given a subgraph
H of G, for an edge e = xy we define distH (e, v) = min{distH (x, v), distH (y, v)}. If there is no path
in H from an endpoint of e to v, then we say that distH (e, v) = ∞. We denote the neighborhood
of a vertex v in a graph G as NG (v), and we omit the subscript “G” when the context is clear.
For some U ⊆ V (G), if U ⊆ N (v), then we say that v dominates U . For a subgraph H of G and
a vertex v, if v dominates V (H), then we will frequently instead say that v dominates H. The
circumference of a graph G, c(G), is the length of a longest cycle in G.
Given a family F of graphs, a graph G is said to be F-free if G contains no member of F as an
induced subgraph. If F contains a claw, that is K1,3 , then G is said to be claw-free. We denote by
hv; x, y, zi the claw with central vertex v and leaves x, y, and z; similarly, hv; x1 , x2 , ..., xt i denotes
a copy of K1,t . For the subgraph induced by U ⊆ V (G) we use the notation G[U ].
1
Dept. of Mathematical and Statistical Sciences,
Univ. of Colorado Denver;
email addresses
[email protected],
[email protected],
[email protected],
[email protected].
2
Research partially supported by NSF Grant # 0742434 UCD GK-12 Transforming Experiences Project
3
Research partially supported by a Simons Foundation Collaboration Grant for Mathematicians # 206692
4
Research partially supported by the National Science Foundation grant DMS 08-38434 “EMSW21-MCTP: Research
Experience for Graduate Students.”
1
An r-factor of a graph G is a spanning r-regular subgraph of G. Thus a 2-factor of a graph G is
a collection of cycles that spans G. A graph is hamiltonian if it contains a spanning cycle, which is
a 2-factor with exactly one component. The problem of determining when a graph is hamiltonian
is a classical and widely studied problem in graph theory (cf. [18, 19]).
Aside from the hamiltonian problem, there are many results throughout the literature that give
conditions ensuring the existence of a 2-factor with given properties. Many results of this type are
outlined in [18, 19], and there have been numerous developments since the writing of those surveys
(see, for instance, [1]) .
Let odd(F) denote the number of odd cycles in a 2-factor F of a graph G, and let odd(G) denote
the minimum of odd(F) over all 2-factors F of G. If G does not contain a 2-factor, we will set
odd(G) = ∞. The focus of this paper is the following.
Problem 1. Given k ≥ 0, determine conditions that guarantee a graph G has odd(G) ≤ k.
We note here that determining if a graph has two disjoint perfect matchings, or equivalently
has odd(G) = 0 is NP-complete, even in the case where G is a bridgeless cubic graph. This follows
from the fact [11] that it is NP-hard to determine if a cubic graph has a proper edge-coloring with
three colors. Such a coloring exists if and only if G contains a pair of disjoint perfect matchings.
In addition to the general literature on the structure of 2-factors in graphs, odd(G) arises
when studying several other interesting problems. For instance, results of Huck and Kochol [13],
Häggkvist and McGuinness [8], and Huck [14] showed that a cubic bridgeless graph with odd(G)
at most four has a family C of at most five even subgraphs such that each edge of G lies in exactly
two members of C. As a consequence, such graphs satisfy the Circuit Double Cover Conjecture (cf.
[22, 23]). Understanding the CDC for cubic graphs is crucial given that Seymour [22] noted that a
smallest counterexample to the circuit double cover conjecture must be cubic.
Note that odd(G) is a measure of how far G is from having a pair of edge-disjoint perfect
matchings, as G contains such a pair if and only if odd(G) = 0. Several other related metrics have
been considered. Let B2 (G) = {(B, B 0 ) : B, B 0 are edge-disjoint matchings of G}, P = {(B, B 0 ) ∈
B2 (G) : |E(B)| + |E(B 0 )| is maximum}, and H be a maximum matching chosen from the pairs
of matchings in P . Then, given a maximum matching M in G, µ(G) = |M |/|H| measures how
close G is to having a pair of disjoint matchings of maximum size. Mkrtchyan, Musoyan, and
Tserunyan [15] studied µ(G) and showed the following.
Theorem 1. For any graph G,
µ(G) ≤
5
4
This bound is tight.
Another measure of how close G is to having a pair of disjoint perfect matchings is given by the
ratio of edges covered by a pair of disjoint matchings to the number of vertices in a graph. This
notion was explored by Mkrtchyan, Petrosyan, and Vardanyan in [16].
Theorem 2. Let G be a cubic graph. Then G contains a pair of disjoint matchings covering at
least 45 |V (G)| edges in G.
The problem of determining when a graph has r edge-disjoint perfect matchings for some r ≥ 2
has also been studied in several contexts. Hilton [9] and Zhang and Zhu [25] examined the number
2
of disjoint 1-factors in regular graphs, and recently Hou [12] considered the problem for nearlyregular graphs (those graphs with ∆(G) − δ(G) ≤ 1). Hoffman and Rodger [10] give necessary and
sufficient condition for a complete multipartite graph to contain r edge-disjoint 1-factors.
In this paper, we are generally interested in investigating Problem 1 in the context of forbidden
subgraphs. In Section 2 we examine the stability of odd(G) under the Ryjáček closure for clawfree graphs. In Section 3 we specifically consider the case k = 0 and examine pairs of forbidden
subgraphs that ensure a graph has a pair of disjoint 1-factors. We then present several constructions
and open problems related to Problem 1 in highly connected claw-free graphs.
2
Closure and odd(G) for Claw-Free Graphs
In [20] Ryjáček introduced the following closure concept. In a graph G, call a vertex v eligible if
G[N (v)] is connected but not complete. The graph formed by completing the neighborhood of a
vertex v is the local completion (of G) at v. The closure of a graph G, denoted cl(G), is constructed
by iteratively performing the local completion operation at eligible vertices until none remain.
This closure is an important tool for the study of cycle structure and hamiltonian-type properties
in large part due to the following result from [20].
Theorem 3. Let G be a claw-free graph. Then
1. the closure cl(G) is well-defined,
2. there is a triangle-free graph H such cl(G) = L(H), where L(H) is the line graph of H,
3. c(G) = c(cl(G)).
Given a graph class C, we say that a property π is stable in C provided that every graph G in
C has property π if and only if cl(G) has property π. Thus, Theorem 3 implies that hamiltonicity
(and more generally the property “c(G) = k”) is stable in the class of claw-free graphs.
Of particular interest here, Ryjáček, Saito, and Schelp [21] showed the following.
Theorem 4. If G is a claw-free graph, then G has a 2-factor with at most k cycles if and only if
cl(G) has a 2-factor with at most k cycles.
The main result of this section is the following extension of Theorem 4.
Theorem 5. If G is a claw-free graph, then odd(G) ≤ k if and only if odd(cl(G)) ≤ k.
Proof. Clearly, if odd(G) ≤ k, then odd(cl(G)) ≤ k, so we proceed by demonstrating the converse.
Let G be a claw-free graph and let G = G0 , G1 , . . . , Gt = cl(G) be a sequence of graphs such that
Gi is the graph formed by local completion of Gi−1 at vertex vi . Suppose to the contrary that
odd(Gt ) ≤ k but odd(G) > k, and let i ≥ 1 be the minimum integer such that Gi has a 2-factor C
with at most k odd cycles and denote vi as v.
Let ECi = (E(C) ∩ E(Gi )) \ E(Gi−1 ), so that ECi is the set of edges of the 2-factor C in Gi which
are not in Gi−1 , and observe that the endpoints of these edges are all in the neighborhood of v.
3
Choose C to minimize |ECi | and furthermore, among all such choices of C select an edge e ∈ ECi such
that distC (e, v) is maximum. For each cycle in C, let Cw denote the cycle containing a particular
vertex w and similarly let Cf denote the cycle containing a particular edge f . Let e = u1 u2 , where
u2 is the predecessor of u1 in Ce , and let P be a shortest path between u1 and u2 in Gi−1 [N (v)].
Since Gi−1 is claw-free, P has order at most four; denote the vertices of P as u1 = x1 x2 ...x` = u2
with 3 ≤ ` ≤ 4.
The proof proceeds in cases by considering distC (e, v). Subcases are then based on the length of
P and the location of certain vertices on given cycles in C. In most cases, we draw a contradiction
by finding a 2-factor with at most k odd cycles in Gi that either uses fewer edges in ECi than C, or
violates the pertinent assumption about distC (e, v). We begin by illuminating several useful edges
and non-edges of Gi−1 .
Claim 1. If Cv = Ce , the following hold:
1. u2 v − ∈
/ E(Gi−1 ),
2. u1 v + ∈
/ E(Gi−1 ),
3. v − v + ∈
/ E(Gi−1 ),
4. if u2 6= v + , then u2 v + ∈ E(Gi−1 ).
Proof of Claim. If u2 v − is an edge, then replacing Cv with u2 v − Cv− u1 vCv u2 results in a 2-factor
C 0 such that odd(C 0 ) ≤ k and |ECi 0 | < |ECi |, a contradiction. Similarly, if u1 v + ∈ E(Gi−1 ), replacing
Cv with u1 v + Cv u2 vCv− u1 contradicts the minimality of |ECi |.
Suppose Cv is a triangle, then v − = u1 and v + = u2 , so clearly, v − v + is not an edge of Gi−1 . If
Cv is a 4-cycle, then u1 = v − or u2 = v + , and in either case, from above, v − v + ∈
/ E(Gi−1 ). Lastly
if Cv has more than four vertices and v − v + ∈ E(Gi−1 ), then replacing Cv with v − v + Cv u2 vu1 Cv v −
contradicts the minimality of |ECi |.
Finally, consider hv; u1 , u2 , v + i. By assumption, u1 u2 ∈
/ E(Gi−1 ), and since part (2) holds
u1 v + ∈
/ E(Gi−1 ), thus since Gi−1 is claw-free, u2 v + ∈ E(Gi−1 ).
The following will also be useful throughout the remainder of the proof.
Claim 2. Let f = w1 w2 ∈ C such that Ce 6= Cf . If u1 w1 ∈ E(Gi−1 ) and u2 w2 ∈ E(Gi−1 ), then
there is a cycle C such that V (C) = V (Ce ) ∪ V (Cf ) where e ∈
/ C.
Proof of Claim. Specifically assume that w1 is the predecessor of w2 (without loss of generality) on
Cf , and replace Ce and Cf with w1 u1 Cu1 u2 w2 Cw2 w1 , resulting in a 2-factor C 0 such that |ECi 0 | <
|ECi |.
Suppose Cv 6= Ce , and consider v + , the successor of v on Cv . The claw hv; v + , u1 , u2 i in Gi−1
implies that without loss of generality v + u2 is an edge in Gi−1 , since e = u1 u2 ∈
/ E(Gi−1 ). However,
+
by Claim 2, we can replace Ce and Cv with u2 v Cv vu1 Ce u2 , contradicting the minimality of |ECi |.
Thus Cv = Ce , so we may apply Claim 1 as needed going forward.
We now proceed by considering distC (e, v).
4
Case 1: distC (e, v) ≥ 3.
By Claim 1, u2 v − ∈
/ E(Gi−1 ), u1 v + ∈
/ E(Gi−1 ), and u2 v + ∈ E(Gi−1 ). Considering the claw
hv; u1 , u2 , v − i we see that u1 v − ∈ E(Gi−1 ). We can now replace Cv with Cv1 = u1 Cv v − u1 and
Cv2 = vCv u2 v. If Cv is odd, this results in a 2-factor C 0 with |ECi 0 | < |ECi |. If Cv is even, then Cv1
and Cv2 have the same parity. If both are even, the minimality of |ECi | is contradicted. Otherwise,
replace Cv with u1 Cv vu1 and v + Cv u2 v + , to contradict the minimality of |ECi |.
Case 2: distC (e, v) = 2.
Without loss of generality, assume e = v −− v (−3) = u1 u2 . Note that since distC (e, v) = 2, Claim 1
implies v + v (−3) ∈ E(Gi−1 ).
We also claim that if x2 6= v − , then Cx2 6= Cv . Indeed, suppose otherwise and note that
− − −−
+
x2 Cv v (−3) v + Cv x−
if x−
2 or
2 v or x2 v is an edge of Gi−1 , then replacing Cv with either x2 vv v
+
+ − −−
0
i
i
−
+
(−3)
−
Cv x2 results in a 2-factor C such that |EC 0 | < |EC |. Also,
x2 vv v x2 Cv v v
that
this implies
+ and x+ 6= v (−3) since v cannot be adjacent to x+ or x− . Thus, since x ; v, x− , x+ is not
=
6
v
x−
2
2
2
2
2
2
2
+
− v −− x v and v + C x− x+ C v (−3) v +
∈
E(G
).
However,
exchanging
C
with
vv
x
induced, x−
i−1
v
2
v
2 2 v
2 2
contradicts the minimality of |ECi |. Therefore x2 ∈
/ V (Cv ) when x2 6= v − .
Case 2.1: ` = 3, so that in particular P = v −− x2 v (−3) = u1 x2 u2 .
By Claim 1, u2 = v (−3) is not adjacent to v − , so x2 6= v − , which implies that Cx2 6= Cv .
+ (−3)
−− . If
(−3) , v −− i cannot be induced in G
and x+
Next, hx2 ; x+
i−1 , so consider each of x2 v
2v
2 ,v
+ (−3)
+ −−
+
−−
x2 v
∈ E(Gi−1 ) or x2 v
∈ E(Gi−1 ), then replacing Cv and Cx2 with either x2 Cx2 x2 v Cv v (−3) x+
2
−− C v (−3) x C − x+ results in a 2-factor C 0 with at most k odd cycles such that |E i | < |E i |.
v
or x+
0
v
2
x
2
C
C
2 2
(−3) , v −− i is not induced in G
Thus, since hx2 ; x+
i−1 , P contains two internal vertices.
2 ,v
Case 2.2: Cx2 = Cv .
At the start of Case 2, we showed that if x2 6= v − , then Cv 6= Cx2 , so we can assume x2 = v −
and ` = 4. By Claim 1, x3 =
6 v + , since v − v + ∈
/ E(Gi−1 ).
We now claim that x3 is not on Cv . Indeed, assume otherwise and consider hx3 ; v (−3) , v − , x−
3 i.
(−3) x− C − vv −− v − x C v (−3) resulting in a 2∈
E(G
),
we
may
replace
C
with
v
If v (−3) x−
3 v
i−1
v
3 v
3
factor C 0 with odd(C 0 ) ≤ k such that |ECi 0 | < |ECi |. If v − x−
3 ∈ E(Gi−1 ), replacing Cv with
− + (−3) C − x vv −− v − similarly contradicts the minimality of |E i |. However, from Claim 1,
v − x−
v 3
3 Cv v v
C
−
(−3)
v v
∈
/ E(Gi−1 ). Thus x3 is not on Cv , so Cx3 6= Cv as claimed.
− (−3) . Since v − v (−3) is not an edge, one of x+ v − or x+ v (−3) is an
Consider the claw x3 ; x+
3
3
3 ,v ,v
− ∈ E(G
−− v − x+ C x v (−3) C − v. Otherwise,
edge in Gi−1 . If x+
v
),
replace
C
and
C
with
vv
i−1
v
x3
v
3
3 x3 3
(−3) ∈ E(G
−− v − x C − x+ v (−3) C − v. Both scenarios result
if x+
i−1 ), replace Cv and Cx3 with vv
3 x3 3
v
3v
in a 2-factor C 0 with at most k odd cycles such that |ECi 0 | < |ECi |, contradicting the minimality of
|ECi |.
Case 2.3: Cx3 = Cv .
In this case, we first show that x3 6= v + . Assume otherwise and consider hx2 ; v −− , v + , x−
2 i. If
− x contradicts the
∈ E(Gi−1 ), then replacing Cv and Cx2 with x2 v + Cv v (−3) vv − v −− x−
C
2 x2 2
+ x− C − x v −− v − vv (−3) C − v +
minimality of |ECi |. If v + x−
∈
E(G
),
then
by
replacing
C
and
C
with
v
i−1
v
x
v
2
2
2 x2 2
in C we contradict the minimality of |ECi |. As Claim 1 shows, v −− v + ∈
/ E(Gi−1 ), implying that
+
hx2 ; v −− , v + , x−
2 i is induced, a contradiction, thus x3 6= v .
v −− x−
2
Now that x3 6= v + , we use hv; v − , v (−3) , x2 i to show that v − x2 ∈ E(Gi−1 ) and v − x−
/ E(Gi−1 ).
2 ∈
5
−
− + −
From here, hx2 ; v − , x−
2 , x3 i is used to complete this subcase, as x3 x2 allows us to use hx3 ; x3 , x3 , x2 i
+ −
i
−
to show that x3 x3 ∈ E(Gi−1 ) which can be used to contradict |EC | and x3 v allows us to contradict
x2 ∈
/ Cv .
By Claim 1, v − v (−3) ∈
/ E(Gi−1 ). If x2v (−3) ∈ E(Gi−1 ), then v −− x2 v (−3) contradicts the mini
imality of P . Thus, since v; v − , v (−3) , x2 is not induced, x2 v − ∈ E(Gi−1 ). Also, if v − x−
2 ∈ EC ,
−
−
0
0
then replacing Cv and Cx2 with v − x−
2 Cx2 x2 vCv v results in a 2-factor C with odd(C ) ≤ k such
that distC 0 (e, v) > distC (e, v), a contradiction to the maximality of distC (e, v).
−
−
Since x2 ; v − , x−
2 , x3 is not induced, either x3 x2 or v x3 is an edge in Gi−1 . First, assume that
+ −
− −
+ − −
−− C x v (−3) C − x+
x3 x−
v 3
v 3
2 ∈ E(Gi−1 ). If x3 x2 or x3 x2 is an edge in Gi−1 , then either x3 x2 Cx2 x2 v
− −
− − + (−3) −
0 with at
−
−−
,
resulting
in
a
2-factor
C
C
and
C
or x3 Cv v v
Cv x3 vv v x2 Cx2 x2 x3 can replace
v
x
2
− +
− −
most k odd cycles such that |ECi 0 | < |ECi |. Since x3 ; x+
3 , x3 , x2 is not induced, x3 x3 ∈ E(Gi−1 ).
−
−− C x− x+ C v (−3) x x− contradicts the minimality
But then replacing Cv and Cx2 with x−
3 2
v 3 3 v
2 C x2 x 2 v
of |ECi |. This means that v − x3 ∈ E(Gi−1 ). Since we could have chosen P = v −− v − x3 v (−3) , from
Case 2.2, the minimality of |ECi | is contradicted, completing Case 2.3.
Case 2.4: Cx3 = Cx2 .
−
+ − −−
By Claim 2, x3 ∈
/ {x+
i.
2 , x2 }. Suppose Cx2 is a 4-cycle, and consider the claw hx2 ; x2 , x2 , v
− +
+ −
−−
−−
(−3)
may replace Cv and Cx2 , resulting in a 2x3 x2 x2 x2 v
If x2 x2 ∈ E(Gi−1 ), then v Cv v
−− (respectively
factor C 0 with at most k odd cycles such that |ECi 0 | < |ECi |, a contradiction. If x+
2v
− −−
+
x2 v ) is an edge in Gi−1 , then replace Cv and Cx2 with v −− Cv v (−3) x3 Cx2 x2 v −− (respectively
−− ) to contradict the minimality of |E i |. Thus, we can assume C
v −− Cv v (−3) x3 Cx−2 x−
x2 is not a
2v
C
4-cycle.
− +
+
−− C v (−3) x x v −−
If both x−
v
3 2
2 x2 and x3 x3 are edges in Gi−1 , then replacing Cv and Cx2 with v
− +
+
− +
0
0
i
i
and x2 Cx2 x3 x3 Cx2 x2 x2 yields a 2-factor C with odd(C ) ≤ k and |EC 0 | < |EC |.
+
−
−
+
First assume x−
/ E(Gi−1 ). Since x2 ; x+
2 x2 ∈
2 , x2 , v is not induced, one of x2 v or x2 v is
an
of Gi−1 . Assume without loss of generality that x−
2 v ∈ E(Gi−1 ). Consider the claw
edge
− − +
∈
E(G
/ E(Gi−1 ). If v + x−
v; x2 , v , v . By Claim 1, v − v + ∈
i−1 ), then replacing Cv and Cx2
2
− −
(−3)
−
+
−−
−
0
with vv
Cv v x2 Cx2 x2 v v v results in a 2-factor C with odd(C 0 ) ≤ k such that |ECi 0 | < |ECi |.
−
− −− x to again
If v − x2 ∈ E(Gi−1 ), then replace Cv and Cx2 with vCv v (−3) v and x2 Cx2 x−
2
2v v
− +
i
contradict the minimality of |EC |. However Gi−1 is claw-free, implying that x2 x2 ∈ E(Gi−1 ), so
+
x−
/ E(Gi−1 ).
3 x3 ∈
−
−
+
−
Since x+
/ E(Gi−1 ) and x3 ; x+
3 x3 ∈
3 , x3 , v is not induced, one of x3 v or x3 v is an edge in Gi−1 .
− + ∈
Without loss
of generality, assume that x+
/ E(Gi−1 ).
3 v ∈ E(Gi−1 ). Again, by Claim 1 v v
+ − +
+ −
+ +
Thus, since v; x3 , v , v
is not induced, either x3 v ∈ E(Gi−1 ) or x3 v ∈ E(Gi−1 ). If
−
− +
3 (−3) C − vv −− v − can replace C and C
x+
v
x2 to contradict the minv
3 v ∈ E(Gi−1 ), then v x3 Cx2 x v
+ +
i
imality of |EC |. Otherwise, if x3 v ∈ E(Gi−1 ), replacing Cv and Cx2 with x2 v −− v − vx2 and
− − + +
v + Cv v (−3) x3 Cx−2 x+
contradicts the minimality of |ECi |. This contradiction completes
2 x2 Cx2 x3 v
Case 2.4.
Case 2.5: Cx3 , Cx2 , and Cv are distinct.
−− i to show that x x+ ∈ E(G
In this case, we begin by using hx2 ; x3 , x+
3 2
i−1 ). This edge allows
2 ,v
− + (−3)
us to consider hx3 ; x3 , x2 , v
i. Since this claw is not induced in Gi−1 , we are able to use edges
in Gi−1 to contradict the minimality of |ECi |.
−− cannot be induced, x v −− , x+ v −− , or x x+ is an edge of G
Since x2 ; x3 , x+
3
3 2
i−1 . Note
2 ,v
2
6
that x3 v −− ∈
/ E(Gi−1 ), as otherwise v −− x3 v (−3) would contradict the minimality of P . Suppose
+ −−
x2 v
∈ E(Gi−1 ), and consider hv; x2 , v − , v (−3) i. If x2 v (−3) ∈ E(Gi−1 ), then v −− x2 v (−3) contradicts the minimality of P . By Claim 1 v − v (−3) ∈
/ E(Gi−1 ). This means x2 v − ∈ E(Gi−1 ). However,
+
−
−−
now replace Cv and Cx2 with v x2 Cx2 x2 v v −− and vCv v (−3) v to contradict the minimality of
|ECi |. Therefore x3 x+
2 ∈ E(Gi−1 ).
+ (−3)
+
+ (−3)
Consider hx3 ; x−
i. This cannot be induced, so at least one of x−
, or
3 , x2 , v
3 x2 , x2 v
− (−3)
− +
x3 v
must be an edge of Gi−1 . When x3 x2 ∈ E(Gi−1 ), replace Cv , Cx2 , and Cx3 with
− −
(−3) C − v −− x C − x+ , resulting in a 2-factor C 0 such that |E i | < |E i |. If x+ v (−3) ∈
x+
2 x2 2
v
2 x 3 C x3 x 3 v
2
C
C0
(−3) C − v −− x C − x+ to contradict the minimality of |E i |. Fiv
E(Gi−1 ), replace Cv and Cx2 with x+
2 x2 2
v
2
C
− (−3) − −−
− x+ x C x− ,
(−3) allows us to replace C , C , and C
with
x
v
C
v
x
C
nally, the edge x−
v
3
x3 3
2
x
v
x
v
x2 2
3
2
3
3
again contradicting the minimality of |ECi |. This contradiction completes the proof of Case 2.5,
which also completes Case 2.
Case 3: distC (e, v) = 1.
Without loss of generality, assume e = v − v −− . By Claim 1, v − v + ∈
/ E(Gi−1 ), so x2 6= v + .
Case 3.1: Cx2 = Cv .
If x2 = v ++ , then replacing Cv with v − x2 Cv v −− v + vv − results in a 2-factor C 0 with at most k
odd cycles such that |ECi 0 | < |ECi |, contradicting the minimality of |ECi |. Thus distCv (v, x2 ) ≥ 3.
+ −
Next we show that either hx2 ; x−
2 x2 , v i is induced or there is a 2-factor in Gi−1 which contradicts
+
− − + − −
the minimality of |ECi |. If x−
2 x2 ∈ E(Gi−1 ), then replacing Cv with vx2 v Cv x2 x2 Cv v results
+ −
− −
0
in a 2-factor C such that distC 0 (e, v) > distC (e, v). If x2 v or x2 v is an edge of Gi−1 then
− −
−− v or vC x v − x+ C v −− v contradicts the minimality of |E i |.
replacing
v 2
v with vC
v x 2 v x 2 Cv v
2 v
C
C−
− is induced, a contradiction. Therefore x ∈
,
v
/
V
(C
).
Thus x2 ; x2 x+
2
v
2
Case 3.2: ` = 3, so in particular P = v − x2 v −− = u1 x2 u2 .
−− , v − i is induced in G
By Claim 2, hx2 ; x+
i−1 , so P must have two internal vertices.
2 ,v
Case 3.3: Cx3 = Cv .
+ − . If
First we show that x3 6= v + . Indeed assume otherwise, and consider x2 ; x−
2 ,v ,v
+ − −
− −− C − v + to contradict the
v + x−
v
2 ∈ E(Gi−1 ), then we can replace Cv and Cx2 with v x2 Cx2 x2 v vv
−
i
−
− x− C − x v + C v −− vv −
minimality of |EC |. If v x2 ∈ E(Gi−1 ) then replacing Cv and Cx2 with
v
v
2 x2 2
+ , v − is induced in
contradicts the minimality of |ECi |. By Claim 1 v − v + ∈
/ E(Gi−1 ), so x2 ; x−
,
v
2
Gi−1 .
−
− −
We now use hx2 ; v − , x3 , x−
2 i to show that x2 x3 ∈ E(Gi−1 ). If x2 v ∈ E(Gi−1 ), then replace Cv
− − −− −
and Cx2 with vx2 Cx2 x2 v v Cv v to contradict the maximality of distC (e, v). If x3 v − ∈ E(Gi−1 ),
then v − x3 v −− contradicts the minimality of P . Thus, x−
2 x3 ∈ E(Gi−1 ).
− −
−− , x− i. If x− v −− ∈ E(G
−
−− x−
Finally, consider hx3 ; x−
i−1 ) replacing Cv and Cx2 with x2 Cx2 x2 v Cv v
3 ,v
2
2
2
− −−
0
0
i
i
results in a 2-factor C with odd(C ) ≤ k such that |EC 0 | < |EC |. If x3 v
∈ E(Gi−1 ), then replac− −−
−
−
ing Cv and Cx2 with v −− Cv − x3 x−
similarly contradicts the minimality of |ECi |.
2 Cx2 x2 v Cv x3 v
− −
− − −
−
Finally if x2 x3 ∈ E(Gi−1 ), then vCv x3 x2 Cx2 x2 v Cv− x3 v can replace Cv and Cx2 to contradict
−− , x− i cannot be induced, x ∈
the maximality of distC (e, v). Since hx3 ; x−
3 / Cv .
3 ,v
2
Case 3.4: Cx3 = Cx2 .
+ −
− +
−
First, considering hx2 ; x−
2 , x2 , v i, we show that x2 x2 ∈ E(Gi−1 ). By Claim 2, x3 6= x2 and
+
+
−
x3 6= x2 . Without loss of generality, v − x2 ∈
/ E(Gi−1 ) (respectively v − x2 ), as otherwise, replacing
7
−
Cv and Cx2 with v − x+
2 Cx2 x2 vCv v results in a 2-factor such that Cv = Cx2 , which was considered
− +
in Case 3.1, and so x2 x2 ∈ E(Gi−1 ).
−− i. By the minimality of P , x v −− ∈
Now consider the claw hx3 ; x+
/ E(Gi−1 ). If v −− x+
2
3 , x2 , v
3 ∈
− +
− C v −− results in a 2-factor C 0
x
x
v
x
x
C
C
E(Gi−1 ), then replacing Cv and Cx2 with v −− x+
3
2
v
x
x
2
2 2
2
3
with at most k odd cycles such that |ECi 0 | < |ECi |. Otherwise, if x2 x+
3 ∈ E(Gi−1 ), then replace Cv and
− +
+
−−
−
−
Cx2 with x2 x3 Cx2 x2 x2 Cx2 x3 v Cv v x2 to contradict the minimality of |ECi |. This contradiction
completes the proof of Case 3.4.
Case 3.5: Cx3 , Cx2 and Cv are distinct.
−
− + / E(G
First, consider the claw hx2 ; x+
i−1 ) as otherwise, replacing
2 , v , x3 i. As in Case 3.4, v x2 ∈
+
−
−
Cv and Cx2 with v x2 Cx2 x2 vCv v results in a 2-factor such that Cv = Cx2 , considered in Case
3.1. By the minimality of P , v − x3 ∈
/ E(Gi−1 ). So we have that x3 x+
2 ∈ E(Gi−1 ).
−− ∈
−− i to show that x+ v −− ∈ E(G
/
Next, we use hx3 ; x2 , x+
i−1 ). By the minimality of P , x2 v
3
3 ,v
+
+
+
E(Gi−1 ). If x3 x2 ∈ E(Gi−1 ), replacing Cx2 and Cx3 with x2 x3 Cx3 x3 x2 Cx2 x2 yields a 2-factor such
−− ∈ E(G
that Cx2 = Cx3 , which is Case 3.4. Thus, x+
i−1 ); however, we can now replace Cv ,
3v
+
+
−
−−
−−
Cx2 , and Cx3 with v x3 Cx3 x3 x2 Cx2 x2 v Cv v , resulting in a 2-factor C 0 such that |ECi 0 | < |ECi |.
This contradiction completes the proof of Case 3.5, and the proof of Theorem 5.
3
Disjoint 1-factors
As noted in the introduction, a 2-factor with no odd components is equivalent to a pair of disjoint
1-factors. Going forward we will refer to such a 2-factor as an even 2-factor. This section specifically
considers Problem 1 when k = 0.
Problem 2. Determine conditions in a graph G which guarantee that G has a pair of disjoint
perfect matchings.
In light of this problem, we have the following immediate corollary of Theorem 5 in the case
k = 0.
Corollary 1. Let G be a claw-free graph. Then G contains a pair of disjoint perfect matchings if
and only if cl(G) contains a pair of disjoint perfect matchings.
It is clear that if a graph G has an even 2-factor, then G has a 2-factor, so we will consider
conditions that ensure a graph has a 2-factor. In [5] Faudree, Faudree, and Ryjáček characterized
the pairs of forbidden subgraphs which guarantee that a large enough 2-connected graph has a
2-factor. In the results that follow, the generalized net N (i, j, k) is the graph obtained by associating one endpoint of each of the paths Pi+1 , Pj+1 and Pk+1 with distinct vertices of a triangle.
Following convention, we will let B(i, j) and Zi denote the generalized nets N (i, j, 0) and N (i, 0, 0),
respectively.
Theorem 6. Let X and Y be connected graphs with at least three edges, and let G be a 2-connected
graph of order n ≥ 10. Then, G being {X, Y }-free implies that G has a 2-factor if and only if, up
to the order of the pairs, X = K1,3 and Y is a subgraph of either P7 , Z4 , B(4, 1), or N (3, 1, 1), or
X = K1,4 and Y = P4 .
8
x
w
z
K2t+1
K2ℓ+1
y
C 1 − C 1[x, y]
C2
Figure 1: The family E of 2-connected {K1,4 , P4 }-free graphs that do not contain disjoint 1-factors.
We begin by considering {K1,4 , P4 }-free graphs. Let E be the family of graphs composed of two
odd cliques and an edge wz joined to two vertices x and y where x may or may not be adjacent
to y (see Figure 1). Every graph G in E is {K1,4 , P4 }-free and does not contain a pair of disjoint
1-factors, as the edge wz is in every perfect matching of G. Our next result demonstrates that
graphs in the family E are the only 2-connected {K1,4 , P4 }-free graphs with no even 2-factor. We
require the following lemma from [5].
Lemma 1. If G is a 2-connected {K1,4 , P4 }-free graph of order at least nine, then G has a 2-factor
with at most two components.
Theorem 7. Any 2-connected {K1,4 , P4 }-free graph of even order at least ten contains an even
2-factor or is a member of the family E.
Proof. Let G be a 2-connected {K1,4 , P4 }-free graph of even order n ≥ 10 that does not contain an
even 2-factor. Thus every 2-factor has at least two components and by Lemma 1 there is a 2-factor
F with exactly two components, call them C 1 and C 2 . Since G is 2-connected there are at least
two disjoint edges, e1 and e2 , between C 1 and C 2 . Let e1 = xx0 and e2 = yy 0 , where x, y ∈ C 1 and
x0 , y 0 ∈ C 2 . If xy ∈ E(C 1 ) and x0 y 0 ∈ E(C 2 ), then C 1 and C 2 can be combined into a single cycle
xC 1 yy 0 C 2− x0 x. Since G is P4 -free, G[{x− , x, x0 , x0+ }] is not an induced P4 . If x− x0+ is an edge,
then xC 1 x− x0+ C 2 x0 x is an even 2-factor in G, so without loss of generality, xx0+ ∈ E(G). Similarly,
since G[{x− , x, x0+ , x0++ }] is not an induced P4 , xx0++ ∈ E(G). A similar argument shows that
there is an induced P4 for each edge from x to C 2 unless x dominates C 2 . Now, if xy ∈ E(C 1 ),
then replace C 1 and C 2 with yC 1 xy 0+ C 2 y 0 y, so y − 6= x. For the same reason that x dominates C 2 ,
y must also dominate C 2 . (Note that if y 0 dominated C 1 instead, then we could combine C 1 and
C 2 .)
Let D be the set of vertices in C 1 that dominate C 2 . There can be no two edges f1 = uv ∈ E(C 1 )
and f2 = u0 v 0 ∈ E(C 2 ) such that uu0 ∈ E(G) and vv 0 ∈ E(G) lest C 1 and C 2 be combined, and
there are at least two vertices in D, so |V (C 1 )| ≥ 4 which implies that |V (C 1 )| ≥ 5 since |V (C 1 )|
is odd. If for some vertex v ∈ D, the edge v − v + is in G, then C 1 can be shortened by using the
edge v − v + to skip v, and C 2 can be extended by including v, forming an even 2-factor. Thus no
such edge exists. Since for any pair of vertices v1 and v2 in C 2 , the graph hx; x− , x+ , v1 , v2 i is not
induced, the edge v1 v2 exists. Thus V (C 2 ) forms a clique.
9
Suppose e1 and e2 were chosen such that C 1 (x, y) ∩ D = ∅. Let v be any vertex in C 1 (x, y − )
such that xv is an edge of G. Then the edge xv + is in G since G[{x0 , x, v, v + }] is not an induced P4 .
Now x dominates C 1 (x, y) and similarly y dominates C 1 (x, y). Let i be the smallest integer such
that x(+i) ∈ C 1 (x, y) does not dominate
C 1 (x, y), and let j be the smallest integer such that y (−j) is
(+i)
0
−
not adjacent to x
. Consider x; x , x , x(+i) , y (−j) . If either x− x(+i) or x− y (−j) is an edge, then
x− x(+i) C 1 y − x(+i−1) C 1− xx0 C 2 x0− yC 1 x− or x− y (−j) C 1 y − x(+i) C 1 y (−j−1) x(+i−1) C 1− xx0 C 2 x0− yC 1 x−
is a hamiltonian cycle. If x0 x− ∈ E(G), x0 x− C 1− xx0+ C 2− x is a hamiltonian cycle, and similarly if
x0 x+i ∈ E(G) or x0 y −j ∈ E(G), then x0 x+i C 1− x+ x+i+1 C 1− xx0+ C 2− x0 or x0 y −j C 1− x+ y −j+1 C 1− xx0+ C 2− x0
is a hamiltonian cycle. Consequently, no such i exists and G[C 1 (x, y)] forms a clique. If |D| = 2,
then this argument can be repeated to show that G[C 1 (y, x)] is also a clique.
Suppose |D| ≥ 3 and let z ∈ D such that C 1 (x, z) ∩ D = {y}. As in the case that |D| = 2,
G[C 1 (y, z)] forms a clique.
If there is an edge between C 1 (x, y) and C 1 (y, z), then we may rearrange C 1 such that y − y + is
an edge and remove y from C 1 and add it to C 2 , and therefore no such edge exists. If x− y + ∈ E(G),
b 1 = y + C 1 x− y + and C
b 2 = xC 1 yx0− C 2− x0 x. Consider
then C 1 [x, y] can be pulled into C 2 to form C
the path G[{x−− , x− , x, x+ }]. From before, x− x+ is not an edge, and x−− x+ allows us to replace
b 1 and C
b 2 with x−− x+ C
b 2− xx− C
b 1 x−− , so xx−− must be an edge. Iterating this argument, we get
C
b 1 . However, since z dominates C 2 , zx0 ∈ E(G) and the cycle zx0 C
b 2 xz + C
b1 z
that x dominates C
−
+
spans G and so is an even 2-factor. Thus, if x y ∈ E(G), then |D| = 2.
Now if xy ∈ E(G), then G[{x− , x, y, y + }] is an induced P4 , otherwise consider G[{x, y − , y, y + }].
If xy + ∈ E(G), then hx; x− , x+ , y + , x0 i is induced. Thus, G[{x, y − , y, y + }] is an induced P4 unless
|D| = 2.
As shown above, |C 1 (x, y)| > 0. Since C 1 is an odd cycle, either |C 1 (x, y)| or |C 1 (y, x)| is
even, so without loss of generality, assume that |C 1 (x, y)| is even. If |C 1 (x, y)| ≥ 4, then C 1 [y, x),
C 1 (x, y), and C 2 + x form an even 2-factor. If |C 1 (x, y)| = 2, then G is a member of E.
Turning our attention to connected, but not necessarily 2-connected, graphs, Fujisawa and Saito
[6] showed the following.
Theorem 8. Let F1 and F2 be connected graphs of order at least three. Then there exists a positive
integer n0 such that every connected {F1 , F2 }-free graph of order at least n0 and minimum degree
at least two has a 2-factor if and only if without loss of generality F1 is a subgraph of K1,3 and F2
is a subgraph of Z2 .
We give the following related result for disjoint 1-factors.
Theorem 9. Every connected {K1,3 , Z2 }-free graph of even order with minimum degree δ(G) =
δ ≥ 3 contains an even 2-factor with at most two components.
Proof. Let G be a connected {K1,3 , Z2 }-free graph of even order. Gould and Jacobson [7] showed
that every 2-connected {K1,3 , Z2 }-free graph is hamiltonian, so we need only consider the case
where G has a cut vertex x. Since G is K1,3 -free, x must lie in exactly two blocks, so let B1 and
B2 be the blocks containing x.
Since G is claw-free, the neighborhood of any vertex is connected or two disjoint cliques,
so G[NB1 (x)] and G[NB2 (x)] are cliques. Further, since d(x) ≥ 3, without loss of generality,
10
|NB1 (x)| ≥ 2. Let v1 , v2 ∈ NB1 (x) and u ∈ NB2 (x). Suppose that NB2 (x) 6= V (B2 ), and let
w ∈ V (B2 )\N (x) such that uw ∈ E(G). We know that there is such a w since G is connected. However, G[{x, v1 , v2 , u, w}] is an induced Z2 . Thus NB2 (x) = V (B2 ), and similarly NB1 (x) = V (B1 ).
Finally, suppose there is some third block, B3 , which necessarily cannot contain x. Without loss
of generality, let v ∈ B2 ∩ B3 be a cut vertex of G. Let u ∈ NB3 (v) and x be the only cut vertex of
B1 . This means that since δ(G) ≥ 3, |V (B1 )| ≥ 4 and x is adjacent to v. Now, let x1 , x2 ∈ V (B1 )
such that x1 6= x and x2 6= x. Then G[{x1 , x2 , x, v, u}] is an induced Z2 , a contradiction.
Thus G consists of blocks B1 and B2 , both of which are complete. Since G has even order, one
block has even order and the other odd order. Without loss of generality, let B1 have odd order.
Then B1 − {x} and B2 are even 2-connected {K1,3 , Z2 }-free graphs so are hamiltonian, and this
yields an even 2-factor of G.
Considering either a clique of any size with a pendant edge or the graph obtained by identifying
a vertex in a clique of any size and a vertex of K3 , the minimum degree condition in our result
cannot be weakened in order to still guarantee an even 2-factor in the graph.
3.1
Minimum Degree and Connectivity Conditions
There are a number of results that give minimum degree and connectivity conditions implying a
claw-free graph has a 2-factor. Egawa and Ota [4] and Choudum and Paulraj [3] independently
showed that a claw-free graph with minimum degree at least four has a 2-factor. Subsequently,
Yoshimoto [24] and Aldred, Egawa, Fujisawa, Ota and Saito [2] demonstrated that δ(G) ≥ 3 suffices
when G is 2-connected and claw-free.
We show next that there are no such conditions that ensure a 3-connected claw-free graph of
even order contains an even 2-factor. Let P be the Petersen graph, and define Pδ as the graph
obtained by replacing each vertex of P with a clique of order δ + 1. The three incident edges at
one vertex v of P are now incident with three different vertices in the clique replacing v in Pδ . (See
Figure 2 for P4 .)
Figure 2: The graph P4 .
Theorem 10. For even δ ≥ 4, Pδ is a 3-connected K1,3 -free graph with minimum degree δ and no
even 2-factor.
11
Proof. It is clear that Pδ is 3-connected since P is 3-connected. The neighborhood of any vertex is
either a clique or two disjoint cliques, thus Pδ is K1,3 -free.
Consider a perfect matching M in Pδ , and a perfect matching M 0 that is disjoint from M . Let
FP ⊂ E(Pδ ) be the set of edges that were originally edges of P (these are the edges that are not
completely contained in a single clique of order δ + 1). Each clique is odd, so M must contain
B ⊆ FP where each clique contains one or three vertices of V (B). If V (B) has three vertices in a
single clique C, then there is no M 0 , since Pδ − M has C as an odd component. Therefore, V (B)
contains exactly one vertex in each clique, and corresponds to a perfect matching in P. However,
any perfect matching in P leaves two components, each isomorphic to C5 . The corresponding
components in Pδ are copies of C5 with vertices replaced by odd cliques. These two components
are odd (of order at least 5(δ + 1)), and so M 0 cannot exist. Thus, Pδ does not contain an even
2-factor.
4
Open Problems
Theorem 10 implies that no minimum degree condition on 1-, 2-, and 3-connected claw-free graphs
will ensure the existence of an even 2-factor. For 4-connected claw-free graphs, we look to the
well-known conjecture of Matthews and Sumner [17].
Conjecture 1 (The Matthews Sumner Conjecture). If G is a 4-connected claw-free graph, then G
is hamiltonian.
A hamiltonian cycle in a graph of even order is an even 2-factor. We therefore make the following
conjecture.
Conjecture 2. If G is a 4-connected, claw-free graph of even order, then G has an even 2-factor
if and only if G is hamiltonian.
We also conjecture the following extension of Theorem 3, and in particular Corollary 1.
Conjecture 3. If G is a k-connected, claw-free graph of even order at least 2k, then G has k
disjoint 1-factors if and only if cl(G) has k disjoint 1-factors.
For k ≥ 3, the condition that G be k-connected in Conjecture 3 is necessary. To see this, consider
the graph G obtained by connecting a vertex v to k − 1 vertices in an odd clique of order at least
2k − 1. The closure of G is complete, but v does not have sufficient degree in G to be in k disjoint
1-factors.
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