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Discrete Applied Mathematics
journal homepage:www.elsevier.com/locate/dam
Disjoint cycles in hypercubes with prescribed vertices in
each cycle
✩Cheng-Kuan Lin
a, Jimmy J.M. Tan
a, Lih-Hsing Hsu
b, Tzu-Liang Kung
c,∗ aDepartment of Computer Science, National Chiao Tung University, Hsinchu 30010, Taiwan, ROCbDepartment of Computer Science and Information Engineering, Providence University, Taichung 43301, Taiwan, ROC cDepartment of Computer Science and Information Engineering, Asia University, Taichung 41354, Taiwan, ROC
a r t i c l e i n f o Article history:
Received 6 August 2012
Received in revised form 24 June 2013 Accepted 2 July 2013
Available online 2 August 2013
Keywords: Spanning cycle Hamiltonian cycle Cyclable Hypercube Graph
a b s t r a c t
A graph G is spanning r-cyclable of order t if for any r nonempty mutually disjoint vertex subsets A1,A2, . . . ,Ar of G with|A1∪A2∪ · · · ∪Ar| ≤ t, there exist r disjoint cycles C1,C2, . . . ,Crof G such that C1∪C2∪ · · · ∪Crspans G, and Cicontains Aifor every i. In
this paper, we prove that the n-dimensional hypercube Qnis spanning 2-cyclable of order n−1 for n≥3. Moreover, Qnis spanning k-cyclable of order k if k ≤n−1 for n≥ 2.
The spanning r-cyclability of a graph G is the maximum integer t such that G is spanning r-cyclable of order k for k=r,r+1, . . . ,t but is not spanning r-cyclable of order t+1. We also show that the spanning 2-cyclability of Qnis n−1 for n≥3.
© 2013 Elsevier B.V. All rights reserved.
1. Introduction
For those graph definitions and notations not defined here, we follow the standard terminology given in [12]. A pair of two sets G
=
(
V,
E)
is a graph if V is a finite set and E is a subset of{
(
a,
b) | (
a,
b)
is an unordered pair of elements of V}
. We say that V=
V(
G)
is the vertex set, and E=
E(
G)
is the edge set. Two vertices u andv
are adjacent if(
u, v) ∈
E. The neighborhood of vertex u in G, denoted by NbdG(
u)
, is the set{
v ∈
V|
(
u, v) ∈
E}
. The degree of u in G, denoted by degG(
u)
, is|
NbdG(
u)|
. A path is a sequence of adjacent vertices, written as⟨
v
0, v
1, . . . , v
m⟩
, in which all the verticesv
0, v
1, . . . , v
maredistinct except that possibly
v
0=
v
m.A cycle of a graph G is a path with at least three vertices such that the first vertex is the same as the last one. A hamiltonian cycle is a spanning cycle in a graph. Until the 1970s, the interest in hamiltonian cycles had long been centered on their relationship to the 4-color problem. Recently, some refined conditions for a graph to be hamiltonian were proposed by researchers [8,17,18], and the study of hamiltonian cycles in general graphs has been fueled by the issue of computational complexity and practical applications. Furthermore, a number of variations were developed and research efforts have been dedicated to pancyclicity [4,9], super spanning connectivity [1,6,19,20], k-ordered hamiltonicity [17], and hamiltonian decomposition [2,21,22] among many other areas. In particular, hamiltonian cycles are a major requirement to design effective interconnection networks [12,14,25,26].
There are several directions of research based on the hamiltonian property. One direction involves the spanning property of cycles. For example, a 2-factor of a graph G is a spanning 2-regular subgraph of G; that is, G has a 2-factor if it can be
✩This work was supported in part by the National Science Council of the Republic of China under Contracts 97-2221-E-126-001-MY3 and
101-2221-E-468-018.
∗Corresponding author. Fax: +886 4 23305737.
E-mail address:tlkung@asia.edu.tw(T.-L. Kung).
0166-218X/$ – see front matter©2013 Elsevier B.V. All rights reserved.
a
b
Fig. 1. Illustration forExamples 1and2.
decomposed into several disjoint cycles. This notion can be applied to identify faulty units in a multiprocessor system. In particular, Fujita and Araki [7] proposed a three-round adaptive diagnosis algorithm by decomposing the hypercube into a fixed number of disjoint cycles such that the length of each cycle is not too small. The other direction addresses the cyclability of a graph G. Let S be a subset of V
(
G)
. Then, S is cyclable in G if there exists a cycle C of G such that S⊆
V(
C)
. Many results of cyclability are known [3,5,11,13,23]. In this paper, we study a new property which is a mixture of these two directions.Now, we extend the concept behind hamiltonian graphs and consider two or more cycles spanning a whole graph. Let
A1
,
A2, . . . ,
Ar be mutually disjoint nonempty vertex subsets of a graph G. Then G is cyclable with respect to A1,
A2, . . . ,
Arif there exist mutually disjoint cycles C1
,
C2, . . . ,
Cr of G such that Cicontains Aifor every i. Obviously, a graph is unlikelyto be cyclable with respect to any r mutually disjoint vertex subsets if r
≥
2. For example, G cannot be cyclable with respect to A1= {
u, v}
and A2=
V(
G) − {
u, v}
for any two vertices u, v
of G. To make this notion more reasonable, we impose one restriction on the order of A1∪
A2· · · ∪
Ar. To be precise, G is r-cyclable of order t if it is cyclable with respectto A1
,
A2, . . . ,
Ar for any r nonempty mutually disjoint subsets A1,
A2, . . . ,
Ar of V(
G)
such that|
A1∪
A2∪ · · ·
Ar| ≤
t. Inaddition, if C1
∪
C2∪ · · · ∪
Crspans G, then G is spanning r-cyclable of order t. Here we have two parameters r and t. We canfix one of them and find the optimal value for the other. The (spanning) r-cyclability of G is t if G is (spanning) r-cyclable of order k for k
=
r,
r+
1, . . . ,
t but is not (spanning) r-cyclable of order t+
1. On the other hand, the (spanning) cyclability of G of order t is r if G is (spanning) k-cyclable of order t for k=
1,
2, . . . ,
r but is not (spanning)(
r+
1)
-cyclable of ordert. According to the presented notion, the problem of finding hamiltonian cycles focuses on r
=
1. It is also noticed that not only is the set of disjoint spanning cycles of G a 2-factor, but also each cycle contains a designated vertex subset. Rather than 2-factors, the number of disjoint cycles is controlled. We give two examples to clarify the proposed notion.Example 1. Fig. 1(a) depicts the Petersen graph. Since the Petersen graph is not hamiltonian, it is not spanning 1-cyclable of any order. However, it is 1-cyclable of order 9. To see that the Petersen graph is spanning 2-cyclable of order 2, we assume that A1
= {
1}
and A2= {
i}
for i̸=
1. We set C1= ⟨
1,
2,
3,
4,
5,
1⟩
and C2= ⟨
6,
8,
10,
7,
9,
6⟩
if i∈ {
6,
7,
8,
9,
10}
; we setC1
= ⟨
1,
5,
4,
9,
6,
1⟩
and C2= ⟨
2,
3,
8,
10,
7,
2⟩
if i∈ {
2,
3}
; we set C1= ⟨
1,
2,
3,
8,
6,
1⟩
and C2= ⟨
4,
5,
10,
7,
9,
4⟩
ifi
∈ {
4,
5}
. Then C1and C2are two disjoint spanning cycles with A1⊂
V(
C1)
and A2⊂
V(
C2)
, respectively.Example 2. Let G be the graph shown inFig. 1(b). Obviously, G is hamiltonian. Thus, it is spanning 1-cyclable of order 10. However, as an example, it is not 2-cyclable with respect to A1
= {
i}
and A2= {
i+
5}
for i=
0,
1,
2,
3,
4. As a result, G is not spanning 2-cyclable of order 2.In this paper, we limit ourself by considering the n-dimensional hypercube Qn as the underlying graph and study its
spanning 2-cyclability. We have the following results: (1) for n
≥
3, Qnis spanning 2-cyclable of order n−
1; (2) Qnisspanning k-cyclable of order k if k
≤
n−
1 for n≥
2.2. Properties of hypercubes
Let u
=
unun−1. . .
u2u1be an n-bit binary string. The Hamming weight of u, denoted byw(
u)
, is the number of indicesi, 1
≤
i≤
n, such that ui=
1. Let u=
unun−1. . .
u2u1and v=
v
nv
n−1. . . v
2v
1be two n-bit binary strings. The Hammingdistance h
(
u,
v)
between u and v is the number of different bits in the corresponding strings. The n-dimensional hypercube, denoted by Qnfor n≥
1, consists of all n-bit binary strings as its vertices, and two vertices u and v are adjacent if and only ifh
(
u,
v) =
1. Obviously, Qnis a bipartite graph with bipartition W= {
u∈
V(
Qn) | w(
u)
is even}
and B= {
u∈
V(
Qn) | w(
u)
is odd
}
. For i=
0,
1, let Qindenote the subgraph of Qninduced by
{
u=
unun−1. . .
u2u1|
un=
i}
. Obviously, Qniis isomorphicto Qn−1with n
≥
2. For any vertex u=
unun−1. . .
u2u1of Qn, we use(
u)
jto denote the bit uj, where 1≤
j≤
n. Moreover,we use
(
u)
kto denote the vertex v=
v
n
v
n−1. . . v
2v
1with ui=
v
ifor 1≤
i̸=
k≤
n andv
k=
1−
uk.The hypercube Qn is one of the most popular interconnection networks for parallel computer/communication
systems [16]. In the following, we discuss some properties of the hypercube that will be used in this paper. First,Theorem 1states that Qnis hamiltonian laceable and hyper-hamiltonian laceable.
Theorem 1 ([10,25]). Assume that n is any positive integer with n
≥
2. Then there exists a hamiltonian path of Qnjoining anytwo vertices from different partite sets. Moreover, there exists a hamiltonian path of Qn
− {
x}
joining y to z if x is in one partiteset whereas y and z are in the other partite set.
In particular,Lemmas 1and2indicate that Qn
− {
w,
b}
remains hamiltonian laceable whenever w and b are vertices indifferent partite sets.
Lemma 1 ([24]). Let n be any positive integer with n
≥
4. Let W and B form the bipartition of Qn. Assume that x and w are anytwo different vertices in W , whereas y and b are any two different vertices in B. Then there exists a hamiltonian path of Qn
−{
w,
b}
joining x and y.
Lemma 2 ([14]). Let n be any positive integer with n
≥
4. Assume that w and b are any two adjacent vertices of Qn, and F isany edge subset of Qn
− {
w,
b}
with|
F| ≤
n−
3. Then there exists a hamiltonian path of(
Qn− {
w,
b}
) −
F joining any twovertices from different partite sets.
Theorem 2generalizes the fault-tolerance of hamiltonian laceability for Qn, andTheorem 3gives two types of
2-disjoint-path cover in Qn.
Theorem 2 ([24]). Assume that n
≥
3. Let Fvbe a union of fvdisjoint pairs of adjacent vertices in Qn, and let Febe a set consistingof feedges in Qnwith fv
+
fe≤
n−
3. Then there exists a hamiltonian path of Qn−
(
Fv∪
Fe)
joining any two vertices fromdifferent partite sets. Moreover, there exists a hamiltonian path of Qn
−
(
Fv∪
Fe∪ {
x}
)
joining y and z if x is in one partite set,and y
,
z are in the other partite set.Theorem 3 ([15]). Let n be any positive integer with n
≥
4. Let W and B form the bipartition of Qn. Assume that x and w areany two different vertices in W , y and b are any two different vertices in B. There are two disjoint paths P1and P2in Qnsuch
that
(
1)
P1is a path of length 2n−1−
1 joining x and y,(
2)
P2is a path of length 2n−1−
1 joining w and b, and(
3)
P1∪
P2spansQn. Moreover, there are two disjoint paths P3and P4in Qnsuch that
(
1)
P3is a path joining x and w,(
2)
P4is a path joining yand b, and
(
3)
P3∪
P4spans Qn.In the rest of this section, we apply the results introduced above to proveLemmas 3and4, which specify 2-disjoint-path covers in Qnthat are able to contain the prescribed vertices. The two lemmas will be used in the proof ofLemma 5, which is
a key result presented in the next section for deriving the spanning 2-cyclability of Qn.
Lemma 3. Let W and B form the bipartition of Qnwith n
≥
4. Suppose that x and u are two different vertices in W , whereas yand v are two different vertices in B. Let S be any nonempty subset of V
(
Qn) − {
x,
y,
u,
v}
with|
S| ≤
n−
3. Then there are twodisjoint paths P1and P2such that
(
1)
P1joins x to y,(
2)
P2joins u to v,(
3)
S⊆
P1, and(
4)
P1∪
P2spans Qn.Proof. We prove this lemma by induction on n. We describe inAppendix Athat this lemma holds for n
=
4. Since Qnisvertex-transitive and edge-transitive, we assume, without loss of generality, that x is in Qn0, and y is in Qn1. For i
∈ {
0,
1}
, we set Wi=
W∩
V(
Qni)
, Bi=
B∩
V(
Qni)
, and Si=
S∩
V(
Qni)
. We have the following cases.Case 1:
|
S0| ≥
1 and|
S1| ≥
1. Thus,|
S0| ≤
n−
4 and|
S1| ≤
n−
4.Subcase 1.1: Both u and v are in Qi
nfor some i
∈ {
0,
1}
. Without loss of generality, we assume that both u and v are in Qn0.Since
|
B0| =
2n−2> (
n−
3) ≥ |
S0∪ {
v}|
for n≥
5, we can choose any vertex b from B0−
(
S0∪ {
v}
)
. By induction, there are two disjoint paths R1and R2in Qn0such that (1) R1joins x to b, (2) R2joins u to v, (3) S0⊆
R1, and (4) R1∪
R2spans Qn0. By Theorem 1, there is a hamiltonian path H of Q1n joining
(
b)
nto y. We set P1= ⟨
x,
R1,
b, (
b)
n,
H,
y⟩
and P2=
R2. Obviously,P1and P2form the desired paths. SeeFig. 2(a).
Subcase 1.2: u is in Q0
n, and v is in Qn1. We set T
= {
p∈
V(
Qn0) | (
p)
n∈
S1}
. Obviously,|
S0∪
T| ≤ |
S0| + |
T| = |
S0| + |
S1| =
|
S| ≤
n−
3. Since|
B0−
(
S0∪
T)| ≥ |
B0| − |
S0∪
T| ≥
2n−2−
(
n−
3) ≥
2 for n≥
5, we can choose two distinct vertices b1and b2in B0−
(
S0∪
T)
. By induction, there are two disjoint paths R1and R2in Qn0such that (1) R1joins x to b1, (2) R2 joins u to b2, (3) S0⊆
R1, and (4) R1∪
R2spans Qn0. Moreover, there are two disjoint paths H1and H2in Qn1such that (1)H1joins
(
b1)
nto y, (2) H2joins(
b2)
nto v, (3) S1⊆
H1, and (4) H1∪
H2spans Qn1. We set P1= ⟨
x,
R1,
b1, (
b1)
n,
H1,
y⟩
andP2
= ⟨
u,
R2,
b2, (
b2)
n,
H2,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 2(b).Subcase 1.3: u is in Q1
n, and v is in Qn0. We set T
= {
p∈
V(
Qn0) | (
p)
n∈
S1}
. Similar to that shown in Subcase 1.2, we have|
B0−
(
S0∪
T∪ {
(
u)
n}
)| ≥
1 and|
W0−
(
S0∪
T∪ {
x, (
y)
n}
)|
ge1. Thus, there exists at least one vertex b in B0−
(
S0∪
T∪ {
(
u)
n}
)
, and there exists at least one vertex w in W0−
(
S0∪
T∪ {
x, (
y)
n}
)
. By induction, there are two disjoint paths R1and R2inQn0such that (1) R1joins x to b, (2) R2joins w to v, (3) S0
⊆
R1, and (4) R1∪
R2spans Qn0. Moreover, there are two disjointpaths H1and H2in Qn1such that (1) H1joins
(
b)
nto y, (2) H2joins u to(
w)
n, (3) S1⊆
H1, and (4) H1∪
H2spans Qn1. We setP1
= ⟨
x,
R1,
b, (
b)
n,
H1,
y⟩
and P2= ⟨
u,
H2, (
w)
n,
w,
R2,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 2(c).Case 2: Either
|
S0| =
0 or|
S1| =
0. Without loss of generality, we assume that|
S0| =
0.Subcase 2.1: Both u and v are in Qn0. Let b be any vertex in B0
− {
v}
. ByTheorem 3, there are two disjoint paths R1and R2 in Qn0such that (1) R1joins x to b, (2) R2joins u to v, and (3) R1∪
R2spans Qn0. ByTheorem 1, there is a hamiltonian patha
b
c
Fig. 2. Illustration for Case 1 ofLemma 3.
a
d
e
b
c
Fig. 3. Illustration for Case 2 ofLemma 3.
H of Qn1joining
(
b)
nto y. We set P1= ⟨
x,
R1,
b, (
b)
n,
H,
y⟩
and P2=
R2. Obviously, P1and P2form the desired paths. SeeFig. 3(a).
Subcase 2.2: Both u and v are in Qn1. Since
|
W1|
>
degQn1(
v) =
n−
1>
n−
2≥ |
S∪ {
u}|
, there exists a vertex w inW1
−
(
S∪ {
u}
)
such that(
v,
w) ∈
E(
Qn)
. Since|
B1| =
2n−2>
n−
3≥ |
S1∪ {
(
x)
n}|
for n≥
5, there exists a vertex b inB1
−
(
S1∪ {
(
x)
n}
)
. ByTheorem 2, there exists a hamiltonian path H of Qn1− {
u,
v,
w}
joining b to y. ByTheorem 3, there aretwo disjoint paths R1and R2in Qn0such that (1) R1joins x to
(
b)
n, (2) R2joins(
u)
nto(
w)
n, and (3) R1∪
R2spans Qn0. We setP1
= ⟨
x,
R1, (
b)
n,
b,
H,
y⟩
and P2= ⟨
u, (
u)
n,
R2, (
w)
n,
w,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 3(b).Subcase 2.3: u is in Q0
n, and v is in Qn1. Obviously, there exists a vertex w1in W1
−
S1such that(
v,
w1) ∈
E(
Qn1)
. Let w2bea vertex in W1
− {
w1}
. ByTheorem 2, there exists a hamiltonian path H of Qn1− {
v,
w1}
joining w2to y. ByTheorem 3, thereare two disjoint paths R1and R2in Qn0such that (1) R1joins x to
(
w2)
n, (2) R2joins u to(
w1)
n, and (3) R1∪
R2spans Qn0. Weset P1
= ⟨
x,
R1, (
w2)
n,
w2,
H,
y⟩
and P2= ⟨
u,
R2, (
w1)
n,
w1,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 3(c).Subcase 2.4: u is in Q1
n, and v is in Qn0.
Suppose that
(
u,
v) ∈
E(
Qn)
. Let w be any vertex in W0. ByTheorem 1, there exists a hamiltonian path R1of Qn0− {
v}
joining x to w, and there exists a hamiltonian path R2of Qn1
− {
u}
joining(
w)
nto y. We set P1= ⟨
x,
R1,
w, (
w)
n,
R2,
y⟩
andP2
= ⟨
u,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 3(d).Suppose that
(
u,
v) ̸∈
E(
Qn)
. Let w be any vertex in W0− {
x, (
y)
n}
. ByTheorem 3, there exist two disjoint paths R1and R2 in Q0nsuch that (1) R1joins x to w, (2) R2joins
(
u)
nto v, and (3) R1∪
R2spans Qn0. ByTheorem 1, there exists a hamiltonianpath H of Q1
n
− {
u}
joining(
w)
nto y. We set P1= ⟨
x,
R1,
w, (
w)
n,
H,
y⟩
and P2= ⟨
u, (
u)
n,
R2,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 3(e).Lemma 4. Let W and B form the bipartition of Qnwith n
≥
5. Let p, x, and y be three different vertices in W , and let q, u,and v be three different vertices in B such that
{
(
p,
q), (
x,
u), (
x,
v)} ⊂
E(
Qn)
. Then there exist two disjoint paths P1and P2inQn
− {
p,
q}
such that(
1)
P1joins x to y,(
2)
P2joins u to v, and(
3)
P1∪
P2spans Qn− {
p,
q}
.Proof. Since n
≥
5, there exists an integer 1≤
k≤
n such that q̸=
(
p)
k, u̸=
(
x)
k, and v̸=
(
x)
k. By the symmetricproperty of Qn, we can assume k
=
n. Without loss of generality, we consider that both p and q are in Qn0. For i∈ {
0,
1}
, weset Wi
=
W∩
V(
Qni)
and Bi=
B∩
V(
Qni)
. Note that{
x,
u,
v} ⊂
V(
Qni)
for some i∈ {
0,
1}
. We have the following cases.Case 1:
{
x,
u,
v} ⊂
V(
Q0n
)
and y∈
V(
Qn1)
. ByTheorem 2, there exists a hamiltonian path R of Qn0− {
p,
q,
x}
joining uand v. ByTheorem 1, there exists a hamiltonian path H of Qn1joining
(
x)
nand y. We set P1
= ⟨
x, (
x)
n,
H,
y⟩
and P2=
R. Obviously, P1and P2form the required paths. SeeFig. 4(a).Case 2: y
∈
V(
Qn0)
and{
x,
u,
v} ⊂
V(
Qn1)
. Since|
B0| =
2n−2>
2, there exists a vertex b in B0− {
q, (
x)
n}
. ByTheorem 2, there exists a hamiltonian path R of Qn0− {
p,
q}
joining b and y. ByTheorem 3, there exist two disjoint paths H1and H2ina
b
c
d
Fig. 4. Illustration forLemma 4.
Q1
n such that (1) H1joins x and
(
b)
n, (2) H2joins u to v, and (3) H1∪
H2spans Qn1. We set P1= ⟨
x,
H1, (
b)
n,
b,
R,
y⟩
andP2
=
H2. Obviously, P1and P2form the required paths. SeeFig. 4(b).Case 3:
{
x,
y,
u,
v} ⊂
V(
Qn0)
. ByTheorem 2, there exists a hamiltonian path R of Qn0− {
p,
q,
u}
joining x and y. Without loss of generality, we write R= ⟨
x,
R1,
w,
v,
z,
R2,
y⟩
. ByTheorem 1, there exist two disjoint paths H1and H2in Qn1such that(1) H1joins
(
w)
nand(
z)
n, (2) H2joins(
u)
nto(
v)
n, and (3) H1∪
H2spans Qn1. We set P1= ⟨
x,
R1,
w, (
w)
n,
H1, (
z)
n, z,
R2,
y⟩
and P2= ⟨
u, (
u)
n,
H2, (
v)
n,
v⟩
. Obviously, P1and P2form the required paths. SeeFig. 4(c).Case 4:
{
x,
y,
u,
v} ⊂
V(
Q1n
)
. Obviously, either u̸=
(
p)
nor v̸=
(
p)
n. Without loss of generality, we assume that u̸=
(
p)
n.Since degQ1
n
(
v) >
3, there exists a vertex z in W1− {
x,
y, (
q)
n
}
such that(
z,
v) ∈
E(
Qn
)
. ByTheorem 2, there exists ahamiltonian path H of Q1
n
− {
u,
v,
z}
joining x and y, and there exists a hamiltonian R of Qn0− {
p,
q}
joining(
u)
nand(
z)
n.We set P1
=
H and P2= ⟨
u, (
u)
n,
R, (
z)
n,
z,
v⟩
. Obviously, P1and P2form the required paths. SeeFig. 4(d). 3. Two disjoint cycles span hypercubesA bipartite graph G, with bipartition W and B, is called 2-disjoint-path-coverable of order t if for any
{
x,
u} ⊂
W ,{
y, v} ⊂
B, and any two disjoint subsets A1,
A2of V(
G) − {
x,
y,
u, v}
with|
A1∪
A2| ≤
t, there exists two disjoint pathsP1and P2of G such that (1) P1joins x and y, (2) P2joins u and
v
, (3) A1⊆
P1, (4) A2⊆
P2, and (5) P1∪
P2spans G. The following lemma is the key result to derive a tight lower bound of spanning 2-cyclability of Qn. Our proof idea is basedon constructing two disjoint paths that can span Qnand cover any two disjoint vertex subsets with the sum of orders not
exceeding n
−
3. The proof will be divided into various cases, each of which may consist of a number of subcases. To stress the main contribution of this paper, we thus defer those tedious details toAppendix Bfor the sake of clarity.Lemma 5. Suppose that n
≥
3. Then, Qnis 2-disjoint-path-coverable of order n−
3.The following theorem holds directly fromLemma 5.
Theorem 4. Assume that n
≥
4. Let A1and A2be any two disjoint vertex subsets of Qnwith|
A1∪
A2| ≤
n−
1. Then there existtwo disjoint cycles C1and C2of Qnsuch that
(
1)
A1⊆
C1(
2)
A2⊆
C2, and(
3)
C1∪
C2spans Qn.Proof. Without loss of generality, we consider
|
A1∪
A2| =
n−
1. There are two cases as follows.Case 1: Both A1 and A2are nonempty. Thus,
|
A1| ≤
n−
2 and|
A2| ≤
n−
2. Since|
A1| + |
A2| =
n−
1≥
3, we may assume, without loss of generality, that|
A1| ≥
2. Let u be a vertex in A2. Since degQn(
u) =
n>
n−
2≥ |
A1|
,there exists a vertex v in NbdQn
(
u) −
A1. (Note that it is possible that v is in A2.) Let x and x′be any two distinct vertices in A1. Since|
(
NbdQn(
x) ∪
NbdQn(
x′
)) − {
x,
x′}| ≥
2n
−
2>
n≥ |
A1∪
A2∪ {
v}|
for n≥
4, there exists a vertex y in(
NbdQn(
x) ∪
NbdQn(
x′
)) − (
A1
∪
A2∪ {
v}
)
. Without loss of generality, we assume that y∈
NbdQn(
x)
. Let A′
1
=
A1− {
x}
andA′2
=
A2− {
u,
v}
. Obviously,|
A′1∪
A′
2
| ≤
n−
3. ByLemma 5, there exist two disjoint paths P1and P2in Qnsuch that (1) P1 joins x and y, (2) P2joins u and v, (3) A1⊆
V(
P1)
, (4) A2⊆
V(
P2)
, and (5) P1∪
P2spans Qn. We set C1= ⟨
x,
P1,
y,
x⟩
andC2
= ⟨
u,
P2,
v,
u⟩
. Obviously, C1and C2form the required cycles in Qn.Case 2: A1or A2is empty. We can assume that A1is empty. First, we consider n
≥
5. Obviously, there exists a cycle C1of length 4 in Qnsuch that V(
C1) ∩
A2= ∅
. ByTheorem 2, there exists a hamiltonian cycle C2of Qn−
V(
C1)
. Then, we haveA2
⊆
C2.On the other hand, we consider n
=
4. Since Q4 is both vertex-symmetric and edge-symmetric, we assume that|
A2∩
V(
Q4i)| =
1 and|
A2∩
V(
Q1−i
4
)| =
2 with i∈ {
0,
1}
. For convenience, let A2∩
V(
Q4i) = {
s}
. Obviously, there exists a cycle C1of length 4 in Q4inot containing s. Moreover, Q4i−
V(
C1)
is a cycle of length 4, denoted by⟨
s,
t,
u,
v,
s⟩
. Then, we can find a hamiltonian path P of Q41−ijoining(
s)
4and(
t)
4. As a result, C2
= ⟨
s, (
s)
4,
P, (
t)
4,
t,
u,
v,
s⟩
and C1form the requested cycles.According toTheorem 4, Qnis spanning 2-cyclable of order n
−
1 for n≥
4. For Q3, let A1= {
x}
and A2= {
u}
, where x and u are different vertices of Q3. Since Q3is vertex-symmetric and edge-symmetric, we assume that x is in Q30, and u is inQ1
3. Clearly, both Q 0
3and Q31are isomorphic to Q2, which is a cycle of length 4. Thus, Q3is spanning 2-cyclable of order 2. We summarize the first main result of this paper as follows.
Corollary 1. The n-cube Qnis spanning 2-cyclable of order n
−
1 for n≥
3.To study the generalized spanning k-cyclability of Qnfor k
≥
3, we argue by induction that Qnis spanning k-cyclableof order k if k
≤
n−
1. Trivially, Q2is spanning 1-cyclable of order 1. As the inductive hypothesis, we assume that Qn−1 is spanning r-cyclable of order r for r≤
n−
2 with n≥
3. Let A= {
u1,
u2, . . . ,
uk}
consist of any k vertices of Qnwithk
≤
n−
1. By the symmetric property of Qn, we may assume that u1is in Qn0, and ukis in Qn1. We set Ai=
A∩
V(
Qni)
fori
∈ {
0,
1}
. Then, A is partitioned into two nonempty subsets A0and A1. Let t= |
A0|
. Without loss of generality, we may assume that ui∈
A0if 1≤
i≤
t, and ui∈
A1if t<
i≤
k. Note that Qniis isomorphic to Qn−1for i=
0,
1. By induction, there exist t disjoint cycles C1,
C2, . . . ,
Ctof Qn0such that uiis in Cifor 1≤
i≤
t and C1∪
C2∪ · · · ∪
Ctspans Qn0, and thereexist k
−
t disjoint cycles Ct+1,
Ct+2, . . . ,
Ckof Qn1such that uiis in Cifor t+
1≤
i≤
k and Ct+1∪
Ct+2∪ · · · ∪
Ckspans Qn1.As a result, C1
,
C2, . . . ,
Ckform k disjoint cycles of Qnsuch that uiis in Cifor 1≤
i≤
k and C1∪
C2∪ · · · ∪
Ckspans Qn. Forclarity, this result is summarized below.
Theorem 5. The n-cube Qnis spanning k-cyclable of order k if k
≤
n−
1 for n≥
2.We give an example to indicate that Qn is not spanning n-cyclable of order n. Let u be any vertex of Qn, and let
{
u1,
u2, . . . ,
un}
be the set of vertices adjacent to u. We set A= {
u1,
u2, . . . ,
un−1} ∪ {
u}
. Obviously,|
A| =
n. SincedegQn−{u1,u2,...,un−1}
(
u) =
1, there is no cycle of G− {
u1,
u2, . . . ,
un−1}
containing u. Thus, we cannot find n cycles C1,
C2, . . . ,
Cnof Qnsuch that uiis in Cifor 1≤
i≤
n−
1, and u is in Cn.4. Concluding remarks
In this paper we proved that Qnis spanning 2-cyclable of order n
−
1 for n≥
3. Now we show an example to indicatethat Qnis not 2-cyclable of order n. Let u and v be any two adjacent vertices of Qn. We set A1
=
NbdQn(
u)−{
v}
and A2= {
u}
.Obviously,
|
A1| + |
A2| =
n. Since degQn−A1(
u) =
1, there is no cycle of G−
A1containing A2. Thus, the spanning 2-cyclabilityof Qnis n
−
1 for n≥
3, and this result is optimal. Furthermore, we proved that Qnis spanning k-cyclable of order k ifk
≤
n−
1 for n≥
2.For possible future directions with our result, we first conjecture that Qnis spanning k-cyclable of order n
−
1 for everyk
≤
n−
1 and n≥
3. As we allowed A1or A2to be empty set in the statement ofTheorem 4, we indeed have a stronger conjecture: assume that n≥
4. Let A1,
A2, . . . ,
Akbe k disjoint vertex subsets of Qnwith|
A1∪
A2∪ · · · ∪
Ak| ≤
n−
1 andk
≤
n−
1, there exist k disjoint cycles C1,
C2, . . . ,
Ckof Qnsuch that (1) Aiis in Cifor 1≤
i≤
k, and (2) C1∪
C2∪ · · · ∪
Ckspans Qn. Notice that the statement is not always true for n
=
3. For counterexample, let A1= {
000,
111}
and A2= ∅
. Then the length of any cycle containing A1is at least 6. Thus, we cannot find two disjoint cycles C1and C2of Q3such that (1) Aiisin Cifor 1
≤
i≤
2, and (2) C1∪
C2spans Q3.Acknowledgments
We would like to express the most immense gratitude to the anonymous reviewers and the editor for their comments and suggestions. We thank also the Editor-in-Chief for his kindly effort in handling this submission.
Appendix A. Q4is 2-disjoint-path-coverable of order one
We prepare the following lemma in advance.
Lemma 6. Let p and q be any two adjacent vertices of Q3. Let u and v be any two nonadjacent vertices of Q3
− {
p,
q}
such thatthey are in different partite sets. Then there exists a hamiltonian path of Q3
− {
p,
q}
joining u and v.Proof. Since Q3 is vertex-symmetric and edge-symmetric, we assume that p
=
000 and q=
001. We have{
u,
v} ∈
{
011,
100}
, {
101,
010}
. Clearly, both
⟨
011,
010,
110,
111,
101,
100⟩
and⟨
101,
100,
110,
111,
011,
010⟩
are hamiltonian paths of Q3− {
p,
q}
.Recall that W and B form the bipartition of Q4. Let A1
= {
z}
and A2= ∅
, where z is any vertex of Q4− {
x,
y,
u,
v}
. SinceQ4is vertex-symmetric and edge-symmetric, we assume that u
=
0000 and v∈ {
0001,
0111}
.Case 1:
{
x,
y,
z} ⊂
V(
Q14
)
. ByTheorem 1, there exists a hamiltonian path P1of Q41joining x and y, and there exists a hamiltonian path P2of Q40joining u and v.Table 1
The vertex b and paths R1and R2.
R1 R2 x=0011,z=0101 ⟨0011,0001,0101,0100=b⟩ ⟨0000,0010,0110,0111⟩ x=0011,z=0110 ⟨0011,0010,0110,0100=b⟩ ⟨0000,0001,0101,0111⟩ x=0101,z=0011 ⟨0101,0001,0011,0010=b⟩ ⟨0000,0100,0110,0111⟩ x=0101,z=0110 ⟨0101,0100,0110,0010=b⟩ ⟨0000,0001,0011,0111⟩ x=0110,z=0011 ⟨0110,0010,0011,0001=b⟩ ⟨0000,0100,0101,0111⟩ x=0110,z=0101 ⟨0110,0100,0101,0001=b⟩ ⟨0000,0010,0011,0111⟩ Table 2 The path P1. x y P1 0011 0001 ⟨0011,0010,0110,0100,0101,0001⟩ 0011 0010 ⟨0011,0001,0101,0100,0110,0010⟩ 0101 0001 ⟨0101,0100,0110,0010,0011,0001⟩ 0101 0100 ⟨0101,0001,0011,0010,0110,0100⟩ 0110 0010 ⟨0110,0100,0101,0001,0011,0010⟩ 0110 0100 ⟨0110,0010,0011,0001,0101,0100⟩
Case 2: Either
{
x} ⊂
V(
Q40), {
y,
z} ⊂
V(
Q41)
or{
y} ⊂
V(
Q40), {
x,
z} ⊂
V(
Q41)
. Without loss of generality, we only consider that{
x} ⊂
V(
Q04
)
and{
y,
z} ⊂
V(
Q41)
. Let b∈
B∩
V(
Q40) − {
v}
. ByTheorem 3, there exist two disjoint paths R1and R2of Q40 such that (1) R1joins x and b, (2) R2joins u and v, and (3) R1∪
R2spans Q40. ByTheorem 1, there exists a hamiltonian pathH of Q41joining
(
b)
4and y. Then, we set P1
= ⟨
x,
R1,
b, (
b)
4,
H,
y⟩
and P2=
R2.Case 3:
{
z} ⊂
V(
Q40), {
x,
y} ⊂
V(
Q41)
. Since degQ04
(
z) =
3>
2, we can choose a vertex s of Q0
4
− {
(
x)
4, (
y)
4,
u,
v}
such that(
s,
z) ∈
E(
Q4)
. Note that both(
x)
4and v are in B, and both(
y)
4and u are in W . Let{
w,
b} = {
s,
z}
such that w∈
W and b∈
B. ByTheorem 3, there exist two disjoint paths R1and R2of Q41such that (1) R1joins x and(
w)
4, (2) R2joins(
b)
4 and y, and (3) R1∪
R2spans Q41. Then, P1is set to be⟨
x,
R1, (
w)
4,
w,
b, (
b)
4,
R2,
y⟩
. ByLemma 6, there exists a hamiltonian path P2of Q40− {
w,
b}
joining u and v.Case 4:
{
x,
y} ⊂
V(
Q04
), {
z} ⊂
V(
Q41)
. ByTheorem 3, there exist two disjoint paths R1and R2of Q40such that (1) R1joins x and y, (2) R2joins u and v, and (3) R1∪
R2spans Q40. We write R1as⟨
x,
H1,
w,
y⟩
. ByTheorem 1, there exists a hamiltonian path H2of Q41joins(
w)
4and(
y)
4. We set P1= ⟨
x,
H1,
w, (
w)
4,
H2, (
y)
4,
y⟩
and P2=
R2.Case 5:
{
x,
z} ⊂
V(
Q40), {
y} ⊂
V(
Q41)
.Subcase 5.1: Suppose that z
∈
B. ByTheorem 3, there exist two disjoint paths R1and R2of Q40such that (1) R1joins x and z, (2) R2joins u and v, and (3) R1∪
R2spans Q40. ByTheorem 1, there exists a hamiltonian path H of Q1
4 joining
(
z)
4and y. We set P1= ⟨
x,
R1,
z, (
z)
4,
H,
y⟩
and P2=
R2.Subcase 5.2: Suppose that z
∈
W and v=
0001. ByTheorem 1, there exists a hamiltonian path R of Q40− {
v}
joiningx and u. We write R as
⟨
x,
R′,
b,
u⟩
. Similarly, there exists a hamiltonian path H of Q14 joining
(
b)
4and y. Then we setP1
= ⟨
x,
R′,
b, (
b)
4,
H,
y⟩
and P2= ⟨
u,
v⟩
.Subcase 5.3: Suppose that z
∈
W and v=
0111. We have{
x,
z} ⊂ {
0011,
0101,
0110}
. We set a vertex b and paths R1and R2according toTable 1. ByTheorem 1, there exists a hamiltonian path H of Q41joining(
b)
4 and y. Then,
P1
= ⟨
x,
R1,
b, (
b)
4,
H,
y⟩
and P2=
R2are the requested paths.Case 6:
{
x,
y,
z} ⊂
V(
Q40)
.Subcase 6.1: v
=
0001. ByTheorem 1, there exists a hamiltonian path R of Q04
− {
v}
. We write R as⟨
x,
R1,
w,
y,
R2,
b,
u⟩
. Similarly, there exists a hamiltonian path H of Q41joining(
w)
4and(
b)
4. We set P1
= ⟨
x,
R1,
w, (
w)
4,
H, (
b)
4,
b,
rev(
R2),
y⟩
and P2= ⟨
u,
v⟩
, where rev(
R2)
is the reverse path of R2.Subcase 6.2: v
=
0111.(i)
(
x,
y) ̸∈ {(
0011,
0100), (
0101,
0010), (
0110,
0101)}
. We set P1according toTable 2. Obviously, P1is a hamiltonian path of Q40− {
u,
v}
. By Theorem 1, there exists a hamiltonian path H of Q14 joining
(
u)
4 and(
v)
4. Then, we set P2 as⟨
u, (
u)
4,
H, (
v)
4,
v⟩
.(ii)
(
x,
y) ∈ {(
0011,
0100), (
0101,
0010), (
0110,
0101)}
. We set R1and R2according toTable 3. Clearly, R1∪
R2spansQ40, and we can write R2as
⟨
u,
R′2,
w,
v⟩
. ByTheorem 1, there exists a hamiltonian path H of Q41joins(
w)
4and(
v)
4. Then we set P1=
R1and P2= ⟨
u,
R′2,
w, (
w)
4,
H, (
v)
4,
v⟩
.Appendix B. Proof ofLemma 5
To prove that Qnis 2-disjoint-path-coverable of order n
−
3, we prepare four propositions as follows. In the rest of thispaper, we continue using W and B to denote the bipartition of Qn. For convenience, we also call W and B partite sets of white
Table 3
The paths R1and R2.
R1 R2 x=0011,y=0100,z∈ {0001,0101} ⟨0011,0001,0101,0100⟩ ⟨0000,0010,0110,0111⟩ x=0011,y=0100,z∈ {0010,0110} ⟨0011,0010,0110,0100⟩ ⟨0000,0001,0101,0111⟩ x=0101,y=0010,z∈ {0001,0011} ⟨0101,0001,0011,0010⟩ ⟨0000,0100,0110,0111⟩ x=0101,y=0010,z∈ {0100,0110} ⟨0101,0100,0110,0010⟩ ⟨0000,0001,0011,0111⟩ x=0110,y=0101,z∈ {0100,0101} ⟨0110,0100,0101,0001⟩ ⟨0000,0010,0011,0111⟩ x=0110,y=0101,z∈ {0010,0011} ⟨0110,0010,0011,0001⟩ ⟨0000,0100,0101,0111⟩
a
b
c
Fig. 5. Illustration forProposition 1.
Proposition 1. Let W and B form the bipartition of Qnwith n
≥
7. Suppose that x and u are any two different vertices in W ,whereas y and v are any two different vertices in B. Furthermore, suppose that x
∈
V(
Qn0)
, y∈
V(
Qn1)
, and y̸=
(
u)
n. Let A0 1and A0
2be any two disjoint nonempty subsets of V
(
Qn0) − {
x,
y,
u,
v}
, and let A11and A12be any two disjoint nonempty subsets ofV
(
Qn1) − {
x,
y,
u,
v}
such that|
A01| + |
A11| + |
A02| + |
A12| =
n−
3. Assume that Qn−1is 2-disjoint-path-coverable of order n−
4.Then, there exist two disjoint paths P1and P2such that
(
1)
P1joins x to y,(
2)
P2joins u to v,(
3)
A10∪
A11⊆
P1,(
4)
A02∪
A12⊆
P2,and
(
5)
P1∪
P2spans Qn.Proof. Obviously,
|
Aji| ≤
n−
6 for i∈ {
1,
2}
and j∈ {
0,
1}
, and|
A11
| + |
A12| + |{
y}| ≤
n−
4. We have the following two cases.Case 1: Both u and v are in Qnjfor some j
∈ {
0,
1}
. Without loss of generality, we assume that j=
0. Since|
V(
Qn0)| =
2n−1
>
n(
n−
4) + (
n−
3) =
n2−
3n−
3≥
n|
A11∪
A12∪ {
y}| + |
A01∪ {
x,
u,
v}|
and 2n−2>
n−
3 for n≥
7, there exists a vertex p in V(
Q0n
) − (
A01∪ {
x,
u,
v}
)
such that(
t)
n̸∈
A11∪
A12∪ {
y}
for every t∈
NbdQn0(
p) ∪ {
p}
, and there exists a blackvertex b in V
(
Q0n
) − (
A02∪ {
v,
p}
)
such that(
b)
n̸∈
A12. Since Qn−1is 2-disjoint-path-coverable of order n−
4, there are two disjoint paths R1and R2in Qn0such that (1) R1joins x to b, (2) R2joins u to v, (3) A01⊆
R1, (4) A02∪ {
p} ⊆
R2, and (5) R1∪
R2 spans Q0n. Without loss of generality, we write R2as
⟨
u,
R2,1,
p,
q,
R2,2,
v⟩
. Again, there are two disjoint paths H1and H2inQn1such that (1) H1joins
(
b)
nto y, (2) H2joins(
p)
nto(
q)
n, (3) A11⊆
H1, (4) A12⊆
H2, and (5) H1∪
H2spans Qn1. We setP1
= ⟨
x,
R1,
b, (
b)
n,
H1,
y⟩
and P2= ⟨
u,
R2,1,
p, (
p)
n,
H2, (
q)
n,
q,
R2,2,
v⟩
. Obviously, P1and P2form the desired paths. SeeFig. 5(a).
Case 2: u is in Qnj, and v is in Q1
−j
n for j
∈ {
0,
1}
. On the one hand, we assume that j=
0; that is, u is in Qn0, and v is in Qn1.Since 2n−2
>
n−
4 for n≥
7, there exists a black vertex b1in V
(
Qn0) −
A0
2such that
(
b1)
n̸∈
A12, and there exists a black vertex b2in V(
Qn0) − (
A01∪ {
b1}
)
such that(
b2)
n̸∈
A11. Since Qn−1is 2-disjoint-path-coverable of order n−
4, there are two disjoint paths R1and R2in Qn0such that (1) R1joins x to b1, (2) R2joins u to b2, (3) A01⊆
R1, (4) A02⊆
R2, and (5) R1∪
R2spansQ0
n; and there are two disjoint paths H1and H2in Qn1such that (1) H1joins
(
b1)
nto y, (2) H2joins(
b2)
nto v, (3) A11⊆
H1, (4) A12
⊆
H2, and (5) H1∪
H2spans Qn1. We set P1= ⟨
x,
R1,
b1, (
b1)
n,
H1,
y⟩
and P2= ⟨
u,
R2,
b2, (
b2)
n,
H2,
v⟩
. Obviously, P1 and P2form the desired paths. SeeFig. 5(b).On the other hand, if j
=
1, then u is in Qn1, and v is in Qn0. Since 2n−2>
n−
3 for n≥
7, there exists a black vertex b in V(
Q0n
) − (
A02∪ {
(
u)
n,
v}
)
such that(
b)
n̸∈
A12, and there exists a white vertex w in V(
Qn0) − (
A10∪ {
x, (
y)
n}
)
such that(
w)
n̸∈
A11. Similarly, there exist disjoint paths R1,
R2,
H1,
H2joining x to b, w to v,(
b)
nto y, and u to(
w)
n, respectively, such that (1) A01
⊆
R1, A02⊆
R2, A11⊆
H1, A12⊆
H2, (2) R1∪
R2spans Qn0, and (3) H1∪
H2spans Qn1. We set P1= ⟨
x,
R1,
b, (
b)
n,
H1,
y⟩
and P2= ⟨
u,
H2, (
w)
n,
w,
R2,
v⟩
. SeeFig. 5(c).Proposition 2. Let W and B form the bipartition of Qnwith n
≥
6. Suppose that x and u are any two different vertices in W ,whereas y and v are any two different vertices in B. Furthermore, suppose that x
∈
V(
Q0n
)
, y∈
V(
Qn1)
, and y̸=
(
u)
n. Let A01andA02be any two disjoint nonempty subsets of V
(
Q0n
) − {
x,
y,
u,
v}
, and let A11be any nonempty subset of V(
Qn1) − {
x,
y,
u,
v}
such that
|
A01| + |
A11| + |
A02| =
n−
3. Assume that Qn−1is 2-disjoint-path-coverable of order n−
4. Then, there exist two disjointpaths P1and P2such that
(
1)
P1joins x to y,(
2)
P2joins u to v,(
3)
A01∪
A11⊆
P1,(
4)
A02⊆
P2, and(
5)
P1∪
P2spans Qn.Proof. We consider the following three cases.
Case 1: Both u and v are in Q0
n. Since 2n
−2
>
n−
4≥ |
A02