Matchings Extend to Perfect Matchings on
Hypercube Networks
Y-Chuang Chen
∗†Kun-Lung Li
Abstract—In this work, we investigate in the prob-lem of perfect matchings with prescribed matchings in the n-dimensional hypercube network Qn. We obtain the following contributions: For any arbitrary match-ing with at most n−1 edges, it can be extended to a perfect matching of Qn for n≥1. Furthermore, for any arbitrary non-forbidden matching with n edges, it also can be extended to a perfect matching of Qn for n ≥ 1. It is shown by J. Fink in 2007 that any arbitrary perfect matching of the n-dimensional hy-percubeQn,n≥2, can be extended to a Hamiltonian cycle. Therefore, it leads to a further result that for any arbitrary non-forbidden matching with at mostn edges, it can be extended to a Hamiltonian cycle of Qn for n≥2.
Keywords: matching, perfect matching, Hamiltonian cycle, hypercube
1
Introduction
The underlying topology of an interconnection network
is usually modeled as a graph G = (V, E), in which
the vertex set V(G) represents processors and the edge
setE(G) represents connections between processors. We
use graphs and networks interchangeably. For the fun-damental graph definitions and notations we follow [2].
G= (V, E) is a simple graph ifV is a finite set andE is
a subset of {(a, b)|(a, b) is an unordered pair ofV}. We
say that V is the vertex set and E is the edge set. The
neighborhood ofv,N(v), is{x|(v, x)∈E}. Two vertices
are adjacent if (a, b) ∈ E. A path is a sequence of
ad-jacent edges (v0, v1),(v1, v2),· · ·,(vm−1, vm), written as
hv0, v1,· · ·, vmi, in which all the vertices v0, v1,· · ·, vm
are distinct. Acycleis a path with at least three vertices
such that the first vertex is the same as the last one. Hence, the length of a path or cycle is the number of
edges in the path or cycle. Letuandvbe two vertices in
a graph, thedistance fromuto v, denoted bydist(u, v),
is the minimum length of any path fromutov. A cycle
is a Hamiltonian cycle if it traverses every vertex of G
exactly once [4, 5].
∗This work was supported in part by the National Science
Coun-cil of the Republic of China under Contract NSC 98-2221-E-159-015.
†Corresponding author: Y-Chuang Chen, Department of
Information Management, Ming Hsin University of Science and Technology, Hsin Feng, Hsinchu 304, Taiwan, R.O.C. Email:[email protected]
A matching M in graph G is a set of pairwise
non-adjacent edges, that is, every vertex is incident with at
most one edge of M. A vertex is matched if it is
in-cident to an edge in the matching M. Otherwise the
vertex is unmatched. A maximum matchingis a
match-ing that contains the largest possible number of edges.
The matching number of a graph is the size of a
maxi-mum matching. Aperfect matchingis a matching which
matches all vertices of the graph. That is, every ver-tex of the graph is incident to exactly one edge of the
matching. A matching is a forbidden matching if every
vertex of N(v) is matched for some unmatched vertex
v∈V(G). Otherwise, it is a non-forbidden matching. A
perfect matching is also aminimum-size edge cover. For
any graph G without isolated vertices, the sum of the
edge covering number and the matching number equals
to the number of vertices. If the graph G has a perfect
matching, then both of the edge cover number and the
matching number are equal to|V(G)|/2 [11].
Perfect matchings can be applied to network
disclo-sure attacks. The Perfect Matching Disclosure Attack
performs a high rate of success when tracing messages sent through a threshold mix in arbitrary scenarios [12]. Another related work concerning perfect matchings is
matching preclusion. A matching M in graph G is
called an almost perfect matching ifM contains exactly
(|V(G)| −1)/2 edges. An edge setF in graphGis called
a matching preclusion set if G\F has neither a perfect
matching nor an almost perfect matching. If a network has a large size of matching preclusion set, each vertex (process) in this network will has a specific partner in the event of edge (link) faults and the network will be robust. On the side, matching preclusion is also related
to the problem ofconnectivityof networks [3, 10].
Thehypercubeis a popular network because of its
attrac-tive properties, including regularity, symmetry, small di-ameter, strong connectivity, recursive construction, par-titionability, and relatively low link complexity [1, 9]. It
has been widely used in parallel systems, such as the
Con-nection machines, Symult S-series, Intel iPSC, iPSC/2,
and SGI Origin 2000 [6, 13, 14]. The formal
defini-tion of an n-dimensional hypercube is given as follows.
Each vertex v in Qn can be distinctly labeled by a
bi-nary n-bit string, v = vnvn−1· · ·v1. For 1 ≤ i ≤ n,
we use vi to denote the binary string v
The Qn consists of all n-bit binary strings
represent-ing its vertices. Two vertices u and v are adjacent if
and only if v = ui with some i. An n-dimensional
hypercube Qn can be constructed from two identical
(n−1)-dimensional hypercubes,Qi,n0−1andQi,n1−1, for some
1 ≤ i ≤ n, where V(Qi,n−01) = {vnvn−1· · ·v1|vi = 0}
and V(Qi,n−11) ={vnvn−1· · ·v1|vi = 1}. The vertex set
of Qn is V(Qn) = Qi,n0−1 ∪Q
i,1
n−1, and the edge set is
E(Qn) = E(Qi,n0−1)∪E(Qi,n1−1)∪P where P is a set of
edges connecting the vertices ofQi,n0−1andQi,n−11in a one
to one fashion. Ann-dimensional hypercube can be
rep-resented as Qn =Qi,n−01
L
Qi,n1−1 withi∈1,2,· · ·, n. We
need some more terms, let v be a vertex of Qi,n−01 and
Qi,n−11, we use v0 to denote the corresponding vertex ofv
inQi,n−11andQni,0−1, respectively. That is, (v, v0) is an edge
of Qn, one ofu, v is inQni,−01, and the other is inQi,n1−1.
We notice that the hypercube Qn is vertex-symmetric
and edge-symmetric forn≥1.
Some results on perfect matchings of hypercubes are dis-cussed in [7, 8, 10]. In this paper, we shall show that
for any arbitrary matching with at most n−1 edges of
the hypercube Qn for n ≥ 1, it can be extended to a
perfect matching of Qn. Furthermore, for any arbitrary
matching with exactly n edges of the hypercube Qn for
n ≥1, it can be extended to a perfect matching ofQn,
except it is a forbidden matching of Qn. It has shown
that any perfect matching of the hypercube Qn, n ≥2,
can be extended to a Hamiltonian cycle [7]. As a result, we have a further result that any non-forbidden matching
with at mostnedges of the hypercubeQn,n≥2, can be
extended to a Hamiltonian cycle.
The rest of this paper is organized as follows. Section 2 shows the perfect matchings with prescribed matchings
in then-dimensional hypercubeQnforn≥1. In Section
3, we give the conclusion remarks.
2
Perfect Matchings of Hypercubes
Lemma 1 For the 3-dimensional hypercube Q3, let M
be an arbitrary matching with at most two edges in the
hypercubeQ3. Then, the matchingM can be extended to
a perfect matching ofQ3.
Proof. First, suppose that M contains exactly zero or
one edge, it is clear thatM can be extended to a perfect
matching of Q3. Now, suppose that M has exactly two
edges. Figure 1.(a) shows all the non-isomorphic cases
thatM contains exactly two edges in the hypercube Q3.
Figure 1.(b) shows the perfect matchings of Q3. As a
result, the proof of this lemma is complete. ♦
Theorem 1 Assume that n ≥ 1 is an integer. Let M
be an arbitrary matching with at mostn−1 edges in the
(a)
(b)
Figure 1: Perfect matchings of the hypercube Q3.
hypercubeQn. Then, the matchingM can be extended to
a perfect matching ofQn.
Proof. We prove it by induction on n. Suppose that
n = 1,2, it is clear that for any matching with at most
n−1 edges, it can be extended to a perfect matching of
Qn. Suppose thatn= 3, by Lemma 1, for any matching
with at most two edges, it can be extended to a perfect
matching ofQ3.
Assume that the theorem is true forn, which means that
for any arbitrary matchingM with at mostn−1 edges,
M can be extended to a perfect matching of Qn. Now,
we shall show that in the hypercube Qn+1, let M0 be
an arbitrary matching with m ≤ n edges in Qn+1, the
matching M0 can be extended to a perfect matching of
Qn+1.
By the definition of hypercubes, for each edge in Qn+1,
the labels of two endpoint vertices are different with
exactly one bit. Since M0 contains at most n edges,
there exists one dimensioni∈1,2,· · ·, n+ 1 inQn+1 =
Qi,0
n
L Qi,1
n such that each edge of M0 is distributed in
Qi,0
n or Qi,n1. Let M0 =M0∩Qni,0 andM1 =M0∩Qi,n1.
That is to say that M0 has m
0 = |M0| edges in Qi,n0,
m1 =|M1| edges in Qni,1, andm0+m1 = m. We may
without loss of generality assume that m0 ≥ m1. We
divide the proof into the following two cases.
Case 1: m0≥m1>0.
Notice that m0 < n and m1 < n since m0 ≥ m1 > 0.
See Figure 2.(a). By inductive hypothesis, M0 can be
extended to a perfect matching of Qi,0
n , say P M0, and
M1 can be extended to a perfect matching ofQi,n1, say
P M1. See Figure 2.(b). Therefore, P M0 ∪P M1 is a
perfect matching ofQn+1.
Case 2: m1= 0.
That is, all the n edges of M0 are distributed in Qi,0
n .
Suppose that m0 < n, by inductive hypothesis, M0
can be extended to a perfect matching of Qi,0
n , and so
M0 can be extended to a perfect matching of Q
n+1.
Q
n
i,0 Q
n i,1
Q
n+1
: elements of M` Q
n
i,0 Q
n i,1
Q
n+1
(a) (b)
{
m0
e
d
g
es
{
m1e
d
g
es
{
{
{
2
n
-1
-m0
e
d
g
es
{
2
n
-1
-m1
e
d
g
e
s
m1
e
d
g
es
m0
e
d
g
es
Figure 2: Case 1: m0≥m1>0.
: elements of M`
(a) Qni,0 Qni,1
Qn+1
{
{
Qni,0 Qni,1 Qn+1
{
n-1e
d
g
es
{
{
2n
-1-2
e
d
g
e
s
(b)
2
n
-1-(
n
+
1
)
e
d
g
e
s
a
b d c
a
b d c
a'
b' d' c'
n
-1
e
d
g
es
2
n
-1-(
n
+
1
)
e
d
g
e
s
{
2n
-1-2
e
d
g
e
s
a'
b' d' c'
Figure 3: Case 2: m1= 0.
Let (a, b) be an edge in M0. By inductive
hypothe-sis, M0 \ {(a, b)} can be extended to a perfect
match-ing of Qi,0
n , say P M0, since m−1 ≤ n−1. If P M0
contains the edge (a, b), it is clear that M0 can be
ex-tended to a perfect matching of Qn+1. Here, we may
consider that P M0 do not contain the edge (a, b). Let
(a, c) and (b, d) be two edges ofP M0. By inductive
hy-pothesis withn≥3,{(a0, c0),(b0, d0)}can be extended to
a perfect matching of Qi,1
n , say P M1. See Figure 3.(a).
Therefore, (P M0∪P M1∪ {(a, b),(a0, b0),(c, c0),(d, d0)})\
{(a, c),(b, d),(a0, c0),(b0, d0)} is a perfect matching of
Qn+1 as Figure 3.(b), and this theorem follows. ♦
Lemma 2 For the 3-dimensional hypercubeQ3, letM be
an arbitrary matching with exactly three edges in the
hy-percube Q3. Then, the matching M can be extended to a
perfect matching ofQ3, exceptM is a forbidden matching
of Q3.
Proof. Figure 4.(a) shows all the non-isomorphic cases
that M contains exactly three edges in the hypercube
Q3. Figure 4.(b) shows the perfect matchings ofQ3with
the matchingM. As a result, the proof of this lemma is
complete. ♦
Lemma 3 For the 3-dimensional hypercubeQ3, letM be
an arbitrary matching with at most one edge in the
hy-percube Q3, andx, y be any two unmatched vertices with
(a)
(b)
a forbidden matching
Figure 4: Perfect matchings of the hypercube Q3.
(a)
(b)
x
y
x y x
y
y x
x
y
x y x
y
y x
Figure 5: Perfect matchings of the hypercubeQ3\ {x, y}.
dist(x, y)is odd. Then, the matchingM can be extended
to a perfect matching ofQ3\ {x, y}.
Proof. Firstly, suppose that|M|= 0, it is clear that for
any two unmatched vertices x, y with dist(x, y) is odd,
there is a perfect matching ofQ3\ {x, y}. Now, suppose
that |M|= 1. Figure 5.(a) shows all the non-isomorphic
cases thatM contains exactly one edge andx, y be any
two unmatched vertices withdist(x, y) is odd in the
hy-percubeQ3. Figure 5.(b) shows the perfect matchings of
Q3\ {x, y} with the matching M. Hence, the proof of
this lemma is complete. ♦
Lemma 4 Assume that n ≥3 is an integer. Let M be
an arbitrary matching with at mostn−2 edges in the
hy-percubeQn, and x, y be any two unmatched vertices with
dist(x, y)is odd. Then, the matchingM can be extended
to a perfect matching ofQn\ {x, y}.
Proof. We prove it by induction on n. Suppose that
n = 3, by Lemma 3, for any matching with at most
one edge and x, y be any two unmatched vertices with
dist(x, y) is odd, it can be extended to a perfect
match-ing ofQ3\ {x, y}.
Assume that the theorem is true forn, which means that
for any arbitrary matching M with at most n−2 edges
odd, M can be extended to a perfect matching of Qn\
{x, y}. Now, we shall show that in the hypercubeQn+1,
let M0 be an arbitrary matching with m ≤n−1 edges
in Qn+1 and x, y be any two unmatched vertices with
dist(x, y) is odd, the matching M0 can be extended to a
perfect matching ofQn+1\ {x, y}.
By the definition of hypercubes, for each edge in Qn+1,
the labels of two endpoint vertices are different with
ex-actly one bit. Since M0 contains at most n−1 edges,
there exists one dimension i∈1,2,· · ·, n+ 1 of Qn+1 =
Qi,0
n
L Qi,1
n such that each edge of M0 is distributed in
Qi,0
n or Qi,n1. Let M0 =M0∩Qni,0 andM1 =M0∩Qi,n1.
That is to say that M0 has m
0 = |M0| edges in Qi,n0,
m1 =|M1| edges in Qni,1, andm0+m1 = m. We may
without loss of generality assume thatm0 ≥m1 and
di-vide the proof into the following three cases.
Case 1: Both xand y are in Qi,1
n .
By Theorem 1,M0can be extended to a perfect matching
ofQi,0
n , sayP M0, sincem0≤m≤n−1. Sincem0≥m1
andm≤n−1,m1≤n−2. By inductive hypothesis,M1
can be extended to a perfect matching of Qi,1
n \ {x, y},
sayP M0
1. Therefore, P M0∪P M10 is a perfect matching
ofQn+1\ {x, y}.
Case 2: Both xand y are in Qi,0
n .
Case 2.1: m0≤n−2.
By inductive hypothesis,M0can be extended to a perfect
matching ofQi,0
n \ {x, y}, sayP M00. By Theorem 1, M1
can be extended to a perfect matching ofQi,1
n , sayP M1.
Therefore, P M0
0∪P M1 is a perfect matching of Qn+1\
{x, y}.
Case 2.2: m0=n−1.
Let (a, b) be an edge of M0, by inductive hypothesis,
M0\ {(a, b)} can be extended to a perfect matching of
Qi,0
n \ {x, y}, say P M00, since m0−1 = n−2. Firstly,
suppose that (a, b) ∈ P M0
0. By Theorem 1, Qi,n1 has
a perfect matching, say P M1. Then, P M0
0∪P M1 is a
perfect matching of Qn+1 \ {x, y}. Now suppose that
(a, b) ∈/ P M0
0. We may let {(a, c),(b, d)} ⊆ P M00.
By Theorem 1, {(a0, c0),(b0, d0)} can be extended to a
perfect matching of Qi,1
n , say P M1, as Figure 6.(a).
Then, (P M0
0 ∪ P M1 ∪ {(a, b),(a0, b0),(c, c0),(d, d0)}) \
{(a, c),(b, d),(a0, c0),(b0, d0)} is a perfect matching of
Qn+1\ {x, y}as Figure 6.(b).
Case 3: x is inQi,0
n and y is inQi,n1.
Case 3.1: m0≤n−2.
InQn+1, sincen≥3, there exists an edge (c, c0) between
Qi,0
n andQi,n1such thatcis an unmatched vertex ofQi,n0,
c 6= x, dist(c, x) is odd, c0 is an unmatched vertex of
Qi,1
n , c06=y, anddist(c0, y) is odd. See Figure 7.(a). By
inductive hypothesis, M0 can be extended to a perfect
matching of Qi,0
n \ {x, c}, say P M00, and M1 can be
ex-tended to a perfect matching of Qi,1
n \ {y, c0}, sayP M10.
Therefore, P M0
0∪P M10 ∪ {(c, c0)} is a perfect matching
(a)
Qni,0 Q Qni,1 n+1 (b) Qni,0
Qni,1 Qn+1
{
n-2ed
g
e
s
a
b
x
y c
d a'
b' c'
d'
{
2
n
-1-(
n
+
1
)
e
d
g
e
s
{
2n
-1-
2
e
d
g
e
s
{
n-2 edg
e
s
a
b
x
y c
d a'
b' c'
d'
{
n-1{
2-
2
e
d
g
e
s
2
n
-1-(
n
+
1
)
e
d
g
e
s
Figure 6: Case 2.2: m0=n−1.
(a) (b)
Qni,0 Qni,1 Qn+1
Qni,0 Q
n i,1 Qn+1
{
m0e
d
g
e
s
x y
c c'
{
m0
e
d
g
es
x y
{
2
n
-1-(
m0
+
1)
e
d
g
es
c c'
{
m1ed
g
es
{
m1ed
g
es
{
2
n
-1-(
m1
+
1)
e
d
g
es
Figure 7: Case 3.1: m0≤n−2.
ofQn+1\ {x, y}as Figure 7.(b).
Case 3.2: m0=n−1.
By Theorem 1,M0can be extended to a perfect matching
of Qi,0
n , say P M0. Let (x, c) be an edge of P M0. By
inductive hypothesis,Qi,1
n \{c0, y}has a perfect matching,
say P M0
1, since dist(c0, y) is odd. Therefore, (P M0∪
P M0
1∪ {(c, c0)})\ {(c, x)}is a perfect matching ofQn+1\
{x, y}. ♦
In the following theorem, we shall show that for any
ar-bitrary non-forbidden matching withm≤nedges in the
hypercubeQn,M can be extended to a perfect matching
ofQn, exceptM is a forbidden matching ofQn.
Theorem 2 Assume that n ≥ 1 is an integer. Let M
be an arbitrary matching with at most nedges in the
hy-percube Qn. Then, the matchingM can be extended to a
perfect matching ofQn, exceptM is a forbidden matching
of Qn.
Proof. We prove it by induction on n. Suppose that
n = 1,2, it is clear that for any matching with at most
nedges, it can be extended to a perfect matching ofQn.
Suppose thatn= 3, by Lemma 1 and 2, for any matching
with at most three edges, it can be extended to a perfect
matching ofQ3, except it is a forbidden matching of Q3.
Assume that the theorem is true forn, which means that
for any arbitrary matching M with at mostnedges, M
is a forbidden matching of Qn. Now, we shall show that
in the hypercubeQn+1, letM0 be an arbitrary matching
withm≤n+ 1 edges inQn+1, the matching M0 can be
extended to a perfect matching of Qn+1, exceptM0 is a
forbidden matching ofQn+1. By Theorem 1, M0 can be
extended to a perfect matching ofQn+1ifm≤n. Hence,
we may only consider that m = n+ 1 in the following
proof of this theorem.
By the definition of hypercubes, the labels of two
end-point vertices of each edge in Qn+1 is exactly different
with one bit. SinceM0 containsn+ 1 edges, there exists
one dimensioni∈ {1,2,· · ·, n+1}ofQn+1=Qi,n0
L Qi,1
n
such that at most one edge of M0 is distributed between
Qi,0
n andQi,n1. LetM0=M0∩Qni,0,M1=M0∩Qi,n1, and
Mc=M0\(M0∪M1). That is that there arem0=|M0|
edges inQi,0
n ,m1=|M1|edges inQni,1,mc=|Mc|= 0 or
1 edge between Qi,0
n and Qi,n1, andm0+m1+mc =m.
We may without loss of generality assume thatm0≥m1.
Hence,m1≤ b(n+ 1)/2c ≤n−1 for n≥3. We divide
the proof into the following three cases.
Case 1: m0=n+ 1.
Assume thatM0 ={(ai, bi)|i = 1,2,· · ·, n+ 1}. We let
M0
1 ={(a0i, b0i)|i = 1,2,· · ·, n+ 1} and P = {(c, c0)|c ∈
(V(Qi,0
n )\{ai, bi|i= 1,2,· · ·, n+1})}. Then,M0∪M10∪P
is a perfect matching ofQn+1.
Case 2: m0=n.
Firstly, suppose that mc = 1. Assume that M0 =
{(ai, bi)|i = 1,2,· · ·, n}. We let M10 = {(a0i, b0i)|i =
1,2,· · ·, n} and P = {(c, c0)|c ∈ (V(Qi,0
n )\ {ai, bi|i =
1,2,· · ·, n})}. Then,M0∪M10∪P is a perfect matching
ofQn+1. Now, suppose thatm1= 1. We divide this case
into two subcases.
Case 2.1: M0 is a non-forbidden matching of Qi,n0.
By inductive hypothesis,M0can be extended to a perfect
matching of Qi,0
n , say P M0. By another, since m1 = 1
and n ≥ 3, M1 can be extended to a perfect matching
of Qi,1
n , say P M1. Therefore, P M0∪P M1 is a perfect
matching ofQn+1.
Case 2.2: M0 is a forbidden matching of Qi,n0.
Let M0 = {(a1, b1),(a2, b2),· · ·,(an, bn)} and c be the
common neighbor ofa1, a2,· · ·, an in Qi,n0. Sincem1= 1
andn≥3, among the (a1, b1),(a2, b2),· · ·,(an, bn), there
exists a matching element, say (an, bn), and a
neigh-bor unmatched vertex of bn, say d, in Qi,n0 such that
d0 is an unmatched vertex. Here, it is clear that c0
is also an unmatched vertex, otherwise M is a
forbid-den matching of Qn+1. By inductive hypothesis, since
(M0∪ {(bn, d)})\ {(an, bn)} is a non-forbidden
match-ing of Qi,0
n with n edges, it can be extended to a
per-fect matching of Qi,0
n , say P M0. By Lemma 4, M1
can be extended to a perfect matching of Qi,1
n \ {c0, d0},
say P M0
1. See Figure 8.(a). Therefore, (P M0∪P M10 ∪
{(an, bn),(c, c0),(d, d0)})\{(an, c),(bn, d)}forms a perfect
Q
n
i,0 Q
n i,1
Qn+1
{
(a) 2 n -1 -( n + 1 ) e d g e s{
2 n -1-2 e dg es{
n edges
c d a n bn a 1 b1 a 2 b2 c' d' Qn
i,0 Q
n i,1 Q n+1
{
(b) 2 n -1 -( n + 1 ) e d ge s{
2 n -1-2 e d g es{
n e dges
c d an bn a1 b1 a2 b2 c' d'
Figure 8: Case 2.2: M0 is a forbidden matching ofQi,n0.
(a)
Qni,0 Q
n i,1 Qn+1
{
m0e d g e s
{
{
m1e d g e s
{
2 n -1-( m1 + 1 ) e d g e s 2 n -1-( m0 + 1 ) e d g e s c an c' a'n (b)Qni,0
Qni,1 Qn+1
{
m0e d g e s
{
{
m1 e d g e s{
2 n -1-( m1 + 1 ) e d g e s 2 n -1-( m0 + 1 ) e d g e s c an c' a'nFigure 9: Case 3.1: an anda0n are unmatched vertices.
matching ofQn+1as Figure 8.(b).
Case 3: m0≤n−1.
Firstly, suppose that mc = 0. M0 can be extended to
a perfect matching of Qi,0
n , sayP M0. By another, since
m1 ≤n−1 for n≥3,M1 can be extended to a perfect
matching ofQi,1
n , sayP M1. Therefore,P M0∪P M1 is a
perfect matching of Qn+1. Now, suppose that mc = 1.
Let Mc = {(c, c0)} where c is in Qi,n0. Then, in Qi,n0,
among thenneighbors ofc, denoted bya1, a2,· · ·, an, if
there exists at least an unmatched vertex, say an, such
that M0∪ {(c, an)} is a non-forbidden matching ofQi,n0
anda0
n is also an unmatched vertex as Figure 9.(a), this
shall be discussed in the following Case 3.1. Otherwise,
ai or a0i is matched in Qn+1 fori= 1,2,· · ·, nas Figure
10.(a), this shall be discussed in Case 3.2.
Case 3.1: an and a0n are unmatched vertices.
Then, by inductive hypothesis,M0∪ {(c, an)} can be
ex-tended to a perfect matching of Qi,0
n , say P M0. Since
m0 ≥ m1 and mc = 1, m1 ≤ b(n+1)2−1c ≤ n−2 for
n≥3. Hence,M1∪ {(c0, a0n)}can be extended to a
per-fect matching ofQi,1
n , sayP M1. See Figure 9.(a).
There-fore, (P M0∪P M1∪ {(c, c0),(an, an0)})\ {(c, an),(c0, a0n)}
is a perfect matching ofQn+1 as Figure 9.(b).
Case 3.2: ai or a0i is matched in Qn+1 for i =
1,2,· · ·, n.
Since m0 ≥ m1, m0 ≤ n − 1, and m0 +
m1 = n, we may without loss of generality
as-sume that M0 = {(a1, b1),(a2, b2),· · ·,(ai, bi)} and
M1 = {(a0i+1, d0i+1),(a0i+2, d0i+2),· · ·,(a0n, d0n)} where i ∈
(a)
Qn
i,0 Q
n i,1
Qn+1
(b) Qn
i,0
Qn i,1
Qn+1
c
g a1
b1
a2
b2 bi
ai
c' a 'n
{
2
n
-1-(
n
+1
)
ed
g
e
s
g'
{
2
n
-1-(
n
-i
)-1
e
d
g
e
s
c
g a1
b1
a2
b2 bi
ai
c' a 'n
{
2
n
-1-(
n
+
1
)
ed
g
es
g'
{
2
n
-1-(
n
-i
)-1
e
d
g
e
s
a'i+1
a'i+2 a'i+2 a'i+1
d 'n d 'i+2 d'i+1
an
bn
bi+1
ai+1 an
bn
bi+1
ai+1
d'n d'i+2 d'i+1
Figure 10: Case 3.2: ai or a0i is matched in Qn+1 for
i= 1,2,· · ·, n.
be a forbidden matching of Qi,0
n , note the c is
the common neighbor of a1, a2,· · ·, an. Then,
{(c, a1),(a2, b2),(a3, b3),· · ·,(an, bn)} is a non-forbidden
matching ofQi,0
n withnedges, and it can be extended to
a perfect matching ofQi,0
n , sayP M0. Let (b1, g)∈P M0.
On account ofdist(c, g) = 3 and inQi,1
n , the distance ofc0
and each of the matched vertices ofQi,1
n is at most 2, sog0
is unmatched. By Lemma 4,M1can be extended to a
per-fect matching ofQi,1
n \{c0, g0}, sayP M10, sincem1≤n−2
and dist(c0, g0) is odd. See Figure 10.(a). Therefore,
(P M0∪P M10∪ {(a1, b1),(c, c0),(g, g0)})\ {(a1, c),(b1, g)}
is a perfect matching ofQn+1. See Figure 10.(b). ♦
It has shown in [7] that every perfect matching of then
-dimensional hypercube withn≥2 can be extended to a
Hamiltonian cycle. Consequently, we have the following corollary.
Corollary 1 Assume that n ≥ 2 is an integer. Let M
be an arbitrary matching with at most nedges in the
hy-percube Qn. Then, the matchingM can be extended to a
Hamiltonian cycle ofQn, exceptM is a forbidden
match-ing of Qn.
3
Conclusion Remarks
Perfect matchings can be applied to network disclosure attacks, such as the Perfect Matching Disclosure Attack. Perfect matching problem is also applicable to connec-tivity of networks. In this paper, we assign an arbitrary
non-forbidden matching with at mostn edges to form a
perfect matching or a Hamiltonian cycle on the
hyper-cubeQn forn≥2.
An open problem is that on the hypercubeQn, if we
as-sign a matching with more thannedges, how can we
re-strict the prescribed matching to obtain a perfect match-ing or a Hamiltonian cycle.
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