Prescribed matchings extend to Hamiltonian cycles in hypercubes with faulty edges

0.19 MB·English

by  Fan Wang

Prescribed matchings extend to Hamiltonian cycles in
hypercubes with faulty edges∗
Fan Wang, Heping Zhang†
3 School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P. R. China
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E-mail addresses: [email protected], [email protected]
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4 Abstract
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] Ruskey and Savage asked the following question: Does every matching of Qn for
O
n ≥ 2 extend to a Hamiltonian cycle of Q ? J. Fink showed that the question is
C n
. true for every perfect matching, and solved the Kreweras’ conjecture. In this paper
h
t
a we consider the question in hypercubes with faulty edges. We show that every
m
matching M of at most 2n−1 edges can be extended to a Hamiltonian cycle of Q
[ n

for n≥ 2. Moreover, we can prove that when n ≥ 4 and M is nonempty this result
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v still holds even if Q has at most n−1−⌈|M|⌉ faulty edges with one exception.
1 n 2
3
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Key words: Hypercube; Hamiltonian cycle; Matching; Edge fault tolerance
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.
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1 Introduction
1
:
v
i
X The n-dimensional hypercube Q is one of the most popular and efficient interconnec-
n
r
a tion networks. There is a large amount of literature on graph-theoretic properties of
hypercubes as well as on their applications in parallel computing (e.g., see [9, 12]).
It is well known that Q is Hamiltonian for every n ≥ 2. This statement dates back
n
to 1872 [8]. Since then, the research on Hamiltonian cycles in hypercubes satisfying
certain additional properties has received considerable attention. The applications in
parallel computing inspired the study of Hamiltonian cycles in hypercubes with faulty
edges [2, 16, 18]. Dvoˇr´ak [4] showed that any set of at most 2n−3 edges of Q (n ≥ 2)
n
that induces vertex-disjoint paths is contained in a Hamiltonian cycle. Wang et al. [18]
∗This work is supported by NSFC (grant no. 61073046).
†Corresponding author.
proved that this result still holds even if Q has some faulty edges, see Lemma 2.8 below.
n
More details about this topic see [5, 15].
Kreweras [11]conjectured that every perfect matching of Q for n ≥ 2canbeextended
n
to a Hamiltonian cycle of Q . In [6, 7] Fink solved this conjecture by proving a stronger
n
result. Ruskey and Savage [14] asked the following question: Does every matching of Q
n
for n ≥ 2 extend to a Hamiltonian cycle of Q ? Fink [6] pointed out that the statement is
n
true for n = 2,3,4. The result in [4] implied that every matching of at most 2n−3 edges
can be extended to a Hamiltonian cycle of Q . Vandenbussche and West [17] showed that
n
every k-suitable matching of at most k(n−k) + (k−1)(k−2) edges for 1 ≤ k ≤ n−3 and
2
every induced matching can be extended to a perfect matching of Q , so can be extended
n
to a Hamiltonian cycle of Q .
n
In this paper, we consider Ruskey and Savage’s question in faulty hypercubes and
obtain the following main results: every matching M of at most 2n − 1 edges can be
extended to a Hamiltonian cycle of Q for n ≥ 2; when n ≥ 4 and M is nonempty this
n
result still holds even if Q has at most n − 1 − ⌈|M|⌉ faulty edges with one exception.
n 2
The rest of this paper is organized as follows. In Section 2 we introduce some necessary
definitions and preliminaries. In Section 3 and Section 4 we discuss bases of induction of
the two main theorems. The main results are stated and proved in Section 5.
2 Definitions and preliminaries
The terminology and notation used in this paper but undefined below can be found in [1].
As usual, the vertex set and edge set of a graph G are denoted by V(G) and E(G). For a
setE ⊆ E(G), letG−E denotethegraphwithverticesV(G)andedgesE(G)\E. Thedis-
tancebetweentwo verticesuandv isthenumber ofedgesinashortest pathbetween uand
v in G, denoted by d (u,v), with the subscript being omitted when the context is clear.
G
For any two edges uv,xy, d(uv,xy) = min{d(u,x),d(u,y),d(v,x),d(v,y)}. Throughout
the paper, n always denotes a positive integer while [n] denotes the set {1,2,...,n}.
The n-dimensional hypercube Q is a graph whose vertex set consists of all binary
n
strings of length n, with two vertices being adjacent whenever the corresponding strings
differ in exactly one position. An edge in Q is called an i-dimensional edge if its
n
endvertices differ in the ith position. The set of all i-dimensional edges of Q is denoted
n
by E . For any given j ∈ [n], let Q0 and Q1 be two (n−1)-dimensional subcubes of
i n−1 n−1
Q induced by all the vertices with the jth positions being 0 and 1, respectively. Since
n
Q − E = Q0 ∪ Q1 , we say that Q is decomposed into two (n − 1)-dimensional
n j n−1 n−1 n
subcubes Q0 and Q1 by E . Any vertex u ∈ V(Q0 ) has in Q1 a unique neighbor,
n−1 n−1 j n−1 n−1
denoted by u . Similarly, any vertex v ∈ V(Q1 ) has in Q0 a unique neighbor, denoted
1 n−1 n−1
by v . For any edge e = uv ∈ E(Q0 ), e denotes the edge u v ∈ E(Q1 ). For given
0 n−1 1 1 1 n−1
F ⊆ E(Q ), let F = F ∩E(Qδ ) for δ ∈ {0,1}.
n δ n−1
Lemma 2.1. [19] For n ≥ 2, let e and f be two disjoint edges in Q . Then Q can be
n n
decomposed into two (n−1)-dimensional subcubes such that one contains e and the other
contains f.
Let us recall the following classical result, originally proved by Havel in [10].
Lemma 2.2. [10] Let n ≥ 1 and x,y ∈ V(Q ) be such that d(x,y) is odd. Then there
n
exists a Hamiltonian path between x and y in Q .
n
Lemma 2.3. [4] For n ≥ 2, let x,y ∈ V(Q ) and e ∈ E(Q ) such that d(x,y) is odd and
n n
e 6= xy. Then there is a Hamiltonian path of Q between x and y passing through edge e.
n
Lemma 2.4. [3] For n ≥ 3, let u,v ∈ V(Q ) and F ⊆ E(Q ) such that d(u,v) is odd
n n
and |F| ≤ 1. Then there exists a Hamiltonian path in Q −F between u and v.
n
A path with endvertices u and v is denoted by P . We say that paths {P }k are
uv i i=1
spanning paths of a graph G if {V(P )}k partitions V(G).
i i=1
Lemma 2.5. [4] For n ≥ 2, let x,y,u,v be pairwise distinct vertices of Q such that
n
both d(x,y) and d(u,v) are odd. Then (i) there exist spanning paths P ,P of Q ; (ii)
xy uv n
moreover, in the case when d(x,y) = 1, path P can be chosen such that P = xy, unless
xy xy
n = 3, d(u,v) = 1 and d(xy,uv) = 2.
Note that for any edge e ∈ E(Q ) there exists a unique edge f ∈ E(Q ) such that
3 3
d(e,f) = 2.
A forest is linear if each component of it is a path.
Lemma 2.6. [4] For n ≥ 2, let E ⊆ E(Q ) with |E| ≤ 2n − 3. Then there exists a
n
Hamiltonian cycle of Q passing through E if and only if the subgraph induced by E is a
n
linear forest.
Lemma 2.7. [13] Let n ≥ 3 and F ⊆ E(Q ) with |F| ≤ n − 2. Then any edge of
n
E(Q )\F lies on a Hamiltonian cycle of Q −F.
n n
Lemma 2.8. [18] For n ≥ 2, let F ⊆ E(Q ), E ⊆ E(Q ) \ F with 1 ≤ |E| ≤ 2n −3,
n n
|F| ≤ n−2−⌊|E|⌋. If the subgraph induced by E is a linear forest, then all edges of E
2
lie on a Hamiltonian cycle of Q −F.
n
3 Base of induction of the first main theorem
A set of edges in a graph G is called a matching if no two edges have a point in common.
A matching is perfect if it covers all of V(G).
Lemma 3.1. [6, 7] Forevery perfectmatching M of K(Q ) there exists a perfectmatching
n
R of Q , n ≥ 2, such that M ∪R is a Hamiltonian cycle of K(Q ), where K(Q ) is the
n n n
complete graph on the vertices of the hypercube Q .
n
Lemma 3.2. Let M be a matching of Q and u,v ∈ V(Q ) such that u ∈/ V(M) and
3 3
d(u,v) is odd. Then there exists a Hamiltonian path of Q between u and v passing through
3
M.
Proof. Since u ∈/ V(M), we have |M| ≤ 3, there are nine possibilities for M and u up to
isomorphism, see Figure 1. Observe that when |M| ≤ 2 we can extend M to a matching
′ ′ ′
M of size 3 which satisfying u ∈/ V(M ). If the conclusion holds for M , then it also holds
for M, so we can assume that |M| = 3. Then there is a Hamiltonian path of Q between
3
u and v passing through M, see Figure 2.
u
u u u u
(1) (2) (3) (4) (5)
u u u u
(6) (7) (8) (9)
Fig. 1. Nine possibilities for M and u of Lemma 3.2 with the edges of M highlighted.
v v
v
v
v
u u u v u u u
v
v
v
v
u v u u u u v u
Fig. 2. Hamiltonian paths in Lemma 3.2 between u and v passing through M.
The following lemma is the base of induction of Theorem 5.1.
Lemma 3.3. Every matching of Q can be extended to a Hamiltonian cycle of Q for
n n
n = 2,3,4.
Proof. Let M be a matching of Q . By Lemma 3.1, every perfect matching can be
n
extended a Hamiltonian cycle of Q for n ≥ 2, so we only need to consider the case that
n
M is not perfect. If n = 2,3, then |M| ≤ 2n− 3 and therefore by Lemma 2.6, there is
a Hamiltonian cycle of Q containing M. If n = 4, since |M| ≤ 7, there exists j ∈ [4]
n
such that |M ∩ E | ≤ 1. Decompose Q into Q0 and Q1 by E such that M is not
j 4 3 3 j 1
perfect. First find a Hamiltonian cycle C of Q0 containing M . If M ∩E = {uu }, then
0 3 0 j 1
select a neighbor v of u on C . Since M is a matching, uv ∈/ M. If M ∩ E = ∅, then
0 j
let uv ∈ E(C ) \M such that u ∈/ V(M ). Since u ∈/ V(M ) and d(u ,v ) is odd, by
0 0 1 1 1 1 1 1
Lemma 3.2 there is a Hamiltonian path P of Q1 passing through M . Then the desired
u1v1 3 1
Hamiltonian cycle of Q is induced by edges of (E(C )∪E(P )∪{uu ,vv })\{uv}.
4 0 u1v1 1 1
4 Base of induction of the second main theorem
Lemma 4.1. Let F = {f} ⊆ E(Q ) and M be a matching of Q −F with |M| = 2. Then
3 3
there exists a Hamiltonian cycle containing M in Q − F except that (Q ,M,F) is the
3 3
case (b) or (c) on Figure 3 up to isomorphism.
(b) (c)
Fig. 3. Two counterexamples of Lemma 4.1 with the edges of M highlighted and the
edge of F dotted.
Proof. There are three non-isomorphic matchings of size 2 in Q , see Figure 4(1)(5)(9).
3
In Figure 4(2), if f is one of the dotted edges, then there is a Hamiltonian cycle containing
M in Q −F. By exhausting all possibilities for {M,f}, as it is presented on Figure 4,
3
we can verify the conclusion holds.
Lemma 4.2. Let M be a matching of Q with |M| = 3. Then there exists at most one
3
edge e in E(Q )\M such that M ∪{e} is not contained in any Hamiltonian cycle of Q .
3 3
(1) (2) (3) (4)
(5) (6) (7) (8)
(9) (10) (11)
Fig. 4. Three non-isomorphic matchings of size 2 in Q and Hamiltonian cycles in
3
Lemma 4.1.
Proof. There are three non-isomorphic matchings of size 3 in Q . By exhausting all
3
possibilities for {M,e}, we can find a Hamiltonian cycle containing M ∪{e} except that
{M,e} is as on Figure 5(1), see Figure 5.
e
(1)
Fig. 5. Hamiltonian cycles of Q containing M ∪{e} in Lemma 4.2 with one exception.
3
Lemma 4.3. Let F ⊆ E(Q ) and M be a matching of Q −F with |M| = 4 and |F| = 1. If
4 4
there exists j ∈ [4] such that M∩E = ∅, then there exists a Hamiltonian cycle containing
j
M in Q −F except that (Q ,M,F) is the case (a) on Figure 6 up to isomorphism.
4 4
(a)
Fig. 6. The counterexample (Q ,M,F) with the edges of M highlighted and the edge of
4
F dotted, where the edges between the two copies of Q are omitted.
3
Proof. Decompose Q into Q0 and Q1 by E . By symmetry, we may assume that |M | ≥
4 3 3 j 0
|M |, moreover, we can choose j such that |M | ≤ 3. Let F = {f}.
1 0
Case 1. |M | = 3. Let M = {xy}.
0 1
Subcase 1.1. f ∈ E(Q0). Apply Lemma 2.6 to obtain a Hamiltonian cycle C of Q0
3 0 3
containing M . If f ∈/ E(C ), let uv ∈ E(C ) \ M such that u v 6= xy; if f ∈ E(C )
0 0 0 0 1 1 0
and f 6= xy, let uv = f. By Lemma 2.3, there exists a Hamiltonian path P passing
1 u1v1
through xy in Q1. Then the desired Hamiltonian cycle in Q −F is induced by edges of
3 4
(E(C )∪E(P ) ∪{uu ,vv })\{uv}.
0 u1v1 1 1
If f ∈ E(C ) and f = xy, then f = x y . Since |E(C ) \(M ∪{e ∈ E(C ) | e = f
0 1 0 0 0 0 0
or e is adjacent to f})| ≥ 8 − (3 + 3) > 1, there exists an edge uv ∈ E(C ) \ M such
0 0
that d(uv,f) = 1, then d(u v ,xy) = 1. By Lemma 2.5, there exist spanning paths
1 1
P = xy,P of Q1. Then the desired Hamiltonian cycle in Q −F is induced by edges
xy u1v1 3 4
of (E(C )∪E(P ) ∪{x x,y y,xy,uu ,vv })\{f,uv}.
0 u1v1 0 0 1 1
Subcase 1.2. f ∈/ E(Q0). Since |F | ≤ 1, by Lemma 2.7, xy lies on a Hamiltonian
3 1
cycle C in Q1 − F . Since |E(C )| − (|M | + |M |) = 4 > 2 + 1, by Lemma 4.2 there
1 3 1 1 0 1
exists an edge uv ∈ E(C ) \ M such that u v ∈/ M , f ∈/ {u u,v v} and M ∪ {u v }
1 1 0 0 0 0 0 0 0 0
is contained in some Hamiltonian cycle C of Q0. Then the desired Hamiltonian cycle in
0 3
Q −F is induced by edges of (E(C )∪E(C )∪{u u,v v})\{u v ,uv}.
4 0 1 0 0 0 0
Case 2. |M | = 2
0
Since |M | = |M | = 2, by symmetry, we may assume that f ∈ E ∪ E(Q0). If
0 1 j 3
f ∈ E , apply Lemma 2.6 to find a Hamiltonian cycle C of Q0 containing M . Since
j 0 3 0
|E(C )\(M ∪{e ∈ E(C ) | e is adjacent to f or e ∈ M })| ≥ 8−(2+2+2) = 2, there
0 0 0 1 1
exists anedge uv ∈ E(C )\M such thatu v ∈/ M and f ∈/ {uu ,vv }. If f ∈ E(Q0) and
0 0 1 1 1 1 1 3
(Q0,M ,F ) is not the case (b) or (c), by Lemma 4.1 there exists a Hamiltonian cycle C
3 0 0 0
containing M in Q0 −F , and let uv ∈ E(C )\M such that u v ∈/ M . If (Q0,M ,F )
0 3 0 0 0 1 1 1 3 0 0
is (b) or (c) and f ∈/ M , since M ∪{f} is a linear forest of size 3, by Lemma 2.6 there
1 1 0
is a Hamiltonian cycle C of Q0 containing M ∪{f} and let uv = f, then u v ∈/ M .
0 3 0 1 1 1
For the above three cases, since M ∪ {u v } is a linear forest of size 3, by Lemma
1 1 1
2.6 there is a Hamiltonian cycle C containing M ∪ {u v } in Q1. Then the desired
1 1 1 1 3
HamiltoniancycleinQ −F isinducedbyedgesof(E(C )∪E(C )∪{uu ,vv })\{uv,u v }.
4 0 1 1 1 1 1
If (Q0,M ,F ) is the case (b) and f ∈ M , then let M \ {f } = {e}. Since M
3 0 0 1 1 1 1
is a matching, we have e ∈ {e ,e ,e ,e ,e ,e ,e }, see Figure 7. Then we can find a
1 2 3 4 5 6 7
′ ′′
Hamiltonian cycle C or C containing M in Q −F except that (Q ,M,F) is (a).
4 4
If (Q0,M ,F ) is the case (c) and f ∈ M , then every edge of E(Q1)\{f } which is
3 0 0 1 1 3 1
not adjacent to f lies in a Hamiltonian cycle C1 or C2 in Q −F, see Figure 8. Hence
1 4
C1 or C2 is the desired Hamiltonian cycle in Q −F.
4
e
1
e
e 2 e
3 5 e
e e 6
4 7
′ ′′
Fig. 7. The two Hamiltonian cycles C and C in Q
4
Fig. 8. The two Hamiltonian cycles C1 and C2 in Q
4
Lemma 4.4. Let F ⊆ E(Q ) and M be a matching of Q −F with |M| = 4 and |F| = 1.
4 4
If |M ∩ E | = 1 for any i ∈ [4], then there exists a Hamiltonian cycle containing M in
i
Q −F.
4
Proof. Since |M ∩ E | = 1 for any i ∈ [4] and |F| = 1, we can choose j ∈ [4] such that
i
|M∩E | = 1 = |F∩E |. Decompose Q into Q0 and Q1 by E such that |M | ≥ |M |. Let
j j 4 3 3 j 0 1
M∩E = {uu } and F = F∩E = {f}. Apply Lemma 2.6 to find a Hamiltonian cycle C
j 1 j 0
of Q0 containing M . Select a neighbor v of u on C such that vv 6= f. Then M ∪{u v }
3 0 0 1 1 1 1
forms a linear forest of size at most 2 and therefore, using Lemma 2.6 again, there is a
Hamiltonian cycle C of Q1 containing M ∪{u v }. Then the desired Hamiltonian cycle
1 3 1 1 1
in Q −F is induced by edges of (E(C )∪E(C )∪{uu ,vv })\{uv,u v }.
4 0 1 1 1 1 1
Lemma 4.5. Let F ⊆ E(Q ) and M be a matching of Q −F with |M| = 4 and |F| = 1.
4 4
Then there exists a Hamiltonian cycle containing M in Q −F except that (Q ,M,F) is
4 4
the case (a) on Figure 6 up to isomorphism.
In Lemma 4.5, since |M| = 4, there exists j ∈ [4] such that M∩E = ∅ or |M∩E | = 1
j i
for any i ∈ [4], so Lemma 4.5 holds by Lemma 4.3 and 4.4.
Lemma 4.6. Let F = {f,g} ⊆ E(Q ) and M = {e,h} be a matching of Q −F. Then
4 4
there exists a Hamiltonian cycle containing M in Q −F.
4
Proof. Apply Lemma 2.1 to decompose Q into Q0 and Q1 by E such that e ∈ E(Q0)
4 3 3 j 3
and h ∈ E(Q1). By symmetry, we may assume that |F | ≥ |F |, then |F | ≤ 1.
3 0 1 1
Case 1. F = ∅. If |F | ≤ 1, by Lemma 2.7, e lies on a Hamiltonian cycle C in
1 0 0
Q0 −F . Since |E(C )| = 8 > 2+4, let uv ∈ E(C )\{e} such that {uu ,vv }∩F = ∅
3 0 0 0 1 1
and u v 6= h. If F = F, since at least one of f and g is not h, we may assume that
1 1 0 1 1
f 6= h. By Lemma 2.7, e lies on a Hamiltonian cycle C in Q0 −{g}. If f ∈/ E(C ), let
1 0 3 0
uv ∈ E(C )\{e} such that u v 6= h; if f ∈ E(C ), let uv = f, note that u v = f 6= h.
0 1 1 0 1 1 1
Next apply Lemma 2.3 to find a Hamiltonian path P of Q1 passing through h.
u1v1 3
Then the desired Hamiltonian cycle in Q −F is induced by edges of (E(C )∪E(P )∪
4 0 u1v1
{uu ,vv })\{uv}.
1 1
Case 2. |F | = 1. Then |F | = |F | = 1 and F ∩E = ∅. By Lemma 2.7, e lies on a
1 0 1 j
Hamiltonian cycle C in Q0 −F and h lies on a Hamiltonian cycle C in Q1 −F . Since
0 3 0 1 3 1
|E(C )\{e}|−|E(Q1)\E(C )| = 7−4 > 1, there has to be an edge uv ∈ E(C )\{e}
0 3 1 0
such that u v ∈ E(C )\{h}. Then the desired Hamiltonian cycle in Q −F is induced
1 1 1 4
by edges of (E(C )∪E(C )∪{uu ,vv })\{uv,u v }.
0 1 1 1 1 1
Using Lemmas 2.8, 3.3, 4.5 and 4.6, we obtain,
Lemma 4.7. Let F ⊆ E(Q ) and M be a matching of Q − F with 1 ≤ |M| ≤ 6 and
4 4
|F| ≤ 3−⌈|M|⌉. Then there exists a Hamiltonian cycle containing M in Q −F except
2 4
that (Q ,M,F) is the case (a) on Figure 6 up to isomorphism.
4
Lemma 4.8. Let F ⊆ E(Q ) and M be a matching of Q − F with 1 ≤ |M| ≤ 8 and
5 5
|F| ≤ 4 − ⌈|M|⌉. Then there exists j ∈ [5] such that |E ∩ (M ∪ F)| ≤ 1 and none
2 j
of (Q0,M ,F ) and (Q1,M ,F ) is the case (a) on Figure 6 up to isomorphism, where
4 0 0 4 1 1
Q0 ∪Q1 = Q −E .
4 4 5 j
Proof. If F = ∅ or |M| ≤ 3, since |M ∪F| ≤ |M|+4−⌈|M|⌉ ≤ 8, we can select j ∈ [5]
2
such that |E ∩(M ∪F)| ≤ 1. Moreover, none of (Q0,M ,F ) and (Q1,M ,F ) is (a). If
j 4 0 0 4 1 1
|M| = 4, then |F| ≤ 2 and |M ∪F| ≤ 6. First select j ∈ [5] such that |E ∩(M ∪F)| ≤ 1.
j
If one of (Q0,M ,F ) and (Q1,M ,F ) is the case (a), then we can choose j ∈ [5] such
4 0 0 4 1 1 0
that|F∩E | ≤ 1,M∩E = ∅ andthetwo subcubes decomposed by E bothcontaintwo
j0 j0 j0
edges of M. Hence j or j satisfies the lemma. So it suffices to prove the case 5 ≤ |M| ≤ 6
0
and |F| = 1.
Since |M|+|F| ≤ 7, there exists i ∈ [5] such that |E ∩(M ∪F)| ≤ 1. Decompose Q
i 5
into two subcubes Q0i and Q1i by E . If none of (Q0i,M ,F ) and (Q1i,M ,F ) is the case
4 4 i 4 0 0 4 1 1
(a), then the conclusion holds. Otherwise, assume that (Q0i,M ,F ) is (a). Without loss
4 0 0
of generality, assume that i = 1 and (M ∪F) ⊆ E , then |(M ∪F)∩(E ∪E ∪E )| ≤ 2.
0 2 3 4 5
Hence, at least two of E ,E and E satisfy |E ∩ (M ∪ F)| ≤ 1, where j ∈ {3,4,5}.
3 4 5 j
Assume that |E ∩ (M ∪ F)| ≤ 1 and |E ∩ (M ∪ F)| ≤ 1. We claim that j = 4 or
4 5
j = 5 satisfies the lemma. If j = 4 does not satisfy the lemma, then one of the two
subcubes decomposed from Q by E is the case (a), see Figure 9. Then the two subcubes
5 4
decomposed from Q by E both contain three edges of M, so none of them is the case
5 5
(a), hence j = 5 satisfies the lemma. The proof is complete.
E E
2 2
E
4
E
1
E
4
E
5
Fig. 9. (Q ,M,F) satisfying one of the two subcubes decomposed by E is the case (a)
5 i
where i = 1,4.
5 Main results
Theorem 5.1. For n ≥ 2, let M be a matching of Q with |M| ≤ 2n− 1. Then there
n
exists a Hamiltonian cycle of Q containing M.
n
Proof. We prove the theorem by induction on n. The theorem holds for n = 2,3,4 by
Lemma 3.3. Assume that the theorem holds for n−1(≥ 4), we are to show it holds for
n(≥ 5). Since |M| ≤ 2n−1, there exists j ∈ [n] such that |M ∩E | ≤ 1. Decompose Q
j n
into Q0 and Q1 by E such that |M | ≥ |M |, then |M | ≤ n−1.
n−1 n−1 j 0 1 1
Claim 1. If there exists a Hamiltonian cycle C containing M in Q0 , then we can
0 0 n−1
construct a Hamiltonian cycle of Q containing M.
n
If M ∩E = {uu }, select a neighbor v of u on C . Since M is a matching, we have
j 1 0
{uv,u v }∩M = ∅. If M ∩E = ∅, since |E(C )|−(|M |+|M |) ≥ 2n−1 −(2n−1) ≥ 1
1 1 j 0 0 1