Fault-free Hamiltonian cycles in balanced hypercube with conditional edge faults

1.4 MB·

by  Pingshan Li

Fault-free Hamiltonian cycles in balanced hypercubes
with conditional edge faults
Pingshan Li Min Xu∗
Sch. Math. Sci. & Lab. Math. Com. Sys., Beijing Normal University, Beijing, 100875, China
Abstract
The balanced hypercube, BH , is a variant of hypercube Q . Zhou et al. [Inform. Sci. 300
n n
7 (2015) 20-27] proposed an interesting problem that whether there is a fault-free Hamiltonian cycle
1 in BH with each vertex incident to at least two fault-free edges. In this paper, we consider this
n
0 problem and show that each fault-free edge lies on a fault-free Hamiltonian cycle in BH after no
n
2
more than 4n−5 faulty edges occur if each vertex is incident with at least two fault-free edges for

n all n≥2. Our result is optimal with respect to the maximum number of tolerated edge faults.
a
J Key words: Balanced hypercubes; Hypercubes; Hamiltonian cycles; Fault-tolerance.

2
1

1 Introduction
]
O
C In the field of parallel and distributed systems, interconnection networks are an important research
. area. Typically,thetopologyofanetworkcanberepresentedasagraphinwhichtheverticesrepresent
h
t processors and the edges represent communication links.
a
m For graph definitions and notations, we follow [1]. A graph G consists of a vertex set V(G) and
[ an edge set E(G), where an edge is an unordered pair of distinct vertices of G. A graph G is called

bipartiteifitsvertexsetcanbepartitionedintotwopartsV ,V suchthateveryedgehasoneendpoint
1 1 2
v in V and one in V . A vertex v is a neighbor of u if (u,v) is an edge of G, and N (u) denotes all
1 2 G
6
the neighbors of u in G. A path P of length (cid:96) from x to y is a finite sequence of distinct vertices
1
2 (cid:104)v0,v1,··· ,v(cid:96)(cid:105) such that x = v0,y = v(cid:96), and (vi,vi+1) ∈ E for 0 ≤ i ≤ (cid:96)−1. We also denote the path
3 P as(cid:104)v ,v ,··· ,v ,Q,v ,v ,··· ,v (cid:105), whereQisthepath(cid:104)v ,v ,··· ,v (cid:105). AcycleC oflength(cid:96)+1
0 1 i j j+1 (cid:96) i i+1 j
0
is a closed path(cid:104)v ,v ,··· ,v ,v (cid:105).
. 0 1 (cid:96) 0
1
A path (resp., cycle) is called a Hamiltonian path (resp., cycle) if it contains all the vertices of
0
7 G. A graph G is said to be Hamiltonian if there is a Hamiltonian cycle. A graph G is said to be
1 Hamiltonian connected if there is a Hamiltonian path between any two vertices of G. A bipartite
:
v graph is Hamiltonian laceable if there is a Hamiltonian path between any two vertices in different
i
X bipartite sets.
r The Hamiltonian property is one of the major requirements in designing network topologies since
a
a topology structure containing Hamiltonian paths or cycles can efficiently simulate many algorithms
that are designed on linear arrays or rings [16].
It is important to consider fault-tolerance in networks since faults may occur in real networks.
Two fault models have been studied in many well-known networks, one is the random fault model,
whichmeansthatfaultsmayoccuranywherewithoutanyrestriction, see, forexample, [13,14,18,24].
The other is the conditional fault model, which assumes that the fault distribution is limited. For
example, some studies on two or more non-faulty edges incident to each vertex can be found in
[2, 7, 9, 10, 15, 19, 21].
The hypercube network has been proved to be one of the most popular interconnection networks
because it possesses many excellent properties such as a recursive structure, regularity, and symmetry.
∗Corresponding author.
E-mail address: [email protected] (M. Xu).
1
The balanced hypercube, proposed by Huang and Wu [8], is a hypercube variant. Similar to hyper-
cubes, balanced hypercubes are bipartite graphs [8] that are vertex-transitive [17] and edge-transitive
[26]. Balanced hypercubes are superior to hypercubes in that they have a smaller diameter than
hypercubes[17].
The balanced hypercube, BH , has been studied by many researchers in recent years. Xu et al.
n
[20] proved that BH is edge-bipancyclic and Hamiltonian laceable. Yang [22] proved that BH is bi-
n n
panconnected. Yang[23]alsodemonstratedthatthesuperconnectivityofBH is4n−4andthesuper
n
edge-connectivity of BH is 4n−2 for n ≥ 2. Lu¨ et al. [12] proved that BH is hyper-Hamiltonian
n n
laceable. Cheng et al. [3] proved that BH is (n−1)-vertex-fault-tolerant edge-bipancyclic. Hao et al.
n
[6] showed that there is a fault-free Hamiltonian path between any two adjacent vertices in BH with
n
2n−2 faulty edges. Cheng et al. [5] proved that BH is 2n−3 edge-fault-tolerant 6-edge-bipancyclic
n
for all n ≥ 2. Zhou et al. [25] proved that BH is 2n−2 edge-fault-tolerant Hamiltonian laceable,
n
and they proposed an interesting problem that whether there is a fault-free Hamiltonian cycle in BH
n
with each vertex incident to at least two fault-free edges. In this paper, we consider this problem and
show that each fault-free edge lies on a fault-free Hamiltonian cycle in BH after no more than 4n−5
n
faulty edges occur if each vertex is incident with at least two fault-free edges for all n ≥ 2. Our result
is optimal with respect to the maximum number of tolerated edge faults.
The rest of this paper is organized as follows. In Section 2, we introduce two equivalent definitions
of balanced hypercubes and discuss some of their properties. In section 3, we introduce some lemmas
used in the proof of the main result and prove the main result. Finally, we conclude this paper and
give an example to show that our result is optimal in Section 4 .
2 Balanced hypercubes
Wu and Huang [8] present two equivalent definitions of BH as follows:
n
Definition 2.1 An n-dimensional balanced hypercube BH has 22n vertices, each labeled by an n-
n
bit string (a ,a ,··· ,a ), where a ∈ {0,1,2,3} for all 0 ≤ i ≤ n − 1. An arbitrary vertex
0 1 n−1 i
(a ,a ,··· ,a ,a , a ,··· ,a ) is adjacent to the following 2n vertices:
0 1 i−1 i i+1 n−1
(1) ((a ±1) mod 4,a ,··· ,a ,a ,a ,··· ,a ) where 1 ≤ i ≤ n−1,
0 1 i−1 i i+1 n−1
(2) ((a0±1) mod 4,a1,··· ,ai−1,(ai+(−1)a0) mod 4,ai+1,··· ,an−1) where 1 ≤ i ≤ n−1.
In BH , the first coordinate a of vertex (a ,a ,··· ,a ) is called the inner index, and the sec-
n 0 0 1 n−1
ond coordinate a (1 ≤ i ≤ n − 1) is called the i-dimension index. From the definition, we have
i
N ((a ,a ,··· ,a )) = N ((a +2,a ,··· ,a )). Figure 1 shows two balanced hypercubes of
BHn 0 1 n−1 BHn 0 1 n−1
dimensional one and two.
Briefly, we assume that ‘+,−’ for the coordinate of a vertex is an operation with mod 4 in the
remainder of the paper. Let X = {(a ,a ,··· ,a ,a ,a ,··· ,a ) | a ∈ {0,1,2,3},0 ≤ k ≤
j,i 0 1 j−1 j j+1 n−1 k
n−1,a = i} for 1 ≤ j ≤ n−1 and i ∈ {0,1,2,3} and let BHj,i = BH [X ]. Then, BH can be
j n−1 n j,i n
dividedintofourcopies: BHj,0 ,BHj,1 ,BHj,2 , andBHj,3 whereBHj,i ∼= BH fori = 0,1,2,3
n−1 n−1 n−1 n−1 n−1 n−1
[3]. We use BHi to denote BHn−1,i for i = 0,1,2,3 .
n−1 n−1
Definition 2.2 An n-dimensional balanced hypercube BH can be constructed recursively as follows:
n
1. BH is a cycle of length four with vertex-set {0,1,2,3}.
1
2. BH is a construct from four copies of BH : BH0 ,BH1 ,BH2 , and BH3 . Each
n n−1 n−1 n−1 n−1 n−1
vertex (a ,a ,···, a ,i) has two extra adjacent vertices:
0 1 n−2
(1) In BHi+1 : (a ±1,a ,··· ,a ,i+1) if a is even.
n−1 0 1 n−2 0
(2) In BHi−1 : (a ±1,a ,··· ,a ,i−1) if a is odd.
n−1 0 1 n−2 0
2
Figure 1: Illustration of BH and BH
1 2
Since BH is a bipartite graph, V(BH ) can be divided into two disjoint parts. Obviously, the
n n
vertex-set V = {a = (a ,a ,··· ,a ) | a ∈ V(BH ) with a odd} and V = {a = (a ,a ,··· ,a ) |
1 0 1 n−1 n 0 2 0 1 n−1
a ∈ V(BH ) with a even} form the desired partition. We use black nodes to denote the vertices in
n 0
V and white nodes to denote the vertices in V .
1 2
Let (u,v) be an edge of BH . If u and v differ only with regard to the inner index, then (u,v)
n
is said to be a 0-dimension edge. If u and v differ not only in terms of the inner index but also with
regard to the i-dimension index, then (u,v) is called the i-dimension edge. We use ∂D (0 ≤ d ≤ n−1)
d
to denote the set of all d-dimension edges.
There are some known properties about BH .
n
Lemma 2.3 ([17, 26]) The balanced hypercube BH is vertex-transitive and edge-transitive.
n
AbipartitegraphGisk-fault-tolerantHamiltonianlaceableifG−F remainsHamiltonianlaceable
for F ⊆ V(G)∪E(G) with |F| ≤ k. A bipartite graph G is k-edge-fault-tolerant Hamiltonian laceable
if G−F remains Hamiltonian laceable for F ⊆ E(G) with |F| ≤ k. Zhou et al. obtained the following
result.
Lemma 2.4 ([25]) The balanced hypercube BH is (2n−2)-edge-fault-tolerant Hamiltonian laceable
n
for n ≥ 2.
Lemma 2.5 ([11]) Let n ≥ 2. Then, BH − ∂D has four components, and each component is
n 0
isomorphic to BH .
n−1
The above lemma shows that one can divide BH into four BH s by deleting ∂D for any d ∈
n n d
{0,1,··· ,n−1}. The four components of BH −∂D are BHj,0 , BHj,1 , BHj,2 , and BHj,3 for
n j n−1 n−1 n−1 n−1
1 ≤ j ≤ n−1. Forconvenience, weuseBH0,0 ,BH0,1 ,BH0,2 , andBH0,3 todenotethecomponents
n−1 n−1 n−1 n−1
of BH −∂D throughout this paper.
n 0
Lemma 2.6 ([4]) Let U,V be two distinct bipartitions of BH , u1,u2 be two different vertices in U,
n
and v1,v2 be two different vertices in V. Then, there are two disjoint paths P,Q such that (1) P joins
u1 and v1, and Q joins u2 and v2; (2) V(P ∪Q) = V(BH ).
n
A graph G is hyper-Hamiltonian laceable if it is Hamiltonian laceable and, for an arbitrary vertex
v in V where i ∈ {0,1}, there is a Hamiltonian path in G−v joining any two different vertices in
i
V . Lu¨ et al. obtained the following result.
1−i
Lemma 2.7 ([12]) The balanced hypercube BH is hyper-Hamiltonian laceable for n ≥ 1.
n
3
3 Main result
In this section, we will give the proof of the main result. First, we introduce some lemmas that will
be used in the proof of the main result.
Definition 3.1 Suppose that G is a graph and F ⊆ E(G). A vertex u is called i-rescuable in G if
|N (u)| = i.
G−F
Lemma 3.2 Let n ≥ 3,F ⊆ E(BH ) with |F| = 4n−5 and δ(BH −F) ≥ 2. Then, at least one
n n
statement holds in the following (1) and (2).
(1) There is an integer m in {0,1,··· ,n−1} such that |F ∩∂D | ≥ 3 and δ(BH −∂D ) ≥ 2.
m n m
(2) There are two integers m,m(cid:48) in {0,1,··· ,n−1} such that |F ∩∂D | ≥ 2 and |F ∩∂D | ≥ 2.
m m(cid:48)
Furthermore, there is no isolated vertex and no more than one 1-rescuable vertex in BH − ∂D
n m
(resp., BH −∂D ).
n m(cid:48)
Proof: Let F be a faulty set in E(BH ) with |F| = 4n−5 and δ(BH −F) ≥ 2. Suppose that u
n n
is k-rescuable, and v is t-rescuable in BH where k ≤ t ≤ min{m | x is m-rescuable in BH , where
n n
x ∈ V(BH )\{u,v}}.
n
Ifk ≥ 4,thereisanintegerm ∈ {0,1,··· ,n−1}suchthat|F∩∂D | ≥ 3because|F| = 4n−5 > 2n.
m
Each vertex is at least 2-rescuable in BH − ∂D for k ≥ 4. Thus, statement (1) holds. In the
n m
following, we assume that 2 ≤ k ≤ 3 and suppose that the dimensions of fault-free edges incident
with u are i ,i ,··· ,i and the dimensions of fault-free edges incident with v are j ,j ,··· ,j where
1 2 k 1 2 t
{i ,i ,··· ,i } and {j ,j ,··· ,j } may be multisets.
1 2 k 1 2 t
Case 1: k = 2.
Subcase 1.1: t = 2.
In this subcase, |F ∩{(u,x) | x ∈ N (u)}| = |F ∩{(v,x) | x ∈ N (v)}| = 2n−2. Since
BHn BHn
2(2n−2)−(4n−5) = 1, we obtain (u,v) ∈ F, and there is no faulty edge in BH −{u,v}. Note
n
that there is no triangle in BH . For an arbitrary vertex x in V(BH )\{u,v}, x is incident with
n n
no more than one faulty edge, which means that x is at least 3-rescuable in BH − ∂D for all
n i
i ∈ {0,1,··· ,n−1} as 2(n−1)−1 ≥ 3. Note that for 4n−5 > 2n, there is an m ∈ {0,1,··· ,n−1}
such that |F ∩∂D | ≥ 3.
m
If m ∈/ {i ,i ,j ,j }, then u and v are 2-rescuable in BH −∂D . Therefore, statement (1) holds.
1 2 1 2 n m
If m ∈ {i ,i ,j ,j }, without loss of generality, let i = m. Note that each vertex in BH is
1 2 1 2 1 n
incident with two i-dimension edges for i ∈ {0,1,··· ,n−1}. The integer m appears no more than
once in the multiset {i ,i ,j ,j }. Thus, u is 1-rescuable, and v is 2-rescuable in BH −∂D . Note
1 2 1 2 n m
that for n ≥ 3, according to the pigeonhole principle, there is an integer m(cid:48) ∈ {0,1,··· ,n−1}\{m}
such that m(cid:48) appears no more than once in the multiset {i ,j ,j }. Thus, |F ∩∂D | ≥ 2. Without
2 1 2 m(cid:48)
loss of generality, we can assume that u is 1-rescuable and v is 2-rescuable in BH −∂D . Therefore,
n m(cid:48)
statement (2) holds.
Subcase 1.2: t = 3.
First, we prove that each vertex in V(BH )\{u,v} is at least 2-rescuable in BH −∂D for all
n n i
i ∈ {0,1,··· ,n−1}.
If (u,v) ∈/ F, there are 4n−5 faulty edges incident with u or v. This means there is no faulty edge
in BH −{u,v}. Thus, each vertex in V(BH )\{u,v} is incident with no more than two faulty edges.
n n
That means it is at least 2-rescuable in BH −∂D for all i ∈ {0,1,··· ,n−1} since 2(n−1)−2 ≥ 2.
n i
If (u,v) ∈ F, there are 4n−6 faulty edges incident with u or v. This means that there is no more
thanonefaultyedgeinBH −{u,v}. SincethereisnotriangleinBH ,eachvertexinV(BH )\{u,v}
n n n
isincidentwithnomorethantwofaultyedges. Thatimpliesthatitisatleast2-rescuableinBH −∂D
n i
for all i ∈ {0,1,··· ,n−1}, as 2(n−1)−2 ≥ 2.
Next, we complete the proof of subcase 1.2.
If there is an m ∈ {0,1,··· ,n−1} such that m ∈/ {i ,i ,j ,j ,j }, then |F ∩∂D | ≥ 3 and u and
1 2 1 2 3 m
v are at least 2-rescuable in BH −∂D . Therefore, statement (1) holds.
n m
Now, we assume that {0,1,··· ,n−1} ⊆ {i ,i ,j ,j ,j }. If n ≥ 4 or n = 3 and i = i , there
1 2 1 2 3 1 2
are two distinct integers m,m(cid:48) in {0,1,··· ,n−1}\{i ,i }. Note that u is incident with two faulty
1 2
4
m (resp., m(cid:48))-dimension edges and u is 2-rescuable, v is at least 1-rescuable in BH −∂D (resp.,
n m
BH −∂D ). Therefore, statement (2) holds.
n m(cid:48)
If n = 3 and i (cid:54)= i , without loss of generality, let i = 0,i = 1. Then 2 ∈ {j ,j ,j }. Thus, u
1 2 1 2 1 2 3
is 2-rescuable, and v is at least 1-rescuable in BH −∂D . Let F(cid:48) be the set of faulty edges incident
n 2
with u or v. Then |F(cid:48)| ≥ 4n−6 = 6 and |F ∩∂D | ≥ |F(cid:48)∩∂D | ≥ 2. Moreover, 2 ≤ |F(cid:48)∩∂D | ≤ 3.
2 2 2
Further, we obtain |F(cid:48) ∩ ∂D | ≥ 2 or |F(cid:48) ∩ ∂D | ≥ 2, as |F(cid:48)| ≥ 4n − 6 = 6. Without loss of
0 1
generality, we can assume that |F(cid:48)∩∂D | ≥ 2. Since u is incident with exactly one faulty 0-dimension
0
edge, v is incident with at least one faulty 0-dimension edge. Thus, 0 appears no more than once
in the multiset {j ,j ,j }. Note that u is 1-rescuable, v is at least 2-rescuable in BH −∂D and
1 2 3 n 0
|F ∩∂D | ≥ |F(cid:48)∩∂D | ≥ 2. Let m = 0,m(cid:48) = 2. Therefore, statement (2) holds.
0 0
Subcase 1.3: t ≥ 4.
Sincet ≥ 4,eachvertexinBH −{u}isatleast2-rescuableinBH −∂D foralli ∈ {0,1,··· ,n−1}.
n n i
Suppose that there are two distinct integers m,m(cid:48) ∈ {0,1,··· ,n−1}\{i ,i }. Note that u is incident
1 2
with two faulty m (resp., m(cid:48))-dimension faulty edges, and u is 2-rescuable in BH − ∂D (resp.,
n m
BH −∂D ). Then, statement (2) holds.
n m(cid:48)
If there aren’t two distinct integers m,m(cid:48) ∈ {0,1,··· ,n−1}\{i ,i }, then n = 3,i (cid:54)= i and
1 2 1 2
|F| = 4n−5 = 7. Withoutlossofgenerality,wecanassumethati = 0,i = 1. Then,|F∩∂D | ≥ 2. If
1 2 2
|F ∩∂D | ≥ 3, then u is 2-rescuable in BH −∂D . Therefore, statement (1) holds. If |F ∩∂D | = 2,
2 n 2 2
then u is 2-rescuable in BH − ∂D . According to the pigeonhole principle, |F ∩ ∂D | ≥ 3 or
n 2 0
|F ∩∂D | ≥ 3. Without loss of generality, we can assume that |F ∩∂D | ≥ 3. Thus, u is 1-rescuable
1 0
in BH −∂D . Let m = 0,m(cid:48) = 2. Therefore, statement (2) holds.
n 0
Case 2: k = 3.
Subcase 2.1: t = 3.
(a) There is a 3-rescuable vertex w in BH where w ∈ V(BH )\{u,v}.
n n
Supposethatthedimensionsoffault-freeedgesincidentwithw ares ,s ,s . Notethat3(2n−3)−
1 2 3
(4n−5) = 2n−4 ≥ 2 and there is no triangle in BH . We obtain n = 3 and |E(BH [{u,v,w}])| =
n n
|F ∩E(BH [{u,v,w}])| = 2. Without loss of generality, let (u,v),(v,w) ∈ F. Since (4×3−5)−(6×
n
3−11) = 0, F ⊆ {e | e is incident to a vertex in {u,v,w}}. Hence, each vertex in V(BH )\{u,v,w}
n
is incident with no more than two faulty edges. This means that each vertex in V(BH )\{u,v,w} is
n
at least 2-rescuable in BH −∂D for all i ∈ {0,1,··· ,n−1}.
n i
Note that 4n−5 = 7 > 2n. There is an integer m ∈ {0,1,··· ,n−1} such that |F ∩∂D | ≥ 3.
m
Without loss of generality, we can assume that m = 0. Since each vertex is incident with exactly two
0-dimension edges, 0 appears no more than three times in the multiset {i ,i ,i ,j ,j ,j ,s ,s ,s }.
1 2 3 1 2 3 1 2 3
If 0 does not appear twice in the multiset {i ,i ,i }, {j ,j ,j } and {s ,s ,s }, then u,v,w are
1 2 3 1 2 3 1 2 3
at least 2-rescuable in BH −∂D . Thus, statement (1) holds.
n 0
Otherwise, without loss of generality, assume that i = i = 0,i = 1. Then, u is 1-rescuable, and
1 2 3
v and w are at least 2-rescuable in BH −∂D . Moreover, u is 2-rescuable in both BH −∂D and
n 0 n 1
BH −∂D .
n 2
If 2 appears four times in the multiset {j ,j ,j ,s ,s ,s }, then |F ∩∂D | ≥ 2. Note that v,w are
1 2 3 1 2 3 1
at least 2-rescuable in BH −∂D . Then, let m = 0,m(cid:48) = 1. Statement (2) holds.
n 1
If 2 appears no more than three times in the multiset {j ,j ,j ,s ,s ,s }, then one of the vertices
1 2 3 1 2 3
in {v,w} is at least 2-rescuable, and another is at least 1-rescuable in BH −∂D . Let m = 0,m(cid:48) = 2.
n 2
Statement (2) holds.
(b) Suppose that each vertex in V(BH )\{u,v} is at least 4-rescuable in BH .
n n
Note that each vertex in V(BH ) \ {u,v} is at least 2-rescuable in BH − ∂D for all i ∈
n n i
{0,1,··· ,n−1}. Let F(cid:48) be the set of faulty edges incident with u or v. Then, |F(cid:48)| ≥ 4n−7.
If there is an integer m ∈ {0,1,··· ,n−1} such that |F(cid:48) ∩∂D | ≥ 3, then m appears no more
m
than once in the multiset {i ,i ,i ,j ,j ,j }. Thus, u (resp., v) is at least 2-rescuable in BH −∂D .
1 2 3 1 2 3 n m
Statement (1) holds.
Otherwise, n = 3 and |F(cid:48) ∩ ∂D | ≤ 2 for all i = 0,1,2. Then, there are two distinct integers
i
m,m(cid:48) ∈ {0,1,2} such that (2) holds. In fact, there are two integers m,m(cid:48) ∈ {0,1,2} such that
|F(cid:48) ∩∂D | = 2 and |F(cid:48) ∩∂D | = 2, as |F(cid:48)| ≥ 4n−7 = 5. Thus, m (resp., m(cid:48)) appears no more
m m(cid:48)
5
than twice in the multiset {i ,i ,i ,j ,j ,j }. Hence, one vertex in {u,v} is at least 2-rescuable and
1 2 3 1 2 3
another is at least 1-rescuable in BH −∂D (resp., BH −∂D ). Statement (2) holds.
n m n m(cid:48)
Subcase 2.2: t ≥ 4.
Note that for t ≥ 4, each vertex in BH − {u} is at least 2-rescuable in BH − ∂D for all
n n i
i ∈ {0,1,··· ,n−1}. Since 4n−5 > 2n, there is an integer m such that |F ∩∂D | ≥ 3. If m appears
m
no more than once in the multiset {i ,i ,i }, then u is at least 2-rescuable in BH −∂D . Statement
1 2 3 n m
(1) holds. If m appears twice in the multiset {i ,i ,i }, without loss of generality, let i = i = m.
1 2 3 1 2
Then, u is 1-rescuable in BH −∂D and let m(cid:48) be an integer in {0,1,··· ,n−1}\{m,i }. Then u is
n m 3
incident with two faulty m(cid:48)-dimension edges and u is 3-rescuable in BH −∂D . Hence, statement
n m(cid:48)
(2) holds.
Owing to the above discussion, the lemma holds. (cid:50)
Lemma 3.3 Let n ≥ 3,F ⊆ E(BH ) with |F| ≤ 4n−5. Then, for any vertex u in BHj,i , there
n n−1
is a fault-free j-dimension edge (v,w) such that (u,v) ∈ E(BHj,i ). Moreover, if u is 1-rescuable in
n−1
BH −∂D , then (u,v) can be a faulty edge.
n j
Proof: Each vertex in V(BH ) is incident with two i-dimension edges for all i ∈ {0,1,··· ,n−1}.
n
Thus, for any vertex u ∈ BHj,i , its degree is 2n − 2 in BHj,i . There is at least one fault-free
n−1 n−1
j-dimension edge (v,w) such that (u,v) ∈ E(BHj,i ) for 2(2n−2)−(4n−5) ≥ 1. If u is incident
n−1
with only one fault-free edge in BHj,i , then u is incident with 2n−3 faulty edges in BHj,i . Thus,
n−1 n−1
|F ∩∂D | ≤ 2n−2. Since 2(2n−3)−(2n−2) = 2n−4 ≥ 2, there is at least one fault-free edge (w,v)
j
such that (u,v) ∈ F.
(cid:50)
Lemma 3.4 Let F ⊆ E(BH ) with |F| ≤ 3 and δ(BH −F) ≥ 2. Then, each edge in BH −F lies
2 2 2
on a fault-free Hamiltonian cycle.
Proof: The proof is rather long, so we provide it in Appendix A. (cid:50)
Theorem 3.5 Let F ⊆ E(BH ) with |F| ≤ 4n−5 and δ(BH −F) ≥ 2. Then, each edge in BH −F
n n n
lies on a fault-free Hamiltonian cycle.
Proof: We prove this theorem by induction on n. By Lemma 3.4, the theorem holds for n = 2.
Assume that this is true for 2 ≤ k ≤ n−1. By Lemma 3.2, one may partite BH along dimension
n
j, 0 ≤ j ≤ n − 1, into four BH s, denoted by BHj,0 ,BHj,1 ,BHj,2 and BHj,3 , such that
n−1 n−1 n−1 n−1 n−1
|F ∩∂D | ≥ 2 and there is no isolated vertex and no more than one 1-rescuable vertex in BH −∂D .
j n j
Let Fi = F ∩BHj,i and e = (u,v) be an arbitrary edge in BH −F. We must show that there is a
n−1 n
Hamiltonian cycle in BH −F that contains e.
n
Case 1: e ∈ E(BH )−∂D .
n j
Without loss of generality, we can assume that e ∈ BHj,0 .
n−1
Subcase 1.1: |Fi| ≤ 4n−9 for all i = 0,1,2,3.
Subcase 1.1.1: δ(BH −F −∂D ) = 1.
n j
By Lemma 3.2, there is exactly one 1-rescuable vertex in BH −∂D , say, w. Without loss of
n j
generality, we can assume that w is a white vertex.
Subcase 1.1.1.1: w ∈ BHj,0 .
n−1
Since δ(BH −F) ≥ 2, w is incident with at least one fault-free j dimension edge, say, (w,b1).
n
By Lemma 3.3, there is a fault-free j-dimension edge (b0,a3) such that (w,b0) ∈ F. By induction,
there is a Hamiltonian cycle C in BHj,0 −F +(w,b0) that contains (u,v). Since w is incident with
0 n−1
exactly two edges in BHj,0 −F +(w,b0), (w,b0) ∈ E(C ). We represent C as (cid:104)u,H ,w,b0,H(cid:48),u(cid:105).
n−1 0 0 0 0
Note that w is incident with 2n − 3 faulty edges in BHj,0 , so we have |F0| ≥ 2n − 3. Hence
n−1
|F1|+|F2|+|F3| = |F|−|F0|−|F ∩∂D | ≤ 4n−5−(2n−3)−2 = 2n−4. Let (a1,b2),(a2,b3) be
j
fault-free j-dimension edges. By Lemma 2.4, there is a fault-free Hamiltonian path H in BHj,i that
i n−1
6
Figure 2: Illustration for subcase 1.1.1.1 of theorem 3.5
joins ai and bi for i = 1,2,3. Then, the cycle C = (cid:104)u,H ,w,b1,H ,a1,b2,H ,a2,b3,H ,a3,b0,H(cid:48),u(cid:105)
0 1 2 3 0
(see figure 2) is the desired cycle.
Subcase 1.1.1.2: w ∈ BHj,1 (or BHj,3 ).
n−1 n−1
Since 4n−5−(2n−3)−2 = 2n−4, |Fi| ≤ 2n−4 for all i = 0,2,3.
Let X = {x0 | (x0,b1) ∈ ∂D \F, where x0 ∈ BHj,0 and (w,b1) ∈ F}. Since 2(2n−3)−[(4n−
j n−1
5)−(2n−3)] = 2n−4 ≥ 2, we can obtain |X| ≥ 1. Moreover, if |X| = 1, there are 2n−2 faulty edges
between BHj,0 and BHj,1 .
n−1 n−1
Subcase 1.1.1.2.1: X = {u}.
Byinduction,thereisafault-freeHamiltoniancycleC inBHj,0 thatcontains(u,v)andaHamil-
0 n−1
toniancycleC inBHj,1 −F+(w,b1)thatcontains(w,b1). SupposethatN (u) = {v,b0}. LetH =
1 n−1 C0 0
C −(u,b0),H = C −(w,b1). Notethat|X| = 1, sothereare2n−2faultyj-dimensionedgesbetween
0 1 1
BHj,0 and BHj,1 . Thus, F ∩E(BHj,0 ,BHj,3 ) = ∅. Let (b0,a3),(a2,b3) be fault-free j-dimension
n−1 n−1 n−1 n−1
edges. Note that |Fi| ≤ 2n−4 for i = 2,3. By Lemma 2.4, there is a Hamiltonian path H of BHj,i
i n−1
that joins ai and bi for i = 2,3. Hence, the cycle C = (cid:104)u,b1,H ,w,b2,H ,a2,b3,H ,a3,b0,H ,u(cid:105) (see
1 2 3 0
figure 3) is the desired cycle.
Subcase 1.1.1.2.2: X (cid:54)= {u}.
(a) |F ∩∂D | ≤ 3.
j
Let a0 ∈ X \ {u} and (w,b2) be a fault-free j-dimension edge. By induction, there is a fault-
free Hamiltonian cycle C in BHj,0 that contains (u,v). Suppose that N (a0) = {b0,d0}. Note
0 n−1 C0
|F ∩∂D | ≤ 3; we can see that b0 or d0 is incident with one fault-free j-dimension edge. Without
j
loss of generality, we can assume that b0 is incident with one fault-free j-dimension edge, say, (b0,a3).
We can represent C as (cid:104)u,H ,a0,b0,H(cid:48),u(cid:105). Let (a2,b3) be a fault-free j-dimension edge. Note that
0 0 0
|Fi| ≤ (4n−5)−(2n−3)−2 = 2n−4 for i = 2,3. By Lemma 2.4, there is a fault-free Hamiltonian
path H in BHj,i that joins ai and bi for i = 2,3. By induction, there is a fault-free Hamiltonian
i n−1
cycle C in BHj,1 that contains (w,b1) as |F1| ≤ 4n − 9 and δ(BHj,1 − F + (w,b1)) ≥ 2. Let
1 n−1 n−1
H = C −(w,b1). Then, the cycle C = (cid:104)u,H ,a0,b1,H ,w,b2,H ,a2,b3,H ,a3,b0,H(cid:48),u(cid:105) (see figure
1 1 0 1 2 3 0
3) is the desired cycle.
(b) |F ∩∂D | ≥ 4.
j
Since [2(n − 1) − 2] + 4 + (2n − 3) = 4n − 3 > 4n − 5, each vertex is at least 3-rescuable in
BHj,0 . Let (u,α) be a fault-free edge in BHj,0 where α (cid:54)= v. Let (F0)(cid:48) = F0 ∪ T where T =
n−1 n−1
{(u,x) | x ∈ N −{v,α}}. Then, |(F0)(cid:48)| ≤ 2n−4+[4n−5−(2n−3)−4] = 4n−10, and
BHj,0(u)n−1
δ(BHj,0 −F(cid:48)) ≥ 2.
n−1 0
Let a0 ∈ X\{u}. By Lemma 3.3, there is a fault-free j-dimension edge (b0,a3) such that (a0,b0) ∈
E(BHj,0 ). By induction, there is a Hamiltonian cycle C in (BHj,0 −(F0)(cid:48))∪(a0,b0) that contains
n−1 0 n−1
7
Figure 3: Illustration for Subcase 1.1.1.2 of theorem 3.5
(a0,b0). Since u is incident with exactly two fault-free edges in BHj,0 −(F0)(cid:48), e = (u,v) ∈ E(C ).
n−1 0
We represent C as (cid:104)u,H ,a0,b0,H(cid:48),u(cid:105). Also by induction, there is a Hamiltonian cycle in BHj,1 −
0 0 0 n−1
F +(w,b1) that contains (w,b1). Let H = C −(w,b1) and (a2,b3) be a fault-free j-dimension edge.
1 1
Note that |Fi| ≤ 2n−4 for i = 2,3. By Lemma 2.4, there is a Hamiltonian path H in BHj,i that
i n−1
joins ai and bi for i = 2,3. Then, the cycle C = (cid:104)u,H ,a0,b1,H ,w,b2,H ,a2,b3,H ,a3,b0,H(cid:48),u(cid:105)
0 1 2 3 0
(see figure 3) is the desired cycle.
Subcase 1.1.1.3: w ∈ BHj,2 .
n−1
Since δ(BH −F) ≥ 2, w is incident with a fault-free j-dimension edge (w,b3). By Lemma 3.3,
n
there is a fault-free j-dimension edge (a1,b2) such that (w,b2) ∈ F. By induction, there is a fault-free
Hamiltonian cycle C in BHj,0 that contains (u,v). We represent it as (cid:104)c1,c2,··· ,c22n−2,c1(cid:105) with
0 n−1
c1 = u,c22n−2 = v. Let M = {(c1,c2),··· ,(c2i−1,c2i),··· ,(c22n−2−1,c22n−2)}. Therefore, M is a set of
22n−3 mutually disjoint edges. There is an edge (c2i−1,c2i) in M such that c2i−1 (resp., c2i) is incident
with a fault-free j-dimension edge, say, (c2i−1,b1) (resp., (c2i,a3)) because 2·22n−3 > 4n−5. We can
represent C as (cid:104)u,H ,c2i−1,c2i,H(cid:48),u(cid:105).
0 0 0
Figure 4: Illustration for subcase 1.1.1.3 of theorem 3.5
8
Note that |Fi| ≤ 2n − 4 for i = 1,3. By Lemma 2.4, there is a fault-free Hamiltonian path
H in BHj,i that joins ai and bi for i = 1,3. By induction, there is a Hamiltonian cycle C
i n−1 2
in BHj,2 − F + (w,b2) that contains (w,b2). Let H = C − (w,b2). Then, the cycle C =
n−1 2 2
(cid:104)u,H ,c2i−1,b1,H ,a1,b2,H ,w,b3,H ,a3,c2i,H(cid:48),u(cid:105) (see figure 4) is the desired cycle.
0 1 2 3 0
Case 1.1.2: δ(BH −F −∂D ) ≥ 2.
n j
Since 3(2n−4)−(4n−7) = 2n−5 > 0, there is an integer i ∈ {1,2,3} such that |Fi| ≤ 2n−4.
Without loss of generality, we can assume that |F1| ≤ 2n − 4. By induction, there is a fault-free
Hamiltonian cycle C in BHj,0 that contains (u,v). Similar to the analysis of subcase 1.1.1.3, we
0 n−1
can represent C as (cid:104)u,H ,c2i−1,c2i,H(cid:48),u(cid:105) where (c2i−1,b1),(c2i,a3) are fault-free j-dimension edges.
0 0 0
By Lemma 3.3, there is a fault-free j-dimension edge (b3,a2) such that (b3,a3) ∈ BHj,3 . In addition,
n−1
by Lemma 3.3, there is a fault-free j-dimension edge (b2,a1) such that (a2,b2) ∈ E(BHj,2 ). By
n−1
induction, there is a Hamiltonian cycle C in (BHj,i −F)∪(ai,bi) that contains (ai,bi) for i = 2,3.
i n−1
Let H = C −(ai,bi) for i=2, 3. By Lemma 2.4, there is a fault-free Hamiltonian path H in BHj,1
i i 1 n−1
thatjoinsa1 andb1. Hence, thecycle(cid:104)u,H ,c2i−1,b1,H ,a1,b2,H ,a2,b3,H ,a3,c2i,H(cid:48),u(cid:105)(seefigure
0 1 2 3 0
5) is the desired cycle.
Figure 5: Illustration for subcase 1.1.2 of theorem 3.5
Subcase 1.2: There is an integer i ∈ {0,1,2,3} such that |Fi| = 4n−8.
Ifδ(BHj,i −Fi) ≥ 2,thereareatleasttwovertex-disjointfaultyedgesfor(4n−8)−[2(n−1)−2] =
n−1
2n−4 ≥ 1. Note that |F ∩∂D | ≤ 3. Then, there is a faulty edge (ui,vi) such that ui is incident with
j
a fault-free j-dimension edge and vi is incident with a fault-free j-dimension edge.
Suppose that δ(BHj,i − Fi) = 1. Note that there is no more than one 1-rescuable vertex in
n−1
BHj,i . Let ui be the 1-rescuable vertex in BHj,i . Since δ(BH −F) ≥ 2, ui is incident with a
n−1 n−1 n
fault-free j-dimension edge. By Lemma 3.3, there is a fault-free j-dimension edge (vi,x) such that
(ui,vi) ∈ Fi. Obviously, δ(BHj,i −Fi+(ui,vi)) ≥ 2.
n−1
Owing to the above discussion, there is an edge (ui,vi) ∈ Fi such that (1) ui is incident with a
fault-free j-dimension edge; (2) vi is incident with a fault-free j-dimension edge; (3) δ(BHj,i −Fi+
n−1
(ui,vi)) ≥ 2, where |Fi| = 4n−8 for i ∈ {0,1,2,3}.
Subcase 1.2.1: |F0| = 4n−8.
By induction, there is a Hamiltonian cycle C that contains (u,v) in BHj,0 − F0 + (u0,v0).
0 n−1
Obviously, |F ∩E(C )| ≤ 1. If F ∩E(C ) = 1, let (a0,b0) = (u0,v0). If |F ∩E(C )| = 0, let (a0,b0) be
0 0 0
an edge such that a0 (resp., b0) is incident with a fault-free j-dimension edge. We can represent the
cycleC as(cid:104)u,H ,a0,b0,H(cid:48),u(cid:105)for2·22n−3 > 4n−5,(a0,b0)exists. Let(a0,b1),(a1,b2),(a2,b3),(a3,b0)
0 0 0
be fault-free j-dimension edges. By Lemma 2.4, there is a fault-free Hamiltonian path H that joins
i
(ai,bi) in BHj,i for i = 1,2,3. The cycle C = (cid:104)u,H ,a0,b1,H ,a1,b2,H ,a2,b3,H ,a3,b0,H(cid:48),u(cid:105) (see
n−1 0 1 2 3 0
figure 6) is the desired cycle.
9
Figure 6: Illustration for subcase 1.2.1 of theorem 3.5
Subcase 1.2.2: |F1| = 4n−8 (or |F3| = 4n−8).
In this subcase, |F0| ≤ 1. By induction, there is a Hamiltonian cycle C in BHj,1 −F1+(u1,v1)
1 n−1
that contains (u1,v1), where u1 (resp., v1) is incident with a fault-free j-dimension edge. We can
assume that (u1,v1) = (a1,b1) and (b1,a0), (a1,b2) are fault-free j-dimension edges incident with a1
or b1. Let H = C −(a1,b1). Since |F0| ≤ 1 and |F ∩∂D | ≤ 3−|F0|, we can choose two fault-free
1 1 j
edges(a0,b0),(a0,d0) ∈ BHj,0 −F0 suchthatb0 isincidentwithafault-freej-dimensionedge(b0,a3).
n−1
Particularly, if a0 = u, let d0 = v. Let (F0)(cid:48) = F0∪T where T = {(a0,x) | x ∈ N (a0)\{b0,d0}}.
BHj,0
n−1
Then, |(F0)(cid:48)| ≤ 2n−3 ≤ 4n−9 and δ(BHj,0 −(F0)(cid:48)) ≥ 2. By induction, there is a Hamiltonian
n−1
cycle C in BHj,0 that contains (u,v). Since a0 is incident with exactly two fault-free edges in
0 n−1
BHj,0 −(F0)(cid:48), we have (a0,b0) ∈ E(C ). We represent C as (cid:104)u,H ,a0,b0,H(cid:48),u(cid:105). Let (a2,b3) be the
n−1 0 0 0 0
fault-free j-dimension edge. By Lemma 2.4, there is a fault-free Hamiltonian path H in BHj,i that
i n−1
joins ai and bi for i = 2,3. Hence, the cycle C = (cid:104)u,H ,a0,b1,H ,a1,b2,H ,a2,b3,H ,a3,b0,H(cid:48),u(cid:105)
0 1 2 3 0
(see figure 7) is the desired cycle.
Figure 7: Illustration for subcase 1.2.2 of theorem 3.5
Subcase 1.2.3: |F2| = 4n−8.
By induction, there is a Hamiltonian in BHj,2 −F2+(u2,v2) that contains (u2,v2) and u2 (resp.,
n−1
v2) is incident with a fault-free j-dimension edge. We can assume that (u2,v2) = (a2,b2) and (a2,b3),
(a1,b2) are fault-free j-dimension edges. Let H = C −(a2,b2). By induction, there is a fault-free
2 2
10