DOI 10.1007/s10878-012-9528-1
On the maximum number of fault-free mutually
independent Hamiltonian cycles in the faulty hypercube
Tzu-Liang Kung· Cheng-Kuan Lin · Lih-Hsing Hsu
Published online: 29 June 2012
© Springer Science+Business Media, LLC 2012
Abstract Hsieh and Yu (2007) first claimed that an injured n-dimensional hypercube
Qncontains (n− 1 − f )-mutually independent fault-free Hamiltonian cycles, where
f ≤ n − 2 denotes the total number of permanent edge-faults in Qn for n≥ 4, and edge-faults can occur everywhere at random. Later, Kueng et al. (2009a) presented a formal proof to validate Hsieh and Yu’s argument. This paper aims to improve this mentioned result by showing that up to (n− f )-mutually independent fault-free Hamiltonian cycles can be embedded under the same condition. Let F denote the set of f faulty edges. If all faulty edges happen to be incident with an identical vertex s, i.e., the minimum degree of the survival graph Qn−F is equal to n−f , then Qn−F contains at most (n− f )-mutually independent Hamiltonian cycles starting from s. From such a point of view, the presented result is optimal. Thus, not only does our improvement increase the number of mutually independent fault-free Hamiltonian cycles by one, but also the optimality can be achieved.
Keywords Interconnection network· Graph · Hypercube · Fault tolerance ·
Hamiltonian cycle
This work is supported in part by the National Science Council of the Republic of China under Contract NSC 98-2218-E-468-001-MY3.
T.-L. Kung (
)Department of Computer Science and Information Engineering, Asia University, Wufeng, Taichung 41354, Taiwan
e-mail:[email protected]
C.-K. Lin
Department of Computer Science, National Chiao Tung University, Hsinchu 30010, Taiwan
L.-H. Hsu
Department of Computer Science and Information Engineering, Providence University, Shalu, Taichung 43301, Taiwan
1 Introduction
In many parallel computer systems, processors are connected on the basis of
inter-connection networks such as meshes, hypercubes, star graphs, bubble-sort networks,
etc. For the sake of simplicity, the underlying topology of an interconnection net-work is usually represented by a graph, whose vertices and edges correspond to pro-cessors and connection links, respectively. Hence, we use the terms, graph and net-work, interchangeably. Throughout this paper, we concentrate on loopless undirected graphs. Some important graph-theory notations and definitions are introduced below. For those not defined here, we follow the standard terminology given by Bondy and Murty (2008).
A graph G consists of a nonempty vertex set V (G) and an edge set E(G), which is a subset of{(u, v) | (u, v) is an unordered pair of elements in V (G)}. Two vertices
uand v of G are adjacent if (u, v)∈ E(G). The neighborhood of vertex v in graph
G, denoted by NG(v), is defined as{u | u ∈ V (G), (v, u) ∈ E(G)}. A graph H is a
subgraph of G if V (H )⊆ V (G) and E(H ) ⊆ E(G). Let S be a nonempty subset of
V (G). The subgraph induced by S is the maximal subgraph of G with vertex set S that contains precisely those edges of G joining two vertices in S. We use G− S to denote the subgraph of G induced by V (G)− S. Analogously, let F be a nonempty subset of E(G). We use G− F to denote the subgraph of G with vertex set V (G) and edge set E(G)− F . The degree of a vertex u in G is the number of edges incident with u. A graph G is k-regular if the degree of every vertex is equal to k. A graph G is bipartite if its vertex set can be partitioned into two disjoint partite sets V0(G)and
V1(G)such that every edge joins a vertex in V0(G)and a vertex in V1(G).
A path P of length k, k≥ 1, from vertex x to vertex y in a graph G is an or-dered sequence of distinct verticesv1, v2, . . . , vk+1 such that v1= x, vk+1= y, and (vi, vi+1)∈ E(G) for every 1 ≤ i ≤ k. Moreover, a path of length 0, con-sisting of a single vertex x, is denoted by x. For convenience, we write P as
v1, v2, . . . , vi, R, vj, . . . , vk+1, where i ≤ j, if R = vi, . . . , vj is a part of P . The ith vertex of P is denoted by P (i); i.e., P (i)= vi. We use (P ) to denote the length of P . To emphasize the start and end vertices of P , we also write P as
P[x, y]. A cycle is a path with at least three vertices such that the last vertex is
adjacent to the first one. For clarity, a cycle of length k, k≥ 3, is represented by
v1, v2, . . . , vk, v1. A path (respectively, cycle) in the graph G is a Hamiltonian path (respectively, Hamiltonian cycle) of G if it traverses every vertex of G. A bipartite graph is Hamiltonian laceable (Simmons1978) if there exists a Hamiltonian path joining any two vertices that are in different partite sets. Moreover, a Hamiltonian laceable graph H is hyper-Hamiltonian laceable (Lewinter and Widulski1997) if for any vertex v∈ Vi(H )with i∈ {0, 1}, there exists a Hamiltonian path in H − {v} joining any two vertices of V1−i(H ).
The n-dimensional hypercube (or n-cube for short), n≥ 1, is one of the most popular network topologies discovered for parallel and distributed computation. Not only is it ideally suited to both special-purpose and general-purpose tasks, but it can efficiently simulate many other networks (Leighton1992). Thus, many attrac-tive properties of hypercubes have been extensively addressed by researchers (Ak-ers and Krishnameurthy 1989; Castañeda and Gotchev 2010; Chang et al. 2004;
Dvoˇrák and Koubek2009, 2010; Fink and Gregor2011; Johnsson and Ho 1989; Kueng et al.2009b; Kung et al.2009; Leighton1992; Leu and Kuo1999; Tsai et al. 2002; Yang et al.1994). The formal definition of an n-cube is given below. For the sake of clarity, let a boldface letter u denote an n-bit binary string bn· · · bi· · · b1. For 1≤ i ≤ n, we use (u)i to denote the binary string bn· · · ¯bi· · · b1. Moreover, we use (u)i to denote the ith bit bi of u. The Hamming weight of u, denoted by
wH(u), is |{i | (u)i = 1, 1 ≤ i ≤ n}|. The n-cube Qn contains 2n vertices, each of which is labeled by an n-bit binary string. For the purpose of notation consis-tency, its vertices are also denoted by boldface letters in the rest of this paper. Two vertices u and v of Qn are adjacent if and only if v= (u)i for some i, and edge
(u, (u)i)is called i-dimensional. Clearly, Qn is a bipartite graph with partite sets
V0(Qn)= {u ∈ V (Qn)| wH(u) is even} and V1(Qn)= {u ∈ V (Qn)| wH(u) is odd}. Sun et al. (2006) first addressed the problem of finding mutually independent Hamiltonian cycles on the n-cube for n≥ 3. Later, Hsieh and Yu (2007) claimed that an injured n-cube contains (n− 1 − f )-mutually independent fault-free Hamil-tonian cycles, and Kueng et al. (2009a) gave a formal proof to validate this claim, where f ≤ n − 2 is the total number of permanent edge-faults that can occur ev-erywhere at random to injure the n-cube. To be precise, we have to introduce the definition of mutually independent Hamiltonian cycles in advance. Let G be a graph with N vertices. A Hamiltonian cycle C of G is represented byu1, u2, . . . , uN, u1, where u1is referred to as the start vertex of C. Naturally every vertex of C can serves as the start one. Two Hamiltonian cycles of G, namely C1= u1, u2, . . . , uN, u1 and C2= v1, v2, . . . , vN, v1, are internally independent if u1= v1and ui = vi for 2≤ i ≤ N. A set {C1, C2, . . . , Cm} of m Hamiltonian cycles of G is m-mutually
in-dependent if and only if any two of them are internally inin-dependent for m≥ 2. The
concept of mutually independent Hamiltonian cycles can be applied in many different areas like those introduced in Hsieh and Yu (2007), Kueng et al. (2008), Kung et al. (2011), Lin et al. (2012), Shih et al. (2010a,2010b), Su et al. (2011a,2011b), Sun et al. (2006). This paper aims to improve the mentioned result (Hsieh and Yu2007; Kueng et al.2009a) by showing that Qn has up to (n− f )-mutually independent fault-free Hamiltonian cycles, starting from any vertex, when f ≤ n − 2 edges are faulty.
The rest of this paper is organized as follows. The basic properties of hypercubes are introduced in Sect.2. Our main theorem is presented in Sect. 3. Finally, some concluding remarks are given in Sect.4.
2 Preliminaries
By definition, the n-cube Qn has a recursive construction; that is, it can be decom-posed into two (n− 1)-dimensional hypercubes. Let Qd,jn denote the subgraph of
Qn induced by {u ∈ V (Qn)| (u)d = j} for 1 ≤ d ≤ n and j ∈ {0, 1}. Obviously,
Qd,jn is isomorphic to Qn−1. Then the d-partition of Qndecomposes Qnalong the
dth dimension into Qd,0n and Qd,1n . The set of crossing edges between Qd,0n and
all d-dimensional edges in Qn. It is known that Qn is vertex-transitive and edge-transitive (Saad and Shultz1988).
A Hamiltonian graph G is said to be f -edge-fault-tolerant Hamiltonian if G− F remains Hamiltonian for every F ⊆ E(G) with |F | ≤ f . A Hamiltonian laceable graph G is said to be f -edge-fault-tolerant Hamiltonian laceable if G− F re-mains Hamiltonian laceable for every F ⊆ E(G) with |F | ≤ f . Similarly, a hyper-Hamiltonian laceable graph G is said to be f -edge-fault-tolerant hyper-hyper-Hamiltonian
laceable if G− F remains hyper-Hamiltonian laceable for every F ⊆ E(G) with
|F | ≤ f .
Lemma 1 (Tsai et al. 2002) Let n≥ 3. Then Qnis (n−2)-edge-fault-tolerant
Hamil-tonian and (n− 2)-edge-fault-tolerant Hamiltonian laceable.
Lemma 2 (Tsai et al. 2002) Let n≥ 3. Then Qnis (n− 3)-edge-fault-tolerant
hyper-Hamiltonian laceable.
Lemma 3 (Sun et al. 2006) Let n≥ 4. Suppose that x and y are two arbitrary vertices
in different partite sets of Qn. Then Qn− {x, y} is Hamiltonian laceable. The proof of the next lemma is presented in AppendixA.
Lemma 4 Let n≥ 4. Suppose that x and y are any two adjacent vertices in Qn. Then
Qn− {x, y} is (n − 3)-edge-fault-tolerant Hamiltonian laceable.
Two Hamiltonian paths of a graph G, represented by P1= u1, u2, . . . , u|V (G)| and P2 = v1, v2, . . . , v|V (G)|, are internally independent if u1= v1, u|V (G)| =
v|V (G)|, and ui = vi for every 1 < i <|V (G)|; P1 and P2 are fully independent if
ui= vi for every 1≤ i ≤ |V (G)|. A set {Pi| 1 ≤ i ≤ m} of m Hamiltonian paths of
Gare m-mutually fully independent if m= 1 or its any two Hamiltonian paths are fully independent for m≥ 2.
Lemma 5 (Sun et al. 2006) Let Qnbe an n-cube for n≥ 2. Suppose that {(wi,bi)∈
E(Qn)| wi ∈ V0(Qn),bi ∈ V1(Qn),1≤ i ≤ n − 1} consists of n − 1 distinct edges
with no shared endpoints. Then Qn contains (n− 1)-mutually fully independent
Hamiltonian paths joining wi and bifor 1≤ i ≤ n − 1.
The next lemma plays an important role in deriving the main result of this paper and was proved by Kueng et al. (2009a). To make the paper self-contained, its proof is given in AppendixB.
Lemma 6 (Kueng et al.2009a) Let F be a set of f edges in Qn, where n≥ 3 and
f ≤ n − 2. Suppose that A = {(wi,bi)∈ E(Qn)| wi ∈ V0(Qn),bi∈ V1(Qn),1≤
i≤ n − 1 − f } consists of n − 1 − f distinct edges with no shared endpoints.
Then Qn− F contains (n − 1 − f )-mutually fully independent Hamiltonian paths
3 Mutually independent Hamiltonian cycles
The mutually independent Hamiltonicity of a graph G, denoted byIHC(G), is de-fined as the maximum integer m such that for any vertex v∈ V (G), there exist m-mutually independent Hamiltonian cycles of G starting from v.
Theorem 1 (Sun et al. 2006)IHC(Qn)= n−1 if n ≤ 3, and IHC(Qn)= n if n ≥ 4. The last lemma and theorem show that Qncontains (n−f )-mutually independent fault-free Hamiltonian cycles, where f denotes the total number of faulty edges in
Qnfor f ≤ n − 2 and n ≥ 4.
Lemma 7 Suppose that F denotes any set of f edges in Q4. Then Q4− F contains
(4− f )-mutually independent Hamiltonian cycles starting from any vertex if f ≤ 2.
Proof Let s be any vertex of Q4in Vi(Q4)for any i∈ {0, 1}. If f = 0, then Theo-rem1has ensured that Q4has 4-mutually independent Hamiltonian cycles starting from s. Thus, we consider f ∈ {1, 2} only. Because the hypercube is edge-transitive, we assume that F contains a 4-dimensional edge. Then Q4can be partitioned along the fourth dimension into Q4,04 and Q4,14 . For the sake of convenience, we define some notations first: F0= F ∩E(Q4,04 ), F1= F ∩E(Q4,14 ), Fc= F ∩ Ec4, f0= |F0|,
f1= |F1|, fc= |Fc|, and δ = 4 − f . Since fc≥ 1, we have f0≤ 1 and f1≤ 1.
Case 1: f = 2. Thus, δ = 2. Without loss of generality, we assume that s is in Q4,04 .
Subcase 1.1: (s, (s)4)is fault-free. Since Q4has eight 4-dimensional edges, we can find two vertex-disjoint edges (u, (u)4)and (v, (v)4), both of which are fault-free such that u∈ V1−i(Q4,04 )and v∈ Vi(Q4,14 ). By Lemma1, Q3is 1-edge-fault-tolerant Hamiltonian laceable. Hence, there exist two Hamiltonian paths H and T of Q4,04 −
F0joining pairs s, u and (v)4,s, respectively. Similarly, there exist two Hamiltonian paths P and R of Q4,14 −F1joining pairs (u)4, (s)4and (s)4,v, respectively. Let C1=
s, H, u, (u)4, P , (s)4,s and C
2= s, (s)4, R,v, (v)4, T ,s. Then C1and C2are 2-mutually independent fault-free Hamiltonian cycles starting from s. See Fig.1(a).
Subcase 1.2: (s, (s)4)is faulty.
Condition 1.2.1: f0= f1= 0. Let x ∈ Vi(Q4,04 )− {s} such that (x, (x)4)is fault-free, and let b be a neighbor of s in Q4,04 such that edges (s, b) and (b, (b)4) are fault-free. By Lemma 2, Q3 is hyper-Hamiltonian laceable. Thus, there exists a Hamiltonian path P of Q4,04 − {b} joining x and s. By Lemma 1, there exists a Hamiltonian path R of Q4,14 joining (b)4 and (x)4. For clarity, P and R are rep-resented byx, P,v, s and (b)4, R,y, (x)4, respectively, where v (respectively,
y) is a neighbor of s (respectively, (x)4). Because Q4,14 is 1-edge-fault-tolerant Hamiltonian laceable, there exists a Hamiltonian path H of Q4,14 − {((x)4,y)}
join-ing (x)4 and (b)4. Then we set C1= s, b, (b)4, R,y, (x)4,x, P,v, s and C2=
s, v, rev(P),x, (x)4, H, (b)4,b, s, where rev(P)is the reverse of P. As a result,
C1and C2are 2-mutually independent fault-free Hamiltonian cycles starting from s. See Fig.1(b).
Fig. 1 Illustration for Lemma7
Table 1 2-mutually independent fault-free Hamiltonian cycles, C1and C2
F C1and C2 {(0, 8), (8, 12)} or {(0, 8), (10, 14)} or {(0, 8), (13, 15)} or {(0, 8), (10, 11)} 0, 1, 5, 4, 6, 7, 3, 11, 15, 14, 12, 13, 9, 8, 10, 2, 00, 2, 10, 8, 9, 11, 15, 14, 12, 13, 5, 1, 3, 7, 6, 4, 0 {(0, 8), (8, 10)} or {(0, 8), (12, 14)} or {(0, 8), (11, 15)} or {(0, 8), (12, 13)} 0, 1, 3, 2, 6, 7, 5, 13, 15, 14, 10, 11, 9, 8, 12, 4, 00, 4, 12, 8, 9, 13, 15, 14, 10, 11, 3, 1, 5, 7, 6, 2, 0 {(0, 8), (8, 9)} or {(0, 8), (9, 10)} 0, 4, 5, 1, 3, 7, 6, 14, 15, 11, 9, 13, 12, 8, 10, 2, 0 0, 2, 10, 8, 12, 14, 15, 11, 9, 13, 5, 4, 6, 7, 3, 1, 0 {(0, 8), (9, 13)} 0, 4, 6, 2, 3, 7, 5, 13, 15, 11, 10, 14, 12, 8, 9, 1, 0 0, 1, 9, 11, 15, 14, 10, 8, 12, 13, 5, 7, 3, 2, 6, 4, 0 {(0, 8), (14, 15)} 0, 2, 6, 4, 5, 7, 3, 11, 15, 13, 12, 14, 10, 8, 9, 1, 0 0, 1, 9, 8, 10, 11, 15, 13, 12, 14, 6, 2, 3, 7, 5, 4, 0
Condition 1.2.2: f0= 0 and f1= 1. With symmetry, we assume that s = 0000. Table1shows that there are 2-mutually independent fault-free Hamiltonian cycles starting from 0000, in which all binary strings are decimalized for the purpose of saving space.
Condition 1.2.3: f0= 1 and f1= 0. By symmetry, we assume that s = 0000. Table2shows that there are 2-mutually independent fault-free Hamiltonian cycles starting from 0000, in which all binary strings are decimalized for the purpose of saving space.
Table 2 2-mutually independent fault-free Hamiltonian cycles, C1and C2 F C1and C2 {(0, 4), (0, 8)} or {(2, 6), (0, 8)} or {(5, 7), (0, 8)} or {(1, 3), (0, 8)} or {(2, 3), (0, 8)} 0, 1, 5, 4, 6, 7, 3, 11, 15, 14, 12, 13, 9, 8, 10, 2, 00, 2, 10, 14, 15, 13, 12, 8, 9, 11, 3, 7, 6, 4, 5, 1, 0 {(0, 2), (0, 8)} or {(4, 6), (0, 8)} or {(3, 7), (0, 8)} or {(4, 5), (0, 8)} 0, 1, 3, 2, 6, 7, 5, 13, 15, 14, 10, 11, 9, 8, 12, 4, 00, 4, 12, 14, 15, 11, 10, 8, 9, 13, 5, 7, 6, 2, 3, 1, 0 {(0, 1), (0, 8)} or {(1, 5), (0, 8)} or {(6, 7), (0, 8)} 0, 4, 5, 7, 3, 1, 9, 8, 12, 13, 15, 11, 10, 14, 6, 2, 0 0, 2, 6, 14, 10, 8, 12, 13, 15, 11, 9, 1, 3, 7, 5, 4, 0 Table 3 3-mutually independent fault-free Hamiltonian cycles, C1, C2, and C3
s C1, C2, C3 00002= 010 0, 1, 5, 4, 6, 7, 3, 11, 15, 14, 12, 13, 9, 8, 10, 2, 0 0, 2, 10, 14, 15, 13, 12, 8, 9, 11, 3, 1, 5, 7, 6, 4, 0 0, 4, 6, 7, 3, 2, 10, 14, 12, 8, 9, 11, 15, 13, 5, 1, 0 01002= 410 4, 0, 1, 3, 2, 6, 7, 15, 11, 10, 14, 12, 8, 9, 13, 5, 4 4, 5, 13, 15, 11, 10, 14, 12, 8, 9, 1, 0, 2, 3, 7, 6, 4 4, 6, 2, 10, 14, 12, 8, 9, 13, 15, 11, 3, 7, 5, 1, 0, 4 01102= 610 6, 2, 0, 1, 5, 7, 3, 11, 9, 13, 15, 14, 10, 8, 12, 4, 6 6, 4, 12, 14, 10, 8, 9, 13, 15, 11, 3, 2, 0, 1, 5, 7, 6 6, 7, 3, 11, 15, 13, 12, 14, 10, 8, 9, 1, 5, 4, 0, 2, 6 01112= 710 7, 3, 1, 5, 4, 0, 2, 10, 11, 15, 13, 9, 8, 12, 14, 6, 7 7, 5, 13, 15, 14, 10, 11, 9, 8, 12, 4, 6, 2, 0, 1, 3, 7 7, 6, 4, 0, 2, 3, 1, 5, 13, 9, 8, 12, 14, 10, 11, 15, 7
Case 2: f = 1. Because the hypercube is edge-transitive, we assume that F = {(0000, 1000)}. With symmetry, we consider s ∈ {0000, 0100, 0110, 0111} only.
Ta-ble3shows that there are 3-mutually independent fault-free Hamiltonian cycles start-ing from s, in which all binary strstart-ings are decimalized for the purpose of savstart-ing space.
Theorem 2 Let n≥ 4. Suppose that F denotes any set of f edges in Qn. Then
Qn− F contains (n − f )-mutually independent Hamiltonian cycles starting from
any vertex if f≤ n − 2.
Proof Let s be any vertex of Qn. If f = 0, then Theorem1has ensured that Qnhas n-mutually independent Hamiltonian cycles starting from s. Thus, we consider 1≤ f ≤
n− 2 below. Without loss of generality, we assume that F contains a d-dimensional
edge for 1≤ d ≤ n. Then, Qncan be partitioned along the dth dimension into Qd,0n and Qd,1n . Furthermore, we assume that s∈ V0(Qd,0n ). For the sake of convenience, we define some notations in advance: F0= F ∩ E(Qd,0n ), F1= F ∩ E(Qd,1n ), Fc=
F ∩ Ec, f0= |F0|, f1= |F1|, fc = |Fc|, and δ = n − f . Since fc≥ 1, we have
Fig. 2 Illustration for
Subcase 1.1 of Theorem2
The proof proceeds by induction on n. Lemma 7 is the induction basis. For
n≥ 5, the inductive hypothesis is that the theorem statement holds for Qk, 4≤ k ≤
n− 1. Since f0≤ f − 1 ≤ n − 3, Qd,0n − F0 has (n− 1 − f0)-mutually indepen-dent Hamiltonian cycles starting from s. Without loss of generality, these Hamil-tonian cycles can be represented by s, vi, Hi,wi,bi,s for 1 ≤ i ≤ n − 1 − f0. Let Fx= Fc∩ {(wi, (wi)d), (bi, (bi)d)| 1 ≤ i ≤ n − 1 − f0} and fx= |Fx|. Ob-viously, there exist at least n− 1 − f0− fx integers in{1, 2, . . . , n − 1 − f0}, say 1, 2, . . . , n− 1 − f0− fx, such that (wi, (wi)d)and (bi, (bi)d)are fault-free for all
i∈ {1, 2, . . . , n − 1 − f0− fx}. For ease of presentation, let δ0= n − 1 − f0− fxand
δ1= n − 2 − f1.
Case 1: (s, (s)d)is faulty. Thus, we have fx≤ fc−1 and δ0= n−1−(f0+fx)≥
n− 1 − (f0+ fc− 1) ≥ n − f = δ. Since δ1= n − 2 − f1≥ n − 2 − (f − 1) = δ − 1, we consider the following two subcases.
Subcase 1.1: Suppose that δ1≥ δ. By Lemma6, Qd,1n − F1has δ-mutually fully independent Hamiltonian paths Pi, 1≤ i ≤ δ, joining (wi)d and (bi)d. Let Ci =
s, vi, Hi,wi, (wi)d, Pi, (bi)d,bi,s for each i ∈ {1, 2, . . . , δ}. Then {C1, C2, . . . , Cδ} is a set of δ-mutually independent Hamiltonian cycles starting from s in Qn− F . See Fig.2for illustration.
Subcase 1.2: Suppose that δ1= δ − 1. Hence, we have f1= f − 1 and Fc=
{(s, (s)d)}; that is, Qd,0
n is fault-free. By Lemma6, Qd,1n − F1has (δ− 1)-mutually fully independent Hamiltonian paths Pi, 1≤ i ≤ δ − 1, joining (wi)d and (bi)d. Let
Ci= s, vi, Hi,wi, (wi)d, Pi, (bi)d,bi,s for 1 ≤ i ≤ δ − 1.
Condition 1.2.1: f = n − 2. Thus, we have δ = 2. Let x be a vertex in V1(Qd,1n ) such that dQn(x, (w1)d)≥ 4. Since dQn((s)d, (w1)d)= 2, we have x = (s)d. Further-more, since Fc= {(s, (s)d)}, edge (x, (x)d)is fault-free. By Lemma 1, Qd,1n − F1 has a Hamiltonian path R joining (v2)d and x. By Lemma3, Qd,0n − {s, v2} has a Hamiltonian path T joining (x)d to b2. Let C2= s, v
2, (v2)d, R,x, (x)d, T ,b2,s. Then{C1, C2} is a set of 2-mutually independent Hamiltonian cycles starting from s in Qn− F . See Fig.3(a).
Condition 1.2.2: f≤ n − 3. Let Yn= {u ∈ V (Qd,1n )| dQd,1
n (u, P1(3))= 3} ∪ {u ∈
V (Qd,1n )| dQd,1
n (u, P2(3))= 3}. Then, we have |Yn| ≥
n−1 3 + 1 > δ for n = 5, and|Yn| ≥ n−1 3
> δ for n≥ 6. Thus, we can choose a vertex y in Yn such that
y /∈ {Pi(2)| 1 ≤ i ≤ δ − 1} ∪ {(vδ)d}. Without loss of generality, we assume that
d
Qd,n1(y, P1(3))= 3. If (y, (s)
d)is a fault-free edge, then let z= (s)d; otherwise, let
z∈ NQd,1
n (y)− {(wi)
d | 1 ≤ i ≤ δ − 1} such that (y, z) is a fault-free edge. Since
|NQd,1
n (y)− {z}| = n − 2 > |{Pi(3)| 2 ≤ i ≤ δ − 1}| + f1= (n − f − 2) + (f − 1) =
n−3, we can find a vertex x in NQd,1
n (y)−{z}−{Pi(3)| 2 ≤ i ≤ δ −1} such that (x, y) is a fault-free edge. It follows from Lemma4that Qd,1n − {x, y} is (n −
4)-edge-fault-Fig. 3 Illustration for
Subcase 1.2 of Theorem2
tolerant Hamiltonian laceable. Since f1= f −1 ≤ n−4, Qd,1n −{x, y} has a fault-free Hamiltonian path R joining (vδ)dand z. By Lemma3, Qd,0n − {s, vδ} has a Hamilto-nian path T joining (x)dto bδ. Let Cδ= s, vδ, (vδ)d, R,z, y, x, (x)d, T ,bδ,s. Then
{C1, C2, . . . , Cδ} is a set of δ-mutually independent Hamiltonian cycles starting from
s in Qn− F . See Fig.3(b).
Case 2: (s, (s)d)is fault-free.
Subcase 2.1: Suppose that f0 + fx ≤ f − 1 and f1 ≤ f − 2. Thus, we have δ0 ≥ δ and δ1 ≥ δ. By Lemma 6, Qd,1n − F1 has δ-mutually fully inde-pendent Hamiltonian paths Pi, 1≤ i ≤ δ, joining (wi)d and (bi)d. Let Ci =
s, vi, Hi,wi, (wi)d, Pi, (bi)d,bi,s for each i ∈ {1, 2, . . . , δ}. Then {C1, C2, . . . , Cδ} is a set of δ-mutually independent Hamiltonian cycles starting from s in Qn− F .
Subcase 2.2: Suppose that f0+ fx= f or f1= f − 1. Thus, we have δ0= δ − 1 or δ1= δ − 1. It follows from Lemma6that Qd,1n − F1has (δ− 1)-mutually fully independent Hamiltonian paths Pi, 1≤ i ≤ δ − 1, joining (wi)dand (bi)d.
Condition 2.2.1: f ≤ n − 3. Let Ci = s, vi, Hi,wi, (wi)d, Pi, (bi)d,bi,s for 1≤ i ≤ δ−1. Obviously, we can choose two vertices x and xin V0(Qd,1n )−{(bδ)d}−
{Pi(2)| 1 ≤ i ≤ δ −1} such that (x, (x)d) /∈ F and (x, (x)d) /∈ F . Since |{(wi)d| 1 ≤
i≤ δ − 1} ∪ {(s)d}| + f1≤ (n − f ) + (f − 1) = n − 1 < n < |NQd,1
n (x)∪ NQd,n1(x
)|,
there exists at least one vertex y in NQd,1
n (x)∪ NQd,n1(x
)such that (x, y) or (x,y)
is fault-free, and y /∈ {(wi)d| 1 ≤ i ≤ δ − 1} ∪ {(s)d}. Without loss of generality, we assume that y∈ NQd,1
n (x). Since f1≤ f − 1 ≤ n − 4, it follows from Lemma2that
Qd,1n −F1−{x} has a Hamiltonian path R joining (s)dand y, and Qd,0n −F0−{s} has a Hamiltonian path T joining (x)d and bδ. Let Cδ= s, (s)d, R,y, x, (x)d, T ,bδ,s.
Fig. 4 Illustration for
Condition 2.2.1 of Theorem2
Fig. 5 Illustration for
Condition 2.2.2 of Theorem2
Then{C1, C2, . . . , Cδ} is a set of δ-mutually independent Hamiltonian cycles starting from s in Qn− F . See Fig.4for illustration.
Condition 2.2.2: f = n − 2. Thus, δ = 2. Let x ∈ V0(Qd,1n )− {(b1)d} such that (x, (x)d) is fault-free. By Lemma 1, Qd,1n − F1 has a Hamiltonian path P joining (b1)d and (s)d. Similarly, Qd,1
n − F1 has a Hamiltonian path R joining
(s)d and x, and Qd,0n − F0 has a Hamiltonian path T joining (x)d and s. Let
C1= s, v1, H1,w1,b1, (b1)d, P1, (s)d,s and C2= s, (s)d, R,x, (x)d, T ,s. Then
{C1, C2} is a set of 2-mutually independent Hamiltonian cycles starting from s in
Qn− F . See Fig.5for illustration.
The proof is completed.
4 Conclusion
In this paper, we improve the result of finding mutually independent fault-free Hamil-tonian cycles in a faulty hypercube, as previously addressed by Hsieh and Yu (2007) and Kueng et al. (2009a). Let F denote the set of f faulty edges in n-cube Qn. Then we show that Qn− F has (n − f )-mutually independent Hamiltonian cycles starting from any vertex if f ≤ n − 2. When all faulty edges happen to be incident with an identical vertex s, i.e., the minimum degree of the survival graph Qn− F is equal to
n− f , then Qn− F contains no more than (n − f )-mutually independent Hamilto-nian cycles starting from s. From such a point of view, the presented result is optimal.
Appendix A: Proof of Lemma4
Proof The proof proceeds by induction on n. Our computer program verifies that
Q4− {x, y} is 1-edge-fault-tolerant Hamiltonian laceable. Please refer to the data reported by Kung (2012). For n≥ 5, we assume that (x, y) is r-dimensional with 1≤ r ≤ n. Let F ⊂ E(Qn− {x, y}) with 1 ≤ |F | ≤ n − 3. If there exists a d-partition of Qn, d∈ {1, 2, . . . , n} − {r}, such that at least one faulty edge is d-dimensional, then Qn is partitioned along the dth dimension; otherwise, every faulty edge is r-dimensional, and Qn is partitioned into Qd,0n and Qd,1n with any d∈ {1, 2, . . . , n} −
{r}. Without loss of generality, we assume that (x, y) is in Qd,0
n . Then, the inductive hypothesis is that Qd,0n − {x, y} is (n − 4)-edge-fault-tolerant Hamiltonian laceable for n≥ 5.
Let Fi = F ∩ E(Qd,in )for i∈ {0, 1}. The following two cases show that Qn−
{x, y} − F is Hamiltonian laceable.
Case 1:|F0| ≤ n − 4. Let s ∈ Vi(Qn)and t∈ V1−i(Qn)for any i∈ {0, 1}. Suppose that both s and t are in Qd,0n . By the inductive hypothesis, Qd,0n −{x, y}−
F0 is Hamiltonian laceable. Hence, there exists a Hamiltonian path P in Qd,0n −
{x, y} − F0joining s and t. Clearly, there are (P )2 vertex-disjoint edges on P . Since
(P ) 2 =
2n−1−3
2 = 2n−2− 1 > n − 3 ≥ |F | for n ≥ 5, there exists an edge (u, v) on P such that{(u, (u)d), (v, (v)d)} ∩ F = ∅. Accordingly, P can be represented ass, P1,u, v, P2,t. By Lemma1, there exists a Hamiltonian path R in Qd,1n − F1 joining (u)dand (v)d. Thens, P
1,u, (u)d, R, (v)d,v, P2,t is a Hamiltonian path of
Qn− {x, y} − F . See Fig.6(a).
Suppose that both s and t are in Qd,1n . Obviously, it follows from Lemma1that there exists a Hamiltonian path R in Qd,1n − F1 joining s and t. Since (R)2 =
2n−1−1
2 = 2
n−2> n− 1 ≥ |{x, y}| + |F | for n ≥ 5, there exists an edge (u, v) on R such that{(v)d, (u)d} ∩ {x, y} = ∅ and {(u, (u)d), (v, (v)d)} ∩ F = ∅.
There-fore, R can be represented ass, R1,u, v, R2,t. By the inductive hypothesis, there exists a Hamiltonian path P in Qd,0n − {x, y} − F0 joining (u)d and (v)d. Then
s, R1,u, (u)d, P , (v)d,v, R2,t is a Hamiltonian path of Qn− {x, y} − F . See Fig.6(b) for illustration.
Suppose that s is in Qd,0n and t is in Qd,1n . Let b∈ V1−i(Qd,0n )− {x, y} such that
(b, (b)d) /∈ F . It follows from the inductive hypothesis that there exists a Hamiltonian path P in Qd,0n − {x, y} − F0joining s and b. By Lemma1, there exists a Hamilto-nian path R in Qd,1n − F1 joining (b)d and t. As a result,s, P, b, (b)d, R,t is a Hamiltonian path in Qn− {x, y} − F . See Fig.6(c) for illustration.
Case 2:|F0| = n − 3. This case implies that every faulty edge is r-dimensional. Similarly, let s∈ Vi(Qn)and t∈ V1−i(Qn)for any i∈ {0, 1}.
Subcase 2.1: Both s and t are in Qd,0n . Let (u, v)∈ F and F= F −{(u, v)}. By the inductive hypothesis, there exists a Hamiltonian path P in Qd,0n − {x, y} − Fjoining
s and t. If (u, v)∈ E(P ), then path P can be represented as s, P1,u, v, P2,t; other-wise, P is written ass, T1,p, q, T2,t, where (p, q) is any edge of P . Clearly, Qd,1n has Hamiltonian paths R and H joining pairs (u)d, (v)dand (p)d, (q)d, respectively. Then eithers, P1,u, (u)d, R, (v)d,v, P2,t or s, T1,p, (p)d, H, (q)d,q, T2,t is a Hamiltonian path of Qn− {x, y} − F . See Fig.7(a) for illustration.
Fig. 6 Illustration for Case 1 of Lemma4
Fig. 7 Illustration for Case 2 of Lemma4
Subcase 2.2: Both s and t are in Qd,1n . Since|F0| = |F | = n − 3 ≥ 2 for n ≥ 5, let (u, v) be a faulty edge such that{(u)d, (v)d} = {s, t}. It is noticed that (u, v) is
an r-dimensional edge. Without loss of generality, we assume that u∈ Vi(Qd,0n ), and let b∈ {1, 2, . . . , n} − {d, r}. By the inductive hypothesis, there exists a fault-free Hamiltonian path P in Qd,0n − {x, y} joining v and u.
Condition 2.2.1: (v)d= s. By Lemma 2, Qd,1n − {s} has a Hamiltonian path R joining (u)d and t. Hence, s, v, P, u, (u)d, R,t is a Hamiltonian path of Qn−
{x, y} − F . See Fig.7(b).
Condition 2.2.2: (u)d= t. Lemma2ensures that Qd,1
n − {t} has a Hamiltonian path H joining s and (v)d. Hence, s, H, (v)d,v, P , u, t is a Hamiltonian path of Qn− {x, y} − F . See Fig.7(c).
Condition 2.2.3: (v)d = s and (u)d = t. Let F= {((u)d, ((u)d)k)| 1 ≤ k ≤ n, k= d, k = r, k = b}. Then it follows from Lemma1that Qd,1n − Fhas a Hamil-tonian path R joining s and t. By the hypercube’s definition, edge ((u)d, (v)d) is on path R. Accordingly, R can be represented as either s, R1, (u)d, (v)d, R2,t
or s, R1, (v)d, (u)d, R2,t. As a result, either s, R1, (u)d,u, P , v, (v)d, R2,t or
s, R1, (v)d,v, P , u, (u)d, R2,t is a Hamiltonian path of Qn− {x, y} − F . See Fig.7(d) for illustration.
Subcase 2.3: s is in Qd,0n and t is in Qd,1n . Because every faulty edge is r-dimensional, we can find a faulty edge (u, v) such that t /∈ {(u)d, (v)d}. Let b ∈ V1−i(Qd,0n )− {x, y} such that b is not incident to any faulty edge. We denote
F− {(u, v)} by F. By the inductive hypothesis, there exists a Hamiltonian path P in
Qd,0n − {x, y} − Fjoining s and b.
Condition 2.3.1: (u, v) /∈ E(P ). Then, Lemma1ensures that there exists a Hamil-tonian path R in Qd,1n joining (b)d and t. Thus,s, P, b, (b)d, R,t is a Hamiltonian path of Qn− {x, y} − F . See Fig.7(e).
Condition 2.3.2: (u, v)∈ E(P ). Then, path P can be represented as s, P1,u, v,
P2,b. By the inductive hypothesis, there exists a Hamiltonian path H in Qd,1n −
{(u)d, (v)d} joining (b)d and t. Therefore,s, P
1,u, (u)d, (v)d,v, P2,b, (b)d, H,t is a Hamiltonian path of Qn− {x, y} − F . See Fig7(f).
Appendix B: Proof of Lemma6
Proof The proof proceeds by induction on n. Suppose that f= 0. Then this case
fol-lows from Lemma5. Suppose that f = n−2. Then we have δ = n − 1 − (n − 2) = 1. By Lemma1, Qn− F has a Hamiltonian path joining any two vertices in different partite sets. Moreover, this theorem is trivial for Q3, as the induction basis. In what follows, we consider 1≤ f ≤ n − 3 and n ≥ 4. The inductive hypothesis is that the theorem statement is true for Qn−1.
Since δ+ f = n − 1 < n, there must exist an integer d in {1, 2, . . . , n} such that
A∪ F contains no d-dimensional edges. Since Qnis edge-transitive, we can assume
d= n. Then we partition Qninto{Qn,0n , Qn,1n } along the nth dimension. Thus, each edge of A∪ F is in either Qn,0n or Qn,1n . For the sake of convenience, we define some notations to be used later: F0= F ∩ E(Qn,0n ), F1= F ∩ E(Qn,1n ), Fc = F ∩ Ecn,
f0= |F0|, f1= |F1|, and δ = n − 1 − f .
Let r0= |{(wi,bi)∈ E(Qn,0n )| 1 ≤ i ≤ δ}| and r1= |{(wi,bi)∈ E(Qn,1n )| 1 ≤
i≤ δ}|. Clearly, r0+ r1= δ. Without loss of generality, we assume {(w1,b1), . . . ,
(wr0,br0)} ⊂ E(Q
n,0
n ) if r1= 0, and {(wr0+1,br0+1), . . . , (wδ,bδ)} ⊂ E(Qn,1n ) if
r1>0. Since n− 1 = δ + f = r0+ r1+ f0+ f1, we have ri + fj ≤ n − 1 for any i, j∈ {0, 1}. Then we have to take the following cases into account.
Case 1: Suppose that ri+ fj≤ n − 2 for any i, j ∈ {0, 1}. Since r0+ f0≤ n − 2,
r0≤ n − 2 − f0= (n − 1) − 1 − f0. By the inductive hypothesis, Qn,0n − F0 has
r0-mutually fully independent Hamiltonian paths Hi[wi,bi], 1 ≤ i ≤ r0, if r0>0. Obviously, Hi[wi,bi] can be represented as wi, Hi,ui,bi, where ui is some vertex adjacent to bi. Similarly, Qn,1n − F1has r1-mutually fully independent Hamiltonian paths Hi[wi,bi] = wi, Hi,ui,bi, r0+ 1 ≤ i ≤ δ, if r1>0.
For r0 > 0, we construct r0 paths in Qn,1n − F1 to incorporate the previ-ously established r0 paths of Qn,0n − F0. Since r0+ f1≤ n − 2, we have r0 ≤
Fig. 8 Illustration for Case 1 of
Lemma6, particularly when 0 < r0< δ
fully independent Hamiltonian paths R1[(u1)n, (b1)n], . . . , Rr0[(ur0)
n, (b r0)
n]. Sim-ilarly, Qn,0n − F0 also contains r1-mutually fully independent Hamiltonian paths
Rr0+1[(ur0+1)
n, (b
r0+1)n], . . . , Rδ[(uδ)n, (bδ)n] if r1 > 0. Accordingly, we set
Pi[wi,bi] = wi, Hi,ui, (ui)n, Ri, (bi)n,bi for every 1 ≤ i ≤ δ. Thus, {P1, P2, . . . ,
Pδ} turns out to be a set of δ-mutually fully independent Hamiltonian paths in
Qn− F . See Fig.8for illustration.
Case 2: Suppose that ri+fi= n−1 for some i ∈ {0, 1}. Without loss of generality, we assume that r0+ f0= n − 1. Since r0= n − 1 − f0≥ n − 1 − f = δ, we must have r0= δ and f0= f ≤ n − 3. It is noticed that r0− 1 = δ − 1 = n − 2 − f =
(n− 1) − 1 − f0. By the inductive hypothesis, Qn,0n − F0 has (r0− 1)-mutually fully independent Hamiltonian paths Hi[wi,bi], 2 ≤ i ≤ r0. Again, Hi[wi,bi] can be represented aswi, Hi,ui,bi, where ui is some vertex adjacent to bi.
Subcase 2.1: Suppose n= 4. Thus, we have r0= 2. By Lemma 2, Q4,04 − F0 has a Hamiltonian path H1[w1,b1] = w1,u1, H1, (b1)j,b1, where u1 is a ver-tex adjacent to w1, and j is some integer of {1, 2, 3}. Let X = {((u1)4, (u2)4)}. Similarly, there exist two Hamiltonian paths R1[(w1)4, (u1)4] and R2[(u2)4, (b2)4] in Q4,14 − X. Obviously, we have R1(7) = R2(1) and R1(8) = R2(2). Then we set P1[w1,b1] = w1, (w1)4, R1, (u1)4,u1, H1, (b1)j,b1 and P2[w2,b2] =
w2, H2,u2, (u2)4, R2, (b2)4,b2. Consequently, P1and P2are 2-mutually fully in-dependent Hamiltonian paths in Q4− F . See Fig.9(a) for illustration.
Subcase 2.2: Suppose n≥ 5. We first consider f0 ≤ n − 4. By the induc-tive hypothesis, Qn,1n has (r0− 1)-mutually fully independent Hamiltonian paths
Ri[(ui)n, (bi)n], 2 ≤ i ≤ r0. Then we can choose an integer j in{1, 2, . . . , n − 1} satisfying conditions (b1)j = w1and ((b1)j)n∈ {R/ i(2n−1− 1) | 2 ≤ i ≤ r0}. Since
r0= n − 1 − f ≤ n − 2, such an integer exists. By Lemma 2, Qn,0n − F0− {b1} has a Hamiltonian path H1[w1, (b1)j] = w1,u1, H1, (b1)j, where u1is some ver-tex adjacent to w1. By Lemma3, there exists a Hamiltonian path R1[(w1)n, (u1)n] in Qn,1n − {(b1)n, ((b1)j)n}. Then we set P1[w1,b1] = w1, (w1)n, R1, (u1)n,u1, H1,
(b1)j, ((b1)j)n, (b1)n,b1 and Pi[wi,bi] = wi, Hi,ui, (ui)n, Ri, (bi)n,bi for 2 ≤
i≤ r0. As a result,{P1, P2, . . . , Pr0} turns out to be a set of r0-mutually fully
inde-pendent Hamiltonian paths in Qn− F . See Fig.9(b).
Next, we consider f0= n − 3. Thus, we have r0= 2. By Lemma1, Qn,0n − F0 has a Hamiltonian path H1[w1,b1] = w1,u1, H1, (b1)j,b1, where u1 is a ver-tex adjacent to w1, and j is some integer of {1, 2, . . . , n − 1}. By Lemma 3, there exists a Hamiltonian path R1[(w1)n, (u1)n] in Qnn,1 − {(b1)n, ((b1)j)n}.
Fig. 9 Illustration for Case 2 of
Lemma6
By the inductive hypothesis, Qn,1n − {((b2)n, ((b1)j)n)} has a Hamiltonian path
R2[(u2)n, (b2)n]. Obviously, we have R2(2n−1 − 1) = ((b1)j)n. Again, we set
P1[w1,b1] = w1, (w1)n, R1, (u1)n,u1, H1, (b1)j, ((b1)j)n, (b1)n,b1 and
P2[w2,b2] = w2, H2,u2, (u2)n, R2, (b2)n,b2. Hence, P1 and P2 are fully inde-pendent Hamiltonian paths in Qn− F . Also see Fig.9(b).
Case 3: Suppose that ri + f1−i = n − 1 for some i ∈ {0, 1}. Without loss of generality, we assume r1+ f0= n − 1. Since r1= n − 1 − f0≥ n − 1 − f = δ, we have r1 = δ and F0= F . By the inductive hypothesis, Qn,1n has (r1− 1)-mutually fully independent Hamiltonian paths Hi[wi,bi] = wi, Hi,ui,bi, where
ui is some vertex adjacent to bi for 1≤ i ≤ r1− 1. Since r1 − 1 = δ − 1 =
n− 2 − f = (n − 1) − 1 − f0, Qn,0n − F0 has (r1− 1)-mutually fully indepen-dent Hamiltonian paths Ri[(ui)n, (bi)n], 1 ≤ i ≤ r1− 1. Then we set Pi[wi,bi] =
wi, Hi,ui, (ui)n, Ri, (bi)n,bi for 1 ≤ i ≤ r1− 1. Next, we choose a vertex v of
V0(Qn,0n )and construct a Hamiltonian path Rr1[(wr1)
n,v] in Qn,0
n − F0 such that
v= Ri(2) and Rr1(2
n−1− 1) = (u
i)nfor every 1≤ i ≤ r1− 1. How can we do that? We distinguish the following subcases.
Subcase 3.1: Suppose that n= 5 or f > 1. Obviously, vertices (u1)n, . . . , (ur1−1)n have at most (r1−1)(n−1) neighboring vertices in Qn,0n . Since|V0(Qn,0n )| = 2n−2>
(r1−1)(n−1) = (n−2−f )(n−1), we can choose a vertex v other than neighbors of
(u1)n, . . . , (ur1−1)
n. Obviously, we have v= R
i(2) for 1≤ i ≤ r1− 1. By Lemma1, there exists a Hamiltonian path Rr1[(wr1)
n,v] in Qn,0
n −F0. Because v is not adjacent to any vertex of {(u1)n, . . . , (ur1−1)n}, we have Rr1(2
n−1− 1) = (u
i)n for every 1≤ i ≤ r1− 1. By Lemma 2, there exists a Hamiltonian path Hr1[(v)
n,b r1] in Qn,1n − {wr1}. Then we set Pr1 = wr1, (wr1) n, R r1,v, (v) n, H r1,br1. Consequently,
Fig. 10 Illustration for Case 3
of Lemma6
{P1, P2, . . . , Pr1} is a set of r1-mutually fully independent Hamiltonian paths in
Qn− F . Fig.10illustrates this subcase.
Subcase 3.2: Suppose that n= 5 and f = 1. Accordingly, we have r1= 3.
Condition 3.2.1: Vertices (u1)n and (u2)n have at least one common neighbor. Since |V0(Qn,0n )| = 2n−2= 8 > 7 = (r1− 1)(n − 1) − 1, we still can choose a vertex v from V0(Qn,0n ) other than neighbors of (u1)n and (u2)n. Obviously, we have v= Ri(2) for 1≤ i ≤ r1− 1. By Lemma1, there exists a Hamiltonian path
Rr1[(wr1)
n,v] of Qn,0
n − F0such that Rr1(2
n−1− 1) = (u
i)nfor every 1≤ i ≤ r1− 1. By Lemma2, there exists a Hamiltonian path Hr1[(v)
n,b r1] in Q n,1 n − {wr1}. Sim-ilarly, we set Pr1= wr1, (wr1) n, R r1,v, (v) n, H
r1,br1. Then, {P1, P2, P3} turns out
be a set of 3-mutually fully independent Hamiltonian paths in Qn− F . Also see Fig.10.
Condition 3.2.2: Vertices (u1)n and (u2)nhave no common neighbors. Then we set the vertex v to be the one adjacent to (u1)nand not identical to R1(2). Obviously, we have v= Ri(2) for 1≤ i ≤ r1− 1. By Lemma1, Qn,0n − F0− {(v, (u1)n)} re-mains Hamiltonian laceable. Thus, there exists a Hamiltonian path Rr1[(wr1)
n,v] of
Qn,0n − F0− {(v, (u1)n)} such that Rr1(2
n−1− 1) = (u
i)n for every 1≤ i ≤ r1− 1. By Lemma2, there exists a Hamiltonian path Hr1[(v)
n,b r1] in Q n,1 n − {wr1}. Simi-larly, we set Pr1 = wr1, (wr1) n, R r1,v, (v) n, H r1,br1. Then, {P1, P2, P3} is a set of
3-mutually fully independent Hamiltonian paths in Qn− F . Also see Fig.10.
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