DOI 10.1007/s11227-008-0206-0
The super spanning connectivity and super spanning
laceability of the enhanced hypercubes
Chung-Hao Chang· Cheng-Kuan Lin ·
Jimmy J.M. Tan· Hua-Min Huang · Lih-Hsing Hsu
Published online: 18 April 2008
© Springer Science+Business Media, LLC 2008
Abstract A k-container C(u, v) of a graph G is a set of k disjoint paths between u
and v. A k-container C(u, v) of G is a k∗-container if it contains all vertices of G. A graph G is k∗-connected if there exists a k∗-container between any two distinct vertices of G. Therefore, a graph is 1∗-connected (respectively, 2∗-connected) if and only if it is Hamiltonian connected (respectively, Hamiltonian). A graph G is super spanning connected if there exists a k∗-container between any two distinct vertices of G for every k with 1≤ k ≤ κ(G) where κ(G) is the connectivity of G. A bipartite graph G is k∗-laceable if there exists a k∗-container between any two vertices from different partite set of G. A bipartite graph G is super spanning laceable if there exists a k∗-container between any two vertices from different partite set of G for every k with 1≤ k ≤ κ(G). In this paper, we prove that the enhanced hypercube
Qn,mis super spanning laceable if m is an odd integer and super spanning connected if otherwise.
Keywords Folded hypercubes· Enhanced hypercubes · Hamiltonian connected ·
Hamiltonian laceable· Super spanning connected · Super spanning laceable
C.-H. Chang (
)The Division of General Education, Ming Hsin University of Science and Technology, Hsinchu, Taiwan 304, China
e-mail:chchang@must.edu.tw
C.-K. Lin· J.J.M. Tan
Department of Computer Science, National Chiao Tung University, Hsinchu, Taiwan 300, China
H.-M. Huang
Department of Mathematics, National Central University, Chung-Li, Taiwan 320, China
L.H. Hsu
Department of Computer Science and Information Engineering, Providence University, Taichung, Taiwan 433, China
1 Introduction
In this paper, a network is represented as a loopless undirected graph. For the graph definitions and notations, we basically follow [3]. G= (V, E) is a graph if V is a finite set and E is a subset of{(u, v) | (u, v) is an unordered pair of V }. We say that V is the vertex set and E is the edge set. Two vertices u and v are adjacent if
(u, v)∈ E. The degree dG(u) of a vertex u of G is the number of edges incident with u. A path is a sequence of vertices represented byv0,v1, . . . ,vk with no re-peated vertex and (vi,vi+1)is an edge of G for all 0≤ i ≤ k − 1. We also write the path P= v0, . . . ,vk as v0, . . . ,vi, Q,vj, . . . ,vk, where Q is a path from vito vj. We use P−1to denote the pathvk,vk−1, . . . ,v1,v0. The length of a path P , l(P ), is the number of edges in P . A path is a Hamiltonian path if it contains all vertices of G. A graph G is Hamiltonian connected if there exists a Hamiltonian path join-ing any two distinct vertices of G. A cycle is a closed pathv0, v1, . . . , vk, v0 where
v0, v1, . . . , vk is a path with k ≥ 2. A Hamiltonian cycle of G is a cycle that tra-verses every vertex of G exactly once. A graph is Hamiltonian if it has a Hamiltonian cycle.
The connectivity of G, κ(G), is the minimum number of vertices whose removal leaves the remaining graph disconnected or trivial. It follows from Menger’s theorem [16] that there are k internal vertex-disjoint paths joining any two distinct vertices when k≤ κ(G). A k-container of a graph G between u and v is a set of k inter-nal vertex-disjoint paths between u and v. Connectivity and container are impotent concepts to measure the fault tolerant of a networks [5,9].
In this paper, we are interested in some special type of containers. A k-container of G between u and v is a k∗-container if it contains all vertices of G. A graph
Gis k∗-connected if there exists a k∗-container between any two distinct vertices. A 1∗-connected graph except K1 and K2 is 2∗-connected. Thus, the concept of
k∗-connected graph is a hybrid concept of connectivity and Hamiltonicity. The study of k∗-connected graph is motivated by the globally 3∗-connected graphs proposed by Albert, Aldred, and Holton [2]. A globally 3∗-connected graph is a cubic graph that is w∗-connected for all 1≤ w ≤ 3. Recently, Lin et al. [12] proved that the pan-cake graph Pn is w∗-connected for any w with 1≤ w ≤ n − 1 if and only if n = 3. The spanning connectivity of a graph G, κ∗(G), is the largest integer k such that G is w∗-connected for all 1≤ w ≤ k if G is 1∗-connected graph. There are some in-teresting results of spanning connectivity [8,13–15]. A graph G is super spanning connected if κ∗(G)= κ(G). Obviously, the complete graph Kn is super spanning connected. Lin et al. [12] studied the n-dimensional pancake graph Pnis super span-ning connected if and only if n= 3. Tsai et al. [18] studied the recursive circulant graphs G(2m,4) is super-connected if and only if m= 2.
A graph G is bipartite if its vertex set can be partitioned into two subsets V0and
V1such that every edge joins vertices of V0and V1. A bipartite graph is k∗-laceable graph if there exists a k∗-container between any two vertices from different partite sets. Note that a 1∗-laceable graph is also known as a Hamiltonian laceable graph. Moreover, a bipartite graph is 2∗-laceable if and only if it is a Hamiltonian graph and all 1∗-laceable graphs except K1 and K2 are 2∗-laceable. A Hamiltonian laceable graph G with partition V0, V1is hyper-Hamiltonian laceable if we remove any vertex
v from a partite set, say V0, there is a Hamiltonian path of G− {v} joining any two vertices in the other partite set V1. If G is a 1∗-laceable graph, we define the spanning laceablility of a bipartite graph G, κ∗(G), to be the largest integer k such that G is w∗-laceable for all 1≤ w ≤ k. A bipartite graph G is super spanning laceable if κ∗(G)= κ(G). Recently, Chang et al. [4] proved that the hypercube graph Qn is super spanning laceable. All bipartite hypercube-like graphs are super spanning laceable [14]. The n-dimensional star graph Snis super spanning laceable if and only if n= 3 [12].
Graph containers do exist in engineering designed information and telecommuni-cation networks or in biological and neural systems ([1,9] and their references). The study of w-container and their w∗-container plays a pivotal role in the design and the implementation of parallel routing and efficient information transmission in a large scale networking systems. In biological informatics and neural informatics, the exis-tence of a w∗-container signifies the effects on the signal transduction system and the reactions in metabolic pathways.
Among all interconnection networks proposed in the literature, the hypercubes Qn is one of the most popular topologies [10]. Let u= u1u2· · · un−1unbe an n-bit binary strings. The hamming weight of u, denoted by w(u), is defined to be the number of
i such that ui = 1. The n-dimensional hypercube Qn consists of all n-bit binary strings as its vertices and two vertices u= u1u2· · · un−1unand v= v1v2· · · vn−1vn are adjacent if and only if u and v differ by exactly one bit, i.e.,ni=1|ui− vi| = 1. Obviously, Qn is a bipartite graph with bipartition W= {u | w(u) is even} and B =
{u | w(u) is odd}. For convenience, the vertices in W are referred as even vertices
and the vertices in B are referred as odd vertices.
Some variations of hypercubes structures have been reported in the literature, for instance, the folded hypercubes FQn by El-Amawy and Latifi [6] and enhanced hypercubes Qn,m (2≤ m ≤ n) by Tzeng NF and Wei S [19]. The folded hyper-cubes FQn is obtained from a hypercubes Qn with add on edges defined by join-ing any vertex u= u1u2· · · un−1un to ¯u = ¯u1¯u2· · · ¯un−1¯un, where ¯ui = 1 − ui is the complement of ui. The enhanced hypercube Qn,m is obtained from a hyper-cubes Qn with add on edges defined by joining any vertex u= u1u2· · · un−1un to
(u)c= ¯u
1¯u2· · · ¯umum+1um+2· · · un−1un. Obviously, FQn= Qn,nand FQnand Qn,m are (n+ 1)-regular. Moreover, FQn is a bipartite graph if and only if n is odd and
Qn,mis a bipartite graph if and only if m is odd.
The rest of the paper is organized as follows. In the next section, we prove some new spanning properties of the hypercubes Qn. In Sect.3, we prove that the folded hypercubes FQn is super spanning laceable if n is an odd integer and super span-ning connected if otherwise. In Sect.4, we prove that the enhanced hypercubes Qn,m is super spanning laceable if m is an odd integer and super spanning connected if otherwise. In the final section, we give our concluding remark.
2 The super spanning laceability of hypercubes
In this section, we review some known results and prove a new theorem. Let u=
the k-th neighbor of u and use (u)k to denote uk. We set Qin−1be the subgraph of
Qn induced by{u ∈ V (Qn)| (u)n= i} for i = 0, 1. Obviously, Qin−1is isomorphic to Qn−1for i= 0, 1. It is well known that Qn is vertex transitive. Furthermore, the permutation on the coordinates of Qnand the componentwise complement operations are graph isomorphisms. Readers can refer reference [7,10] for a survey about the properties of hypercubes. We have the following lemmas:
Lemma 1 [11] Qnis hyper-Hamiltonian laceable if and only if n≥ 2.
Lemma 2 [4] Qnis super spanning laceable for any positive integer n.
Chang et al. [4] proved that the following two paths spanning property of hyper-cube.
Lemma 3 [4] Assume that n≥ 2. Let x1and x2be two distinct even vertices of Qn and y1and y2be two distinct odd vertices of Qn. Then there exist two paths P1and
P2of Qnsuch that (1) Pi joins xi and yi for 1≤ i ≤ 2 and (2) P1∪ P2spans Qn.
Lemma 4 [17] Qn− {x, y} is Hamiltonian laceable if x is an even vertex, y is an odd vertex of Qn, and n≥ 4.
There is another version of Menger theorem on k-connected graphs, called k-fan version. Let G be a graph. Let x be a vertex in G and S= {y1,y2, . . . ,yk} be a set of k vertices not containing x. An (x, S)-fan is a set of disjoint paths{P1, P2, . . . , Pk} such that Pi is a path joining x to yi for 1≤ i ≤ k. The k-fan version Menger’s theorems states that there exists an (x, S)-fan of G between any vertex x and any k set S not containing x with 1≤ k ≤ κ(G). With this observation, we define a spanning fan is a fan that spans G. The following theorem states that there exists a spanning (x, S)-fan,
{P1, P2, . . . , Pk}, of Qnbetween any vertex x and S= {y1,y2, . . . ,yk} with yk being the only vertices in{y1,y2, . . . ,yk} in the partite set not containing x and 1 ≤ k ≤ n. The requirement that yk is the only vertex in{y1,y2, . . . ,yk} in the partite set not containing x is needed just because Qnis a bipartite graph with the same number of vertices in both partite sets.
Theorem 1 Assume that k≤ n and x is a vertex of Qn. Let U = {y1,y2, . . . ,yk} be a subset of V (Qn)− {x} with yi= yj for every i= j and yk is the only vertex in
{y1,y2, . . . ,yk} such that ykand x are in different partite set. Then there is a spanning
(x, U )-fan of Qn.
Proof By Lemma2, this statement is holds on every Qnif k= 1. Suppose that k = 2 and n≥ 2. By Lemma2, there is a Hamiltonian path P= y1, R1,x, R2,y2 of Qn joining y1 to y2. We set P1= x, R−11 ,y1 and P2= x, R2,y2. Then P1 and P2 forms the required paths. Thus, we assume that 3≤ k ≤ n, and this theorem is true for Qn−1. Since Qn is vertex transitive, we assume that x= 0n. Thus, x is an even vertex and x∈ Q0n−1. We have the following cases:
Fig. 1 Illustration for Theorem1
Case 1: (yk)i = 0 for some 1 ≤ i ≤ n. Since Qn is edge transitive, we assume that
(yk)n= 0. Thus, yk ∈ Q0n−1. For 0≤ j ≤ 1, we set Uj = {yi | yi ∈ Qjn−1for 1≤
i≤ k}. Without loss of generality, we assume that U0= {ym+1,ym+2, . . . ,yk} ⊆
Q0n−1and U1= {y1,y2, . . . ,ym} ⊆ Q1n−1for some 0≤ m ≤ k − 1.
Subcase 1.1: m= 0. Let ˜U = U0− {yk−1}. Obviously, | ˜U| = k − 1. By induc-tion, there is a spanning (x, ˜U )-fan,{R1, R2, . . . , Rk−1}, of Q0n−1. Without loss of generality, we assume that yk−1∈ Rk−1 where Rk−1 is joining x to yt for some
t∈ {1, 2, . . . , k − 2, k}. We can write Rk−1as x, H1,yk−1,z, H2,yt. (Note that z = yk if l(H2)= 0.) By Lemma 2, there is a Hamiltonian path W of Q1n−1joining
(x)n to (z)n. We set Pi = Ri for every 1≤ i ≤ k − 2, Pk−1= x, H1,yk−1, and
Pk= x, (x)n, W, (z)n,z, H2,yt. Then {P1, P2, . . . , Pk} forms a set of required paths of Qn. See Fig.1(a) for an illustration for k= 6 and t = 6.
Subcase 1.2: m= 1. Thus, y1∈ Q1n−1. By induction, there is a spanning (x, U0)-fan,
{R1, R2, . . . , Rk−1}, in Q0n−1such that Ri joins x to yi+1for every 1≤ i ≤ k − 1. By Lemma2, there is a Hamiltonian path W of Q1n−1joining (x)nto y1. We set P1=
x, (x)n, W,y
1 and Pi= Ri−1 for every 2≤ i ≤ k. Then {P1, P2, . . . , Pk} forms a spanning (x, U )-fan of Qn. See Fig.1(b) for an illustration for k= 6.
Subcase 1.3: m= 2. We have {y1,y2} ⊆ Q1n−1. Since there are 2n−2even vertices in
Q0n−1and 2n−2− |U0∪ {x}| − 1 = 2n−2− (k − 2) ≥ 2n−2− n + 2 ≥ 1 if n ≥ 3, we can choose an even vertex u in Q0n−1− (U0∪ {x}). By induction, there is a spanning
(x, U0∪ {u})-fan, {R1, R2, . . . , Rk−1} of Q0n−1such that (1) Ri joins x to yi+2for every 1≤ i ≤ k − 2 and (2) Rk−1joins x to u. By Lemma3, there exist two disjoint paths S1and S2of Q1n−1such that (1) S1joins (u)nto y1, (2) S2joins (x)nto y2, and (3) S1∪S2spans Q1n−1. We set P1= x, Rk−1,u, (u)n, S1,y1, P2= x, (x)n, S2,y2, and Pi= Ri−2for every 3≤ i ≤ k. Then {P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig.1(c) for an illustration for k= 6.
Subcase 1.4: 3≤ m ≤ k − 2. We have k ≥ 5. Hence, n ≥ 5. Since m ≥ 3 and k ≤ n, |U0− {yk}| = k − m − 1 ≤ k − 4 ≤ n − 4.
We claim that there exists an even vertex u in Q1n−1− U1 such that (yi)n∈/
NQ1
n−1(u) for every m+ 1 ≤ i ≤ k − 1. Such claim holds because (n − 1)|U0−
{yk}| + |U1| ≤ (n − 1)(n − 4) + (n − 2) ≤ (n − 1)(n − 3) − 1 < 2n−2for all n≥ 5. Since m≤ k − 2 and k ≤ n, m + 1 ≤ n − 1. By induction, there is a spanning
(u, U1∪ {(x)n})-fan, {W1, W2, . . . , Wm+1} in Q1n−1 such that (1) Wi joins u to yi for every 1≤ i ≤ m and (2) Wm+1 joins u to (x)n. We write Wi as u, vi, Wi ,yi for every 1≤ i ≤ m − 1. Since u is an even vertex in Q1n−1, vi is an odd ver-tex in Q1n−1 and (vi)n is an even vertex in Q0n−1 for every 1≤ i ≤ m − 1. Let
˜
U0= U0∪ {(vi)n|1 ≤ i ≤ m − 1}. Obviously, | ˜U0| = (k − m) + (m − 1) = k − 1. By induction, there is a spanning (x, ˜U0)-fan, {R1, R2, . . . , Rk−1}, of Q0n−1 such that (1) Ri joins x to (vi)n for every 1≤ i ≤ m − 1 and (2) Ri joins x to yi+1for every m≤ i ≤ k − 1. We set Pi = x, Ri, (vi)n,vi, Wi ,yi for every 1 ≤ i ≤ m − 1,
Pm= x, (x)n, Wm−1+1,u, Wm,ym, and Pi = Ri−1 for every m+ 1 ≤ i ≤ k. Then
{P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig.1(d) for an illustration for k= 6 and m = 3.
Subcase 1.5: m= k − 1 and k − 1 ≥ 3. Let ˜U1= (U1− {y1}) ∪ {(x)n}. Obviously,
| ˜U1| = k − 1. By induction, there is a spanning (y1, ˜U1)-fan,{W1, W2, . . . , Wk−1}, in
Q1n−1such that (1) W1joins y1to (x)n and (2) Wi joins y1to yi for every 2≤ i ≤
k− 1. We write Wi asy1,vi, Wi ,yi for every 2 ≤ i ≤ k − 1. Since y1is an even vertex in Q1n−1, vi is an odd vertex in Q1n−1 and (vi)n is an even vertex in Q0n−1 for every 2≤ i ≤ k − 1. Let ˜U0= {yk} ∪ {(vi)n|2 ≤ i ≤ k − 1}. Obviously, | ˜U0| =
k− 1. By induction, there is a spanning (x, ˜U0)-fan,{R1, R2, . . . , Rk−1}, in Q0n−1 such that (1) R1joins x to yk and (2) Ri joins x to (vi)nfor every 2≤ i ≤ k − 1. We set P1= x, (x)n, W−1
1 ,y1, Pi = x, Ri, (vi)n,vi, Wi ,yi for every 2 ≤ i ≤ k − 1, and Pk= R1. Then{P1, P2, . . . , Pk} forms a (x, U)-fan of Qn. See Fig.1(e) for an illustration for k= 6.
Case 2: (yk)i= 1 for every 1 ≤ i ≤ n. Obviously, n is odd with n ≥ 3 and yk∈ Q1n−1. Since Qnis edge transitive, we assume that U0= {y1,y2, . . . ,ym} ⊆ Q0n−1and U1=
{ym+1,ym+2, . . . ,yk} ⊆ Q1n−1for some 1≤ m ≤ k − 2.
Subcase 2.1: m= k − 2. We have {yk−1,yk} ⊆ Q1n−1. Let H be a Hamiltonian path of Q1n−1joining yk−1to yk. We write H asyk−1, H1,u, (x)n, H2,yk. Since
(x)n is an odd vertex, u is an even vertex and (u)n is an odd vertex in Q0n−1. (Note that yk−1= u if l(H1)= 0 or (x)n= yk if l(H2)= 0.) By induction, there is a spanning (x, U0∪ {(u)n})-fan, {R1, R2, . . . , Rk−1} in Q0n−1 such that (1) Ri joins x to yi for 1≤ i ≤ k − 2 and (2) Rk−1joins x to (u)n. We set Pi = Ri for 1≤ i ≤ k − 2, Pk−1= x, Rk−1, (u)n,u, H1−1,yk−1, and Pk = x, (x)n, H2,yk. Then{P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig. 1(f) for an il-lustration for k= 6.
Subcase 2.2: m= k − 3. We have n ≥ 5 and {yk−2,yk−1,yk} ⊆ Q1n−1. Since m+ 1 ≤
n− 2 < 2n−2, we can pick an even vertex z∈ Q0n−1− ({yi| 1 ≤ i ≤ k − 3} ∪ {x}). By Lemma3, there exist two disjoint paths S1and S2of Q1n−1such that (1) S1joins (z)n to yk−2, (2) S2joins (x)nto yk−1, and (3) S1∪ S2spans Q1n−1. Obviously, yk∈ Sifor some 1≤ i ≤ 2.
Subcase 2.2.1: yk∈ S1. We write S1as(z)n, H1,yk,u, H2,yk−2. Obviously, u is an even vertex and (u)nis an odd vertex in Q0
n−1. Let ˜U0= U0∪ {z, (u)n}. Obviously,
| ˜U0| = k − 1. By induction, there is a spanning (x, ˜U0)-fan, {R1, R2, . . . , Rk−1}, in Q0n−1 such that (1) Ri joins x to yi for 1≤ i ≤ k − 3, (2) Rk−2 joins x to z, and (3) Rk−1 joins x to (u)n. We set Pi = Ri for 1≤ i ≤ k − 3, Pk−2=
x, Rk−1, (u)n,u, H2,yk−2, Pk−1= x, (x)n, S2,yk−1, and Pk = x, Rk−2,z, (z)n,
H1,yk. Then {P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig.1(g) for an illustration for k= 6.
Subcase 2.2.2: yk ∈ S2. Similar to Subcase 2.2.1, there is a spanning (x, U )-fan of Qn.
Subcase 2.3: 1≤ m ≤ k − 4. We have k ≥ 5. Moreover, n ≥ 5. Since m ≤ k − 4 and k≤ n, |U0| = m ≤ k − 4 ≤ n − 4.
We claim that there exists an even vertex u in Q1n−1− U1 such that (yi)n∈/
NQ1
n−1(u) for every 1≤ i ≤ m. Such claim holds because (n − 1)|U0| + |U1− {yk}| =
(n− 1)m + (k − m) − 1 = (n − 2)m + k − 1 ≤ (n − 2)(n − 4) + n − 1 = (n − 1)(n −
4)+ 3 < 2n−2for all n≥ 5.
Let ˜U1= (U1− {yk}) ∪ {(x)n}. Obviously, | ˜U1| = k − m. By induction, there is a spanning (u, ˜U1)-fan,{Wm+1, Wm+2, . . . , Wk}, in Q1n−1joining u to ˜U1 such that (1) Wi joins u to yi for every m+ 1 ≤ i ≤ k − 1 and (2) Wk joins u to (x)n. We write Wi asu, vi, Wi ,yi for every m + 1 ≤ i ≤ k − 1. Since u is an even vertex in Q1n−1, vi is an odd vertex in Q1n−1and (vi)n is an even vertex in Q0n−1for every
m+ 1 ≤ i ≤ k − 2.
Subcase 2.3.1: yk ∈ Wk. We write Wk as u, H1,z, yk, H2, (x)n. Since yk is an odd vertex in Q1n−1, z is an even vertex in Q1n−1, and (z)n is an odd vertex in Q0n−1. Let ˜U0= U0∪ {(vi)n|m + 1 ≤ i ≤ k − 2} ∪ {(z)n}. Obviously, | ˜U0| =
m+ (k − m − 2) + 1 = k − 1. By induction, there is a spanning (x, ˜U0)-fan,
{R1, R2, . . . , Rk−1}, in Q0n−1such that (1) Ri joins x to yi for 1≤ i ≤ m, (2) Ri joins
x to (vi)nfor every m+ 1 ≤ i ≤ k − 2, and (3) Rk−1joins x to (z)n. We set Pi = Ri for every 1≤ i ≤ m, Pi = x, Ri, (vi)n,vi, Wi ,yi for every m + 1 ≤ i ≤ k − 2,
Then{P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig. 1(h) for an il-lustration for k= 6 and m = 2.
Subcase 2.3.2: yk∈ Wi for some 1≤ i ≤ k − 1. Without loss of generality, we as-sume that yk∈ Wk−1. We write Wk−1asu, vk−1, H1,yk,z, H2,yk−1. Since ykis an odd vertex in Q1n−1, z is an even vertex in Q1n−1and (z)nis an odd vertex in Q0n−1. Let ˜U0= U0∪ {(vi)n|m + 1 ≤ i ≤ k − 2} ∪ {(z)n}. Obviously, | ˜U0| = m + (k − m − 2)+ 1 = k − 1. By induction, there is a spanning (x, ˜U0)-fan,{R1, R2, . . . , Rk−1}, in Q0n−1 such that (1) Ri joins x to yi for every 1≤ i ≤ m, (2) Ri joins x to
(vi)n for every m+ 1 ≤ i ≤ k − 2, and (3) Rk−1 joins x to (z)n. We set Pi = Ri for every 1≤ i ≤ m, Pi = x, Ri, (vi)n,vi, Wi ,yi for every m + 1 ≤ i ≤ k − 2,
Pk−1= x, Rk−1, (z)n,z, H2,yk−1, and Pk= x, (x)n, Wk−1,u, vk−1, H1,yk. Then
{P1, P2, . . . , Pk} forms a spanning (x, U)-fan of Qn. See Fig.1(i) for an illustration
for k= 6 and m = 2.
3 The super spanning properties of folded hypercubes
Let u= u1u2· · · un−1un be a vertex of FQn. The c-neighbor of u in FQn, (u)c, is
¯u1¯u2· · · ¯un. Note that (u)c and u are of the same parity if and only if n is an even integer. Let Ec= {(u1u2· · · un,¯u1¯u2· · · ¯un)| u1u2· · · un∈ V (FQn)}. By definition, the n-dimensional folded hypercube FQnis obtained from Qnby adding Ec. Let f be a function on V (FQn)defined by f (u)= u if (u)n= 0 and f (u) = ((u)c)n if otherwise. The following theorem can be proved easily.
Theorem 2 The function f is an isomorphism of FQninto itself.
Let FQjn−1be the subgraph of FQn induced by{v ∈ V (FQn)| (v)n= j} for 0 ≤
j≤ 1. Obviously, FQjn−1is isomorphic to Qn−1for 0≤ j ≤ 1.
Lemma 5 Let x be an even vertex and y be an odd vertex of FQn for any positive integer n≥ 2. Then there exists a k∗-container of FQn between x and y for every 1≤ k ≤ n + 1.
Proof Since FQ2is isomorphic to the complete graph K4, this statement holds for
n= 2. Suppose that n ≥ 3. Since Qn is a spanning subgraph of FQn, by Lemma2, there exists a k∗-container between x and y for every 1≤ k ≤ n. Thus, we only need to construct an (n+ 1)∗-container of FQn between x and y. Since FQn is vertex transitive, we assume that x= 0n∈ V (FQ0n−1).
Case 1: y∈ FQ0n−1. We have the following cases:
Subcase 1.1: n= 3. Without loss of generality, we assume that y = 100. We set P1= 000, 001, 101, 100, P2= 000, 010, 110, 100, P3= 000, 100, and P4=
000, 111, 011, 100. Then {P1, P2, P3, P4} forms a 4∗-container of FQ3between x and y.
Fig. 2 Illustration for Lemma5
Subcase 1.2: n≥ 4. Since FQ0n−1is isomorphic to Qn−1, by Lemma2, there is an
(n− 1)∗-container{P1, P2, . . . , Pn−1} of FQ0n−1between x and y.
Subcase 1.2.1: (x)c= (y)n. Obviously, (x)c and (y)c are of different parity. Since FQ1n−1is isomorphic to Qn−1, by Lemma3, there exist two disjoint paths S1and S2 of FQ1n−1such that (1) S1joins (x)nto (y)n, (2) S2joins (x)c to (y)c, and (3) S1∪
S2spans FQ1n−1. We set Pn= x, (x)n, S1, (y)n,y and Pn+1= x, (x)c, S2, (y)c,y. Then{P1, P2, . . . , Pn+1} forms an (n + 1)∗-container of FQnbetween x and y. See Fig.2(a) for illustration for n= 5.
Subcase 1.2.2: (x)c= (y)n. Then (y)c= (x)nand n is even.
Suppose that n= 4. We have x = 0000 and y = 1110. We set P1= 0000, 0001, 1110, P2= 0000, 0010, 0110, 1110, P3= 0000, 0100, 0101, 0111, 0011, 1011, 1001, 1101, 1100, 1110, P4 = 0000, 1000, 1010, 1110, and P5 = 0000, 1111, 1110. Then {P1, P2, P3, P4, P5} forms a 5∗-container of FQ4between x and y.
Since 2n−1− 2 ≥ 3(n − 1) for n ≥ 6, there is one path Pi in{P1, P2, . . . , Pn−1} such that I (Pi)≥ 3. Without loss of generality, we may assume that I (Pn−1)≥ 3. We write Pn−1 as x, u, v, H, y where u is an odd vertex and v is an even ver-tex. By Lemma 4, there is a Hamiltonian path W of Q1n−1− {(x)n, (y)n} joining (u)n to (v)n. We set P
n−1= x, u, (u)n, W, (v)n,v, H, y, Pn= x, (x)n= (y)c,y, and Pn+1= x, (x)c= (y)n,y. Then {P
1, P2, . . . , Pn−2, Pn −1, Pn, Pn+1} forms an
(n+ 1)∗-container of FQnbetween x and y. See Fig.2(b) for illustration for n= 6.
Case 2: y∈ FQ1n−1. We have the following cases:
Subcase 2.1: n is odd and y∈ {(x)n, (x)c}. By Theorem2, we only consider that
y= (x)c. By Lemma2, there is an n∗-container{P
1, P2, . . . , Pn} of Qn between x and y. We set Pn+1= x, y. Then {P1, P2, . . . , Pn+1} forms an (n + 1)∗-container of FQnbetween x and y.
Subcase 2.2: n is odd and y /∈ {(x)c, (x)n}. Since y ∈ FQ1n−1 and y is an odd ver-tex, we have y= (x)c or y= (x)n if n= 3. Thus, n ≥ 5. Since there are 2n−2 even vertices in FQ0n−1 and 2n−2≥ n − 1 for n ≥ 5, we can choose (n − 4) dis-tinct even vertices u1, u2, . . . ,un−4in FQ0n−1− {x, (y)c, (y)n} such that (ui)n= (x)c for 1≤ i ≤ n − 4. Let v be an odd vertex of FQ0n−1 and let U0= {ui|1 ≤ i ≤
n− 4} ∪ {(y)c, (y)n,v}. Obviously, |U
0| = n − 1. By Theorem 1, there is a span-ning (x, U0)-fan,{R1, R2, . . . , Rn−1}, in FQ0n−1 such that (1) Ri joins x to ui for 1≤ i ≤ n − 4, (2) Rn−3joins x to (y)c, (3) Rn−2joins x to (y)n, and (4) Rn−1joins
x to v. Let U1= {(ui)n|1 ≤ i ≤ n − 4} ∪ {(x)c, (x)n, (v)n}. Obviously, |U1| = n − 1. By Theorem1, there is a spanning (y, U1)-fan,{H1, H2, . . . , Hn−1}, in FQ1n−1such that (1) Hi joins (ui)n to y for 1≤ i ≤ n − 4, (2) Hn−3joins (x)c to y, (3) Hn−2 joins (x)n to y, and (4) Hn−1 joins (v)n to y. We set Pi = x, Ri,ui, (ui)n, Hi,y for 1≤ i ≤ n − 4, Pn−3= x, Rn−3, (y)c,y, Pn−2= x, Rn−2, (y)n,y, Pn−1=
x, Rn−1,v, (v)n, Hn−1,y, Pn= x, (x)c, Hn−3,y, and Pn+1= x, (x)n, Hn−2,y. Then{P1, P2, . . . , Pn+1} forms an (n + 1)∗-container of FQnbetween x and y. See Fig.2(c) for illustration for n= 5.
Subcase 2.3: n is even with n≥ 4 and y = (x)n. Since there are 2n−2even vertices in FQ0n−1and 2n−2≥ n − 1 for n ≥ 4, we can choose (n − 2) distinct even vertices
u1, u2, . . . ,un−2in FQ0n−1− {x}. Let U0= {ui|1 ≤ i ≤ n − 2} ∪ {(y)c}. Obviously,
|U0| = n − 1. By Theorem1, there is a spanning (x, U0)-fan,{R1, R2, . . . , Rn−1}, in FQ0n−1such that (1) Rijoins x to uifor 1≤ i ≤ n−2 and (2) Rn−1joins x to (y)c. Let
U1= {(ui)n|1 ≤ i ≤ n−2}∪{(x)c,}. Obviously, |U1| = n−1. By Theorem1, there is a spanning (y, U1)-fan,{H1, H2, . . . , Hn−1}, in FQ1n−1such that (1) Hi joins (ui)nto
y for 1≤ i ≤ n − 2 and (2) Hn−1joins (x)cto y. We set Pi = x, Ri,ui, (ui)n, Hi,y for 1≤ i ≤ n − 2, Pn−1= x, Rn−1, (y)c,y, Pn= x, (x)c, Hn−1,y, and Pn+1=
x, y = (x)n. Then {P
1, P2, . . . , Pn+1} forms an (n + 1)∗-container of FQnbetween
x and y. See Fig.2(d) for illustration for n= 6.
Subcase 2.4: n is even with n≥ 4 and y = (x)n. Since there are 2n−2even vertices in FQ0n−1and 2n−2≥ n − 1 for n ≥ 4, we can choose (n − 3) distinct even vertices u1,
u2, . . . ,un−3in FQ0n−1− {x, (y)n} such that (ui)n= (x)nfor 1≤ i ≤ n − 3. Let U0=
{ui|1 ≤ i ≤ n − 3} ∪ {(y)n, (y)c} and let U1= {(ui)n|1 ≤ i ≤ n − 3} ∪ {(x)n, (x)c}. Obviously,|U0| = |U1| = n−1. Similar to Subcase 2.2, there is an (n+1)∗-container
of FQnbetween x and y.
Theorem 3 FQnis super spanning laceable if n is an odd integer and FQnis super spanning connected if n is an even integer.
Proof Since FQ1is isomorphic to Q1, this statement holds for n= 1. By Lemma5, this statement holds if n is odd and n≥ 3. Thus, we assume that n is even. Let x and y be any two different vertices of FQn. We need to find a k∗-container of FQnbetween
x and y for 1≤ k ≤ n + 1. Without loss of generality, we assume that x is an even
vertex. By Lemma5, this statement holds if y is an odd vertex. Thus, we assume that
y is an even vertex. Without loss of generality, we assume that (x)n= 0 and (y)n= 1. Let f be the function on V (FQn)defined by f (u)= u if (u)n= 0 and f (u) = ((u)c)n if otherwise. By Theorem 2, f is an isomorphism from FQn into itself. In other
words, we still get FQnif we relabel all the vertices u with f (u). However, f (x)= x is an even vertex and f (y)= ((y)c)n is an odd vertex. By Lemma5, there exists a
k∗-container of FQn between f (x) and f (y) for every 1≤ k ≤ n + 1. Thus, there exists a k∗-container of FQnbetween x and y for every 1≤ k ≤ n + 1. This theorem
is proved.
4 The super spanning properties of enhanced hypercubes
Let u= u1u2· · · un−1un be a vertex of Qn,m. Similar to before, c-neighbor of u in Qn,m, (u)c, is ¯u1¯u2· · · ¯umum+1um+2· · · un−1un. Note that (u)c and u are of the same parity if and only if m is even. Let Ec= {(u1u2· · · un,¯u1¯u2· · · ¯umum+1um+2· · ·
un−1un)|u1u2· · · un∈ V (Qn,m)}. By definition, the n-dimensional enhanced hyper-cube Qn,mis obtained from Qn by adding Ec. Obviously, Qn,mis FQ
n if m= n. We use Qjn,mto denote the subgraph of Qn,minduced by{v ∈ V (Qn,m)| (v)n= j} for 0≤ j ≤ 1. Moreover, we use Qijn,m to denote the subgraph of Qn,minduced by
{v ∈ V (Qn,m)|(v)n−1= i and (v)n= j} for 0 ≤ i, j ≤ 1.
Lemma 6 Let x and y be any two distinct vertices of Qjn,m with n− m ≥ 1 for some j . Suppose that there is a k∗-container of Qjn,m between x and y and there is an 1∗-container of Q1n,m−j between (x)nand (y)n. Then there is a (k+ 1)∗-container of Qn,mbetween x and y.
Proof Let{P1, P2, . . . , Pk} be a k∗-container of Qjn,m between x and y and W be a Hamiltonian path of Q1n,m−j joining (x)n to (y)n. Set Pk+1= x, (x)n, W, (y)n,y. Then{P1, P2, . . . , Pk+1} forms a (k + 1)∗-container of Qn,mbetween x and y.
Lemma 7 Let x be an even vertex and y be an odd vertex of Qn,n−1for any positive integer n≥ 3. Then there exists a k∗-container of Qn,n−1between x and y for every 1≤ k ≤ n + 1.
Proof Since Qn is a spanning subgraph of Qn,n−1, by Lemma2, there exists a k∗ -container of Qn,n−1 between x and y for every 1≤ k ≤ n. Thus, we only need to construct an (n+ 1)∗-container of Qn,n−1between x and y. Without loss of general-ity, we assume that x∈ Q00n,n−1. We have the following cases:
Case 1: y∈ Q00n,n−1∪ Q10n,n−1. Since Qn,n00−1∪ Q10n,n−1= Q0n,n−1 is isomorphic to FQn−1, by Lemma5, there exists an n∗-container of Q0n,n−1between x and y. Since
Q01n,n−1∪ Q11n,n−1= Q1n,n−1 is isomorphic to FQn−1, by Lemma 5, there exists a Hamiltonian path of Q1n,n−1 joining (x)n to (y)n. By Lemma 6, there exists an
(n+ 1)∗-container of Qn,n−1between x and y.
Case 2: y∈ Q01n,n−1. Suppose that n= 3. We have x = 000 and y = 001. We set
P1= 000, 001, P2= 000, 010, 011, 001, P3= 000, 100, 101, 001, and P4=
000, 110, 111, 001. Then {P1, P2, P3, P4} forms a 4∗-container of Q3,2 between
Now, we consider n≥ 4. Since Q00n,n−1∪ Q01n,n−1 is isomorphic to Qn−1, by Lemma2, there exists an (n− 1)∗-container{P1, P2, . . . , Pn−1} of Q00n,n−1∪ Q01n,n−1 joining x to y. Obviously, (x)c and (y)c are different parity. Note that (x)n−1 is an odd vertex and (y)n−1 is an even vertex. By Lemma 3, there exist two dis-joint paths S1and S2 of Q10n,n−1∪ Q11n,n−1 such that (1) S1 joins (x)n−1to (y)n−1, (2) S2 joins (x)c to (y)c, and (3) S1∪ S
2 spans Q10n,n−1∪ Q11n,n−1. We set Pn=
x, (x)n−1, S
1, (y)n−1,y and Pn+1= x, (x)c, S2, (y)c,y. Then {P1, P2, . . . , Pn+1} forms an (n+ 1)∗-container of Qn,mbetween x and y. See Fig.3(a) for illustration.
Case 3: y∈ Q11n,n−1. Suppose that n= 3. We have x = 000 and y = 111. We set P1= 000, 100, 101, 111, P2= 000, 010, 011, 111, P3= 000, 001, 111, and
P4= 000, 110, 111. Then {P1, P2, P3, P4} forms a 4∗-container of Q3,2between x and y.
Now, we consider n≥ 4. Since y is adjacent to (n − 2) even vertices in Q11n,n−1, we can choose an even vertex z∈ Q11n,n−1 which is a neighbor of y such that
(z)n= (x)c and (z)n= (x)n−1. Let v= (z)n. Obviously, v is an odd vertex. Since
Q00n,n−1∪ Q10n,n−1= Q0n,n−1is isomorphic to FQn−1, by Lemma5, there exists an
n∗-container{R1, R2, . . . , Rn} of Q00n,n−1∪ Q10n,n−1between x and v. Since v is adja-cent to n vertices in Q00n,n−1∪ Q10n,n−1, by relabeling, we can write Riasx, R i,ui,v for 1≤ i ≤ n − 3, write Rn−2asx, Rn −2, (y)n,v, write Rn−1asx, R n−1, (v)c,v, and write Rn asx, R n, (v)n−1,v. Let A = {(ui)n| 1 ≤ i ≤ n − 3}. Obviously, A is a set of (n− 3) odd vertices of Q11n,n−1. Since Q11n,n−1 is isomorphic to Qn−2, by Theorem1, there is a spanning (y, A∪ {z})-fan, {H1, H2, . . . , Hn−2} in Q11n,n−1such that (1) Hi joins (ui)n to y for 1≤ i ≤ n − 3 and (2) Hn−2 joins z to y. We set
Pi= x, Ri ,ui, (ui)n, Hi,y for 1 ≤ i ≤ n − 3 and Pn−2= x, Rn −2, (y)n,y. Suppose that (n−1) is an odd integer. We set Pn−1= x, R n−1, (v)c,v, z, Hn−2,y. Since Q01n,n−1is isomorphic to Qn−2, by Lemma3, there exist two disjoint paths S1 and S2of Q01n,n−1such that (1) S1joins ((v)n−1)nto (y)c, (2) S2joins (x)nto (y)n−1, and (3) S1∪ S2 spans Q01n,n−1. Let Pn= x, Rn , (v)n−1, ((v)n−1)n, S1, (y)c,y, and
Pn+1= x, (x)n, S2, (y)n−1,y. Then {P1, P2, . . . , Pn+1} forms an (n+1)∗-container of Qn,n−1between x and y. See Fig.3(b) for illustration.
Suppose that (n− 1) is an even integer. We set Pn−1 = x, R n, (v)n−1,v, z,
Hn−2,y. Suppose that (y)c= (x)n. By Lemma1, there exists a Hamiltonian path
Sof Q01n,n−1− {(x)n} joining ((v)c)n to (y)n−1. Set Pn= x, R
n−1, (v)c, ((v)c)n, S,
(y)n−1,y and Pn+1 = x, (x)n = (y)c,y. Then {P1, P2, . . . , Pn+1} forms an
(n+ 1)∗-container of Qn,n−1 between x and y. See Fig.3(c) for illustration. Thus, we assume that (y)c= (x)n. By Lemma 3, there exist two disjoint paths S1 and
S2 of Q01n,n−1 such that (1) S1 joins ((v)c)n to (y)c, (2) S2 joins (x)n to (y)n−1, and (3) S1∪ S2 spans Q01n,n−1. Let Pn = x, Rn −1, (v)c, ((v)c)n, S1, (y)c,y and
Pn+1= x, (x)n, S2, (y)n−1,y. Then {P1, P2, . . . , Pn+1} forms an (n+1)∗-container
Fig. 3 Illustration for Lemma7
Lemma 8 Let x be an even vertex and y be an odd vertex of Qn,m for any two positive integers n≥ m ≥ 2. Then there exists a k∗-container of Qn,mbetween x and
y for every 1≤ k ≤ n + 1.
Proof Since Q2,2is isomorphic to complete graph K4, this statement holds for n= 2. Suppose that n≥ 3.
Since Qnis a spanning subgraph of Qn,m, by Lemma2, there exists a k∗-container of Qn,mbetween x and y for every 1≤ k ≤ n. Thus, we only need to construct an
(n+ 1)∗-container of Qn,mbetween x and y. Without loss of generality, we assume that x∈ Q00n,m. We prove our claim by induction on t= n − m. The induction bases are t= 0 and 1. By Lemma5, our claim holds for t= 0. With Lemma7, our claim holds for t= 1. Consider t ≥ 2 and assume that our claim holds for (t − 1). We have the following cases:
Case 1: y∈ Q00n,m∪Q10n,m. Since Q00n,m∪Q10n,mis isomorphic to Qn−1,m, by induction, there exists an n∗-container of Q00n,m∪ Q10n,m between x and y. Since Q01n,m∪ Q11n,m is isomorphic to Qn−1,m, by induction, there is a Hamiltonian path of Q01n,m∪ Q11n,m joining (x)nto (y)n. Thus, by Lemma6, there exists an (n+ 1)∗-container of Qn,m between x and y.
Fig. 4 Illustration for Lemma8
Case 2: y∈ Q01n,m. Note that Q01n,m and Q10n,m are symmetric with respect to Qn,m and Q00n,m∪ Q01n,mis isomorphic to Qn−1,m. Similar to Case 1, there is an (n+ 1)∗ -container of Qn,mbetween x and y.
Case 3: y∈ Q11n,m. Since y is adjacent to (n− 1) vertices in Q11n,m, we can choose a neighbor z of y in Q11n,m such that z= (y)c and (z)n= (x)n−1. Let v= (z)n. Ob-viously, v is an odd vertex. Since Q00
n,m∪ Q10n,mis isomorphic to Qn−1,m, by induc-tion, there exists an n∗-container{R1, R2, . . . , Rn} of Q00n,m∪ Q10n,m joining x to v. Since v is adjacent to n vertices in Q00n,m∪ Q10n,m, by relabeling, we can write Ri as x, Ri ,ui,v for 1 ≤ i ≤ n − 2, write Rn−1 as x, R n−1, (y)n,v, and write Rn asx, R n, (v)n−1,v. Since Q11n,mis isomorphic to Qn−2,m, by induction, there ex-ists an (n− 1)∗-container{H1, H2, . . . , Hn−1} of Q11n,m joining z to y. Since y is adjacent to (n− 1) vertices in Q11n,m and (z, y)∈ E(Q11n,m), one of these paths is
z, y. Without loss of generality, we assume that Hi= z, (ui)n, Hi ,y for 1 ≤ i ≤
n− 2 and Hn−1= z, y. We set Pi = x, Ri ,ui, (ui)n, Hi ,y for 1 ≤ i ≤ n − 2,
Pn−1= x, Rn −1, (y)n,y, and Pn= x, Rn , (v)n−1,v, z, y. Since Q01n,m is isomor-phic to Qn−2,m, by induction, there exists a Hamiltonian path W in Q01n,mjoining (x)n to (y)n−1. We set Pn+1= x, (x)n, W, (y)n−1,y. Then {P1, P2, . . . , Pn+1} forms an
(n+ 1)∗-container of Qn,mbetween x and y. See Fig.4for illustration.
Lemma 9 Qn,n−1 is 1∗-connected and 2∗-connected if n is an odd integer with
n≥ 3.
Proof Since any 1∗-connected graph with more than 3 vertices is 2∗-connected. Thus, we only need to show Qn,n−1is 1∗-connected. Suppose that x and y are two distinct vertices of Qn,n−1. Without loss of generality, we assume that x∈ Q0n,n−1.
Suppose that y∈ Q0n,n−1. By Theorem 3, there exists a Hamiltonian path R=
x, v, R ,y in Q0
n,n−1joining x to y and there exists a Hamiltonian path H in Q 1 n,n−1 joining (x)n to (v)n. Set P = x, (x)n, H, (v)n,v, R ,y. Thus, P forms a Hamil-tonian path in Qn,n−1joining x to y. See Fig.5(a) for illustration.
Suppose that y∈ Q1n,n−1. Note that there are (2n−1− 1) vertices in Q0
n,n−1− {x} and 2n−1− 1 ≥ 3 for n ≥ 3. We can pick a vertex z in Q0n,n−1such that (z)n= y. By
Fig. 5 Illustration for Lemma9
Theorem3, there exists a Hamiltonian path R in Q0n,n−1joining x to z and there exists a Hamiltonian path H in Q1n,n−1joining (z)nto y. Set P= x, R, z, (z)n, H,y. Thus,
P forms a Hamiltonian path in Qn,n−1joining x to y. See Fig.5(b) for illustration.
Lemma 10 Q3,2is super spanning connected.
Proof Let x and y be any two different vertices of Q3,2. By Lemma 9, Q3,2 is 1∗-connected and 2∗-connected. Hence, we need to construct a 3∗-container and a 4∗-container between x and y. Without loss of generality, we assume that x= 000. By Lemma7, this statement holds if y is an odd vertex. Thus, we assume that y is an even vertex. We list all possible cases as follows:
y 3∗-container 4∗-container 110 000, 010, 110 000, 100, 110 000, 001, 011, 101, 111, 110 000, 010, 110 000, 100, 110 000, 001, 011, 101, 111, 110 000, 110 011 000, 010, 011 000, 100, 101, 001, 011 000, 110, 111, 011 000, 001, 011 000, 010, 011 000, 100, 101, 011 000, 110, 111, 011 101 000, 001, 011, 101 000, 010, 110, 111, 101 000, 100, 101 000, 001, 101 000, 010, 011, 101 000, 100, 101 000, 110, 111, 101 Lemma 11 Suppose that n≥ 3 is an odd integer. Let x and y be any two different
even vertices of Qn,n−1. Then there exists a k∗-container of Qn,n−1between x and y for every 1≤ k ≤ n + 1.
Proof By Lemma10, this statement holds for Q3,2. Thus, we assume that n≥ 5. By Lemma9, Qn,n−1is 1∗-connected and 2∗-connected. Thus, we need to construct a
k∗-container between x and y for every 3≤ k ≤ n + 1. Without loss of generality, we assume that x∈ Q00n,n−1.
Case 1: (y)n= 0. Since Q00n,n−1∪ Q10n,n−1 is isomorphic to FQn−1, by Theorem3, there exists a (k− 1)∗-container of Q00n,n−1∪ Q10n,n−1 between x and y for every 2≤ k − 1 ≤ n. By Lemma6, there is a k∗-container of Qn,n−1between x and y for every 3≤ k ≤ n + 1.
Case 2: (y)n= 1. Since y is an even vertex, |{i | i = n and (y)i = 1}| is odd. With-out loss of generality, we assume that (y)n−1= 1. Thus, y ∈ Q11n,n−1. We have the following cases:
Subcase 2.1: n≤ k ≤ n + 1. Since y is adjacent to (n − 2) vertices in Q11n,n−1, we can choose a neighbor z of y in Q11n,n−1 such that (z)n = (x)n−1. Let v=
(z)n. Obviously, v is an even vertex. By Theorem3, there exists an n∗-container
{R1, R2, . . . , Rn} of Q00n,n−1∪ Q10n,n−1between x and v. Since v is adjacent to n ver-tices in Q00n,m∪Q10n,m, by relabeling, we can write Riasx, R i,ui,v for 1 ≤ i ≤ n−3, write Rn−2 asx, Rn −2, (v)c,v, write Rn−1as x, Rn −1, (y)n,v, and write Rn as
x, R
n, (v)n−1,v. Let A = {(ui)n|1 ≤ i ≤ n − 3}. Obviously, A is a set of (n − 3) even vertices of Q11n,n−1.
Subcase 2.1.1: k= n + 1. Since Qn−2 is a spanning sbgraph of Q11n,n−1, by The-orem1, there is a spanning (y, A∪ {z})-fan, {H1, H2, . . . , Hn−2}, in Q11n,n−1 such that (1) Hi joins (ui)n to y for 1≤ i ≤ n − 3 and (2) Hn−2 joins z to y. We set
Pi= x, Ri ,ui, (ui)n, Hi,y for 1 ≤ i ≤ n − 3, Pn−2= x, R n−2, (v)c,v, z, Hn−2,y, and Pn−1= x, Rn −1, (y)n,y.
Suppose that (y)n−1= (x)n. Note that Qn−2 is a spanning subgraph of Q01n,n−1. By Lemma1, there exists a Hamiltonian path S of Q01n,n−1− {(x)n} joining ((v)n−1)n
to (y)c. We set Pn = x, R n, (v)n−1, ((v)n−1)n, S, (y)c,y and Pn+1= x, (x)n=
(y)n−1,y. Then {P1, P2, . . . , Pn+1} forms an (n + 1)∗-container of Qn,n−1between
x and y. See Fig.6(a) for illustration.
Now, we consider (y)n−1= (x)n. Since Qn
−2is a spanning subgraph of Q01n,n−1, by Lemma3, there exist two disjoint paths S1 and S2 of Q01n,n−1 such that (1) S1 joins ((v)n−1)n to (y)n−1, (2) S2joins (x)n to (y)c, and (3) S1∪ S
2 spans Q01n,n−1. Let Pn= x, R n, (v)n−1, ((v)n−1)n, S
1, (y)n−1,y and Pn+1= x, (x)n, S2, (y)c,y. Then{P1, P2, . . . , Pn+1} forms an (n + 1)∗-container of Qn,n−1 between x and y. See Fig.6(b) for illustration.
Subcase 2.1.2: k= n. Obviously, A ∪ {((v)n−1)n} is a set of (n − 2) even vertices
of Q01n,n−1∪ Q11n,n−1. Since Qn−1 is a spanning subgraph of Q01n,n−1∪ Q11n,n−1, by Theorem1, there is a spanning (y, A∪ {z, ((v)n−1)n})-fan, {H
1, H2, . . . , Hn−1}, in
Q01n,n−1∪ Q11n,n−1such that (1) Hi joins (ui)nto y for 1≤ i ≤ n − 3, (2) Hn−2joins z to y, and (3) Hn−1joins ((v)n−1)nto y. We set Pi= x, R
i,ui, (ui)n, Hi,y for 1 ≤
i≤ n−3, Pn−2= x, Rn −2, (v)c,v, z, Hn−2,y, Pn−1= x, R n−1, (y)n,y, and Pn=
x, R
n, (v)n−1, ((v)n−1)n, Hn−1,y. Then {P1, P2, . . . , Pn} forms an n∗-container of
Qn,n−1between x and y. See Fig.6(c) for illustration.
Subcase 2.2: 3≤ k ≤ n − 1. Let v be an even vertex of Q00n,n−1∪ Q10n,n−1such that
Fig. 6 Illustration for
Lemma11
by Theorem3, there exists a k∗-container{R1, R2, . . . , Rk} of Q00n,n−1∪ Q10n,n−1 be-tween x and v. We write Ri= x, Ri ,ui,v for 1 ≤ i ≤ k. Let A = {u1,u2, . . . ,uk}. Since k≥ 3, at most one vertex of A is an even vertex. Without loss of general-ity, we assume that{u1,u2, . . . ,uk−1} is a set of (k − 1) odd vertices. Obviously,
{(ui)n| 1 ≤ i ≤ k − 1} is a set of (k − 1) even vertices of Q01n,n−1∪ Q11n,n−1. Since
Qn−1is a spanning subgraph of Q01n,n−1∪ Q11n,n−1, by Theorem1, there is a spanning
(y,{(ui)n| 1 ≤ i ≤ k − 1} ∪ {(v)n})-fan, {H1, H2, . . . , Hk}, of Q01n,n−1∪ Q11n,n−1such that (1) Hi joins (ui)n to y for 1≤ i ≤ k − 1 and (2) Hk joins (v)n to y. We set
Pi = x, R i,ui, (ui)n, Hi,y for 1 ≤ i ≤ k − 1 and Pk = x, R k,uk,v, (v)n, Hk,y. Then{P1, P2, . . . , Pk} forms a k∗-container of Qn,n−1between x and y. See Fig.6(d)
for illustration.
Lemma 12 Suppose that n≥ 3 and m is even. Let x and y be any two different even
vertices of Qn,m. Then there exists a Hamiltonian path P of Qn,mbetween x and y. Proof For the fixed number m, we prove this statement by induction on t= n − m. Suppose that x and y be any two different even vertices of Qn,m. By Lemma10, this statement holds for t= 1. Consider t ≥ 2 and assume that our claim holds for (t − 1). Without loss of generality, we assume that x∈ Q0
n,m.
Suppose that y∈ Q0n,m. Since Q0n,mis isomorphic to Qn−1,m, by induction, there exists a Hamiltonian path R= x, v, R ,y in Q0n,mjoining x to y and there exists a Hamiltonian path H in Q1n,mjoining (x)nto (v)n. Set P = x, (x)n, H, (v)n,v, R ,y. Thus, P forms a Hamiltonian path in Qn,mjoining x to y.
Suppose that y∈ Q1n,m. We pick an even vertex z in Q0n,m such that z= x. By induction, there exists a Hamiltonian path R in Q0n,m joining x to z. Obvi-ously, (z)nis an odd vertex of Q1
n,m. Since Qn−1is a spanning subgraph of Q1n,m, by Lemma 2, there exists a Hamiltonian path H in Q1n,m joining (z)n to y. Set
P = x, R, z, (z)n, H,y. Thus, P forms a Hamiltonian path in Qn,mjoining x to y.
Lemma 13 Suppose that n≥ 3 and m is even. Let x and y be any two different even
vertices of Qn,m. Then there exists a k∗-container of Qn,mbetween x and y for every 1≤ k ≤ n + 1.
Proof By Lemma10, this statement holds for n= 3. Suppose that n ≥ 4. We claim that there exists a k∗-container between x and y for every 1≤ k ≤ n + 1. By Lemma12, this statement holds for k= 1. Note that Qn is a spanning subgraph of Qn,mand Qn is Hamiltonian. So, this statement holds for k= 2. We claim that there exists a k∗-container between x and y for every 3≤ k ≤ n + 1. Without loss of generality, we assume that x∈ Q00n,m. We prove our claim by induction on t= n − m. The induction bases are t= 0 and 1. By Theorem3, our claim holds for t= 0. With Lemma11, this statement holds for t= 1. Consider t ≥ 2 and assume that this state-ment holds for (t− 1). We have the following cases.
Case 1: y∈ Q00n,m∪Q10n,m. Since Q00n,m∪Q10n,mis isomorphic to Qn−1,m, by induction, there exists a (k− 1)∗-container{P1, P2, . . . , Pk−1} of Q00n,m∪ Q10n,mbetween x and
y for every 2≤ k − 1 ≤ n. By Lemma6, there exists a k∗-container of Qn,mbetween
x and y.
Case 2: y∈ Q01n,m. Note that Q01n,mand Q10n,mare symmetric with respect to Qn,mand
Q00n,m∪ Q01n,mis isomorphic to Qn−1,m. Similar to Case 1, there is a k∗-container of
Qn,mbetween x and y for every 3≤ k ≤ n + 1.
Case 3: y∈ Q11n,m.
Subcase 3.1: 3≤ k ≤ n. Let v be an even vertex of Q00n,m∪ Q10n,msuch that v= x and
(y)nis not a neighbor of v. By induction, there exists a k∗-container{R1, R2, . . . , Rk} of Q00n,m∪ Q10n,m between x and v. We write Ri = x, Ri ,ui,v for 1 ≤ i ≤ k. Let
A= {u1,u2, . . . ,uk}. Since k ≥ 3, at most one vertex of A is an even vertex. With-out loss of generality, we assume that {u1,u2, . . . ,uk−1} is a set of (k − 1) odd vertices. Obviously, {(ui)n| 1 ≤ i ≤ k − 1} is a set of (k − 1) even vertices of
Q01n,m∪ Q11n,m. By Theorem1, there is a spanning (y,{(ui)n| 1 ≤ i ≤ k − 1} ∪ {(v)n })-fan,{H1, H2, . . . , Hk}, of Qn,m01 ∪ Q11n,msuch that (1) Hi joins (ui)nto y for 1≤ i ≤
k−1 and (2) Hkjoins (v)nto y. We set Pi= x, R i,ui, (ui)n, Hi,y for 1 ≤ i ≤ k −1 and Pk= x, Rk ,uk,v, (v)n, Hk,y. Then {P1, P2, . . . , Pk} forms a k∗-container of
Qn,mbetween x and y.
Subcase 3.2: k= n + 1. Since y is adjacent to (n − 1) vertices in Q11n,m, we can choose a neighbor z of y in Q11n,msuch that z= (y)cand (z)n= (x)n−1. Let v= (z)n. Obviously, both v and ((v)n−1)n are even vertices. By induction, there exists an
n∗-container{R1, R2, . . . , Rn} of Q00n,m∪ Q10n,m between x and v. Since v is adja-cent to n vertices in Q00n,m∪ Q10n,m, by relabeling, we can write Ri asx, R i,ui,v for
