On the Spanning w -Wide Diameter of the Star Graph*
Cheng-Kuan Lin
Institute of Computer Science and Engineering, National Chiao Tung University, Hsinchu, Taiwan 30010, Republic of China
Hua-Min Huang
Department of Mathematics, National Central University, Chungli, Taiwan 32001, Republic of China
D. Frank Hsu
Department of Computer and Information Science, Fordham University, New York, NY 10023
Lih-Hsing Hsu
Department of Computer Science and Information Engineering, Providence University, Taichung, Taiwan 43301, Republic of China
Let u and v be any two distinct nodes of an undirected graph G, which is k -connected. A container C(u, v ) between u and v is a set of internally disjoint paths {P1, P2,. . . , Pw} between u and v where 1 ≤ w ≤ k . The width of C(u, v) is w and the length of C(u, v) {written as l[C(u, v)]} is max{l(Pi) | 1 ≤ i ≤ w }. A w -container C(u, v) is a container with width w . The w -wide distance between u and v , dw(u, v ), is min{l(C(u, v)) | C(u, v) is a w -container}. A w -container C(u, v) of the graph G is a w∗-container if every node of G is incident with a path in C(u, v). That means that the w -container C(u, v) spans the whole graph. Let Snbe the n-dimensional star graph with n ≥ 5. It is known that Sn is bipartite. In this article, we show that, for any pair of distinct nodes u and v in different partite sets of Sn, there exists an (n − 1)∗-container C(u, v) and the (n − 1)-wide distance d(n−1)(u, v ) is less than or equal to n!
n−2+ 1. In addition, we also show the existence of a 2∗-container C(u, v) and the 2-wide distance d2(u, v) is bounded above byn2!+ 1. ©2006 Wiley Periodicals, Inc. NETWORKS, Vol. 48(4), 235–249 2006
Keywords: diameter; hamiltonian; hamiltonian laceable; star graphs
Received May 2005; accepted July 2006
Correspondence to: L.-H. Hsu; e-mail: lhhsu@cis.nctu.edu.tw
*Preliminary results of this article were presented at the ISPAN’04 confer-ence and an extended abstract can be found in [20].
DOI 10.1002/net.20135
Published online in Wiley InterScience (www.interscience.wiley. com).
©2006 Wiley Periodicals, Inc.
1. BASIC DEFINITIONS
An interconnection network connects the processors of parallel computers. Its architecture can be represented as a graph, in which the nodes correspond to processors and the edges correspond to connections. Hence, we use graphs and networks interchangeably. There are many mutually conflict-ing requirements in designconflict-ing the topology for computer networks. The n-cube is one of the most popular topolo-gies [17]. The n-dimensional star network Snwas proposed in [1] as “an attractive alternative to the n-cube” topology for interconnecting processors in parallel computers. Since its introduction, the network Snhas received considerable atten-tion. Akers and Krishnamurthy [1] showed that the star graphs are node transitive and edge transitive. Jwo et al. [15] showed that the star graphs are bipartite. Star graphs are able to embed grids [15]: trees [3, 5, 8], and hypercubes [22]. Cycle embed-dings and path embedembed-dings are studied in [10–13, 15, 18, 23]. The diameter and fault diameters of star graphs were com-puted in [1, 16, 24]. Some interesting properties of star graphs are studied in [7, 9, 19].
For graph definitions and notation we follow [4]. 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 node set and E is the edge set. A graph G is vertex tran-sitive if there is an isomorphism f from G into itself such that f(u) = v for any two nodes u and v of G. A graph G is edge transitive if there is an isomorphism f from G into itself such that f((u, v)) = (x, y) for any two edges (u, v) and (x, y) of G. For a node u in graph G, NG(u) denotes the
neighbor-hood of u, which is the set{v | (u, v) ∈ E}. For any node u of V , we denote the degree of u by degG(u) = |NG(u)|.
A graph G is k-regular if degG(u) = k for all nodes u in G. Two nodes u and v are adjacent if(u, v) ∈ E. A path P is a sequence of adjacent nodes, written as v1, v2,. . . , vk, in which the nodes v1, v2,. . . , vkare distinct except that possibly
v1= vk. We use P−1to denote the pathvk, vk−1,. . . , v2, v1.
Let I(P) = V(P) − {v1, vk} be the set of the internal nodes of P. A set of paths {P1, P2,. . . , Pk} is internally
node-disjoint (abbreviated as node-disjoint) if I(Pi) ∩ I(Pj) = ∅ for any i = j. The length of a path Q, l(Q), is the number of edges in Q. We also write the path v1, v2,. . . , vk as v1, Q1, vi, vi+1,. . . , vj, Q2, vt,. . . , vk, where Q1is the path
v1, v2,. . . , vi and Q2is the pathvj, vj+1,. . . , vt. Hence, it is possible to write a path asv1, Q, v1, v2,. . . , vk if l(Q) = 0. We use d(u, v) to denote the distance between u and v, that is, the length of a shortest path joining u and v. The diameter of a graph G, D(G), is defined as max{d(u, v) | u, v ∈ V}. A path is a hamiltonian path if it contains all nodes of G. A graph G is hamiltonian connected if for any two dis-tinct nodes of G there is a hamiltonian path of G between them. A cycle is a path with at least three nodes such that the first node is the same as the last node. A hamilto-nian cycle of G is a cycle that traverses every node 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 nodes whose removal leaves the remaining graph discon-nected or trivial. It follows from Menger’s Theorem [21] that there are k internally node-disjoint paths joining any two distinct nodes u and v when k ≤ κ(G). A container C(u, v) between two distinct nodes u and v in G is a set of internally disjoint paths {P1, P2,. . . , Pr} between u and v. The width of C(u, v) is r. A w-container is a container of width w. The length of C(u, v) = {P1,. . . , Pr}, l(C(u, v)), is max{l(Pi) | 1 ≤ i ≤ r}. The w-wide distance between u and
v, dw(u, v), is min{l(C(u, v)) | C(u, v) is a w-container}. The
w-diameter of G, Dw(G), is max{dw(u, v) | u, v ∈ V, u = v}. In particular, the wide diameter of G is Dκ(G)(G). The wide diameter is used to measure the performance of multipath communication in networks [14].
In this article, we are interested in a specific type of con-tainer. We say that a w-container C(u, v) is a w∗-container if every node of G is incident with a path in C(u, v). A graph G is w∗-connected if there exists a w∗-container between any two distinct nodes u and v. Obviously, a graph G is 1∗-connected if and only if it is hamiltonian connected. Moreover, a graph G is 2∗-connected if it is hamiltonian. The study of w∗-connected graphs is motivated by the globally 3∗-connected graphs pro-posed by Albert et al. [2]. A globally 3∗-connected graph is a 3-regular 3∗-connected graph. Assume that a graph G is w∗-connected. Obviously, w ≤ κ(G). A graph G is super spanning connected if G is w∗-connected for any w with 1≤ w ≤ κ(G). In [19,26], some families of graphs are proved to be super spanning connected.
Graph containers do exist in engineering designed infor-mation and telecommunication networks or in biological and neural systems ([1, 14] and its references). The study of w-container, w-wide distance, and their w∗-versions
plays a pivotal role in design and implementation of parallel routing and efficient information transmission in large-scale networking systems. In biological informat-ics and neuroinformatinformat-ics, the existence and structure of a w∗-container signifies the cascade effect in the sig-nal transduction system and the reaction in a metabolic pathway.
A graph G is bipartite if its node set can be partitioned into two subsets V1and V2such that every edge joins nodes
between V1and V2. Let G be a bipartite graph with bipartition V1and V2such that|V1| ≥ |V2|. Suppose that there exists
a w∗-container C(u, v) = {P1, P2,. . . , Pw} in G joining u to
v with u, v ∈ V1. Obviously, the number of nodes in Pi is 2ti+ 1 for some integer ti. There are ti− 1 nodes of Piin V1
other than u and v, and tinodes of Piin V2. As a consequence,
|V1| =
w
i=1(ti− 1) + 2 and |V2| =
w
i=1ti. Therefore, any bipartite graph G withκ(G) ≥ 3 is not w∗-connected for any w, 3≤ w ≤ κ(G).
Let G be a w∗-laceable bipartite graph with bipartite node sets V1 and V2 and |V1∪ V2| ≥ 2. From the above
dis-cussion,|V1| = |V2|. For this reason, a bipartite graph is w∗-laceable if there exists a w∗-container between any two nodes from different partite sets for some w, 1≤ w ≤ κ(G). A 1∗-laceable graph is also known as a hamiltonian laceable graph [25]. Moreover, a graph G is 2∗-laceable if and only if it is hamiltonian. All 1∗-laceable graphs except K1and K2
are 2∗-laceable. A bipartite graph G is super spanning lace-able if G is i∗-laceable for all 1≤ i ≤ κ(G). Recently, Chang et al. [6] proved that the n-dimensional hypercube Qnis super spanning laceable for every positive integer n. It was proved in [19] that the n-dimensional star graph Snis super spanning laceable if and only if n= 3.
We also define the w∗-laceable distance between any two nodes u and v from different partite sets, dsL
w(u, v), as min{l(C(u, v)) | C(u, v) is a w∗-container}. The w∗L-diameter of G, denoted by DsL
w(G), is defined as max{dwsL(u, v) | u and v are nodes from different partite sets}. In particular, the spanning wide diameter of G is DsL
κ(G)(G). In this arti-cle, we evaluate the spanning wide diameter of Sn and the 2∗L-diameter of S
n.
FIG. 2. Illustration for Lemma 3.
In Section 2, we give the definition of star graphs and introduce some basic properties of star graphs. In Section 3, we prove some hamiltonian path properties of star graphs. Then we discuss the spanning wide diameter of Sn in Section 4. In Section 5, we discuss the 2∗L-spanning diameter of Sn.
2. STAR GRAPHS AND THEIR PROPERTIES
Assume that n ≥ 2. We use n to denote the set {1, 2, . . . , n}, where n is a positive integer. A permuta-tion on n is a sequence of n distinct elements ui∈ n,
u1u2. . . ui. . . un. An inversion of u1u2. . . ui. . . unis a pair
(i, j) such that ui < ujand i > j. An even permutation is a permutation with an even number of inversions, and an odd permutation is a permutation with an odd number of inver-sions. The n-dimensional star network, denoted by Sn, is a graph with the node set V(Sn) = {u1u2. . . un | ui ∈ n and ui = uj for i = j}. The edges are specified as follows:
u1u2. . . ui. . . unis adjacent to v1v2. . . vi. . . vnby an edge in dimension i with 2≤ i ≤ n if vj = uj for j /∈ {1, i}, v1= ui and vi = u1. By definition, Snis an(n−1)-regular graph with
n! nodes. Moreover, it is node transitive and edge transitive [1]. The star graphs S2, S3, and S4are shown in Figure 1 for
an illustration.
We use boldface to denote nodes in Sn. Hence, u1, u2,. . . ,
un denotes a sequence of nodes in Sn. We use e to denote the element 12. . . n. It is known that Snis a bipartite graph with one partite set containing those nodes corresponding to odd permutations and the other partite set containing those nodes corresponding to even permutations. We will use white nodes to represent those even permutation nodes and use black nodes to represent those odd permutation nodes. Let
u= u1u2. . . unbe any node of the star graph Sn. We say that
uiis the i-th coordinate of u, denoted by(u)i, for 1≤ i ≤ n. By the definition of Sn, there is exactly one neighbor v of u such that u and v are adjacent through an i-dimensional edge with 2 ≤ i ≤ n. For this reason, we use (u)i to denote the unique i-neighbor of u. Obviously,((u)i)i= u. For 1 ≤ i ≤ n, let Sn{i}denote the subgraph of Sninduced by those nodes u with(u)n= i. Obviously, Sncan be decomposed into n sub-graphs Sn{i}, 1≤ i ≤ n, and each Sn{i}is isomorphic to Sn−1. Thus, the star graph can be constructed recursively. Obvi-ously, u∈ S{(u)n}
n and(u)n∈ Sn{(u)1}. Let I ⊆ n. We use
SInto denote the subgraph of Sninduced by∪i∈IV
Sn{i}
. For 1≤ i = j ≤ n, we use Ei,jto denote the set of edges between S{i}n and S{j}n . For 1≤ i = j ≤ n, we use S{(i,j)}n to denote the subgraph of Sninduced by those nodes u with(u)n−1 = i and
(u)n= j. Obviously, S{(i,j)}n = Sn{(j,i)}and S{(i,j)}n is isomorphic to Sn−2.
Lemma 1 ([23]). Assume that n≥ 4. |Ei,j| = (n − 2)! for
any 1≤ i = j ≤ n. Moreover, there are (n−2)!2 edges joining black nodes of Sn{i}to white nodes of S{j}n .
Lemma 2 ([1]). Let u and v be any two distinct nodes of Snwith d(u, v) ≤ 2. Then (u)1= (v)1. Moreover,{((u)i)1|
2≤ i ≤ n − 1} = n − {(u)1,(u)n} if n ≥ 3.
3. HAMILTONIAN PATHS OF STAR GRAPHS Theorem 1 ([11]). Snis hamiltonian laceable if and only
if n= 3.
Theorem 2 ([18]). Suppose that w is any black node of Sn
with n≥ 4. Then there is a hamiltonian path P of Sn− {w}
between any two distinct white nodes u and v.
Lemma 3. Let n≥ 5 and I = {a1, a2,. . . , ar} be a subset
ofn for some r ∈ n. Assume that u is a white node in S{a1} n
and v is a black node in S{ar}
n . Then there is a hamiltonian
path P of SnIjoining u to v.
Proof. Let x1 = u and yr = v. Obviously, Sn{ai} is isomorphic to Sn−1 for every i ∈ r. By Theorem 1, this result holds on r = 1. Suppose that r ≥ 2. By Lemma 1, there are(n − 2)!/2 ≥ 3 edges joining black nodes of S{ai}
n to white nodes of S{ai+1}
n for every i ∈ r − 1. For every
i∈ r −1, we can choose a black node yi∈ Sn{ai}and a white node xi+1 ∈ S{ani+1} such that(yi, xi+1) ∈ Eai,ai+1. By Theo-rem 1, there is a hamiltonian path Hiof Sn{ai}joining xito yi for every i∈ r. Then x1, H1, y1, x2, H2, y2,. . . , xr, Hr, yr forms the desired hamiltonian path of SnI joining u to v. See
Figure 2 for an illustration. ■
Lemma 4. Assume that r and s are any two adjacent nodes of Snwith n ≥ 4. Then, for any white node u in Sn− {r, s}
and for any i ∈ n, there exists a hamiltonian path P of Sn − {r, s} joining u to some black node v of Sn− {r, s}
with(v)1= i.
Proof. Because Snis node transitive and edge transitive, we assume that r = e and s = (e)2. Obviously, both e and (e)2 are in S{n}
n . We prove this lemma by induction on n. Suppose that n= 4. The required hamiltonian paths of S4−
1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 4312, 2314, 3214, 4213, 2413, 1423, 3421, 2431, 4231, 3241, 1243 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 4312, 2314, 3214, 4213, 1243, 3241, 4231, 2431, 3421, 1423, 2413 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 3412, 2413, 1423, 3421 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 2431, 3421, 1423, 2413, 3412, 4312, 2314, 3214, 4213, 1243, 3241, 4231 1423, 3421, 4321, 2341, 1342, 4312, 3412, 2413, 4213, 3214, 2314, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243 1423, 3421, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413 1423, 3421, 2431, 4231, 3241, 1243, 4213, 2413, 3412, 1432, 4132, 3142, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 4312, 2314, 3214 1423, 3421, 2431, 4231, 3241, 1243, 2143, 3142, 4132, 1432, 3412, 2413, 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123 2143, 3142, 4132, 1432, 3412, 2413, 1423, 4123, 3124, 1324, 4321, 3421, 2431, 4231, 3241, 2341, 1342, 4312, 2314, 3214, 4213, 1243 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 3124, 4123, 1423, 3421, 4321, 2341, 1342, 4312, 3412, 2413 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 3124, 4123, 1423, 2413, 3412, 4312, 1342, 2341, 4321, 3421 2143, 3142, 4132, 1432, 2431, 3421, 4321, 2341, 1342, 4312, 3412, 2413, 1423, 4123, 3124, 1324, 2314, 3214, 4213, 1243, 3241, 4231 2314, 3214, 4213, 1243, 3241, 4231, 2431, 3421, 1423, 2413, 3412, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432 2314, 3214, 4213, 1243, 3241, 4231, 2431, 1432, 4132, 3142, 2143, 4123, 3124, 1324, 4321, 3421, 1423, 2413, 3412, 4312, 1342, 2341 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421, 2431, 4231, 3241, 1243, 4213, 3214 2314, 3214, 4213, 1243, 3241, 4231, 2431, 3421, 1423, 2413, 3412, 1432, 4132, 3142, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 4312 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 4321, 3421, 1423, 2413, 3412, 1432, 4132, 3142, 2143, 4123, 3124, 1324 2431, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 4312, 1342, 2341, 4321, 3421, 1423, 2413 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421 2431, 3421, 4321, 1324, 3124, 4123, 1423, 2413, 3412, 1432, 4132, 3142, 2143, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 3241, 4231 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 2341, 1342, 4312, 3412, 2413, 1423, 3421, 4321, 1324, 2314, 3214, 4213, 1243 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 4321, 3421, 1423, 2413, 3412, 4312, 1342, 2341 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 4321, 2341, 1342, 4312, 3412, 2413, 1423, 3421 3124, 4123, 2143, 3142, 4132, 1432, 2431, 3421, 1423, 2413, 3412, 4312, 1342, 2341, 4321, 1324, 2314, 3214, 4213, 1243, 3241, 4231 3241, 4231, 2431, 1432, 4132, 3142, 1342, 2341, 4321, 3421, 1423, 2413, 3412, 4312, 2314, 3214, 4213, 1243, 2143, 4123, 3124, 1324 3241, 4231, 2431, 1432, 4132, 3142, 2143, 1243, 4213, 3214, 2314, 1324, 3124, 4123, 1423, 3421, 4321, 2341, 1342, 4312, 3412, 2413 3241, 4231, 2431, 1432, 4132, 3142, 2143, 1243, 4213, 3214, 2314, 1324, 3124, 4123, 1423, 2413, 3412, 4312, 1342, 2341, 4321, 3421 3241, 1243, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 3142, 4132, 1432, 3412, 4312, 2314, 3214, 4213, 2413, 1423, 3421, 2431, 4231 3412, 2413, 1423, 3421, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432 3412, 2413, 1423, 3421, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341 3412, 2413, 1423, 3421, 4321, 1324, 3124, 4123, 2143, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 3241, 4231, 2431, 1432, 4132, 3142 3412, 2413, 1423, 3421, 2431, 1432, 4132, 3142, 1342, 4312, 2314, 3214, 4213, 1243, 2143, 4123, 3124, 1324, 4321, 2341, 3241, 4231 4132, 3142, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 4312, 2314, 3214, 4213, 1243, 3241, 4231, 2431, 3421, 1423, 2413, 3412, 1432 4132, 3142, 2143, 4123, 3124, 1324, 4321, 3421, 1423, 2413, 3412, 1432, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341 4132, 3142, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 4312, 2314, 3214, 4213, 1243, 3241, 4231, 2431, 1432, 3412, 2413, 1423, 3421 4132, 3142, 2143, 4123, 3124, 1324, 2314, 3214, 4213, 1243, 3241, 4231, 2431, 1432, 3412, 2413, 1423, 3421, 4321, 2341, 1342, 4312 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421, 2431, 4231, 3241, 1243 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 1423, 3421, 2431, 4231, 3241, 1243, 2143, 3142, 4132, 1432, 3412, 2413 4213, 3214, 2314, 4312, 1342, 2341, 4321, 1324, 3124, 4123, 2143, 1243, 3241, 4231, 2431, 3421, 1423, 2413, 3412, 1432, 4132, 3142 4213, 3214, 2314, 4312, 1342, 2341, 3241, 1243, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 4123, 3124, 1324, 4321, 3421, 2431, 4231 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421, 2431, 4231, 3241, 2341, 1342, 4312, 2314, 3214, 4213, 1243 4321, 1324, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421, 2431, 4231, 3241, 1243, 4213, 3214, 2314, 4312, 1342, 2341 4321, 1324, 3124, 4123, 2143, 1243, 4213, 3214, 2314, 4312, 1342, 2341, 3241, 4231, 2431, 3421, 1423, 2413, 3412, 1432, 4132, 3142 4321, 1324, 3124, 4123, 2143, 1243, 3241, 2341, 1342, 3142, 4132, 1432, 3412, 4312, 2314, 3214, 4213, 2413, 1423, 3421, 2431, 4231
Assume that this result holds in Skfor every 4≤ k ≤ n−1. We have the following cases:
Case 1. u ∈ Sn{n}. By induction, there is a hamiltonian path P of S{n}n − {e, (e)2} joining u to a black node x with
(x)1 = n − 1. Note that (x)nis a white node of S{n−1}n . We choose a black node v in Sn−2n with(v)1= i. By Lemma 3,
there is a hamiltonian path Q of Sn−1n joining (x)n to v.
Thenu, P, x, (x)n, Q, v forms the desired hamiltonian path of Sn− {e, (e)2} joining u to v with (v)1= i. See Figure 3(a)
for an illustration.
Case 2. u ∈ Sn{k}for some k∈ n − 1. By Lemma 1, there are(n − 2)!/2 ≥ 3 edges joining black nodes of Sn{k}to white nodes of Sn{n}. We can choose a white node x∈ Sn{n}−{e, (e)2} with(x)1= k. By Theorem 1, there is a hamiltonian path P
FIG. 4. Illustration for Theorem 3.
of Sn{k}joining u to the black node(x)n. By induction, there is a hamiltonian path Q of Sn{n} − {e, (e)2} joining x to a black node y with(y)1∈ n − 1 − {k}. We choose a black
node v in Sn−1−{k,(y)1}
n with(v)1 = i. By Lemma 3, there
exists a hamiltonian path R of Sn−1−{k}n joining the white node(y)nto v. Thenu, P, (x)n, x, Q, y,(y)n, R, v forms the desired hamiltonian path of Sn− {e, (e)2} joining u to v with
(v)1= i. See Figure 3(b) for an illustration. ■ Theorem 3. Let n≥ 5 and I = {a1, a2,. . . , ar} be a subset
ofn for some r ∈ n. Then SnIis hamiltonian laceable.
Proof. Let u be a white node and v be a black node of SIn. By Lemma 3, this theorem holds on either r = 1 or r ≥ 2 and (u)n = (v)n. Thus, we assume that r ≥ 2 and (u)n= (v)n. Without loss of generality, we assume that(u)n= (v)n= a1.
Case 1. (v)n∈ SI
n. Without loss of generality, we assume that(v)n∈ S{ar}
n . By Theorem 2, there is a hamiltonian path
P of S{a1}
n − {v} joining u to a white node x with (x)1 = a2. By Lemma 3, there is a hamiltonian path Q of SnI−{a1} joining the black node (x)n to the white node (v)n. Then u, P, x, (x)n, Q,(v)n, v forms the desired hamiltonian path of SnIjoining u to v. See Figure 4(a) for an illustration. Case 2. (u)n/∈ SI
n and(v)n/∈ SIn. We can choose a white node y with y being a neighbor of v in S{a1}
n and(y)1= ar.
Obviously, y= u. By Lemma 4, there is a hamiltonian path P of S{a1}
n − {v, y} joining u to a black node x with (x)1= a2.
By Lemma 3, there is a hamiltonian path Q of SI−{a1} n joining the white node (x)n to the black node (y)n. Then u, P, x, (x)n, Q,(y)n, y, v is the desired hamiltonian path of
SInjoining u to v. See Figure 4(b) for an illustration. ■
Theorem 4. Assume that r and s are two adjacent nodes of Snwith n≥ 5. Then Sn− {r, s} is hamiltonian laceable.
Proof. Because Snis node transitive and edge transitive, we assume that r = e and s = (e)2. Obviously, both e and (e)2 are in S{n}
n . Let u be a white node and v be a black node of Sn− {e, (e)2}. We want to find a hamiltonian path of
Sn− {e, (e)2} joining u to v.
Case 1. u, v ∈ Sn{n}. By Lemma 4, there is a hamiltonian path
P of S{n}n − {e, (e)2} joining u to a black node y with (y)1= 1.
We write P as u, Q1, x, v, Q2, y. (Note that l(Q1) = 0 if u = x and l(Q2) = 0 if v = y.) By Theorem 3, there is a
hamiltonian path R of Snn−1joining the black node(x)nto the white node(y)n. Thenu, Q
1, x,(x)n, R,(y)n, y,(Q2)−1, v is
the desired hamiltonian path of Sn− {e, (e)2} joining u to v. See Figure 5(a) for an illustration.
Case 2. u, v ∈ Sn{k} for some k ∈ n − 1. By Theo-rem 1, there is a hamiltonian path P of Sn{k}joining u to v. By Lemma 1, there are(n−2)!/2 ≥ 3 edges joining white nodes of S{k}n to black nodes of S{n}n . We can choose a white node x of Sn{k}, with(x)nbeing a black node of Sn{n}− {e, (e)2}. We write P asu, Q1, x, y, Q2, v. (Note that l(Q1) = 0 if u = x
and l(Q2) = 0 if v = y.) Because d(x, y) = 1, by Lemma 2, (y)1 ∈ n − 1 − {k}. By Lemma 4, there is a hamiltonian
path R of S{n}n − {e, (e)2} joining (x)nto a white node z with
(z)1∈ n − 1 − {k}. By Theorem 3, there is a hamiltonian
path T of Sn−1−{k}n joining the black node(z)nto the white node(y)n. Thenu, Q1, x,(x)n, R, z,(z)n, T ,(y)n, y, Q2, v is
the desired hamiltonian path of Sn− {e, (e)2} joining u to v. See Figure 5(b) for an illustration.
Case 3. u ∈ S{n}n and v∈ S{k}n for some k ∈ n − 1. By Lemma 4, there is a hamiltonian path P of S{n}n − {e, (e)2} joining u to a black node x with(x)1∈ n−1. By Theorem 3,
there is a hamiltonian path Q of Sn−1n joining the white node
(x)nto v. Thenu, P, x, (x)n, Q, v is the desired hamiltonian path of Sn− {e, (e)2} joining u to v. See Figure 5(c) for an illustration.
Case 4. u ∈ Sn{k}and v∈ Sn{l}with k, l, and n being distinct. By Lemma 1, there are(n − 2)!/2 ≥ 3 edges joining black nodes of S{k}n to white nodes of Sn{n}. We choose a black node
x of Sn{k}with(x)nbeing a white node of S{n}n − {e, (e)2}. By Theorem 1, there is a hamiltonian path P of Sn{k}joining u to x. By Lemma 4, there is a hamiltonian path Q of Sn{n}− {e, (e)2} joining(x)nto a black node y with(y)
1∈ n − 1 − {k}. By
Theorem 3, there is a hamiltonian path R of Sn−1−{k}n joining
the white node(y)nto v. Thenu, P, x, (x)n, Q, y,(y)n, R, v is the desired hamiltonian path of Sn−{e, (e)2} joining u to v.
See Figure 5(d) for an illustration. ■
Lemma 5. Assume that n≥ 5. Suppose that p and q are two different white nodes of Sn, and r and s are two different
black nodes of Sn. Then there exist two disjoint paths P1and P2such that (1) P1joins p to r, (2) P2joins q to s, and (3) P1∪ P2spans Sn.
Proof. Because Sn is edge transitive, we assume that
p∈ S{n}n and q∈ S{n−1}n . Suppose that r∈ S{i}n and s∈ S{j}n . Case 1. i, j ∈ n − 2 with i = j. By Theorem 3, there is a hamiltonian path P1of Sn{i,n}joining p to r. Again, there is a hamiltonian path P2of Snn−1−{i}joining q to s. Then P1and P2are the desired paths.
Case 2. i, j ∈ n−2 with i = j. We can choose a white node
x with x being a neighbor of s in S{i}n and(x)1∈ n − 1 − {i}.
By Lemma 4, there is a hamiltonian path Q of Sn{i}− {s, x} joining r to a white node y with(y)1= n. By Theorem 3, there
is a hamiltonian path P of Sn{n}joining p to the black node(y)n. Moreover, there is a hamiltonian path R of Snn−1−{i}joining
q to the black node(x)n. Then P1 = p, P, (y)n, y, Q−1, r
and P2= q, R, (x)n, x, s are the desired paths.
Case 3. Either (i = n and j ∈ n −1), or (i ∈ n−{n −1} and j = n − 1). By symmetry, we assume that i = n and j∈ n − 1. By Theorem 3, there is a hamiltonian path P1of S{n}n joining p to r. Moreover, there is a hamiltonian path P2
of Snn−1joining q to s. Then P1and P2are the desired paths.
Case 4. Either (i = n − 1 and j ∈ n − 2), or (i ∈ n − 2 and j = n). By symmetry, we assume that i = n − 1 and j ∈ n − 2. By Lemma 1, there exist (n − 2)!/2 ≥ 3 edges joining white nodes of Sn{n−1} to black nodes of Sn{n}. We can choose a white node x in S{n−1}n − {q} with (x)1= n. By Theorem 3, there is a hamiltonian path R of Sn{n−1} joining q to r. We write R asq, R1, y, x, R2, r. By Theorem 3,
there is a hamiltonian path P of Sn{n}joining p to the black node
(x)n. Because d(x, y) = 1, by Lemma 2, (y)n∈ Sn−2
n . By
Theorem 3, there exists a hamiltonian path Q of Snn−2joining the white node(y)nto s. Then P1= p, P, (x)n, x, R2, r and P2= q, R1, y,(y)n, Q, s are the desired paths.
Case 5. i = n−1 and j = n. By Theorem 3, there is a hamil-tonian path Q of Sn{n}joining p to s. Again, there is a hamilto-nian path R of S{n−1}n joining q to r. We choose a white node
x∈ S{n}n with(x)1= n − 1. We write Q as p, Q1, x, y, Q2, s
and write R as q, R1, w,(x)n, R2, r. Obviously, y is a
black node and w is a white node. Because d(x, y) = 1, by Lemma 2, (y)1∈ n − 2. Because d((x)n, w) = 1, by
Lemma 2, (w)1∈ n − 2. By Theorem 3, there exists a
hamiltonian path W of Snn−2 joining the black node(w)n to the white node(y)n. Then P1= p, Q1, x,(x)n, R2, r and P2= q, R1, w,(w)n, W ,(y)n, y, Q2, s are the desired paths.
Case 6. Either i = j = n or i = j = n − 1. By sym-metry, we assume that i = j = n. By Theorem 3, there is a hamiltonian path P of Sn{n} joining p to s. We can write
P asp, R1, r, x, R2, s. By Theorem 3, there is a
hamilto-nian path Q of Snn−1joining q to the black node(x)n. Then
P1= p, R1, r and P2= q, Q, (x)n, x, R2, s are the desired
paths. ■
4. THE(n − 1)∗L-DIAMETER OF Sn
Let u be a node of Snwith n≥ 4 and let m be any integer with 3 ≤ m ≤ n. We set Fm(u) = {(u)i | 3 ≤ i ≤ m} ∪ {((u)i)i−1| 3 ≤ i ≤ m}.
Lemma 6. Assume that u is a white node of Snand j∈ n
with n≥ 4. Then there is a hamiltonian path P of Sn− Fn(u)
joining u to some black node v with(v)1= j.
Proof. We prove this lemma by induction on n. Because Sn is node transitive, we assume that u = e. Suppose that
n = 4. The required hamiltonian paths of S4− F4(e) are
listed below:
j= 1 1234, 2134, 3124, 4123, 1423, 3421, 2431, 1432, 4132, 3142, 2143, 1243, 4213, 2413, 3412, 4312, 1342, 2341, 4321, 1324 j= 2 1234, 2134, 4132, 3142, 1342, 4312, 3412, 1432, 2431, 3421, 1423, 2413, 4213, 1243, 2143, 4123, 3124, 1324, 4321, 2341 j= 3 1234, 2134, 4132, 1432, 2431, 3421, 1423, 4123, 3124, 1324, 4321, 2341, 1342, 4312, 3412, 2413, 4213, 1243, 2143, 3142 j= 4 1234, 2134, 3124, 1324, 4321, 2341, 1342, 3142, 4132, 1432, 2431, 3421, 1423, 4123, 2143, 1243, 4213, 2413, 3412, 4312
Assume that this statement holds on any Sk for every 4≤ k ≤ n−1. We have Fn(e) = Fn−1(e)∪{(e)n,((e)n)n−1}. By induction, there is a hamiltonian path P of S{n}n − Fn−1(e) joining e to a black node x with(x)1 = 1. By Lemma 4,
there is a hamiltonian path Q of S{1}n − {(e)n,((e)n)n−1} joining the white node (x)n to a black node y with (y)1 = 2. We can choose a black node z of Sn−1−{1}n with (z)1 = j. By Theorem 3, there exists a hamiltonian
path R of Sn−1−{1}n joining the white node (y)n to z.
Thene, P, x, (x)n, Q, y,(y)n, R, z is a desired hamiltonian
path. ■
Lemma 7. Let u = u1u2u3u4 be any white node of S4. There exist three paths P1, P2, and P3such that (1) P1joins u to the black node u2u4u1u3with l(P1) = 7, (2) P2joins u to the white node u3u4u1u2with l(P2) = 8, (3) P3joins u to the white node u4u1u3u2with l(P3) = 8, and (4) P1∪P2∪P3 spans S4.
Proof. Because S4 is node transitive, we assume that u= 1234. Then we set P1= 1234, 3214, 4213, 1243, 2143, 4123, 1423, 2413, P2= 1234, 4231, 3241, 2341, 4321, 3421, 2431, 1432, 3412, and P3= 1234, 2134, 3124, 1324, 2314, 4312, 1342, 3142, 4132.
Obviously, P1, P2, and P3are the desired paths. ■
Lemma 8. Let u= u1u2u3u4be any white node of S4. Let i1i2i3be a permutation of u2, u3, and u4. There exist four
paths P1, P2, P3, and P4of S4such that (1) P1joins u to a white node w with(w)1= i1and l(P1) = 2, (2) P2joins u to a white node x with(x)1= i2and l(P2) = 2, (3) P3joins u to a black node y with(y)1= i3and l(P3) = 19, (4) P4joins u to a black node z with z= y, (z)1 = i3, and l(P4) = 19, (5) P1∪ P2∪ P3spans S4, (6) P1∪ P2∪ P4spans S4, (7) V(P1) ∩ V(P2) ∩ V(P3) = {u}, and (8) V(P1) ∩ V(P2) ∩ V(P4) = {u}.
Proof. Because S4 is node transitive, we assume that u = 1234. Because u = 1234, we have {i1, i2} ⊂ {2, 3, 4}
and i3 ∈ {2, 3, 4} − {i1, i2}. Without loss of generality, we
suppose that i1 < i2. The required four paths are listed
below. i1= 2 P1= 1234, 4231, 2431 i2= 3 P2= 1234, 2134, 3124 i3= 4 P3= 1234, 3214, 2314, 1324, 4321, 3421, 1423, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 4213, 1243, 3241, 2341, 1342, 4312 P4= 1234, 3214, 2314, 1324, 4321, 3421, 1423, 2413, 4213, 1243, 3241, 2341, 1342, 4312, 3412, 1432, 4132, 3142, 2143, 4123 i1= 2 P1= 1234, 4231, 2431 i2= 4 P2= 1234, 3214, 4213 i3= 3 P3= 1234, 2134, 3124, 4123, 2143, 1243, 3214, 2314, 1342, 4312, 2314, 1324, 4321, 3421, 1423, 2413, 3412, 1432, 4132, 3142 P4= 1234, 2134, 3142, 1324, 2314, 4312, 1342, 3142, 4132, 1432, 3412, 2413, 1423, 4123, 2143, 1243, 3241, 2341, 4321, 3421 i1= 3 P1= 1234, 2134, 3124 i2= 4 P2= 1234, 3214, 4213 i3= 2 P3= 1234, 4231, 3241, 1243, 2143, 4123, 1423, 2413, 3412, 4312, 2314, 1324, 4321, 3421, 2431, 1432, 4132, 3142, 1342, 2341 P4= 1234, 4231, 3241, 1243, 2143, 4123, 1423, 3421, 2431, 1432, 4132, 3142, 1342, 2341, 4321, 1324, 2314, 4312, 3412, 2413
Thus, this statement is proved. ■
Lemma 9. Assume that n≥ 5 and i1i2. . . in−1is an
(n−1)-permutation onn. Let u be any white node of Sn. Then there
exist(n−1) paths P1, P2,. . . , Pn−1of Snsuch that (1) P1joins u to a black node y1with(y1)1= i1and l(P1) = n(n−2)!−1, (2) Pjjoins u to a white node yjwith(yj)1= ijand l(Pj) =
n(n − 2)! for every 2 ≤ j ≤ n − 1, (3) ∪n−1
j=1Pjspans Sn, and
(4)∩nj=1−1V(Pj) = {u}.
Proof. The proof of this lemma is rather tedious. The authors strongly suggest the reader skim over the proof first and comprehend the details later.
Because Sn is node transitive, we assume that u = e. Without loss of generality, we suppose that i2< i3< · · · < in−1.
Case 1. n = 5. Hence, n(n − 2)! = 30. We have i2= 4, i3≥
2, and i4 ≥ 3. We set x1 = (e)5and xi = ((xi−1)i)5 for every 2≤ i ≤ 4, and x5= ((x4)3)5. Note that xi is a black node in S{i}n for every i ∈ 4 and x5is a black node in Sn{1}. Obviously, x1= x5. We set H = e, x1,(x1)2, x2,(x2)3, x3, (x3)4, x
4,(x4)3, x5.
Case 1.1. i1= 3. We have i2= 4, i3= 3, and i4= 1. Let u1= 24135, u2= 41325, and u3= 34125. We set W1= e = 12345, 32145, 42135, 12435, 21435, 41235, 14235, 24135= u1, W2= e = 12345, 21345, 31245, 13245, 23145, 43125, 13425, 31425, 41325= u2, and W3= e = 12345, 42315, 32415, 23415, 43215, 34215, 24315, 14325, 34125= u3. Obviously, W1∪ W2∪ W3spans S5{5}and V(Wi) ∩ V(Wj) = {e} for every i, j ∈ 3 with i = j. By Lemma 4, there exists a hamiltonian path Q1 of S5{2}− {x2,(x2)3} joining the white node (u1)5 to a black node y1 with (y1)1 = i1. Again, there exists a hamiltonian path Q2 of S5{4}− {x4,(x4)3} joining the black node (u2)5 to a white node y2 with (y2)1 = i2. Moreover, there exists a
hamil-tonian path Q3of S5{3}− {x3,(x3)4} joining the black node (u3)5to a white node y
3 with(y3)1 = i3. Similarly, there
exists a hamiltonian path Q4 of S{1}5 − {x1,(x1)2} joining the black node x5 to a white node y4 with (y4)1 = i4.
We set
P1= e, W1, u1,(u1)5, Q1, y1, P2= e, W2, u2,(u2)5, Q2, y2, P3= e, W3, u3,(u3)5, Q3, y3, and P4= e, H, x5, Q4, y4.
FIG. 6. Illustration of case 1.
Obviously, l(P1) = 29 and l(Pi) = 30 for every 2 ≤ i ≤ 4. Apparently, P1, P2, P3, and P4 are the desired paths. See
Figure 6(a) for an illustration.
Case 1.2. i1= 3. We have i2= 4, i3= 1, and i4= 2. Let u1= 31425, u2= 42135, and u3= 21435. We set W1= e = 12345, 21345, 41325, 14325, 34125, 43125, 13425, 31425= u1, W2= e = 12345, 32145, 23145, 13245, 31245, 41235, 14235, 24135, 42135= u2, and W3= e = 12345, 42315, 24315, 34215, 43215, 23415, 32415, 12435, 21435= u3. Obviously, W1∪W2∪W3spans S{5}5 and V(Wi)∩V(Wj) = {e} for every i, j∈ 3 with i = j. By Lemma 4, there exists a hamiltonian path Q1 of S5{3}− {x3,(x3)4} joining the white node (u1)5 to a black node y1 with (y1)1 = i1. Again,
there exists a hamiltonian path Q2 of S5{4}− {x4,(x4)3} joining the black node (u2)5 to a white node y2 with (y2)1 = i2. Moreover, there exists a hamiltonian path Q3of S{1}5 − {x1,(x1)2} joining the black node x5to a white node y3 with(y3)1= i3. Similarly, there exists a hamiltonian path Q4
of S5{2}− {x2,(x2)3} joining the black node (u3)5to a white node y4with(y4)1= i4. We set
P1= e, W1, u1,(u1)5, Q1, y1, P2= e, W2, u2,(u2)5, Q2, y2, P3= e, H, x5, Q3, y3, and P4= e, W3, u3,(u3)5, Q4, y4.
Obviously, l(P1) = 29 and l(Pi) = 30 for every 2 ≤ i ≤ 4. Apparently, P1, P2, P3, and P4 are the desired paths. See
Figure 6(b) for an illustration.
Case 2. n ≥ 6. Because n − 1 ≥ 5, we have ik = k + 2 for every 2 ≤ k ≤ n − 4, in−3 = 1, in−2 = 2, and
in−1= 3. We set uj= (e)j+2and vj= ((e)j+2)j+1for every j∈ n − 4. Thus, uj is a black node in S{(n−1,n)}n and vj is a white node in Sn{(n−1,n)} for every j ∈ n − 4. Note that
Fn−2(e) = {uj| j ∈ n − 4} ∪ {vj| j ∈ n − 4}.
By Lemma 6, there is a hamiltonian path P of S{(n−1,n)}n −
Fn−2(e) joining e to a black node x1 with (x1)1= 2. We
recursively set xj as the unique neighbor of (xj−1)n−1 in Sn{(j,n)} with(xj)1 = j + 1 for every 2 ≤ j ≤ n − 4, and
we set xn−3as the unique neighbor of(xn−4)n−1in Sn{(n−3,n)} with(xn−3)1 = n − 1. It is easy to see that xj is a black node for 1 ≤ j ≤ n − 3 and {(xj)n−1, x
j+1} ⊂ Sn{(j+1,n)}for 1≤ j ≤ n − 4. We construct Pj for every 1≤ j ≤ n − 1 as follows:
1. j∈ n − 4 − {1}. By Lemma 4, there is a hamiltonian path Tj of Sn{(j+1,n)}− {(xj)n−1, xj+1} joining the black node (vj)n−1 to a white node zj with(zj)1 = j + 2.
Again, there is a hamiltonian path Tj of Sn{j+2}joining the black node(zj)nto a white node yjwith(yj)1= ij. Then we set Pjase, uj, vj,(vj)n−1, Tj, zj,(zj)n, Tj , yj. Obviously, l(Pj) = n(n − 2)!.
2. j = n − 3. We choose a white node yn−3 in Sn{1} with(yn−3)1 = in−3. Note that there are((n − 3)!/2) edges joining some black nodes of Sn{(n−2,n)} to some white nodes of S{1}n and there are((n − 3)!/2) edges joining some white nodes of Sn{(n−2,n)} to some black nodes of S{1}n . We choose a white node r in Sn{1} with (r)nbeing a black node in S{(n−2,n)}
n and choose a black node s in S{1}n with(s)nbeing a white node in S{(n−2,n)}n . By Lemma 5, there exist two disjoint paths H1 and
H2 of S{1}n such that (1) H1 joins (e)n to r, (2) H2
joins s to yn−3, and (3) H1∪ H2 spans S{1}n . By The-orem 3, there is a hamiltonian path H of Sn{(n−2,n)} joining the black node (r)n to the white node (s)n. We set Pn−3 ase, (e)n, H1, r,(r)n, H,(s)n, s, H2, yn−3.
Obviously, l(Pn−3) = n(n − 2)!.
3. j= n − 1. By Lemma 4, there is a hamiltonian path Q1
of Sn{(2,n)}− {(x1)n−1, x2} joining the black node (v1)n−1 to a white node q with(q)1= 3. Again, there is a
hamil-tonian path Q2 of Sn{3} joining the black node(q)n to a white node yn−1 with(yn−1)1 = in−1. We set Pn−1 as e, u1, v1,(v1)n−1, Q1, q,(q)n, Q2, yn−1. Obviously,
l(Pn−1) = n(n − 2)!.
4. We construct P1and Pn−2 dependent on whether i1 =
n− 1 or not. We set L as x1,(x1)n−1, x2,(x2)n−1,. . . , xn−4,(xn−4)n−1, xn−3,(xn−3)n. By Theorem 1, there is a hamiltonian path W of Sn{(1,n)}joining the black node (e)n−1to a white node p with(p)
1= 2.
Suppose that i1= n − 1. By Theorem 1, there is a
hamiltonian path R of Sn{n−1} joining the white node
(xn−3)n to a black node y1 with (y1)1= i1. Again, there
exists a hamiltonian path Z of S{2}n−1 joining the black node (p)n to a white node yn−2 with (yn−2)1 = in−2. We set P1 as e, P, x1, L,(xn−3)n, R, y1 and Pn−2 as e, (e)n−1, W , p,(p)n, Z, y
n−2. Obviously, l(P1) = n(n −
2)!−1 and l(Pn−2) = n(n−2)!. Apparently, P1, P2,. . . , Pn−1 are the desired paths. See Figure 7(a) for an illustration for the case n= 7.
FIG. 7. Illustration of case 3 with n= 7.
Suppose that i1 = n − 1. Note that in−2 = n − 1. Because(xn−3)nis a white node in S{n−1}n with((xn−3)n)1= n and ((xn−3)n)n = n − 1, there is a black node z in S{n−1}n such that z is the unique neighbor of (xn−3)n with (z)1= 2. Because ((xn−3)n)n−1= n − 3, we have
(z)n−1 = n − 3. Note that (z)n is a white node in S{2}n with ((z)n)n−1 = n − 3. Because (p)n is a black node in S{2}n with ((p)n)1 = n and ((p)n)n = 2, there is a white node t in S{2}n such that t is the unique neighbor of (p)n with (t)1 = n − 1. Because (p)n−1 = 1 and
(p)n= n, we have ((p)n)n−1 = 1 and ((p)n)1= n. Because
((p)n)
n−1= 1, we have (t)n−1= 1. Because ((z)n)n−1= n − 3 and (t)n−1= 1, we have (z)n= t. By Theorem 4, there is a hamiltonian path W1 of Sn{2}− {(p)n, t} join-ing (z)n to a black node y1n with (y1)1= i1. Again,
there is a hamiltonian path W2 of Sn{n−1}− {(xn−3)n, z} joining (t)n to a white node yn−2 with (yn−2)1= in−2. We set P1 as e, P, x1, L,(xn−3)n, z,(z)n, W1, y1 and Pn−2 as e, (e)n−1, W , p,(p)n, t,(t)n, W2, yn−2. Obviously, l(P1) = n(n − 2)! − 1 and l(Pn−2) = n(n − 2)!. Apparently,
P1, P2,. . . , Pn−1are the desired paths. See Figure 7(b) for an
illustration for the case n= 7. ■
TABLE 1. All Hamiltonian Cycles in S4.
1234, 4231, 3241, 1243, 4213, 3214, 2314, 1324, 4321, 2341, 1342, 4312, 3412, 2413, 1423, 3421, 2431, 1432, 4132, 3142, 2143, 4123, 3124, 2134, 1234 1234, 4231, 3241, 1243, 4213, 2413, 3412, 1432, 2431, 3421, 1423, 4123, 2143, 3142, 4132, 2134, 3124, 1324, 4321, 2341, 1342, 4312, 2314, 3214, 1234 1234, 4231, 3241, 1243, 2143, 4123, 3124, 2134, 4132, 3142, 1342, 2341, 4321, 1324, 2314, 4312, 3412, 1432, 2431, 3421, 1423, 2413, 4213, 3214, 1234 1234, 4231, 3241, 2341, 1342, 4312, 2314, 1324, 4321, 3421, 2431, 1432, 3412, 2413, 1423, 4123, 3124, 2134, 4132, 3142, 2143, 1243, 4213, 3214, 1234 1234, 4231, 3241, 2341, 1342, 3142, 2143, 1243, 4213, 3214, 2314, 4312, 3412, 2413, 1423, 4123, 3124, 1324, 4321, 3421, 2431, 1432, 4132, 2134, 1234 1234, 4231, 3241, 2341, 4321, 3421, 2431, 1432, 4132, 3142, 1342, 4312, 3412, 2413, 1423, 4123, 2143, 1243, 4213, 3214, 2314, 1324, 3124, 2134, 1234 1234, 4231, 2431, 1432, 3412, 4312, 2314, 3214, 4213, 2413, 1423, 3421, 4321, 1324, 3124, 4123, 2143, 1243, 3241, 2341, 1342, 3142, 4132, 2134, 1234 1234, 4231, 2431, 1432, 4132, 2134, 3124, 1324, 4321, 3421, 1423, 4123, 2143, 3142, 1342, 2341, 3241, 1243, 4213, 2413, 3412, 4312, 2314, 3214, 1234 1234, 4231, 2431, 1432, 4132, 3142, 2143, 1243, 3241, 2341, 1342, 4312, 3412, 2413, 4213, 3214, 2314, 1324, 4321, 3421, 1423, 4123, 3124, 2134, 1234 1234, 4231, 2431, 3421, 1423, 2413, 3412, 1432, 4132, 2134, 3124, 4123, 2143, 3142, 1342, 4312, 2314, 1324, 4321, 2341, 3241, 1243, 4213, 3214, 1234 1234, 4231, 2431, 3421, 1423, 4123, 3124, 1324, 4321, 2341, 3241, 1243, 2143, 3142, 1342, 4312, 2314, 3214, 4213, 2413, 3412, 1432, 4132, 2134, 1234 1234, 4231, 2431, 3421, 4321, 2341, 3241, 1243, 4213, 2413, 1423, 4123, 2143, 3142, 1342, 4312, 3412, 1432, 4132, 2134, 3124, 1324, 2314, 3214, 1234 1234, 3214, 4213, 1243, 3241, 4231, 2431, 1432, 3412, 2413, 1423, 3421, 4321, 2341, 1342, 4312, 2314, 1324, 3124, 4123, 2143, 3142, 4132, 2134, 1234 1234, 3214, 4213, 2413, 3412, 4312, 2314, 1324, 3124, 4123, 1423, 3421, 4321, 2341, 1342, 3142, 2143, 1243, 3241, 4231, 2431, 1432, 4132, 2134, 1234 1234, 3214, 4213, 2413, 1423, 4123, 2143, 1243, 3241, 4231, 2431, 3421, 4321, 2341, 1342, 3142, 4132, 1432, 3412, 4312, 2314, 1324, 3124, 2134, 1234 1234, 3214, 2314, 4312, 1342, 3142, 4132, 1432, 3412, 2413, 4213, 1243, 2143, 4123, 1423, 3421, 2431, 4231, 3241, 2341, 4321, 1324, 3124, 2134, 1234 1234, 3214, 2314, 1324, 4321, 3421, 2431, 4231, 3241, 2341, 1342, 4312, 3412, 1432, 4132, 3142, 2143, 1243, 4213, 2413, 1423, 4123, 3124, 2134, 1234 1234, 3214, 2314, 1324, 3124, 4123, 2143, 1243, 4213, 2413, 1423, 3421, 4321, 2341, 3241, 4231, 2431, 1432, 3412, 4312, 1342, 3142, 4132, 2134, 1234
Using depth first search, we list all hamiltonian cycles in S4in Table 1.
Lemma 10. DsL
3(S4) = 15.
Proof. Let u be any white node in S4, and let v be any
black node in S4. Because S4is node transitive, we assume
that u = 1234. Suppose that d(u, v) = 1. Because S4 is
edge transitive, we assume that v= 2134. Let {P1, P2, P3} be
a 3∗-container joining u to v. Because S4is 3-regular, one of
three paths, say P3, isu, v. Thus, P1∪ P−12 forms a
hamil-tonian cycle of S4not using the edge (u, v). From Table 1,
we obtain dsL
3(u, v) = 15. Thus, D
sL
3(S4) ≥ 15. Suppose that d(u, v) = 1. Then v ∈ {1324, 1243, 1432, 2413, 2341, 3142, 4123, 4312, 3421}. We find the following set of 3∗-containers
of S4between u= 1234 and v: P1= 1234, 4231, 3241, 1243, 2143, 4123, 3124, 1324 C1234, 1324 P2= 1234, 2134, 4132, 3142, 1342, 2341, 4321, 1324 P3= 1234, 3214, 4213, 2413, 1423, 3421, 2431, 1432, 3412, 4312, 2314, 1324 P1= 1234, 2134, 4132, 1432, 3412, 2413, 4213, 1243 C(1234, 1243) P2= 1234, 3214, 2314, 4312, 1342, 3142, 2143, 1243 P3= 1234, 4231, 2431, 3421, 1423, 4123, 3124, 1324, 4321, 2341, 3241, 1243 P1= 1234, 3214, 2314, 1324, 4321, 3421, 2431, 1432 C(1234, 1432) P2= 1234, 4231, 3241, 2341, 1342, 4312, 3412, 1432 P3= 1234, 2134, 3124, 4123, 1423, 2413, 4213, 1243, 2143, 3142, 4132, 1432 P1= 1234, 3214, 4213, 2413 C(1234, 2413) P2= 1234, 4231, 2431, 3421, 4321, 2341, 3241, 1243, 2143, 4123, 1423, 2413 P3= 1234, 2134, 3124, 1324, 2314, 4312, 1342, 3142, 4132, 1432, 3412, 2413 P1= 1234, 4231, 3241, 2341 C(1234, 2341) P2= 1234, 3214, 2314, 4312, 3412, 2413, 4213, 1243, 2143, 3142, 1342, 2341 P3= 1234, 2134, 4132, 1432, 2431, 3421, 1423, 4123, 3124, 1324, 4321, 2341 P1= 1234, 2134, 4132, 3142 C(1234, 3142) P2= 1234, 4231, 3241, 2341, 4321, 3421, 2431, 1432, 3412, 4312, 1342, 3142 P3= 1234, 3241, 2341, 4321, 3421, 1423, 4123, 2143, 1243, 4213, 2413, 3412 P1= 1234, 2134, 3124, 4123 C(1234, 4123) P2= 1234, 3214, 4213, 2413, 3412, 4312, 2314, 1324, 4321, 3421, 1423, 4123 P3= 1234, 4231, 2431, 1432, 4132, 3142, 1342, 2341, 3241, 1243, 2143, 4213 P1= 1234, 3214, 2314, 4312 C(1234, 4312) P2= 1234, 2134, 4132, 3142, 2143, 4123, 3124, 1324, 4321, 2341, 1342, 4312 P3= 1234, 4231, 3241, 1243, 4213, 2413, 1423, 3421, 2431, 1432, 3412, 4312 P1= 1234, 4231, 2431, 3421 C(1234, 3421) P2= 1234, 2134, 3124, 4123, 2143, 3142, 4132, 1432, 3412, 2413, 1423, 3421 P3= 1234, 3214, 4213, 1243, 3241, 2341, 1342, 4312, 2314, 1324, 4321, 3421
From this table, dsL
3 (u, v) ≤ 15 if d(u, v) = 1. Hence, DsL
3(S4) = 15. ■
Lemma 11. DsL
n−1(Sn) ≥ nn−2! + 1 = (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 if n ≥ 5.
Proof. Let u and v be two adjacent nodes of Sn. Obviously, u and v are in different partite sets. Let {P1, P2,. . . , Pn−1} be any (n − 1)∗-container of Snjoining u to v. Obviously, one of these paths isu, v. Thus, max{l(Pi) | 1≤ i ≤ n −1} ≥ nn!−2−2+1 = n−2n! −n−22 +1 = nn−2! +1. Hence, dsL n−1(u, v) ≥ n! n−2+ 1 and D sL n−1(Sn) ≥ nn−2! + 1. ■ Lemma 12. DsL 4(S5) ≤ 41.
Proof. Let u be any white node and v be any black node of S5. Obviously, d(u, v) is odd.
Case 1. d(u, v) = 1. Because S5 is node transitive and
edge transitive, we may assume that u = e = 12345 and v= (e)5= 52341. By Lemma 7, there exist three paths P1, P2, and P3 of S5{5} such that (1) P1 joins 12345 to the
black node 24135 with l(P1) = 7, (2) P2 joins 12345 to
the white node 34125 with l(P2) = 8, (3) P3joins 12345 to
the white node 41325 with l(P3) = 8, and (4) P1∪ P2∪ P3
spans S{5}5 . Similarly, there exist three paths Q1, Q2, and Q3
of S5{1} such that (1) Q1 joins 52341 to the white node
24531 with l(Q1) = 7, (2) Q2 joins 52341 to the black
node 34521 with l(Q2) = 8, (3) Q3joins 52341 to the black
node 45321 with l(Q3) = 8, and (4) Q1∪ Q2∪ Q3spans S5{1}. By Theorem 1, there is a hamiltonian path R1of S5{2}
joining the white node 54132 to the black node 14532, there is a hamiltonian path R2of S{3}5 joining the black node 54123
to the white node 14523, and there is a hamiltonian path R3
of S5{4}joining the black node 51324 to the white node 15324. Then we set T1= e = 12345, P1, 24135, 54132, R1, 14532, 24531, (Q1)−1, 52341= (e)5, T2= e = 12345, P2, 34125, 54123, R2, 14523, 34521, (Q2)−1, 52341= (e)5, T3= e = 12345, P3, 41325, 51324, R3, 15324, 45321, (Q3)−1, 52341= (e)5, and T4= e = 12345, 52341 = (e)5.
Obviously,{T1, T2, T3, T4} is a 4∗-container of S5between e
and(e)5. Moreover, l(T1) = 39, l(T2) = l(T3) = 41, and l(T4) = 1. Thus, ds4L(e, (e)5) ≤ 41.
Case 2. d(u, v) ≥ 3. Because d(u, v) ≥ 3, there is i ∈ {2, 3, 4, 5} such that (u)i= (v)iand{(u)i,(v)i} ∩ {(u)1,(v)1} = ∅.
Without loss of generality, we assume that(u)5= (v)5and
{(u)5,(v)5} ∩ {(u)1,(v)1} = ∅. Moreover, we assume that (u)5= 5, (v)5= 4, (u)1= 1, and (v)1= 5. Because (u)1=
1 and(u)5= 5, we have {(u)2,(u)3,(u)4} = {2, 3, 4}.
Subcase 2.1. (v)1 = 1. We have {(v)2,(v)3,(v)4} =
{2, 3, 5}. By Lemma 8, there exist four paths P1, P2, P3,
and P4 of S5{5} such that (1) P1 joins u to a white node w with (w)1 = 2 and l(P1) = 2, (2) P2 joins u to a
white node x with (x)1 = 3 and l(P1) = 2, (3) P3 joins u to a black node y with (y)1 = 4 and l(P3) = 19, (4) P4 joins u to a black node z = y with (z)1 = 4 and l(P4) = 19, (5) P1∪ P2∪ P3spans S{5}5 , (6) P1∪ P2∪ P4
spans S5{5}, (7) V(P1) ∩ V(P2) ∩ V(P3) = {u}, and (8) V(P1) ∩ V(P2) ∩ V(P4) = {u}.
Similarly, there exist four paths Q1, Q2, Q3, and Q4of S{4}5
such that (1) Q1joins v to a black node p with(p)1= 2 and l(Q1) = 2, (2) Q2joins v to a black node q with(q)1= 3
and l(Q2) = 2, (3) Q3joins v to a white node r with(r)1= 5
and l(Q3) = 19, (4) Q4joins v to a white node s = r with (s)1 = 5 and l(Q4) = 19, (5) Q1∪ Q2∪ Q3 spans S5{4},
(6) Q1∪Q2∪Q4spans S5{4}, (7) V(Q1)∩V(Q2)∩V(Q3) = {v},
and (8) V(Q1) ∩ V(Q2) ∩ V(Q4) = {v}.
By Lemma 1, there are exactly three edges joining some black nodes of S{5}5 to some white nodes of S{4}5 . By the pigeon-hole principle, at least one node in{y, z} is adjacent to a node in{r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1be the hamiltonian path of S{1}5 joining
the black node(u)5to the white node(v)5, T2be the
hamil-tonian path of S5{2}joining the black node(w)5to the white node(p)5, and T3be the hamiltonian path of S5{3}joining the
black node(x)5to the white node(q)5. We set H1= u, (u)5, T1,(v)5, v,
H2= u, P1, w,(w)5, T2,(p)5, p, Q−11 , v, H3= u, P2, x,(x)5, T3,(q)5, q, Q−12 , v, and H4= u, P3, y, r, Q−13 , v.
Obviously, {H1, H2, H3, H4} is a 4∗-container of S5
between u and v. Moreover, l(H1) = 25, l(H2) = l(H3) =
29, and l(H4) = 39. Thus, d4sL(u, v) ≤ 41.
Subcase 2.2. (v)1= a ∈ {2, 3}. We have {(v)2,(v)3,(v)4}
= {1, 2, 3, 5} − {a}. Let b be the only element in {2, 3} − {a}. By Lemma 8, there exist four paths P1, P2, P3, and P4of S5{5}
such that (1) P1joins u to a white node w with(w)1= a and l(P1) = 2, (2) P2joins u to a white node x with(x)1 = b
and l(P2) = 2, (3) P3joins u to a black node y with(y)1= 4
and l(P3) = 19, (4) P4joins u to a black node z = y with (z)1 = 4 and l(P4) = 19, (5) P1∪ P2 ∪ P3 spans S{5}5 ,
(6) P1∪P2∪P4spans S5{5}, (7) V(P1)∩V(P2)∩V(P3) = {u},
and (8) V(P1) ∩ V(P2) ∩ V(P4) = {u}.
Again, there exist four paths Q1, Q2, Q3, and Q4of S5{4}
such that (1) Q1joins v to a black node p with(p)1= 1 and l(Q1) = 2, (2) Q2joins v to a black node q with(q)1 = b
and l(Q2) = 2, (3) Q3joins v to a white node r with(r)1= 5
and l(Q3) = 19, (4) Q4joins v to a white node s= r with (s)1= 5 and l(Q4) = 19, (5) Q1∪ Q2∪ Q3spans S{4}5 , (6) Q1∪ Q2∪ Q4spans S5{4}, (7) V(Q1) ∩ V(Q2) ∩ V(Q3) = {v},
and (8) V(Q1) ∩ V(Q2) ∩ V(Q4) = {v}.
By Lemma 1, there are exactly three edges joining some black nodes of S5{5} to some white nodes of S5{4}. By the pigeon-hole principle, at least one node in{y, z} is adjacent to a node in{r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1be the hamiltonian path of S5{1}joining
the black node(u)5to the white node(p)5, T2be the
hamil-tonian path of S{a}5 joining the black node(w)5to the white node(v)5, and T3be the hamiltonian path of S{b}5 joining the
black node(x)5to the white node(q)5. We set H1= u, (u)5, T1,(p)5, p, Q−11 , v, H2= u, P1, w,(w)5, T2,(v)5, v,
H3= u, P2, x,(x)5, T3,(q)5, q, Q−12 , v, and H4= u, P3, y, r, Q3−1, v.
Obviously, {H1, H2, H3, H4} is a 4∗-container of S5
between u and v. Moreover, l(H1) = l(H2) = 27, l(H3) =
29, and l(H4) = 39. Thus, d4sL(u, v) ≤ 41. ■
Lemma 13. dsL
n−1(u, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 = nn−2! + 1 for every n ≥ 6.
Proof. Let u be any white node and v be any black node of Sn. Obviously, d(u, v) is odd.
Case 1. d(u, v) = 1. Because the star graph is node tran-sitive and edge trantran-sitive, we may assume that u = e and
