The Spanning Diameter of the Star Graphs
The Spanning Diameter of the Star Graphs Cheng-Kuan Lin and Hua-Min Huang Department of Mathematics National Central University Chung-Li, Taiwan D. F. Hsu Department of Computer and Information Science Fordham University 113 West 60th Street New York, NY 10458, USA Lih-Hsing Hsu Department of Computer and Information Engineering Ta Hwa Institute of Technology, Hsinchu, Taiwan Abstract Assume that u and v are any two distinct vertices of different partite sets of the n-dimensional star graph Sn with n ≥ 5. We prove that there are (n − 1) internally n−1 disjoint paths P1 , P2 , . . . , Pn−1 joining u to v such that ∪i=1 Pi spans Sn and l(Pi ) ≤ n! (n−1)!+2(n−2)!+2(n−3)!+1 = n−2 +1. We also prove that there are two internally disjoint paths Q1 and Q2 joining u to v such that Q1 ∪ Q2 spans Sn and l(Qi ) ≤ n! 2 +1 for i = 1, 2.
Keywords: diameter, hamiltonian, hamiltonian laceable, star graphs.
1
Basic Definitions
For the graph definition and notation we 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. For a vertex u, N (u) denotes the neighborhood of u which is the set {v | (u, v) ∈ E}. For any vertex x of V , deg G (x) denotes its degree in G. We use δ(G) to denote min{degG (x) | x ∈ V }, and we use ∆(G) to denote max{degG (x) | x ∈ V }. Two vertices u and v are adjacent if (u, v) ∈ E. A path is represented by hv0 , v1 , . . . , vk i. The length of a path Q, l(Q), is the number of edges in Q. We also write the path hv0 , v1 , . . . , vk i as hv0 , Q1 , vi , vi+1 , . . . , vj , Q2 , vt , . . . , vk i, where Q1 is the path hv0 , v1 , . . . , vi i and Q2 is the path hvj , vj+1 , . . . , vt i. Hence, it is possible to write a path as hv0 , v1 , Q, v1 , v2 , . . . , vk i if l(Q) = 0. 1
We use d(u, v) to denote the distance between u and v, i.e., the length of the 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 vertices of G. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two distinct vertices of G. A cycle is a path with at least three vertices such that the first vertex is the same as the last vertex. A hamiltonian cycle of G is a cycle that traverses 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 [11] that there are k internally vertex-disjoint (abbreviated as disjoint) paths joining any two distinct vertices u and v when k ≤ κ(G). A container C(u, v) between two distinct vertices u and v in G is a set of internally disjoint paths {P1 , P2 , . . . , Pk } between u and v. The width of C(u, v) is k. A w-container is a container of width w. The length of a C(u, v) = {P1 , . . . , Pk }, l(C(u, v)), is max{l(Pi ) | 1 ≤ i ≤ k}. 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 6= v}. In particular, the wide diameter of G is Dκ(G) (G). The wide diameter is an important measure for interconnection networks [7]. In this paper, we are interested in a specifies type of containers. We say that a wcontainer C(u, v) is a w ∗ -container if every vertex 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 vertices 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 graph is motivated by the globally 3∗ -connected graphs proposed by Albert, Aldred, and Holton [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) ≤ δ(G) ≤ ∆(G). A graph G is super spanning connected if G is w ∗ -connected for any w with 1 ≤ w ≤ κ(G). In [10], it is proved that the pancake graph Pn is super spanning connected if and only if n 6= 3. A graph G is bipartite if its vertex set can be partitioned into two subsets V1 and V2 such that every edge joins vertices between V1 and V2 . Let G be a k-connected bipartite graph with bipartition V1 and V2 such that | V1 |≥| V2 |. Suppose that there exists a k ∗ -container C(u, v) = {P1 , P2 , . . . , Pk } in a bipartite graph G joining u to v with u, v ∈ V1 . Since the length of Pi = 2ki + 1 is an odd integer for all 1 ≤ i ≤ k, there are ki − 1 vertices of Pi in V1 P other than u and v, and ki vertices of Pi in V2 . As a consequence, | V1 |= ki=1 (ki − 1) + 2 P and | V2 |= ki=1 ki . Therefore, any bipartite graph G with κ(G) ≥ 3 is not k ∗ -connected. For this reason, we define that a bipartite graph is k ∗ -laceable if there exists a k ∗ -container between any two vertices from different partite sets. Obviously, any k ∗ -laceable graph with k ≥ 1 is a bipartite graph with the equal size of bipartition. An 1∗ -laceable graph is also known as hamiltonian laceable graph. Moreover, a bipartite graph is 2 ∗ -laceable if and only if it is hamiltonian. Obviously, all 1∗ -laceable graphs except K1 and K2 are 2∗ -laceable. A k-regular bipartite graph G is super spanning laceable if G is i∗ -laceable for all 1 ≤ i ≤ κ(G). It was proved in [10] that the star graph Sn is super spanning laceable if and only if n 6= 3.
2
We also define the w ∗ -lacable distance between any two vertices u and v from different partite sets, dswL (u, v), is min{l(C(u, v)) | C(u, v) is a w ∗ -container}. The w ∗L -diameter of G, denoted by DwsL (G), as max{dswL (u, v) | u and v are vertices from different partite sets}. sL In particular, the wide spanning diameter of G is Dκ(G) (G). In this paper, we prove that sL n! Dκ(Sn ) (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1 = n−2 + 1 and D2sL (Sn ) = n! + 1. 2 In Section 2, we give the definition of the star graphs and introduce some basic properties sL n! of star graphs. Then we prove that Dκ(S (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1 = n−2 +1 n) sL n! in Section 3. In Section 4, we prove that D2 (Sn ) = 2 + 1. We conclude the paper in the final section.
2
The star graphs and some notation conventions
Let n be a positive integer. We use hni to denote the set {1, . . . , n}. The n-dimensional star graph, denoted by Sn , is a graph with the vertex set V (Sn ) = {u1 . . . un | ui ∈ hni and ui 6= uj for i 6= j}. The adjacent is defined as follows: u1 . . . ui . . . un is adjacency to v1 . . . vi . . . vn through an edge of dimension i with 2 ≤ i ≤ n if vj = uj for j ∈ / {1, i}, v1 = ui , and vi = u1 . The star graphs S2 , S3 , and S4 are illustration in Figure 1. The star graphs are an important family of interconnection networks proposed by Akers and Krishnameurthy [1]. Some interesting properties of star graphs are studied [4, 5, 8, 12, 13]. It is know that the connectivity of Sn is n − 1. We use bold face to denote vertices in Sn . Hence u1 , u2 , . . . , un is denotes a sequence of vertices in Sn .
a 12
e
21
1324
f
S2 g
123 321
312
231
c
2134
3142
d
1423
4213
1234
4132
1342
4123
1243
4231
1432
4312
2143
b
213
3124
2314 3214
2413
b
c
3241
2431
2341
3421 4321
d
3412
e
f
g
a
132 S3
S4
Figure 1: The star graphs S2 , S3 , and S4 . By definition, Sn is an (n − 1)-regular graph with n! vertices. Moreover, it is vertex transitive and edge transitive. We use e to denote the element 12 . . . n. It is known that Sn 3
is a bipartite graph with one partite set containing those vertices corresponds to odd permutations and the other partite set containing those vertices corresponds to even permutations. We will use white vertices to represent those even permutation vertices and use black vertices to represent those odd permutation vertices. Let u = u1 u2 . . . un be any vertex of the star graph Sn . We say that ui is 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 {i} unique i-neighbor of u. Obviously, ((u)i )i = u. For 1 ≤ i ≤ n, let Sn denote the subgraph of Sn induced by those vertices u with (u)n = i. Obviously, Sn can be decomposed into n {i} {i} subgraph Sn , 1 ≤ i ≤ n, and each Sn is isomorphic to Sn−1 . Thus, the star graph can be constructed recursively. Let I ⊆ hni. We use SnI to denote the subgraph of Sn induced by {i} {j} ∪i∈I V (Sn{i} ). For 1 ≤ i 6= j ≤ n, we use E i,j to denote the set of edges between Sn and Sn . {(i,j)} For 1 ≤ i 6= j ≤ n, we use Sn to denote the subgraph of Sn induced by those vertices u {(i,j)} {(j,i)} {(i,j)} with (u)n−1 = i and (u)n = j. Obviously, Sn 6= Sn and Sn is isomorphic Sn−2 . Obviously, we have the following lemmas. Lemma 1 | E i,j |= (n − 2)! for any 1 ≤ i 6= j ≤ n. Moreover, there are (n − 2)!/2 edges {i} {j} joining black vertices of Sn to white vertices of Sn . Lemma 2 Assume that u and v are any two distinct vertices of Sn with 1 ≤ d(u, v) ≤ 2. Then (u)1 6= (v)1 . Theorem 1 [8] Sn is hamiltonian laceable if and only if n 6= 3. Theorem 2 [8] For any black vertex w and any two distinct white vertices u, v of S n with n ≥ 4. There exists a hamiltonian path of Sn − {w} joining u to v. Lemma 3 Assume that n ≥ 5 and I = {i1 , i2 , . . . , im } is a nonempty subset of hni. Let u be a white vertex and v be a black vertex of SnI with (u)n = i1 and (v)n = im . Then there exists a hamiltonian path of SnI joining u to v. {i }
Proof. Obviously, Sn j is isomorphic to Sn−1 for all 1 ≤ j ≤ m. By Theorem 1, this lemma is ture on m = 1. Thus, we assume that m ≥ 2. We set that x1 as u and set that ym as v. By Lemma 1, | E ij ,ij+1 |= (n − 2)!. We choose (yj , xj+1 ) ∈ E ij ,ij+1 with yj is a black vertex and xj+1 is a white vertex for 1 ≤ j ≤ m − 1. By Theorem 1, there is a hamiltonian path {i } Pj of Sn j joining xj to yj for all 1 ≤ j ≤ m. Then hx1 , P1 , y1 , x2 , P2 , y2 , . . . , xm , Pm , ym i forms a hamiltonian path of SnI joining u to v. 2 Lemma 4 Assume that r and s are any two adjacent vertices of Sn with n ≥ 4. Then there exists a hamiltonian path of Sn − {r, s} joining any white vertex u to a black vertex v with (v)1 = i for any i ∈ hni. Proof. Since Sn is vertex transitive and edge transitive, we assume that r = e and s = (e)2 . {n} Obviously, r, s ∈ Sn . We prove this lemma by induction on n. The required hamiltonian paths of S4 − {1234, 2134} are listed below. 4
(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324) (3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134) (3124)(2134)(4132)(1432)(2431)(3421)(4321)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(4312)(3412)(2413)(1423)(4123)(2143)(3142) (3124)(2134)(4132)(1432)(2431)(3421)(4321)(1432)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(3142)(2143)(4123)(1423)(2413)(3412)(4312) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (4132)(3124)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(2413)(1423)(4123)(2143)(3142)(1342)(4312) (1342)(4312)(3412)(2413)(1423)(4123)(2143)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432) (1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(2143)(4123)(1423)(2413) (1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142) (1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312) (2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324) (2314)(3214)(4213)(1243)(3241)(2341)(4321)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413) (2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214) (2314)(3214)(4213)(1243)(3241)(2341)(4321)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(1423)(2413)(3412)(4312)(1342)(3142)(2143)(4123) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413) (4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(3142)(1342)(4312)(3412)(3412)(1423)(4123) (1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413)(4213)(3241)(2314)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413) (1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214) (1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123) (2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (2143)(3142)(1342)(4312)(3412)(2413)(1423)(4123)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134) (2143)(1243)(3241)(2341)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(3124)(2134)(4132)(3142) (2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123) (3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243) (3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341) (3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(2143)(4123)(1423)(2413)(3412)(4312)(1342)(3142) (3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(2143)(3142)(1342)(4312)(3412)(2413)(1423)(4123) (4321)(3421)(2431)(1432)(3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324) (4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341) (4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421) (4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(3142)(2143)(4123)(1324)(2413)(3412)(4321) (2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(3412)(4312)(1342)(3142)(2143)(4123)(1423)(2413) (2431)(1432)(3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(4321)(3421) (2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(3412)(2413)(1423)(4123)(2143)(3142)(1342)(4312) (3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(4321)(3421)(2431)(1432) (3412)(4312)(1342)(3142)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134) (3412)(4312)(1342)(3132)(2143)(1243)(3241)(2341)(4321)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(3124)(2134)(4132)(1432)(2431)(3421) (3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)
Assume that this lemma is hold on Sk for all 4 ≤ k < n. We have the following cases: {n}
{n}
Case 1: u ∈ Sn . Obviously, Sn is isomorphism to Sn−1 . By induction, there exists a {n} hamiltonian path P of Sn − {r, s} joining u to a black vertex x with (x)1 = n − 1. Let v hn−2i is a black vertex of Sn with (v)1 = i. By Lemma 3, there exists a hamiltonian path Q of hn−1i Sn joining the white vertex (x)n to v. Then hu, P, x, (x)n , Q, vi forms the desired path. {k}
Case 2: u ∈ Sn for some k 6= n. By Lemma 1, there exists (n − 2)!/2 ≥ 3 edges {n} {k} joining black vertices of Sn to white vertices of Sn . We can choose a black vertex x of {k} {n} {k} Sn such that (x)n is a white vertex of Sn − {r, s}. Since Sn is isomorphism to Sn−1 , {k} by Theorem 1, there exists a hamiltonian path P of Sn joining u to x. By induction, {n} there exists a hamiltonian path Q of Sn − {r, s} joining (x)n to a black vertex y with hn−1i−{k,r} (y)1 = r ∈ hn − 1i − {k}. Let v is a black vertex of Sn with (v)1 = i. By Lemma 3, hn−1i−{k} there exists a hamiltonian path R of Sn joining the white vertex (y)n to v. Obviously, hu, P, x, (x)n , Q, y, (y)n , R, vi forms the desired path. 2 Theorem 3 Assume that n ≥ 5 and I = {i1 , i2 , . . . , im } is a nonempty subset of hni. Then SnI is hamiltonian laceable. Proof. Suppose that u is a white vertex and v is a black vertex of SnI . By Lemma 3, this theorem is true on either m = 1, or m ≥ 2 and (u)n 6= (v)n . Thus, we assume that m ≥ 2 and (u)n = (v)n . Without loss of generality, we assume that (u)n = (v)n = i1 . 5
Case 1: (v)1 ∈ I. Without loss of generality, we assume that (v)1 = im . Obviously, {i } {i } Sn 1 isomorphic to Sn−1 . By Theorem 2, there exists a hamiltonian path P of Sn 1 − {v} joining u to a white vertex x such that (x)1 = i2 . By Lemma 3, there exists a hamiltonian I−{i } path Q of Sn 1 joining the black vertex (x)n to the white vertex (v)n . Then the path hu, P, x, (x)n , Q, (v)n , vi forms a hamiltonian path of SnI joining u to v. Case 2: (u)1 ∈ / I and (v)1 ∈ / I. We can choose a white vertex y be a neighbor of v in {i1 } Sn with (y)n = im . Obviously, y 6= u. By Lemma 4, there exists a hamiltonian path P of {i } Sn 1 − {v, y} joining u to a black vertex x such that (x)n = i2 . By Lemma 3, there exists a I−{i } hamiltonian path Q of Sn 1 joining the white vertex (x)n to the black vertex (y)n . Then the path hu, P, x, (x)n , Q, (y)n , y, vi forms a hamiltonian path of SnI joining u to v. 2 Theorem 4 Assume that n ≥ 5. Sn −{r, s} is hamiltonian laceable for any adjacent vertices r and s. Proof. Since the Sn is vertex transitive and edge transitive, we assume that r = e and {n} s = (e)2 . Obviously, r, s ∈ Sn . Suppose that u is a white vertex and v is a black vertex of Sn − {r, s}. We want to find a hamiltonian path of Sn − {r, s} joining u to v. {n}
{n}
Case 1: u, v ∈ Sn . By Lemma 4, there exists a hamiltonian path P of Sn − {r, s} joining u to a black vertex y with (y)1 ∈ hn − 1i. Obviously, P can be write as hu, Q1 , x, v, Q2 , yi. (Note that l(Q1 ) = 0 if u = x and l(Q2 ) = 0 if v = y.) By Theorem 3, there exists a hn−1i hamiltonian path R of Sn joining the black vertex (x)n to the white vertex (y)n . Then hu, Q1 , x, (x)n , R, (y)n , y, (Q2 )−1 , vi forms a desired hamiltonian path. {k}
Case 2: u, v ∈ Sn for some k 6= n. By Theorem 1, there exists a hamiltonian path P of {k} Sn joining u to v. By Lemma 1, there exists (n − 2)!/2 ≥ 3 edges joining black vertices {k} {n} {k} of Sn to white vertices of Sn . We can choose a black vertex x of Sn with (x)1 = n and (x)n ∈ / {r, s}. Obviously, P can be write as hu, Q1 , x, y, Q2 , vi. (Note that l(Q1 ) = 0 if u = x and l(Q2 ) = 0 if v = y.) Obviously, d(x, y) = 1. By Lemma 2, (x)1 6= (y)1 . By {n} Lemma 4, there exists a hamiltonian path R of Sn − {r, s} joining the white vertex (x)n to a black vertex z with (z)1 ∈ hn − 1i − {k}. By Theorem 3, there exists a hamiltonian hn−1i−{k} path T of Sn joining the black vertex (z)n to the white vertex (y)n . Obviously, hu, Q1 , x, (x)n , R, z, (z)n , T, (y)n , y, Q2 , vi forms a desired hamiltonian path. {n}
{k}
Case 3: u ∈ Sn and v ∈ Sn for some k 6= n. By Lemma 4, there exists a hamiltonian {n} path P of Sn − {r, s} joining u to a black vertex x with (x)1 ∈ hn − 1i. By Theorem hn−1i 3, there exists a hamiltonian path Q of Sn joining the white vertex (x)n to v. Then hu, P, x, (x)n , Q, vi forms a desired hamiltonian path. {k}
{l}
Case 4: u ∈ Sn and v ∈ Sn with k, l, and n are distinct. By Lemma 1, there exists {k} {n} (n − 2)!/2 ≥ 3 edges joining black vertices of Sn to white vertices of Sn , we can choose {k} a black vertex x of Sn with (x)1 = n and (x)n ∈ / {r, s}. By Theorem 1, there exists a {k} hamiltonian path P of Sn joining u to x. By Lemma 4, there exists a hamiltonian path Q 6
{n}
of Sn − {r, s} joining the white vertex (x)n to a black vertex y with (y)1 ∈ hn − 1i − {k}. hn−1i−{k} By Lemma 3, there exists a hamiltonian path R of Sn joining the white vertex (y)n to v. Then hu, P, x, (x)n , Q, y, (y)n , R, vi forms a desired hamiltonian path. 2 Lemma 5 Assume that n ≥ 4. Let u be a white vertex of Sn and Fn = {(u)i | 3 ≤ i ≤ n} ∪ {((u)i )i−1 | 3 ≤ i ≤ n}. Then there exists a hamiltonian path of Sn − Fn joining u to a black vertex v with (v)1 = j for any j ∈ hni. Proof. We proved this lemma by induction on n. Since Sn is vertex transitive, we assume that u = e. The required hamiltonian paths of S4 − F4 are listed below. (1234)(2134)(3124)(4123)(1423)(3421)(2431)(1432)(4132)(3142)(2143)(1243)(4213)(2413)(3412)(4312)(1342)(2341)(4321)(1324) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(1423)(2413)(4213)(1243)(2143)(4123)(3124)(1324)(4321)(2341) (1234)(2134)(4132)(1432)(2431)(3421)(1423)(4123)(3124)(1324)(4321)(2341)(1342)(4312)(3412)(2413)(4213)(1243)(2143)(3142) (1234)(2134)(3124)(1324)(4321)(2341)(1342)(3142)(4132)(1432)(2431)(3421)(1423)(4123)(2143)(1243)(4213)(2413)(3412)(4312)
{n}
Assume that the lemma is true for any Sk with 4 ≤ k ≤ n − 1. Obviously, Sn is {n} isomorphic to Sn−1 . By induction, there exists a hamiltonian path P of Sn − Fn−1 joining {1} e to a black vertex x with (x)1 = 1. By Lemma 4, there exists a hamiltonian path Q of Sn − {(e)n , ((e)n )n−1 } joining the white vertex (x)n to a black vertex y with (y)1 = 2. We can hn−1i−{1} with (z)1 = i. By Theorem 3, there exists a hamiltonian choose a black vertex z of Sn hni−{1,n} path R of Sn joining the white vertex (y)n to z. Then he, P, x, (x)n, Q, y, (y)n , R, zi forms a desired hamiltonian path. 2 Lemma 6 Assume that n ≥ 5. Suppose that p and q are two different white vertices of S n , and r and s are two different black vertices of Sn . Then there exist two disjoint paths P1 and P2 such that (1) P1 joins p to r, (2) P2 joins q to s, and (3) P1 ∪ P2 spans Sn . Proof. Without loss of generality, we assume that (p)n = n and (q)n = n − 1. Let (r)n = i and (s)n = j. {i,n}
Case 1: i, j ∈ hn − 2i and i 6= j. By Theorem 3, there exists a hamiltonian path P1 of Sn hn−1i−{i} joining p to r. Again, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 form the desired paths. Case 2: i, j ∈ hn − 2i and i = j. We can choose a white vertex x be the neighborhood of r {i} hni−{i,n−1} in Sn with (x)1 6= n − 1. By Theorem 3, there exists a hamiltonian path P of Sn joining p to the black vertex (x)n . By Lemma 4, there exists a hamiltonian path Q of {i} Sn − {r, x} joining s to a black vertex y with (y)1 = n − 1. By Theorem 1, there exists a {n−1} hamiltonian path Q of Sn joining q to the black vertex (y)n . Then P1 = hp, P, (x)n , x, ri and P2 = hq, R, (y)n , y, Q−1 , si form the desired paths. Case 3: either i ∈ hn − 2i and j = n − 1, or i = n and j ∈ hn − 2i. By symmetric, we assume that i ∈ hn − 2i and j = n − 1. By Theorem 3, there exists a hamiltonian path P1 hni−{n−1} {n−1} of Sn joining p to r. By Theorem 1, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 are the desired paths. 7
Case 4: either i = n − 1 and j ∈ hn − 2i, or i ∈ hn − 2i and j = n. By symmetric, we assume that i = n − 1 and j ∈ hn − 2i. By Theorem 1, there exists a hamiltonian path R of {n−1} {n} Sn joining q to r. We can choose a white vertex x ∈ R with (x)n ∈ Sn . We can write {n} R as hq, R1 , y, x, R2 , ri. By Theorem 1, there exists a hamiltonian path W of Sn joining p to the black vertex (x)n . We set P1 as hp, W, (x)n , x, R2 , ri. Obviously, y is a black vertex hn−2i and (y)1 ∈ hn − 2i. By Theorem 3, there exists a hamiltonian path Q of Sn joining the n n white vertex (y) to s. We set P2 as hq, R1 , y, (y) , Q, si. Then P1 and P2 forms the desired paths. {n}
Case 5: i = n and j = n − 1. By Theorem 1, there exists a hamiltonian path P1 of Sn hn−1i joining p to r. By Theorem 3, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 are the desired paths. {n}
Case 6: i = n−1 and j = n. We can choose a white vertex x ∈ Sn −{s} with (x)1 = n−1. {n} By Theorem 1, there exists a hamiltonian path R of Sn joining p to s. Again, there exists {n−1} a hamiltonian path Q of Sn joining q to r. We can write R as hp, R1 , x, y, R2 , si, n and write Q as hq, Q1 , w, (x) , Q2 , ri. Obviously, y is a black vertex and w is a white vertex. Since d(x, y) = 1, by Lemma 2, (y)1 6= (x)1 = n − 1. Since d((x)n , w) = 1, by hn−1i Lemma 2, (w)1 6= ((x)n )1 = n. By Theorem 3, there exists a hamiltonian path W of Sn joining the black vertex (w)n to the white vertex (y)n . Then P1 = hp, R1 , x, (x)n , Q2 , ri and P2 = hq, Q1 , w, (w)n, W, (y)n , y, R2 ri form a desired paths. Case 7: either i = j = n or i = j = n − 1. By symmetric, we assume that i = j = n. By {n} Theorem 1, there exists a hamiltonian path P of Sn joining p to s. We can write P as hp, R1 , r, x, R2 , si. We set P1 as hp, R1 , ri. Obviously, x is a white vertex and (x)1 ∈ hn − 1i. hn−1i By Theorem 3, there exists a hamiltonian path Q of Sn joining q to the black vertex (x)n . Then P1 and P2 = hq, Q, (x)n , x, R2 , qi are the desired paths. 2
3
The (n − 1)∗L -diameter of Sn
Lemma 7 Let u = u1 u2 u3 u4 be any white vertex of S4 . There exist three paths P1 , P2 , and P3 such that (1) P1 joins u to the black vertex u2 u4 u1 u3 with l(P1 ) = 7, (2) P2 joins u to the white vertex u3 u4 u1 u2 with l(P2 ) = 8, (3) P3 joins u to the white vertex u4 u1 u3 u2 with l(P3 ) = 8, and (4) P1 ∪ P2 ∪ P3 spans S4 . Proof.
Since S4 is vertex transitive, we assume that u = 1234. Then we set P1 = h1234, 3214, 4213, 1243, 2143, 4123, 1423, 2413i, P2 = h1234, 4231, 3241, 2341, 4321, 3421, 2431, 1432, 3412i, and P3 = h1234, 2134, 3124, 1324, 2314, 4312, 1342, 3142, 4132i.
Obviously, P1 , P2 , and P3 forms the desired paths.
8
2
Lemma 8 Let u = u1 u2 u3 u4 be any white vertex of S4 . Let i1 i2 i3 be a permutation of u2 , u3 , and u4 . There exist four paths P1 , P2 , P3 , and P4 of S4 such that (1) P1 joins u to a white vertex w with (w)1 = i1 and l(P1 ) = 2, (2) P2 joins u to a white vertex x with (x)1 = i2 and l(P2 ) = 2, (3) P3 joins u to a black vertex y with (y)1 = i3 and l(P3 ) = 19, (4) P4 joins u to a black vertex z with z 6= y, (z)1 = i3 , and l(P4 ) = 19, (5) P1 ∪ P2 ∪ P3 spans S4 , and (6) P1 ∪ P2 ∪ P4 spans S4 . Proof. Since S4 is vertex transitive, we assume that u = 1234. Without loss of generality, we suppose that i1 < i2 . The required four paths are listed below. P1 P2 P3 P4
= = = =
(1234)(2134)(3124) (1234)(3214)(4213) (1234)(4231)(3241)(1243)(2143)(4123)(1423)(2413)(3412)(4312)(2314)(1324)(4321)(3421)(2431)(1432)(4132)(3142)(1342)(2341) (1234)(4231)(3241)(1243)(2143)(4123)(1423)(3421)(2431)(1432)(4132)(3142)(1342)(2341)(4321)(1324)(2314)(4312)(3412)(2413)
P1 P2 P3 P4
= = = =
(1234)(4231)(2431) (1234)(3214)(4213) (1234)(2134)(3124)(4123)(2143)(1243)(3214)(2314)(1342)(4312)(2314)(1324)(4321)(3421)(1423)(2413)(3412)(1432)(4132)(3142) (1234)(2134)(3142)(1324)(2314)(4312)(1342)(3142)(4132)(1432)(3412)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)
P1 P2 P3 P4
= = = =
(1234)(4231)(2431) (1234)(2134)(3124) (1234)(3214)(2314)(1324)(4321)(3421)(1423)(4123)(2143)(3142)(4132)(1432)(3412)(2413)(4213)(1243)(3241)(2341)(1342)(4312) (1234)(3214)(2314)(1324)(4321)(3421)(1423)(2413)(4213)(1243)(3241)(2341)(1342)(4312)(3412)(1432)(4132)(3142)(2143)(4123)
2 Lemma 9 Assume that n ≥ 5 and i1 i2 . . . in−1 is an (n − 1)-permutation of hni. Let u be any white vertex of Sn . Then there exist (n − 1) paths P1 , P2 , . . . , Pn−1 of Sn such that (1) P1 is joins u to a black vertices y1 with (y1 )1 = i1 and l(P1 ) = n(n − 2)! − 1, (2) Pj is joins u to a white vertices yi with (yj )1 = ij and l(Pj ) = n(n − 2)! for 2 ≤ j ≤ n − 1, and (3) n−1 Pj spans Sn . ∪j=1 Proof. Since Sn is vertex transitive, we assume that u = e. Without loss of generality, we suppose that i2 < i3 < . . . < in−1 . Case 1: n = 5. Obviously, i2 6= 4, i3 = 6 1, and i4 6= 1. Let x1 = (e)5 , x2 = ((x1 )2 )5 , x3 = ((x2 )3 )5 , and x4 = ((x3 )4 )5 . Assume that i1 = 3. We set W1 = h12345, 32145, 42135, 12435, 21435, 41235, 14235, 24135i, W2 = h12345, 21345, 31245, 13245, 23145, 43125, 13425, 31425, 41325i, and W3 = h12345, 42315, 32415, 23415, 43215, 34215, 24315, 14325, 34125i. Let u1 = 24135, u2 = 41325, u3 = 34125, and u4 = ((x4 )3 )5 . Obviously, xi is a black vertex {i} {2} of S5 . By Lemma 4, there exists a hamiltonian path Q1 of S5 − {x2 , (x2 )3 } joining the white vertex (u1 )5 to a black vertex y1 with (y1 )1 = i1 . Again, there exists a hamiltonian {4} path Q2 of S5 − {x4 , (x4 )3 } joining the black vertex (u2 )5 to a white vertex y2 with {3} (y2 )1 = i2 . Moreover, there exists a hamiltonian path Q3 of S5 − {x3 , (x3 )4 } joining the black vertex (u3 )5 to a white vertex y3 with (y3 )1 = i3 . And, there exists a hamiltonian path 9
{2}
Q4 of S5 We set
− {x2 , (x2 )3 } joining the black vertex (u4 )5 to a white vertex y4 with (y4 )1 = i4 . P1 P2 P3 P4
= = = =
he, W1 , u1 , (u1 )5 , Q1 , y1 i, he, W2 , u2 , (u2 )5 , Q2 , y2 i, he, W3 , u3 , (u3 )5 , Q4 , y4 i. and he, x1 , (x1 )2 , x2 , (x2 )3 , x3 , (x3 )4 , x4 , (x4 )3 , u4 , Q3 , y3 i.
Obviously, l(P1 ) = 29, and l(Pi ) = 30 for 2 ≤ i ≤ 4. Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. Assume that i1 6= 3. Obviously, i3 6= 3. We set W1 = h12345, 21345, 41325, 14325, 34125, 43125, 13425, 31425i, W2 = h12345, 32145, 23145, 13245, 31245, 41235, 14235, 24135, 42135i, and W3 = h12345, 42315, 24135, 34215, 43215, 23415, 32415, 12435, 21435i. Let u1 = 31425, u2 = 42135, u3 = 21435, and u4 = ((x4 )3 )5 . Obviously, xi is a black vertex {i} {3} of S5 . By Lemma 4, there exists a hamiltonian path Q1 of S5 − {x3 , (x3 )4 } joining the white vertex (u1 )5 to a black vertex y1 with (y1 )1 = i1 . Again, there exists a hamiltonian {4} path Q2 of S5 − {x4 , (x4 )3 } joining the black vertex (u2 )5 to a white vertex y2 with {1} (y2 )1 = i2 . Moreover, there exists a hamiltonian path Q3 of S5 − {x1 , (x1 )2 } joining the black vertex (u4 )5 to a white vertex y3 with (y3 )1 = i3 . And, there exists a hamiltonian path {2} Q4 of S5 − {x2 , (x2 )3 } joining the black vertex (u3 )5 to a white vertex y4 with (y4 )1 = i4 . We set P1 P2 P3 P4
= = = =
he, W1 , u1 , (u1 )5 , Q1 , y1 i, he, W2 , u2 , (u2 )5 , Q2 , y2 i, he, x1 , (x1 )2 , x2 , (x2 )3 , x3 , (x3 )4 , x4 , (x4 )3 , u4 , Q3 , y3 i. and he, W3 , u3 , (u3 )5 , Q4 , y4 i.
Obviously, l(P1 ) = 29, and l(Pi ) = 30 for 2 ≤ i ≤ 4. Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. Case 2: n ≥ 6. We set that xj = (e)j and yj = ((e)j )j−1 for 3 ≤ j ≤ n − 2. Obviously, uj is a black vertex and vj is a white vertex for 3 ≤ j ≤ n − 2. {(n−1,n)}
Assume that i1 6= n − 3. By Lemma 5, there exists a hamiltonian path R of Sn − Fn−2 joining e to a black vertex x1 with (x1 )1 = 2. We recursively set xj as the unique {(j,n)} neighborhood of (xj−1 )n−1 in Sn with (xj )1 = j + 1 for 2 ≤ j ≤ n − 5, and we set {(n−4,n)} xn−4 as the unique neighborhood of (xn−5 )n−1 in Sn with (xn−4 )1 = n − 1. It is {(j+1,n)} easy to see that xj is a black vertex for 1 ≤ j ≤ n − 4 and {(xj )n−1 , xj+1 } ∈ Sn for 1 ≤ j ≤ n − 5. Choose xn−3 as any black vertex in Snn−1 with (xn−3 )1 = n and 10
{n−1}
(xn−3 )n−1 = n − 3. By Theorem 1, there exists a hamiltonian path R0 of Sn joinn ing the white vertex (xn−4 ) to xn−3 . Then we set the black vertex y1 as the unique {(n−3,n)} neighborhood of the white vertex (xn−3 )n in Sn with (y1 )1 = i1 . We set P1 as n−1 n−1 n−1 he, R, x1 , (x1 ) , x2 , (x2 ) , . . . , xn−5 , (xn−5 ) , xn−4 , (xn−4 )n , R0 , xn−3 , (xn−3 )n , y1 i. Obviously, l(P1 ) = n(n − 2)! − 1. {1}
By Theorem 1, there exists a hamiltonian path H of Sn joining the black vertex (e)n to the a white vertex z with (z)1 = n and (z)n−2 = n − 2. Obviously, there exists a hamiltonian {(n−2,n)} path H 0 of Sn joining the black vertex (z)n to a white vertex y2 with (y2 )1 = a2 . We n set P2 as he, (e) , H, z, (z)n , H 0 , y2 i. Obviously, l(P2 ) = n(n − 2)!. {(1,n)}
By Theorem 1, there exists a hamiltonian path W of Sn joining the black vertex n−1 (e) to a white vertex r with (r)1 = n − 1. Moreover, there exists a hamiltonian path W 0 {n−1} of Sn joining the black vertex (r)n to a white vertex y3 with (y3 )1 = a3 . We set P3 as n−1 he, (e) , W, r, (r)n , W 0 , y3 i. Obviously, l(P3 ) = n(n − 2)!. Let i be any index with 4 ≤ i ≤ n − 2. By Lemma 4, there exists a hamiltonian {(i−2,n)} path Qi of Sn − {(xi−3 )n−1 , xi−2 } joining the black vertex (vi−1 )n−1 to a white ver{i−2} tex zi with (zi )1 = i − 2. Again, there exists a hamiltonian path Q0i of Sn joinn ing the black vertex (zi ) to a white vertex yi with (yi )1 = ai . Then we set Pi as he, ui−1 , vi−1 , (vi−1 )n−1 , Qi , zi , (zi )n , Q0i , yi i. Obviously, l(Pi ) = n(n − 2)!. {(n−3,n)}
By Lemma 4, there exists a hamiltonian path R of Sn − {(xn−3 )n , y1 } joining the black vertex (vn−3 )n−1 to a white vertex r with (r)1 = n − 2. Obviously, there ex{n−2} ists a hamiltonian path R0 of Sn joining the black vertex (r)n to a white vertex yn−1 with (yn−1 )1 = an−1 . Hence we set Pn−1 as he, un−1 , vn−1 , (vn−1 )n−1 , R, r, (r)n, R0 , yn−1 i. Obviously, l(Pn−1 ) = n(n − 2)!. Apparently, P1 , P2 , . . . , Pn−1 form the desired paths. See Figure 2 for illustration for the case n = 6. {(n−1,n)}
Assume that i1 = n − 3. By Lemma 5, there exists a hamiltonian path R of Sn − Fn−2 joining e to a black vertex x1 with (x1 )1 = n − 3. We recursively set xj as the unique {(n−j−1,n)} neighborhood of (xj−1 )n−1 in Sn with (xj )1 = n − j − 1 for 2 ≤ j ≤ n − 5, and we set {(3,n)} xn−4 as the unique neighborhood of (xn−5 )n−1 in Sn with (xn−4 )1 = n−1. It is easy to see {n−j−2,n} n−1 that xj is a black vertex for 1 ≤ j ≤ n − 4 and {(xj ) , xj+1 } ∈ Sn for 1 ≤ j ≤ n − 5. {n−1} 0 By Theorem 1, there exists a hamiltonian path R of Sn joining the white vertex (xn−4 )n to a black vertex xn−3 with (xn−3 )1 = n and (xn−3 )n−1 = 2. Then we set the black vertex y1 {(2,n)} as the unique neighborhood of the white vertex (xn−3 )n in Sn with (y1 )1 = i1 . We set P1 n−1 n−1 n−1 as he, R, x1 , (x1 ) , x2 , (x2 ) , . . . , xn−5 , (xn−5 ) , xn−4 , (xn−4 )n , R0 , xn−3 , (xn−3 )n , y1 i. Obviously, l(P1 ) = n(n − 2)! − 1. {1}
By Theorem 1, there exists a hamiltonian path H of Sn joining the black vertex (e)n to a white vertex z with (z)1 = n and (z)n−2 = n − 2. Again, there exists a hamiltonian 11
S6{6} e
S6{(5,6)}
S6{5}
S6 (x2)6
x1
{(2,6)}
x2 (x1)6
u3 v3 u4 v4
S6{1}
S6{(4,6)} y2
(e)6
(v 3)5
P1
(z)6
z4
S6{(3,6)}
x3
(x3)6
(v4)5
z
P2
S6{(1,6)}
(e)5 r
y1 z5
S6{2}
S6{4}
(z4)6
P4 y4
(r)6
S6{3}
P3
y5
(z 5)6
y3
P5
Figure 2: Illustration for the case 1 of Lemma 9.
{(n−2,n)}
path H 0 of Sn joining the black vertex (z)n to a white vertex y2 with (y2 )1 = a2 . We set P2 as he, (e)n , H, z, (z)n , H 0 , y2 i. Obviously, l(P2 ) = n(n − 2)!. {(1,n)}
joining the black vertex By Theorem 1, there exists a hamiltonian path W of Sn (e)n−1 to a white vertex r with (r)1 = n − 1. Moreover, there exists a hamiltonian path W 0 {n−1} of Sn joining the black vertex (r)n to a white vertex y3 with (y3 )1 = i3 . We set P3 as he, (e)n−1 , W, r, (r)n , W 0 , y3 i. Obviously, l(P3 ) = n(n − 2)!. Let j be any index with 4 ≤ j ≤ n − 2. By Lemma 4, there exists a hamiltonian {(j−1,n)} − {(xn−j−1 )n−1 , xn−j } joining the black vertex (vj )n−1 to a white path Qj of Sn {j−1} joinvertex zj with (zj )1 = j − 1. Again, there exists a hamiltonian path Q0j of Sn n ing the black vertex (zj ) to a white vertex yj with (yj )1 = ij . Then we set Pj as he, uj , vj , (vj )n−1 , Qj , zj , (zj )n , Q0j , yj i. Obviously, l(Pj ) = n(n − 2)!. {(2,n)}
By Lemma 4, there exists a hamiltonian path R of Sn − {(xn−3 )n , y1 } joining the n−1 black vertex (v3 ) to a white vertex r with (r)1 = n − 2. Obviously, there exists a {n−2} 0 joining the black vertex (r)n to a white vertex yn−1 with hamiltonian path R of Sn (yn−1 )1 = in−1 . Hence we set Pn−1 as he, u3 , v3 , (v3 )n−1 , R, r, (r)n , R0 , yn−1 i. Obviously, l(Pn−1 ) = n(n − 2)!. 12
Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. See Figure 3 for illustration for the case n = 6. The lemma is proved. 2 {6}
S6
e
{(5,6)}
S6
x1
{(2,6)} 6
S
{5} 6
S
x3
y1
(x3)6
u3 {(4,6)}
v3 u4 v4
S6
{1}
S6
y2
(e)6
(v3)5
P1
(z)6
z
P2
z4 (x2)6
{(3,6)}
S6
x2
{(1,6)}
(x1)6
S6
(e)5 r
(v4)5 z5
{2}
S6
{4}
S6
(z4)6 {3}
(r)6
S6
P4 y4
y5
(z5)6
y3
P3
P5
Figure 3: Illustration for the case 2 of Lemma 9.
sL Lemma 10 Dn−1 (Sn ) ≥
n! n−2
+ 1 = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1.
Proof. Let u and v are two adjacent vertices of Sn . Let P1 , P2 , . . . , Pn−1 be any (n − 1)∗ container of Sn joining u to v. Obviously, one of these path is hu, vi. Thus, max{l(Pi ) | 1 ≤ sL n! n! n! L i ≤ n − 1} ≥ n−2 + 1. Hence, dsn−1 (u, v) ≥ n−2 + 1 and Dn−1 (Sn ) ≥ n−2 + 1. 2 Lemma 11 D4sL (S5 ) ≤ 41. Proof. odd.
Let u be any white vertex and v be any black vertex of S5 . Obviously, d(u, v) is
Case 1: d(u, v) = 1. Since the S5 is vertex transitive and edge transitive, we may assume {5} that u = e = 12345 and v = (e)5 = 52341. By Lemma 7, there exist P1 , P2 , and P3 of S5 such that (1) P1 joins 12345 to the black vertex 24135 with l(P1 ) = 7, (2) P2 joins 12345 to the white vertex 34125 with l(P2 ) = 8, (3) P3 joins 12345 to the white vertex 41325 with {5} {1} l(P3 ) = 8, and (4) P1 ∪ P2 ∪ P3 spans S4 . Similarly, there exist Q1 , Q2 , and Q3 of S5 13
such that (1) Q1 joins 52341 to the white vertex 24531 with l(Q1 ) = 7, (2) Q2 joins 52341 to the black vertex 34521 with l(Q2 ) = 8, (3) Q3 joins 52341 to the black vertex 45321 with {1} l(Q3 ) = 8, and (4) Q1 ∪ Q2 ∪ Q3 spans S4 . By Theorem 1, there is a hamiltonian path {2} R1 of S5 joining the white vertex 54132 to the black vertex 14532, there is a hamiltonian {3} path R2 of S5 joining the black vertex 34125 to the white vertex 41324, and there is a {4} hamiltonian path R3 of S5 joining the black vertex 51324 to the white vertex 15424. Then we set T1 T2 T3 T4
= = = =
h12345, P1, 24135, 54132, R1, 14532, 24531, (Q1)−1 , 52341i, h12345, P2, 34125, 54123, R2, 14523, 34521, (Q2)−1 , 52341i, h12345, P3, 41325, 51324, R3, 15324, 45321, (Q3)−1 , 52341i, and h12345, 52341i.
Obviously, {T1 , T2 , T3 , T4 } forms a 4∗ -container of S5 between 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. Since d(u, v) ≥ 3, we may assume that u = e, (v)5 = 4, and (v)1 ∈ {1, 2, 3}. {5}
Subcase 2.1: (v)1 = 1. By Lemma 8, there exist four paths P1 , P2 , P3 , and P4 of S5 such that (1) P1 joins e to a white vertex w with (w)1 = 2 and l(P1 ) = 2, (2) P2 joins e to a white vertex x with (x)1 = 3 and l(P1 ) = 2, (3) P3 joins e to a black vertex y with (y)1 = 4 and l(P3 ) = 19, (4) P4 joins e to a black vertex z 6= y with (z)1 = 4 and l(P4 ) = 19, (5) {5} {5} P1 ∪ P2 ∪ P3 spans S5 , and (6) P1 ∪ P2 ∪ P4 spans S5 . {4}
Similarly, there exist four paths Q1 , Q2 , Q3 , and Q4 of S5 such that (1) Q1 joins v to a black vertex p with (p)1 = 2 and l(Q1 ) = 2, (2) Q2 joins v to a black vertex q with (q)1 = 3 and l(Q2 ) = 2, (3) Q3 joins v to a white vertex r with (r)1 = 5 and l(Q3 ) = 19, (4) P4 joins {4} v to a white vertex s 6= r with (s)1 = 5 and l(Q4 ) = 19, (5) Q1 ∪ Q2 ∪ Q3 spans S5 , and {4} (6) Q1 ∪ Q2 ∪ Q4 spans S5 . {5}
By Lemma 1, there are exactly three edges joining a black vertices of S5 to a white {4} vertices of S5 . By pigeon-hole principle, at least one vertex in {y, z} is adjacent to a vertex in {r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1 be the {1} hamiltonian path of S5 joining the black vertex (e)5 to the white vertex (v)5 , T2 be the {2} hamiltonian path of S5 joining the black vertex (w)5 to the white vertex (p)5 , and T3 be {3} the hamiltonian path of S5 joining the black vertex (x)5 to the white vertex (q)5 . We set H1 H2 H3 H4
= = = =
he, (e)5 , T1 , (v)5 , vi, he, P1 , w, (w)5, T2 , (p)5 , p, Q−1 1 , vi, 5 5 −1 he, P2 , x, (x) , T3 , (q) , q, Q2 , vi, and he, P3 , y, r, Q−1 3 , vi. 14
Obviously, {H1 , H2 , H3 , H4 } forms a 4∗ -container of S5 between e and v. Moreover, l(H1 ) = 25, l(H2 ) = l(H3 ) = 29, and l(H4 ) = 39. Thus, ds4L (e, v) ≤ 41. Subcase 2.2: (v)1 = a ∈ {2, 3}. Let b be the only element in {2, 3} − {a}. By Lemma 8, {5} there exist four paths P1 , P2 , P3 , and P4 of S5 such that (1) P1 joins e to a white vertex w with (w)1 = a and l(P1 ) = 2, (2) P2 joins e to a white vertex x with (x)1 = b and l(P2 ) = 2, (3) P3 joins e to a black vertex y with (y)1 = 4 and l(P3 ) = 19, (4) P4 joins e to a black {5} vertex z 6= y with (z)1 = 4 and l(P4 ) = 19, (5) P1 ∪ P2 ∪ P3 spans S5 , and (6) P1 ∪ P2 ∪ P4 {5} spans S5 . {4}
Again, there exist four paths Q1 , Q2 , Q3 , and Q4 of S5 such that (1) Q1 joins v to a black vertex p with (p)1 = 1 and l(Q1 ) = 2, (2) Q2 joins v to a black vertex q with (q)1 = b and l(Q2 ) = 2, (3) Q3 joins v to a white vertex r with (r)1 = 5 and l(Q3 ) = 19, (4) Q4 joins {4} v to a white vertex s 6= r with (s)1 = 5 and l(Q4 ) = 19, (5) Q1 ∪ Q2 ∪ Q3 spans S5 , and {4} (6) Q1 ∪ Q2 ∪ Q4 spans S5 . {5}
By Lemma 1, there are exactly three edges joining a black vertices of S5 to a white {4} vertices of S5 . By pigeon-hole principle, at least one vertex in {y, z} is adjacent to a vertex in {r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1 be the {1} hamiltonian path of S5 joining the black vertex (e)5 to the white vertex (p)5 , T2 be the {a} hamiltonian path of S5 joining the black vertex (w)5 to the white vertex (v)5 , and T3 be {b} the hamiltonian path of S5 joining the black vertex (x)5 to the white vertex (q)5 . We set H1 H2 H3 H4
= = = =
he, (e)5 , T1 , (p)5 , p, Q−1 1 , vi, 5 he, P1 , w, (w) , T2 , (v)5 , vi, he, P2 , x, (x)5 , T3 , (q)5 , q, Q−1 2 , vi, and −1 he, P3 , y, r, Q3 , vi.
Obviously, {H1 , H2 , H3 , H4 } forms a 4∗ -container of S5 between e and v. Moreover, l(H1 ) = l(H2 ) = 27, l(H3 ) = 29, and l(H4 ) = 39. Thus, ds4L (e, v) ≤ 41. 2 L Lemma 12 dsn−1 (u, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 =
Proof. odd.
n! n−2
+ 1 for n ≥ 6.
Let u be any white vertex and v be any black vertex of Sn . Obviously, d(u, v) is
Case 1: d(u, v) = 1. Since the star graph is vertex transitive and edge transitive, we may assume that u = e and v = (e)n . {n}
By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 2 and P1 = n(n − 2)! − 1, (2) Pi joins e to a white vertices {n} n−2 Pi spans Sn . xi with (xi )1 = i + 1 and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪i=1 {1} And there exist n − 2 paths Q1 , Q2 , . . . , Qn−2 of Sn such that (1) Q1 joins (e)n to a white vertices y1 with (y1 )1 = 2 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins (e)n to a black vertices yi 15
{1}
n−2 with (yi )1 = i + 1 and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪i=1 Qi spans Sn . By {2} Theorem 1, there exists a hamiltonian path R1 of Sn joining the white vertex (x1 )n to the {i+1} black vertex (y1 )n and there exists a hamiltonian path Ri of Sn joining the black vertex n n (xi ) to the white vertex (yi ) for 2 ≤ i ≤ n − 3. n n We set Pi = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , (e) i for 1 ≤ i ≤ n−2 and Pn−1 = he, (e) i. Then P1 , P2 , . . . , Pn−1 forms an (n − 1)∗ -container between e and (e)n . Obviously, l(P1 ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! − 1, l(Pi ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 for 2 ≤ i ≤ n − 2, L and l(Pn−1 ) = 1. Hence dsn−1 (e, (e)n ) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1.
Case 2: d(u, v) ≥ 3. Since d(u, v) ≥ 3, we may assume that u = e, (v)n = n − 1, and (v)1 6= n. {n}
Subcase 2.1: (v)1 = 1. By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 1 and l(P1 ) = n(n − 2)! − 1, (2) Pi joins e to a white vertex xi with (xi )1 = i and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and {n} {n−1} n−2 (3) ∪i=1 Pi spans Sn . Again there exist (n − 2) paths Q1 , Q2 , . . . , Qn−2 of Sn such that (1) Q1 joins v to a white vertex y1 with (y1 )1 = 1 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins v to a black vertex yi with (yi )1 = i and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪n−2 i=1 Qi {n−1} spans Sn . {1}
By Lemma 6, there exist two disjoint paths H1 and H2 of Sn such that (1) H1 joins the white vertex (x1 )n to the black vertex (y1 )n , (2) H2 joins the white vertex (e)n to the black {1} vertex (v)n , and (3) H1 ∪ H2 spans Sn . For 2 ≤ i ≤ n − 2, there exists a hamiltonian path {i} Ri of Sn joining the black vertex (xi )n to the white vertex (yi )n . We set T1 = he, P1 , x1 , (x1 )n , H1 , (y1 )n , y1 , Q−1 j , vi, Ti = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , vi for 2 ≤ i ≤ n − 2, and n n Tn−1 = he, (e) , H1 , (v) , vi. Obviously, {T1 , T2 , . . . , Tn−1 } forms an (n − 1)∗ -container. Moreover, l(Ti ) ≤ (n − 1)! + L 2(n − 2)! + 2(n − 3)! + 1. Thus, dsn−1 (e, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1. Subcase 2.2: (v)1 = t 6= 1, n. By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of {n} Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 1 and l(P1 ) = n(n − 2)! − 1, (2) Pi joins e to a white vertex xi with (xi )1 = i and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, {n} {n−1} n−2 Pi spans Sn . Again there exist (n − 2) paths Q1 , Q2 , . . . , Qn−2 of Sn such and (3) ∪i=1 that (1) Q1 joins v to a white vertex y1 with (y1 )1 = 1 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins v to a black vertex yi with (yi )1 = i and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) {n−1} n−2 ∪i=1 Qi spans Sn . {t}
Let the black vertex w be the unique neighbor of the white vertex (v)n of Sn with {1} (w)1 = 1. By Lemma 6, there exist two disjoint paths H1 and H2 of Sn such that (1) H1 joins the white vertex (x1 )n to the black vertex (y1 )n , (2) H2 joins the white vertex (e)n to 16
{1}
the black vertex (w)n , and (3) H1 ∪ H2 span Sn . {t}
By Theorem 4, there exists a hamiltonian path Rt of Sn joining the black vertex (xt )n {i} to the white vertex (yt )n . By Theorem 1, there exists a hamiltonian path Ri of Sn joining the black vertex (xi )n to the white vertex (yi )n for 2 ≤ i ≤ n − 2 with i 6= t. We set T1 = he, P1 , x1 , (x1 )n , H1 , (y1 )n , y1 , Q−1 j , vi, Ti = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , vi for 2 ≤ i ≤ n − 2, and n n n Tn−1 = he, (e) , H2 , (w) , w, (v) , vi. Obviously, {T1 , T2 , . . . , Tn−1 } forms an (n − 1)∗ -container. Moreover, l(Ti ) ≤ (n − 1)! + L 2(n − 2)! + 2(n − 3)! + 1. Thus, dsn−1 (e, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1. 2 It is easy to check that = 1 and = 3. Using computer, we have that D3sL (S4 ) = 15 by brute force. By Lemmas 10, 11, and 12, we have the following theorem: D1sL (S2 )
D2sL (S3 )
sL Theorem 5 Dn−1 (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! for n ≥ 5.
4
The 2∗L -diameter Sn
Lemma 13 Assume that u be any white vertex of S4 . Let a and b are two distinct elements of h4i. There exist two paths P1 and P2 of S4 such that (1) P1 joins u to a black vertex x with (x)1 = a and l(P1 ) = 7, (2) P2 joins u to a white vertex y with (y)1 = b and l(P2 ) = 16, and (3) P1 ∪ P2 spans S4 . Proof. Without loss of generality, we may assume that u = e. The required two paths are listed below. P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2
= = = = = = = = = = = = = = = = = = = = = = = =
(1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(3421)(2341)(3241)(4231)(2431) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(4231)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(2431)(4231)(3241)(2341)(4321) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(4231)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(2314)(1324)(3124)(2134)(4132)(3142)(1342) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(2134)(3124)(1324)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(4321)(3421)(2431)(4231)(3241) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(2134)(3124)(1324)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(3241)(4231)(2431)(3421)(4321) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(4231)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(3124)(1324)(2314)(3214)(4213)(2413)(1423) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(3421)(2341)(3241)(4231)(2431) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(2431)(4231)(3241)(2341)(4321) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(1342)(2341)(3241)(1243)(4213)(2413)(3412)(1432)(4132)(3142)(2143)(4123)(1423) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(3241)(1243)(4213)(2413)(1423)(4123)(2143) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(1342)(2341)(3241)(1243)(4213)(2413)(1423)(4123)(2143)(3142)(4132)(1432)(3412)
2
17
Theorem 6 3 sL 15 D2 (Sn ) = n! 2
if n = 3, if n = 4, and + 1 if n ≥ 5.
D2sL (S3 )
Proof. It is easy to check that = 3. Using computer, we have that D2sL (S4 ) = 15 by brute force. Thus, we assume that n ≥ 5. Let u be a white vertex and v be a black vertex of Sn . Let P1 and P2 be any two 2∗ -container of Sn joining u to v. Obviously, + 1. Hence, ds2L (u, v) ≥ n!2 + 1 and D2sL (Sn ) ≥ n!2 + 1. Hence we only max{l(P1 ), l(P2 )} ≥ n! 2 need to show that ds2L (u, v) ≤ n!2 + 1. Without loss of generality, we assume that u = e and {n−1} v ∈ Sn . {5}
Case 1: n = 5. By Lemma 13, there exist two paths H1 and H2 of S5 such that (1) H1 joins e to a black vertex x with (x)1 = 1 and l(H1 ) = 7, (2) H2 joins e to a white vertex {5} y with (y)1 = 3 and l(H2 ) = 16, and (3) H1 ∪ H2 spans S5 . Again, there exist two paths {4} T1 and T2 of S5 such that (1) T1 joins v to a white vertex p with (p)1 = 2 and l(T1 ) = 7, {4} (2) T2 joins v to a black vertex q with (q)1 = 3 and l(T2 ) = 16, and (3) T1 ∪ T2 spans S5 . {1,2} By Lemma 3, there exists a hamiltonian path W of S5 joining the white vertex (x)5 to {3} the black vertex (p)5 . Moreover, there exists a hamiltonian path Z of S5 joining the black vertex (y)5 to the white vertex (q)5 . We set L1 = he, H1 , x, (x)n , W, (p)n , p, T1−1 , vi and L2 = he, H2 , y, (y)n , Z, (q)n , q, T2−1 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = 61 and l(L2 ) = 59. Hence + 1. ds2L (u, v) ≤ n! 2 {n−1}
Case 2: n ≥ 6 is even. Let x be any neighborhood of v in Sn with (x)1 6= n. Obviously, x is a white vertex. Let j be any index in hni − {(x)1 , n − 1}. Obviously, (e)j is a black vertex. Let a1 , a2 , . . . , an−2 be a permutation of hn − 2i such that a1 = j and (x)1 = a3 . By {n} Theorem 2, there exists a hamiltonian path P of Sn − {(e)j } joining e to a white vertex p with (p)1 = a1 and (p)2 = an−2 . By Lemma 4, there exists a hamiltonian path H of {a } Sn 1 − {(p)n , ((p)n )2 } joining the white vertex ((e)j )n to a black vertex y with (y)1 = a3 . Let H = {a1 , a2 , . . . , a n2 } and H 0 = hn − 2i − H. By Lemma 3, there exists a hamiltonian H−{a }
1 path T of Sn joining the white vertex (y)n to the black vertex (x)n . By Theorem 1, {a } there exists a hamiltonian path H 0 of Sn n−2 joining the black vertex (((p)n )2 )n to a white vertex q with (q)1 = n − 3 and (q)n 6= v. By Theorem 2, there exists a hamiltonian path {n−1} T 0 of Sn − {x} joining the black vertex (q)n to v. We set
L1 = he, (e)j , ((e)j )n , H, y, (y)n, T, (x)n , x, vi and L2 = he, P, p, (p)n, ((p)n )2 , (((p)n )2 )n , H 0 , q, (q)n , T 0 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = n!2 + 1 and l(L2 ) = Hence ds2L (u, v) ≤ n!2 + 1. See Figure 4 for illustration the case n = 6. 18
n! 2
− 1.
4
(((p) 6)2)6
e (e)
S6{5}
S6{a }
S6{6} j
q
(q) 6
v x
p
S6{a }
S6{a }
1
S6{a }
3
2
(p) 6 ((p) 6)2 ((e) j)6
y
(y) 6
(x) 6
Figure 4: Illustration for the case 2 of Theorem 6.
{n−1}
Case 3: n ≥ 7 is odd. Let x be any neighborhood of v in Sn with (x)1 6= n. Obviously, x is a white vertex. Let j be any index in hni − {(x)1 , n − 1}. Obviously, (e)j is a black vertex. Let a1 , a2 , . . . , an−2 be a permutation of hn − 2i such that a1 = j and {n} (x)1 = a2 . By Theorem 2, there exists a hamiltonian path P of Sn − {(e)j } joining e to a white vertex p with (p)1 = a1 and (p)2 = an−3 . By Lemma 4, there exists a hamiltonian {a } path R of Sn 1 − {(p)n , ((p)n )2 } joining the white vertex ((e)j )n to a black vertex y with n−1
−1
{(a ,a )}
2 (y)1 = an−2 and (y)n−1 = a2 . Let X be the subgraph of Sn induced by ∪i=2 Sn i n and {(a ,a )} {(n−1,a )} {(n,a )} n i n n−3 ∪ Sn ∪ Sn n . By Lemma 3, Y be the subgraph of Sn induced by ∪i= n−1 Sn 2
there exists a hamiltonian path W of X joining the white vertex (y)n to a black vertex z with (z)1 = a3 . Let H = {a1 , a2 , . . . , a n−1 } and H 0 = hn − 3i − H. Again, there exists a H−{a }
2
1 hamiltonian path T of Sn joining the white vertex (z)n to the black vertex (x)n . By {H 0 } Theorem 3, there exists a hamiltonian path R0 of Sn joining the black vertex (((p)n )2 )n to a white vertex q with (q)1 = an−2 and (q)n−1 = n. By Lemma 3, there exists a hamiltonian path W 0 of Y joining the white vertex (q)n to a black vertex r with (r)1 = n − 1 and r 6= v. {n−1} By Theorem 2, there exists a hamiltonian path T 0 of Sn − {x} joining the black vertex n (r) ) to the black vertex v. We set
L1 = he, (e)j , ((e)j )n , R, y, (y)n , W, z, (z)n , T, (x)n , x, vi and L2 = he, P, p, (p)n , ((p)n )2 , (((p)n )2 )n , R0 , q, (q)n , W 0 , r, (r)n , T 0 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = n!2 + 1 and l(L2 ) = Hence ds2L (u, v) ≤ n!2 + 1. See Figure 5 for illustration the case n = 7.
19
n! 2
− 1. 2
S7{7}
4
(((p) 7)2)7
e (e)
S7{6}
S7{a } j
p
q
S7{a }
v
5
(q)
x
7
r
S7{a }
(r)7
S7{a }
1
3
(p) 7
2
z
(y) 7 7 2
((p) ) ((e) j)7
S7{a }
(z)n
y
(x) 7
Figure 5: Illustration for the case 3 of Theorem 6.
5
Conclusion
sL n! In this paper, we prove that Dn−1 (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 = n−2 + 1 and sL sL n! n! D2 (Sn ) = 2 + 1 for n ≥ 5. Actually, we prove that d2 (u, v) = 2 + 1 for any two vertices u and v from different bipartite set of Sn .
Recently, we have proved that Sn is super spanning laceable [10]. Hence we will study for 3 ≤ k ≤ n−1. We conjecture that DksL (Sn ) = n! +1 for n ≥ 5 and 2 ≤ k ≤ n−2. k
DksL (Sn )
References [1] S. B. Akers and B. Krishnameurthy, ”A group-theoretic model for symmetric interconnection networks,” IEEE Trans. Comput., pp. 555–566, 1989. [2] M. Albert, R.E.L. Aldred, and D. Holton, “On 3*-connected graphs,” Australasian Journal of Combinatorics, 24, 193-208, 2001. [3] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, North Holland, New York, (1980). [4] A. Bouabdallah, M. C. Heydemann, J. Opatmy, and D. Sotteau, “Embedding complete binary trees into star and pancake graphs”, Theoory Comput. Syst., pp. 279–305, 1998. [5] K. Day and A. Tripathi, ”A comparative study of topological properties,” IEEE Trans. Parallel Distrib. Syst., vol. 5, no. 1, pp. 31–38, 1994. [6] L. Gargano, U. Vaccaro, and A. Vozella, “Fault tolerant routing in the star and pancake interconnection networks,” Inform Process. Lett., pp. 315–320, 1993. 20
[7] D. Frank Hsu, “On container width and length in graphs, groups and networks”, IEICE Trans, Fundamentals, vol. E77-A, no. 4, pp. 668-680, 1994. [8] Sun-Yuan Hsieh, Gen-Huey Chen, and Chin-Wen Ho, “Hamiltonian-laceability of star graphs”, Networks, vol. 36, no. 4, pp. 225-232, 2000. [9] Tseng-Kuei Li, Jimmy J. M. Tan, and Lih-Hsing Hsu, “Hamiltonian laceability on edge fault star graph”, Submitted. [10] Cheng-Kuan Lin, Hua-Min Huang, and Lih-Hsing Hsu, “The super connectivity of the pancake graphs and gtar graphs”, Submitted. [11] K. Menger, “Zur allgemeinen Kurventheorie”, Fund. Math., 10, pp. 95–115, 1927. [12] K. Qiu, H. Meijer, and S. G. Alk, ”Decomposing a star graph into disjoint cycles,” Information Processing Letters, vol. 39, no. 4, pp. 125–129, 1991. [13] Y. Rouskov and P. K. Srimani, ”Fault diameter of star graphs,” Information Processing Letters, vol. 48, no. 5, 1993.
21
Keywords: diameter, hamiltonian, hamiltonian laceable, star graphs.
1
Basic Definitions
For the graph definition and notation we 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. For a vertex u, N (u) denotes the neighborhood of u which is the set {v | (u, v) ∈ E}. For any vertex x of V , deg G (x) denotes its degree in G. We use δ(G) to denote min{degG (x) | x ∈ V }, and we use ∆(G) to denote max{degG (x) | x ∈ V }. Two vertices u and v are adjacent if (u, v) ∈ E. A path is represented by hv0 , v1 , . . . , vk i. The length of a path Q, l(Q), is the number of edges in Q. We also write the path hv0 , v1 , . . . , vk i as hv0 , Q1 , vi , vi+1 , . . . , vj , Q2 , vt , . . . , vk i, where Q1 is the path hv0 , v1 , . . . , vi i and Q2 is the path hvj , vj+1 , . . . , vt i. Hence, it is possible to write a path as hv0 , v1 , Q, v1 , v2 , . . . , vk i if l(Q) = 0. 1
We use d(u, v) to denote the distance between u and v, i.e., the length of the 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 vertices of G. A graph G is hamiltonian connected if there exists a hamiltonian path joining any two distinct vertices of G. A cycle is a path with at least three vertices such that the first vertex is the same as the last vertex. A hamiltonian cycle of G is a cycle that traverses 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 [11] that there are k internally vertex-disjoint (abbreviated as disjoint) paths joining any two distinct vertices u and v when k ≤ κ(G). A container C(u, v) between two distinct vertices u and v in G is a set of internally disjoint paths {P1 , P2 , . . . , Pk } between u and v. The width of C(u, v) is k. A w-container is a container of width w. The length of a C(u, v) = {P1 , . . . , Pk }, l(C(u, v)), is max{l(Pi ) | 1 ≤ i ≤ k}. 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 6= v}. In particular, the wide diameter of G is Dκ(G) (G). The wide diameter is an important measure for interconnection networks [7]. In this paper, we are interested in a specifies type of containers. We say that a wcontainer C(u, v) is a w ∗ -container if every vertex 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 vertices 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 graph is motivated by the globally 3∗ -connected graphs proposed by Albert, Aldred, and Holton [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) ≤ δ(G) ≤ ∆(G). A graph G is super spanning connected if G is w ∗ -connected for any w with 1 ≤ w ≤ κ(G). In [10], it is proved that the pancake graph Pn is super spanning connected if and only if n 6= 3. A graph G is bipartite if its vertex set can be partitioned into two subsets V1 and V2 such that every edge joins vertices between V1 and V2 . Let G be a k-connected bipartite graph with bipartition V1 and V2 such that | V1 |≥| V2 |. Suppose that there exists a k ∗ -container C(u, v) = {P1 , P2 , . . . , Pk } in a bipartite graph G joining u to v with u, v ∈ V1 . Since the length of Pi = 2ki + 1 is an odd integer for all 1 ≤ i ≤ k, there are ki − 1 vertices of Pi in V1 P other than u and v, and ki vertices of Pi in V2 . As a consequence, | V1 |= ki=1 (ki − 1) + 2 P and | V2 |= ki=1 ki . Therefore, any bipartite graph G with κ(G) ≥ 3 is not k ∗ -connected. For this reason, we define that a bipartite graph is k ∗ -laceable if there exists a k ∗ -container between any two vertices from different partite sets. Obviously, any k ∗ -laceable graph with k ≥ 1 is a bipartite graph with the equal size of bipartition. An 1∗ -laceable graph is also known as hamiltonian laceable graph. Moreover, a bipartite graph is 2 ∗ -laceable if and only if it is hamiltonian. Obviously, all 1∗ -laceable graphs except K1 and K2 are 2∗ -laceable. A k-regular bipartite graph G is super spanning laceable if G is i∗ -laceable for all 1 ≤ i ≤ κ(G). It was proved in [10] that the star graph Sn is super spanning laceable if and only if n 6= 3.
2
We also define the w ∗ -lacable distance between any two vertices u and v from different partite sets, dswL (u, v), is min{l(C(u, v)) | C(u, v) is a w ∗ -container}. The w ∗L -diameter of G, denoted by DwsL (G), as max{dswL (u, v) | u and v are vertices from different partite sets}. sL In particular, the wide spanning diameter of G is Dκ(G) (G). In this paper, we prove that sL n! Dκ(Sn ) (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1 = n−2 + 1 and D2sL (Sn ) = n! + 1. 2 In Section 2, we give the definition of the star graphs and introduce some basic properties sL n! of star graphs. Then we prove that Dκ(S (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1 = n−2 +1 n) sL n! in Section 3. In Section 4, we prove that D2 (Sn ) = 2 + 1. We conclude the paper in the final section.
2
The star graphs and some notation conventions
Let n be a positive integer. We use hni to denote the set {1, . . . , n}. The n-dimensional star graph, denoted by Sn , is a graph with the vertex set V (Sn ) = {u1 . . . un | ui ∈ hni and ui 6= uj for i 6= j}. The adjacent is defined as follows: u1 . . . ui . . . un is adjacency to v1 . . . vi . . . vn through an edge of dimension i with 2 ≤ i ≤ n if vj = uj for j ∈ / {1, i}, v1 = ui , and vi = u1 . The star graphs S2 , S3 , and S4 are illustration in Figure 1. The star graphs are an important family of interconnection networks proposed by Akers and Krishnameurthy [1]. Some interesting properties of star graphs are studied [4, 5, 8, 12, 13]. It is know that the connectivity of Sn is n − 1. We use bold face to denote vertices in Sn . Hence u1 , u2 , . . . , un is denotes a sequence of vertices in Sn .
a 12
e
21
1324
f
S2 g
123 321
312
231
c
2134
3142
d
1423
4213
1234
4132
1342
4123
1243
4231
1432
4312
2143
b
213
3124
2314 3214
2413
b
c
3241
2431
2341
3421 4321
d
3412
e
f
g
a
132 S3
S4
Figure 1: The star graphs S2 , S3 , and S4 . By definition, Sn is an (n − 1)-regular graph with n! vertices. Moreover, it is vertex transitive and edge transitive. We use e to denote the element 12 . . . n. It is known that Sn 3
is a bipartite graph with one partite set containing those vertices corresponds to odd permutations and the other partite set containing those vertices corresponds to even permutations. We will use white vertices to represent those even permutation vertices and use black vertices to represent those odd permutation vertices. Let u = u1 u2 . . . un be any vertex of the star graph Sn . We say that ui is 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 {i} unique i-neighbor of u. Obviously, ((u)i )i = u. For 1 ≤ i ≤ n, let Sn denote the subgraph of Sn induced by those vertices u with (u)n = i. Obviously, Sn can be decomposed into n {i} {i} subgraph Sn , 1 ≤ i ≤ n, and each Sn is isomorphic to Sn−1 . Thus, the star graph can be constructed recursively. Let I ⊆ hni. We use SnI to denote the subgraph of Sn induced by {i} {j} ∪i∈I V (Sn{i} ). For 1 ≤ i 6= j ≤ n, we use E i,j to denote the set of edges between Sn and Sn . {(i,j)} For 1 ≤ i 6= j ≤ n, we use Sn to denote the subgraph of Sn induced by those vertices u {(i,j)} {(j,i)} {(i,j)} with (u)n−1 = i and (u)n = j. Obviously, Sn 6= Sn and Sn is isomorphic Sn−2 . Obviously, we have the following lemmas. Lemma 1 | E i,j |= (n − 2)! for any 1 ≤ i 6= j ≤ n. Moreover, there are (n − 2)!/2 edges {i} {j} joining black vertices of Sn to white vertices of Sn . Lemma 2 Assume that u and v are any two distinct vertices of Sn with 1 ≤ d(u, v) ≤ 2. Then (u)1 6= (v)1 . Theorem 1 [8] Sn is hamiltonian laceable if and only if n 6= 3. Theorem 2 [8] For any black vertex w and any two distinct white vertices u, v of S n with n ≥ 4. There exists a hamiltonian path of Sn − {w} joining u to v. Lemma 3 Assume that n ≥ 5 and I = {i1 , i2 , . . . , im } is a nonempty subset of hni. Let u be a white vertex and v be a black vertex of SnI with (u)n = i1 and (v)n = im . Then there exists a hamiltonian path of SnI joining u to v. {i }
Proof. Obviously, Sn j is isomorphic to Sn−1 for all 1 ≤ j ≤ m. By Theorem 1, this lemma is ture on m = 1. Thus, we assume that m ≥ 2. We set that x1 as u and set that ym as v. By Lemma 1, | E ij ,ij+1 |= (n − 2)!. We choose (yj , xj+1 ) ∈ E ij ,ij+1 with yj is a black vertex and xj+1 is a white vertex for 1 ≤ j ≤ m − 1. By Theorem 1, there is a hamiltonian path {i } Pj of Sn j joining xj to yj for all 1 ≤ j ≤ m. Then hx1 , P1 , y1 , x2 , P2 , y2 , . . . , xm , Pm , ym i forms a hamiltonian path of SnI joining u to v. 2 Lemma 4 Assume that r and s are any two adjacent vertices of Sn with n ≥ 4. Then there exists a hamiltonian path of Sn − {r, s} joining any white vertex u to a black vertex v with (v)1 = i for any i ∈ hni. Proof. Since Sn is vertex transitive and edge transitive, we assume that r = e and s = (e)2 . {n} Obviously, r, s ∈ Sn . We prove this lemma by induction on n. The required hamiltonian paths of S4 − {1234, 2134} are listed below. 4
(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324) (3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134) (3124)(2134)(4132)(1432)(2431)(3421)(4321)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(4312)(3412)(2413)(1423)(4123)(2143)(3142) (3124)(2134)(4132)(1432)(2431)(3421)(4321)(1432)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(3142)(2143)(4123)(1423)(2413)(3412)(4312) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (4132)(3124)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142) (4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(2413)(1423)(4123)(2143)(3142)(1342)(4312) (1342)(4312)(3412)(2413)(1423)(4123)(2143)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432) (1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(2143)(4123)(1423)(2413) (1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142) (1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312) (2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324) (2314)(3214)(4213)(1243)(3241)(2341)(4321)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413) (2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214) (2314)(3214)(4213)(1243)(3241)(2341)(4321)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(1423)(2413)(3412)(4312)(1342)(3142)(2143)(4123) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413) (4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214) (4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(3142)(1342)(4312)(3412)(3412)(1423)(4123) (1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413)(4213)(3241)(2314)(1324)(3124)(2134)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413) (1423)(4123)(2143)(3142)(1342)(4312)(3412)(2413)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214) (1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123) (2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243) (2143)(3142)(1342)(4312)(3412)(2413)(1423)(4123)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134) (2143)(1243)(3241)(2341)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(3124)(2134)(4132)(3142) (2143)(1243)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123) (3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243) (3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(4321)(2341) (3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(2143)(4123)(1423)(2413)(3412)(4312)(1342)(3142) (3241)(2341)(4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(2143)(3142)(1342)(4312)(3412)(2413)(1423)(4123) (4321)(3421)(2431)(1432)(3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324) (4321)(3421)(2431)(1432)(3412)(4312)(1342)(3142)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(2143)(1243)(3241)(2341) (4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421) (4321)(3421)(2431)(1432)(4132)(2134)(3124)(1324)(2314)(3214)(4213)(1243)(3241)(2341)(1342)(3142)(2143)(4123)(1324)(2413)(3412)(4321) (2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(3412)(4312)(1342)(3142)(2143)(4123)(1423)(2413) (2431)(1432)(3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(4321)(3421) (2431)(3421)(4321)(2341)(3241)(1243)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(1432)(3412)(2413)(1423)(4123)(2143)(3142)(1342)(4312) (3412)(4312)(1342)(2341)(3241)(1243)(2143)(3142)(4132)(2134)(3124)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(4321)(3421)(2431)(1432) (3412)(4312)(1342)(3142)(4132)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134) (3412)(4312)(1342)(3132)(2143)(1243)(3241)(2341)(4321)(1324)(2314)(3214)(4213)(2413)(1423)(4123)(3124)(2134)(4132)(1432)(2431)(3421) (3412)(1432)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124)(2134)(4132)(3142)(1342)(4312)
Assume that this lemma is hold on Sk for all 4 ≤ k < n. We have the following cases: {n}
{n}
Case 1: u ∈ Sn . Obviously, Sn is isomorphism to Sn−1 . By induction, there exists a {n} hamiltonian path P of Sn − {r, s} joining u to a black vertex x with (x)1 = n − 1. Let v hn−2i is a black vertex of Sn with (v)1 = i. By Lemma 3, there exists a hamiltonian path Q of hn−1i Sn joining the white vertex (x)n to v. Then hu, P, x, (x)n , Q, vi forms the desired path. {k}
Case 2: u ∈ Sn for some k 6= n. By Lemma 1, there exists (n − 2)!/2 ≥ 3 edges {n} {k} joining black vertices of Sn to white vertices of Sn . We can choose a black vertex x of {k} {n} {k} Sn such that (x)n is a white vertex of Sn − {r, s}. Since Sn is isomorphism to Sn−1 , {k} by Theorem 1, there exists a hamiltonian path P of Sn joining u to x. By induction, {n} there exists a hamiltonian path Q of Sn − {r, s} joining (x)n to a black vertex y with hn−1i−{k,r} (y)1 = r ∈ hn − 1i − {k}. Let v is a black vertex of Sn with (v)1 = i. By Lemma 3, hn−1i−{k} there exists a hamiltonian path R of Sn joining the white vertex (y)n to v. Obviously, hu, P, x, (x)n , Q, y, (y)n , R, vi forms the desired path. 2 Theorem 3 Assume that n ≥ 5 and I = {i1 , i2 , . . . , im } is a nonempty subset of hni. Then SnI is hamiltonian laceable. Proof. Suppose that u is a white vertex and v is a black vertex of SnI . By Lemma 3, this theorem is true on either m = 1, or m ≥ 2 and (u)n 6= (v)n . Thus, we assume that m ≥ 2 and (u)n = (v)n . Without loss of generality, we assume that (u)n = (v)n = i1 . 5
Case 1: (v)1 ∈ I. Without loss of generality, we assume that (v)1 = im . Obviously, {i } {i } Sn 1 isomorphic to Sn−1 . By Theorem 2, there exists a hamiltonian path P of Sn 1 − {v} joining u to a white vertex x such that (x)1 = i2 . By Lemma 3, there exists a hamiltonian I−{i } path Q of Sn 1 joining the black vertex (x)n to the white vertex (v)n . Then the path hu, P, x, (x)n , Q, (v)n , vi forms a hamiltonian path of SnI joining u to v. Case 2: (u)1 ∈ / I and (v)1 ∈ / I. We can choose a white vertex y be a neighbor of v in {i1 } Sn with (y)n = im . Obviously, y 6= u. By Lemma 4, there exists a hamiltonian path P of {i } Sn 1 − {v, y} joining u to a black vertex x such that (x)n = i2 . By Lemma 3, there exists a I−{i } hamiltonian path Q of Sn 1 joining the white vertex (x)n to the black vertex (y)n . Then the path hu, P, x, (x)n , Q, (y)n , y, vi forms a hamiltonian path of SnI joining u to v. 2 Theorem 4 Assume that n ≥ 5. Sn −{r, s} is hamiltonian laceable for any adjacent vertices r and s. Proof. Since the Sn is vertex transitive and edge transitive, we assume that r = e and {n} s = (e)2 . Obviously, r, s ∈ Sn . Suppose that u is a white vertex and v is a black vertex of Sn − {r, s}. We want to find a hamiltonian path of Sn − {r, s} joining u to v. {n}
{n}
Case 1: u, v ∈ Sn . By Lemma 4, there exists a hamiltonian path P of Sn − {r, s} joining u to a black vertex y with (y)1 ∈ hn − 1i. Obviously, P can be write as hu, Q1 , x, v, Q2 , yi. (Note that l(Q1 ) = 0 if u = x and l(Q2 ) = 0 if v = y.) By Theorem 3, there exists a hn−1i hamiltonian path R of Sn joining the black vertex (x)n to the white vertex (y)n . Then hu, Q1 , x, (x)n , R, (y)n , y, (Q2 )−1 , vi forms a desired hamiltonian path. {k}
Case 2: u, v ∈ Sn for some k 6= n. By Theorem 1, there exists a hamiltonian path P of {k} Sn joining u to v. By Lemma 1, there exists (n − 2)!/2 ≥ 3 edges joining black vertices {k} {n} {k} of Sn to white vertices of Sn . We can choose a black vertex x of Sn with (x)1 = n and (x)n ∈ / {r, s}. Obviously, P can be write as hu, Q1 , x, y, Q2 , vi. (Note that l(Q1 ) = 0 if u = x and l(Q2 ) = 0 if v = y.) Obviously, d(x, y) = 1. By Lemma 2, (x)1 6= (y)1 . By {n} Lemma 4, there exists a hamiltonian path R of Sn − {r, s} joining the white vertex (x)n to a black vertex z with (z)1 ∈ hn − 1i − {k}. By Theorem 3, there exists a hamiltonian hn−1i−{k} path T of Sn joining the black vertex (z)n to the white vertex (y)n . Obviously, hu, Q1 , x, (x)n , R, z, (z)n , T, (y)n , y, Q2 , vi forms a desired hamiltonian path. {n}
{k}
Case 3: u ∈ Sn and v ∈ Sn for some k 6= n. By Lemma 4, there exists a hamiltonian {n} path P of Sn − {r, s} joining u to a black vertex x with (x)1 ∈ hn − 1i. By Theorem hn−1i 3, there exists a hamiltonian path Q of Sn joining the white vertex (x)n to v. Then hu, P, x, (x)n , Q, vi forms a desired hamiltonian path. {k}
{l}
Case 4: u ∈ Sn and v ∈ Sn with k, l, and n are distinct. By Lemma 1, there exists {k} {n} (n − 2)!/2 ≥ 3 edges joining black vertices of Sn to white vertices of Sn , we can choose {k} a black vertex x of Sn with (x)1 = n and (x)n ∈ / {r, s}. By Theorem 1, there exists a {k} hamiltonian path P of Sn joining u to x. By Lemma 4, there exists a hamiltonian path Q 6
{n}
of Sn − {r, s} joining the white vertex (x)n to a black vertex y with (y)1 ∈ hn − 1i − {k}. hn−1i−{k} By Lemma 3, there exists a hamiltonian path R of Sn joining the white vertex (y)n to v. Then hu, P, x, (x)n , Q, y, (y)n , R, vi forms a desired hamiltonian path. 2 Lemma 5 Assume that n ≥ 4. Let u be a white vertex of Sn and Fn = {(u)i | 3 ≤ i ≤ n} ∪ {((u)i )i−1 | 3 ≤ i ≤ n}. Then there exists a hamiltonian path of Sn − Fn joining u to a black vertex v with (v)1 = j for any j ∈ hni. Proof. We proved this lemma by induction on n. Since Sn is vertex transitive, we assume that u = e. The required hamiltonian paths of S4 − F4 are listed below. (1234)(2134)(3124)(4123)(1423)(3421)(2431)(1432)(4132)(3142)(2143)(1243)(4213)(2413)(3412)(4312)(1342)(2341)(4321)(1324) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432)(2431)(3421)(1423)(2413)(4213)(1243)(2143)(4123)(3124)(1324)(4321)(2341) (1234)(2134)(4132)(1432)(2431)(3421)(1423)(4123)(3124)(1324)(4321)(2341)(1342)(4312)(3412)(2413)(4213)(1243)(2143)(3142) (1234)(2134)(3124)(1324)(4321)(2341)(1342)(3142)(4132)(1432)(2431)(3421)(1423)(4123)(2143)(1243)(4213)(2413)(3412)(4312)
{n}
Assume that the lemma is true for any Sk with 4 ≤ k ≤ n − 1. Obviously, Sn is {n} isomorphic to Sn−1 . By induction, there exists a hamiltonian path P of Sn − Fn−1 joining {1} e to a black vertex x with (x)1 = 1. By Lemma 4, there exists a hamiltonian path Q of Sn − {(e)n , ((e)n )n−1 } joining the white vertex (x)n to a black vertex y with (y)1 = 2. We can hn−1i−{1} with (z)1 = i. By Theorem 3, there exists a hamiltonian choose a black vertex z of Sn hni−{1,n} path R of Sn joining the white vertex (y)n to z. Then he, P, x, (x)n, Q, y, (y)n , R, zi forms a desired hamiltonian path. 2 Lemma 6 Assume that n ≥ 5. Suppose that p and q are two different white vertices of S n , and r and s are two different black vertices of Sn . Then there exist two disjoint paths P1 and P2 such that (1) P1 joins p to r, (2) P2 joins q to s, and (3) P1 ∪ P2 spans Sn . Proof. Without loss of generality, we assume that (p)n = n and (q)n = n − 1. Let (r)n = i and (s)n = j. {i,n}
Case 1: i, j ∈ hn − 2i and i 6= j. By Theorem 3, there exists a hamiltonian path P1 of Sn hn−1i−{i} joining p to r. Again, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 form the desired paths. Case 2: i, j ∈ hn − 2i and i = j. We can choose a white vertex x be the neighborhood of r {i} hni−{i,n−1} in Sn with (x)1 6= n − 1. By Theorem 3, there exists a hamiltonian path P of Sn joining p to the black vertex (x)n . By Lemma 4, there exists a hamiltonian path Q of {i} Sn − {r, x} joining s to a black vertex y with (y)1 = n − 1. By Theorem 1, there exists a {n−1} hamiltonian path Q of Sn joining q to the black vertex (y)n . Then P1 = hp, P, (x)n , x, ri and P2 = hq, R, (y)n , y, Q−1 , si form the desired paths. Case 3: either i ∈ hn − 2i and j = n − 1, or i = n and j ∈ hn − 2i. By symmetric, we assume that i ∈ hn − 2i and j = n − 1. By Theorem 3, there exists a hamiltonian path P1 hni−{n−1} {n−1} of Sn joining p to r. By Theorem 1, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 are the desired paths. 7
Case 4: either i = n − 1 and j ∈ hn − 2i, or i ∈ hn − 2i and j = n. By symmetric, we assume that i = n − 1 and j ∈ hn − 2i. By Theorem 1, there exists a hamiltonian path R of {n−1} {n} Sn joining q to r. We can choose a white vertex x ∈ R with (x)n ∈ Sn . We can write {n} R as hq, R1 , y, x, R2 , ri. By Theorem 1, there exists a hamiltonian path W of Sn joining p to the black vertex (x)n . We set P1 as hp, W, (x)n , x, R2 , ri. Obviously, y is a black vertex hn−2i and (y)1 ∈ hn − 2i. By Theorem 3, there exists a hamiltonian path Q of Sn joining the n n white vertex (y) to s. We set P2 as hq, R1 , y, (y) , Q, si. Then P1 and P2 forms the desired paths. {n}
Case 5: i = n and j = n − 1. By Theorem 1, there exists a hamiltonian path P1 of Sn hn−1i joining p to r. By Theorem 3, there exists a hamiltonian path P2 of Sn joining q to s. Then P1 and P2 are the desired paths. {n}
Case 6: i = n−1 and j = n. We can choose a white vertex x ∈ Sn −{s} with (x)1 = n−1. {n} By Theorem 1, there exists a hamiltonian path R of Sn joining p to s. Again, there exists {n−1} a hamiltonian path Q of Sn joining q to r. We can write R as hp, R1 , x, y, R2 , si, n and write Q as hq, Q1 , w, (x) , Q2 , ri. Obviously, y is a black vertex and w is a white vertex. Since d(x, y) = 1, by Lemma 2, (y)1 6= (x)1 = n − 1. Since d((x)n , w) = 1, by hn−1i Lemma 2, (w)1 6= ((x)n )1 = n. By Theorem 3, there exists a hamiltonian path W of Sn joining the black vertex (w)n to the white vertex (y)n . Then P1 = hp, R1 , x, (x)n , Q2 , ri and P2 = hq, Q1 , w, (w)n, W, (y)n , y, R2 ri form a desired paths. Case 7: either i = j = n or i = j = n − 1. By symmetric, we assume that i = j = n. By {n} Theorem 1, there exists a hamiltonian path P of Sn joining p to s. We can write P as hp, R1 , r, x, R2 , si. We set P1 as hp, R1 , ri. Obviously, x is a white vertex and (x)1 ∈ hn − 1i. hn−1i By Theorem 3, there exists a hamiltonian path Q of Sn joining q to the black vertex (x)n . Then P1 and P2 = hq, Q, (x)n , x, R2 , qi are the desired paths. 2
3
The (n − 1)∗L -diameter of Sn
Lemma 7 Let u = u1 u2 u3 u4 be any white vertex of S4 . There exist three paths P1 , P2 , and P3 such that (1) P1 joins u to the black vertex u2 u4 u1 u3 with l(P1 ) = 7, (2) P2 joins u to the white vertex u3 u4 u1 u2 with l(P2 ) = 8, (3) P3 joins u to the white vertex u4 u1 u3 u2 with l(P3 ) = 8, and (4) P1 ∪ P2 ∪ P3 spans S4 . Proof.
Since S4 is vertex transitive, we assume that u = 1234. Then we set P1 = h1234, 3214, 4213, 1243, 2143, 4123, 1423, 2413i, P2 = h1234, 4231, 3241, 2341, 4321, 3421, 2431, 1432, 3412i, and P3 = h1234, 2134, 3124, 1324, 2314, 4312, 1342, 3142, 4132i.
Obviously, P1 , P2 , and P3 forms the desired paths.
8
2
Lemma 8 Let u = u1 u2 u3 u4 be any white vertex of S4 . Let i1 i2 i3 be a permutation of u2 , u3 , and u4 . There exist four paths P1 , P2 , P3 , and P4 of S4 such that (1) P1 joins u to a white vertex w with (w)1 = i1 and l(P1 ) = 2, (2) P2 joins u to a white vertex x with (x)1 = i2 and l(P2 ) = 2, (3) P3 joins u to a black vertex y with (y)1 = i3 and l(P3 ) = 19, (4) P4 joins u to a black vertex z with z 6= y, (z)1 = i3 , and l(P4 ) = 19, (5) P1 ∪ P2 ∪ P3 spans S4 , and (6) P1 ∪ P2 ∪ P4 spans S4 . Proof. Since S4 is vertex transitive, we assume that u = 1234. Without loss of generality, we suppose that i1 < i2 . The required four paths are listed below. P1 P2 P3 P4
= = = =
(1234)(2134)(3124) (1234)(3214)(4213) (1234)(4231)(3241)(1243)(2143)(4123)(1423)(2413)(3412)(4312)(2314)(1324)(4321)(3421)(2431)(1432)(4132)(3142)(1342)(2341) (1234)(4231)(3241)(1243)(2143)(4123)(1423)(3421)(2431)(1432)(4132)(3142)(1342)(2341)(4321)(1324)(2314)(4312)(3412)(2413)
P1 P2 P3 P4
= = = =
(1234)(4231)(2431) (1234)(3214)(4213) (1234)(2134)(3124)(4123)(2143)(1243)(3214)(2314)(1342)(4312)(2314)(1324)(4321)(3421)(1423)(2413)(3412)(1432)(4132)(3142) (1234)(2134)(3142)(1324)(2314)(4312)(1342)(3142)(4132)(1432)(3412)(2413)(1423)(4123)(2143)(1243)(3241)(2341)(4321)(3421)
P1 P2 P3 P4
= = = =
(1234)(4231)(2431) (1234)(2134)(3124) (1234)(3214)(2314)(1324)(4321)(3421)(1423)(4123)(2143)(3142)(4132)(1432)(3412)(2413)(4213)(1243)(3241)(2341)(1342)(4312) (1234)(3214)(2314)(1324)(4321)(3421)(1423)(2413)(4213)(1243)(3241)(2341)(1342)(4312)(3412)(1432)(4132)(3142)(2143)(4123)
2 Lemma 9 Assume that n ≥ 5 and i1 i2 . . . in−1 is an (n − 1)-permutation of hni. Let u be any white vertex of Sn . Then there exist (n − 1) paths P1 , P2 , . . . , Pn−1 of Sn such that (1) P1 is joins u to a black vertices y1 with (y1 )1 = i1 and l(P1 ) = n(n − 2)! − 1, (2) Pj is joins u to a white vertices yi with (yj )1 = ij and l(Pj ) = n(n − 2)! for 2 ≤ j ≤ n − 1, and (3) n−1 Pj spans Sn . ∪j=1 Proof. Since Sn is vertex transitive, we assume that u = e. Without loss of generality, we suppose that i2 < i3 < . . . < in−1 . Case 1: n = 5. Obviously, i2 6= 4, i3 = 6 1, and i4 6= 1. Let x1 = (e)5 , x2 = ((x1 )2 )5 , x3 = ((x2 )3 )5 , and x4 = ((x3 )4 )5 . Assume that i1 = 3. We set W1 = h12345, 32145, 42135, 12435, 21435, 41235, 14235, 24135i, W2 = h12345, 21345, 31245, 13245, 23145, 43125, 13425, 31425, 41325i, and W3 = h12345, 42315, 32415, 23415, 43215, 34215, 24315, 14325, 34125i. Let u1 = 24135, u2 = 41325, u3 = 34125, and u4 = ((x4 )3 )5 . Obviously, xi is a black vertex {i} {2} of S5 . By Lemma 4, there exists a hamiltonian path Q1 of S5 − {x2 , (x2 )3 } joining the white vertex (u1 )5 to a black vertex y1 with (y1 )1 = i1 . Again, there exists a hamiltonian {4} path Q2 of S5 − {x4 , (x4 )3 } joining the black vertex (u2 )5 to a white vertex y2 with {3} (y2 )1 = i2 . Moreover, there exists a hamiltonian path Q3 of S5 − {x3 , (x3 )4 } joining the black vertex (u3 )5 to a white vertex y3 with (y3 )1 = i3 . And, there exists a hamiltonian path 9
{2}
Q4 of S5 We set
− {x2 , (x2 )3 } joining the black vertex (u4 )5 to a white vertex y4 with (y4 )1 = i4 . P1 P2 P3 P4
= = = =
he, W1 , u1 , (u1 )5 , Q1 , y1 i, he, W2 , u2 , (u2 )5 , Q2 , y2 i, he, W3 , u3 , (u3 )5 , Q4 , y4 i. and he, x1 , (x1 )2 , x2 , (x2 )3 , x3 , (x3 )4 , x4 , (x4 )3 , u4 , Q3 , y3 i.
Obviously, l(P1 ) = 29, and l(Pi ) = 30 for 2 ≤ i ≤ 4. Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. Assume that i1 6= 3. Obviously, i3 6= 3. We set W1 = h12345, 21345, 41325, 14325, 34125, 43125, 13425, 31425i, W2 = h12345, 32145, 23145, 13245, 31245, 41235, 14235, 24135, 42135i, and W3 = h12345, 42315, 24135, 34215, 43215, 23415, 32415, 12435, 21435i. Let u1 = 31425, u2 = 42135, u3 = 21435, and u4 = ((x4 )3 )5 . Obviously, xi is a black vertex {i} {3} of S5 . By Lemma 4, there exists a hamiltonian path Q1 of S5 − {x3 , (x3 )4 } joining the white vertex (u1 )5 to a black vertex y1 with (y1 )1 = i1 . Again, there exists a hamiltonian {4} path Q2 of S5 − {x4 , (x4 )3 } joining the black vertex (u2 )5 to a white vertex y2 with {1} (y2 )1 = i2 . Moreover, there exists a hamiltonian path Q3 of S5 − {x1 , (x1 )2 } joining the black vertex (u4 )5 to a white vertex y3 with (y3 )1 = i3 . And, there exists a hamiltonian path {2} Q4 of S5 − {x2 , (x2 )3 } joining the black vertex (u3 )5 to a white vertex y4 with (y4 )1 = i4 . We set P1 P2 P3 P4
= = = =
he, W1 , u1 , (u1 )5 , Q1 , y1 i, he, W2 , u2 , (u2 )5 , Q2 , y2 i, he, x1 , (x1 )2 , x2 , (x2 )3 , x3 , (x3 )4 , x4 , (x4 )3 , u4 , Q3 , y3 i. and he, W3 , u3 , (u3 )5 , Q4 , y4 i.
Obviously, l(P1 ) = 29, and l(Pi ) = 30 for 2 ≤ i ≤ 4. Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. Case 2: n ≥ 6. We set that xj = (e)j and yj = ((e)j )j−1 for 3 ≤ j ≤ n − 2. Obviously, uj is a black vertex and vj is a white vertex for 3 ≤ j ≤ n − 2. {(n−1,n)}
Assume that i1 6= n − 3. By Lemma 5, there exists a hamiltonian path R of Sn − Fn−2 joining e to a black vertex x1 with (x1 )1 = 2. We recursively set xj as the unique {(j,n)} neighborhood of (xj−1 )n−1 in Sn with (xj )1 = j + 1 for 2 ≤ j ≤ n − 5, and we set {(n−4,n)} xn−4 as the unique neighborhood of (xn−5 )n−1 in Sn with (xn−4 )1 = n − 1. It is {(j+1,n)} easy to see that xj is a black vertex for 1 ≤ j ≤ n − 4 and {(xj )n−1 , xj+1 } ∈ Sn for 1 ≤ j ≤ n − 5. Choose xn−3 as any black vertex in Snn−1 with (xn−3 )1 = n and 10
{n−1}
(xn−3 )n−1 = n − 3. By Theorem 1, there exists a hamiltonian path R0 of Sn joinn ing the white vertex (xn−4 ) to xn−3 . Then we set the black vertex y1 as the unique {(n−3,n)} neighborhood of the white vertex (xn−3 )n in Sn with (y1 )1 = i1 . We set P1 as n−1 n−1 n−1 he, R, x1 , (x1 ) , x2 , (x2 ) , . . . , xn−5 , (xn−5 ) , xn−4 , (xn−4 )n , R0 , xn−3 , (xn−3 )n , y1 i. Obviously, l(P1 ) = n(n − 2)! − 1. {1}
By Theorem 1, there exists a hamiltonian path H of Sn joining the black vertex (e)n to the a white vertex z with (z)1 = n and (z)n−2 = n − 2. Obviously, there exists a hamiltonian {(n−2,n)} path H 0 of Sn joining the black vertex (z)n to a white vertex y2 with (y2 )1 = a2 . We n set P2 as he, (e) , H, z, (z)n , H 0 , y2 i. Obviously, l(P2 ) = n(n − 2)!. {(1,n)}
By Theorem 1, there exists a hamiltonian path W of Sn joining the black vertex n−1 (e) to a white vertex r with (r)1 = n − 1. Moreover, there exists a hamiltonian path W 0 {n−1} of Sn joining the black vertex (r)n to a white vertex y3 with (y3 )1 = a3 . We set P3 as n−1 he, (e) , W, r, (r)n , W 0 , y3 i. Obviously, l(P3 ) = n(n − 2)!. Let i be any index with 4 ≤ i ≤ n − 2. By Lemma 4, there exists a hamiltonian {(i−2,n)} path Qi of Sn − {(xi−3 )n−1 , xi−2 } joining the black vertex (vi−1 )n−1 to a white ver{i−2} tex zi with (zi )1 = i − 2. Again, there exists a hamiltonian path Q0i of Sn joinn ing the black vertex (zi ) to a white vertex yi with (yi )1 = ai . Then we set Pi as he, ui−1 , vi−1 , (vi−1 )n−1 , Qi , zi , (zi )n , Q0i , yi i. Obviously, l(Pi ) = n(n − 2)!. {(n−3,n)}
By Lemma 4, there exists a hamiltonian path R of Sn − {(xn−3 )n , y1 } joining the black vertex (vn−3 )n−1 to a white vertex r with (r)1 = n − 2. Obviously, there ex{n−2} ists a hamiltonian path R0 of Sn joining the black vertex (r)n to a white vertex yn−1 with (yn−1 )1 = an−1 . Hence we set Pn−1 as he, un−1 , vn−1 , (vn−1 )n−1 , R, r, (r)n, R0 , yn−1 i. Obviously, l(Pn−1 ) = n(n − 2)!. Apparently, P1 , P2 , . . . , Pn−1 form the desired paths. See Figure 2 for illustration for the case n = 6. {(n−1,n)}
Assume that i1 = n − 3. By Lemma 5, there exists a hamiltonian path R of Sn − Fn−2 joining e to a black vertex x1 with (x1 )1 = n − 3. We recursively set xj as the unique {(n−j−1,n)} neighborhood of (xj−1 )n−1 in Sn with (xj )1 = n − j − 1 for 2 ≤ j ≤ n − 5, and we set {(3,n)} xn−4 as the unique neighborhood of (xn−5 )n−1 in Sn with (xn−4 )1 = n−1. It is easy to see {n−j−2,n} n−1 that xj is a black vertex for 1 ≤ j ≤ n − 4 and {(xj ) , xj+1 } ∈ Sn for 1 ≤ j ≤ n − 5. {n−1} 0 By Theorem 1, there exists a hamiltonian path R of Sn joining the white vertex (xn−4 )n to a black vertex xn−3 with (xn−3 )1 = n and (xn−3 )n−1 = 2. Then we set the black vertex y1 {(2,n)} as the unique neighborhood of the white vertex (xn−3 )n in Sn with (y1 )1 = i1 . We set P1 n−1 n−1 n−1 as he, R, x1 , (x1 ) , x2 , (x2 ) , . . . , xn−5 , (xn−5 ) , xn−4 , (xn−4 )n , R0 , xn−3 , (xn−3 )n , y1 i. Obviously, l(P1 ) = n(n − 2)! − 1. {1}
By Theorem 1, there exists a hamiltonian path H of Sn joining the black vertex (e)n to a white vertex z with (z)1 = n and (z)n−2 = n − 2. Again, there exists a hamiltonian 11
S6{6} e
S6{(5,6)}
S6{5}
S6 (x2)6
x1
{(2,6)}
x2 (x1)6
u3 v3 u4 v4
S6{1}
S6{(4,6)} y2
(e)6
(v 3)5
P1
(z)6
z4
S6{(3,6)}
x3
(x3)6
(v4)5
z
P2
S6{(1,6)}
(e)5 r
y1 z5
S6{2}
S6{4}
(z4)6
P4 y4
(r)6
S6{3}
P3
y5
(z 5)6
y3
P5
Figure 2: Illustration for the case 1 of Lemma 9.
{(n−2,n)}
path H 0 of Sn joining the black vertex (z)n to a white vertex y2 with (y2 )1 = a2 . We set P2 as he, (e)n , H, z, (z)n , H 0 , y2 i. Obviously, l(P2 ) = n(n − 2)!. {(1,n)}
joining the black vertex By Theorem 1, there exists a hamiltonian path W of Sn (e)n−1 to a white vertex r with (r)1 = n − 1. Moreover, there exists a hamiltonian path W 0 {n−1} of Sn joining the black vertex (r)n to a white vertex y3 with (y3 )1 = i3 . We set P3 as he, (e)n−1 , W, r, (r)n , W 0 , y3 i. Obviously, l(P3 ) = n(n − 2)!. Let j be any index with 4 ≤ j ≤ n − 2. By Lemma 4, there exists a hamiltonian {(j−1,n)} − {(xn−j−1 )n−1 , xn−j } joining the black vertex (vj )n−1 to a white path Qj of Sn {j−1} joinvertex zj with (zj )1 = j − 1. Again, there exists a hamiltonian path Q0j of Sn n ing the black vertex (zj ) to a white vertex yj with (yj )1 = ij . Then we set Pj as he, uj , vj , (vj )n−1 , Qj , zj , (zj )n , Q0j , yj i. Obviously, l(Pj ) = n(n − 2)!. {(2,n)}
By Lemma 4, there exists a hamiltonian path R of Sn − {(xn−3 )n , y1 } joining the n−1 black vertex (v3 ) to a white vertex r with (r)1 = n − 2. Obviously, there exists a {n−2} 0 joining the black vertex (r)n to a white vertex yn−1 with hamiltonian path R of Sn (yn−1 )1 = in−1 . Hence we set Pn−1 as he, u3 , v3 , (v3 )n−1 , R, r, (r)n , R0 , yn−1 i. Obviously, l(Pn−1 ) = n(n − 2)!. 12
Apparently, P1 , P2 , . . . , Pn−1 forms the desired paths. See Figure 3 for illustration for the case n = 6. The lemma is proved. 2 {6}
S6
e
{(5,6)}
S6
x1
{(2,6)} 6
S
{5} 6
S
x3
y1
(x3)6
u3 {(4,6)}
v3 u4 v4
S6
{1}
S6
y2
(e)6
(v3)5
P1
(z)6
z
P2
z4 (x2)6
{(3,6)}
S6
x2
{(1,6)}
(x1)6
S6
(e)5 r
(v4)5 z5
{2}
S6
{4}
S6
(z4)6 {3}
(r)6
S6
P4 y4
y5
(z5)6
y3
P3
P5
Figure 3: Illustration for the case 2 of Lemma 9.
sL Lemma 10 Dn−1 (Sn ) ≥
n! n−2
+ 1 = (n − 1)! + 2(n − 2)! + 2(n − 2)! + 1.
Proof. Let u and v are two adjacent vertices of Sn . Let P1 , P2 , . . . , Pn−1 be any (n − 1)∗ container of Sn joining u to v. Obviously, one of these path is hu, vi. Thus, max{l(Pi ) | 1 ≤ sL n! n! n! L i ≤ n − 1} ≥ n−2 + 1. Hence, dsn−1 (u, v) ≥ n−2 + 1 and Dn−1 (Sn ) ≥ n−2 + 1. 2 Lemma 11 D4sL (S5 ) ≤ 41. Proof. odd.
Let u be any white vertex and v be any black vertex of S5 . Obviously, d(u, v) is
Case 1: d(u, v) = 1. Since the S5 is vertex transitive and edge transitive, we may assume {5} that u = e = 12345 and v = (e)5 = 52341. By Lemma 7, there exist P1 , P2 , and P3 of S5 such that (1) P1 joins 12345 to the black vertex 24135 with l(P1 ) = 7, (2) P2 joins 12345 to the white vertex 34125 with l(P2 ) = 8, (3) P3 joins 12345 to the white vertex 41325 with {5} {1} l(P3 ) = 8, and (4) P1 ∪ P2 ∪ P3 spans S4 . Similarly, there exist Q1 , Q2 , and Q3 of S5 13
such that (1) Q1 joins 52341 to the white vertex 24531 with l(Q1 ) = 7, (2) Q2 joins 52341 to the black vertex 34521 with l(Q2 ) = 8, (3) Q3 joins 52341 to the black vertex 45321 with {1} l(Q3 ) = 8, and (4) Q1 ∪ Q2 ∪ Q3 spans S4 . By Theorem 1, there is a hamiltonian path {2} R1 of S5 joining the white vertex 54132 to the black vertex 14532, there is a hamiltonian {3} path R2 of S5 joining the black vertex 34125 to the white vertex 41324, and there is a {4} hamiltonian path R3 of S5 joining the black vertex 51324 to the white vertex 15424. Then we set T1 T2 T3 T4
= = = =
h12345, P1, 24135, 54132, R1, 14532, 24531, (Q1)−1 , 52341i, h12345, P2, 34125, 54123, R2, 14523, 34521, (Q2)−1 , 52341i, h12345, P3, 41325, 51324, R3, 15324, 45321, (Q3)−1 , 52341i, and h12345, 52341i.
Obviously, {T1 , T2 , T3 , T4 } forms a 4∗ -container of S5 between 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. Since d(u, v) ≥ 3, we may assume that u = e, (v)5 = 4, and (v)1 ∈ {1, 2, 3}. {5}
Subcase 2.1: (v)1 = 1. By Lemma 8, there exist four paths P1 , P2 , P3 , and P4 of S5 such that (1) P1 joins e to a white vertex w with (w)1 = 2 and l(P1 ) = 2, (2) P2 joins e to a white vertex x with (x)1 = 3 and l(P1 ) = 2, (3) P3 joins e to a black vertex y with (y)1 = 4 and l(P3 ) = 19, (4) P4 joins e to a black vertex z 6= y with (z)1 = 4 and l(P4 ) = 19, (5) {5} {5} P1 ∪ P2 ∪ P3 spans S5 , and (6) P1 ∪ P2 ∪ P4 spans S5 . {4}
Similarly, there exist four paths Q1 , Q2 , Q3 , and Q4 of S5 such that (1) Q1 joins v to a black vertex p with (p)1 = 2 and l(Q1 ) = 2, (2) Q2 joins v to a black vertex q with (q)1 = 3 and l(Q2 ) = 2, (3) Q3 joins v to a white vertex r with (r)1 = 5 and l(Q3 ) = 19, (4) P4 joins {4} v to a white vertex s 6= r with (s)1 = 5 and l(Q4 ) = 19, (5) Q1 ∪ Q2 ∪ Q3 spans S5 , and {4} (6) Q1 ∪ Q2 ∪ Q4 spans S5 . {5}
By Lemma 1, there are exactly three edges joining a black vertices of S5 to a white {4} vertices of S5 . By pigeon-hole principle, at least one vertex in {y, z} is adjacent to a vertex in {r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1 be the {1} hamiltonian path of S5 joining the black vertex (e)5 to the white vertex (v)5 , T2 be the {2} hamiltonian path of S5 joining the black vertex (w)5 to the white vertex (p)5 , and T3 be {3} the hamiltonian path of S5 joining the black vertex (x)5 to the white vertex (q)5 . We set H1 H2 H3 H4
= = = =
he, (e)5 , T1 , (v)5 , vi, he, P1 , w, (w)5, T2 , (p)5 , p, Q−1 1 , vi, 5 5 −1 he, P2 , x, (x) , T3 , (q) , q, Q2 , vi, and he, P3 , y, r, Q−1 3 , vi. 14
Obviously, {H1 , H2 , H3 , H4 } forms a 4∗ -container of S5 between e and v. Moreover, l(H1 ) = 25, l(H2 ) = l(H3 ) = 29, and l(H4 ) = 39. Thus, ds4L (e, v) ≤ 41. Subcase 2.2: (v)1 = a ∈ {2, 3}. Let b be the only element in {2, 3} − {a}. By Lemma 8, {5} there exist four paths P1 , P2 , P3 , and P4 of S5 such that (1) P1 joins e to a white vertex w with (w)1 = a and l(P1 ) = 2, (2) P2 joins e to a white vertex x with (x)1 = b and l(P2 ) = 2, (3) P3 joins e to a black vertex y with (y)1 = 4 and l(P3 ) = 19, (4) P4 joins e to a black {5} vertex z 6= y with (z)1 = 4 and l(P4 ) = 19, (5) P1 ∪ P2 ∪ P3 spans S5 , and (6) P1 ∪ P2 ∪ P4 {5} spans S5 . {4}
Again, there exist four paths Q1 , Q2 , Q3 , and Q4 of S5 such that (1) Q1 joins v to a black vertex p with (p)1 = 1 and l(Q1 ) = 2, (2) Q2 joins v to a black vertex q with (q)1 = b and l(Q2 ) = 2, (3) Q3 joins v to a white vertex r with (r)1 = 5 and l(Q3 ) = 19, (4) Q4 joins {4} v to a white vertex s 6= r with (s)1 = 5 and l(Q4 ) = 19, (5) Q1 ∪ Q2 ∪ Q3 spans S5 , and {4} (6) Q1 ∪ Q2 ∪ Q4 spans S5 . {5}
By Lemma 1, there are exactly three edges joining a black vertices of S5 to a white {4} vertices of S5 . By pigeon-hole principle, at least one vertex in {y, z} is adjacent to a vertex in {r, s}. Without loss of generality, we assume that y is adjacent to r. Let T1 be the {1} hamiltonian path of S5 joining the black vertex (e)5 to the white vertex (p)5 , T2 be the {a} hamiltonian path of S5 joining the black vertex (w)5 to the white vertex (v)5 , and T3 be {b} the hamiltonian path of S5 joining the black vertex (x)5 to the white vertex (q)5 . We set H1 H2 H3 H4
= = = =
he, (e)5 , T1 , (p)5 , p, Q−1 1 , vi, 5 he, P1 , w, (w) , T2 , (v)5 , vi, he, P2 , x, (x)5 , T3 , (q)5 , q, Q−1 2 , vi, and −1 he, P3 , y, r, Q3 , vi.
Obviously, {H1 , H2 , H3 , H4 } forms a 4∗ -container of S5 between e and v. Moreover, l(H1 ) = l(H2 ) = 27, l(H3 ) = 29, and l(H4 ) = 39. Thus, ds4L (e, v) ≤ 41. 2 L Lemma 12 dsn−1 (u, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 =
Proof. odd.
n! n−2
+ 1 for n ≥ 6.
Let u be any white vertex and v be any black vertex of Sn . Obviously, d(u, v) is
Case 1: d(u, v) = 1. Since the star graph is vertex transitive and edge transitive, we may assume that u = e and v = (e)n . {n}
By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 2 and P1 = n(n − 2)! − 1, (2) Pi joins e to a white vertices {n} n−2 Pi spans Sn . xi with (xi )1 = i + 1 and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪i=1 {1} And there exist n − 2 paths Q1 , Q2 , . . . , Qn−2 of Sn such that (1) Q1 joins (e)n to a white vertices y1 with (y1 )1 = 2 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins (e)n to a black vertices yi 15
{1}
n−2 with (yi )1 = i + 1 and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪i=1 Qi spans Sn . By {2} Theorem 1, there exists a hamiltonian path R1 of Sn joining the white vertex (x1 )n to the {i+1} black vertex (y1 )n and there exists a hamiltonian path Ri of Sn joining the black vertex n n (xi ) to the white vertex (yi ) for 2 ≤ i ≤ n − 3. n n We set Pi = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , (e) i for 1 ≤ i ≤ n−2 and Pn−1 = he, (e) i. Then P1 , P2 , . . . , Pn−1 forms an (n − 1)∗ -container between e and (e)n . Obviously, l(P1 ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! − 1, l(Pi ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 for 2 ≤ i ≤ n − 2, L and l(Pn−1 ) = 1. Hence dsn−1 (e, (e)n ) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1.
Case 2: d(u, v) ≥ 3. Since d(u, v) ≥ 3, we may assume that u = e, (v)n = n − 1, and (v)1 6= n. {n}
Subcase 2.1: (v)1 = 1. By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 1 and l(P1 ) = n(n − 2)! − 1, (2) Pi joins e to a white vertex xi with (xi )1 = i and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and {n} {n−1} n−2 (3) ∪i=1 Pi spans Sn . Again there exist (n − 2) paths Q1 , Q2 , . . . , Qn−2 of Sn such that (1) Q1 joins v to a white vertex y1 with (y1 )1 = 1 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins v to a black vertex yi with (yi )1 = i and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) ∪n−2 i=1 Qi {n−1} spans Sn . {1}
By Lemma 6, there exist two disjoint paths H1 and H2 of Sn such that (1) H1 joins the white vertex (x1 )n to the black vertex (y1 )n , (2) H2 joins the white vertex (e)n to the black {1} vertex (v)n , and (3) H1 ∪ H2 spans Sn . For 2 ≤ i ≤ n − 2, there exists a hamiltonian path {i} Ri of Sn joining the black vertex (xi )n to the white vertex (yi )n . We set T1 = he, P1 , x1 , (x1 )n , H1 , (y1 )n , y1 , Q−1 j , vi, Ti = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , vi for 2 ≤ i ≤ n − 2, and n n Tn−1 = he, (e) , H1 , (v) , vi. Obviously, {T1 , T2 , . . . , Tn−1 } forms an (n − 1)∗ -container. Moreover, l(Ti ) ≤ (n − 1)! + L 2(n − 2)! + 2(n − 3)! + 1. Thus, dsn−1 (e, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1. Subcase 2.2: (v)1 = t 6= 1, n. By Lemma 9, there exist (n − 2) paths P1 , P2 , . . . , Pn−2 of {n} Sn such that (1) P1 joins e to a black vertex x1 with (x1 )1 = 1 and l(P1 ) = n(n − 2)! − 1, (2) Pi joins e to a white vertex xi with (xi )1 = i and l(Pi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, {n} {n−1} n−2 Pi spans Sn . Again there exist (n − 2) paths Q1 , Q2 , . . . , Qn−2 of Sn such and (3) ∪i=1 that (1) Q1 joins v to a white vertex y1 with (y1 )1 = 1 and l(Q1 ) = n(n − 2)! − 1, (2) Qi joins v to a black vertex yi with (yi )1 = i and l(Qi ) = n(n − 2)! for 2 ≤ i ≤ n − 2, and (3) {n−1} n−2 ∪i=1 Qi spans Sn . {t}
Let the black vertex w be the unique neighbor of the white vertex (v)n of Sn with {1} (w)1 = 1. By Lemma 6, there exist two disjoint paths H1 and H2 of Sn such that (1) H1 joins the white vertex (x1 )n to the black vertex (y1 )n , (2) H2 joins the white vertex (e)n to 16
{1}
the black vertex (w)n , and (3) H1 ∪ H2 span Sn . {t}
By Theorem 4, there exists a hamiltonian path Rt of Sn joining the black vertex (xt )n {i} to the white vertex (yt )n . By Theorem 1, there exists a hamiltonian path Ri of Sn joining the black vertex (xi )n to the white vertex (yi )n for 2 ≤ i ≤ n − 2 with i 6= t. We set T1 = he, P1 , x1 , (x1 )n , H1 , (y1 )n , y1 , Q−1 j , vi, Ti = he, Pi , xi , (xi )n , Ri , (yi )n , yi , Q−1 i , vi for 2 ≤ i ≤ n − 2, and n n n Tn−1 = he, (e) , H2 , (w) , w, (v) , vi. Obviously, {T1 , T2 , . . . , Tn−1 } forms an (n − 1)∗ -container. Moreover, l(Ti ) ≤ (n − 1)! + L 2(n − 2)! + 2(n − 3)! + 1. Thus, dsn−1 (e, v) ≤ (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1. 2 It is easy to check that = 1 and = 3. Using computer, we have that D3sL (S4 ) = 15 by brute force. By Lemmas 10, 11, and 12, we have the following theorem: D1sL (S2 )
D2sL (S3 )
sL Theorem 5 Dn−1 (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! for n ≥ 5.
4
The 2∗L -diameter Sn
Lemma 13 Assume that u be any white vertex of S4 . Let a and b are two distinct elements of h4i. There exist two paths P1 and P2 of S4 such that (1) P1 joins u to a black vertex x with (x)1 = a and l(P1 ) = 7, (2) P2 joins u to a white vertex y with (y)1 = b and l(P2 ) = 16, and (3) P1 ∪ P2 spans S4 . Proof. Without loss of generality, we may assume that u = e. The required two paths are listed below. P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2 P1 P2
= = = = = = = = = = = = = = = = = = = = = = = =
(1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(3421)(2341)(3241)(4231)(2431) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(4231)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(1423)(2413)(4213)(3214)(2314)(1324)(3124) (1234)(2134)(4132)(3142)(1342)(4312)(3412)(1432) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(2431)(4231)(3241)(2341)(4321) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(4231)(3241)(2341)(4321)(3421)(2431)(1432)(3412)(4312)(2314)(1324)(3124)(2134)(4132)(3142)(1342) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(2134)(3124)(1324)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(4321)(3421)(2431)(4231)(3241) (1234)(3214)(4213)(1243)(2143)(4123)(1423)(2413) (1234)(2134)(3124)(1324)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(3241)(4231)(2431)(3421)(4321) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(4231)(2431)(3421)(4321)(2341)(3241)(1243)(2143)(4123)(3124)(1324)(2314)(3214)(4213)(2413)(1423) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(3421)(2341)(3241)(4231)(2431) (1234)(2134)(4132)(1432)(3412)(4312)(1342)(3142) (1234)(3214)(2314)(1324)(3124)(4123)(2143)(1243)(4213)(2413)(1423)(3421)(2431)(4231)(3241)(2341)(4321) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(1342)(2341)(3241)(1243)(4213)(2413)(3412)(1432)(4132)(3142)(2143)(4123)(1423) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(3412)(1432)(4132)(3142)(1342)(2341)(3241)(1243)(4213)(2413)(1423)(4123)(2143) (1234)(2134)(3124)(1324)(4321)(3421)(2431)(4231) (1234)(3214)(2314)(4312)(1342)(2341)(3241)(1243)(4213)(2413)(1423)(4123)(2143)(3142)(4132)(1432)(3412)
2
17
Theorem 6 3 sL 15 D2 (Sn ) = n! 2
if n = 3, if n = 4, and + 1 if n ≥ 5.
D2sL (S3 )
Proof. It is easy to check that = 3. Using computer, we have that D2sL (S4 ) = 15 by brute force. Thus, we assume that n ≥ 5. Let u be a white vertex and v be a black vertex of Sn . Let P1 and P2 be any two 2∗ -container of Sn joining u to v. Obviously, + 1. Hence, ds2L (u, v) ≥ n!2 + 1 and D2sL (Sn ) ≥ n!2 + 1. Hence we only max{l(P1 ), l(P2 )} ≥ n! 2 need to show that ds2L (u, v) ≤ n!2 + 1. Without loss of generality, we assume that u = e and {n−1} v ∈ Sn . {5}
Case 1: n = 5. By Lemma 13, there exist two paths H1 and H2 of S5 such that (1) H1 joins e to a black vertex x with (x)1 = 1 and l(H1 ) = 7, (2) H2 joins e to a white vertex {5} y with (y)1 = 3 and l(H2 ) = 16, and (3) H1 ∪ H2 spans S5 . Again, there exist two paths {4} T1 and T2 of S5 such that (1) T1 joins v to a white vertex p with (p)1 = 2 and l(T1 ) = 7, {4} (2) T2 joins v to a black vertex q with (q)1 = 3 and l(T2 ) = 16, and (3) T1 ∪ T2 spans S5 . {1,2} By Lemma 3, there exists a hamiltonian path W of S5 joining the white vertex (x)5 to {3} the black vertex (p)5 . Moreover, there exists a hamiltonian path Z of S5 joining the black vertex (y)5 to the white vertex (q)5 . We set L1 = he, H1 , x, (x)n , W, (p)n , p, T1−1 , vi and L2 = he, H2 , y, (y)n , Z, (q)n , q, T2−1 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = 61 and l(L2 ) = 59. Hence + 1. ds2L (u, v) ≤ n! 2 {n−1}
Case 2: n ≥ 6 is even. Let x be any neighborhood of v in Sn with (x)1 6= n. Obviously, x is a white vertex. Let j be any index in hni − {(x)1 , n − 1}. Obviously, (e)j is a black vertex. Let a1 , a2 , . . . , an−2 be a permutation of hn − 2i such that a1 = j and (x)1 = a3 . By {n} Theorem 2, there exists a hamiltonian path P of Sn − {(e)j } joining e to a white vertex p with (p)1 = a1 and (p)2 = an−2 . By Lemma 4, there exists a hamiltonian path H of {a } Sn 1 − {(p)n , ((p)n )2 } joining the white vertex ((e)j )n to a black vertex y with (y)1 = a3 . Let H = {a1 , a2 , . . . , a n2 } and H 0 = hn − 2i − H. By Lemma 3, there exists a hamiltonian H−{a }
1 path T of Sn joining the white vertex (y)n to the black vertex (x)n . By Theorem 1, {a } there exists a hamiltonian path H 0 of Sn n−2 joining the black vertex (((p)n )2 )n to a white vertex q with (q)1 = n − 3 and (q)n 6= v. By Theorem 2, there exists a hamiltonian path {n−1} T 0 of Sn − {x} joining the black vertex (q)n to v. We set
L1 = he, (e)j , ((e)j )n , H, y, (y)n, T, (x)n , x, vi and L2 = he, P, p, (p)n, ((p)n )2 , (((p)n )2 )n , H 0 , q, (q)n , T 0 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = n!2 + 1 and l(L2 ) = Hence ds2L (u, v) ≤ n!2 + 1. See Figure 4 for illustration the case n = 6. 18
n! 2
− 1.
4
(((p) 6)2)6
e (e)
S6{5}
S6{a }
S6{6} j
q
(q) 6
v x
p
S6{a }
S6{a }
1
S6{a }
3
2
(p) 6 ((p) 6)2 ((e) j)6
y
(y) 6
(x) 6
Figure 4: Illustration for the case 2 of Theorem 6.
{n−1}
Case 3: n ≥ 7 is odd. Let x be any neighborhood of v in Sn with (x)1 6= n. Obviously, x is a white vertex. Let j be any index in hni − {(x)1 , n − 1}. Obviously, (e)j is a black vertex. Let a1 , a2 , . . . , an−2 be a permutation of hn − 2i such that a1 = j and {n} (x)1 = a2 . By Theorem 2, there exists a hamiltonian path P of Sn − {(e)j } joining e to a white vertex p with (p)1 = a1 and (p)2 = an−3 . By Lemma 4, there exists a hamiltonian {a } path R of Sn 1 − {(p)n , ((p)n )2 } joining the white vertex ((e)j )n to a black vertex y with n−1
−1
{(a ,a )}
2 (y)1 = an−2 and (y)n−1 = a2 . Let X be the subgraph of Sn induced by ∪i=2 Sn i n and {(a ,a )} {(n−1,a )} {(n,a )} n i n n−3 ∪ Sn ∪ Sn n . By Lemma 3, Y be the subgraph of Sn induced by ∪i= n−1 Sn 2
there exists a hamiltonian path W of X joining the white vertex (y)n to a black vertex z with (z)1 = a3 . Let H = {a1 , a2 , . . . , a n−1 } and H 0 = hn − 3i − H. Again, there exists a H−{a }
2
1 hamiltonian path T of Sn joining the white vertex (z)n to the black vertex (x)n . By {H 0 } Theorem 3, there exists a hamiltonian path R0 of Sn joining the black vertex (((p)n )2 )n to a white vertex q with (q)1 = an−2 and (q)n−1 = n. By Lemma 3, there exists a hamiltonian path W 0 of Y joining the white vertex (q)n to a black vertex r with (r)1 = n − 1 and r 6= v. {n−1} By Theorem 2, there exists a hamiltonian path T 0 of Sn − {x} joining the black vertex n (r) ) to the black vertex v. We set
L1 = he, (e)j , ((e)j )n , R, y, (y)n , W, z, (z)n , T, (x)n , x, vi and L2 = he, P, p, (p)n , ((p)n )2 , (((p)n )2 )n , R0 , q, (q)n , W 0 , r, (r)n , T 0 , vi. Obviously, L1 and L2 forms a 2∗ -container. Moreover, l(L1 ) = n!2 + 1 and l(L2 ) = Hence ds2L (u, v) ≤ n!2 + 1. See Figure 5 for illustration the case n = 7.
19
n! 2
− 1. 2
S7{7}
4
(((p) 7)2)7
e (e)
S7{6}
S7{a } j
p
q
S7{a }
v
5
(q)
x
7
r
S7{a }
(r)7
S7{a }
1
3
(p) 7
2
z
(y) 7 7 2
((p) ) ((e) j)7
S7{a }
(z)n
y
(x) 7
Figure 5: Illustration for the case 3 of Theorem 6.
5
Conclusion
sL n! In this paper, we prove that Dn−1 (Sn ) = (n − 1)! + 2(n − 2)! + 2(n − 3)! + 1 = n−2 + 1 and sL sL n! n! D2 (Sn ) = 2 + 1 for n ≥ 5. Actually, we prove that d2 (u, v) = 2 + 1 for any two vertices u and v from different bipartite set of Sn .
Recently, we have proved that Sn is super spanning laceable [10]. Hence we will study for 3 ≤ k ≤ n−1. We conjecture that DksL (Sn ) = n! +1 for n ≥ 5 and 2 ≤ k ≤ n−2. k
DksL (Sn )
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[7] D. Frank Hsu, “On container width and length in graphs, groups and networks”, IEICE Trans, Fundamentals, vol. E77-A, no. 4, pp. 668-680, 1994. [8] Sun-Yuan Hsieh, Gen-Huey Chen, and Chin-Wen Ho, “Hamiltonian-laceability of star graphs”, Networks, vol. 36, no. 4, pp. 225-232, 2000. [9] Tseng-Kuei Li, Jimmy J. M. Tan, and Lih-Hsing Hsu, “Hamiltonian laceability on edge fault star graph”, Submitted. [10] Cheng-Kuan Lin, Hua-Min Huang, and Lih-Hsing Hsu, “The super connectivity of the pancake graphs and gtar graphs”, Submitted. [11] K. Menger, “Zur allgemeinen Kurventheorie”, Fund. Math., 10, pp. 95–115, 1927. [12] K. Qiu, H. Meijer, and S. G. Alk, ”Decomposing a star graph into disjoint cycles,” Information Processing Letters, vol. 39, no. 4, pp. 125–129, 1991. [13] Y. Rouskov and P. K. Srimani, ”Fault diameter of star graphs,” Information Processing Letters, vol. 48, no. 5, 1993.
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