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Panpositionable Hamiltonicity of the Alternating

Group Graphs

Yuan-Hsiang Teng and Jimmy J. M. Tan

Department of Computer Science, National Chiao Tung University, Hsinchu City, Taiwan 300, Republic of China

Lih-Hsing Hsu

Department of Computer Science and Information Engineering, Providence University, Taichung County, Taiwan 433, Republic of China

The alternating group graph AGn is an interconnection network topology based on the Cayley graph of the alternating group. There are some interesting results concerning the hamiltonicity and the fault tolerant hamil-tonicity of the alternating group graphs. In this article, we propose a new concept called panpositionable hamil-tonicity. A hamiltonian graph G is panpositionable if for any two different vertices x and y of G and for any integer l satisfying d(x, y) ≤ l ≤ |V (G)| − d(x, y), there exists a hamiltonian cycle C of G such that the relative distance between x , y on C is l . We show that AGn is panposi-tionable hamiltonian if n≥ 3.©2007 Wiley Periodicals, Inc. NETWORKS, Vol. 50(2), 146–156 2007

Keywords: alternating group graph; hamiltonian; hamiltonian connected; panpositionable hamiltonian

1. INTRODUCTION

Network topology is usually represented by a graph where the vertices represent processors and the edges represent the links between processors. For graph definitions and nota-tion we follow Ref. [6]. Let G = (V, E) be a graph, where

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 of G. Two vertices u and v are adjacent if (u, v) ∈ E. A path P is represented by v0, v1, v2,. . . , vk.

The length of a path P is the number of edges in P, denoted by L(P). We sometimes write the path v0, v1, v2,. . . , vk as

v0, P1, vi, vi+1,. . . , vj, P2, vt,. . . , vk, where P1 is the path

v0, v1,. . . , vi and P2 is the path vj, vj+1,. . . , vt. It is

Received January 2006; accepted November 2006

Correspondence to: J. M. Tan; e-mail: jmtan@cs.nctu.edu.tw

Contract grant sponsor: National Science Council of the Republic of China; Contract grant number: NSC 94-2213-E-009-138

DOI 10.1002/net.20184

Published online in Wiley InterScience (www.interscience.wiley. com).

©2007 Wiley Periodicals, Inc.

possible to write a pathv0, v1, P, v1, v2,. . . , vk if L(P) = 0.

We use dG(u, v), or simply d(u, v) if there is no ambiguity,

to denote the distance between u and v in a graph G, i.e., the length of a shortest path joining u and v in G. A cycle is a path with at least three vertices such that the first vertex is the same as the last one. We use dC(u, v) and DC(u, v) to denote

the shorter and the longer distance between u and v on a cycle

C of G, respectively. It is possible that DC(u, v) = dC(u, v)

if the lengths of the two disjoint paths joining u and v in

C are equal. A path is a hamiltonian path if its vertices are

distinct and span V . A graph G is hamiltonian connected if there exists a hamiltonian path joining any two vertices of G. A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A graph G is hamiltonian if there exists a hamiltonian cycle in G. The hamiltonian properties are impor-tant aspects of designing an interconnection network. Many related works have appeared in the literature [1, 5, 7].

We propose a new concept called panpositionable hamil-tonicity. A hamiltonian graph G is panpositionable if for any two different vertices x and y of G and for any integer l satisfy-ing d(x, y) ≤ l ≤ |V(G)|−d(x, y), there exists a hamiltonian cycle C of G such that the relative distance between x, y on C is l; more precisely, dC(x, y) = l if l ≤ |V(G)|2  or

DC(x, y) = l if l > |V(G)|2 . Given a hamiltonian cycle C, if

dC(x, y) = l, we have DC(x, y) = |V(G)| − dC(x, y).

There-fore, a graph is panpositionable hamiltonian if for any integer

l with d(x, y) ≤ l ≤ |V(G)|2 , there exists a hamiltonian cycle

C of G with dC(x, y) = l. One trivial example, the complete

graph Knwith n≥ 3, is panpositionable.

There are several requirements in designing a good topol-ogy for an interconnection network, such as connectivity and hamiltonicity. The hamiltonian property is one of the major requirements in designing an interconnection network. The hamiltonian property is fundamental to the deadlock-free routing algorithms of distributed systems [8, 9]. A high-reliability network design can be based on constructing a hamiltonian cycle in an interconnection network. Similar

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FIG. 1. AG3and AG4.

to the importance of hamiltonicity for the communication between processors in an interconnection network, panposi-tionable hamiltonicity allows more flexible communication in a hamiltonian network. The panpositionable hamiltonian property inherits the hamiltonian property and advances it further. The concept is interesting and useful in the study of interconnection networks.

The alternating group graph [7] was proposed by Jwo et al. The interconnection scheme is based on the Cayley graph of the set of all even permutations. Cheng and Lipman investi-gated some properties of this family of graphs in Refs. [2–4]. Letn = {1, 2, . . . , n}. Then p = p1p2. . . pnis a permutation

of elements inn if pi= pjfor all i= j. Suppose that piand

pjare two different symbols in p with i< j; then the pair (i, j)

is an inversion if pi> pj. A permutation is even if the number

of inversions is even. Let g+i and gi be two operations such that pg+i and pgi are permutations obtained from permuta-tion p by rotating the symbols in posipermuta-tion 1, 2, and i from left to right and from right to left, respectively. For example, we have 1234g+4 = 4132 and 1234g4 = 2431. The n-dimensional alternating group graph AGnis the graph(Vn, En), where Vnis

the set of all even permutations, and En= {(p, q) | p, q ∈ Vn,

and q = pgi+ or q = pgi for i = 3, 4, . . . , n}. Figure 1 illustrates AG3 and AG4. An alternating group graph AGn

is a regular graph of degree 2(n − 2) with n2! vertices, and

AGnis vertex symmetric and edge symmetric [7]. The

diam-eter of AGnis3(n−2)2 . There are some studies concerning

hamiltonicity of the alternating group graph. Jwo et al. [7] showed that the alternating group graph is hamiltonian and hamiltonian connected. Chang et al. [1] showed that AGnis

(n − 2)-vertex fault tolerant hamiltonian and (n − 3)-vertex

fault tolerant hamiltonian connected if n ≥ 4. A graph G is panconnected if there exists a path of length l joining any two different vertices x and y with d(x, y) ≤ l ≤ |V(G)| − 1. A graph G is pancyclic if it contains a cycle of length l for each l satisfying 3 ≤ l ≤ |V(G)|. Chang et al. also proved that AGnis panconnected and pancyclic for all n≥ 3 [1].

In this article, we study the panpositionable hamiltonic-ity of the alternating group graph AGn. We show that the

alternating group graph AGnis panpositionable hamiltonian

for all n ≥ 3. In the following section, we discuss some basic properties of the alternating group graph. In Section 3,

we prove our main theorem. In the final section, we present our conclusion and explain some relationship between the panpositionable hamiltonian property and the panconnected property.

2. SOME PROPERTIES OF THE ALTERNATING GROUP GRAPHS

For each n ≥ 3, let Vn[i] = {p | p = p1p2. . . pn and

pn = i}. It is the set of all vertices with the rightmost

posi-tion being i. Let AGn[i] denote the subgraph of AGninduced

by Vn[i]. It is easy to see that each AGn[i] is isomorphic

to AGn−1. Thus, AGn can be recursively constructed from

n copies of AGn−1. Each AGn[i] represents a subcomponent

of AGn. Let I be a subset of{1, 2, . . . , n}. We use AGn(I) to

denote the subgraph of AGninduced byi∈IVn[i]. We call

AGn(I) an incomplete alternating group graph if |I| < n. We

observe that each AGn[i] can be recursively decomposed into

its smaller subcomponents. We use Eni,j to denote the set of

edges between AGn[i] and AGn[j]. Let F be a faulty set, which

may include faulty edges, faulty vertices, or both. The good edge set GEni,j(F) is the set of edges (u, v) ∈ Eni,j such that

{u, v, (u, v)} ∩ F = ∅. We need some basic properties of the alternating group graphs. The following proposition follows directly from the definition of the alternating group graphs.

Proposition 1. Let n be an integer with n ≥ 5, and let i and j be two distinct elements ofn. Suppose that H is one subcomponent of AGn[j] with the (n − 1)th position being h

and the nth position being j for some h∈ n − {i, j}. Then

|Ei,j

n| = (n−2)!, and the number of edges between AGn[i] and

H is(n−3)!. Moreover, if (u, v) and (u , v ) are distinct edges

in Eni,j, then{u, v}



{u , v } = ∅, and (u, u ) ∈ E(AG

n[i]) if

and only if(v, v ) ∈ E(AGn[j]).

Let i∈ n, and let u be a vertex in AGn[i]. We say that

v is a neighbor of u if v is adjacent to u. We use NI(u) to

denote the set of all neighbors of u, which are in AGn(I).

Particularly, we use N(u) as an abbreviation of Nn−{i}(u). We call vertices in N(u) the outer neighbors of u. Obviously, |Ni(u)| = 2(n − 3) and |N(u)| = 2. We say that vertex u is

adjacent to subcomponent AGn[j] if u has an outer neighbor

in AGn[j]. Then, we define the adjacent subcomponent AS(u)

of u as{j | u is adjacent to AGn[j]}. We have the following

proposition:

Proposition 2. Suppose that n≥ 4 and i ∈ n. Let u and v be two distinct vertices in AGn[i].

(a) If d(u, v) = 1, then |AS(u) ∩ AS(v)| = 1 and AS(u) = AS(v).

(b) If d(u, v) = 2, then AS(u) = AS(v).

Proof. Let u = u1u2. . . un, v = v1v2. . . vn, and un =

vn = i. If d(u, v) = 1, we have v = ugck for some 3 ≤

k ≤ n − 1 and c ∈ {+, −}. Without loss of generality, we

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and uk = v1if v= ug+k. Let ug+n and ugn be the two outer

neighbors of u. Let vg+n and vgn be the two outer neighbors of v. Then, AS(u) = {u1, u2} and AS(v) = {v1, v2}. Thus AS(u) ∩ AS(v) = {u1} = {v2} and |AS(u) ∩ AS(v)| = 1. It is

obvious that AS(u) = AS(v).

If d(u, v) = 2, there exists a vertex w ∈ V(AGn[i]) such

that d(u, w) = d(w, v) = 1, because u, v ∈ AGn[i]. Let w =

w1w2. . . wn. If u= wg+k and v= wgk for 3 ≤ k ≤ n − 1,

then AS(u) = {w1, wk} and AS(v) = {w2, wk}. If u = wgcxand

v= wgcyfor some x = y, 3 ≤ x, y ≤ n − 1, and c ∈ {+, −},

then wx∈ AS(u) and wx /∈ AS(v). Thus AS(u) = AS(v). The

statement follows. ■

Chang et al. studied the fault hamiltonicity and fault hamil-tonian connectivity of the alternating group graphs in Ref. [1]. The result is listed as follows.

Theorem 1. [1] Alternating group graphs AGn, n≥ 4, are

n− 2 vertex-fault tolerant hamiltonian and n − 3 vertex-fault tolerant hamiltonian connected.

The above theorem states that with up to n− 2 faulty vertices AGn still has a hamiltonian cycle, and with up to

n− 3 faulty vertices AGnis still hamiltonian connected. The

following lemmas consider the hamiltonian connectivity of the subgraphs AGn(I) of the alternating group graphs AGn.

Lemma 1. Suppose that I ⊆ {1, 2, 3, 4} with |I| ≥ 2. If x∈ V(AG4[i]), y ∈ V(AG4[j]), and (x, y) is an edge between AG4[i] and AG4[j] with i = j ∈ I, then there is a hamiltonian path of AG4(I) joining x and y.

Proof. The alternating group graph AG4is known to be

edge symmetric. Without loss of generality, we may consider that x= 3241 and y = 1342, which are two adjacent vertices of AG4 in Figure 1. If I = {1, 2}, then 3241, 4321, 2431,

4132, 3412, 1342 forms a hamiltonian path of AG4(I) from x to y. If I = {1, 2, 3}, then 3241, 2431, 4321, 1423, 2143,

4213, 3412, 4132, 1342 forms a hamiltonian path of AG4(I)

from x to y. If I = {1, 2, 3, 4}, then by Theorem 1, AG4is

hamiltonian connected. Hence the lemma follows. ■

Lemma 2. Suppose that

1. n≥ 5,

2. I⊆ n with |I| ≥ 2, 3. F⊆ V(AGn) ∪ E(AGn), and

4. AGn[l] − F is hamiltonian connected for each l ∈ I and

|F| ≤ 2n − 7.

Then, for any x ∈ V(AGn[i]) and y ∈ V(AGn[j]) with

i= j ∈ I, there is a hamiltonian path of AGn(I) − F joining

x and y.

Proof. Consider that |F| ≤ 2n − 7. Suppose that |GEi1,i2

n (F)| < 3 for some i1, i2 ∈ n. Since |Eni1,i2| =

(n − 2)! ≥ 2(n − 2), this implies that |F| > 2n − 7. We get

a contradiction. Hence we have|GEi1,i2

n (F)| ≥ 3. We prove

this lemma by induction on|I|. Suppose that |I| = 2, and

I = {i, j} for some i, j. Since |GEi,jn(F)| ≥ 3, there exists an

edge(u, v) ∈ GEni,j(F) such that u = x ∈ V(AGn[i]) and

v= y ∈ V(AGn[j]). By the assumption that each AGn[l] − F

is hamiltonian connected, there is a hamiltonian path P1of AGn[i]−F from x to u and a hamiltonian path P2of AGn[j]−F

from v to y. Thusx, P1, u, v, P2, y forms a hamiltonian path

of AGn(I) − F from x to y.

Assume that the result is true for all I with 2≤ |I | < |I|. Thus there is an i ∈ I with i = i, j. Since |GEni ,j(F)| ≥

3, there exists an edge (u, v) ∈ GEni ,j(F) such that u ∈

V(AGn[i ]) and v = y ∈ V(AGn[j]). Then, there is a

hamilto-nian path P1of AGn(I −{j})−F from x to u and a hamiltonian

path P2of AGn[j] − F from v to y. Thus x, P1, u, v, P2, y

forms a hamiltonian path of AGn(I) − F from x to y. Hence

the lemma follows. ■

Jwo et al. [7] presented a shortest path routing algorithm for the alternating group graph AGn, and gave some

charac-terizations of the minimum length path between two arbitrary vertices in AGn. With this algorithm, we can find a minimum

length path between any two distinct vertices of AGnas stated

in the following proposition.

Proposition 3. [7] Let i, j ∈ n and i = j. Suppose that

u= u1u2· · · unand v= v1v2· · · vnare two vertices in AGn.

(a) If un= vn= i, then u and v belong to the same

subcom-ponent AGn[i]. A shortest path from u to v in AGnis completely

contained in AGn[i]. That is, d(u, v) = dAGn[i](u, v).

(b) If un = i, vn = j, and vx = i for some x ∈ {1, 2},

there exists a vertex s ∈ V(AGn[i]) adjacent to v such that

u, P, s, v is the minimum length path between u and v in

AGn, where P is a path completely contained in AGn[i]. That

is, d(u, v) = dAGn[i](u, s) + 1 = d(u, s) + 1.

(c) If un= i, vn = j, and vx = i for some x ∈ {3, 4, . . . ,

n− 1}, there exist vertices s ∈ V(AGn[i]) and t ∈ V(AGn[j]),

where t is adjacent to v, and(s, t) ∈ Ei,jn, such thatu, P, s, t, v

is the minimum length path between u and v in AGn, where P

is a path completely contained in AGn[i]. That is, d(u, v) =

dAGn[i](u, s) + 2 = d(u, s) + 2.

Example. Suppose that u and v are two vertices in

AG5. If u = 12345 and v = 21435, then u ∈

V(AG5[5]) and v ∈ V(AG5[5]). A shortest path from u to v is 12345 g + 3 −→ 31245 g−4 −→ 14235 g+3 −→ 21435, and case

(a) holds. If u = 12345 and v = 15432, then u ∈ V(AG5[5]) and v ∈ V(AG5[2]). A shortest path from u to v is 12345 g + 3 −→ 31245 g−4 −→ 14235 g+3 −→ 21435 g5− −→ 15432,

and case (b) holds. If u = 12345 and v = 34512, then u ∈ V(AG5[5]) and v ∈ V(AG5[2]). A shortest path from u to v is 12345 g − 4 −→ 24315 g−5 −→ 45312 g3+ −→ 34512, and case (c) holds.

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3. PANPOSITIONABLE HAMILTONICITY OF THE ALTERNATING GROUP GRAPHS

In this section, we prove that the alternating group graph

AGnis panpositionable hamiltonian for all n≥ 3. The basic

idea is to study AG3and AG4first, and then to prove the result

for n≥ 5 by induction on n.

Lemma 3. Alternating group graphs AGnare

panposition-able hamiltonian for n= 3, 4.

Proof. For n= 3, since AG3is isomorphic to the

com-plete graph K3, the result holds for n= 3 trivially. Now, we

consider THE case n= 4. Let u and v be any two vertices of

AG4in Figure 1.

The alternating group graph is known to be vertex sym-metric and edge symsym-metric, and to have diameter3(n−2)2  [7]. The diameter of AG4 is 3. Without loss of generality,

to prove this lemma it is enough to consider u = 1234

and v = 3124 for d(u, v) = 1; u = 1234 and v =

4321 for d(u, v) = 2; u = 1234 and v = 2143 for

d(u, v) = 3. For each l ∈ {d(u, v), d(u, v) + 1, . . . ,|V(AG4)| 2 },

we construct a hamiltonian cycle HC of AG4 such that dHC(u, v) = l. These hamiltonian cycles HC are listed

below.

d(u, v) dHC(u, v) The cycle HC

1 1 1234, 3124, 4321, 2431, 3241, 2143, 1423, 4213, 2314, 3412, 1342, 4132, 1234 1 2 1234, 2314, 3124, 4321, 1423, 4213, 3412, 4132, 1342, 2143, 3241, 2431, 1234 1 3 1234, 2431, 4321, 3124, 1423, 4213, 2143, 3241, 1342, 4132, 3412, 2314, 1234 1 4 1234, 2431, 4321, 1423, 3124, 2314, 3412, 4213, 2143, 3241, 1342, 4132, 1234 1 5 1234, 2314, 4213, 2143, 1423, 3124, 4321, 2431, 3241, 1342, 3412, 4132, 1234 1 6 1234, 2431, 4132, 1342, 3241, 4321, 3124, 1423, 2143, 4213, 3412, 2314, 1234 2 2 1234, 3124, 4321, 3241, 2431, 4132, 3412, 1342, 2143, 1423, 4213, 2314, 1234 2 3 1234, 3124, 1423, 4321, 2431, 3241, 2143, 1342, 4132, 3412, 4213, 2314, 1234 2 4 1234, 2314, 3124, 1423, 4321, 2431, 3241, 1342, 2143, 4213, 3412, 4132, 1234 2 5 1234, 3124, 2314, 4213, 1423, 4321, 2431, 3241, 2143, 1342, 3412, 4132, 1234 2 6 1234, 2314, 3412, 4213, 1423, 3124, 4321, 2431, 3241, 2143, 1342, 4132, 1234 3 3 1234, 3124, 1423, 2143, 3241, 4321, 2431, 4132, 1342, 3412, 4213, 2314, 1234 3 4 1234, 3124, 4321, 3241, 2143, 1423, 4213, 2314, 3412, 1342, 4132, 2431, 1234 3 5 1234, 3124, 4321, 2431, 3241, 2143, 1423, 4213, 2314, 3412, 1342, 4132, 1234 3 6 1234, 2431, 3241, 4321, 3124, 1423, 2143, 1342, 4132, 3412, 4213, 2314, 1234

Thus the lemma holds. ■

In the following lemma, we show that there exist two ver-tex disjoint paths spanning all the vertices in an incomplete alternating group graph. We need the lemma later in our main theorem. One may skip the proof temporarily, and come back to it later.

Lemma 4. Suppose that n ≥ 5, I ⊆ n with |I| ≥ 2, x1 ∈ V(AGn[i1]) and x2 ∈ V(AGn[i2]) with i1 = i2 ∈ I. Then, for any pair of distinct vertices(y1, y2) in V(AGn(I)),

there exist two disjoint paths, one joining x1and yifor some

i ∈ {1, 2}, and the other joining x2and yj with i = j, such

that these two paths span all the vertices in AGn(I).

Proof. Let i1, i2,. . . , i|I|be|I| distinct indices of n. We

prove this lemma by finding two disjoint paths P1and P2in AGn(I) such that P1joins x1and yi, and P2joins x2and yjwith

i = j. Moreover, P1and P2span all the vertices in AGn(I).

According to the location of y1and y2, we have the following

cases:

Case 1. Suppose that y1 and y2 are located in different

subcomponents, and x1and x2are not both adjacent to yifor

every i∈ {1, 2}.

Subcase 1.1. Suppose that x1, x2, yi, and yj are located

in four different subcomponents. We assume that yi

V(AGn[i3]) and yj∈ V(AGn[i4]) with |I| ≥ 4. See Figure 2a

for an illustration. By Lemma 2, we can find a hamiltonian path P1from x1to yiin AGn({i1, i3}). Similarly, we can find a

hamiltonian path P2from x2to yjin AGn(I − {i1, i3}).

There-fore, P1and P2are two disjoint paths spanning all the vertices

in AGn(I).

Subcase 1.2. Suppose that one of y1, y2 and one of x1, x2are located in the same subcomponent. Without loss of generality, we may assume that x1and yi are located in the

same subcomponent, x2 and yj are located in different

sub-components, yi ∈ V(AGn[i1]), and yj ∈ V(AGn[i3]) with

|I| ≥ 3. See Figure 2b for an illustration. By Theorem 1, since AGn[i1] is hamiltonian connected, we can find a

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FIG. 2. Illustrations for Lemma 4.

find a hamiltonian path P2 from x2 to yj in AGn(I − {i1}).

Therefore, P1and P2are two disjoint paths spanning all the

vertices in AGn(I).

Subcase 1.3. Suppose that x1and yiare located in the same

subcomponent for some i∈ {1, 2}, and x2and yjare located

in the same subcomponent with i= j. We assume that yi

V(AGn[i1]) and yj∈ V(AGn[i2]) with |I| ≥ 2. See Figure 2c

for an illustration. Without loss of generality, we may assume that i= 1 and j = 2. By Theorem 1, since AGn[i1] is

hamil-tonian connected, we can find a hamilhamil-tonian path P1from y1

to x1in AGn[i1]. If |I| ≥ 3, since |N(y2)| = 2, we can find

an edge(y2, y 2) ∈ Ei2,j

n such that y 2 ∈ V(AGn[j]) for some

j∈ I−{i1, i2}. By Lemma 2, we can find a hamiltonian path P2

from y 2to x2in AGn(I −{i1})−{y2}. Let P2= y2, y 2, P 2, x2.

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y2to x2in AGn[i2]. Let P2= y2, P 2, x2. Therefore, P1and P2are two disjoint paths spanning all the vertices in AGn(I).

Case 2. Suppose that yi and yj are located in the same

subcomponent, and x1and x2are not both adjacent to yi for

every i∈ {1, 2}.

Subcase 2.1. Suppose that y1, y2∈ V(AGn[i1]) or y1, y2∈ V(AGn[i2]) with |I| ≥ 2. See Figure 2d for an illustration.

Without loss of generality, we consider the former case and assume that i= 1 and j = 2. By Theorem 1, AGn[i1] − {y2}

is hamiltonian connected, hence we can find a hamiltonian path P1from y1 to x1 in AGn[i1] − {y2}. If |I| ≥ 3, since

|N(y2)| = 2, we can find an edge (y2, y 2) ∈ E

i1,j

n such that

y 2 ∈ V(AGn[j]) for some j ∈ I − {i1, i2}. By Lemma 2, we

can find a hamiltonian path P2 from y2 to x2in AGn(I −{i1}).

If|I| = 2, there exists an edge (y2, y 2) ∈ Eni1,i2such that y 2∈ V(AGn[i2]). By Theorem 1, there is a hamiltonian path P 2

from y 2to x2in AGn[i2]. Let P2= y2, y 2, P2 , x2. Therefore, P1and P2are two disjoint paths spanning all the vertices in AGn(I).

Subcase 2.2. Suppose that y1, y2 ∈ V(AGn[i3]). Without

loss of generality, we consider two subcases:

Subcase 2.2.1. Suppose there exists some xi∈ AS(y1) for

i ∈ {1, 2} with |I| ≥ 3. See Figure 2e for an illustration.

Without loss of generality, we may assume that i= 1. Since

x1 ∈ AS(y1), we can find an edge (y1, y 1) ∈ Ei1,i3

n such that

y 1 ∈ V(AGn[i1]) and x1 = y 1. By Theorem 1, we can find

a hamiltonian path P 1 from y1 to x1 in AGn[i1]. Let P1 =

y1, y 1, P1 , x1. Let y 2 = y1 ∈ V(AGn[i3]). By Theorem 1,

since AGn[i3] − {y1} is hamiltonian connected, we can find

a hamiltonian path P 2 from y2 to y 2 in AGn[i3] − {y1}. If

|I| ≥ 4, since |N(y

2)| = 2, we can find an edge (y 2, y 2) ∈ Ei3,j

n such that y2 ∈ V(AGn[j]) for some j ∈ I − {i1, i2, i3}.

By Lemma 2, we can find a hamiltonian path P 2 from y 2 to x2 in AGn(I − {i1, i3}). If |I| = 3, there exists an edge (y

2, y 2) ∈ E

i3,i2

n such that y 2 ∈ V(AGn[i2]). By Theorem

1, there is a hamiltonian path P 2from y 2 to x2in AGn[i2].

Let P2= y2, P 2, y 2, y2 , P 2, x2. Therefore, P1and P2are two

disjoint paths spanning all the vertices in AGn(I).

Subcase 2.2.2. Suppose that x1, x2 /∈ AS(y1)∪AS(y2) with

|I| ≥ 4. See Figure 2f for an illustration. Since |N(y1)| =

2, we can find an edge (y1, y 1) ∈ Ei1,j1

n such that y 1

V(AGn[j1]) for some j1∈ I −{i1, i2, i3}. By Lemma 2, we can

find a hamiltonian path P 1from y 1to x1in AGn({i1, j1}). Let P1= y1, y 1, P1 , x1. Let y2 ∈ V(AGn[i3]) and y 2∈ Ni3(y1).

By Proposition 2, we have AS(y1) = AS(y 2). By Theorem 1, since AGn[i3] − {y1} is hamiltonian connected, we can find a

hamiltonian path P 2from y2to y 2in AGn[i3]−{y1}. If |I| ≥ 5,

since|N(y 2)| = 2, we can find an edge (y2 , y 2) ∈ Ei3,j2

n such

that y 2 ∈ V(AGn[j2]) for some j2 ∈ I − {i1, i2, i3, j1}. By

Lemma 2, we can find a hamiltonian path P2 from y 2to x2in AGn(I − {i1, i3, j1}). If |I| = 4, since |N(y2 )| = 2, we can

find an edge(y 2, y 2) ∈ Ei3,i2

n such that y2 ∈ V(AGn[i2]). Since AGn[i2] is hamiltonian connected, there is a hamiltonian path P 2from y 2to x2in AGn[i2]. Let P2= y2, P 2, y 2, y2 , P 2, x2.

Therefore, P1and P2are two disjoint paths spanning all the

vertices in AGn(I).

Case 3. Suppose that x1and x2are adjacent to yifor some

i∈ {1, 2}. Without loss of generality, we assume that i = 1.

If y2 ∈ V(AGn[i1]), let P1 = x1, y1. Suppose that F =

{x1, y1}. By Lemma 2, we can find a hamiltonian path P2from x2to y2in AGn(I) − F. If y2 /∈ V(AGn[i1]), let P1= x2, y1.

Suppose that F= {x2, y1}. By Lemma 2, we can find a

hamil-tonian path P2from x1to y2in AGn(I)−F. Therefore, P1and

P2are two disjoint paths spanning all the vertices in AGn(I).

Thus the lemma follows. ■

We now prove our main result.

Theorem 2. Alternating group graphs AGn are

panposi-tionable hamiltonian if n≥ 3.

Proof. We prove this theorem by induction on n. By Lemma 3, AG3 and AG4 are panpositionable hamiltonian.

Suppose that the result holds for AGn−1for some n≥ 5. We

observe that AGncan be recursively constructed from n copies

of AGn−1, and each AGn−1is panpositionable hamiltonian by

the inductive hypothesis, for all n ≥ 5. Let s and t be two distinct vertices of AGn. Then for each l ∈ {d(s, t), d(s, t) +

1, d(s, t) + 2, . . . ,|V(AGn)|

2 }, we shall find a hamiltonian cycle

of AGnsuch that the distance between s and t on the cycle is l.

The idea of the proof is to expand the path between s and t to various lengths by inserting one or more subcomponents of

AGn−1. We achieve this purpose by our expanding algorithm

described below, and we can construct a path connecting s and t with the length of the path being l for any integer l with

d(s, t) ≤ l ≤ |V(AGn)|

2 .

Case 1. Suppose that s and t belong to the same subcompo-nent AGn[i]. There will be two subcases in Case 1; Figure 3a

and b illustrate Subcase 1.1 and Subcase 1.2, respectively. See Figure 3a first. Suppose that s, t ∈ V(AGn[i]) for some

i∈ n. By Proposition 3, d(s, t) = dAGn[i](s, t). Since AGn[i] is isomorphic to AGn−1, by the inductive hypothesis, for each

l0∈ {d(s, t), d(s, t)+1, d(s, t)+2, . . . , |V(AGn[i])|−d(s, t)},

we can construct a hamiltonian cycle HCi of AGn[i] such

that the distance between s and t on the cycle is l0. Let u and v be the two neighbors of t on HCi. Let HCi =

s, LP, u, t, v, RP, s, and let P0 = s, LP, u, t. Without loss

of generality, let L(P0) = l0. By Proposition 2, d(t, u) = 1, so we have|AS(t) ∩ AS(u)| = 1. This means that we can find a subcomponent AGn[j1] for which j1 ∈ n − {i},

such that there exist two disjoint edges (u, p1) and (t, q1) in Ei,j1

n . By Proposition 1, (p1, q1) ∈ E(AGn[j1]). Since

|N(t)| = 2, we can find a subcomponent AGn[ht] different

from AGn[i] and AGn[j1], and a vertex t ∈ V(AGn[ht]) such

that(t, t ) ∈ Ei,ht

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FIG. 3. Theorem 2, case 1.

d(t, v) ≤ 2 and hence AS(t) ⊇ {j1, ht}, AS(t) = AS(v), and

|N(v)| = 2. So we can find another subcomponent AGn[hv]

and a vertex v ∈ V(AGn[hv]) such that (v, v ) ∈ Eni,hv for

some hv∈ n−{i, j1, ht}. By Lemma 2, there exists a

hamil-tonian path HP of AGn(n − {i}) joining t and v . Thus

s, P0, t, t , HP, v , v, RP, s forms a hamiltonian cycle, and for

each l0∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . , |V(AGn[i])| −

d(s, t)}, the distance between s and t on the cycle is l0. Now we present an algorithm called st-expansion to expand the path P0between s and t to various lengths. We

describe the details as follows.

We can insert one subcomponent of AGn[j1] into P0 as

follows. See Figure 4a for an illustration. Because p1and q1

are adjacent, we may regard them as in the same subcompo-nent of AGn[j1], say C. The subcomponent C is isomorphic

to AGn−2. By Theorem 1, there is a hamiltonian path HP1of C joining p1and q1with L(HP1) = |V(AGn−2)| − 1. We can

insert more than one subcomponent of AGn[j1] into P0as

fol-lows. See Figure 4b for an illustration. We regard p1and q1as

in different subcomponents of AGn[j1]. By Lemma 1 if n = 5

and by Lemma 2 if n > 5, there is a hamiltonian path HP1 joining p1and q1with L(HP1) = m|V(AGn−2)| − 1, where

m is the number of subcomponents of AGn[j1] we wanted to

insert. Thus we can construct a path HP1between p1and q1

such that L(HP1) = I1|V(AGn−2)| − 1 for each integer I1

with 1≤ I1≤ n −1. Let P1= s, LP, u, p1, HP1, q1, t. Thus

we have L(P1) = l0+ I1|V(AGn−2)| = l0+I1(n−2)!2 . Since

d(s, t) ≤ l0 ≤ |V(AGn[i])| − d(s, t), we have I1(n−2)!2 +

d(s, t) ≤ L(P1) ≤ I1(n−2)! 2 +(n−1)!2 − d(s, t). For each 1 ≤ I1≤ n−1,(I1−1)(n−2)! 2 +(n−1)!2 − d(s, t) ≥ I1(n−2)! 2 + d(s, t)

if n≥ 5. Therefore, for each l1∈ {d(s, t), d(s, t)+1, d(s, t)+

2,. . . , (n − 1)! − d(s, t)}, we can construct a path P1from s to t such that the distance between s and t on the path is l1.

Similar as above, we can expand the path between s and

t more. For each 2 ≤ x ≤ n2, let ux−1 and tx−1 be two

adjacent vertices on HPx−1, where HPx−1 is a hamiltonian

path of AGn[jx−1] joining px−1and qx−1. By Propositions 1

and 2, there exist two distinct edges(ux−1, px) and (tx−1, qx)

in Ejx−1,jx

n for some jx∈ n − {i, ht, hv, j1,. . . , jx−1} such that

(px, qx) ∈ E(AGn[jx]). See Figure 4c for an illustration. We

can insert one subcomponent of AGn[jx] into P0as follows.

Because px and qx are adjacent, we may regard them as in

the same subcomponent of AGn[jx], say C. The

subcompo-nent C is isomorphic to AGn−2. By Theorem 1, there is a

hamiltonian path HPxof C joining pxand qxwith L(HPx) =

|V(AGn−2)| − 1. We can insert more than one subcomponent

of AGn[jx] into P0as follows. We regard pxand qxas in

dif-ferent subcomponents of AGn[jx]. By Lemma 1 if n = 5 and

by Lemma 2 if n> 5, there is a hamiltonian path HPx

join-ing pxand qxwith L(HPx) = m|V(AGn−2)| − 1, where m is

the number of subcomponents of AGn[jx] we wanted to insert.

Thus, we can construct a path HPxbetween pxand qxsuch that

L(HPx) = Ix|V(AGn−2)|−1 for each integer Ixwith 1≤ Ix

n−1. Let Px= s, LP, u, p1,. . . , px, HPx, qx,. . . , q1, t. Thus

we have L(Px) = l0+ (x − 1)|V(AGn−1)| + Ix|V(AGn−2)| =

l0+(x−1)(n−1)!2 +Ix(n−2)!

2 . Since d(s, t) ≤ l0≤ |V(AGn[i])|− d(s, t), we have (x−1)(n−1)!2 + Ix(n−2)! 2 + d(s, t) ≤ L(Px) ≤ Ix(n−2)! 2 + x(n−1)! 2 − d(s, t). For each 1 ≤ Ix ≤ n − 1, (Ix−1)(n−2)! 2 + x(n−1)! 2 −d(s, t) ≥ Ix(n−2)! 2 +(x−1)(n−1)!2 +d(s, t)

if n≥ 5. Therefore, for each lx∈ {d(s, t), d(s, t)+1, d(s, t)+

2,. . . ,(x+1)(n−1)!2 − d(s, t)}, we can construct a path Pxfrom

s to t such that the distance between s and t on the path is lx

by using st-expansion. Notice that the maximal value of lxis (n

2+1)(n−1)!

2 −d(s, t), which is greater than

n!

4, and|V(AG

n)|

2 =

n!

4. Hence for any integer l with d(s, t) ≤ l ≤ |V(AG

n)|

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FIG. 4. An illustration of st-expansion.

can construct a path joining s and t with the length of the path being l. We will use st-expansion for the remaining cases of the proof.

To construct a hamiltonian cycle, we consider the follow-ing two subcases:

Subcase 1.1. All the vertices of AGn({j1,. . . , jx}) are on

the path Px for some 1 ≤ x ≤ n2. See Figure 3a

for an illustration. By Lemma 2, there is a hamiltonian path HP of AGn(n − {i, j1,. . . , jx}) joining t and v

in which t ∈ V(AGn[ht]) and v ∈ V(AGn[hv]). Thus

s, Px, t, t , HP, v , v, RP, s forms a hamiltonian cycle, and

for each l∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . ,|V(AGn)|

2 }, the

distance between s and t on the cycle is l.

Subcase 1.2. Not all the vertices of AGn({j1,. . . , jx}) are

on the path Pxfor some 1≤ x ≤ n2. See Figure 3b for an

illustration. Then we can find two adjacent vertices y and z in AGn[jx], which are not on the path Px. Let F ⊆ V(Px).

By Proposition 1 and Proposition 2, there exist two dis-tinct edges (y, y ) ∈ Ejx,hy

n and (z, z ) ∈ Enjx,hz such that

y = t ∈ V(AGn[hy]) and z = v ∈ V(AGn[hz]),

respec-tively. If AGn[jx] − F is isomorphic to AGn−2, by Theorem 1,

there is a hamiltonian path HP from y to z in AGn[jx] − F.

If AGn[jx] − F contains more than one subcomponent of

AGn[jx], by Lemma 1 if n = 5, and by Lemma 2 if n > 5,

there is a hamiltonian path HP from y to z in AGn[jx] − F. By

Lemma 4, there exist two disjoint paths DP1and DP2, such

that DP1joins t and y , and DP2joins v and z . Moreover, the

two paths span all of the vertices in AGn(n − {i, j1,. . . , jx}).

Thuss, Px, t, t , DP1, y , y, HP, z, z , DP2, v , v, RP, s forms

a hamiltonian cycle, and for each l ∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . ,|V(AGn)|

2 }, the distance between s and t on

the cycle is l.

Now, we consider the case in which s and t belong to differ-ent subcompondiffer-ents of AGn. Suppose that s∈ V(AGn[i]) and

t ∈ V(AGn[j]) for any i = j ∈ n. By Proposition 3, there

exists a minimum length path connecting s and t with the form s, MP, t , t or s, MP, t , t , t, where MP is a path in AGn[i], t ∈ V(AGn[i]), and t ∈ V(AGn[j]). That is, d(s, t) =

dAGn[i](s, t )+1 = d(s, t )+1 or d(s, t) = dAGn[i](s, t )+2 =

d(s, t ) + 2. Thus we have the following cases:

Case 2. Suppose that s and t belong to different subcom-ponents of AGn, and the minimum length path connecting s

and t has the forms, MP, t , t. Then d(s, t) = d(s, t ) + 1. See Figure 5a for an illustration. Since AGn[i] is

isomor-phic to AGn−1, by the inductive hypothesis, for each l0 ∈

{d(s, t ), d(s, t )+1, d(s, t ) + 2, . . . , |V(AGn[i])|−d(s, t )},

we can construct a hamiltonian cycle HCi of AGn[i] such

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FIG. 5. Theorem 2, Case 2 and Case 3.

u and v be the two neighbors of t on HCi. Let HCi =

s, LP, u, t , v, RP, s, and let P0 = s, LP, u, t , t. Without

loss of generality, let L(P0) = l0+ 1. By Proposition 2,

d(t , u) = 1, so we have |AS(t ) ∩ AS(u)| = 1. This means that we can find a subcomponent AGn[j1] in which j1 ∈ n − {i}. If t /∈ V(AGn[j1]), by using st -expansion,

the proof is the same as Case 1 but we replace vertex t in Case 1 with vertex t in this case. So we consider the case in which t ∈ V(AGn[j1]), that is, j1 = j. Let q1 = t.

There exist two disjoint edges (u, p1) and (t , q1) in Ei,j1

n .

By Proposition 1,(p1, q1) ∈ E(AGn[j1]). By Proposition 2, d(t , v) ≤ 2 hence AS(t ) = {j1}, and AS(t ) = AS(v). Since |N(t )| = 2, we can find a subcomponent AGn[ht], and a

vertex t ∈ V(AGn[ht]) such that (t , t ) ∈ Eni,htfor some ht

n − {i, j1}. Since |N(v)| = 2 and AS(t ) = AS(v), we can

find a subcomponent AGn[hv], and a vertex v ∈ V(AGn[hv])

such that(v, v ) ∈ Ei,hv

n for some hv ∈ n − {i, j1, ht}. By

Lemma 2, there exists a hamiltonian path HP of AGn(n−{i})

joining t and v . Thuss, P0, t, HP, v , v, RP, s forms a

hamil-tonian cycle, and for each l0 ∈ {d(s, t), d(s, t) + 1, d(s, t) +

2,. . . , |V(AGn[i])| − d(s, t) + 1}, the distance between s and

t on the cycle is l0.

By using st -expansion, for any integer l with d(s, t ) ≤

l|V(AGn)|

2 , we can construct a path joining s and t with

the length of the path being l . Therefore, for any integer l with d(s, t) ≤ l ≤ |V(AGn)|

2 , we can construct a path joining s

and t with the length of the path being l.

To construct a hamiltonian cycle, we consider the follow-ing two subcases:

Subcase 2.1. All the vertices of AGn({j1,. . . , jx}) are on

the path Px for some 1 ≤ x ≤ n2. By Lemma 2, there is

a hamiltonian path HP of AGn(n − {i, j1,. . . , jx}) joining

t and v where t ∈ V(AGn[ht]) and v ∈ V(AGn[hv]). Thus

s, Px, t, t , t , HP, v , v, RP, s forms a hamiltonian cycle, and

for each l∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . ,|V(AGn)|

2 }, the

distance between s and t on the cycle is l.

Subcase 2.2. Not all the vertices of AGn({j1,. . . , jx}) are

on the path Px for some 1 ≤ x ≤ n2. See Figure 5a

for an illustration. Then, we can find two adjacent ver-tices y and z in AGn(jx), which are not on the path Px. Let

F ⊆ V(Px). By Proposition 1 and Proposition 2, there exist

two distinct edges (y, y ) ∈ Ejx,hy

n and(z, z ) ∈ Ejnx,hz such

that y = t ∈ V(AGn[hy]) and z = v ∈ V(AGn[hz]),

respectively. If AGn[jx] − F is isomorphic to AGn−2, by

Theorem 1, there is a hamiltonian path HP from y to z in AGn[jx] − F. If AGn[jx] − F contains more than one

subcomponent of AGn[jx], by Lemma 1 if n = 5, and

by Lemma 2 if n > 5, there is a hamiltonian path HP from y to z in AGn[jx] − F. By Lemma 4, there exist

two disjoint paths DP1 and DP2, such that DP1 joins t

and y , and DP2 joins v and z . Moreover, the two paths

span all of the vertices in AGn(n − {i, j1,. . . , jx}). Thus

s, Px, t, t , t , DP1, y , y, HP, z, z , DP2, v , v, RP, s forms a

hamiltonian cycle, and for each l ∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . ,|V(AGn)|

2 }, the distance between s and t on

the cycle is l.

Case 3. Suppose that s and t belong to different subcompo-nents of AGn, and the minimum length path connecting s and

t has the forms, MP, t , t , t. Then d(s, t) = d(s, t ) + 2. See Figure 5b for an illustration. Since AGn[i] is

isomor-phic to AGn−1, by the inductive hypothesis, for each l0 ∈

{d(s, t ), d(s, t )+1, d(s, t )+2, . . . , |V(AG

n[i])|−d(s, t )},

we can construct a hamiltonian cycle HCi of AGn[i] such

that the distance between s and t on the cycle is l0. Let u and v be the two neighbors of t on HCi. Let HCi =

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s, LP, u, t , v, RP, s, and let P0 = s, LP, u, t , t , t.

With-out loss of generality, let L(P0) = l0+ 2. By Proposition 2,

d(t , u) = 1, so we have |AS(t ) ∩ AS(u)| = 1. This means that we can find a subcomponent AGn[j1] for which j1∈ n − {i}. If t, t /∈ V(AGn[j1]), by using st -expansion,

the proof is the same as Case 1 but we replace vertex

t in Case 1 with vertex t in this case. So we consider the case in which t, t ∈ V(AGn[j1]), that is, j1 = j.

There exist two disjoint edges (u, p1) and (t , t ) in Ei,j1

n .

By Proposition 1,(p1, t ) ∈ E(AGn[j1]). By Proposition 2, d(t , v) ≤ 2 hence AS(t ) = {j1}, and AS(t ) = AS(v). Since |N(t )| = 2, we can find a subcomponent AGn[ht], and a

vertex t∈ V(AGn[ht]) such that (t , t) ∈ Eni,htfor some ht

n − {i, j1}. Since |N(v)| = 2 and AS(t ) = AS(v), we can

find a subcomponent AGn[hv], and a vertex v ∈ V(AGn[hv])

such that(v, v ) ∈ Ei,hv

n for some hv ∈ n − {i, j1, ht}. Let

F ⊆ V(AGn) and F = {t∗}. By Lemma 2, there exists a

hamiltonian path HP of AGn(n − {i}) joining t and v . Thus

s, P0, t, HP, v , v, RP, s forms a hamiltonian cycle, and for

each l0∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . , |V(AGn[i])| −

d(s, t) + 1}, the distance between s and t on the cycle is l0. Now we modify the st-expansion slightly to expand the path P0between s and t to various lengths. We describe the

details as follows.

Suppose that n≥ 6. See Figure 4d for an illustration. We can insert one subcomponent of AGn[j1], which is isomorphic

to AGn−2, into P0as follows. Because d(p1, t) = 2, which

is less than the diameter of AGn−2, and by the symmetric

property of the alternating group graph, we may regard p1

and t as in the same subcomponent of AGn[j1], say C. By

Lemma 2, there is a hamiltonian path HP1of C− Fjjoining

p1 and t with L(HP1) = |V(AGn−2)| − 2. Let C∗ be the

m subcomponents of AGn[j1] we wanted to insert into P0,

where m is the number of subcomponents of AGn[j1]. We

regard p1 and t as in different subcomponents of AGn[j1].

By Lemma 2, there is a hamiltonian path HP1 of C− Fj

joining p1 and t with L(HP1) = m|V(AGn−2)| − 2. Thus,

we can construct a path HP1 between p1 and t such that L(HP1) = I1|V(AGn−2)| − 2 for each integer I1 with 1 ≤

I1 ≤ n − 1. Let P1 = s, LP, u, p1, HP1, t. Thus, we have L(P1) = l0+ I1|V(AGn−2)| − 2 = l0+ I1(n−2)!2 − 2. Since

d(s, t)−2 ≤ l0≤ |V(AGn[i])|−d(s, t)+2, we haveI1(n−2)!2 +

d(s, t)−4 ≤ L(P1) ≤ I1(n−2)! 2 +(n−1)!2 −d(s, t). For each 1 ≤ I1≤ n−1,(I1−1)(n−2)! 2 +(n−1)!2 −d(s, t) ≥ I1(n−2)! 2 +d(s, t)−4

if n≥ 6. Therefore, for each l1∈ {d(s, t), d(s, t)+1, d(s, t)+

2,. . . , (n − 1)! − d(s, t)}, we can construct a path P1from

s to t such that the distance between s and t on the path

is l1. Then, similar to the st-expansion described in Case 1,

we can expand the path between s and t such that for each

lx∈ {d(s, t), d(s, t) + 1, d(s, t) + 2, . . . ,(x+1)(n−1)!2 − d(s, t)},

we can construct a path Pxfrom s to t such that the distance

between s and t on the path is lx. Hence for any integer l with

d(s, t) ≤ l ≤ |V(AGn)|

2 , we can construct a path joining s and t with the length of the path being l.

For n= 5, that is, AG5, we have d(s, t) = 4 in this case.

As described above,s, LP, u, t , t, HP, v , v, RP, s forms a

hamiltonian cycle, and for each l0∈ {4, 5, 6, . . . , 12}, the

dis-tance between s and t on the cycle is l0. Let Fj ⊆ V(AGn[j1])

and Fj = {t }. By Theorem 1, we can find a

hamilto-nian path HP1of AGn[j1] − Fj joining p1and t. Let P1 =

s, LP, u, p1, HP1, t. We have 11 ≤ L(P1) ≤ 19. Therefore,

for each l1 ∈ {4, 5, 6, . . . , 19}, we can construct a path P1

from s to t such that the distance between s and t on the path is l1in AG5. Let u1and t1be two adjacent vertices on HP1.

Then, for each l2∈ {4, 5, 6, . . . ,|V(AG2 5)|}, we can construct a

path P2from s to t such that the distance between s and t on

the path is l2by u1t1-expansion.

To construct a hamiltonian cycle, the proof is the same as that given in Subcase 2.1 and Subcase 2.2 by replacing vertex

t in Case 2 with vertex t∗in this case.

Hence the theorem is proved. ■

4. CONCLUDING REMARKS

In this article, we have proposed a new concept called panpositionable hamiltonicity. We now explain some rela-tionship between panpositionable hamiltonicity and pancon-nectivity. A hamiltonian graph G is panpositionable if for any two different vertices x and y of G and for any inte-ger l satisfying d(x, y) ≤ l ≤ |V(G)| − d(x, y), there exists a hamiltonian cycle C of G such that the relative distance between x, y on C is l. A graph G is panconnected if there exists a path of length l joining any two different vertices

x and y with d(x, y) ≤ l ≤ |V(G)| − 1. If G is

panpo-sitionable hamiltonian, it is clear that there exists a path of length l joining any two different vertices x and y with

d(x, y) ≤ l ≤ |V(G)| − d(x, y). If we can find that G

con-tains a path of length l joining any two different vertices x and y with|V(G)| − d(x, y) + 1 ≤ l ≤ |V(G)| − 1, then we can conclude that G is panconnected. By Theorem 1, the fault tolerant hamiltonian properties of the alternating group graph AGn, there exists a path of length l joining any two

different vertices x and y with n2! − 4 ≤ l ≤ n2!− 1 in AGn

if n≥ 4 [1]. Therefore, we can obtain the following known result as a corollary.

Corollary 1. [1] Alternating group graphs AGn are

pan-connected for all n≥ 4.

We give an example to show that a panconnected graph G is not necessarily panpositionable hamiltonian. Let

n, s1, s2,. . . , sr be integers with 1 ≤ s1 < s2 < · · · < sr.

The circulant graph C(n; s1, s2,. . . , sr) is a graph with

ver-tex set {0, 1, . . . , n − 1}. Two vertices i and j are adjacent if and only if i − j = ±sk (mod n) for some k where

1 ≤ k ≤ r. We can check that C(n; 1, 2) is panconnected by brute force for n∈ {5, 6, 7, 8, 9, 10}. However, C(10; 1, 2) is not panpositionable hamiltonian. Figure 6 shows the struc-ture of C(10; 1, 2). Consider vertex 0 and vertex 2, with

d(0, 2) = 1. We prove by contradiction that C(10; 1, 2) does

not contain a hamiltonian cycle HC with dHC(0, 2) = 5.

(11)

FIG. 6. The circulant graph C(10; 1, 2).

of C(10; 1, 2) with dHC(0, 2) = 5. There are three

possi-ble paths, P1 = 0, 8, 9, 1, 3, 2, P2 = 0, 9, 1, 3, 4, 2, and P3 = 0, 1, 3, 5, 4, 2, of length 5 joining vertex 0 and

ver-tex 2. If HC contains P1, then the edges(0, 1), (0, 2), (0, 9)

cannot belong to HC. If HC contains P2 or P3, then the

edges (2, 0), (2, 1), (2, 3) cannot belong to HC. Hence for

n = 10, there does not exist any hamiltonian cycle in C(10; 1, 2) such that the distance on the cycle between

ver-tex 0 and verver-tex 2 is 5. So C(10; 1, 2) is not panpositionable hamiltonian. In fact, the circulant graph C(n; 1, 2) is pan-connected for every n ≥ 5, but it is not panpositionable hamiltonian for some values of n. Therefore, the panposi-tionable hamiltonian property is a stronger property for an interconnection network. Future work will try to find the panpositionable hamiltonicity of other interconnection net-works and some relationships between these hamiltonian-like concepts.

Acknowledgments

The authors thank the anonymous referees for their helpful suggestions and comments which, remove some ambiguities and improve presentation.

REFERENCES

[1] J.M. Chang, J.S. Yang, Y.L. Wang, and Y. Cheng, Pan-connectivity, fault-tolerant hamiltonicity and hamiltonian-connectivity in alternating group graphs, Networks 44 (2004), 302–310.

[2] E. Cheng and M. Lipman, Fault tolerant routing in split-stars and alternating group graphs, Congr Numer 139 (1999), 21–32.

[3] E. Cheng and M. Lipman, Vulnerability issues of star graphs, alternating group graphs and split-stars: Strength and toughness, Discr Appl Math 118 (2002), 163–179.

[4] E. Cheng, M. Lipman, and H.A. Park, Super-connectivity of star graphs, alternating group graphs and split-stars, Ars Combin 59 (2001), 107–116.

[5] J. Fan, Hamilton-connectivity and cycle-embedding of the Möbius cubes, Inform Process Lett 82 (2002), 113–117. [6] F. Harary, Graph theory, Addison-Wesley, Reading,

Mas-sachusetts, 1994.

[7] J.S. Jwo, S. Lakshmivarahan, and S.K. Dhall, A new class of interconnection networks based on the alternating group, Networks 23 (1993), 315–326.

[8] X. Lin and P.K. McKinley, Deadlock-free multicast worm-hole routing in 2-D mesh multicomputers, IEEE Trans Parallel Distrib Syst 5 (1994), 793–804.

[9] Y.C. Tseng, M.H. Yang, and T.Y. Juang, Achieving fault-tolerant multicast in injured wormhole-routed tori and meshes based on Euler path construction, IEEE Trans Com-put 48 (1999), 1282–1296.

數據

FIG. 1. AG 3 and AG 4 .
FIG. 2. Illustrations for Lemma 4.
FIG. 3. Theorem 2, case 1.
FIG. 4. An illustration of st-expansion.
+3

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