J. Parallel Distrib. Comput. 63 (2003) 1277–1287

ARTICLE IN PRESS

$Grant

China. Gra�Fax: +8

E-mail a

0743-7315/$

doi:10.1016

Embedding kðn � kÞ edge-disjoint spanning trees inarrangement graphs$

Chin-Tsai Lin�

Department of Information Management, Kun-Shan University of Technology, 949 Da Wan Rd., Yung-Kang City, Tainan Hsien 710, Taiwan, ROC

Received 29 August 2002; revised 6 May 2003

Abstract

The arrangement graphs are a class of generalized star graphs. In this paper we construct a graph that consists of the maximum

number of directed edge-disjoint spanning trees in an arrangement graph. The paths that connect the common root node to any

given node through different spanning trees are node-disjoint, and the lengths of these paths differ from the shortest possible lengths

by a small additive constant. This graph can be used to derive fault-tolerant algorithms for broadcasting and scattering problems

without prior knowledge of the faulty elements of the network.

r 2003 Elsevier Inc. All rights reserved.

Keywords: Edge-disjoint spanning trees; Node-disjoint paths; Alternating group graphs; Arrangement graphs; Fault tolerance

1. Introduction

A suitable interconnection network is an importantpart of a multicomputer or multiprocessor system. Thisnetwork is usually modeled by a symmetric graph, wherethe nodes denote the processing elements and the edgesdenote the communication channels. One of the mostefficient interconnection networks is that of binary n-cubes [12]. Another family of regular graphs, called stargraphs [1,2], has been extensively studied. Subsequently,the arrangement graph was proposed as a generalizationof the star graph in an attempt to solve the scalabilityproblem of the star graph topology, while preserving itsattractive features [6]. An (n; k)-arrangement graph,denoted by An;k; is specified by the two integers, n and k;where 1pkpn � 1: Its vertices correspond to thearrangements of k elements chosen out of the setf1; 2;y; ng; and its edges connect two vertices thatdiffer in exactly one dimension. The ðn; kÞ-arrangement

graphs are regular in degree kðn � kÞ; size n!ðn�kÞ!;

diameter I3k=2m; and average distance [4] Hk þ kðk�2Þn

;

where Hk ¼ Ski¼11=i denotes the kth Harmonic number

[14]. It is also known that An;k can be decomposed into n

Sponsor: National Science Council of the Republic of

nt No.: NSC 90-2213-E-168-012.

86-6-273-2726.

ddress: ctlin@mail.ksut.edu.tw.

- see front matter r 2003 Elsevier Inc. All rights reserved.

/S0743-7315(03)00107-2

subgraphs An�1;k�1 by fixing each different element inone particular position 1 to k [6]. If k is made equal ton � 1; the star graph will be obtained. The alternatinggroup graph is also obtained if k is made equal to n � 2[4,13]. Some research on the arrangement graph can befound in [3–8,11,15]. By tuning the two parameters n

and k; we can have a more suitable size when designingan interconnection network.

Efficient parallel algorithms for multiprocessors ormulticomputers must make effective use of the under-lying communication channels. In this paper, a schemeis introduced for constructing on the ðn; kÞ-arrangementgraph kðn � kÞ directed edge-disjoint spanning trees

(EDSTs). This is the maximum number of edge-disjointspanning trees that can be constructed on the ðn; kÞ-arrangement graph, since the degree is kðn � kÞ: Thedepth of the EDSTs graph differs from the minimumpossible depth by a small constant. The root nodeof EDSTs has kðn � kÞ node-disjoint paths to each oneof the other nodes, and one path through each of thekðn � kÞ edge-disjoint spanning trees. Similar graphshave been previously constructed on other interconnec-tion networks, such as the binary hypercube [12], thecube-connected-cycles (CCC) [10], and the star graphs[9]. The construction of the EDSTs graph is equivalentto the problem of exploiting the node-disjoint pathsbetween one node and all the other n!

ðn�kÞ! � 1 nodes ofthe ðn; kÞ-arrangement graph. Using the EDSTs graph

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C.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–12871278

we can derive fault-tolerant algorithms for the broad-casting and scattering problems, as proposed byFragopoulou and Akl [9].

The remainder of the paper is organized as follows.Section 2 introduces the notations and definitions thatare used here. Section 3 presents the scheme ofembedding kðn � kÞ directed edge-disjoint spanningtrees. Conclusions are drawn in Section 4.

14

13

12

24 34

23

32

43

53 52

54

Fig. 1. A5;2:

1

8 9

3 4 5 6 72

Fig. 2. Position-held-by relationships between elements for v =

6531792 in A9;7:

2. Notations and definitions

Let n and k be two integers satisfying 1pkpn � 1;and let us denote /nS ¼ f1; 2;y; ng and /kS ¼f1; 2;y; kg: Let Pn

k be the set of arrangements of k

elements out of the n elements of /nS: The k elementsof an arrangement v are denoted as v1v2yvk:

Definition 1. The ðn; kÞ-arrangement graph An;k ¼ðV ;EÞ is given by:

V ¼fv1v2yvkjvi in /nS and viavj for

iajg ¼ Pnk; and

E ¼fðv; uÞjv and u in V ; and for some i in

/kSviaui; and vj ¼ uj for jaig:

For example, A5;2 is shown in Fig. 1.The routing in arrangement graphs is based on a cycle

representation for the set of arrangements described asfollows [6]. Let v ¼ v1v2yvk be an arrangement in Pn

k

and let EXTðvÞ ¼ /nS� fv1; v2;y; vkg: We refer to theelements of EXTðvÞ as the external elements of v: Thespecial node (arrangement) 12yk is called the identitynode, and is denoted Ik: The elements of EXTðIkÞ arecalled the foreign elements, and they are the elementsk þ 1; k þ 2;y; n: A tuple C ¼ ðx1; x2;y; xtÞ is calleda p-cycle for some node v; if for any i between 1 andt � 1; xipk and the position xi in v is held by xiþ1; andthe position xt is held by x1 if xtpk: The cycle C iscalled an internal cycle if all of its elements x1; x2;y;and xt are smaller than or equal to k: On the other hand,C is called a hybrid cycle1 if it contains one foreignelement. Any node v of An;k can be represented by a setof internal cycles and hybrid cycles such that eachhybrid cycle contains exactly one foreign element, whichis always written as the last element in the cycle.Therefore, the first element of a hybrid cycle is alwaysnon-foreign and is external to v: For example, in A9;7 thenode v ¼ 6531792 can be represented by the two cyclesð2; 5; 7Þ and ð4; 1; 6; 9Þ because position 1 is held by 6,position 2 is held by 5, and so on (see Fig. 2). Note thatthe external foreign elements, such as 8, are regarded asif they were at their correct positions with respect to Ik

even though the corresponding positions do not really

1Hybrid cycles are called external cycles in [6].

exist. The cycle ð2; 5; 7Þ is an internal cycle since all itselements are less than or equal to k ¼ 7; while the cycleð4; 1; 6; 9Þ is hybrid since it has the foreign element 9.Elements 3 and 8 are omitted in the cycle notationbecause 3 is at its correct position with respect to Ik; and8 is an external foreign element. We refer to the internalelements that are at their correct positions with respectto Ik as the invariant elements.

Definition 2. Consider a node q ¼ q1q2yqk of An;k: Letgq be the bijection from /nS to itself:

gqðlÞ ¼xði mod tÞþ1 if l is in a p-cycle ðx1; x2;y; xtÞ

of q; and l ¼ xi;

l otherwise:

8><>:

We define Fq; the translation with respect to q; as thebijection from Pn

k to itself: For all node v ¼ v1v2yvk inPn

k;

FqðvÞ ¼ gqðv1Þgqðv2ÞygqðvkÞ:The reverse translation with respect to q; denoted byF�1

q ; is g�1q ðv1Þg�1

q ðv2Þyg�1q ðvkÞ; where g�1

q is the inversefunction of gq:

ARTICLE IN PRESSC.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–1287 1279

Clearly, Fq and F�1q are automorphisms on An;k: In

An;k; for two adjacent nodes v and u different in vi andui; FqðvÞ and FqðuÞ are different in gqðviÞ and gqðuiÞ: Forexample, in A7;4 if q is 3652 ¼ ð1; 3; 5Þð4; 2; 6Þ;Fqð1357Þ ¼ 3517; Fqð1457Þ ¼ 3217; F�1

q ð1357Þ ¼ 5137;and F�1

q ð1457Þ ¼ 5637:Therefore, the routing between two arbitrary nodes v

and u in An;k can be translated by F�1u to the routing

between node F�1u ðvÞ and the identity node Ik ¼ 12yk:

That is achieved by sorting each p-cycle of F�1u ðvÞ to

move each of its misplaced elements to its correctposition of Ik:

We now define the rotation operation that isimportant for constructing the kðn � kÞ directed edge-disjoint spanning trees.

Definition 3. Let us define a bijection r from the setf1; 2;y; ng to itself, as follows:

rðiÞ ¼i if i4k;

ði mod kÞ þ 1 otherwise:

�

(Note that r maps foreign elements to themselves, and itmaps the elements in f1; 2;y; kg as follows:1-2-?-k � 1-k-1Þ: The rotation of a node v

of An;k is defined to be the node:

RðvÞ ¼ rðvkÞrðv1Þrðv2Þyrðvk�1Þ

or equivalently RðvÞ ¼ v0 so that v0rðiÞ ¼ rðviÞ; 1pipk:

This means that the position of each element of v andthe element itself are mapped through r: For example, inA7;4; Rð4152Þ ¼ 3125 and Rð6543Þ ¼ 4651: By rotationof a network, we mean that rotation is applied to eachnode of the network. The rotation operation is anautomorphism on An;k that has the following twoproperties.

(1)

It maps the identity node 12yk to itself. As anextension to this, nodes v and RðvÞ are always at thesame distance from the identity node.

(2)

Let ðv; uÞ be an edge, where u differs from v byreplacing vi with ui: Then ðRðvÞ;RðuÞÞ is an edge andRðuÞ differs from RðvÞ by replacing RðvÞrðiÞ ¼ rðviÞwith RðuÞrðiÞ ¼ rðuiÞ:

The conjugation operation is also important forconstructing the kðn � kÞ directed edge-disjoint span-ning trees.

Definition 4. The conjugation of a node v of An;k withrespect to the permutation ði; jÞ; iaj and k þ1pi; jpn; denoted by ði; jÞ 3 v 3 ði; jÞ; is the node derivedfrom v by mapping both the position of each element ofv and the element itself through ði; jÞ:

Since both i and j are foreign elements, the nodeði; jÞ 3 v 3 ði; jÞ is derived from v by replacing i with j and

vice versa if they occur. For example, the conjugation ofv ¼ 6543 of A7;4 with respect to ð5; 6Þ is 5643. It can beverified that the conjugation operation preserves thecycle structures, except that the roles of the foreignelements are exchanged, and it is also an automorphismon An;k: (In fact, the rotation operation R can beconsidered as the conjugation with respect to r:)

We also use the following notations for a node v:

ðx; yÞ 3 v

¼ neighbor node of v that is derived from v

by replacing x with y for some x infv1; v2;y; vkg and y in EXTðvÞ;

Cx

¼ p-cycle that contains element x; eðCxÞ ¼ external non-foreign (first) element in Cx

if Cx is hybrid (0 otherwise),

f ðCxÞ ¼ foreign element in Cx if Cx is hybrid

(0 otherwise),

hðxÞ ¼ position held by element x if xeEXTðvÞ

(0 otherwise).

According to the positions held by the foreign elements,we can enforce a unique cyclic order among the hybridcycles as follows.

Definition 5. If the cycle structure of v contains a hybridcycle Cj ¼ ðx1; x2;y; xt; jÞ and at least one other hybridcycle, then zð jÞ is the first position in v; among hð jÞ þ1;y; k; 1;y; hð jÞ � 1; that is held by a foreign elementother than j: Otherwise, zð jÞ ¼ 0: For example, for nodev ¼ 57638 ¼ ð2; 7Þð4; 3; 6Þð1; 5; 8Þ of A9;5 we have zð7Þ ¼hð6Þ ¼ 3; zð6Þ ¼ hð8Þ ¼ 5; and zð8Þ ¼ hð7Þ ¼ 2:

For some node v containing at least two hybrid cycles,the cyclic order contributes a scheme to construct a setof node-disjoint shortest paths from v to Ik: if Cj is oneof the hybrid cycles, then we derive a path by firstsorting Czð jÞ; then sorting the p-cycles other than Cj andCzð jÞ; and finally sorting Cj: For example, the shortestpath from v ¼ 57638 of A9;5 to Ik that ends with sorting(1, 5, 8) is 57638-5

%2638-526

%48-52

%348-

%12348-

1234%5: This path sorts ð2; 7Þ; ð4; 3; 6Þ; and ð1; 5; 8Þ in

order. Similarly, we have another two shortest paths,which end with sorting ð2; 7Þ and ð4; 3; 6Þ; respectively:57638-576

%48-57

%348-

%17348-1734

%5-1

%2345; 57638

-%17638-1763

%5-1

%2635-126

%45-12

%345: It can be

verified that these paths are node-disjoint to one anotherexcept the ends.

The following definition is the trick to derive a consistentscheme to construct node-disjoint shortest paths.

Definition 6. For each node v (excluding the identitynode), we define w as follows:

(1)

If the cycle structure of v contains a hybrid cycleCj ¼ ðx1;x2;y;xt; jÞ; t41; then wð jÞ ¼ x1; i.e., theexternal non-foreign element in Cj: For example, for

ARTICLE IN PRESSC.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–12871280

node v ¼ 5143 ¼ ð2; 1; 5Þð3; 4Þ of A7;4 we havevwð5Þ ¼ v2 ¼ 1:

(2)

If the cycle structure of v contains a hybrid cycleCj ¼ ðx; jÞ and at least one other hybrid cycle, thenwð jÞ ¼ x: For example, for node v ¼ 1365 ¼ð4; 5Þð2; 3; 6Þ of A7;4 we have vwð5Þ ¼ v4 ¼ 5:

(3)

If the cycle structure of v contains a unique hybridcycle ðx; jÞ and at least one internal cycle, then wð jÞis the first position in v; among x þ1;y; k; 1;y; x � 1; that belongs to an internalcycle. For example, if node v ¼ 5243 ¼ ð1; 5Þð3; 4Þof A7;4; then we have vwð5Þ ¼ v3 ¼ 4:

(4)

Otherwise, let wð jÞ ¼ 0 and vwð jÞ ¼ 0 (for dummypurpose). For example, if v ¼ 1534 ¼ ð2; 5Þ of A7;4;then we have wð5Þ ¼ vwð5Þ ¼ wð6Þ ¼ vwð6Þ ¼ wð7Þ ¼vwð7Þ ¼ 0:

The idea behind wð jÞ is to indicate the position tocorrect first for the shortest paths from v to Ik that fix j

at position hð jÞ of each node except node Ik (the lasthop is via node ðhð jÞ; jÞ 3 Ik). If there are at least twohybrid cycles in v; Cases (1) and (2) relate to the shortestpaths that start with sorting Czð jÞ as described above.Case (3) relates to the shortest paths that start withsorting the internal cycle containing wð jÞ and end withsorting the hybrid cycle ðx; jÞ; which have a dual shortestpath starting with sorting the hybrid cycle ðx; jÞ andending with sorting the internal cycle containing wð jÞ:For example, in A7;4 the shortest path from node 5243via node 5234 to Ik is 5243-52

%13-521

%4-52

%34-

%1234:

Its dual shortest path is 5243-%1243-12

%53-125

%4-

12%34: If the cycle structure of v contains a unique hybrid

cycle and at least one internal cycle, then the shortestpaths of Case (1) contain a subpath of Case (3) after theelements x1; x2;y; and xt�1 are corrected with respectto Ik: For example, in A7;4 the shortest path from node5143 via node 5234 to Ik is 5143-5

%243-52

%13-521

%4-

52%34-

%1234; which contains the above shortest path

from 5243 via 5234 to 1234:

Definition 7. A directed edge ðv; uÞ in An;k is of type ði; jÞif F�1

v ðuÞ ¼ ði; jÞ 3 Ik; 1pipk and k þ 1pjpn:

3. The edge-disjoint spanning trees

In this section we construct the kðn � kÞ directededge-disjoint spanning trees graph, rooted at the identitynode of An;k: The spanning tree rooted at node ði; jÞ 3 Ik

is denoted by Tj

i ; and each spanning tree includes allnodes of An;k except the identity node. The reverse-direction spanning tree of T

ji is a rendezvous result of

the disjoint paths whose last hop is via node ði; jÞ 3 Ik

from each one of the other nodes to the identity node.By joining all T

ji ’s with Ik as the common root, we

obtain the EDSTs graph. Since An;k is symmetric, the

EDSTs graph can be easily translated into a graphrooted at any other node.

We now describe an algorithm, Parentðv;Tj

i Þ; forgiven node v (excluding the identity node and itsneighbors), that computes the parent node of v in eachone of the spanning tree T

ji for 1pipk and k þ

1pjpn: In this algorithm, pj

i ðvÞ denotes the parent ofnode v in spanning tree T

ji :

For example, the EDSTs graph on A5;2 is shown inFig. 3. The spanning trees T

ji ; 1pip3 and 4pjp5; of

A5;3 are shown in Figs. 4 and 5. Table 1 shows the parentnodes (as well as the corresponding statements) of v ¼234AB7C98 ¼ ð1; 2; 3; 4;AÞð5;BÞð6; 7;CÞð8; 9Þ in A13;9

through different spanning trees.In the algorithm, statements (9) and ð90Þ can be

combined; and they are separated merely for thefollowing analysis. In order to clarify this idea, wemay express v ¼ v1v2yvk by the form v1v2yvkfyg;which has v followed by EXTðvÞ:

Lemma 1. Let v ¼ v1v2yvk be a node of An;k and x be an

external element of v: In An;k all the paths from v

satisfying the following conditions are node-disjoint to one

another except node v:

(1)

the lengths of the paths are 3 or less, (2) the paths start at different neighbors of v and end at a

node of the form yfvi;yg; 1pipk; and

(3)

all the external elements in EXTðvÞ � fxg are

external with respect to any node in the paths

(including the end nodes).

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(1)

(1)

(4)(7)(7)

(9’)

(1)(1) (1) (1) (1) (1) (1)

(1) (3) (3)(2)

(1)

12

52

1525 51 21

(1)

(1)

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(1)

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(1)

(1)

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(1) (3) (3)(2)

(1)

12

(1)

(1)

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(1) (3) (3)(2)

(1)

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(1)

(1)

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(9’)

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(1) (3) (3)(2)

(1)

12

(1)

(1)

(4)(7)(7)

(9’)

(1)(1)(1)

(1) (1) (1) (1)

(1) (3) (3)(2)

(1)

T

T

TT

T42

32 45 43 41

35 53 23 13 31

54 34 24

1452

42 32 54 53 51

34 24 14 43 23 13 41 31

45 35 25

15

13

15 14 53 43 23

54 51 52 45 41 42 25 24 21

35 34 31

3214

15 13 54 24

53 51 52 35 31 32 25 23

45 41

4215

14 13 45 35 25

43 41 42 34 31 32 24 23 21

54 53 51

52

31

32

41

42

25

1 2

T

Fig. 3. EDSTs graph on A5;2:

C.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–1287 1281

Proof. Assume there exist two such paths coincident at

some node. The subgraph of An;k having all the elementsin EXTðvÞ � fxg external with respect to any node is astar graph Akþ1;k: It is known that any minimum cycle ina star graph is of length 6. Therefore, the two paths mustbe both of length 3, and form a cycle of length 6. Sinceany pair of farthest nodes in a minimum cycle of a stargraph have identical external element(s), the coincidentend node cannot be of the form yfvi;yg; whichcontradicts the second condition. &

Theorem 1. The Parentðv;Tj

i Þ algorithm defines a span-

ning tree, rooted at the node ði; jÞ 3 Ik: The kðn � kÞspanning trees constructed by the Parent algorithm have

the following properties:

(1)

If all the edges of the spanning trees are directed from

parent to children nodes, these spanning trees are

edge-disjoint. Consequently, the EDSTs graph with

node 12yk as the common root contains all edges of

An;k except those edges that are directed towards

node 12yk:

(2) The identity node has kðn � kÞ disjoint paths to any

other node of An;k; such that one path through each

one of the kðn � kÞ spanning trees. The lengths of

these paths differ from the shortest possible lengths

by a small additive constant.

(3) The depth of the EDSTs graph is less than or equal to

I3ðk � 1Þ=2mþ 5; which is optimal to within a small

additive constant. This constant is less than or equal

to 3.

(4) Each spanning tree can be derived from its preceding

one by the application of a rotation or conjugation.

From the properties of the rotation and conjugation

operations, we conclude that all the spanning trees

are isomorphic.

Proof. We prove each of the properties separately.

(1)

It is sufficient to prove that the parents of a node v

in different spanning trees are different neighbors ofv: In other words, we have to prove each one of then � k external elements moves evenly to k differentpositions through k different spanning trees. Node v

must belong to one of the following three cases:

(i) v does not contain any hybrid cycle. That is,

all foreign elements are external. Hence,p

ji ðvÞ ¼ v1yvi�1jviþ1yvk; for 1pipk and

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(6)

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(7)(4)(4)(6)(9’)(4)(7)(7)(1)(1)(9’)(4)

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(3)

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(2)

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413425453523

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314

145 345 142 154 254 214 134

(8) (9’) (9’) (9’) (9’)(8)

(7)(4)(4)(6)(9’)(4)(7)(7)(1)(1)(9’)(4)

(1)(1)(1)(1)(5)(4)(7)

(1) (1) (3)(2)

(1)(4)(1)(1)(7)(1)(1)

(3)

(1)(1)

(3)(1) (1)

412 521 321 451 431213415513253 452 125153

235 543 215 315 243 432 512 312 524 324 341 251 351 231 531352 135152

(1)(3)

(1)

(1)(3)

(2) (2)

(9’)

(7)

(1)

T 41

435

Fig. 4. Spanning trees T41 ;T4

2 ; and T43 of A5;3:

C.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–12871282

k þ 1pjpn: Clearly, the parents of node v inthe kðn � kÞ spanning trees are all different.

(ii)

v contains exactly one hybrid cycle Cj; where j

is foreign. Clearly, any external foreignelement f is at position i of p

fi ðvÞ for

1pipk: We now show the external non-foreign element eðCjÞ is at a different positionof p

ji ðvÞ for 1pipk: Let l be the position of

element eðCjÞ in pj

i ðvÞ: Table 2 summarizes allthe possible cases of i and such v: In this table,the cases of v correspond to the cases of thehybrid cycle of length 2, 3, and more,respectively. It can be verified that in each

case of v there is a one-to-one correspondencebetween i’s and l’s. Therefore, eðCjÞ is at adifferent position of p

ji ðvÞ for 1pipk:

(iii)

v contains at least two hybrid cycles. AssumeCj1 ;Cj2 ;y;Cjs are the hybrid cycles of v;where j1; j2;y; js are foreign elements andzð jtÞ is in Cjðt mod sÞþ1

for 1ptps: Clearly, anyexternal foreign element f of v is at position i

of pfi ðvÞ for 1pipk: Without loss of general-

ity, given l; 1plpk; we show in Table 3 thereexists a pair, i and j; where jAf j1; j2;y; jsg;such that eðCjtÞ is in position l of p

ji ðvÞ:

Therefore, all the pj

i ðvÞ’s are different.

ARTICLE IN PRESS

534

(1)(1)(3)

(6)

(2)

(2)

123

(8) (9’) (9’) (9’) (9’)(8)

(7)(4)(4)(6)(9’)(4)(7)(7)(1)(1)(9’)(4)

(1)(1)(1)(1)(5)(4)(7)

(1) (1) (3)(2)

(1)(4)(1)(1)(7)(1)(1)

(3)

(1)(1)

(3)(1) (1)(1)

(3)(1)

(1)(3) (2) (2)

(9’)

(7)

(1)

231

213

T35

125

124 425 145 325 135

423 421 415 143 142 245 324 345 321 315 134 132 435 235

412 413 243 241 524 341 342 521 215 314 312 154 152 532 431 432 234

541 523 214 514 354 512 153 531 452 251 254 534

543 542 513 453 451 351 253

352

(1)(1)(3)

(2)

(2)

123

(8) (9’) (9’) (9’) (9’)(8)

(7)(4)(4)(6)(9’)(4)(7)(7)(1)(1)(9’)(4)

(1)(1)(1)(1)(5)(4)(7)

(1) (1) (3)(2)

(1)(4)(1)(1)(7)(1)(1)

(3)

(1)(1)

(3)(1) (1)

312

(1)(3)

(1)

(1)(3) (2) (2)

(9’)

(7)

(1)

213

(6)

T25

153

143 154 453 152 253

124 134 354 423 413 451 142 452 132 352 243 254 251

314 324 421 431 145 432 412 135 351 342 543 513 215 234 214 231 241

435 125 321 341 345 542 315 523 235 514 531 541 245

425 415 325 524 534 532 521

512

(1)(1)(3)

(6)

(2)

(2)

123

132(8) (9’) (9’) (9’) (9’)

(8)

(7)(4)(4)(6)(9’)(4)(7)(7)(1)(1)(9’)(4)

(1)(1)(1)(1)(5)(4)(7)

(1) (1) (3)(2)

(1)(4)(1)(1)(7)(1)(1)

(3)

(1)(1)

(3)(1) (1)

213

231

(1)(3)

(1)

(1)(3) (2) (2)

(9’)

(7)

(1)

T15

523

423 543 524 513 521

143 243 542 124 413 514 512 421 321 541 531

342 142 134 234 453 214 314 253 532 412 312 425 325 351 241 341 431

254 153 432 452 415 352 125 251 345 235 435 451

154 354 152 145 245 215 135

315

324

Fig. 5. Spanning trees T51 ; T5

2 ; and T53 of A5;3:

Table 1

Parent nodes of v ¼ 234AB7C98 in A13:9

pj

i j ¼ A j ¼ B j ¼ C j ¼ D

i ¼ 1 ð4Þ: 234%1B7C98 ð8Þ: 23

%5AB7C98 ð8Þ: 23

%6AB7C98 ð1Þ: D34AB7C98

i ¼ 2 ð6Þ: 23%1AB7C98 ð9Þ:

%534AB7C98 ð9Þ:

%634AB7C98 ð1Þ: 2D4AB7C98

i ¼ 3 ð90Þ: 2%14AB7C98 ð9Þ: 2

%54AB7C98 ð9Þ: 2

%64AB7C98 ð1Þ: 23DAB7C98

i ¼ 4 ð3Þ: 234A%57C98 ð7Þ: 234A

%17C98 ð7Þ: 234AB7

%198 ð1Þ: 234DB7C98

i ¼ 5 ð7Þ: 234%5B7C98 ð3Þ: 234AB

%6C98 ð7Þ: 234AB7

%598 ð1Þ: 234AD7C98

i ¼ 6 ð8Þ: 234AB%1C98 ð8Þ: 234AB

%5C98 ð4Þ: 234AB7

%698 ð1Þ: 234ABDC98

i ¼ 7 ð7Þ: 234%6B7C98 ð7Þ: 234A

%67C98 ð3Þ:

%134AB7C98 ð1Þ: 234AB7D98

i ¼ 8 ð90Þ: 234AB7C9%1 ð90Þ: 234AB7C9

%5 ð90Þ: 234AB7C9

%6 ð1Þ: 234AB7CD8

i ¼ 9 ð90Þ: 234AB7C%18 ð90Þ: 234AB7C

%58 ð90Þ: 234AB7C

%68 ð1Þ: 234AB7C9D

C.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–1287 1283

ARTICLE IN PRESS

Table 2

Position held by element eðCjÞ in pj

i ðvÞ

v: eðCjÞ ¼ hð jÞ v: vwð jÞ ¼ hð jÞ v: eðCjÞahð jÞ and vwð jÞahð jÞ

Tj

i : i ¼ hð jÞ (2): l ¼ hðvwð jÞÞ (2): l ¼ hðvwð jÞÞ (2): l ¼ hðvwð jÞÞT

ji : i ¼ eðCjÞ same as i ¼ hð jÞ (4): l ¼ hð jÞ (4): l ¼ hð jÞ

Tj

i : i ¼ vwð jÞ (5): l ¼ hð jÞ same as i ¼ hð jÞ (6): l ¼ hðhð jÞÞT

ji : i ¼ other ð90Þ: l ¼ hðiÞ ð90Þ: l ¼ hðiÞ ð90Þ: l ¼ hðiÞ

C.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–12871284

Table 3

(2) ði; jÞ pair corresponding to l; 1plpk ðbðtÞ ¼ ðt � 2 mod sÞ þ 1Þ

Conditions of l stmt : ði; jÞ

l is invariant or in an internal cycle ð90Þ: ðvl ; jtÞ

We explain how the path from each node v to theidentity node is created through each one of thespanning trees. Let us consider T

ji : If vi ¼ j; then

the parent is that derived by either statement (2) or(3), and which has the properties:

l is neither invariant nor external, ð9Þ: ðvl ; jtÞ

and vl ; vv ef j1; j2;y; jsg (a)

l

l ¼ hð jtÞ (i.e., vl ¼ jt) ð4Þ: ðeðCjt Þ; jtÞl ¼ eðCjt Þ ð3Þ: ðhð jbðtÞÞ; jbðtÞÞ

Element j is fixed at position i: Hence, if theparent of p

ji ðvÞ exists, it is also derived from

pj

i ðvÞ by applying either statement (2) or (3).

l ¼ hðhð jtÞÞ (i.e., vvl

¼ jtÞ ð6Þ: ðveðCjtÞ; jtÞ

0 0

(b)

l ¼ hð jt0 Þ; t at; and 1pt ps (i.e., vl ¼ jt0 ) (7): ðhð jtÞ; jt0 Þl ¼ eðC Þ; t0at; and 1pt0ps (9): ðv ; j Þ

If f is an external foreign element of v; f is alsoan external foreign element of p

ji ðvÞ:

jt0 l t

l ¼ hðhð j ÞÞ; t0at; and 1pt0ps (i.e., v ¼ j 0 ) (8): ðeðC Þ; j Þ

0 v t j 0 t (c)

t l t

pj

i ðvÞ is closer to the identity node than v; that is, v

has a path of minimum length to Ik through Tj

i :

If j is an external element of v; then the parent of v isthat derived by statement (1), in which j is at position i:Otherwise, element j is first made external, and thenmoves immediately to position i along the path. By acareful trace of the Parent algorithm, the possiblesequences of statements applied before element j hasbecome external are as follows: (Please consider hybridcycles of length 2, 3, and more, respectively.)

It can be observed that any external foreign elementof v is always external through the parts of the paths,any external non-foreign element of v other than eðCjÞ isalways external through the parts of the paths if the firststatement applied is (4), (5), (6), or (90), and any externalnon-foreign element of v other than eðCiÞ and eðCjÞ isalways external through the parts of the paths if the firststatement applied is either (8) or (9).

To prove the paths are node-disjoint, we nowdistinguish among different of nodes.

(i) v does not contain any hybrid cycle. The paths arenode-disjoint since j is fixed at position i by T

ji ; while

the other foreign elements remain external.(ii) v contains exactly one hybrid cycle Cf ; where f is

foreign. Let us first consider the path from v to Ik

through Tj

i ; jaf : Within the path, foreign element j isfixed at position i; and if f is external with respect to anode, then f is also external with respect to its parentnode. Thus, these paths through T

ji ’s are node-disjoint

to one another for k þ 1pjpn; jaf ; and 1pipk: Theyare also node-disjoint to the paths through T

fi ’s for

1pipk; since j is external with respect to any node inthe path through T

fi : We now show that the paths

ARTICLE IN PRESSC.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–1287 1285

through Tfi ’s, for 1pipk; are also node-disjoint to one

another. The last parts of these paths are node-disjointbecause f is fixed at position i and the other foreignelements are always external. Therefore, we have toprove that the first parts of the paths are also node-disjoint. It can be observed that the paths must startwith the statement (4), (5), (6), or ð90Þ; and the first partsof the paths satisfy the conditions of Lemma 1.Therefore, these paths are node-disjoint to one another.Furthermore, the lengths of these paths differ from theminimum possible length by a small additive constantbecause the part of each path from node yf f ;yg to Ik

is a shortest path through the subgraph of An;k that fixesf at position i:

(iii) v contains at least two hybrid cycles. AssumeCj1 ;Cj2 ;y;Cjs are the hybrid cycles of v as mentionedabove. By the same way as in (ii), for k þ 1pjpn andjef j1; j2;y; jsg the paths from v to Ik through T

ji ’s

are node-disjoint to one another and to the other paths.The last parts of the other paths are node-disjoint to oneanother because jt; 1ptps; is fixed at position i; and theother foreign elements will be always external once theyhave been made external through the parts of the paths.We have to prove that the first parts of the paths are alsonode-disjoint. These paths must start by applyingstatement (4), (6), (7), (8), (9), or ð90Þ: By Lemma 1,for 1ptps and 1pipk; the first parts of paths throughT

jti ’s which start by applying statement (4), (6), or ð90Þ

are node-disjoint to one another. The paths that start byapplying statement (8) are node-disjoint to the otherpaths because they distinguish themselves by makinghð f ðCiÞÞ always external and eðCjtÞ fixed at positionhðhð f ðCiÞÞÞ through the first parts of the paths. Thepaths that start by applying statement (9) are node-disjoint to the other paths because they distinguishthemselves by making i always external and eðCjtÞ fixedat position hðiÞ through the first parts of the paths(herein i is in some other hybrid cycle than Cjt). Also,the parent nodes of v by statement (7) are not one of thenodes in the other paths. Therefore, they are all node-disjoint. Similarly, the lengths of these paths differ fromthe minimum possible length by a small additiveconstant.

(3)

From part 2 of this lemma, we note that the depthof each spanning tree (from node ði; jÞ 3 vÞ is lessthan or equal to the diameter of An�1;k�1 plus 4,I3ðk � 1Þ=2mþ 4: Consequently, the depth of theEDSTs graph (from node 12yk) is less than orequal to I3ðk � 1Þ=2mþ 5; which is the diameter ofAn;k plus 4 (if k is odd) or 3 (if k is even). A lowerbound for the depth of the EDSTs graph is posedby the fault diameter of An;k: The fault diameter ofAn;k is the diameter of the remaining graph when anarbitrary set of kðn � kÞ � 1 nodes are removedfrom An;k and is more than the fault free diameter of

An;k except k ¼ 1 (complete graphs). Therefore,the depth of the EDSTs graph is optimal towithin a small constant, which is less than or equalto 3.

(4)

According to the definition of the rotation opera-tion, when we say that each spanning tree can becyclically obtained as a rotation of its precedingone, which is equivalent to saying that spanning treeT

j

rðiÞ can be obtained as a rotation of spanningtree T

ji ; 1pipk and k þ 1pjpn: According to the

definition of the conjugation operation, when wesay that each spanning tree can be obtained as aconjugation of the preceding one, which is equiva-lent to saying that spanning tree T

j2i can be

obtained as a conjugation of spanning tree Tj1

i for3pipn; k þ 1pj1; j2pn; and j1aj2:

We first prove that edge ðv; pj1

i ðvÞÞ belongs tospanning tree T

j1i if and only if edge

(ð j1; j2Þ 3 v 3 ð j1; j2Þ; ð j1; j2Þ 3 pj1

i ðvÞ 3 ð j1; j2Þ) belongsto spanning tree T

j2i : From the properties of the

conjugation operation, ð j1; j2Þ 3 v 3 ð j1; j2Þ preservesthe cycle structure of v; except that j1 is replaced byj2; and vice versa. Consequently, from the defini-tions of wð jÞ; zð jÞ; eðCjÞ; f ðCjÞ; and hð jÞ it can beverified that the roles of wð j1Þ; zð j1Þ; eðCj1Þ;f ðCj1Þ; and hð j1Þ are replaced by wð j2Þ; zð j2Þ;eðCj2Þ; f ðCj2Þ; and hð j2Þ respectively and vice versa.From these we conclude that node v derives itsparent in T

j1i from a specific statement (1) to ð90Þ of

Parentðv;Tj1

i Þ if and only if node ð j1; j2Þ 3 v 3 ð j1; j2Þderives its parent from the same statement ofParentðð j1; j2Þ 3 v 3 ð j1; j2Þ; T

j2i Þ:

We now prove that if edge ðv; pj

i ðvÞÞ belongs tospanning tree T

ji ; then edge ðRðvÞ;Rðp j

i ðvÞÞÞ be-longs to spanning tree T

j

rðiÞ: From the properties ofthe rotation operation, the rotation of a node v isnode RðvÞ ¼ v0; so that v0rðiÞ ¼ rðviÞ: Furthermore, ifvi ¼ j then v0rðiÞ ¼ j; and if xAEXTðvÞ thenrðxÞAEXTðv0Þ: In addition, from the definitions ofwð jÞ; zð jÞ; eðCjÞ; f ðCjÞ; and hð jÞ it can be verifiedthat v0wð jÞ ¼ rðvwð jÞÞ; and zð jÞ; eðCjÞ; f ðCjÞ; andhð jÞ of v0 are equal to rðzð jÞÞ; rðeðCjÞÞ; rð f ðCjÞÞ;and rðhð jÞÞ of v respectively. From these weconclude that if node v derives its parent in T

ji by

a specific statement (1) to ð90Þ of Parent(v; Tj

i ),then node v0 derives its parent by the same statementof Parentðv0;T

j

rðiÞÞ: It can be verified for nodesthat derive their parents through each differentstatement of the Parent algorithm, that if node v isconnected to its parent in T

ji by replacing vm

then node v0 is connected to its parent in Tj

rðiÞ byreplacing v0rðmÞ:

Since both the rotation operation and theconjugation operation are automorphisms on An;k;all spanning trees, which are T

ji ’s for 1pipk and

k þ 1pjpn; are isomorphic. &

ARTICLE IN PRESSC.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–12871286

The Parent algorithm is a generalization of thatproposed in [9]; If k ¼ n � 1; they derive the sameEDSTs graph for An;k:

The EDSTs graphs can be used not only to derivefault-tolerant algorithms for broadcasting and scatter-ing problems similar to that in [9], but also to optimizeboth the start-up time and the transmission time in thebroadcasting problems [16].

However, one thing should be mentioned about thetranslation operation on the arrangement graphs.Consider a multi-node broadcasting or scattering algo-rithm. Under the all-port store-and-forward commu-nication model [12] n!

ðn�kÞ!kðn � kÞ directed edges areavailable on An;k for message transmission at each timestep. Since the algorithm proceeds symmetrically fromall nodes of the network, messages originating at eachone of the n!

ðn�kÞ! nodes of An;k are transmitted through atmost kðn � kÞ directed edges at each step. Let EiðIkÞ bethe set of kðn � kÞ directed edges on which messagesoriginating at the identity node are transmitted at timestep i of the algorithm. Let EiðqÞ be the set of kðn � kÞedges on which messages originating at node q aretransmitted at step i of the algorithm and are obtainedfrom EiðIkÞ using the translation operation with respectto q: Since the translation operation may not preservethe types of edges, there may be conflicts during theexecuting of the algorithm if no regulation is provided.Nevertheless, in an alternating group graph, if eachnode is represented by an even permutation and thetranslation of a node v ¼ v1v2yvn with respect to somenode q is redefined as the function composition q � v ¼qv1

qv2yqvn ; then the translation operation will preserve

the types of edges. The following lemma guarantees thatno conflicts arise during the execution of the algorithmson the alternating group graphs and it can be proved ina similar way to [9] for the star graph.

Theorem 2. At each time step i of the multi-node

broadcasting or scattering algorithm, if the kðn � kÞdirected edges in EiðIkÞ are all of different types and k ¼

Table 4

The lower bounds and the number of time steps performed on An;n�2 for degr

Problem Lower bounds

Message transmissions

Single-node broadcasting Mðn!2� 1Þ

Fault-tolerant single-node broadcasting Mð2n � 4Þðn!2� 1Þ

Fault-tolerant multi-node broadcasting Mð2n � 4Þðn!2� 1Þn!

2

Fault-tolerant single-node scattering Mð2n � 4Þtn

Fault-tolerant multi-node scattering Mð2n � 4Þn!2tn

n � 1 or n � 2; then the sets of directed edges EiðqÞ; where

q ranges over all nodes of An;k; are disjoint.

Proof. Assume two different edges ðv; uÞaðv0; u0ÞAEiðIkÞ for some i; and take the edges ðFsðvÞ;FsðuÞÞAEiðsÞ and ðFs0 ðv0Þ;Fs0 ðu0ÞÞAEiðs0Þ which are obtained byðv; uÞ and ðv0; u0Þ; respectively, under translation withrespect to two different nodes of An;k; s and s0:Also assume that ðFsðvÞ;FsðuÞÞ ¼ ðFs0 ðv0Þ;Fs0 ðu0ÞÞ:Let typeðaÞ denote the type of an edge a: Sincethe type of an edge is preserved under translation, weconclude that typeððv; uÞÞ ¼ typeððFsðvÞ;FsðuÞÞÞ ¼typeððFs0 ðv0Þ;Fs0 ðu0ÞÞÞ ¼ typeððv0; u0ÞÞ which contradictsour assumption that ðv; uÞ and ðv0; u0Þ are two differentedges of EiðIkÞ: &

4. Concluding remarks

This paper shows a scheme to construct kðn � kÞedge-disjoint spanning trees in An;k: The scheme isdifferent from that developed by Chen et al. [3] in thefollowing aspects:

(1)

ee of f

The number of constructed edge-disjoint spanningtrees is equal to kðn � kÞ; which is maximum, ratherthan 2ðn � kÞ:

(2)

All the spanning trees have a common root.Furthermore, the paths from a node to the rootthrough different spanning trees are node-disjoint.

(3)

The depth of the EDSTs graph is less than or equalto I3ðk � 1Þ=2mþ 5 rather than 2k � 1: If wereverse the direction of the edges adjacent to nodeIk; the depth of any spanning tree will be less thanor equal to I3ðk � 1Þ=2mþ 4:

The scheme can also be used to construct a maximumset of node-disjoint paths between any two nodes. Inmany cases, the node-disjoint paths are different fromthose proposed by Day and Tripathi [5].

ault tolerance 2n � 5 (average distance tn ¼ n þ Hn þ 7n� 1

n�1� 6)

Time steps performed

Time steps

J Mð2n�4Þnþ I3ðn�2Þ

2mþ 1 J M

ð2n�4Þnþ I2ðn�2Þ2

mþ 3

M þ I3ðn�2Þ2

m M þ I3ðn�2Þ2

mþ 3

Mðn!2� 1Þ Mðn!

2� 1Þ

Mðn!2� 1Þ Mðn!

2� 1Þ

Mtn OðMtnÞ

ARTICLE IN PRESSC.-T. Lin / J. Parallel Distrib. Comput. 63 (2003) 1277–1287 1287

With the translation operation, these spanning treescan be used to derive optimal or asymptotically optimalfault-tolerant algorithms on the alternating groupgraphs for multi-node broadcasting and scatteringproblems without prior knowledge of the faultyelements of the network, as necessary for star graphs[9]. Table 4 summarizes the lower bounds and thenumber of time steps performed to broadcast or scatterM messages from each source node on the alternatinggroup graph An;n�2: How to use the spanning trees toderive general conflict-free and fault-tolerant algorithmson the arrangement graphs for multi-node broadcastingand scattering problems is a topic for future research.

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Chin-Tsai Lin received the Master of

Science degree and Ph.D. degree in

computer science and information engi-

neering from Taiwan University in 1989

and 1996, respectively. He is currently an

Assistant Professor of Information Man-

agement at the Kun-Shan University of

Technology, Taiwan. From 1999 to

2001, he worked as an Assistant Profes-

sor of Information Management at Da-

Yeh University, Taiwan. His research

interests include network quality of service, parallel processing,

and interconnection networks.