On ring embedding in hypercubes with faulty nodes and links

ELSEVIER Information Processing Letters 68 (1998) 207-214

Informqtion ~~~~ryx3

On ring embedding in hypercubes with faulty nodes and links

Abhijit Sengupta ’

Department of Computer Science, University of South Carolina, Columbia, SC 29208, USA

Received 17 March 1997; received in revised form 13 May 1998

Communicated by C. Morgan

Abstract

In this paper, we show that in an n-dimensional hypercuhe Qn with fn faulty nodes and fe faulty edges, such that f,+f,<n-l,aringoflength2” - 2f,, can be embedded avoiding the faulty elements when fn B 0 or fe -c n - 1. When fn = 0 and fe = n - 1, if all the faulty edges are not incident on the same node, a Hamiltonian cycle can be embedded avoiding the faulty elements when n 2 4. For a Q3, however, if f,, = 0 and fe = 2, a Hamiltonian cycle might not exist even when all faulty edges are not incident on the same node. We show that a ring of size 6 can be embedded in that case. When f,, = 0 and fe = n - 1, if all the faulty edges are incident on the same node, clearly a Hamiltonian cycle cannot exist and we show that a ring of size 2” - 2 can he embedded. This generalizes a recent result of Tseng (1996) where the number of edge faults were assumed not to exceed n - 4. 0 1998 Elsevier Science B.V. All rights reserved.

Keywords: Interconnection networks; Node and edge faults; Hypercube

1. Introduction

In this paper we consider the problem of embedding a ring in a hypercube interconnection network when there are node and/or edge faults. It is known that [2, 41 when there are only edge faults, an n-dimensional hypercube remains Hamiltonian with up to n - 2 edge faults. Most of the earlier works consider only faulty edges [3-51 or only faulty nodes [ 1,7]. In a recent paper [6], considering both node and edge faults, it has been shown that in an n-dimensional hypercube, if there are fn node faults and fe edge faults,

fn+fe<n-l> fe6ne4,

a ring of length at least 2n - 2fn can be embedded avoiding the faulty elements. This paper generalizes

’ Email: [email protected].

the result of [6] (where the number of edge faults were assumed not to exceed n - 4) and shows that as long as fn+fe<n-l,aringoflength2”-2f,canbe embedded avoiding the faulty elements when fn > 0 orf,cn-l.Whenf,=Oandf,=n-1,ifall the faulty edges are not incident on the same node, we show that a Hamiltonian cycle can be embedded for Q - n’s with n > 4. For a Q3, however, if fn = 0 and fe = 2, a Hamiltonian cycle might not exist even when all faulty edges are not incident on the same node. We find that distribution of faulty edges and show that a ring of size 6 can be embedded in that case. When

fa=O and fe=n-1,

if all the faulty edges are incident on the same node, obviously a Hamiltonian cycle cannot exist and we show that a ring of size 2” - 2 (which clearly is the longest ring) can be embedded in that case.

0020-0190/98/$ - see front matter 0 1998 Elsevier Science B.V. All rights reserved

PII: SOOZO-0190(98)00159-S

208 A. Sengupta /Information Processing Letters 68 (1998) 207-214

2. Definitions and notations

An n-dimensional binary hypercube Qn is a graph with 2n nodes. Assuming each node is denoted by an n-bit vector, two nodes are adjacent iff their vector representations differ exactly in one bit position; if that bit position is i then the edge between the nodes is called the ith dimensional edge. Each bit position is called a dimension of Qn. Let 2 = { 1,2, . . . , n} represent the set of dimensions of a Q,. An n-bit vector having only i th bit 1 and everything else 0 will

be denoted by ei. @ will denote bitwise vector sum in modulo 2. For any node uj in a Qn, assuming uj gives the binary vector representation of the node, the i-dimensional edge incident on uj will be denoted by

vj i uj @ ei or by the pair of nodes (uj , uj @ ei ) . The node uj $ ei will be called the neighbor of uj along the dimension i and will be denoted by Ni (uj). For any two nodes vi and uj , let H(vi , vj) denote the set of dimensions in which the vector representations of vi and uj differ. For an edge (vi, Vi) and for a dimension t, such that (vi, vj) is not in dimension t, the edge (N,(vi), Nt(vj)) will be denoted by Zt((vi, uj)). For a set of edges S, let

It(S) = {Me) I e E S}.

For any X c Z, let Qn,x denote the edge-induced subgraph of Q,, induced by only the edges in dimensions appearing in X . It is simple to observe that Qn, x is a disconnected graph with each connected compo-

nent being a Qlxl . Suppose we form a graph G’ where each component of Qn,x is a node in G’ and two nodes representing components ci and cj of Qn,x are adjacent in G’ iff there is a node u in ci and a node v in cj such that u and v are adjacent in Q,,. Clearly, G’ is a

Qn_lxl and we will denote G’ by K(Q,,x). K(Qn,x) is a condensation of Qn obtained from Q, by squeez-

ing the set of nodes in a component of Qn,x into a single node and replacing parallel edges by a single edge. For some t E Z, let X = Z - It}. Then the Qn can be viewed as two Q,,xs having t-dimensional edges between the nodes in them. We will refer to these two Qn_ 1 s as being formed by splitting the Qn along the dimension t .

Any string of symbols chosen from 2 will be called a dimension sequence. For any node vi and a dimension sequence f, let waZk(vi ,t) denote the walk traversed in Qn as one starts from vi and traverses the

edges in the dimensions appearing in c and also in the same order as they appear in e. For example, if

t; = ~1x2~3, then w&(ui, c) denotes the walk

Vj XI Vj Cl3 ex, x2 Vj @ e,, @ ex2

3. Hamiltonian property of faulty hypercube

For any permutation n = x1x2. . . xn on Z, let a(n) denote the string s,, * x, (‘*’ is the concatenation operation), where sj , 2 < j < n, is defined recursively assj =sj-1 *xj * sj-1 andsr =xl.Clearly,o(n) is a string of length 2”. It is known [2] that for any vi and n, wuZk(vi , (T(X)) is a Hamiltonian cycle (HC) in Qn, which we will denote, in short, by C (vi, n). Assuming vc denote the node represented by the zero binary vector, C(vo, n) will be referred to as the canonical HC for the permutation n and will be denoted, in short, by C(n). It is simple to see that the following

properties hold for every C (vi, n) .

Property 1. For a permutation n = ~1x2 . . .x,, every

edge in dimension x1 appears in the HC given by

C(vi,n)foranychoiceofvi.

For a permutation YC = ~1x2 . _ x,, and some vi, suppose we construct C(vi , n) and consider any node va in the C(vi , IT). Of the two edges incident on v,, one is in dimension xl. Let z, be the dimension of the other edge incident on v,. Let Nz(v,) = vb, N,, (v,) = v,,, N,, (vg) = VJ (see Fig. 1). It is simple to see that N, (v,) = vg and the edges (vY , v,), (v,, up) and (vg, us) are three consecutive edges of C(vi , n). Hence, if the edge (v,, up) is faulty and/or any or both the nodes v, and VP are faulty, the faulty

nodes and/or the edge can be skipped using the edge (vr , IQ) in dimension z, provided this edge and its endpoints are fault-free. We will refer this to “standard node and/or edge skipping” and the edge ( vY , US),

Z

Fig. 1. Standard node and/or edge skipping.

A. Sengupta /Information Processing L.etters 68 (1998) 207-214 209

which is in the same dimension as that of (vu, VP), will be called the bridge of skip or the bridge around vu or up. Clearly for every edge (v,, vb) in a C (vi, n), there is a bridge around u, in the same dimension as (v,, up) unless (u,, VP) is in dimension x1.

hqmty 2. An edge Uj _ uk appears in C(Ui , n) iff the edge Uj @ ui _ uk @3 ui appears in C(X).

Given a node uj in a Q,,, let Zj denote the set of bit positions where uj has 1. Given an i-dimensional edge (uj, uj @ ei) in a Qn and a permutation n = x1.72.. . xn, it has been shown [4] that the edge appears in the canonical HC C(rr) iff one of the following conditions hold.

(Cl) xt = i,

(C2) x2 = i andxt E Zj, (C3) if xk = i, 2 -c k < n, then xk-1 E Zj and

h,x2,..., Xk-2) n zj = 0,

(C4) ifx,=i,theneitherZj=PIorZj={x,_i}. Given a set of faulty edges Se, if we can find a permutation n such that none of the above conditions holds for every edge in Se then we have found an HC avoiding the faulty edges. It has been shown [4] that in a Qn, if only edges could be faulty and there are at most n - 2 faulty edges, then there exist numerous canonical HCs in the Qn. Given a set of faulty edges S, , 1 Se ) < n - 2, the following Algorithm A, determines a node u and a permutation n such that the HC C(u, n) does not contain any edge from Se.

Algorithm A.

Given: A set of faulty edges Se with 1 S, 1 < n - 2 in a Q,,.

To find: A node u and a permutation K on Z such that

no edge of S, appears in C(u, n). Method: Let I Se I = t and Z,, Zb, respectively be the

set of fault-free and faulty dimensions.

LetZg={z~,z2,...,zg}andZb={Y1,Y2,...,Yb}.

Henceb<tandg+b=n. Case 1: b = t, that is, all the faulty edges are in

distinct dimensions. Assume that ui yi vi @ eYi is the faulty edge in dimension yi, 1 < i < b.

Stepl.l:Setn +ZIZ2...Zg_lyly2...ybZg.

Step 1.2: Form the vector representation of the node u as follows. Set u t the zero vector.

forj:=2tobdo set the yj-ltb bit of u equal to the

yj _ 1 th bit of the faulty edge in dimension Yj .

Set the zg_ 1 th bit of v to be equal to the zg_ 1 th bit of the faulty edge in dimension yt .

Case 2: b < t. Thus all the faulty edges are not in distinct dimensions.

Step 2.1: Choose b faulty edges such that these edges are in distinct dimensions.

Step 2.2: Set n t 2122.. .zg-1yly2.. . ybzS. Step 2.3: Find u as in case 1 using edges chosen in

Step 2.1. Step 2.4: Assume that of the remaining t - b faulty

edges, the jth edge is wj 2 wi @ eUj

and is in dimension oj .

forj:=ltot-bdo ifz,-l-jthbitofwj isOthen

u + u @ e+-j.

From the description of the algorithm, the following observations are simple. (01) given two fault-free dimensions y, z E Z, we

can always choose a n = ~1x2.. .n, such that x1 = y and xn = z;

(02) if there are t faulty dimensions, they can be ar-

bitrarily permuted among x,+1, xn-2, . . . , xn-t; (03) any set of b faulty edges in distinct dimensions

can be used in Steps 1.2 and 2.1; (04) Algorithm A sets only IS,] bits of u; the re-

maining n - IS,] bits of u can be chosen arbitrarily-the bit positions that can be cho-

sen arbitrarily are z,th, ybth, zlth, zzth, . . ., z,,-~s,~-~th bits positions of u.

Proof of correctness of Algorithm A. For the choice of n, none of the faulty dimensions appears as the first or the last dimension of n and hence neither (Cl) nor (C4) holds. Assuming 7t = ~1x2 . . .x,, , (C2) and (C3)

imply that the edge uj i Uj @ ei appears in C(n) iff whenxk=i,k#l,k#n, l xk_tth bit of Uj is 1 and

0 Xlth, XTth, . . . . xk-2th bits of Uj are all zeroes (applicable fork > 2 only). Consider Case 1 of Algorithm A. With the cho-

sen n, if possible let us assume that Uj yi uj @ eYj

appears in C(u, n). Then, Uj @ u Yj Uj @ 2, @3eyj ap-


pears in C(n). By construction, the yj_lth bit of v is same as the yj_ 1 th bit of uj . Hence the yj _ 1 th bit of uj @ v is 0 and SO is the yj-tth bit of uj @ v @ eyj. But if the yj_ 1 th bit of uj $ v is 0 then this edge cannot appear in C(n) by (C2). Hence we are done by contradiction.

Consider Case 2 of Algorithm A. With the chosen n, if possible let us assume (for Step 2.4) that wj ‘yi wj @ eaj appears in C(V, n). Then, wj $ v cri wj $ v @ euj appears in C(v, n). By construction, the zjth bit of wj and v are opposite. Hence the zg-t-jth bit of wj @ v is 1 and SO is the z,-1-jth bit of wj $ v @ eaj.

But if the zg-t_jth bit of wj $ v is 1 then this edge cannot appear in C(n) by (C3). Hence we are done by contradiction. q

4. Main result

We now consider the problem of embedding a ring in a faulty Q,, when there are faulty nodes and edges. To avoid the faulty nodes or edges, often the standard node and/or edge skipping will be used. Throughout the paper, Fn and F, will, respectively, denote the set

of faulty nodes and faulty edges in a Qn . Let

IFnl = fn, IFel= fe, IFI=f, where F = F,, U Fe, F being the set of the faults in the Qn. We will assume that f < n - 1 to ensure that the graph obtained by removing the faulty components remains connected. We will show, by case by case analysis, that a ring of length at least 2” - 2fn can be embedded avoiding the faulty nodes and edges.

Case 1: fn = 0 and fe < n - 1. For fe < n - 2, Algorithm A finds an HC in the faulty Qn. So we will consider only the case fe = n - 1. We will show that unless some node is an endpoint of all the faulty edges, there is an HC avoiding Fe for all hypercubes larger

than Q3. Subcase A. All faulty edges are in distinct dimen-

sions. Assume that the edge induced subgraph of Qn, induced by Fe, has maximum matching of size one. Then there is a node such that all n - 1 faulty edges are incident on it. Clearly an HC avoiding the faulty edges cannot exist. However, there exists a cycle containing 2” - 2 nodes avoiding the faulty edges and clearly this is the largest ring that can be embedded (all rings in a Qn must be of even length). The cycle can be con- structed as follows. Choose any edge h E Fe. Form

F,‘=F,-{h}.ThuslF,‘(=n-2.LetJ.beindimen- sion t. Since all edges are in distinct dimensions, t is a fault-free dimension when F,’ is assumed to be the set of faulty edges. Apply Algorithm A using F,’ as the set of faulty edges to find a C(v, n). Clearly the C(v, n) will have the edge h appearing in it. Using the standard edge skipping, the faulty edge )c can be avoided. Note

that such a skipping is possible because the bridge of skip is in the same dimension as that of h and since all faulty edges are in distinct dimensions, the bridge

is fault-free. Assume that the edge induced subgraph of Qn,

induced by F, , has maximum matching of size at least two. We show that an HC can be embedded in the faulty Qn avoiding the faulty edges. Let (v,, VB) and (vv, us) be two matching edges in dimensions t and x, respectively. Hence v,, VP, vy, vg are all distinct. Form

F,‘=F,-{(u,,ug)}.

Let y 4 (t, x} be any dimension in which va and vy

differ. Since v, , VP, vy , vg are all distinct, y exists. With F,’ as the set of faulty edges, t is a fault-free

dimension. Apply Algorithm A using F,’ as the set of faulty edges to find a C(v, n). Choose x,, = t. If y is a fault-free dimension then choose x1 = y and x2 = x. If y is not a fault-free dimension then choose .x,-2 = y and x+1 = x. Using Algorithm A, as we find C( v, n) , note that since xn = t , there are only two t-dimensional edges appearing C(v, n) and these are VI v CB et and v $ e,“_, t v @ e,,_, @ et. Since C(v, n) avoids all the faulty edges of F,‘, the only faulty edge of Fe that can possibly appear in C(v, rr) is one of these two edges. Since all faulty edges are in distinct dimensions, the only faulty edge in dimension x is (v, , v,g ). According to Algorithm A, the construction of the vector v is such that the yth bit of v is equal to the yth bit of v, which is different from the yth bit of vy (since v, and vy differ in yth bit). Hence nei-

ther of the two edges v i v @ et and v @ exn_, L

u @ e,,_, @ et is the only faulty edge in dimension t. Hence the C (v, nt) avoids the faulty set Fe.

Subcase B. All faulty edges are not in distinct dimensions. There are at least two faulty edges in the same dimension, say, t. Split up the Qn into two Qn_ 1 s along the dimension t; let us call them X and Y. For j E (X, Y}, let Sx = {e 1 e is a faulty edge and e E j}. Let FX = SX U It(&) and Fy = Sy U Zt(Sx).

A. Sengupta /Information Processing Letters 68 (1998) 207-214 211

HC in X HC in Y

\ / v 1 Do V t

1

Fig. 2. Combining two HCs.

Fig. 3. A bad Q3.

Since there are at least two faulty edges in dimension t, IFxl, IFyl < n - 3. Since each of X and Y is a en-t, by [4] there is an HC in X avoiding the faulty edges. Let us choose any such HC. In every node of the node sequence forming the HC, if the tth bit is flipped, we clearly obtain an HC in Y that avoids the faulty edges in Y. Let vi and uj be two nodes appearing consecutively along the HC in X. If the t-dimensional edges of Vi and uj are fault-free, then an HC in the Qn can be formed as shown in the Fig. 2. If the number of faulty edges in the dimension t is fewer than half of the total number of edges in dimension t , then such vi and uj will exist such that the t-dimensional edges of ui and uj are fault-free and we are done. There are 2”-’ edges in dimension t and 2+’ > n - 1 for n > 4. Hence for Q4 and larger hypercubes, the HC exists. For Q3, an HC does not exist if there are two faulty edges as shown in Fig. 3. In this case, however, a ring of size 6 exists as can be shown in Fig. 3. For any other two edge faults in the same dimension, an HC exists as shown in Fig. 3 (faulty edges are marked x). The embedded ring is shown in bold lines in Fig. 3.

Case 2: fn = 1 and fe 6 n - 2. We show that when fe = n - 2, a ring of size 2” - 2 can be embedded in a Qn avoiding the faulty node and edges.

Let uj be the faulty node. Let (Y, /? be two fault- free dimensions. Choose y E (a, /I) such that NY (uj)

is the endpoint of at least three fault-free edges. We

1

!“;J = 1 “j

Fig. 4. Fault-free bridge around Uj

first claim that such a y always exists. Let ut = Na (vi) and v2 = Ng(vj). If ~1 has at least three fault-free edges incident on it, then y = cr. If not, then all the

n - 2 faulty edges are incident on ut and none of them has the other endpoint ~2, because, otherwise we

would have had a cycle u1 v2uj ut of length 3 which is impossible. This implies, ~2 is an endpoint of all fault-

free edges. In that case, y = /I. Apply Algorithm A to find an HC C (u, n) in the Q,,

avoiding the faulty edges. While using Algorithm A, choose R = ~1x2 . ..xn suchthatxl=yandx,=z (see (01)) and the dimensions of the faulty edges incident on ut = Ny(uj) appear as x,-1,x,+2,. . . (see (02)) and the faulty edges incident on ut are included among the faulty edges of distinct dimensions (see (03)). We claim that with this choice of xc, in the

C(v, rr) formed by Algorithm A, the faulty node uj can be skipped by the standard skipping technique. We

show that the bridge around uj cannot be a faulty edge. Suppose uj appears in the C(v, n) as shown in the

Fig. 4, that is, uj x uj @ eh appears in C(u, n).

If possible, assume the bridge around uj is faulty. Clearly the bridge is one of the faulty edges incident on ut and h is the dimension of the bridge. In n = x1x2... xn , if h = xi, then since at least three fault-free edges are incident on ~1, and the dimensions of the

faulty edges incident on ut appear as x+1, xn_2, . . . , and the bridge around uj is faulty, we must have i > 2.

According to the construction of C(u, n), xi_tth bit of v = xi_lth bit of ~1 = xi_tth bit of uj, as i > 2.

Hence, xi- 10-1 bit of v @ uj = 0. By condition (C2), the edge v @ uj h u @ uj @ eA cannot appear in C(Z) and this implies, Vj h Vj @ ei cannot appear in C(v, n). Hence the bridge around uj cannot be a faulty edge.

Thus, there is a C(V, n) given by Algorithm A which does not use any faulty edge and the bridge around nj is fault-free. Since there is only one faulty node uj in Qn, the endpoints of the bridge are fault-free and hence the embedding is possible by standard node skipping.


:_______a s 5 52 vs v4 vs Vk

Fig. 5. The HC and the nodes Vj and uk in it.

Case 3: f,, = 2 and fe < n - 3. We show that when fe = n - 3, a ring of size at least 2” - 4 can be embedded in a Qn avoiding the faulty node and edges.

Let a, /3, y be three fault-free dimensions. Let uj and Vk be the two faulty nodes. If h(vj , Vk) < 2 then choose y E {a, /I, v} - H(vj, vk), otherwise choose

any y E Ia, B, ~1. Let {a, B, Y) - IY} = Iz, ~1. Apply

Algorithm A to find an HC C (v , n) in the Qn avoiding the faulty edges. While using Algorithm A, choose ?r = x1x2.. . x,, such that x1 = y, x2 = z and x, = w

(see (01)). We prove that with this choice, the bridges around both Uj and Vk are fault-free and also the endpoints of the bridges are fault-free. We consider the case of Vj , the proof for Vk is identical.

Assume uj appears in the HC as shown in Fig. 5.

Without loss of generality, assume Vj and vk appear in the HC as shown in Fig. 5. By Property 1, all the y-dimensional edges appear in the HC. If possible, assume that the bridge around nj is in dimension h and is faulty. Then, there is a faulty edge in dimension h incident on ~1 = NY (uj) and the edge Vj L Uj @ ek appears in C(v, n). Assume that in n = x1x2.. .xn, h = xi. Then i > 2 (recall that x1 and x2 are fault-free dimensions).

Suppose the faulty edge in dimension h incident on vt was chosen among the edges in distinct dimensions, then according to the construction of C(v, n), xi-1 th bitofv=xi_tthbitofvt=xi_tthbitofvj.Hence, xi-tth bit of v @ Vj = 0. By condition (C3), the edge v @ uj h v @ uj @ eA cannot appear in C(n) and this implies, Vj A. Vj @Ed cannot appear in C(v, n). This is a contradiction.

Now suppose the faulty edge in dimension h incident on vt was not chosen among the edges in distinct dimensions. Then, this edge is considered in the Step 2.4 of Algorithm A, and suppose the xpth bit of v gets set when this edge is being considered. Clearly p < i - 1 and x,th bit of v and vt must be opposite. Since, fe < n - 3, the xtth bit (that is, yth bit) of v is never set in Step 1.2 or in Step 2.4 of Algo-

rithm A, that is, p > 1. From Algorithm A, xpth bit

of v = 1 @ x,th bit of vt = 1 $ x,th bit of uj (since

vi and uj differ in x1 th bit and p > 1). Hence x,th bit

of v @ uj = 1. Since P < i - 1, by condition (C3), the

edge v @ uj 1 v @ vj @ ei cannot appear in C(n) and

this implies, uj h Vj @ eh cannot appear in C(v, n).

This is a contradiction.

Now we show how the nodes Vj and vk can be

skipped by standard node skipping. Since the bridges

around nj and Vk have been shown to be fault-free,

we are done if we can justify that the endpoints of the

bridges are fault-free. Note that if h(Vj, vk) < 2, then

ShCe Y $! H(Vj, vk) but Y E H(Vj, VI), Y E H(Vj, IQ),

and y E H(vj , vq), Vk cannot be either of the nodes

vt , vg , ~4. If Vk = v2 then, the endpoints of the bridge

around Vj are fault-free (Vj and vk are the only faulty

nodes) and hence the standard node skipping will give

Us a ring Of length 2n - 2. If Vk # V:! then, Vk iS Vg

or farther away from vj along the HC and hence, the

standard node slopping will give us a ring of length

2” - 4. On the other hand, if h(vj , Vk) 2 3, using

similar argument, it is simple to see that vk cannot be

either of the nodes vt, ~2, v3, v5 (all these nodes have

Hamming distance less than 3 from Vj). If Vk = 9~4,

then since the bridges around both nj and Vk are fault-

free, the standard node skipping will give us a ring of

length 2” - 2 as shown in Fig. 6. In short, either we

have ring of length 2n - 2 or of length 2” - 4.

Case 4: fe < n - 4. We show that in this case, there

exists a ring of length at least 2’ - 2fn avoiding the

faulty nodes and faulty edges. We will assume that

f,, 2 3, otherwise, we are done by the previous cases. Let Z, and Zb be, respectively the set of fault-free and

faulty dimensions. Clearly, I Zb 1 < fe < n - 4. It might be noted that, given the set of faulty nodes F,, there

existsasetYcZ, lYl>n- fn+l,suchthatevery

component of Qn, r (refer to the definition of Q,,, r on

p. 2) has at most one faulty node. The set Y can be

determined by the following Algorithm B (the proof

A. Sengupta /Information Processing Letters 68 (1998) 207-214 213

Fig. 6. The nodes vj and IJ~ are close in the HC.

of correctness of the algorithm is omitted because of its obviousness).

Algorithm B.

Given: A set of faulty edges F,, with fn=lFnl<n-linaQn.

Tofind:AsetY~Z,IYl>n-f,+l,suchthat every component of Qn, y has at most one faulty node

Method: Stepl:LetF,={ut,u2,...,uf,,}.Findanydimension

j in which ~1 and ~2 differ. Set X t {j}.

Step 2: for j := 3 to f,, do if there is some i, 1 6 i < j - 1,

such that ui and uj appear in the same component of Q,,z_x

then find a dimension k in which ui and uj differ and set X t X U {k}.

Step 3: Set Y t Z - X.

Once the set Y has been computed from F, using Algorithm B, let W = Y - Zb. Clearly, each component of Q,,w has no faulty edge and has at most one faultynode.NotethatlWl=(Y-Zbl>n-f,+l- fe > 2, since fn + fe < n - 1. Choose P g W such that I P( = 2. Thus, each component of Qn,p is a Q2

with at most one faulty node and no faulty edge, while K(Qn,p) is a Qn_2 (with each node of K( Q,,p) is a condensed version of a Q2) and all the faulty edges in the Q,, appear as the edges of K ( Qn, p) in the condensed form as discussed before. Since there are at mostn-4faultyedgesinK(Q,,p)andK(Q,,p)isa Qn_2, an HC in K ( Qn,p), avoiding the faulty edges, can be determined using Algorithm A.

Letco,cl,c2 ,..., c~+_~ be the node sequence of an HC in K ( Qn,p) as determined by Algorithm A. Recall that each cj represents four nodes (connected as a Q2) of the original Qn and each such Q2 has at most one faulty node and no faulty edge. Also the edges between two consecutive such Q2s are faulfree. Consider any two nodes cj and c( j+l) mod 2+2 of K(Q,,p) that appear adjacent in the HC determined

(3

a

b 8

au d e

C f

(iv)

(4 (vi) (vii)

Fig. 7. Traversing two consecutive Q2s.

a

d B OU b e

C f

Fig. 8. Starting/ending the ring formation.

by Algorithm A. The two Q2s represented by cj and C~j+l) mod 2n-2 are shown (from left to right) in Fig. 7 enumerating all the seven possible cases where the dark dot represents a faulty node. The cases (i)-(iv) are the cases when each of the two Q2s has a faulty node, while the cases (v) and (vi) are the cases when only one of the two Q2s has a faulty node and the case (vii) is the case when none of them has a faulty node. The table in Fig. 9 gives, for all the possible cases and for all starting nodes in the Q2 represented by cj, the node

sequence traversed in it and the node of C(j+l) mod p-2

reached after traversal. It is simple to note that for this traversal, in a Qz, if there is a faulty node then at most one node, in addition to the faulty node, might not be included in the traversal, whereas if there is no faulty node in a Q2, all the nodes of that Q2 are included

214 A. Sengupta / Information Processing Letters 68 (1998) 207-214

Case Starting at Node sequence traversed

(i) b bcde

i cde

dcbg

(ii)

(iii)

dcf &f cde

bade ade

dabg

(iv) a b C

W b

i

abcf

bcf Cbg

bcde cde

dcbg

(vi) a b

ii

abcde badcf

c&g d&f

(vii) a b

ii

abcde bcdah caizbg

hcf

Fig. 9. Node sequence of traversal for different cases.

in the traversal. In short, at most fX additional nodes might not be included where fX 6 fn .

Using this traversal, we can embed a ring in a straight forward fashion as follows. Consider any two nodes cj and cu+ij mod ~“-2 of K( Qn,p) such that the Q2 represented by cj has no faulty node in it whereas c(~+~) mod 2n-2 has one faulty node. Since the number of faulty nodes is at most n - 1, such a cj will exist.

Suppose the Q2s represented by cj and cu+lj mod 2”-2 are as shown in Fig. 8 from left to right.

Start from the node e in the Q2 on right. Clearly, in that Q2, the node sequence ef is traversed. If after traversal of the 2n-2 - 1 Q~s, the path ends up in node a or d or c, follow the node sequence adcbe or dcbe or cdabe, respectively to complete the ring. If, however, the path ends up in the node b, follow the node sequence budcf to complete the ring. Thus the total number of nodes fomit not included in the ring is at most fX + fn + 1, where one additional node from Q2s represented by cj and C(j+l) mod 2n-2 might not

be included. Since the length of the ring must be even, fornit 6 fx + fn + 1 6 2 fn + 1 must be even, implying j&t < 2fn. Thus we have formed a ring of length at least 2” - 2 fn .

References

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rings in hypercubes, J. Parallel Distrib. Comput. ll(l991) 63-

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[2] MS. Chen, K.G. Shin, Processor allocation in an N-cube

multiprocessor using gray codes, IEEE Trans. Comput. C-36

(1987) 1396-1407.

[3] S. Latifi, S. Zheng, N. Baghetzadeh, Optimal ring embedding

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puting Symposium, 1992, pp. 178-184.

[4] A. Sen, A. Sengupta, S. Bandyopadhyay, On some topological

properties of hypercube, incomplete hypercube and supercube,

in: Proc. International Parallel Processing Symposium, New

port Beach, April, 1993, pp. 636-642.

[5] Y-C. Tseng, T.-H. Lai, Gn the embedding of a class of regular

graphs in a faulty hypercube, J. Parallel D&rib. Comput. 37

(2) (1996) 200-206.

[6] Y.-C. Tseng, Embedding a ring in a hypercube with both both

faulty links and faulty nodes, Inform. Process. Lett. 59 (1996) 217-222.

[7] P-J. Yang, S.-B. Tien, C.S. Raghavendra, Embedding of multi-

dimensional meshes on to faulty hypercubes, in: Proc. In-

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On ring embedding in hypercubes with faulty nodes and links