Long paths in hypercubes with conditional node-faults
Tz-Liang Kueng
a, Tyne Liang
a,*, Lih-Hsing Hsu
b, Jimmy J.M. Tan
a,1 aDepartment of Computer Science, National Chiao Tung University, 1001 University Road, Hsinchu 30050, Taiwan, ROC
b
Department of Computer Science and Information Engineering, Providence University, 200 Chung Chi Road, Taichung 43301, Taiwan, ROC
a r t i c l e
i n f o
Article history:
Received 12 December 2007
Received in revised form 4 October 2008 Accepted 10 October 2008 Keywords: Interconnection network Hypercube Fault tolerance Conditional fault Linear array Path embedding
a b s t r a c t
Let F be a set of f 6 2n 5 faulty nodes in an n-cube Qnsuch that every node of Qnstill has at least two fault-free neighbors. Then we show that Qn F contains a path of length at least 2n
2f 1 (respectively, 2n 2f 2) between any two nodes of odd (respectively, even) distance. Since the n-cube is bipartite, the path of length 2n
2f 1 (or 2n 2f 2) turns out to be the longest if all faulty nodes belong to the same partite set. As a contribution, our study improves upon the previous result presented by [J.-S. Fu, Lon-gest fault-free paths in hypercubes with vertex faults, Information Sciences 176 (2006) 759–771] where only n 2 faulty nodes are considered.
Ó 2008 Elsevier Inc. All rights reserved.
1. Introduction
In many parallel computer systems, processors are connected on the basis of interconnection networks, referred to as net-works henceforth. Among various kinds of netnet-works, hypercube is one of the most attractive topologies discovered for its suitability in both special-purpose and general-purpose tasks[11]. One important issue to address in hypercubes is how to embed other networks into hypercubes. By definition[11], embedding one guest network G into another host network H is a form of injective mapping,
g
, from the node set of G to the node set of H. A link of G corresponds to a path of H underg
. Often embedding takes cycles, paths, or meshes as guest networks[3–5,19,20]because these architectures are extensively applied in parallel systems.Fault-tolerant embedding in hypercubes has been widely addressed in researches[2,6,7,9,13,15–18]. For example, Latifi et al.[9]proved that an n-dimensional hypercube (or n-cube), Qn, is Hamiltonian even if it has n 2 faulty links. On the other
hand, Tsai et al.[15]showed that Qn(n P 3) is both Hamiltonian laceable and strongly Hamiltonian laceable even if it has
n 2 faulty links. Recently, Tsai and Lai[17]addressed the conditional edge-fault-tolerant edge-bipancyclicity of hyper-cubes. As Tseng [18] showed, a faulty n-cube, containing fe6n 4 faulty links and fv6n 1 faulty nodes with feþ fv6n 1, has a fault-free cycle of length at least 2n 2fv. Furthermore, Fu[6]showed that a fault-free cycle of length at least 2n 2f can be embedded into an n-cube with 1 6 f 6 2n 4 faulty nodes. Fu[7]also proved that a fault-free path of
length at least 2n 2f 1 (or 2n 2f 2) can be embedded to join two arbitrary nodes of odd (or even) distance in an n-cube with f 6 n 2 faulty nodes.
0020-0255/$ - see front matter Ó 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.ins.2008.10.015
*Corresponding author. Tel.: +886 3 5131365; fax: +886 3 5721490. E-mail address:tliang@cs.nctu.edu.tw(T. Liang).
1
This work was supported in part by the National Science Council of the Republic of China under Contract NSC 96-2221-E-009-137-MY3. Contents lists available atScienceDirect
Information Sciences
Basically, the components of a network may fail independently. It is unlikely that all failures would be close to each other. Based on this phenomenon, the conditional node-faults[10]were defined in such a way that each node of a faulty network still has at least g fault-free neighbors. In this paper, we concern that g ¼ 2. More precisely, a network is said to be condi-tionally faulty if and only if every node has at least two fault-free neighbors. Under this premise, we would like to extend Fu’s result[7]by showing that a conditionally faulty n-cube with f 6 2n 5 faulty nodes still contains a fault-free path of length at least 2n 2f 1 (respectively, 2n 2f 2) between any two fault-free nodes of odd (respectively, even) distance. Con-sider a 4-cube with four faulty nodes, 0000, 0011, 1100, and 1111, as shown inFig. 1, in which every node has at least two fault-free neighbors. Then the length of the longest path between nodes 0110 and 1001 is 4 < 24
2 4 2. This is why we concentrate only on f 6 2n 5 faulty nodes.
It is sufficient to assume that every node should have at least two fault-free neighbors while a long path is constructed between every pair of fault-free nodes. Consider the scenario that u is a fault-free node with only one fault-free neighbor, namely v. Then the longest path between u and v happens to be of length 1. To avoid such a degenerate situation, it is nec-essary that, for any pair u;
v
of adjacent nodes, u has some fault-free neighbor other than v, and vice versa. On the other hand, it is also statistically reasonable to require that every node needs to have at least two fault-free neighbors. Suppose, with a random fault model, the probabilities of node failures are identical and independent. Let PrðnÞ denote the probability that every node of the n-cube Qn, containing 2n 5 faulty nodes, is adjacent to at least two fault-free neighbors. Because Qnhas 2nnodes, there are 2n
2n 5
ways to distribute 2n 5 faulty nodes. In the random fault model, all these fault distri-butions have equal probability of occurrence. Clearly, Prð3Þ ¼ 1 and Prð4Þ ¼ 1
24 4 3 24 3 ¼ 31 35, where 2 4 43 is the number of faulty node distributions that there exists some node having three faulty neighbors. When n P 5, the number of faulty node distributions that there exists some node having n faulty neighbors is 2n 2
n
n n 5
. Moreover, the number of faulty node distributions that there exists some node having exactly n 1 faulty neighbors is 2n n
n 1 2n n n 4 . Since 2 n n n 4 P 2 n n n 5
for n P 5, we can derive that
PrðnÞ ¼ 1 Prðsome node has at least n 1 faulty neighborsÞ ¼ 1 2n 2 n n n 5 þ 2n n 1n 2n n n 4 2n 2n 5 P1 2n ð1 þ nÞ 2 n n n 4 2n 2n 5 ¼ 1 2 n ð1 þ nÞ ð2n 2n þ 5Þ Q2n5k ¼ n3k Q2n k ¼ 2nnþ1k ¼ 1 ðn 3Þðn 2Þ 2n n þ 1 n 1 2n n 2n 5 2n 3 n þ 1 2n 2 2n 2n þ 5 2n 1 ,LðnÞ:
Fig. 1. A conditionally faulty Q4with four faulty nodes. Every faulty node is marked by an ‘‘X” symbol. The length of the longest path between nodes 0110
It is not difficult to compute PrðnÞ numerically, such as Prð5Þ ¼ 6157
6293, Prð6Þ ¼ 96965279706503, etc. Since limn!1LðnÞ ¼ 1, PrðnÞ
ap-proaches to 1 as n increases.
The rest of this paper is organized as follows. In Section2, basic definitions and notations are introduced. In Section
3, a partition procedure, named PARTITION, is proposed to divide a conditionally faulty n-cube into two conditionally faulty subcubes. In Section4, we show that a conditionally faulty n-cube with f 6 2n 5 faulty nodes has a fault-free path of length at least 2n 2f 1 (respectively, 2n 2f 2) between any two fault-free nodes of odd (respectively, even) distance. Finally, the conclusion and discussion are presented in Section5.
2. Preliminaries
Throughout this paper, we concentrate on loopless undirected graphs. For the graph definitions, we follow the ones given by Bondy and Murty [1]. A graph G consists of a node set VðGÞ and a link set EðGÞ that is a subset of fðu;
v
Þ j ðu;v
Þ is an unordered pair ofVðGÞg. It is bipartite if its node set can be partitioned into two disjoint partite sets, V0ðGÞ and V1ðGÞ, such that every link joins a node of V0ðGÞ and a node of V1ðGÞ.A path P of length k from node x to node y in a graph G is a sequence of distinct nodes h
v1
;v2
; . . . ;vkþ1
i such that x ¼v1
, y ¼vkþ1
, and ðvi
;viþ1
Þ 2 EðGÞ for every 1 6 i 6 k if k P 1. Moreover, a path of length 0 consisting of a single node x is denoted by hxi. For convenience, we write P as hv1
; . . . ;vi
;Q ;vj
; . . . ;vkþ1
i, where Q ¼ hvi
; . . . ;vj
i. The ith node of P is denoted by PðiÞ; i.e., PðiÞ ¼vi
. A cycle is a path with at least three nodes such that the last node is adjacent to the first one. For clarity, a cycle of length k is represented by hv1
;v2
; . . . ;vk
;v1
i. A path (or cycle) in a graph G is a Hamil-tonian path (or HamilHamil-tonian cycle) if it spans G. A bipartite graph is HamilHamil-tonian laceable[14]if there exists a Hamiltonian path between any two nodes that are in different partite sets. Furthermore, a Hamiltonian laceable graph G is hyper-Hamiltonian laceable [12]if, for any nodev
2 ViðGÞ and i 2 f0; 1g, there exists a Hamiltonian path of G fv
g betweenany two nodes of V1iðGÞ. Later Hsieh et al.[8]introduced strongly Hamiltonian laceability. A Hamiltonian laceable graph
G is strongly Hamiltonian laceable if there exists a path of length jVðGÞj 2 between any two nodes in the same partite set.
Let u ¼ bn. . .bi. . .b1 be an n-bit binary string. For 1 6 i 6 n, we use ðuÞi to denote the binary string bn. . . bi. . .b1.
Moreover, we use ½ui to denote the bit bi of u. The Hamming weight of u, denoted by wHðuÞ, is
jf1 6 j 6 n j ½uj ¼ 1gj. The n-cube Qnconsists of 2 n
nodes and n2n1 links. Each node corresponds to an n-bit binary string. Two nodes, u and v, are adjacent if and only if
v
¼ ðuÞi for some i, and we call the link ðu; ðuÞiÞ i-dimensional. We define dimððu;v
ÞÞ ¼ i ifv
¼ ðuÞi. The Hamming distance between u and v, denoted by hðu;v
Þ, is defined to be j f1 6 j 6 n j ½uj–½v
jg j. Hence two nodes, u and v, are adjacent if and only if hðu;v
Þ ¼ 1. It is well known that Qnis a bipartite graph with partite sets V0ðQnÞ ¼ fu 2 VðQnÞ j wHðuÞ is eveng and V1ðQnÞ ¼ fu 2 VðQnÞ j wHðuÞ is oddg.
A graph G is node-transitive if, for any pair
v1
;v2
of VðGÞ, there exists some automorphisml
:VðGÞ ! VðGÞ such thatl
ðv1
Þ ¼v2
. A graph G is link-transitive if, for any two links e1 ¼ ðu1;v1
Þ and e2 ¼ ðu2;v2
Þ of G, there exists someauto-morphism w : VðGÞ ! VðGÞ such that wðu1Þ ¼ u2and wð
v1
Þ ¼v2
. As introduced in[11], Qnis both node-transitive andlink-transitive. The following two theorems reveal the link-fault-tolerant Hamiltonian laceability of hypercubes. Theorem 1 [15]. Let n P 3. Suppose that F # EðQnÞ is a set of utmost n 2 faulty links. Then Qn F is Hamiltonian laceable and strongly Hamiltonian laceable.
Theorem 2 [15]. Let n P 3. Suppose that F # EðQnÞ is a set of utmost n 3 faulty links. Then Qn F is hyper-Hamiltonian
laceable.
3. Partition of faulty hypercubes
In this section, we show that a conditionally faulty n-cube can be partitioned into two conditionally faulty subcubes if it has 2n 5 or less faulty nodes. First of all, we introduced some notations to be used later. For 1 6 j 6 n and i 2 f0; 1g, let Qj;i
n be a subgraph of Qninduced by fu 2 VðQnÞ j ½uj ¼ ig. Obviously, Q j;i
n is isomorphic to Qn1. Then the node partition of
Qninto subgraphs Q j;0 n and Q
j;1
n is called j-partition. For convenience, we use FðGÞ to denote the set of all faulty nodes in
graph G. For any node u of G, its neighborhood NGðuÞ is defined by NGðuÞ ¼ f
v
2 VðGÞ j ðu;v
Þ 2 EðGÞg. In addition, let NFGðuÞdenote the set NGðuÞ \ FðGÞ.
Suppose Qn, n P 4, is conditionally faulty with f 6 2n 5 faulty nodes. Moreover, suppose u, v, and w are three nodes
of this faulty n-cube, and each of them has only two fault-free neighbors. Then we discuss how the faulty nodes will be distributed conditionally. For simplification, let U ¼ NF
QnðuÞ, V ¼ N
F
Qnð
v
Þ, and W ¼ NF QnðwÞ.
If j V \ W j ¼ 0, then we have f Pj V [ W j ¼ j V j þ j W j ¼ 2n 4, contradicting the requirement that f 6 2n 5. Therefore, j V \ W jP 1 needs to be satisfied. Similarly, we also have j U \ V jP 1 and j U \ W jP 1. Since any two nodes of an n-cube can have utmost two common neighbors, we obtain that j V \ W j; j U \ V j; j U \ W j2 f1; 2g. We first consider the case that at least one of j V \ W j, j U \ V j, and j U \ W j is equal to 1. Without loss of generality, we suppose j V \ W j ¼ 1.
I. First, we concern the case that j V \ W j ¼ j U \ V j ¼ j U \ W j ¼ 1. If j U \ V \ W jP 1, we have 2n 5 P f P j U [ V [ W j ¼ 3ðn 2Þ ð1 þ 1 þ 1Þ þ 1 ¼ 3n 8; i.e., n 6 3. Since n P 4, we only concern j U \ V \ W j ¼ 0. Then we have 2n 5 P f Pj U [ V [ W jP 3ðn 2Þ ð1 þ 1 þ 1Þ ¼ 3n 9; i.e., n 6 4.Fig. 2a depicts a faulty 4-cube with j V \ W j ¼ j U \ V j ¼ j U \ W j ¼ 1 and j U \ V \ W j ¼ 0.Fig. 2b is a cube-styled layout isomorphic toFig. 2a. We can examineFig. 2a in a top-down viewpoint. Since hypercube is node-transitive, we can assume that u ¼ t1. By
link-transitivity, we assume that t4and t5are faulty neighbors of u. Since j U \ V j ¼ 1, we obtain
v
2 ft7;t8;t9;t10g.With-out loss of generality, we assume that
v
¼ t10. Since j U \ W j ¼ j V \ W j ¼ 1 and j U \ V \ W j ¼ 0, we see thatw ¼ t9and V \ W ¼ ft15g. As a consequence, this happens to be the only possibility. However, node t11has only
one fault-free neighbor. Thus it is not conditionally faulty.
II. Secondly, we consider the case that j V \ W j ¼ j U \ V j ¼ 1 and j U \ W j ¼ 2. By the definition of hypercube, we see that j NQnðuÞ \ NQnð
v
Þ \ NQnðwÞ j6 1. Obviously, we have j U \ V \ W j6j NQnðuÞ \ NQnðv
Þ \ NQnðwÞ j. In particular, we claim that j U \ V \ W j ¼ 1. Suppose, by contradiction, that j U \ V \ W j ¼ 0. Then we have U \ V \ W ¼ ðU \ VÞ \ ðU \ WÞ ¼ ;. Since U \ V–; and U \ W–;, we conclude that V \ W ¼ ;. That is, the assumption of j U \ V \ W j ¼ 0 leads to a contradiction between j V \ W j ¼ 1 and V \ W ¼ ;. As a result, j U \ V \ W j is equal to 1. Accordingly, we have 2n 5 P f Pj U [ V [ W j ¼ 3ðn 2Þ ð1 þ 1 þ 2Þ þ 1 ¼ 3n 9; i.e., n 6 4. SeeFig. 2c for illustration. For clarity,Fig. 2d is an isomorphic layout ofFig. 2c. Similarly, we can examineFig. 2c in a top-down viewpoint. By node-transitivity, we assume that u ¼ t1. By link-transitivity, we assume that t4and t5are faultyneigh-bors of u. Since j U \ W j ¼ 2, we have w ¼ t11. Since j V \ W j ¼ j U \ V j ¼ 1 and j U \ V \ W j ¼ 1, we obtain
v
2 ft7;t8;t9;t10g. Without loss of generality, we assume thatv
¼ t10. Then this turns out to be the only possibility.It is noticed that node t8has only two fault-free neighbors.
III. Next, we concern the case that j V \ W j ¼ 1 and j U \ V j ¼ j U \ W j ¼ 2. Similarly, we have j U \ V \ W j ¼ 1. Since ðU \ VÞ [ ðU \ WÞ # U, we have j ðU \ VÞ [ ðU \ WÞ j6j U j. However, we have a contradiction that j ðU \ VÞ [ ðU \ WÞ j ¼ j U \ V j þ j U \ W j j U \ V \ W j ¼ 2 þ 2 1 ¼ 3 > n 2 ¼ j U j if n 6 4. In what follows, we suppose that n P 5. As a consequence, we have 2n 5 P f Pj U [ V [ W j ¼ 3ðn 2Þ ð1 þ 2 þ 2Þ þ 1 ¼ 3n 10; i.e., n ¼ 5. SeeFig. 3a. Again, we examineFig. 3a in a top-down viewpoint. By node-transitivity, we assume that u ¼ t1. By link-transitivity, we assume that t4, t5, and t6are faulty neighbors of u. Since j U \ V j ¼ j U \ W j ¼ 2,
we have f
v
;wg ft14;t15;t16g. Without loss of generality, we assume thatv
¼ t14and w ¼ t16. Since j V \ W j ¼ 1,we have t26RV [ W. Moreover, we have 2n 5 P f Pj V [ W j ¼ j V j þ j W j j V \ W j ¼ ðn 2Þ þ ðn 2Þ 1 Fig. 2. Every faulty node is marked by an ‘‘X” symbol. (a) The Q4with j NFQ4ðuÞ \ N
F Q4ðvÞ j ¼ j N F Q4ðvÞ \ N F Q4ðwÞ j ¼ j N F Q4ðuÞ \ N F Q4ðwÞ j ¼ 1; (b) a layout isomorphic to (a); (c) the Q4with j N
F Q4ðuÞ \ N F Q4ðvÞ j ¼ j N F Q4ðvÞ \ N F Q4ðwÞ j ¼ 1 and j N F Q4ðuÞ \ N F Q4ðwÞ j ¼ 2; (d) a layout isomorphic to (c).
Fig. 3. Every faulty node is marked by an ‘‘X” symbol. Each of u,v, w, and z has only two fault-free neighbors. (a) The Q5with j NFQ5ðvÞ \ N F Q5ðwÞ j ¼ 1 and j NF Q5ðuÞ \ N F Q5ðvÞ j ¼ j N F Q5ðuÞ \ N F Q5ðwÞ j ¼ 2; (b) the Q5with j N F Q5ðuÞ \ N F Q5ðvÞ j ¼ j N F Q5ðvÞ \ N F Q5ðwÞ j ¼ j N F Q5ðuÞ \ N F Q5ðwÞ j ¼ 2.
¼ 2n 5; that is, f ¼ 2n 5 and U # V [ W. Then we have either t202 V or t232 V. Without loss of generality, we
assume that t232 V. Similarly, we can assume that t252 W. As a result, this is the only possibility. It is noted that node
t12¼ z has three faulty neighbors, and j NFQ5ðxÞ j6 2 for each x 2 VðQ5Þ fu;
v
;w; zg.Q
6 w v t10 t2 t3 t4 t5 t6 t7 t8 t9 t11 t 12 t13 t14 t15 t16 t17 t18 t19 t20 t21 t22 t23 t24 t25 t26 t27 t28 t29 t30 t31 t32 t33 t34 t35 t36 t37 t38 t39 t40 t41 t42The links among nodes t
23,... , t
57are omitted.
t43 t44 t45 t46 t47 t48 t49 t50 t51 t52 t53 t54 t55 t56 t57 u = t1 u = t1 w v z t10 t2 t3 t4 t5 t6 t7 t8 t9 t11 t 12 t13 t14 t15 t16 t17 t18 t19 t20 t21 t22 t23 t24 t25 t26 t27 t28 t29 t30 t31 t32 t33 t34 t35 t36 t37 t38 t39 t40 t41 t42
The links among nodes t
23,... , t
57are omitted.
t43 t44 t45 t46 t47 t48 t49 t50 t51 t52 t53 t54 t55 t56 t57
Q
6a
b
Fig. 4. Every faulty node is marked by an ‘‘X” symbol. The Q6 with j N F Q6ðuÞ \ N F Q6ðvÞ j ¼ j N F Q6ðvÞ \ N F Q6ðwÞ j ¼ j N F Q6ðuÞ \ N F Q6ðwÞ j ¼ 2. (a) j NF Q6ðuÞ j ¼ j N F Q6ðvÞ j ¼ j N F Q6ðwÞ j ¼ j N F Q6ðzÞ j ¼ 4 and j N F Q6ðxÞ j6 3 for x 2 VðQ6Þ fu;v;w; zg; (b) j N F Q6ðuÞ j ¼ j N F Q6ðvÞ j ¼ j N F Q6ðwÞ j ¼ 4 and j N F Q6ðxÞ j6 3 for x 2 VðQ6Þ fu;v;wg.
Now we consider the case that j V \ W j ¼ j U \ V j ¼ j U \ W j ¼ 2. Again, we have j U \ V \ W j ¼ 1. Since j ðU \ VÞ [ ðU \ WÞ j6j U j, we still have a contradiction that j ðU \ VÞ [ ðU \ WÞ j ¼ j U \ V j þ j U \ W j j U \ V \ W j ¼ 2 þ 2 1 ¼ 3 > n 2 ¼ j U j if n 6 4. In what follows, we suppose n P 5. Then we have 2n 5 P f P j U [ V [ W j ¼ 3ðn 2Þ ð2 þ 2 þ 2Þ þ 1 ¼ 3n 11; i.e., n 2 f5; 6g. Note that j U [ V [ W j ¼ 4 if n ¼ 5 and j U [ V [ W j ¼ 7 if n ¼ 6. SeeFig. 3b andFig. 4a and b. InFig. 3b, it is not difficult to see that j NF
Q5ðxÞ j6 2 for each x 2 VðQ5Þ fu;
v
;w; zg. We explainFig. 4as follows. By node-transitivity, we assume that u ¼ t1. By link-transitivity, weassume that t4, t5, t6, and t7 are faulty neighbors of u. Since j U \ V j ¼ j U \ W j ¼ 2, we deduce that
f
v
;wg ftij 17 6 i 6 22g. Since j U \ V \ W j ¼ 1, we can assume thatv
¼ t20 and w ¼ t22. Then we havej V \ ft30;t36;t39;t42g j ¼ 2 and j W \ ft32;t38;t41;t42g j ¼ 2. Since j V \ W j ¼ 2, we have V \ W ¼ ft6;t42g. If t392 V and
t412 W, then node t18happens to have only two fault-free neighbors (seeFig 4a); otherwise, we have j NFQ6ðxÞ j6 3 for each x 2 VðQ6Þ fu;
v
;wg (see Fig. 4b, in which nodes t36 and t41, for example, are faulty). Hence these figures cover allpossibilities.
According to the analysis presented earlier, a conditionally faulty n-cube with f 6 2n 5 faulty nodes is likely to contain three or four nodes such that each of them has only two fault-free neighbors. Since 2n 5 6 n 2 for n 6 3, we concentrate only on the case that n P 4. To summarize, we have the following two lemmas.
Lemma 1. Suppose that an n-cube Qnðn P 4Þ is conditionally faulty with f 6 2n 5 faulty nodes. Let u;
v
;w; z 2 VðQnÞ such that j NFQnðuÞ j ¼ j N F Qnðv
Þ j ¼ j N F QnðwÞ j ¼ j N F QnðzÞ j ¼ n 2 and j N FQnðxÞ j6 n 3 for every x 2 VðQnÞ fu;
v
;w; zg. Then the faulty nodes are distributed as illustrated in Figs.2c,3a and b, and4a. In Figs.2c and3a, no dimensions can be used to partition Qnin such a way that both resulting subcubes are conditionally faulty. InFig. 3b andFig. 4a, there exists some dimension j of f1; 2; . . . ; ng such that both Qj;0n and Qj;1
n are conditionally faulty with 2n 7 or less faulty nodes. Proof. In Figs. 2c and 3a, we check, by brute force, that either Qk;0n or Q
k;1
n contains a node with only one fault-free neighbor
for each k 2 f1; 2; . . . ; ng; that is, there does not exist any dimension to partition Qnsuch that both ðn 1Þ-cubes are
condi-tionally faulty. In Figs. 3b and 4a, let j be any integer of f1; 2; . . . ; ng such that ðuÞjis faulty. Then both Qj;0 n and Q
j;1
n are
con-ditionally faulty with 2n 7 or less faulty nodes. h
Lemma 2. Suppose that an n-cube Qnðn P 4Þ is conditionally faulty with f 6 2n 5 faulty nodes. Let u;
v
;w 2 VðQnÞ such thatj NFQnðuÞ j ¼ j N F Qnð
v
Þ j ¼ j N F QnðwÞ j ¼ n 2 and j N FQnðxÞ j6 n 3 for every x 2 VðQnÞ fu;
v
;wg. Then the faulty nodes are dis-tributed as illustrated inFig. 4b. Moreover, there exists some dimension j of f1; 2; . . . ; ng such that both Qj;0n and Qj;1n are condition-ally faulty with 2n 7 or less faulty nodes.Proof. Let j 2 f1; 2; . . . ; ng such that ðuÞj
2 NFQnðuÞ \ N F Qnð
v
Þ \ N F QnðwÞ. Then both Q j;0 n and Q j;1n are conditionally faulty with
2n 7 or less faulty nodes. h
Lemma 3. Suppose that an n-cube Qnðn P 4Þ is conditionally faulty with f 6 2n 5 faulty nodes. Let u and v be two nodes of Qn
such that j NF
QnðuÞ j ¼ j N
F
Qnð
v
Þ j ¼ n 2 and j NF
QnðxÞ j6 n 3 for every x 2 VðQnÞ fu;
v
g. Then there exists some dimension k of f1; 2; . . . ; ng such that both Qk;0n and Qk;1n are conditionally faulty. When n P 5, both Qk;0n and Qk;1n contain 2n 7 or less faulty nodes. Proof. Since j NF QnðuÞ j ¼ j N F Qnðv
Þ j ¼ n 2 and f 6 2n 5, we have j N F QnðuÞ \ N FQnð
v
Þ jP 1. Since any two nodes of Qncan have utmost two common neighbors, we consider the following two cases.Case 1: Suppose that j NFQnðuÞ \ N F
Qnð
v
Þ j ¼ 2. Let i and j be two integers such that fðuÞ i;ðuÞjg ¼ NFQnðuÞ \ N F Qnð
v
Þ. Obviously, we have ðuÞi ¼ ðv
Þjand ðuÞj ¼ ðv
Þi. Then we can partition Qnalong dimension k 2 fi; jg. As a result, both Qk;0n and Qk;1n contain at least n 3 faulty nodes. SeeFig. 5a.Case 2: Suppose that j NF QnðuÞ \ N
F
Qnð
v
Þ j ¼ 1. We claim first that this case holds only for n P 5. By contradiction, we suppose n ¼ 4. Let p and q be two integers such that both ðuÞpand ðuÞqare faulty. Since j NFQnðuÞ \ NF
Qnð
v
Þ j ¼ 1, we havev
–ððuÞpÞq. Thus node ððuÞpÞq happens to have only two fault-free neighbors, which contradicts the assumption that j NFQnðxÞ j6 n 3 for every x 2 VðQnÞ fu;v
g.Let i and j be two integers such that fðuÞig ¼ fð
v
Þjg ¼ NFQnðuÞ \ N F Qnðv
Þ. Since j N F QnðuÞ fðuÞ i g j þ j NFQnðv
Þ fðv
Þ j g j ¼ 2ðn 3Þ > n 2 ¼ j f1; . . . ; ng fi; jg j for n P 5, there exists some dimension k of f1; . . . ; ng fi; jg such that both ðuÞkand ðv
Þkare faulty. As a result, either Qk;0n or Qk;1
n contains exactly two faulty nodes. SeeFig. 5b. In either case, both Qk;0
n and Qk;1n are conditionally faulty. h
Lemma 4. Suppose that an n-cube Qnðn P 4Þ is conditionally faulty with f 6 2n 5 faulty nodes. Let z be a unique node with
exactly n 2 faulty neighbors. Then there exists some dimension j of f1; 2; . . . ; ng such that both Qj;0 n and Q
j;1
n are conditionally
faulty. Except for the case depicted inFig. 5c, both Qj;0n and Qj;1n contain 2n 7 or less faulty nodes if n P 5.
Proof. Since Qnis node-transitive, we assume z ¼ 0n. Since Qnis also link-transitive, we assume that ðzÞ1and ðzÞ2are
fault-free. Because z is a unique node with exactly n 2 faulty neighbors, we have j NF
k 2 f3; . . . ; ng, we have NFQk;0 n ðxÞ # N F QnðxÞ and N F Qk;1nðyÞ # N F QnðyÞ for x 2 VðQ k;0 n Þ fzg and y 2 VðQ k;1 n Þ. Thus we obtain j NFQk;0 n ðxÞ j6j N F QnðxÞ j6 n 3 and j N F Qk;1 n ðyÞ j6j N F QnðyÞ j6 n 3 for x 2 VðQ k;0 n Þ fzg and y 2 VðQ k;1 n Þ. In addition, we have j NFQk;0
n ðzÞ j ¼ ðn 2Þ 1 ¼ n 3 for every k 2 f3; . . . ; ng. Let j be an integer of f3; . . . ; ng. Then both Q
j;0 n and Q
j;1
n are
condi-tionally faulty.
Suppose f 6 2n 6. We see that, for any j 2 f3; . . . ; ng, both Qj;0n and Q j;1
n contain 2n 7 or less faulty nodes. Suppose f ¼ 2n 5. We assume, by contraposition, that either Qj;0
n or Qj;1n contains 2n 6 faulty nodes for any j 2 f3; . . . ; ng. Then, for any x of FðQnÞ fðzÞkj 3 6 k 6 ng, we have ½xj ¼ ½zj for every j 2 f3; . . . ; ng. Hence we have FðQnÞ fðzÞkj 3 6 k 6 ng # fz; ððzÞ1Þ2g. Since j FðQnÞ fðzÞkj 3 6 k 6 ng j ¼ f ðn 2Þ ¼ n 3 6 2 ¼ j fz; ððzÞ1Þ2g j, we derive that n 6 5. That is, if n P 6, there exists some dimension j of f3; . . . ; ng such that both Qj;0
n and Qj;1n are conditionally faulty with 2n 7 or less faulty nodes. Since j FðQnÞ fðzÞkj 3 6 k 6 ng j ¼ 2 for n ¼ 5, nodes z and ððzÞ1Þ2are faulty; that is, FðQ5Þ ¼ fz; ðzÞ3;ðzÞ4;ðzÞ5;ððzÞ1Þ2g, as shown inFig. 5c. Therefore,Fig. 5c happens to be the only possibility that either Qj;0n or Qj;1
n contains 2n 6 faulty nodes for every j 2 f3; . . . ; ng. h
Lemma 5. Suppose that an n-cube Qnðn P 4Þ contains f 6 2n 5 faulty nodes such that every node has at least three fault-free
neighbors. Then there exists some dimension j of f1; . . . ; ng such that both Qj;0 n and Q
j;1
n are conditionally faulty. For n P 5, Q j;0 n and
Qj;1
n contain 2n 7 or less faulty nodes.
Proof. Since every node has at least three fault-free neighbors, every ðn 1Þ-dimensional subcube of Qnis conditionally
faulty. First, we consider the case that f 6 2n 6. Let u and v be two distinct faulty nodes, and let j 2 f1; . . . ; ng such that ½uj–½
v
j. Then both Qj;0n and Qj;1n contain 2n 7 or less faulty nodes.Now we consider the case that f ¼ 2n 5. For n P 5, we claim that there exists some dimension j of f1; . . . ; ng such that j FðQj;0n Þ j6 2n 7 and j FðQj;1nÞ j6 2n 7. For 1 6 k 6 n, we define that qk ¼ 1 if ½uk ¼ ½
v
k for every two distinct faulty nodes u;v
2 FðQnÞ, and qk ¼ 0 otherwise. Let q ¼Pn
k ¼ 1qk. Clearly, all faulty nodes are located in either Q k;0
n or Q
k;1 n if qk ¼ 1. For convenience, let f1 6 k 6 n j qk ¼ 0g ¼ fi1; . . . ;inqg. Then both Qj;0n and Q
j;1
n contain at least one faulty node for j 2 fi1; . . . ;inqg. Suppose, by contradiction, either Qj;0n or Qj;1n contains only one faulty node for every j 2 fi1; . . . ;inqg. For
v
2 FðQnÞ, let Aðv
Þ ¼ f1 6 k 6 n j FðQk;0n Þ ¼ fv
g or FðQk;1
n Þ ¼ f
v
gg. Since Qnis node-transitive, we assume that e ¼ 0nis a faulty node such that j AðeÞ j achieves the maximum of set fj Aðv
Þ jjv
2 FðQnÞg. For convenience, let p ¼ j AðeÞ j. Obviously, we have 1 6 p 6 n q. Moreover, let AðeÞ ¼ fi1; . . . ;ipg. Forv
2 FðQnÞ feg, we see that ½v
k ¼ 1 for each k 2 fi1; . . . ;ipg. Let BðkÞ ¼ fv
2 FðQnÞ feg j ½v
k–½ekg for k 2 fipþ1; . . . ;inqg. Since we assumed, by contradiction, that either Qj;0n or Qj;1 n has only one faulty node for each j 2 fi1; . . . ;inqg, we have j BðjÞ j ¼ 1 for each j 2 fipþ1; . . . ;inqg. Since Qn is link-transitive, we assume that fi1; . . . ;ipg ¼ f1; . . . ; pg and fipþ1; . . . ;inqg ¼ fp þ 1; . . . ; n qg. Then we have ðFðQnÞ fegÞ S
k2fipþ1;...;inqgBðkÞ # f0 np1p
g. Accordingly, we derive that 1 ¼ j f0np1pg jPj ðFðQnÞ fegÞ Sk2fipþ1;...;inqgBðkÞ jPj FðQnÞ j j feg j Pk2fi
pþ1;...;inqgj BðkÞ j ¼ ð2n 5Þ 1 ðn q pÞ; that is, p þ q 6 7 n. Recall that p P 1 and q P 0. Thus, we have n 2 f5; 6g. Now we can identify all faulty nodes according to the values of p, q, and n.
Case 1: Suppose ðn; q; pÞ ¼ ð5; 0; 1Þ. Since p ¼ 1, we have ½
v
1 ¼ 1 for eachv
2 FðQ5Þ feg and j BðjÞ j ¼ 1 for each j 2 f2; 3; 4; 5g. Thus we have FðQ5Þ ¼ f00000; 00011; 00101; 01001; 10001g. Clearly, node 00001 has five faulty neighbors.Case 2: Suppose ðn; q; pÞ ¼ ð5; 0; 2Þ. Similarly, we have FðQ5Þ ¼ f00000; 00111; 01011; 10011; 00011g. Then node 00011 has three faulty neighbors.
Case 3: Suppose ðn; q; pÞ ¼ ð5; 1; 1Þ. We have FðQ5Þ ¼ f00000; 00011; 00101; 01001; 00001g. Again, node 00001 has four faulty neighbors.
Case 4: Suppose ðn; q; pÞ ¼ ð6; 0; 1Þ. We have FðQ6Þ ¼ f000000; 000011; 000101; 001001; 010001; 100001; 000001g. Thus, node 000001 has six faulty neighbors.
In short, node 0np1phas at least n 2 faulty neighbors, which contradicts the requirement that every node has at least three fault-free neighbors. Hence there exists some dimension j of f1; . . . ; ng such that both Qj;0n and Q
j;1
n are conditionally faulty with 2n 7 or less faulty nodes. h
Suppose that Qnis conditionally faulty with utmost 2n 5 faulty nodes. Let F ¼ FðQnÞ. For n P 5, we propose a
proce-dure PARTITION(Qn, F) to determine j-partition of Qnaccording to the following rules: Q5 z Qn u v n-4 faulty neighbors n-4 faulty neighbors k ( )u k ( )v Qn u j ( )u v j ( )v n-4 faulty neighbors n-4 faulty neighbors i ( )u = i ( )v = 3 ( )z 4 ( )z 5 ( )z 1 ( )z 2 ( )z 1 ( )z ( )2 z
Fig. 5. Every faulty node is marked by an ‘‘X” symbol. (a,b) j NF QnðuÞ j ¼ j N
F
QnðvÞ j ¼ n 2 and j N F
QnðxÞ j6 n 3 for x 2 VðQnÞ fu;vg; (c) a faulty node distribution on Q5; (d) a conditionally faulty 4-cube with four faulty nodes.
(1) Suppose that at least three nodes of Qn have exactly n 2 faulty neighbors, respectively. If Qn has its faulty
nodes distributed as shown inFig. 3a, it will be partitioned along dimension j ¼ dimððt1;t5ÞÞ. Then one resulting
subcube has its faulty nodes distributed as inFig. 2b. Otherwise,Lemma 1 and Lemma 2ensure that Qncan be
partitioned along some dimension j such that both Qj;0n and Q j;1
n are conditionally faulty with 2n 7 or less faulty
nodes.
(2) Suppose that there exist exactly two nodes of Qn with n 2 faulty neighbors, respectively. By Lemma 3, there
exists some dimension j of f1; . . . ; ng such that both Qj;0n and Q j;1
n are conditionally faulty with 2n 7 or less faulty
nodes.
(3) Suppose that there is only one node of Qnwith exactly n 2 faulty neighbors. Denote it by z. If the faulty nodes
are distributed as inFig. 5c, we partition Qnalong any dimension j 2 fi j ðzÞiis faultyg. Then one resulting subcube
turns out to have 2n 6 faulty nodes, distributed as in Fig. 5d. Otherwise, we can apply Lemma 4 to choose a dimension j of f1; . . . ; ng such that both Qj;0
n and Q j;1
n are conditionally faulty with 2n 7 or less faulty nodes.
(4) Suppose that every node of Qnhas at least three fault-free neighbors. Obviously, every ðn 1Þ-cube is conditionally faulty.
ByLemma 5, there exists some dimension j of f1; . . . ; ng such that both Qj;0
n and Q j;1
n contain 2n 7 or less faulty nodes.
The following corollary summarizes what is obtained by procedure PARTITION(Qn, F). Also, it is a summary ofLemmas
1–5.
Corollary 1. Suppose that an n-cube Qn ðn P 5Þ is conditionally faulty with f 6 2n 5 faulty nodes. Except for the cases illustrated in Figs.2c,3a, and5c, there exists some dimension j of f1; 2; . . . ; ng such that both Qj;0n and Q
j;1
n are conditionally faulty with 2n 7 or less faulty nodes.
4. Long paths in faulty hypercubes
The following theorem was proved by Fu[7].
Theorem 3 [7]. Let u and v denote two arbitrary fault-free nodes of an n-cube with f 6 n 2 faulty nodes, where n P 3. If hðu;
v
Þ is odd (or even), then there exists a fault-free path of length at least 2n 2f 1 (or 2n 2f 2) between u and v.To improve the above result, we need the following lemma.
Lemma 6. Let z 2 VðQ4Þ, fi; j; p; qg ¼ f1; 2; 3; 4g, and F ¼ fðzÞi;ðzÞj;ðzÞpg. Suppose that s and t are any two nodes of Q4 F such that fs; tg–fz; ðzÞqg. Then Q4 F has a path of length at least 9 or 8 between s and t if hðs; tÞ is odd or even, respectively. Proof. By symmetry, let z ¼ 0000, i ¼ 1, j ¼ 2, p ¼ 3, and q ¼ 4. We partition Q4into Q
4;0 4 and Q 4;1 4 . Then Q 4;1 4 is fault-free and z 2 V0ðQ4;04 Þ.
Case 1: Both s and t are in Q4;0
4 F. Since Q 4;1
4 is fault-free,Theorem 1ensures that Q 4;1
4 contains a path P of length 7 (respectively, 6) between ðsÞ4and ðtÞ4if hðs; tÞ is odd (respectively, even). Thus, hs; ðsÞ4;P; ðtÞ4;ti is a fault-free path of length 9 (respectively, 8) between s and t if hðs; tÞ is odd (respectively, even).
Case 2: Both s and t are in Q4;1
4 . If hðs; tÞ is odd,Theorem 1ensures that Q 4;1
4 fð1101; 1111Þg contains a path P of length 7 between s and t. Clearly, path P does not pass through (1101,1111). Since it spans Q4;14 , we have 1111 2 VðPÞ. Accordingly, link (1110,1111) or (1011,1111) is on P. Thus P can be written as hs; R1;1110; 1111; R2;ti or hs; T1;1011; 1111; T2;ti. As a result, hs; R1;1110; 0110; 0111; 1111; R2;ti or hs; T1;1011, 0011; 0111; 1111; T2;ti is a path of length 9 between s and t. On the other hand, if hðs; tÞ is even, then we consider two cases as follows. Suppose first that s; t 2 V0ðQ4;14 Þ. By Theorem 1, Q4;14 fð1101; 1111Þg contains a path P of length 6 between s and t. Again, link (1110,1111) or (1011,1111) is on P, and thus the desired path can be constructed as above. Suppose that s; t 2 V1ðQ4;14 Þ. ByTheorem 2, Q
4;1
4 f1001g contains a path P of length 6 between s and t. Obviously, link (1110,1111), (1101,1111), or (1011,1111) is on P. Hence the desired path can be constructed similarly.
Case 3: Suppose that s is in Q4;04 F and t is in Q 4;1
4 . First, we consider the case that s–z. If s 2 V0ðQ4Þ, then s is adjacent to node 0111. Clearly, there exists some node v of f0110; 0101; 0011g fsg such that ð
v
Þ4–t. ByTheorem 1, Q4;14 has a path P of length 6 or 7 between ð
v
Þ4and t if hðs; tÞ is odd or even, respectively. Then hs; 0111;v
;ðv
Þ4;P; ti is a fault-free path of length 9 or 10 if hðs; tÞ is odd or even, respectively. If s 2 V1ðQ4Þ, then we have s ¼ 0111. Obviously, there exists some node u of f0110; 0101; 0011g such that ðuÞ4–t. Similarly, Q4;14 has a path T of length 7 (respectively, 6) between ðuÞ
4 and t if hðs; tÞ is odd (respectively, even). Then hs; u; ðuÞ4;T; ti is a fault-free path of length 9 (respectively, 8) if hðs; tÞ is odd (respectively, even).
Next, we consider the case that s ¼ z. If hðs; tÞ is even, it follows fromTheorem 1that Q4;14 has a path H of length 7 between ðsÞ4 ¼ ðzÞ4 and t. Then hs ¼ z; ðzÞ4;H; ti is a fault-free path of length 8. If hðs; tÞ is odd,Theorem 2ensures that Q4;14 f1100g has a path R of length 6 between ðzÞ4 and t. Clearly, node 1111 is on R. Accordingly, link (1111,1110), (1111,1101), or (1111,1011) is on R. For example, path R can be written as hðzÞ4;R1;1111; 1110; R2;ti if ð1111; 1110Þ 2 EðRÞ. Then hs ¼ z; ðzÞ4;R1;1111; 0111; 0110, 1110; R2;ti is a fault-free path of length 9 between s and t. h
Theorem 4. Let F be a set of f 6 3 faulty nodes in Q4such that every node of Q4has at least two fault-free neighbors. Suppose that s and t are two arbitrary nodes of Q4 F. Then Q4 F contains a path of length at least 15 2f (respectively, 14 2f ) between s and t if hðs; tÞ is odd (respectively, even).
WithTheorem 4andLemma 6, we will be able to prove the next theorem.
Theorem 5. Let F be a set of f faulty nodes in Qnðn P 1Þ such that every node of Qnhas at least two fault-free neighbors. Suppose f ¼ 0 if n 2 f1; 2g, and f 6 2n 5 if n P 3. Let s and t be two arbitrary nodes of Qn F. Then Qn F contains a path of length at least 2n 2f 1 (respectively, 2n 2f 2) between s and t if hðs; tÞ is odd (respectively, even).
Proof. The result is trivial for n 2 f1; 2g. When n 2 f3; 4g, the result follows fromTheorem 3orTheorem 4, respectively. In what follows we consider the case that n P 5. Except for the faulty node distribution illustrated inFig. 3a, procedure PAR-TITION(Qn, F) returns j-partition of Qnsuch that both Qj;0n and Q
j;1
n are conditionally faulty. If Q5has its faulty nodes
distrib-uted as inFig. 3a, then PARTITION(Q5, F) returns j-partition of Q5such that one subcube has its faulty nodes distributed as in
Fig. 2b. Accordingly, the proof can be justified by the induction on n. Our inductive hypothesis is that the result holds for
Qn1. For convenience, let F0 ¼ FðQj;0nÞ and F1 ¼ FðQj;1nÞ. Moreover, let f0 ¼ j F0j and f1 ¼ j F1j. Without loss of generality,
we assume that s 2 V0ðQn FÞ.
Case 1: Suppose f062n 7 and f162n 7. Without loss of generality, we assume that f06f1. In particular, for the case illustrated inFig. 3a, Qj;05 is conditionally faulty with f0 ¼ 2 faulty nodes, and Qj;15 is not conditionally faulty with f1 ¼ 3 faulty nodes distributed as inFig. 2b.
Subcase 1.1. Both s and t are in Qj;0
n . By inductive hypothesis, Q j;0
n F0contains a path H0of length L at least 2n1 2f0 1 (respectively, 2n1 2f0 2) between s and t if hðs; tÞ is odd (respectively, even). Clearly, we have j f
v
2 VðQj;1nÞ jj NF Qj;1
nð
v
Þ jP n 2g j6 1. Let A ¼ fðH0ðiÞ; H0ði þ 1ÞÞ j 1 6 i 6 L and i 1ðmod 2Þg be a set of disjoint links on H0. Since j A j ¼ d2Le > f1þ 1 Pj F1[ fv
2 VðQj;1nÞ jj NF Qj;1
nð
v
Þ jP n 2g j for n P 5, there exists an odd integer ^ı, 1 6 ^ı 6 L, such that j F1\ fðH0ð^ıÞÞj;ðH0ð^ı þ 1ÞÞjg j ¼ 0, j NFQj;1 nððH0ð^ıÞÞ j Þ j6 n 3, and j NFQj;1 nððH0ð^ı þ 1ÞÞ jÞ j6 n 3 are satisfied. Let x ¼ H0ð^ıÞ and y ¼ H0ð^ı þ 1Þ. Hence path H0can be written as hs; H00;x; y; H000;ti.
If Qj;1
n is conditionally faulty, our inductive hypothesis asserts that Q j;1
n F1has a path H1of length at least 2n1 2f1 1
between ðxÞjand ðyÞj. Otherwise, the faulty nodes of Qj;1
n are distributed as inFig. 2b. Since both ðxÞ j
and ðyÞjhave two or more fault-free neighbors in Qj;1n,Lemma 6ensures that Q
j;1
n has a fault-free path H1of length at least 2n1 2f1 1 between ðxÞ j
and ðyÞj. Then hs; H0 0;x; ðxÞ
j
;H1;ðyÞj;y; H000;ti is a fault-free path of length at least 2 n
2f 1 (respectively, 2n 2f 2) be-tween s and t if hðs; tÞ is odd (respectively, even). SeeFig. 6a.
Subcase 1.2. Both s and t are in Qj;1n. We consider first that the faulty nodes of Q j;1
5 are distributed as depicted inFig. 2b. Let z denote the node with only one fault-free neighbor r in Qj;15 . Note that f0 ¼ 2 and f1 ¼ 3.
Suppose fs; tg ¼ fz; rg. Then a long path between s and t is constructed as follows. On the one hand, we assume that s ¼ z and t ¼ r. Since j V0ðQj;05Þ F0jPj V0ðQj;05Þ j j F0j ¼ 24 2 > 4 ¼ j F1[ ftg j, there exists some fault-free node x
of V0ðQj;05Þ such that ðxÞ jRF
1[ ftg. By inductive hypothesis, Qj;05 F0has a path H0of length at least 24 2f0 1 between
x y s t
Q
nj,0Q
nj,1H
0H
0 (x)j j (y)H
1 sQ
5j,0Q
5 j,1P
1 tP
2P
3P
4P
5 x1 y1 x2 y2 x3 y3 x4 y4 j (y2) j (x2) j (y1) j (x1) j (y3) j (x3) j (y4) j (x4)H
0 s zQ
5j,0Q
5 j,1 j (z) yH
1 x t j (y) j (y)H
0 s z j yH
1 x = tQ
5j,0Q
5 j,1 x s tH
0 j (x)H
1 j (s)Q
5j,0Q
5 j,1 b s tQ
nj,0Q
nj,1H
1 j (b)H
0 x t sH
0 j (x)H
1Q
5j,0Q
5 j,1 j (t) x y s tQ
nj,0Q
nj,1H
1H
1 j (x) j (y)H
0 (x)jðsÞj and x. By Lemma 6, Qj;1
5 F1 has a path H1 of length at least 24 2f1 2 between ðxÞ
j and t. As a result,
hs; ðsÞj;H0;x; ðxÞj;H1;ti is a fault-free path of length at least 25 2f 1 (seeFig. 6b). On the other hand, we assume that
t ¼ z and s ¼ r. Since j V1ðQj;05Þ F0jPj V1ðQj;05Þ j j F0j ¼ 24 2 > 4 ¼ j F1[ fsg j, there exists some fault-free node x
of V1ðQj;05Þ such that ðxÞ j
RF1[ fsg. Again, the inductive hypothesis asserts that Qj;05 has a fault-free path H0of length at least
24
2f0 1 between x and ðtÞj;Lemma 6asserts that Qj;1has a fault-free path H1of length at least 24 2f1 2 between s
and ðxÞj. Then hs; H1;ðxÞj;x; H0;ðtÞj;ti is a fault-free path of length at least 25 2f 1 (seeFig. 6c).
Suppose fs; tg–fz; rg. ThenLemma 6asserts that Qj;15 F1contains a path H1of length L at least 24 2f1 1 (respectively,
24
2f1 2) between s and t if hðs; tÞ is odd (respectively, even). Let A ¼ fðH1ðiÞ; H1ði þ 1ÞÞ j 1 6 i 6 L and i 1ðmod 2Þg be
a set of disjoint links. Since j A j ¼ dL
2e > 2 ¼ f0, there exists an odd integer ^ı, 1 6 ^ı 6 L, such that
F0\ fðH1ð^ıÞÞj;ðH1ð^ı þ 1ÞÞjg ¼ ;. Let x ¼ H1ð^ıÞ and y ¼ H1ð^ı þ 1Þ. Accordingly, path H1 can be written as hs; H01;x; y; H001;ti.
Again, the inductive hypothesis asserts that Qj;0
5 F0has a path H0of length at least 24 2f0 1 between ðxÞjand ðyÞj. Then
hs; H0 1;x; ðxÞ
j
;H0;ðyÞj;y; H001;ti is a fault-free path of length at least 2
5 2f 1 or 25 2f 2 if hðs; tÞ is odd or even,
respec-tively. SeeFig. 6d.
Now we consider the case that faulty nodes of Qj;1
5 are not distributed as depicted inFig. 2b, or n P 6. Then Q j;1 n is
con-ditionally faulty. By inductive hypothesis, Qj;1
n F1 has a path H1 of length L at least 2n1 2f1 1 (respectively,
2n1 2f1 2) between s and t if hðs; tÞ is odd (respectively, even). Similarly, let A ¼ fðH1ðiÞ; H1ði þ 1ÞÞ j 1 6
i 6 L and i 1ðmod 2Þg be a set of disjoint links. Since j A j ¼ dL
2e > f0 for n P 5, there is a link ðx; yÞ of A such that
F0\ fðxÞj;ðyÞjg ¼ ;. Accordingly, path H1can be written as hs; H01;x; y; H 00
1;ti. By inductive hypothesis, Q j;0
n F0has a path
H0of length at least 2n1 2f0 1 between ðxÞ j
and ðyÞj. Again, hs; H0 1;x; ðxÞ
j
;H0;ðyÞj;y; H001;ti is a fault-free path of length
at least 2n 2f 1 or 2n 2f 2 if hðs; tÞ is odd or even, respectively. SeeFig. 6d. Subcase 1.3. Suppose that s is in Qj;0n and t is in Q
j;1 n . Note that j fx 2 VðQ j;1 n Þ jj N F Qj;1
nðxÞ jP n 2g j6 1. On the one hand, we consider the case that node t has only one fault-free neighbor, denoted by r, in Qj;1n . On this occasion, n is equal to 5. Since j V1ðQj;0nÞ F0jP 2n2 f0>f1þ 2 ¼ j F1[ ft; rg j for n ¼ 5, there exists a fault-free node b of V1ðQj;0n Þ F0 such that ðbÞjRF1[ ft; rg. On the other hand, we consider the case that node t has at least two fault-free neighbors in Qj;1n. Since j V1ðQj;0nÞ F0jP 2n2 f0>f1þ 2 Pj F1j þ j ftg j þ j fx 2 VðQj;1n Þ jj N F Qj;1 nðxÞ jP n 2g jPj F1[ ftg [ fx 2 VðQ j;1 nÞ jj N F Qj;1 nðxÞ jP n 2g j for n P 5, there exists a fault-free node b of V1ðQj;0n Þ F0such that ðbÞjRF1[ ftg [ fx 2 VðQj;1n Þ jj N
F
Qj;1nðxÞ jP n 2g. By inductive hypothesis, Qj;0
n F0has a path H0of length at least 2n1 2f0 1 between s and b. If the faulty nodes of Qj;1n
are distributed as illustrated inFig. 2b,Lemma 6asserts that Qj;1n F1has a path H1of length at least 2n1 2f1 1
(respec-tively, 2n1
2f1 2) between ðbÞ
jand t if hððbÞj
;tÞ is odd (respectively, even); otherwise, the inductive hypothesis asserts that Qj;1
n F1has a path H1of length at least 2n1 2f1 1 (respectively, 2n1 2f1 2) between ðbÞjand t if hððbÞj;tÞ is odd
(respectively, even). Then hs; H0;b; ðbÞj;H1;ti is a fault-free path of length at least 2n 2f 1 (respectively, 2n 2f 2)
be-tween s and t if hðs; tÞ is odd (respectively, even). SeeFig. 6e.
Case 2: Suppose either f0 ¼ 2n 6 or f1 ¼ 2n 6. ByLemmas 1–5, we know that this case may occur while n ¼ 5. More
precisely, the faulty nodes happen to be distributed as illustrated inFig. 5c where z is itself a faulty node with three faulty neighbors. Without loss of generality, we assume that f0 ¼ 4; thus, ðzÞjis a unique faulty node in Qj;15.
Subcase 2.1. Both s and t are in Qj;05. By inductive hypothesis, Qj;05 ðF0 fzgÞ contains a path H0 of length L at least 9 ¼ 24 2 3 1 (respectively, 8 ¼ 24 2 3 2) between s and t if hðs; tÞ is odd (respectively, even).
First, we consider the case that node z is not on H0. Let A ¼ fðH0ðiÞ; H0ði þ 1ÞÞ j 1 6 i 6 L and i 1ðmod 2Þg be a set of
disjoint links on H0. Since j A j ¼ dL2e > 1 ¼ f1, there exists an odd integer ^ı, 1 6 ^ı 6 L, such that both ðH0ð^ıÞÞj and
ðH0ð^ı þ 1ÞÞjare fault-free. Let x ¼ H0ð^ıÞ and y ¼ H0ð^ı þ 1Þ. Hence path H0can be written as hs; H00;x; y; H000;ti. It follows from
inductive hypothesis that Qj;1 5 fðzÞ
j
g has a path H1 of length at least 13 ¼ 24 2 1 1 between ðxÞj and ðyÞj. Then
hs; H00;x; ðxÞ j;H
1;ðyÞj;y; H000;ti is a fault-free path of length at least 23 > 2 5
2 5 1 (respectively, 22 > 25 2 5 2) be-tween s and t if hðs; tÞ is odd (respectively, even).
Now we consider the case that node z is on H0. Since the length of H0 is at least 9, we can write H0 as
hs; H00;x; z; y; H 00
0;ti. Clearly, ðxÞ
j and ðyÞj are fault-free nodes in the same partite set of Qj;1
5. ByTheorem 2, Q
j;1
5 is
hyper-Hamiltonian laceable; thus Qj;15 fðzÞjg has a path H1 of length 14 between ðxÞj and ðyÞj. Then hs; H00;x; ðxÞ j
; H1;ðyÞj;y; H000;ti is a fault-free path of length at least 23 > 2
5
2 5 1 (respectively, 22 > 25 2 5 2) between s and t if hðs; tÞ is odd (respectively, even).
Subcase 2.2. Both s and t are in Qj;15 . For the sake of clarity, we distinguish whether hðs; tÞ is odd or even. Suppose that hðs; tÞ is odd. By inductive hypothesis, Qj;1
5 fðzÞ j
g contains a path H1of length L at least 13 between s and t.
Obviously, we have ðzÞjRVðH1Þ. Consequently, ð
v
Þj–z for anyv
2 VðH1Þ. Let A ¼ fðH1ðiÞ; H1ði þ 1ÞÞ j 1 6 i 6 L andi 1ðmod 2Þg be a set of disjoint links on H1. Since j A j j F0 fzg j ¼ dL2e ðf0 1Þ P 7 ð4 1Þ ¼ 4, there exist four
links of A, namely ðx1;y1Þ, ðx2;y2Þ, ðx3;y3Þ, and ðx4;y4Þ, such that ðxiÞjand ðyiÞ
jare fault-free for all i 2 f1; 2; 3; 4g. Thus path
H1 can be written as hs; P1;x1;y1;P2;x2;y2;P3;x3;y3;P4;x4;y4;P5;ti. Then hs; P1;x1;ðx1Þj;ðy1Þ j ;y1;P2;x2,ðx2Þj;ðy2Þ j ;y2; P3;x3;ðx3Þj;ðy3Þ j ;y3;P4;x4;ðx4Þj;ðy4Þ j
;y4, P5;ti is a fault-free path of length at least 21 ¼ 25 2 5 1 between s and t. See
Fig. 6f.
Suppose that hðs; tÞ is even. If s and ðzÞjbelong to the different partite sets of Qj;1
5,Theorem 2asserts that Q
j;1 5 fðzÞ
j
g has a path H1of length 14 between s and t. Similar to the case that hðs; tÞ is odd, there exist four disjoint links on H1, namely ðx1;y1Þ,
ðx2;y2Þ, ðx3;y3Þ, and ðx4;y4Þ, such that ðxiÞj and ðyiÞ
j are fault-free for all i 2 f1; 2; 3; 4g. Accordingly, we can write
H1 ¼ hs; P1;x1;y1;P2;x2;y2;P3;x3;y3;P4;x4;y4;P5;ti. Then hs; P1;x1, ðx1Þj;ðy1Þ j ;y1, P2;x2;ðx2Þj, ðy2Þ j ;y2, P3;x3;ðx3Þj, ðy3Þ j , y3;P4, x4;ðx4Þj;ðy4Þ j
, y4;P5;ti is a fault-free path of length at least 22 > 25 2 5 2 between s and t. If nodes s and ðzÞjbelong
to the same partite set of Qj;15, then we construct a fault-free path as follows. Since Q j;0
5 is conditionally faulty, we denote
by x any fault-free neighbor of z in Qj;0
5. By inductive hypothesis, Q j;0
5 ðF0 fzgÞ has a path H0 of length at least
9 ¼ 24 2 3 1 between x and z. We can write path H0as hx; H00;y; zi, where y is also a fault-free neighbor of z. Without
loss of generality, let j ¼ 5, fx; yg ¼ fðzÞ1;ðzÞ2g, and X ¼ fððzÞj;ððzÞjÞ3Þ; ððzÞj;ððzÞjÞ4Þg. Since j X j ¼ 2,Theorem 1ensures that Qj;1
5 X is strongly Hamiltonian laceable; hence it has a path H1 of length 14 between s and t. Obviously, both
ððzÞj;ðxÞjÞ and ððzÞj;ðyÞjÞ are on H1, and we can write H1as hs; H01;ðxÞ j ;ðzÞj;ðyÞj;H00 1;ti. Then hs; H 0 1;ðxÞ j ;x; H0 0;y; ðyÞ j ;H00 1;ti is a
fault-free path of length at least 22 > 25 2 5 2 between s and t.
Subcase 2.3. Suppose that s is in Qj;05 and t is in Qj;15 . By inductive hypothesis, Qj;05 ðF0 fzgÞ has a path H0of length at least 9 (respectively, 8) between s and z if hðs; zÞ is odd (respectively, even). Accordingly, path H0can be written as hs; H00;x; y; zi. Since ðzÞjis a unique faulty node in Qj;15, both ðxÞ
j
and ðyÞjare fault-free. If ðyÞj–t, it follows from inductive hypothesis that Qj;1
5 fðzÞ j
g has a path H1of length at least 13 (respectively, 12)
be-tween ðyÞj and t if hððyÞj;tÞ is odd (respectively, even). Then hs; H0 0;x; y; ðyÞ
j
;H1;ti is a path of length at least
21 ¼ 25 2 5 1 (respectively, 20 ¼ 25 2 5 2) between s and t if hðs; tÞ is odd (respectively, even). SeeFig. 6g. Other-wise, if ðyÞj
¼ t, then our inductive hypothesis asserts that Qj;15 fðzÞ j
g has a path H1of length at least 13 between ðxÞjand
ðyÞj. Then hs; H0 0;x; ðxÞ
j
;H1;ðyÞj ¼ ti is a path of length at least 21 ¼ 25 2 5 1 (respectively, 20 ¼ 25 2 5 2)
be-tween s and t if hðs; tÞ is odd (respectively, even). SeeFig. 6h. Therefore the proof is completed. h
5. Conclusion
In this paper, we show that a conditionally faulty n-cube with f 6 2n 5 faulty nodes contains a fault-free path of length at least 2n
2f 1 (respectively, 2n 2f 2) between any two fault-free nodes of odd (respectively, even) distance. When compared with the previous results presented by Fu[7], our results can tolerate almost double that faulty nodes under an additional condition that every node has two or more fault-free neighbors. It has been well grounded that 2n 5 is the max-imum number of faulty nodes tolerable in Qnif n ¼ 4. Yet it is not easy to show that a fault-free path of length at least
2n 2f 1 (or 2n 2f 2) cannot be embedded to connect any two nodes in a conditionally faulty n-cube with f faulty nodes for f P 2n 4 and n P 5. In fact, we conjecture that an n-cube may tolerate more than 2n 5 faulty nodes with re-spect to fault-tolerant path embedding. Therefore, we intend to find, in our future work, the tight upper bound to the number of tolerable faulty nodes.
Acknowledgements
The authors would like to express the most immense gratitude to the anonymous referees and Editor-in-Chief for their insightful and constructive comments. They greatly improve the readability of this paper.
Appendix A. Proof of Theorem 4
In order to proveTheorem 4, we address the following two lemmas in advance.
Lemma 7. Suppose that Q3is conditionally faulty with f 6 2 faulty nodes. Let s and t denote any two fault-free nodes of Q3. Then Q3contains a fault-free path of length at least 7 2f ðrespecti
v
ely; 6 2f Þ between s and t if hðs; tÞ is odd (respectively, even). Proof. If f < 2, this result follows from Theorem 3. Thus we only consider the case that f ¼ 2. For convenience, let F ¼ FðQ3Þ. Since Q3is node-transitive, we assume that node 000 is faulty. To require that every node of Q3has at leasttwo fault-free neighbors, the other faulty node must be one of {001, 010, 100, 111}.
Case 1: One of {001, 010, 100} is faulty. Obviously, each of {001,010,100} is adjacent to 000. Since Q3is link-transitive, we assume that 001 2 F; that is, F ¼ f000; 001g. Then we partition Q3into Q2;03 and Q
2;1
3 . Hence we have F # VðQ 2;0
3 Þ. SeeFig. 7a. Subcase 1.1. Both s and t are in Q2;03 F. Without loss of generality, we assume that s ¼ 101 and t ¼ 100. Obviously, hs ¼ 101; 111; 110; 100 ¼ ti is a fault-free path of length 3 ¼ 7 2 2.
Subcase 1.2. Both s and t are in Q2;13 . If hðs; tÞ is odd, then Q 2;1
3 contains a path of length 3 between s and t. Otherwise, Q 2;1 3
contains a path of length 2 between s and t. Subcase 1.3. Suppose that s is in Q2;03 F and t is in Q
2;1
3 . Without loss of generality, we assume s ¼ 101 and list the required
Case 2: Node 111 is faulty. SeeFig. 7b for illustration.
Subcase 2.1. Both s and t are in Q2;03 f000g. For every possible combination of s and t, we list the required paths inTable 1. Subcase 2.2. Both s and t are in Q2;13 f111g. This subcase is symmetric toSubcase 2.1.
Subcase 2.3. Suppose that s is in Q2;0
3 f000g and t is in Q 2;1
3 f111g. For every possible combination of s and t, we list the
required paths inTable 1.
In summary, Q3 F contains a path of length at least 7 2f (respectively, 6 2f ) between s and t if hðs; tÞ is odd
(respec-tively, even). h
Lemma 8. Let w 2 V0ðQ3Þ and fi; j; kg ¼ f1; 2; 3g. Suppose that b1 and b2 are two arbitrary nodes of V1ðQ3Þ. Then
Q3 fw; ððwÞiÞjg contains a path of length four between b1and b2if and only if fb1;b2g–fðwÞk;ðððwÞiÞjÞkg.
Proof. Since Q3is node-transitive and link-transitive, we assume that w ¼ 000, i ¼ 1, j ¼ 2, and k ¼ 3. SeeFig. 7c. Then
we list all the required paths inTable 1. h
Fig. 7. (a, b) Illustrations forLemma 7; (c) the distribution of faulty nodes indicated inLemma 8.
Table 1
The required paths forLemma 7andLemma 8.
Subcase 1.3ofLemma 7 s ¼ 101 t ¼ 010 hs ¼ 101; 100; 110; 010 ¼ ti t ¼ 011 hs ¼ 101; 111; 011 ¼ ti t ¼ 110 hs ¼ 101; 100; 110 ¼ ti t ¼ 111 hs ¼ 101; 100; 110; 111 ¼ ti Subcase 2.1ofLemma 7 s ¼ 101 t ¼ 001 hs ¼ 101; 100; 110; 010; 011; 001 ¼ ti t ¼ 100 hs ¼ 101; 001; 011; 010; 110; 100 ¼ ti s ¼ 001 t ¼ 100 hs ¼ 001; 011; 010; 110; 100 ¼ ti Subcase 2.3ofLemma 7 s ¼ 001 t ¼ 010 hs ¼ 001; 011; 010 ¼ ti t ¼ 011 hs ¼ 001; 101; 100; 110; 010; 011 ¼ ti t ¼ 110 hs ¼ 001; 011; 010; 110 ¼ ti s ¼ 100 t ¼ 010 hs ¼ 100; 110; 010 ¼ ti t ¼ 011 hs ¼ 100; 110; 010; 011 ¼ ti t ¼ 110 hs ¼ 100; 101; 001; 011; 010; 110 ¼ ti s ¼ 101 t ¼ 010 hs ¼ 101; 100; 110; 010 ¼ ti t ¼ 011 hs ¼ 101; 001; 011 ¼ ti t ¼ 110 hs ¼ 101; 100; 110 ¼ ti Lemma 8 b1 ¼ 001 b2 ¼ 010 hb1 ¼ 001; 101; 100; 110; 010 ¼ b2i b2 ¼ 100 hb1 ¼ 001; 101; 111; 110; 100 ¼ b2i b2 ¼ 111 hb1 ¼ 001; 101; 100; 110; 111 ¼ b2i b1 ¼ 010 b2 ¼ 100 hb1 ¼ 010; 110; 111; 101; 100 ¼ b2i b2 ¼ 111 hb1 ¼ 010; 110; 100; 101; 111 ¼ b2i
Theorem 4. Let F be a set of f 6 3 faulty nodes in Q4such that every node of Q4has at least two fault-free neighbors. Suppose that
s and t are two arbitrary nodes of Q4 F. Then Q4 F contains a path of length at least 15 2f (respectively, 14 2f ) between s
and t if hðs; tÞ is odd (respectively, even).
Proof. If f < 3, this result follows fromTheorem 3. Thus we concentrate only on the case that f ¼ 3. ByLemmas 1–5,Fig. 2c happens to be a unique case that a conditionally faulty Q4with three faulty nodes cannot be partitioned along any dimension
in such a way that both subcubes are conditionally faulty. On this occasion, we partition Q4along an arbitrary dimension j;
otherwise, there exists some dimension j such that both Qj;0 4 and Q
j;1
4 are conditionally faulty.
Case 1: Both Qj;04 and Qj;14 are conditionally faulty. For convenience, let F0 ¼ FðQj;04Þ and F1 ¼ FðQj;14 Þ. Without loss of generality, we assume that f0 ¼ j F0j ¼ 2 and f1 ¼ j F1j ¼ 1. Moreover, we assume s 2 V0ðQ4 FÞ.
Subcase 1.1. Both s and t are in Qj;04 . ByLemma 7, Qj;04 F0contains a path H0of length at least 3 ¼ 7 2f0(respectively, 2 ¼ 6 2f0) between s and t if hðs; tÞ is odd (respectively, even). Obviously, H0can be written as hs ¼ x0;x1;x2;H00;ti. If ðx1Þj is faulty, then ðx0Þjand ðx2Þj are fault-free. ByTheorem 2, Qj;14 is hyper-Hamiltonian laceable. Thus Q
j;1
4 fðx1Þjg has a Hamiltonian path H1between ðx0Þjand ðx2Þj. As a result, hs ¼ x0;ðx0Þj;H1;ðx2Þj;x2;H00;ti is a fault-free path of length at least 15 2f (respectively, 14 2f ) when hðs; tÞ is odd (respectively, even). If ðx1Þjis fault-free, then ðx0Þjor ðx2Þjis fault-free. Suppose, for example, that ðx0Þjis fault-free. ByLemma 7, Qj;14 F1has a fault-free path H1of length at least 7 2f1between ðx0Þjand ðx1Þj. As a result, hs ¼ x0;ðx0Þj;H1;ðx1Þj;x1;x2;H00;ti is a fault-free path of length at least 15 2f (respectively, 14 2f ) when hðs; tÞ is odd (respectively, even).
Subcase 1.2. Both s and t are in Qj;1
4. First, we consider the case that hðs; tÞ is odd. ByLemma 7, Q j;1
4 F1contains a path T1of
length at least 5 ¼ 7 2f1between s and t. Let A ¼ fðT1ðiÞ; T1ði þ 1ÞÞ j 1 6 i 6 5 and i 1ðmod2Þg be a set of disjoint links
on T1. Since j A j ¼ 3 > f0, there exists an odd integer ^ı, 1 6 ^ı 6 5, such that both ðT1ð^ıÞÞjand ðT1ð^ı þ 1ÞÞjare fault-free. Let
w ¼ T1ð^ıÞ and b ¼ T1ð^ı þ 1Þ. Accordingly, T1can be written as hs; T01;w; b; T 00
1;ti. ByLemma 7, Q j;0
4 F0has a path T0of length
at least 7 2f0between ðwÞjand ðbÞj. As a result, hs; T01;w; ðwÞ j;T
0;ðbÞj;b; T001;ti is a fault-free path of length at least 15 2f
between s and t.
Next, we consider the case that hðs; tÞ is even. Hence we have t 2 V0ðQ4 FÞ. Let u denote the faulty node in Qj;14. Then we
distinguish the following two subcases.
Subcase 1.2.1. Suppose that u 2 V1ðQj;14 Þ. By Theorem 2, Q j;1
4 is hyper-Hamiltonian laceable. Thus Q j;1
4 fug has a Hamiltonian path H1 from s to t. Obviously, the length of H1 is equal to 6. Let B ¼ fðH1ðiÞ; H1ði þ 1ÞÞ j 1 6 i 6 6 and i 1ðmod 2Þg be a set of disjoint links on T1. Since j B j ¼ 3 > f0, there exists an odd integer ^ı, 1 6 ^ı 6 6, such that both ðH1ð^ıÞÞjand ðH1ð^ı þ 1ÞÞjare fault-free. Let w ¼ H1ð^ıÞ and b ¼ H1ð^ı þ 1Þ. Thus H1can be written as hs; H0
1;w; b; H001;ti. ByLemma 7, Q j;0
4 F0 has a path H0 of length at least 7 2f0between ðwÞ j
and ðbÞj. As a result, hs; H01;w; ðwÞ
j
;H0;ðbÞj;b; H001;ti is a fault-free path of length at least 14 2f0>14 2f between s and t. Subcase 1.2.2. Suppose that u 2 V0ðQj;14Þ. Since hðs; tÞ is even, it follows fromLemma 7that Q
j;1
4 F1has a path T1of length
at least 6 2f1 ¼ 4 between s and t. If there exists a link ðw; bÞ on T1such that both ðwÞjand ðbÞjare fault-free, then a path of
length at least 14 2f can be constructed in a way similar to that described in Subcase 1.2.1. Otherwise, we have F0\ fðT1ðiÞÞj;ðT1ði þ 1ÞÞjg–; for every i. Then we claim that both ðT1ð2ÞÞjand ðT1ð4ÞÞj are faulty. Since f0 ¼ 2, we see that
j F0\ fðT1ð1ÞÞj;ðT1ð2ÞÞj;ðT1ð3ÞÞjg j ¼ 1 and j F0\ fðT1ð3ÞÞj;ðT1ð4ÞÞj;ðT1ð5ÞÞjg j ¼ 1. Then we have F0\ fðT1ð1ÞÞj;
ðT1ð2ÞÞj;ðT1ð3ÞÞjg ¼ ðF0\ fðT1ð1ÞÞj;ðT1ð2ÞÞjgÞ \ ðF0\ fðT1ð2ÞÞj;ðT1ð3ÞÞjgÞ ¼ fðT1ð2ÞÞjg. Similarly, we have F0\ fðT1ð3ÞÞj;
ðT1ð4ÞÞj;ðT1ð5ÞÞjg ¼ fðT1ð4ÞÞjg. That is, F0 ¼ fðT1ð2ÞÞj;ðT1ð4ÞÞjg. ByLemma 8, Qj;04 F0contains either a path T0of length
4 between ðT1ð1ÞÞj and ðT1ð3ÞÞj or a path R0 of length 4 between ðT1ð3ÞÞj and ðT1ð5ÞÞj. As a result,hs ¼ T1ð1Þ;
ðT1ð1ÞÞj;T0;ðT1ð3ÞÞj;T1ð3Þ; T1ð4Þ; T1ð5Þ ¼ ti or hs ¼ T1ð1Þ; T1ð2Þ; T1ð3Þ; ðT1ð3ÞÞj;R0, ðT1ð5ÞÞj;T1ð5Þ ¼ ti is a fault-free path of
length 8 ¼ 14 2f .
Subcase 1.3. Suppose that s is in Qj;04 and t is in Q j;1
4. Since f0 ¼ 2, we have j V1ðQj;04Þ F0jP 2 ¼ j F1[ ftg j and
j VðQj;04Þ ðF0[ fsgÞ j ¼ 5 >j F1[ ftg j. If hðs; tÞ is odd, we choose a node x of V1ðQj;04Þ F0such that ðxÞjis fault-free; otherwise,
we choose a node x of VðQj;0
4Þ ðF0[ fsgÞ such that ðxÞjRF1[ ftg. ByLemma 7, Qj;04 F0contains a path H0of length at least
7 2f0(respectively, 6 2f0) between s and x when hðs; xÞ is odd (respectively, even). Similarly, Q j;1
4 F1contains a path H1of
length at least 7 2f1 (respectively, 6 2f1) between ðxÞ j
and t when hððxÞj;tÞ is odd (respectively, even). As a result, hs; H0;x; ðxÞj;H1;ti is a fault-free path of length at least 15 2f (respectively, 14 2f ) if hðs; tÞ is odd (respectively, even).
Case 2: Suppose Q4has its faulty nodes distributed as inFig. 2c. To be precise, we assume F ¼ f0000; 0011; 1100g. Then
we partition Q4into Q4;04 and Q 4;1
4 . It is noticed that Q 4;0
4 is not conditionally faulty.
Subcase 2.1. Both s and t are in Q4;04 f0000; 0011g. By Theorem 3, Q 4;0
4 f0000g has a path T0 of length at least 5 (respectively, 4) between s and t if hðs; tÞ is odd (respectively, even).
We consider first that hðs; tÞ is odd. Thus the length of path T0is greater than or equal to 5. Then T0passes through every
node of V0ðQ4;04 Þ f0000g. In particular, the faulty node 0011 is on T0. Hence T0can be written as hs; T00;x; 0011; y; T000;ti. Since
hð0011; 1100Þ ¼ 4, both ðxÞ4and ðyÞ4are fault-free. Since hððxÞ4
;ðyÞ4Þ is even,Theorem 3ensures that Q4;1
4 f1100g has a
path T1of length at least 4 between ðxÞ4and ðyÞ4. As a result, hs; T00;x; ðxÞ 4
;T1;ðyÞ4;y; T000;ti is a fault-free path of length at least
9 ¼ 15 2f .
Next, we consider the case that hðs; tÞ is even. We distinguish whether the faulty node 0011 is on T0. If node 0011 is on T0,
then a path of length at least 8 can be constructed to join s and t in a way similar to that described earlier. Otherwise, there exists a link ðw; bÞ on T0such that both ðwÞ4and ðbÞ4are fault-free. Hence T0can be written as hs; R00;w; b; R
00
0;ti. ByTheorem 3,
Q4;14 f1100g has a path T1of length at least 5 between ðwÞ4and ðbÞ4. Then hs; R00;w; ðwÞ 4
;T1;ðbÞ4;b; R000;ti turns out to be a
fault-free path of length at least 10 > 14 2f .
Subcase 2.2. Suppose that s is in Q4;04 f0000; 0011g and t is in Q4;14 f1100g. ByTheorem 3, Q4;04 f0000g has a path T0of length at least 5 (respectively, 4) between nodes s and 0011 if hðs; 0011Þ is odd (respectively, even). Accordingly, we write T0 as hs; T0
0;x; y; 0011i. Since hð0011; 1100Þ ¼ 4, both ðxÞ 4
and ðyÞ4 is fault-free. On the one hand, we assume ðyÞ4–t. By
Theorem 3, Q4;14 f1100g has a path T1 of length at least 5 (respectively, 4) between ðyÞ4 and t if hððyÞ4;tÞ is odd
(respectively, even). As a result, hs; T00;x; y; ðyÞ 4
;T1;ti is a fault-free path of length at least 9 ¼ 15 2f (respectively, 8 ¼ 14 2f ) if hðs; tÞ is odd (respectively, even). On the other hand, if ðyÞ4 ¼ t, thenTheorem 3ensures that Q4;14 f1100g has a path R1of length at least 5 between ðxÞ4and ðyÞ4. Then hs; T00;x; ðxÞ
4
;R1;ðyÞ4 ¼ ti turns out to be a fault-free path of length at least 9 ¼ 15 2f (respectively, 8 ¼ 14 2f ) if hðs; tÞ is odd (respectively, even).
Subcase 2.3. Both s and t are in Q4;1
4 f1100g. We list the required paths obtained by brute force inTable 2.
Therefore the proof is completed. h
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Table 2
The required paths inSubcase 2.3ofTheorem 4.
s ¼ 1101 t ¼ 1110 hs ¼ 1101; 1001; 0001; 0101; 0100; 0110; 0010; 1010; 1110 ¼ ti t ¼ 1111 hs ¼ 1101; 1001; 0001; 0101; 0100; 0110; 0010; 1010; 1110; 1111 ¼ ti t ¼ 1000 hs ¼ 1101; 0101; 0001; 1001; 1011; 1111; 1110; 1010; 1000 ¼ ti t ¼ 1001 hs ¼ 1101; 0101; 0100; 0110; 1110; 1111; 1011; 1010; 1000; 1001 ¼ ti t ¼ 1010 hs ¼ 1101; 0101; 0100; 0110; 1110; 1111; 1011; 1001; 1000; 1010 ¼ ti t ¼ 1011 hs ¼ 1101; 0101; 0001; 1001; 1000; 1010; 1110; 1111; 1011 ¼ ti s ¼ 1110 t ¼ 1111 hs ¼ 1110; 1010; 1000; 1001; 1101; 0101; 0100; 0110; 0111; 1111 ¼ ti t ¼ 1000 hs ¼ 1110; 0110; 0100; 0101; 0001; 1001; 1011; 1010; 1000 ¼ ti t ¼ 1001 hs ¼ 1110; 0110; 0100; 0101; 1101; 1111; 1011; 1010; 1000; 1001 ¼ ti t ¼ 1010 hs ¼ 1110; 0110; 0100; 0101; 0001; 1001; 1101; 1111; 1011; 1010 ¼ ti t ¼ 1011 hs ¼ 1110; 0110; 0100; 0101; 0001; 1001; 1101; 1111; 1011 ¼ ti s ¼ 1111 t ¼ 1000 hs ¼ 1111; 0111; 0110; 0100; 0101; 0001; 1001; 1011; 1010; 1000 ¼ ti t ¼ 1001 hs ¼ 1111; 0111; 0101; 0100; 0110; 0010; 1010; 1000; 1001 ¼ ti t ¼ 1010 hs ¼ 1111; 0111; 0110; 0100; 0101; 1101; 1001; 1000; 1010 ¼ ti t ¼ 1011 hs ¼ 1111; 0111; 0101; 0100; 0110; 0010; 1010; 1000; 1001; 1011 ¼ ti s ¼ 1000 t ¼ 1001 hs ¼ 1000; 1010; 1110; 0110; 0100; 0101; 1101; 1111; 1011; 1001 ¼ ti t ¼ 1010 hs ¼ 1000; 1001; 1101; 0101; 0100; 0110; 1110; 1111; 1011; 1010 ¼ ti t ¼ 1011 hs ¼ 1000; 1001; 1101; 0101; 0100; 0110; 1110; 1111; 1011 ¼ ti s ¼ 1001 t ¼ 1010 hs ¼ 1001; 1011; 1111; 0111; 0101; 0100; 0110; 1110; 1010 ¼ ti t ¼ 1011 hs ¼ 1001; 1000; 1010; 1110; 0110; 0100; 0101; 1101; 1111; 1011 ¼ ti s ¼ 1010 t ¼ 1011 hs ¼ 1010; 1000; 1001; 1101; 0101; 0100; 0110; 0111; 1111; 1011 ¼ ti
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