Long Paths in Faulty Hypercubes with Conditional Node-faults
7.1 Partition of an n-cube with conditional node-faults
Here we will show that a conditionally faulty n-cube can be partitioned into two conditionally faulty subcubes if it has 2n − 5 or less faulty nodes. Recall that F (G) denotes the set of all faulty elements in a network G. Let u be a node of G. For convenience, we use NGF(u) to denote the set of all faulty neighbors of u; i.e., NGF(u) = NG(u) ∩ F (G).
Suppose Qn, n ≥ 4, is conditionally faulty with f ≤ 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 = NQFn(u), V = NQFn(v), and W = NQFn(w).
If |V ∩ W | = 0, then we have f ≥ |V ∪ W | = |V | + |W | = 2n − 4, contradicting the requirement that f ≤ 2n − 5. Therefore, |V ∩ W | ≥ 1 needs to be satisfied. Similarly, we also have |U ∩ V | ≥ 1 and |U ∩ W | ≥ 1. Since any two nodes of an n-cube can have utmost two common neighbors, we obtain that |V ∩ W |, |U ∩ V |, |U ∩ W | ∈ {1, 2}. We first consider the case that at least one of |V ∩ W |, |U ∩ V |, and |U ∩ W | is equal to 1. Without loss of generality, we suppose |V ∩ W | = 1.
I. Firstly, we concern the case that |V ∩ W | = |U ∩ V | = |U ∩ W | = 1. If |U ∩ V ∩ W | ≥ 1, we have 2n − 5 ≥ f ≥ |U ∪ V ∪ W | = 3(n − 2) − (1 + 1 + 1) + 1 = 3n − 8; i.e., n ≤ 3. Since n ≥ 4, we only concern |U ∩ V ∩ W | = 0. Then we have 2n − 5 ≥ f ≥
|U ∪ V ∪ W | ≥ 3(n − 2) − (1 + 1 + 1) = 3n − 9; i.e., n ≤ 4. Figure 7.2(a) depicts a faulty 4-cube with |V ∩ W | = |U ∩ V | = |U ∩ W | = 1 and |U ∩ V ∩ W | = 0. Figure 7.2(b) is a cube-styled layout isomorphic to Figure 7.2(a). We can examine Figure 7.2(a) in a top-down viewpoint. Since hypercube is node-transitive, we can assume that u = t1. By link-transitivity, we assume that t4 and t5 are faulty neighbors of u. Since |U ∩ V | = 1, we obtain v ∈ {t7, t8, t9, t10}. Without loss of generality, we assume that v = t10. Since
u
As a consequence, this happens to be the only possibility. However, node t11 has only one fault-free neighbor. Thus it is not conditionally faulty.
II. Secondly, we consider the case that |V ∩ W | = |U ∩ V | = 1 and |U ∩ W | = 2. By
See Figure 7.2(c) for illustration. For clarity, Figure 7.2(d) is an isomorphic layout of Figure 7.2(c). Similarly, we can examine Figure 7.2(c) in a top-down viewpoint.
By node-transitivity, we assume that u = t1. By link-transitivity, we assume that t4 and t5 are faulty neighbors of u. Since |U ∩ W | = 2, we have w = t11. Since
|V ∩ W | = |U ∩ V | = 1 and |U ∩ V ∩ W | = 1, we obtain v ∈ {t7, t8, t9, t10}. Without loss of generality, we assume that v = t10. Then this turns out to be the only possibility.
It is noticed that node t8 has only two fault-free neighbors.
III. Next, we concern the case that |V ∩ W | = 1 and |U ∩ V | = |U ∩ W | = 2. Similarly, we have |U ∩V ∩W | = 1. Since (U ∩V )∪(U ∩W ) ⊆ U, we have |(U ∩V )∪(U ∩W )| ≤ |U|.
However, we have a contradiction that |(U ∩V )∪(U ∩W )| = |U ∩V |+|U ∩W |−|U ∩V ∩ W | = 2+2−1 = 3 > n−2 = |U| if n ≤ 4. In what follows, we suppose that n ≥ 5. As a consequence, we have 2n−5 ≥ f ≥ |U ∪V ∪W | = 3(n−2)−(1+2+2)+1 = 3n−10; i.e., n = 5. See Figure 7.3(a). Again, we examine Figure 7.3(a) in a top-down viewpoint.
By node-transitivity, we assume that u = t1. By link-transitivity, we assume that t4, t5, and t6 are faulty neighbors of u. Since |U ∩ V | = |U ∩ W | = 2, we have {v, w} ⊂ {t14, t15, t16}. Without loss of generality, we assume that v = t14and w = t16. Since |V ∩ W | = 1, we have t26 ∈ V ∪ W . Moreover, we have 2n − 5 ≥ f ≥ |V ∪ W | =/
|V | + |W | − |V ∩ W | = (n − 2) + (n − 2) − 1 = 2n − 5; that is, f = 2n − 5 and
Q5
Figure 7.3: Every faulty node is marked by an “X” symbol. Each of u, v, w, and z has only two fault-free neighbors. (a) The Q5 with |NQF5(v) ∩ NQF5(w)| = 1 and |NQF5(u) ∩ NQF5(v)| =
|NQF5(u)∩NQF5(w)| = 2; (b) the Q5 with |NQF5(u)∩NQF5(v)| = |NQF5(v)∩NQF5(w)| = |NQF5(u)∩
NQF5(w)| = 2.
U ⊆ V ∪ W . Then we have either t20 ∈ V or t23 ∈ V . Without loss of generality, we assume that t23 ∈ V . Similarly, we can assume that t25 ∈ W . As a result, this is the only possibility. It is noted that node t12 = z has three faulty neighbors, and
|NQF5(x)| ≤ 2 for each x ∈ V (Q5) − {u, v, w, z}. Figure 7.4 as follows. By node-transitivity, we assume that u = t1. By link-transitivity, we assume that t4, t5, t6, and t7 are faulty neighbors of u. Since |U ∩ V | = |U ∩ W | = 2, we deduce that {v, w} ⊂ {ti | 17 ≤ i ≤ 22}. Since |U ∩ V ∩ W | = 1, we can assume that v = t20 and w = t22. Then we have |V ∩ {t30, t36, t39, t42}| = 2 and |W ∩ {t32, t38, t41, t42}| = 2. Since
|V ∩ W | = 2, we have V ∩ W = {t6, t42}. If t39∈ V and t41 ∈ W , then node t18 happens to have only two fault-free neighbors (see Figure 7.4(a)); otherwise, we have |NQF6(x)| ≤ 3 for each x ∈ V (Q6) − {u, v, w} (see Figure 7.4(b), in which nodes t36 and t41, for example, are faulty). Hence these figures cover all possibilities.
According to the analysis presented earlier, a conditionally faulty n-cube with f ≤ 2n − 5
Q6
v w t10
t2 t3 t4 t5 t6 t7
t8 t9
t11 t12 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
(b)
The links among nodes t23,...,t57 are omitted.
t43 t44 t45 t46 t47 t48 t49 t50 t51 t52 t53 t54 t55 t56 t57
=u t1
Q6 t1=u
v w t10 z
t2 t3 t4 t5 t6 t7
t8 t9
t11 t12 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
(a)
The links among nodes t23,...,t57 are omitted.
t43 t44 t45 t46 t47 t48 t49 t50 t51 t52 t53 t54 t55 t56 t57
Figure 7.4: Every faulty node is marked by an “X” symbol. The Q6 with |NQF6(u)∩NQF6(v)| =
|NQF6(v) ∩ NQF6(w)| = |NQF6(u) ∩ NQF6(w)| = 2. (a) |NQF6(u)| = |NQF6(v)| = |NQF6(w)| =
|NQF6(z)| = 4 and |NQF6(x)| ≤ 3 for x ∈ V (Q6) − {u, v, w, z}; (b) |NQF6(u)| = |NQF6(v)| =
faulty nodes is likely to contain three or four nodes, every of which has only two fault-free neighbors. Since 2n − 5 ≤ n − 2 for n ≤ 3, we concentrate only on the case that n ≥ 4. To summarize, we have the following two lemmas.
Lemma 7.1. Suppose that an n-cube Qn (n ≥ 4) is conditionally faulty with f ≤ 2n−5 faulty nodes. Let u, v, w, z ∈ V (Qn) such that |NQFn(u)| = |NQFn(v)| = |NQFn(w)| = |NQFn(z)| = n−2 and |NQFn(x)| ≤ n−3 for every x ∈ V (Qn)−{u, v, w, z}. Then the faulty nodes are distributed as illustrated in Figure 7.2(c), Figure 7.3(a,b), and Figure 7.4(a). In Figure 7.2(c) and Figure 7.3(a), no dimensions can be used to partition Qn in such a way that both resulting subcubes are conditionally faulty. In Figure 7.3(b) and Figure 7.4(a), there exists some dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
Proof. In Figure 7.2(c) and Figure 7.3(a), we check, by brute force, that either Qk,0n or Qk,1n contains a node with only one fault-free neighbor for each k ∈ {0, 1, . . . , n − 1}; that is, there does not exist any dimension to partition Qn such that both (n − 1)-cubes are conditionally faulty. In Figure 7.3(b) and Figure 7.4(a), let j be any integer of {0, 1, . . . , n − 1} such that (u)j is faulty. Then both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
Lemma 7.2. Suppose that an n-cube Qn (n ≥ 4) is conditionally faulty with f ≤ 2n − 5 faulty nodes. Let u, v, w ∈ V (Qn) such that |NQFn(u)| = |NQFn(v)| = |NQFn(w)| = n − 2 and
|NQFn(x)| ≤ n − 3 for every x ∈ V (Qn) − {u, v, w}. Then the faulty nodes are distributed as illustrated in Figure 7.4(b). Moreover, there exists some dimension j of {0, 1, . . . , n − 1}
such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
Proof. Let j ∈ {0, 1, . . . , n − 1} such that (u)j ∈ NQFn(u) ∩ NQFn(v) ∩ NQFn(w). Then both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
Lemma 7.3. Suppose that an n-cube Qn (n ≥ 4) is conditionally faulty with f ≤ 2n − 5 faulty nodes. Let u and v be two nodes of Qn such that |NQFn(u)| = |NQFn(v)| = n − 2 and
|NQFn(x)| ≤ n − 3 for every x ∈ V (Qn) − {u, v}. Then there exists some dimension k of {0, 1, . . . , n − 1} such that both Qk,0n and Qk,1n are conditionally faulty. When n ≥ 5, both Qk,0n and Qk,1n contain 2n − 7 or less faulty nodes.
Proof. Since |NQFn(u)| = |NQFn(v)| = n − 2 and f ≤ 2n − 5, we have |NQFn(u) ∩ NQFn(v)| ≥ 1.
Since any two nodes of Qncan have utmost two common neighbors, we consider the following two cases.
Case 1: Suppose that |NQFn(u) ∩ NQFn(v)| = 2. Let i and j be two integers such that {(u)i, (u)j} = NQFn(u) ∩ NQFn(v). Obviously, we have (u)i = (v)j and (u)j = (v)i. Then we can partition Qn along dimension k ∈ {i, j}. As a result, both Qk,0n and Qk,1n contain at least n − 3 faulty nodes. See Figure 7.5(a).
Case 2: Suppose that |NQFn(u) ∩ NQFn(v)| = 1. We claim first that this case holds only for n ≥ 5. By contradiction, we suppose n = 4. Let p and q be two integers such that both
Q5 (d) a conditionally faulty 4-cube with four faulty nodes.
(u)p and (u)q are faulty. Since |NQFn(u) ∩ NQFn(v)| = 1, we have v 6= ((u)p)q. Thus node ((u)p)qhappens to have only two fault-free neighbors, which contradicts the assumption that
|NQFn(x)| ≤ n − 3 for every x ∈ V (Qn) − {u, v}.
Let i and j be two integers such that {(u)i} = {(v)j} = NQFn(u) ∩ NQFn(v). Since
|NQFn(u) − {(u)i}| + |NQFn(v) − {(v)j}| = 2(n − 3) > n − 2 = |{0, 1, . . . , n − 1} − {i, j}|
for n ≥ 5, there exists some dimension k of {0, 1, . . . , n − 1} − {i, j} such that both (u)k and (v)k are faulty. As a result, either Qk,0n or Qk,1n contains exactly two faulty nodes. See Figure 7.5(b).
In either case, both Qk,0n and Qk,1n are conditionally faulty.
Lemma 7.4. Suppose that an n-cube Qn (n ≥ 4) is conditionally faulty with f ≤ 2n − 5 faulty nodes. Let z be a unique node with exactly n − 2 faulty neighbors. Then there exists some dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty.
Except for the case depicted in Figure 7.5(c), both Qj,0n and Qj,1n contain 2n − 7 or less faulty nodes if n ≥ 5.
Proof. Since Qn is node-transitive, we assume z = 0n. Since Qn is also link-transitive, we assume that (z)0 and (z)1 are fault-free. Because z is a unique node with exactly n − 2 faulty neighbors, we have |NQFn(x)| ≤ n − 3 for x ∈ V (Qn) − {z}. For every k ∈ {2, 3, . . . , n − 1},
Suppose f = 2n−5. We assume, by contraposition, that either Qj,0n or Qj,1n contains 2n−6 faulty nodes for any j ∈ {2, 3, . . . , n − 1}. Then, for any x of F (Qn) − {(z)k | 2 ≤ k ≤ n − 1},
we have (x)j = (z)j for every j ∈ {2, 3, . . . , n − 1}. Hence we have F (Qn) − {(z)k| 2 ≤ k ≤ n − 1} ⊆ {z, ((z)0)1}. Since |F (Qn) − {(z)k | 2 ≤ k ≤ n − 1}| = f − (n − 2) = n − 3 ≤ 2 = |{z, ((z)0)1}|, we derive that n ≤ 5. That is, if n ≥ 6, there exists some dimension j of {2, 3, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes. Since |F (Qn) − {(z)k| 2 ≤ k ≤ n − 1}| = 2 for n = 5, nodes z and ((z)0)1 are faulty; that is, F (Q5) = {z, (z)2, (z)3, (z)4, ((z)0)1}, as shown in Figure 7.5(c). Therefore, Figure 7.5(c) happens to be the only possibility that either Qj,0n or Qj,1n contains 2n − 6 faulty nodes for every j ∈ {2, 3, . . . , n − 1}.
Lemma 7.5. Suppose that an n-cube Qn (n ≥ 4) contains f ≤ 2n − 5 faulty nodes such that every node has at least three fault-free neighbors. Then there exists some dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty. For n ≥ 5, both Qj,0n and Qj,1n contain 2n − 7 or less faulty nodes.
Proof. Since every node has at least three fault-free neighbors, every (n − 1)-dimensional subcube of Qn is conditionally faulty. First, we consider the case that f ≤ 2n − 6. Let u and v be two distinct faulty nodes, and let j ∈ {0, 1, . . . , n − 1} such that (u)j 6= (v)j. Then both Qj,0n and Qj,1n contain 2n − 7 or less faulty nodes.
Now we consider the case that f = 2n − 5. For n ≥ 5, we claim that there exists some dimension j of {0, 1, . . . , n − 1} such that |F (Qj,0n )| ≤ 2n − 7 and |F (Qj,1n )| ≤ 2n − 7. For 0 ≤ k ≤ n − 1, we define that qk = 1 if (u)k = (v)k for every two distinct faulty nodes u, v ∈ F (Qn), and qk = 0 otherwise. Let q =Pn
k=1qk. Clearly, all faulty nodes are located in either Qk,0n or Qk,1n if qk= 1. For convenience, let {0 ≤ k ≤ n−1 | qk = 0} = {i1, . . . , in−q}.
Then both Qj,0n and Qj,1n contain at least one faulty node for j ∈ {i1, . . . , in−q}.
Suppose, by contradiction, either Qj,0n or Qj,1n contains only one faulty node for every j ∈ {i1, . . . , in−q}. For v ∈ F (Qn), let A(v) = {0 ≤ k ≤ n − 1 | F (Qk,0n ) = {v} or F (Qk,1n ) = {v}}. Since Qn is node-transitive, we assume that e = 0n is a faulty node such that |A(e)|
achieves the maximum of set {|A(v)| | v ∈ F (Qn)}. For convenience, let p = |A(e)|. Obvi-ously, we have 1 ≤ p ≤ n − q. Moreover, let A(e) = {i1, . . . , ip}. For v ∈ F (Qn) − {e}, we see that (v)k = 1 for each k ∈ {i1, . . . , ip}. Let B(k) = {v ∈ F (Qn) − {e} | (v)k 6= (e)k} for k ∈ {ip+1, . . . , in−q}. Since we assumed, by contradiction, that either Qj,0n or Qj,1n has only one faulty node for each j ∈ {i1, . . . , in−q}, we have |B(j)| = 1 for each j ∈ {ip+1, . . . , in−q}.
Since Qnis link-transitive, we assume that {i1, . . . , ip} = {0, . . . , p−1} and {ip+1, . . . , in−q} = {p, . . . , n − q − 1}. Then we have (F (Qn) − {e}) −S
k∈{ip+1,...,in−q}B(k) ⊆ {0n−p1p}. Ac-cordingly, we derive that 1 = |{0n−p1p}| ≥ |(F (Qn) − {e}) − S
k∈{ip+1,...,in−q}B(k)| ≥
|F (Qn)| − |{e}| −P
k∈{ip+1,...,in−q}|B(k)| = (2n − 5) − 1 − (n − q − p); that is, p + q ≤ 7 − n.
Recall that p ≥ 1 and q ≥ 0. Thus, we have n ∈ {5, 6}. Now we can identify all faulty nodes according to the values of p, q, and n.
Case 1: Suppose (n, q, p) = (5, 0, 1). Since p = 1, we have (v)0 = 1 for each v ∈ F (Q5) − {e} and |B(j)| = 1 for each j ∈ {1, 2, 3, 4}. Thus we have F (Q5) = {00000, 00011, 00101, 01001, 10001}. Clearly, node 00001 has five faulty neighbors.
Case 2: Suppose (n, q, p) = (5, 0, 2). Similarly, we have F (Q5) = {00000, 00111, 01011, 10011, 00011}. Then node 00011 has three faulty neighbors.
Case 3: Suppose (n, q, p) = (5, 1, 1). We have F (Q5) = {00000, 00011, 00101, 01001, 00001}. Again, node 00001 has four faulty neighbors.
Case 4: Suppose (n, q, p) = (6, 0, 1). We have F (Q6) = {000000, 000011, 000101, 001001, 010001, 100001, 000001}. Thus, node 000001 has six faulty neighbors.
In short, node 0n−p1p has 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 {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
Suppose that Qnis conditionally faulty with utmost 2n−5 faulty nodes. Let F = F (Qn).
For n ≥ 5, we propose a procedure PARTITION(Qn, F ) to determine j-partition of Qn
according to the following rules:
(1) Suppose that at least three nodes of Qn have exactly n − 2 faulty neighbors, respec-tively. If Qn has its faulty nodes distributed as shown in Figure 7.3(a), it will be partitioned along dimension j = dim((t1, t5)). Then one resulting subcube has its faulty nodes distributed as in Figure 7.2(b). Otherwise, Lemma 7.1 and Lemma 7.2 ensure that Qn can be partitioned along some dimension j such that both Qj,0n and Qj,1n 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, respec-tively. By Lemma 7.3, there exists some dimension j of {0, 1, . . . , n−1} such that both Qj,0n and Qj,1n 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 in Figure 7.5(c), we partition Qn along any dimension j ∈ {i | (z)i is faulty}. Then one resulting subcube turns out to have 2n − 6 faulty nodes, distributed as in Figure 7.5(d). Otherwise, we can apply Lemma 7.4 to choose a dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.
(4) Suppose that every node of Qn has at least three fault-free neighbors. Obviously, every (n − 1)-cube is conditionally faulty. By Lemma 7.5, there exists some dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n contain 2n − 7 or less faulty nodes.
The following corollary summarizes what is obtained by procedure PARTITION(Qn, F ).
Also, it is a summary of Lemmas 7.1−7.5.
Corollary 7.1. Suppose that an n-cube Qn (n ≥ 5) is conditionally faulty with f ≤ 2n − 5 faulty nodes. Except for the cases illustrated in Figure 7.2(c), Figure 7.3(a), and Fig-ure 7.5(c), there exists some dimension j of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty nodes.