Long Paths in Faulty Hypercubes with Conditional Node-faults
7.2 Long path embedding in faulty hypercubes
The following theorem was proved by Fu [23].
Theorem 7.1. [23] Suppose that n ≥ 3. Let u and v denote two arbitrary fault-free nodes of an n-cube with f ≤ n − 2 faulty nodes. If h(u, v) 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 7.6. Let z ∈ V (Q4), {i, j, p, q} = {0, 1, 2, 3}, and F = {(z)i, (z)j, (z)p}. Suppose that s and t are any two nodes of Q4− F such that {s, t} 6= {z, (z)q}. Then Q4− F has a path of length at least 9 or 8 between s and t if h(s, t) is odd or even, respectively.
Proof. By symmetry, let z = 0000, i = 0, j = 1, p = 2, and q = 3. We partition Q4 into Q3,04 and Q3,14 . Then Q3,14 is fault-free and z ∈ V0(Q3,04 ).
Case 1: Both s and t are in Q3,04 − F . Since Q3,14 is fault-free, Theorem 4.2 ensures that Q3,14 contains a path P of length 7 (respectively, 6) between (s)3 and (t)3 if h(s, t) is odd (respectively, even). Thus, hs, (s)3, P, (t)3, ti is a fault-free path of length 9 (respectively, 8) between s and t if h(s, t) is odd (respectively, even).
Case 2: Both s and t are in Q3,14 . If h(s, t) is odd, Theorem 4.2 ensures that Q3,14 − {(1101, 1111)} contains a path P of length 7 between s and t. Clearly, path P does not pass through (1101, 1111). Since it spans Q3,14 , we have 1111 ∈ V (P ). 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 h(s, t) is even, then we consider two cases as follows. Suppose first that s, t ∈ V0(Q3,14 ). By Theorem 4.2, Q3,14 − {(1101, 1111)} 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 ∈ V1(Q3,14 ). By Theorem 4.3, Q3,14 − {1001} 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 Q3,04 − F and t is in Q3,14 . First, we consider the case that s6= z. If s ∈ V0(Q4), then s is adjacent to node 0111. Clearly, there exists some node v of {0110, 0101, 0011} − {s} such that (v)3 6= t. By Theorem 4.2, Q3,14 has a path P of length 6 or 7 between (v)3 and t if h(s, t) is odd or even, respectively. Then hs, 0111, v, (v)3, P, ti is a fault-free path of length 9 or 10 if h(s, t) is odd or even, respectively. If s ∈ V1(Q4), then we
have s = 0111. Obviously, there exists some node u of {0110, 0101, 0011} such that (u)3 6= t.
Similarly, Q3,14 has a path T of length 7 (respectively, 6) between (u)3 and t if h(s, t) is odd (respectively, even). Then hs, u, (u)3, T, ti is a fault-free path of length 9 (respectively, 8) if h(s, t) is odd (respectively, even).
Next, we consider the case that s = z. If h(s, t) is even, it follows from Theorem 4.2 that Q3,14 has a path H of length 7 between (s)3 = (z)3 and t. Then hs = z, (z)3, H, ti is a fault-free path of length 8. If h(s, t) is odd, Theorem 4.3 ensures that Q3,14 − {1100} has a path R of length 6 between (z)3 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 h(z)3, R1, 1111, 1110, R2, ti if (1111, 1110) ∈ E(R). Then hs = z, (z)3, R1, 1111, 0111, 0110, 1110, R2, ti is a fault-free path of length 9 between s and t.
Lemma 7.7. Suppose that Q3 is conditionally faulty with f ≤ 2 faulty nodes. Let s and t denote any two fault-free nodes of Q3. Then Q3 contains a fault-free path of length at least 7 − 2f (respectively, 6 − 2f ) between s and t if h(s, t) is odd (respectively, even).
Proof. If f < 2, this result follows from Theorem 7.1. Thus we only consider the case that f = 2. For convenience, let F = F (Q3). Since Q3 is node-transitive, we assume that node 000 is faulty. To require that every node of Q3 has at least two 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 Q3 is link-transitive, we assume that 001 ∈ F ; that is, F = {000, 001}. Then we partition Q3 into Q1,03 and Q1,13 . Hence we have F ⊆ V (Q1,03 ). See Figure 7.6(a).
Subcase 1.1: Both s and t are in Q1,03 − F . Without loss of generality, we assume that s = 101 and t = 100. Obviously, hs = 101, 111, 110, 100 = ti is a fault-free path of length 3 = 7 − 2 · 2.
Subcase 1.2: Both s and t are in Q1,13 . If h(s, t) is odd, then Q1,13 contains a path of length 3 between s and t. Otherwise, Q1,13 contains a path of length 2 between s and t.
Subcase 1.3: Suppose that s is in Q1,03 − F and t is in Q1,13 . Without loss of generality, we assume s = 101 and list the required path in Table 7.1.
Case 2: Node 111 is faulty. See Figure 7.6(b) for illustration.
Subcase 2.1: Both s and t are in Q1,03 − {000}. For every possible combination of s and t, we list the required paths in Table 7.1.
Subcase 2.2: Both s and t are in Q1,13 − {111}. This subcase is symmetric to Subcase 2.1.
Table 7.1: The required paths for Lemma 7.7 and Lemma 7.8.
Figure 7.6: (a,b) Illustrations for Lemma 7.7; (c) the distribution of faulty nodes indicated in Lemma 7.8.
Subcase 2.3: Suppose that s is in Q1,03 − {000} and t is in Q1,13 − {111}. For every possible combination of s and t, we list the required paths in Table 7.1.
In summary, Q3 − F contains a path of length at least 7 − 2f (respectively, 6 − 2f ) between s and t if h(s, t) is odd (respectively, even).
Lemma 7.8. Let w ∈ V0(Q3) and {i, j, k} = {0, 1, 2}. Suppose that b1 and b2 are two arbitrary nodes of V1(Q3). Then Q3− {w, ((w)i)j} contains a path of length 4 between b1 and b2 if and only if {b1, b2} 6= {(w)k, (((w)i)j)k}.
Proof. Since Q3 is node-transitive and link-transitive, we assume that w = 000, i = 0, j = 1, and k = 2. See Figure 7.6(c). Then we list all the required paths in Table 7.1.
Theorem 7.2. Let F be a set of f ≤ 3 faulty nodes in Q4 such that every node of Q4 has 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 h(s, t) is odd (respectively, even).
Proof. If f < 3, this result follows from Theorem 7.1. Thus we concentrate only on the case that f = 3. By Lemmas 7.1−7.5, Figure 7.2(c) happens to be a unique case that a conditionally faulty Q4 with three faulty nodes cannot be partitioned along any dimension in such a way that both subcubes are conditionally faulty. On this occasion, we partition Q4 along an arbitrary dimension j; otherwise, there exists some dimension j such that both Qj,04 and Qj,14 are conditionally faulty.
Case 1: Both Qj,04 and Qj,14 are conditionally faulty. For convenience, let F0 = F (Qj,04 ) and F1 = F (Qj,14 ). Without loss of generality, we assume that f0 = |F0| = 2 and f1 = |F1| = 1. Moreover, we assume s ∈ V0(Q4− F ).
Subcase 1.1: Both s and t are in Qj,04 . By Lemma 7.7, Qj,04 − F0 contains a path H0 of length at least 3 = 7 − 2f0 (respectively, 2 = 6 − 2f0) between s and t if h(s, t) is odd (respectively, even). Obviously, H0 can be written as hs = x0, x1, x2, H0′, ti. If (x1)j is faulty, then (x0)j and (x2)j are fault-free. By Theorem 4.3, Qj,14 is hyper-hamiltonian laceable. Thus Qj,14 − {(x1)j} has a hamiltonian path H1 between (x0)j and (x2)j. As a result, hs = x0, (x0)j, H1, (x2)j, x2, H0′, ti is a fault-free path of length at least 15 − 2f (respectively, 14 − 2f ) when h(s, t) is odd (respectively, even). If (x1)j is fault-free, then (x0)j or (x2)j is fault-free. Suppose, for example, that (x0)j is fault-free. By Lemma 7.7, Qj,14 − F1 has a fault-free path H1 of length at least 7 − 2f1 between (x0)j and (x1)j. As a result, hs = x0, (x0)j, H1, (x1)j, x1, x2, H0′, ti is a fault-free path of length at least 15 − 2f (respectively, 14 − 2f ) when h(s, t) is odd (respectively, even).
Subcase 1.2: Both s and t are in Qj,14 . First, we consider the case that h(s, t) is odd.
By Lemma 7.7, Qj,14 − F1 contains a path T1 of length at least 5 = 7 − 2f1 between s and t. Let A = {(T1(i), T1(i + 1)) | 1 ≤ i ≤ 5 and i ≡ 1 (mod 2)} be a set of disjoint links on T1. Since |A| = 3 > f0, there exists an odd integer ˆı, 1 ≤ ˆı ≤ 5, such that both (T1(ˆı))j and
(T1(ˆı+1))j are fault-free. Let w = T1(ˆı) and b = T1(ˆı+1). Accordingly, T1 can be written as hs, T1′, w, b, T1′′, ti. By Lemma 7.7, Qj,04 − F0 has a path T0 of length at least 7 − 2f0 between (w)j and (b)j. As a result, hs, T1′, w, (w)j, T0, (b)j, b, T1′′, ti is a fault-free path of length at least 15 − 2f between s and t.
Next, we consider the case that h(s, t) is even. Hence we have t ∈ V0(Q4 − F ). Let u denote the faulty node in Qj,14 . Then we distinguish the following two subcases.
Subcase 1.2.1: Suppose that u ∈ V1(Qj,14 ). By Theorem 4.3, Qj,14 is hyper-hamiltonian laceable. Thus Qj,14 − {u} has a hamiltonian path H1 from s to t. Obviously, the length of H1 is equal to 6. Let B = {(H1(i), H1(i + 1)) | 1 ≤ i ≤ 6 and i ≡ 1 (mod 2)} be a set of disjoint links on T1. Since |B| = 3 > f0, there exists an odd integer ˆı, 1 ≤ ˆı ≤ 6, such that both (H1(ˆı))j and (H1(ˆı + 1))j are fault-free. Let w = H1(ˆı) and b = H1(ˆı + 1). Thus H1 can be written as hs, H1′, w, b, H1′′, ti. By Lemma 7.7, Qj,04 − F0 has a path H0 of length at least 7 − 2f0 between (w)j and (b)j. As a result, hs, H1′, w, (w)j, H0, (b)j, b, H1′′, 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 ∈ V0(Qj,14 ). Since h(s, t) is even, it follows from Lemma 7.7 that Qj,14 − F1 has a path T1 of length at least 6 − 2f1 = 4 between s and t. If there exists a link (w, b) on T1 such that both (w)j and (b)j are 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 ∩ {(T1(i))j, (T1(i + 1))j} 6= ∅ for every i. Then we claim that both (T1(2))j and (T1(4))j are faulty. Since f0 = 2, we see that |F0 ∩ {(T1(1))j, (T1(2))j, (T1(3))j}| = 1 and |F0∩ {(T1(3))j, (T1(4))j, (T1(5))j}| = 1. Then we have F0∩ {(T1(1))j, (T1(2))j, (T1(3))j} = (F0∩ {(T1(1))j, (T1(2))j}) ∩ (F0∩ {(T1(2))j, (T1(3))j}) = {(T1(2))j}. Similarly, we have F0∩ {(T1(3))j, (T1(4))j, (T1(5))j} = {(T1(4))j}. That is, F0 = {(T1(2))j, (T1(4))j}. By Lemma 7.8, Qj,04 − F0 contains either a path T0 of length 4 between (T1(1))jand (T1(3))j or a path R0 of length 4 between (T1(3))j and (T1(5))j. As a result, hs = T1(1), (T1(1))j, T0, (T1(3))j, T1(3), T1(4), T1(5) = ti or hs = T1(1), T1(2), T1(3), (T1(3))j, R0, (T1(5))j, T1(5) = ti is a fault-free path of length 8 = 14 − 2f .
Subcase 1.3: Suppose that s is in Qj,04 and t is in Qj,14 . Since f0 = 2, we have |V1(Qj,04 ) − F0| ≥ 2 = |F1 ∪ {t}| and |V (Qj,04 ) − (F0 ∪ {s})| = 5 > |F1 ∪ {t}|. If h(s, t) is odd, we choose a node x of V1(Qj,04 ) − F0 such that (x)j is fault-free; otherwise, we choose a node x of V (Qj,04 ) − (F0∪ {s}) such that (x)j ∈ F/ 1 ∪ {t}. By Lemma 7.7, Qj,04 − F0 contains a path H0 of length at least 7 − 2f0 (respectively, 6 − 2f0) between s and x when h(s, x) is odd (respectively, even). Similarly, Qj,14 − F1 contains a path H1 of length at least 7 − 2f1
(respectively, 6 − 2f1) between (x)j and t when h((x)j, t) is odd (respectively, even). As a result, hs, H0, x, (x)j, H1, ti is a fault-free path of length at least 15 − 2f (respectively, 14 − 2f ) if h(s, t) is odd (respectively, even).
Case 2: Suppose Q4 has its faulty nodes distributed as in Figure 7.2(c). To be precise, we assume F = {0000, 0011, 1100}. Then we partition Q4 into Q3,04 and Q3,14 . It is noticed that Q3,04 is not conditionally faulty.
Subcase 2.1: Both s and t are in Q3,04 −{0000, 0011}. By Theorem 7.1, Q3,04 −{0000} has a path T0 of length at least 5 (respectively, 4) between s and t if h(s, t) is odd (respectively, even).
We consider first that h(s, t) is odd. Thus the length of path T0 is greater than or equal to 5. Then T0 passes through every node of V0(Q3,04 ) − {0000}. In particular, the faulty node 0011 is on T0. Hence T0 can be written as hs, T0′, x, 0011, y, T0′′, ti. Since h(0011, 1100) = 4, both (x)3 and (y)3 are fault-free. Since h((x)3, (y)3) is even, Theorem 7.1 ensures that Q3,14 − {1100} has a path T1 of length at least 4 between (x)3 and (y)3. As a result, hs, T0′, x, (x)3, T1, (y)3, y, T0′′, ti is a fault-free path of length at least 9 = 15 − 2f .
Next, we consider the case that h(s, t) 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 sand t in a way similar to that described earlier. Otherwise, there exists a link (w, b) on T0
such that both (w)3 and (b)3 are fault-free. Hence T0 can be written as hs, R′0, w, b, R0′′, ti.
By Theorem 7.1, Q3,14 − {1100} has a path T1 of length at least 5 between (w)3 and (b)3. Then hs, R′0, w, (w)3, T1, (b)3, b, R0′′, ti turns out to be a fault-free path of length at least 10 > 14 − 2f .
Subcase 2.2: Suppose that s is in Q3,04 − {0000, 0011} and t is in Q3,14 − {1100}. By Theorem 7.1, Q3,04 − {0000} has a path T0 of length at least 5 (respectively, 4) between nodes s and 0011 if h(s, 0011) is odd (respectively, even). Accordingly, we write T0 as hs, T0′, x, y, 0011i. Since h(0011, 1100) = 4, both (x)3 and (y)3 is fault-free. On the one hand, we assume (y)3 6= t. By Theorem 7.1, Q3,14 − {1100} has a path T1 of length at least 5 (respectively, 4) between (y)3 and t if h((y)3, t) is odd (respectively, even). As a result, hs, T0′, x, y, (y)3, T1, ti is a fault-free path of length at least 9 = 15 − 2f (respectively, 8 = 14 − 2f ) if h(s, t) is odd (respectively, even). On the other hand, if (y)3 = t, then Theorem 7.1 ensures that Q3,14 − {1100} has a path R1 of length at least 5 between (x)3 and (y)3. Then hs, T0′, x, (x)3, R1, (y)3 = ti turns out to be a fault-free path of length at least 9 = 15 − 2f (respectively, 8 = 14 − 2f ) if h(s, t) is odd (respectively, even).
Subcase 2.3: Both s and t are in Q3,14 − {1100}. We list the required paths obtained by brute force in Table 7.2.
Therefore the proof is completed.
With Theorem 7.2 and Lemma 7.6, we will be able to prove the next theorem.
Theorem 7.3. Let F be a set of f faulty nodes in Qn (n ≥ 1) such that every node of Qn
has at least two fault-free neighbors. Suppose f = 0 if n ∈ {1, 2}, and f ≤ 2n − 5 if n ≥ 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 h(s, t) is odd (respectively, even).
Proof. The result is trivial for n ∈ {1, 2}. When n ∈ {3, 4}, the result follows from Theo-rem 7.1 or TheoTheo-rem 7.2, respectively. In what follows we consider the case that n ≥ 5. Except
Table 7.2: The required paths in Subcase 2.3 of Theorem 7.2.
s= 1101 t= 1110 hs = 1101, 1001, 0001, 0101, 0100, 0110, 0010, 1010, 1110 = ti t= 1111 hs = 1101, 1001, 0001, 0101, 0100, 0110, 0010, 1010, 1110, 1111 = ti t= 1000 hs = 1101, 0101, 0001, 1001, 1011, 1111, 1110, 1010, 1000 = ti t= 1001 hs = 1101, 0101, 0100, 0110, 1110, 1111, 1011, 1010, 1000, 1001 = ti t= 1010 hs = 1101, 0101, 0100, 0110, 1110, 1111, 1011, 1001, 1000, 1010 = ti t= 1011 hs = 1101, 0101, 0001, 1001, 1000, 1010, 1110, 1111, 1011 = ti s= 1110 t= 1111 hs = 1110, 1010, 1000, 1001, 1101, 0101, 0100, 0110, 0111, 1111 = ti
t= 1000 hs = 1110, 0110, 0100, 0101, 0001, 1001, 1011, 1010, 1000 = ti t= 1001 hs = 1110, 0110, 0100, 0101, 1101, 1111, 1011, 1010, 1000, 1001 = ti t= 1010 hs = 1110, 0110, 0100, 0101, 0001, 1001, 1101, 1111, 1011, 1010 = ti t= 1011 hs = 1110, 0110, 0100, 0101, 0001, 1001, 1101, 1111, 1011 = ti s= 1111 t= 1000 hs = 1111, 0111, 0110, 0100, 0101, 0001, 1001, 1011, 1010, 1000 = ti
t= 1001 hs = 1111, 0111, 0101, 0100, 0110, 0010, 1010, 1000, 1001 = ti t= 1010 hs = 1111, 0111, 0110, 0100, 0101, 1101, 1001, 1000, 1010 = ti t= 1011 hs = 1111, 0111, 0101, 0100, 0110, 0010, 1010, 1000, 1001, 1011 = ti s= 1000 t= 1001 hs = 1000, 1010, 1110, 0110, 0100, 0101, 1101, 1111, 1011, 1001 = ti t= 1010 hs = 1000, 1001, 1101, 0101, 0100, 0110, 1110, 1111, 1011, 1010 = ti t= 1011 hs = 1000, 1001, 1101, 0101, 0100, 0110, 1110, 1111, 1011 = ti s= 1001 t= 1010 hs = 1001, 1011, 1111, 0111, 0101, 0100, 0110, 1110, 1010 = ti
t= 1011 hs = 1001, 1000, 1010, 1110, 0110, 0100, 0101, 1101, 1111, 1011 = ti s= 1010 t= 1011 hs = 1010, 1000, 1001, 1101, 0101, 0100, 0110, 0111, 1111, 1011 = ti
for the faulty node distribution illustrated in Figure 7.3(a), procedure PARTITION(Qn, F ) returns j-partition of Qn such that both Qj,0n and Qj,1n are conditionally faulty. If Q5 has its faulty nodes distributed as in Figure 7.3(a), then PARTITION(Q5, F ) returns j-partition of Q5 such that one subcube has its faulty nodes distributed as in Figure 7.2(b). Accordingly, the proof can be justified by the induction on n. Our inductive hypothesis is that the result holds for Qn−1. For convenience, let F0 = F (Qj,0n ) and F1 = F (Qj,1n ). Moreover, let f0 = |F0| and f1 = |F1|. Without loss of generality, we assume that s ∈ V0(Qn− F ).
Case 1: Suppose f0 ≤ 2n − 7 and f1 ≤ 2n − 7. Without loss of generality, we assume that f0 ≤ f1. In particular, for the case illustrated in Figure 7.3(a), Qj,05 is conditionally faulty with f0 = 2 faulty nodes, and Qj,15 is not conditionally faulty with f1 = 3 faulty nodes distributed as in Figure 7.2(b).
Subcase 1.1: Both s and t are in Qj,0n . By inductive hypothesis, Qj,0n − F0 contains a path H0 of length L at least 2n−1− 2f0− 1 (respectively, 2n−1− 2f0− 2) between s and t if h(s, t) is odd (respectively, even). Clearly, we have |{v ∈ V (Qj,1n ) | |NF
Qj,1n (v)| ≥ n − 2}| ≤ 1.
Let A = {(H0(i), H0(i + 1)) | 1 ≤ i ≤ L and i ≡ 1 (mod 2)} be a set of disjoint links on H0. Since |A| = ⌈L2⌉ > f1 + 1 ≥ |F1 ∪ {v ∈ V (Qj,1n ) | |NQFj,1
n (v)| ≥ n − 2}| for n ≥ 5, there exists an odd integer ˆı, 1 ≤ ˆı ≤ L, such that |F1 ∩ {(H0(ˆı))j, (H0(ˆı + 1))j}| = 0,
|NF
Qj,1n ((H0(ˆı))j)| ≤ n − 3, and |NQFj,1
n ((H0(ˆı+ 1))j)| ≤ n − 3 are satisfied. Let x = H0(ˆı) and y= H0(ˆı + 1). Hence path H0 can be written as hs, H0′, x, y, H0′′, ti.
If Qj,1n is conditionally faulty, our inductive hypothesis asserts that Qj,1n − F1 has a path
H1 of length at least 2n−1− 2f1− 1 between (x)j and (y)j. Otherwise, the faulty nodes of Qj,1n are distributed as in Figure 7.2(b). Since both (x)j and (y)j have two or more fault-free neighbors in Qj,1n , Lemma 7.6 ensures that Qj,1n has a fault-free path H1 of length at least 2n−1− 2f1 − 1 between (x)j and (y)j. Then hs, H0′, x, (x)j, H1, (y)j, y, H0′′, ti is a fault-free path of length at least 2n− 2f − 1 (respectively, 2n− 2f − 2) between s and t if h(s, t) is odd (respectively, even). See Figure 7.7(a).
Subcase 1.2: Both s and t are in Qj,1n . We consider first that the faulty nodes of Qj,15 are distributed as depicted in Figure 7.2(b). Let z denote the node with only one fault-free neighbor r in Qj,15 . Note that f0 = 2 and f1 = 3.
Suppose {s, t} = {z, r}. Then a long path between s and t is constructed as follows. On the one hand, we assume that s = z and t = r. Since |V0(Qj,05 ) − F0| ≥ |V0(Qj,05 )| − |F0| = 24−2 > 4 = |F1∪{t}|, there exists some fault-free node x of V0(Qj,05 ) such that (x)j ∈ F/ 1∪{t}.
By inductive hypothesis, Qj,05 − F0 has a path H0 of length at least 24 − 2f0 − 1 between (s)j and x. By Lemma 7.6, Qj,15 − F1 has a path H1 of length at least 24− 2f1− 2 between (x)j and t. As a result, hs, (s)j, H0, x, (x)j, H1, ti is a fault-free path of length at least 25− 2f − 1 (see Figure 7.7(b)). On the other hand, we assume that t = z and s = r. Since
|V1(Qj,05 ) − F0| ≥ |V1(Qj,05 )| − |F0| = 24− 2 > 4 = |F1∪ {s}|, there exists some fault-free node xof V1(Qj,05 ) such that (x)j ∈ F/ 1∪ {s}. Again, the inductive hypothesis asserts that Qj,05 has a fault-free path H0 of length at least 24 − 2f0 − 1 between x and (t)j; Lemma 7.6 asserts that Qj,1 has a fault-free path H1 of length at least 24− 2f1− 2 between s and (x)j. Then hs, H1, (x)j, x, H0, (t)j, ti is a fault-free path of length at least 25− 2f − 1 (see Figure 7.7(c)).
Suppose {s, t} 6= {z, r}. Then Lemma 7.6 asserts that Qj,15 − F1 contains a path H1 of length L at least 24 − 2f1 − 1 (respectively, 24 − 2f1 − 2) between s and t if h(s, t) is odd (respectively, even). Let A = {(H1(i), H1(i + 1)) | 1 ≤ i ≤ L and i ≡ 1 (mod 2)} be a set of disjoint links. Since |A| = ⌈L2⌉ > 2 = f0, there exists an odd integer ˆı, 1 ≤ ˆı ≤ L, such that F0 ∩ {(H1(ˆı))j, (H1(ˆı + 1))j} = ∅. Let x = H1(ˆı) and y = H1(ˆı + 1). Accordingly, path H1 can be written as hs, H1′, x, y, H1′′, ti. Again, the inductive hypothesis asserts that Qj,05 − F0 has a path H0 of length at least 24 − 2f0 − 1 between (x)j and (y)j. Then hs, H1′, x, (x)j, H0, (y)j, y, H1′′, ti is a fault-free path of length at least 25−2f −1 or 25−2f −2 if h(s, t) is odd or even, respectively. See Figure 7.7(d).
Now we consider the case that faulty nodes of Qj,15 are not distributed as depicted in Figure 7.2(b), or n ≥ 6. Then Qj,1n is conditionally faulty. By inductive hypothesis, Qj,1n − F1 has a path H1of length L at least 2n−1−2f1−1 (respectively, 2n−1−2f1−2) between s and t if h(s, t) is odd (respectively, even). Similarly, let A = {(H1(i), H1(i+1)) | 1 ≤ i ≤ L and i ≡ 1 (mod 2)} be a set of disjoint links. Since |A| = ⌈L2⌉ > f0 for n ≥ 5, there is a link (x, y) of A such that F0 ∩ {(x)j, (y)j} = ∅. Accordingly, path H1 can be written as hs, H1′, x, y, H1′′, ti.
By inductive hypothesis, Qj,0n − F0 has a path H0 of length at least 2n−1− 2f0− 1 between (x)j and (y)j. Again, hs, H1′, x, (x)j, H0, (y)j, y, H1′′, ti is a fault-free path of length at least 2n− 2f − 1 or 2n− 2f − 2 if h(s, t) is odd or even, respectively. See Figure 7.7(d).
Subcase 1.3: Suppose that s is in Qj,0n and t is in Qj,1n . Note that |{x ∈ V (Qj,1n ) |
|NF
Qj,1n (x)| ≥ n − 2}| ≤ 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
|V1(Qj,0n ) − F0| ≥ 2n−2− f0 > f1 + 2 = |F1∪ {t, r}| for n = 5, there exists a fault-free node b of V1(Qj,0n ) − F0 such that (b)j ∈ F/ 1 ∪ {t, r}. On the other hand, we consider the case that node t has at least two fault-free neighbors in Qj,1n . Since |V1(Qj,0n ) − F0| ≥ 2n−2− f0 >
f1 + 2 ≥ |F1| + |{t}| + |{x ∈ V (Qj,1n ) | |NQFj,1
n (x)| ≥ n − 2}| ≥ |F1 ∪ {t} ∪ {x ∈ V (Qj,1n ) |
|NF
Qj,1n (x)| ≥ n − 2}| for n ≥ 5, there exists a fault-free node b of V1(Qj,0n ) − F0 such that (b)j ∈ F/ 1∪ {t} ∪ {x ∈ V (Qj,1n ) | |NF
Qj,1n (x)| ≥ n − 2}.
By inductive hypothesis, Qj,0n − F0 has a path H0 of length at least 2n−1 − 2f0 − 1 between s and b. If the faulty nodes of Qj,1n are distributed as illustrated in Figure 7.2(b), Lemma 7.6 asserts that Qj,1n −F1 has a path H1 of length at least 2n−1−2f1−1 (respectively, 2n−1− 2f1 − 2) between (b)j and t if h((b)j, t) is odd (respectively, even); otherwise, the inductive hypothesis asserts that Qj,1n − F1 has a path H1 of length at least 2n−1− 2f1 − 1 (respectively, 2n−1 − 2f1 − 2) between (b)j and t if h((b)j, t) is odd (respectively, even).
Then hs, H0, b, (b)j, H1, ti is a fault-free path of length at least 2n− 2f − 1 (respectively, 2n− 2f − 2) between s and t if h(s, t) is odd (respectively, even). See Figure 7.7(e).
Case 2: Suppose either f0 = 2n − 6 or f1 = 2n − 6. By Lemmas 7.1−7.5, we know that this case may occur while n = 5. More precisely, the faulty nodes happen to be distributed as illustrated in Figure 7.5(c) where z is itself a faulty node with three faulty neighbors.
Without loss of generality, we assume that f0 = 4; thus, (z)j is a unique faulty node in Qj,15 . Subcase 2.1: Both s and t are in Qj,05 . By inductive hypothesis, Qj,05 − (F0 − {z}) contains a path H0 of length L at least 9 = 24− 2 · 3 − 1 (respectively, 8 = 24 − 2 · 3 − 2) between s and t if h(s, t) is odd (respectively, even).
First, we consider the case that node z is not on H0. Let A = {(H0(i), H0(i + 1)) | 1 ≤ i ≤ L and i ≡ 1 (mod 2)} be a set of disjoint links on H0. Since |A| = ⌈L2⌉ > 1 = f1, there exists an odd integer ˆı, 1 ≤ ˆı ≤ L, such that both (H0(ˆı))j and (H0(ˆı+1))j are fault-free. Let x= H0(ˆı) and y = H0(ˆı+ 1). Hence path H0 can be written as hs, H0′, x, y, H0′′, ti. It follows from inductive hypothesis that Qj,15 −{(z)j} has a path H1 of length at least 13 = 24−2·1−1 between (x)j and (y)j. Then hs, H0′, x, (x)j, H1, (y)j, y, H0′′, ti is a fault-free path of length at least 23 > 25− 2 · 5 − 1 (respectively, 22 > 25− 2 · 5 − 2) between s and t if h(s, t) 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, H0′, x, z, y, H0′′, ti. Clearly, (x)j and (y)j are fault-free nodes in the same partite set of Qj,15 . By Theorem 4.3, Qj,15 is hyper-hamiltonian laceable; thus Qj,15 − {(z)j} has a path H1 of length 14 between (x)j and (y)j. Then hs, H0′, x, (x)j, H1, (y)j, y, H0′′, ti is a fault-free path of length at least 23 > 25− 2 · 5 − 1 (respectively, 22 > 25− 2 · 5 − 2) between s and t if h(s, t) is odd (respectively, even).
Subcase 2.2: Both s and t are in Qj,15 . For the sake of clarity, we distinguish whether
x
h(s, t) is odd or even.
Suppose that h(s, t) is odd. By inductive hypothesis, Qj,15 − {(z)j} contains a path H1 of length L at least 13 between s and t. Obviously, we have (z)j ∈ V (H/ 1). Conse-quently, (v)j 6= z for any v ∈ V (H1). Let A = {(H1(i), H1(i + 1)) | 1 ≤ i ≤ L and i ≡ 1 (mod 2)} be a set of disjoint links on H1. Since |A| − |F0 − {z}| = ⌈L2⌉ − (f0 − 1) ≥ 7 − (4 − 1) = 4, there exist four links of A, namely (x1, y1), (x2, y2), (x3, y3), and (x4, y4), such that (xi)j and (yi)j are fault-free for all i ∈ {1, 2, 3, 4}. 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, (x1)j, (y1)j, y1, P2, x2, (x2)j, (y2)j, y2, P3, x3, (x3)j, (y3)j, y3, P4, x4, (x4)j, (y4)j, y4, P5, ti is a fault-free path of length at least 21 = 25 − 2 · 5 − 1 between s and t. See Figure 7.7(f).
Suppose that h(s, t) is even. If s and (z)j belong to the different partite sets of Qj,15 , Theorem 4.3 asserts that Qj,15 − {(z)j} has a path H1 of length 14 between s and t. Similar to the case that h(s, t) is odd, there exist four disjoint links on H1, namely (x1, y1), (x2, y2), (x3, y3), and (x4, y4), such that (xi)j and (yi)j are fault-free for all i ∈ {1, 2, 3, 4}. Accord-ingly, we can write H1 = hs, P1, x1, y1, P2, x2, y2, P3, x3, y3, P4, x4, y4, P5, ti. Then hs, P1, x1, (x1)j, (y1)j, y1, P2, x2, (x2)j, (y2)j, y2, P3, x3, (x3)j, (y3)j, y3, P4, x4, (x4)j, (y4)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 (z)j belong to the same partite set of Qj,15 , then we construct a fault-free path as follows. Since Qj,05 is conditionally faulty, we denote by x any fault-free neighbor of z in Qj,05 . By inductive hypothesis, Qj,05 −(F0−{z}) has a path H0 of length at least 9 = 24−2·3−1 between x and z.
We can write path H0 as hx, H0′, y, zi, where y is also a fault-free neighbor of z. Without loss of generality, let j = 4, {x, y} = {(z)0, (z)1}, and X = {((z)j, ((z)j)2), ((z)j, ((z)j)3)}. Since
|X| = 2, Theorem 4.2 ensures that Qj,15 − X is strongly hamiltonian laceable; hence it has a path H1 of length 14 between s and t. Obviously, both ((z)j, (x)j) and ((z)j, (y)j) are on H1, and we can write H1 as hs, H1′, (x)j, (z)j, (y)j, H1′′, ti. Then hs, H1′, (x)j, x, H0′, y, (y)j, H1′′, 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 − (F0 − {z}) has a path H0 of length at least 9 (respectively, 8) between s and z if h(s, z) is odd (respectively, even). Accordingly, path H0 can be written as hs, H0′, x, y, zi.
Since (z)j is a unique faulty node in Qj,15 , both (x)j and (y)j are fault-free.
If (y)j 6= t, it follows from inductive hypothesis that Qj,15 −{(z)j} has a path H1 of length at least 13 (respectively, 12) between (y)j and t if h((y)j, t) is odd (respectively, even).
Then hs, H0′, x, y, (y)j, H1, ti is a path of length at least 21 = 25 − 2 · 5 − 1 (respectively, 20 = 25− 2 · 5 − 2) between s and t if h(s, t) is odd (respectively, even). See Figure 7.7(g).
Otherwise, if (y)j = t, then our inductive hypothesis asserts that Qj,15 − {(z)j} has a path H1 of length at least 13 between (x)j and (y)j. Then hs, H0′, x, (x)j, H1, (y)j = ti is a path of length at least 21 = 25 − 2 · 5 − 1 (respectively, 20 = 25− 2 · 5 − 2) between s and t if h(s, t) is odd (respectively, even). See Figure 7.7(h).
Therefore the proof is completed.