Fault Diameter of Hypercubes
5.2 Shortest paths in faulty hypercubes
n if |F | ≤ n − 2, n + 1 if |F | = n − 1, n + 2 if |F | = 2n − 3.
5.2 Shortest paths in faulty hypercubes
We can improve Theorem 5.1, mentioned earlier, by proving the next three propositions.
Proposition 5.1. Suppose that u and v are any two distinct nodes of Qn with h(u, v) = n.
Let F be a set of 2n − 3 hybrid node-faults and/or link-faults in Qn such that both u and v are fault-free with at least one reachable neighbor. Then dQn−F(u, v) = n.
Proof. It is not difficult to verify that this proposition holds for n = 2. Hence we concern only the case that n ≥ 3. Let Iu = {i1, . . . , ip} be a set of p distinct integers of {0, 1, . . . , n−1}
such that (u)i1, . . . , (u)ip are reachable neighbors of u. Similarly, let Iv = {i′1, . . . , i′q} ⊆ {0, 1, . . . , n−1} be a set of q distinct integers such that (v)i′1, . . . , (v)i′q are reachable neighbors of v. We distinguish the following two cases.
Case 1: Suppose that Iu ∩ Iv 6= ∅. Let j ∈ Iu ∩ Iv. Then we partition Qn into Qj,0n and Qj,1n . For convenience, let F0 = F (Qj,0n ) and F1 = F (Qj,1n ). Since h(u, v) = n, nodes u and v are located in different subcubes. Moreover, we have h(u, (v)j) = n − 1. By the pigeonhole principle, we have |F0| ≤ n − 2 or |F1| ≤ n − 2. Without loss of generality, we
assume that |F0| ≤ n − 2. Moreover, we assume u ∈ V (Qj,0n ). By Lemma 5.1, Qj,0n has at least one fault-free path L of length n − 1 between u and (v)j. Hence hu, L, (v)j, vi forms a fault-free path of length n between u and v.
Case 2: Suppose that Iu∩Iv = ∅. Since |F | = 2n−3, we can conclude that 3 ≤ p+q ≤ n.
Without loss of generality, we assume that p ≥ q. Thus we have p ≥ 2.
Suppose first that n = 3. We have p = 2 and q = 1. Let j ∈ Iv. Without loss of generality, we assume that u ∈ V (Qj,0n ). Obviously, Qj,0n is fault-free, and it has a fault-free path L of length two between u and (v)j. Then hu, L, (v)j, vi is a fault-free path of length 3. See Figure 5.1(a).
Suppose that n ≥ 4. Let j ∈ Iu. Since Iu ∩ Iv = ∅, (u)j is a reachable neighbor of u, whereas (v)j is an unreachable neighbor of v. Again, we assume u ∈ V (Qj,0n ). Let F0 = F (Qj,0n ) and F1 = F (Qj,1n ). If |F1| ≤ n − 2, Lemma 5.1 ensures that Qj,1n has a fault-free path R of length n−1 between (u)j and v. Hence hu, (u)j, R, vi is a fault-free path of length n between u and v. See Figure 5.1(b).
Suppose that |F1| ≥ n − 1. Thus we have |F0| + |Ecj| ≤ n − 2. Let ˜Iv = {k ∈ Iv | ((v)k)j ∈ NQn−F((v)k)}, where NQn−F((v)k) is the set of all reachable neighbors of (v)k.
Subcase 2.1: Suppose that ˆIv 6= ∅. Let k ∈ ˜Iv and Θ be a subgraph of Qn induced by {x ∈ V (Qn) | (x)j = (u)j, (x)k= (u)k}. Then Θ is an (n − 2)-cube inside Qj,0n . Because (v)j is an unreachable neighbor of v and it is outside Θ, there are utmost n − 3 faulty elements in Θ. By Lemma 5.1, Θ has a fault-free path L of length n − 2 between u and ((v)k)j. So hu, L, ((v)k)j, (v)k, vi is a fault-free path of length n. See Figure 5.1(c).
Subcase 2.2: Suppose that ˆIv = ∅. Let k1 ∈ Iv. Since |F | ≤ 2n − 3 and p + q ≤ n, there exists an integer k2 ∈ {0, 1, . . . , n − 1} − {j, k1} such that ((v)k1)k2 is a reachable neighbor of (v)k1 and (((v)k1)k2)j is a reachable neighbor of ((v)k1)k2. Let w = ((v)k1)k2 and Ω be a subgraph of Qn induced by {x ∈ V (Qn) | (x)j = (u)j, (x)k1 = (u)k1, (x)k2 = (u)k2}. Then Ω is an (n − 3)-cube inside Qj,0n . Obviously, (u)k1, (v)j, and ((v)k1)j are unreachable neighbors of u, v, and (v)k1, respectively. Since (u)k1, (v)j, and ((v)k1)j are outside Ω, there are utmost n − 4 faulty elements in Ω. It follows from Lemma 5.1 that Ω has a fault-free path L of length n − 3 between u and (w)j. So hu, L, (w)j, w, (w)k2 = (v)k1, vi is a fault-free path of length n between u and v. See Figure 5.1(d).
In summary, we conclude that dQn−F(u, v) = n, and the proof is completed.
Proposition 5.2. Suppose that u and v are any two distinct nodes of Qn, n ≥ 3. Let F be a set of utmost 2n − 4 hybrid node-faults and/or link-faults in Qn such that both u and v
u
are fault-free with at least one reachable neighbor. Then
dQn−F(u, v) ≤ a set of utmost 2n − 3 hybrid node-faults and/or link-faults in Qn such that both u and v are fault-free with at least one reachable neighbor. Then
dQn−F(u, v) ≤
n + 1 if h(u, v) = n − 1 and n ≥ 2, h(u, v) + 4 if h(u, v) ≤ n − 2 and n ≥ 3.
Proof. For the sake of clarity, we prove Proposition 5.2 and Proposition 5.3 simultaneously.
The proof is by induction on n. Obviously, the result is true for n = 2. As our inductive hypothesis, we assume that the result holds for Qn−1 with n ≥ 3. Since h(u, v) ≤ n − 1, we can partition Qn along some dimension j such that both u and v are in the same subcube. By transitivity, we assume that j = 0. Without loss of generality, we assume that u, v ∈ V (Q0,0n ). For convenience, let F0 = F (Q0,0n ) and F1 = F (Q0,1n ). Then we distinguish two cases.
Case 1: Suppose that |F1| ≤ 2n − 5 = 2(n − 1) − 3. First, we consider the case that both u and v have at least one reachable neighbor in Q0,1n . Then it follows from the inductive hypothesis that dQn−F(u, v) = dQ0,1n −F1(u, v) = n − 1 if h(u, v) = n − 1 for n ≥ 3, dQn−F(u, v) ≤ dQ0,1n −F1(u, v) ≤ n if h(u, v) = n − 2 for n ≥ 3, and dQn−F(u, v) ≤ dQ0,1
n −F1(u, v) ≤ h(u, v) + 4 if h(u, v) ≤ n − 3 for n ≥ 4. See Figure 5.2(a).
Now we consider the case that either u or v has no reachable neighbors in Q0,1n . Thus, we have |F1| ≥ n − 1 and |F0| + |Ec0| ≤ n − 2. Since n − 1 ≤ |F1| ≤ 2n − 5, we have n ≥ 4. Without loss of generality, we assume that u has no reachable neighbors in Q0,1n . Accordingly, (u)0 is the unique reachable neighbor of u.
Suppose first that h(u, v) = n − 1. Since h((u)0, v) = n, it follows from Proposition 5.1 that dQn−F((u)0, v) = n. Let P be a fault-free path of length n between (u)0 and v.
Obviously, we have u /∈ V (P ). Hence hu, (u)0, P, vi turns out to be a fault-free path of length n + 1. See Figure 5.2(b).
Suppose that h(u, v) ≤ n − 2. If (v)0 is a reachable neighbor of v, then it follows from Corollary 5.1 that dQ0,0n −F0((u)0, (v)0) ≤ h((u)0, (v)0) + 2 = h(u, v) + 2 since |F0| ≤ n − 2.
Let R be a shortest path between (u)0 and (v)0 in Q0,0n − F0. Then hu, (u)0, R, (v)0, vi forms a fault-free path of length at most h(u, v) + 4. See Figure 5.2(b). In particular, we have
|F0| ≤ n − 3 if |F | = 2n − 4. Therefore, Q0,0n − F0 has a path R of length n − 2 between (u)0 and (v)0 if h(u, v) = n − 2 and |F | = 2n − 4. As a result, hu, (u)0, R, (v)0, vi is a fault-free path of length n. On the other hand, if (v)0 is an unreachable neighbor of v, then we have (v)0 ∈ F or (v, (v)0) ∈ F . By Lemma 5.1, Q0,0n has n − 1 internally node-disjoint paths L1, . . . , Ln−1 between (u)0 and (v)0. For clarity, Li can be written as h(u)0, L′i, ((v)0)i, (v)0i for 1 ≤ i ≤ n−1. Let Ti = h(u)0, L′i, ((v)0)i, (v)i, vi with 1 ≤ i ≤ n−1. Then {T1, . . . , Tn−1} is a set of n − 1 internally node-disjoint paths between (u)0 and v. We distinguish two subcases.
Subcase 1.1: One of {T1, . . . , Tn−1}, say Ti, is fault-free. Hence hu, (u)0, Ti, vi is a path of length at most h(u, v) + 4 between u and v. See Figure 5.2(d). In particular, we consider the case that h(u, v) = n − 2. Clearly, n − 2 paths of {T1, . . . , Tn−1} are of length n − 1.
When n ≥ 5, u and v have no common neighbors. Since
{(v)0, (v, (v)0)} ∪
n−1[
i=1
{(u)i, (u, (u)i)}
!
∩
n−1[
i=1
V (Ti) ∪ E(Ti)
!
= ∅,
at most n − 3 faults may appear on T1, . . . , Tn−1. Hence there still exists a fault-free path Tk of {T1, . . . , Tn−1} such that ℓ(Tk) = n − 1 if n ≥ 5. Then hu, (u)0, Tk, vi is a fault-free path of length n.
Subcase 1.2: None of {T1, . . . , Tn−1} is fault-free. We claim first that h(u, v) = 2.
Moreover, it is noticed that |F | = 2n − 3 in this subcase. Because T1, . . . , Tn−1 are internally node-disjoint and u has n − 1 unreachable neighbors in Q0,1n , we conclude that Ti, 1 ≤ i ≤ n − 1, contains exactly one faulty element. Since V (Ti) ∩ V (Q0,1n ) = {v, (v)i} for
1 ≤ i ≤ n − 1, there exist two distinct integers t1 and t2, 1 ≤ t1, t2 ≤ n − 1, such that F (Tt1) = {(v)t1} = {(u)t2} and F (Tt2) = {(v)t2} = {(u)t1}. By transitivity, we assume that t1 = n − 1 and t2 = n − 2. Again, Lemma 5.1 ensures that Q0,1n has n − 1 internally node-disjoint paths R1, . . . , Rn−1 of length at most 4 between u and v. For clarity, we can write Ri as hu, R′i, (v)i, vi for 1 ≤ i ≤ n − 1. Thus we have ℓ(Rn−2) = ℓ(Rn−1) = 2 and ℓ(Ri) = 4 for 1 ≤ i ≤ n − 3. Because (v)0 is an unreachable neighbor of v, thus v has a reachable neighbor in Q0,1n , say (v)k with some k ∈ {1, . . . , n − 3}. To be precise, we write Rk = hu, xk, yk, (v)k, vi and Lk = h(u)0, (xk)0, (yk)0, ((v)k)0, (v)0i, where xk is some neighbor of u, and yk is a common neighbor of xk and (v)k.
Subcase 1.2.1: Suppose that ((v)k)0 is an unreachable neighbor of (v)k. Let Sk(1) = h(u)0, (xk)0, (yk)0i and Sk(2) = h(yk)0, yk, (v)ki be two paths. Because Tk has only one faulty element, path Sk(1) is fault-free. Since
V (Sk(2)) ∪ E(Sk(2))
∩S
i6=kV (Ti) ∪ E(Ti)
= ∅, path Sk(2) is also fault-free. Then hu, (u)0, Sk(1), (yk)0, Sk(2), (v)k, vi turns out to be a fault-free path of length 6. See Figure 5.2(e).
Subcase 1.2.2: Suppose that ((v)k)0 is a reachable neighbor of (v)k. Let Θ be a subgraph of Q0,0n induced by {x ∈ V (Q0,0n ) | (x)p = (u)p, p ∈ {1, . . . , n−3}−{k}}. Obviously, Θ is isomorphic to Q3. Then we claim that |F (Θ)| ≤ 2. Since |F0| ≤ n − 2, this claim holds for n = 4 trivially. In what follows, we concern that n ≥ 5. It is easy to see that Lk, Ln−2, and Ln−1 are inside Θ. Moreover, we have (V (Ti) ∪ E(Ti)) ∩ (V (Θ) ∪ E(Θ)) = {(u)0} for i ∈ {1, . . . , n − 3} − {k}. Since Ti contains one faulty element for each 1 ≤ i ≤ n − 1, at least n − 4 faulty elements are outside Θ; i.e., |F (Θ)| ≤ 2. Since h((u)0, ((v)k)0) = 3, it follows from Lemma 5.1 that Θ has a fault-free path S of length 3 between (u)0 and ((v)k)0. As a result, hu, (u)0, S, ((v)k)0, (v)k, vi is a fault-free path of length 6. See Figure 5.2(f).
Case 2: Suppose that |F0| ≥ 2n − 4. Thus, we have |F1| + |Ec0| ≤ 1.
Subcase 2.1: Suppose that (u)0 and (v)0 are reachable neighbors of u and v, respec-tively. Since |F1| ≤ 1, it follows from Lemma 5.1 that Q0,1n has a fault-free path R of length at most h(u, v) + 2 between (u)0 and (v)0. Then hu, (u)0, R, (v)0, vi is a fault-free path of length at most h(u, v) + 4 between u and v. See Figure 5.2(g). Obviously, we have ℓ(R) = h(u, v) if |F | ≤ 2n − 4. Therefore, hu, (u)0, R, (v)0, vi turns out to be a fault-free path of length h(u, v) + 2.
Subcase 2.2: Suppose that (u)0 or (v)0 is an unreachable neighbor of u or v, re-spectively. If |F | ≤ 2n − 4, then |F1| + |Ec0| = 0. Thus we have |F | = 2n − 3 in this case. Since |F1| + |Ec0| ≤ 1, we assume that (u)0 is an unreachable neighbor of u and (v)0 is a reachable neighbor of v. Let (u)k be a reachable neighbor of u with some k ∈ {1, . . . , n − 1}. If (u)k = v, then we have dQn−F(u, v) = h(u, v) = 1. In what follows we assume (u)k 6= v. Since |F1| + |Ec0| ≤ 1, node ((u)k)j is a reachable neigh-bor of (u)k. If (u)k 6= (v)k, then h((u)k, v) = h(u, v) − 1. By Lemma 5.1, Q0,1n has a fault-free path R of length at most h((u)k, v) + 2 = h(u, v) + 1 between ((u)k)0 and (v)0. Then hu, (u)k, ((u)k)0, R, (v)0, vi is a fault-free path of length at most h(u, v) + 4.
u
According to Lemma 5.1 and Propositions 5.1−5.3, we can compute the fault diameter of hypercubes as follows.
Theorem 5.3. Let F be a set of hybrid node-faults and/or link-faults in Qn, n ≥ 3, such that every node of Qn has at least one reachable neighbor. Then the diameter of Qn− F is computed as follows:
Proof. Suppose first that n 6= 4. The results are direct consequences from Lemma 5.1 and
0100 0101
0110 0111
0000 0001
0010
0011
1101
1110 1111
1000
1001
1010 1011
1100
Q
4Figure 5.3: The distance between 0100 and 0111 in the faulty 4-cube is 6.
Propositions 5.1−5.3.
Suppose that n = 4. Applying Lemma 5.1 and Propositions 5.1−5.3, we have D(Q4 − F ) = 4 if |F | ≤ 2, D(Q4− F ) = 5 if |F | = 3, D(Q4− F ) ≤ 6 if |F | = 4, and D(Q4− F ) = 6 if |F | = 5. Let F = {0000, 0101, 0110, (0111, 1111)}. Then dQ4−F(0100, 0111) = 6. See Figure 5.3. Therefore, D(Q4− F ) = 6 if |F | = 4.
The proof is completed.