Paths of Variable Lengths in Hypercubes with Conditional
6.2 Path embedding in faulty hypercubes
P
i∈S|Fi| ≤ n; i.e., n ≤ 5. Then a dimension j of S can be chosen so that both Qj,0n and Qj,1n contain 2n − 7 or less faulty links.
(3.2.1) When n = 5, we claim that |F (Qj,0n )| · |F (Qj,1n )| > 0 for some j ∈ S. Let ei = (bi4. . . bii. . . bi0, bi4. . . ¯bii. . . bi0) be an i-dimensional link of Q5 for i ∈ {0, 1, 2, 3, 4}. Suppose that F = {e0, e1, e2, e3, e4} is a faulty set of Q5 such that |F (Qi,05 )| · |F (Qi,15 )| = 0 for each i ∈ {0, 1, 2, 3, 4}. Then we have b0i = b1i= b2i = b3i= b4i for each i ∈ {0, 1, 2, 3, 4}; i.e., all faulty links are incident with an identical node. This contradicts the assumption that every node is incident to utmost n − 3 faulty links.
(3.2.2) Similarly, there exists an integer j ∈ S such that |F (Qj,04 )| · |F (Qj,14 )| > 0.
In summary, the proposed procedure determines a j-partition of Qn such that both Qj,0n and Qj,1n are conditionally faulty with |F (Qj,0n )| + |F (Qj,1n )| ≤ 2n − 6.
6.2 Path embedding in faulty hypercubes
The following theorem characterizes a property of some shortest paths in a faulty n-cube.
Q5
z ( )zk3
( )zk4
( )zk5
( )zk1
( )zk2
( )zk1
( )k2
Q4
z ( )zk3
( )zk4
( )zk1
( )zk2
( )zk1
( )k2
(a) (b)
Figure 6.1: The distributions of faulty links indicated in (2.2).
Theorem 6.1. Let F be a set of 2n − 5 faulty links in Qn such that every node of Qn− F has at least two neighbors. Moreover, let j be an integer of {0, 1, . . . , n − 1} such that both Qj,0n and Qj,1n are conditionally faulty with 2n − 7 or less faulty links. Suppose that u is a node of Qj,0n , and v is a node of Qj,1n . Then there exists a shortest path P∗ between u and v in Qn− F such that P∗ crosses the dimension j exactly once.
Proof. Since |F (Qj,0n )| + |F (Qj,1n )| ≤ |F | = 2n − 5, we assume that |F (Qj,1n )| ≤ n − 3.
Since (u)j 6= (v)j, every shortest path between u and v crosses the dimension j an odd number of times. If there exists a shortest path between u and v crossing the dimension j exactly once, the proof is done. Thus, we assume that one shortest path between u and v, namely P , crosses the dimension j more than once. Accordingly, the shortest path P can be represented as hu, P0, x1, (x1)j, P1, (x2)j, x2, P2, x3, (x3)j, . . . , xr, (xr)j, Pr, vi with odd integer r ≥ 3. For convenience, let H = h(x1)j, P1, (x2)j, x2, P2, x3, (x3)j, . . . , xr, (xr)j, Pr, vi.
By Corollary 5.1, we have dQj,1
n −F (Qj,1n )((x1)j, v) ≤ h((x1)j, v) + 2. Suppose that R is a shortest path between (x1)j and v in Qj,1n − F (Qj,1n ). Then we have ℓ(H) ≤ ℓ(R). Since r ≥ 3, we have ℓ(H) ≥ h((x1)j, v) + 2 ≥ ℓ(R). As a result, P∗ = hu, P0, x1, (x1)j, R, vi happens to be a shortest path between u and v and it crosses the dimension j exactly once.
Theorem 6.2 is proved in [71].
Theorem 6.2. [71] Let F be a set of n − 2 faulty links in Qn (n ≥ 2). Suppose that u and v are any two different nodes of Qn− F . Then Qn− F contains a path of length l between u and v for every l satisfying dQn−F(u, v) ≤ l ≤ 2n− 1 and 2|(l − dQn−F(u, v)).
As Tsai [66] showed, an n-cube with 2n − 5 conditional link-faults is both hamiltonian laceable and strongly hamiltonian laceable.
Theorem 6.3. [66] Let F be a set of faulty links in Qn (n ≥ 3) such that every node of Qn− F has at least two neighbors. Then Qn− F is both hamiltonian laceable and strongly hamiltonian laceable if |F | ≤ 2n − 5.
The next two lemmas will be applied to prove Theorem 7.3.
Lemma 6.1. [66] Assume that n ≥ 2. Let x and u be two distinct nodes of V0(Qn); let y and v be two distinct nodes of V1(Qn). Then there exist two node-disjoint paths P1 and P2
such that the following conditions are satisfied: (1) P1 joins x to y, (2) P2 joins u to v, and (3) V (P1) ∪ V (P2) = V (Qn).
Lemma 6.2. Let v be any node of Qn (n ≥ 3), and let (w, b) be any link of Qn− {v}. For every odd integer l in the range from 1 to 2n− 3, Qn− {v} has a path of length l between w and b.
Proof. Since Qn is node-transitive, we assume that v = 0n. We prove this lemma by the induction on n. The induction base depends on Q3. With the link-transitivity, the required paths are listed in Table 6.2.
Table 6.2: The paths of variable lengths between w and b in Q3− {000}.
(w, b) = (011, 001) h011, 111, 101, 001i, h011, 111, 110, 100, 101, 001i (w, b) = (011, 111) h011, 001, 101, 111i, h011, 001, 101, 100, 110, 111i (w, b) = (101, 001) h101, 111, 011, 001i, h101, 100, 110, 111, 011, 001i (w, b) = (101, 100) h101, 111, 110, 100i, h101, 111, 011, 010, 110, 100i (w, b) = (101, 111) h101, 100, 110, 111i, h101, 100, 110, 010, 011, 111i
When n ≥ 4, we assume that the result is true for Qn−1. Then we partition Qn along some dimension p other than dim((w, b)). Obviously, v is located in Qp,0n .
Case 1: Suppose that (w, b) is in Qp,0n . By the inductive hypothesis, Qp,0n − {v} has a path of odd length l0 between w and b for any odd integer l0 from 1 to 2n−1− 3. Let H be a path of length 2n−1− 3 between w and b in Qp,0n − {v}. Since 2n−1− 3 > 1, we can represent H as hw, u, H0, bi. By Theorem 6.2, Qp,1n has a path H1 of odd length l1 between (w)p and (b)p for any odd integer l1 from 1 to 2n−1− 1. As a result, hw, (w)p, H1, (u)p, u, H0, bi is a path of odd length 2n−1− 2 + l1, in the range from 2n−1− 1 to 2n− 3.
Case 2: Suppose that (w, b) is in Qp,1n . By Theorem 6.2, Qp,1n has a path of odd length l1
between w and b for any odd integer l1 from 1 to 2n−1−1. Let H be a path of length 2n−1−1 between w and b in Qp,1n . Then we can choose a link (x, y) on H such that v /∈ {(x)p, (y)p}.
Hence, we can represent H as hw, H1′, x, y, H1′′, bi. By the inductive hypothesis, Qp,0n − {v}
has a path H0 of odd length l0 between (x)p and (y)pfor any odd integer l0 from 1 to 2n−1−3.
As a result, hw, H1′, x, (x)p, H0, (y)p, y, H1′′, bi is a path of odd length 2n−1+ l0, in the range from 2n−1+ 1 to 2n− 3.
As Shih [56] showed, any fault-free link of Qn lies on a cycle of even length from 6 to 2n when up to 2n − 5 conditional link-faults may occur.
Theorem 6.4. [56] Let F be a set of 2n − 5 faulty links in Qn such that every node of Qn− F has at least two neighbors. Suppose that u and v are any two adjacent nodes of Qn− F . Then Qn− F contains a path of odd length l between u and v if l is in the range from 1 to 2n− 1 excluding 3.
In the following discussion, we devote to constructing paths between any two nodes with distance greater than 1.
Theorem 6.5. Let F be a set of 2n − 5 faulty links in Qn (n ≥ 3) such that every node of Qn− F has at least two neighbors. Suppose that u and v are two arbitrary nodes of Qn− F with distance d∗ = dQn−F(u, v) ≥ 2. Then Qn− F contains a path of length l between u and v for every integer l satisfying both d∗ ≤ l ≤ 2n− 1 and 2|(l − d∗).
Proof. Applying procedure Partition(Qn, F , u, v), we can determine a j-partition of Qn such that both Qj,0n and Qj,1n are conditionally faulty with |F (Qj,0n )| + |F (Qj,1n )| ≤ 2n − 6. As a result, the proof can proceed by the induction on n. The induction base, depending upon Q3, follows from Theorem 6.2. As our inductive hypothesis, we assume that the result holds for Qn−1 when n ≥ 4.
Case I: Suppose that u and v are in the different partite sets of Qn. Without loss of generality, we assume that u ∈ V0(Qn) and v ∈ V1(Qn). By Theorem 6.3, Qn − F is hamiltonian laceable. Moreover, a shortest path between u and v can be easily obtained by a simple breadth-first search. Therefore, we mainly concentrate on the paths of odd lengths in the range from d∗+ 2 to 2n− 3.
Subcase I.1: Suppose that |F (Qj,0n )| ≤ 2n − 7 and |F (Qj,1n )| ≤ 2n − 7. Without loss of generality, we assume that |F (Qj,0n )| ≥ |F (Qj,1n )|; thus, |F (Qj,1n )| ≤ n − 3.
Subcase I.1.1: Suppose that both u and v are in Qj,0n . By the inductive hypoth-esis, Qj,0n − F (Qj,0n ) contains a path H0 of length 2n−1 − 1 between u and v. Let A = {(H0(i), H0(i + 1)) | 1 ≤ i ≤ 2n−1, i ≡ 1 (mod 2)} be a set of disjoint links on H0. Since
|A| = ⌈2n−12−1⌉ > 2n − 5 for any n ≥ 4, there exists a link (w, b) of A such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free. Hence, H0can be written as hu, H0′, w, b, H0′′, vi.
Since |F (Qj,1n )| ≤ n − 3, it follows from Theorem 6.2 that Qj,1n − F (Qj,1n ) contains a path H1 of odd length l1 between (w)j and (b)j for any odd integer l1 from 1 to 2n−1− 1. As a result, hu, H0′, w, (w)j, H1, (b)j, b, H0′′, vi is a path of odd length 2n−1+ l1, in the range from 2n−1+ 1 to 2n− 1. See Figure 6.2(a) for illustration.
The paths of lengths less than 2n−1+1 can be obtained as follows. By Proposition 5.3, we have d∗ = dQn−F(u, v) ≤ h(u, v) + 4 and dQj,0
n −F (Qj,0n )(u, v) ≤ h(u, v) + 4. By the inductive hypothesis, Qj,0n − F (Qj,0n ) has a path T0 of length l0 between u and v for any odd integer l0 in the range from dQj,0
n −F (Qj,0n )(u, v) to 2n−1− 1. If d∗ = h(u, v) or d∗ = h(u, v) + 4, then dQj,0n −F (Qj,0n )(u, v) = d∗. Otherwise, if d∗ = h(u, v) + 2, then dQj,0n −F (Qj,0n )(u, v) ≤ d∗+ 2.
Subcase I.1.2: Suppose that both u and v are in Qj,1n . Since |F (Qj,1n )| ≤ n − 3, it
Qnj,0 shortest path between u and v in Qn− F that does not cross the dimension j. By inductive hypothesis, Qj,1n − F (Qj,1n ) contains a path T1 of odd length l1 between u and v for each shortest path P∗ between u and v in Qn− F such that P∗ crosses the dimension j exactly once. Thus, P∗ can be represented as hu, P0, x, (x)j, P1, vi, where P0 is a shortest path
joining u to some node x in Qj,0n − F (Qj,0n ), and P1 is a shortest path joining (x)j to v in Qj,1n − F (Qj,1n ). See Figure 6.2(e,f) for illustration.
Subcase I.1.3.1: Suppose that ℓ(P0) > 0 and ℓ(P1) > 0. By Theorem 6.2, Qj,1n −F (Qj,1n ) contains a path T1of length l1 between (x)j and v for each l1 satisfying ℓ(P1) ≤ l1 ≤ 2n−1−1 and 2|(l1− ℓ(P1)). Suppose that ℓ(P0) = 1. It follows from Theorem 6.4 that Qj,0n − F (Qj,0n ) contains a path T0of odd length l0 between u and x for any odd integer l0 in the range from 1 to 2n−1− 1 excluding 3. Suppose that ℓ(P0) > 1. By the inductive hypothesis, Qj,0n − F (Qj,0n ) contains a path T0 of length l0 between u and x for each l0 satisfying ℓ(P0) ≤ l0 ≤ 2n−1− 1 and 2|(l0− ℓ(P0)). As a result, hu, T0, x, (x)j, T1, vi is a path of odd length l0+ l1+ 1, in the range from d∗ to 2n− 3.
Subcase I.1.3.2: Suppose that ℓ(P0) = 0 or ℓ(P1) = 0. Since d∗ = dQn−F(u, u) > 1, we have u 6= x or v 6= (x)j. With symmetry, we assume that ℓ(P0) = 0. By the inductive hypothesis, Qj,1n − F (Qj,1n ) contains a path T1 of even length l1 between (x)j and v for each even integer l1 from ℓ(P1) to 2n−1− 2. As a result, hu = x, (x)j, T1, vi is a path of odd length l1+ 1 in the range from ℓ(P1) + 1 = d∗ to 2n−1− 1.
The paths of odd lengths in the range from 2n−1+ 1 to 2n− 1 are constructed as follows.
Since |V1(Qj,0n )| = 2n−2 > 2n − 5 for n ≥ 4, we can choose a node y from V1(Qj,0n ) such that (y, (y)j) is fault-free. Let R0 be a path joining u to y in Qj,0n − F (Qj,0n ), and R1 be a path joining (y)j to v in Qj,1n − F (Qj,1n ). Similar to Subcase I.1.3.1, H = hu, R0, y, (y)j, R1, vi is a path of any odd length in the range from d′ = dQj,0n −F (Qj,0n )(u, y) + dQj,1n −F (Qj,1n )((y)j, v) + 1 to 2n− 1. By Theorem 5.3, we have d′ ≤ (n + 1) + (n − 1) + 1 ≤ 2n−1+ 1 for n ≥ 4. That is, H can be a path of any odd length in the range from 2n−1+ 1 to 2n− 1.
Subcase I.2: Suppose that |F (Qj,0n )| = 2n − 6 or |F (Qj,1n )| = 2n − 6. Without loss of generality, we assume that |F (Qj,0n )| = 2n − 6. Thus, Qj,1n is fault-free. By procedure Partition(Qn, F , u, v), the faulty links are distributed as shown in Figure 6.1.
Subcase I.2.1: Suppose that both u and v are in Qj,0n . Let (w, b) be a faulty link of Qj,0n such that both (w, (w)j) and (b, (b)j) are fault-free. For convenience, let F0 = F (Qj,0n ) − {(w, b)}. By the inductive hypothesis, Qj,0n − F0 has a path Pl of odd length l between u and v for any odd integer l in the range from dQj,0
n −F0(u, v) to 2n−1− 1. If (w, b) is on Pl, we write Pl as hu, Pl′, w, b, Pl′′, vi and define ePl = hu, Pl′, w, (w)j, (b)j, b, Pl′′, vi.
Otherwise, Plcan be written as hu, Pl′, x, y, Pl′′, vi, where (x, y) is a link on Plsuch that both (x, (x)j) and (y, (y)j) are fault-free. Similarly, we define ePl = hu, Pl′, x, (x)j, (y)j, y, Pl′′, vi.
Then ePl is a path of length l + 2. By Proposition 5.3, we have d∗ = dQn−F(u, v) ≤ h(u, v) + 4 and dQj,0n −F0(u, v) ≤ h(u, v) + 4. First, if d∗ = h(u, v) or d∗ = h(u, v) + 4, then we have d∗ = dQj,0
n −F0(u, v), and thus l ranges from d∗to 2n−1−1. Next, if d∗ = h(u, v)+2 = dQj,0
n −F0(u, v), then l ranges from d∗ to 2n−1−1. Finally, if d∗ = h(u, v) + 2 and dQj,0
n −F0(u, v) = h(u, v) + 4, then l ranges from d∗+ 2 to 2n−1− 1. For the final case, a shortest path between u and v in Qn− F can be constructed by a breadth-first search. In summary, the paths of odd lengths from d∗+ 2 to 2n−1+ 1 are constructed.
By Theorem 6.2, Qj,1n contains a path T1 of length l1 between (w)j and (b)j for each odd integer l1 from 1 to 2n−1− 1. Similarly, Qj,1n contains a path R1 of length l1 between (x)j and (y)j for each odd integer l1from 1 to 2n−1−1. Thus, hu, P2′n−1−1, w, (w)j, T1, (b)j, b, P2′′n−1−1, vi (or hu, P2′n−1−1, x, (x)j, R1, (y)j, y, P2′′n−1−1, vi) is a path of length 2n−1+ l1, in the range from 2n−1+ 1 to 2n− 1.
Subcase I.2.2: Suppose that both u and v are in Qj,1n . Let (w, (w)i) be a faulty link in Qj,0n such that both (w, (w)j) and ((w)i, ((w)i)j) are fault-free. Since d∗ = dQn−F(u, v) > 1, we assume that (w)j is different from u and v. Moreover, since n ≥ 4, we assume that t ∈ {0, 1, . . . , n − 1} − {j, i}. Let X = {((w)j, ((w)j)k) | k /∈ {i, j, t}}. Since |X| = n − 3, our inductive hypothesis ensures that Qj,1n −X contains a path T1 of odd length l1 between u and vfor any odd integer l1 satisfying d∗ ≤ l1 ≤ 2n−1−1. Let T1 denote a path of length 2n−1−1 between u and v in Qj,1n − X. It is noted that ((w)j, ((w)j)i) is on T1. Hence, T1 can be represented as hu, T1′, (w)j, ((w)j)i, T1′′, vi. By Theorem 6.4, Qj,0n − (F (Qj,0n ) − {(w, (w)i)}) contains a path T0 of odd length l0 between w and (w)i for 5 ≤ l0 ≤ 2n−1− 1. As a result, hu, T1′, (w)j, w, T0, (w)i, ((w)j)i, T1′′, vi is a path of odd length 2n−1+ l0, in the range from 2n−1+ 5 to 2n− 1. See Figure 6.3(a) for illustration.
Let T0 denote the longest path between w and (w)i in Qj,0n − (F (Qj,0n ) − {(w, (w)i)}).
Moreover, let A = {(T0(k), T0(k + 1)) | 1 ≤ k ≤ 2n−1, k ≡ 1 (mod 2)} be a set of disjoint links on T0. The paths of lengths 2n−1+ 1 and 2n−1+ 3 can be obtained as follows:
(a) Since |A| = ⌈2n−12−1⌉ > 3 for n ≥ 4, there exists a link (x, y) of A such that both F ∩ {(x, (x)j), (y, (y)j)} = ∅ and {(x)j, (y)j} ∩ {u, v} = ∅ are satisfied. Without loss of generality, we assume that x ∈ V0(Qn). By Lemma 6.1, there exist two node-disjoint paths P1 and P2 in Qj,1n such that (i) P1 joins u to (x)j, (ii) P2 joins (y)j to v, and (iii) V (P1) ∪ V (P2) = V (Qj,1n ). As a result, hu, P1, (x)j, x, y, (y)j, P2, vi is a path of length 2n−1+ 1. See Figure 6.3(b) for illustration.
(b) We write T0 as hw = x0, x1, . . . , x2n−1−1 = (w)ii. Then we can choose a pair of nodes from {{x0, x3}, {x1, x4}, {x2, x5}}, namely {xk, xk+3}, such that both F ∩ {(xk, (xk)j), (xk+3, (xk+3)j)} = ∅ and |{(xk)j, (xk+3)j} ∩ {u, v}| ≤ 1 are satisfied.
(b.1) Suppose that xk ∈ V0(Qn). If |{(xk)j, (xk+3)j} ∩ {u, v}| = 0, Lemma 6.1 en-sures that Qj,1n has two node-disjoint paths P1 and P2 such that (i) P1 joins u to (xk)j, (ii) P2 joins (xk+3)j to v, and (iii) V (P1) ∪ V (P2) = V (Qj,1n ). Hence, hu, P1, (xk)j, xk, xk+1, xk+2, xk+3, (xk+3)j, P2, vi is a path of length 2n−1 + 3. If
|{(xk)j, (xk+3)j}∩{u, v}| = 1, we assume that (xk)j = v. By Theorem 4.3, Qj,1n − {v} has a hamiltonian path H1 joining u to (xk+3)j. Then hu, H1, (xk+3)j, xk+3, xk+2, xk+1, xk, (xk)j = vi is a path of length 2n−1+ 3. See Figure 6.3(c).
(b.2) Suppose that xk ∈ V1(Qn). The required paths can be obtained similarly.
Subcase I.2.3: Suppose that u is in Qj,0n , and v is in Qj,1n . If (u, (u)j) is fault-free, the shortest path between u and v can be of the form hu, (u)j, P1, vi, where P1 is a shortest path joining (u)j to v in Qj,1n . By the inductive hypothesis, Qj,1n contains a path T1 of even length
H0 2n − 6, the j-partition determined by Partition(Qn, F , u, v) guarantees that link (v, (v)j) is fault-free if h(u, v) is odd. (See (2.2) in Section 6.1). Let (w, b) be a faulty link in Qj,0n such that both (w, (w)j) and (b, (b)j) are fault-free. By the inductive hypothesis, Qj,0n − (F (Qj,0n ) − {(w, b)}) contains a path H0 of length 2n−1− 2 between u to (v)j. Three subcases are distinguished.
Subcase I.2.3.1: Suppose that (w, b) is not located on H0. See Figure 6.3(d). We choose a link (x, y) on H0 such that (x, (x)j) and (y, (y)j) are fault-free, and ((x)j, (y)j) is not incident with v. Thus, H0 can be represented as hu, H0′, x, y, H0′′, (v)ji. By Lemma 6.2, Qj,1n − {v} contains a path T1 of odd length l1 between (x)j and (y)j for any odd integer l1 from 1 to 2n−1− 3. Consequently, hu, H0′, x, (x)j, T1, (y)j, y, H0′′, (v)j, vi is a path of odd length 2n−1+ l1, in the range from 2n−1+ 1 to 2n− 3.
Subcase I.2.3.2: Suppose that (w, b) is located on H0, and (w, b) is not incident with (v)j. See Figure 6.3(e). Thus, H0 can be represented as hu, H0′, w, b, H0′′, (v)ji. By Lemma 6.2, Qj,1n − {v} contains a path T1 of odd length l1 between (w)j and (b)j for
1 ≤ l1 ≤ 2n−1− 3. Hence, hu, H0′, w, (w)j, T1, (b)j, b, H0′′, (v)j, vi is a path of odd length 2n−1+ l1, in the range 2n−1+ 1 to 2n− 3.
Subcase I.2.3.3: Suppose that (w, b) is located on H0, and (w, b) is incident with (v)j. See Figure 6.3(f). Let w = (v)j. Thus, H0 can be represented as hu, H0′, b, w = (v)ji.
By Theorem 6.2, Qj,1n contains a path T1 of odd length l1 between (b)j and v for any odd integer l1 satisfying 1 ≤ l1 ≤ 2n−1− 1. Then hu, H0′, b, (b)j, T1, vi is a path of odd length 2n−1+ l1− 2, in the range from 2n−1− 1 to 2n− 3.
Case II: Suppose that u and v belong to the same partite set of Qn− F . Thus, the distance d∗ between u and v is even. Without loss of generality, we assume that u, v ∈ V0(Qn). By Theorem 6.3, Qn− F is strongly hamiltonian laceable. Moreover, a shortest path between u and v can be obtained by a breadth-first search. Hence, we concentrate on the paths of even lengths in the range from d∗+ 2 to 2n− 4.
Subcase II.1: Suppose that |F (Qj,0n )| ≤ 2n − 7 and |F (Qj,1n )| ≤ 2n − 7. Without loss of generality, we assume that |F (Qj,0n )| ≥ |F (Qj,1n )|. Thus, |F (Qj,1n )| ≤ n − 3.
Subcase II.1.1: Suppose that both u and v are in Qj,0n . By the inductive hypothesis, Qj,0n −F (Qj,0n ) has a path H0 of length 2n−1−2 between u and v. Let A = {(H0(i), H0(i+1)) | 1 ≤ i ≤ 2n−1 − 1, i ≡ 1 (mod 2)} be a set of disjoint links on H0. First, suppose that
|F (Qj,0n )| > 0. Since |A| = ⌈2n−12−2⌉ > 2n − 5 − |F (Qj,0n )| for n ≥ 4, there exists a link (w, b) of A such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free. Next, suppose that
|F (Qj,0n )| = 0 and n ≥ 5. Since |A| = ⌈2n−12−2⌉ > 2n − 5, there still exists a link (w, b) of A such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free. Finally, suppose that
|F (Qj,0n )| = 0 and n = 4. If there does not exist any node z of V1(Qj,04 ) such that (z, (z)j) is faulty, there must exist a link (w, b) on H0 such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free. If there exists a node z of V1(Qj,04 ) such that (z, (z)j) is faulty, then it follows from Theorem 4.3 that Qj,04 − {z} has a hamiltonian path, still namely H0, between u and v. Obviously, there also exists a link (w, b) on H0 such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free. In summary, H0 can be written as hu, H0′, w, b, H0′′, vi. Since
|F (Qj,1n )| ≤ n − 3, it follows from Theorem 6.2 that Qj,1n − F (Qj,1n ) contains a path H1 of odd length l1 between (w)j and (b)j for any odd integer l1 satisfying 1 ≤ l1 ≤ 2n−1 − 1. As a result, hu, H0′, w, (w)j, H1, (b)j, b, H0′′, vi is a path of even length in the range from 2n−1 to 2n− 2.
The paths of lengths less than 2n−1 are obtained as follows. By Proposition 5.3, we have d∗ = dQn−F(u, v) ≤ h(u, v) + 4 and dQj,0
n −F (Qj,0n )(u, v) ≤ h(u, v) + 4. By inductive hypothesis, Qj,0n −F (Qj,0n ) has a path T0of length l0 between u and v for any even length from dQj,0
n −F (Qj,0n )(u, v) to 2n−1−2. If d∗ = h(u, v) or d∗ = h(u, v)+4, then dQj,0
n −F (Qj,0n )(u, v) = d∗. If d∗ = h(u, v) + 2, then dQj,0
n −F (Qj,0n )(u, v) ≤ d∗+ 2.
Subcase II.1.2: Suppose that both u and v are in Qj,1n . Since |F (Qj,1n )| ≤ n−3, it follows from Lemma 5.1 that d∗ ≤ h(u, u) + 2. Thus, Qn− F has a shortest path between u and v that does not cross the dimension j. By the inductive hypothesis, Qj,1n − F (Qj,1n ) contains a
path T1 of length l1 between u and v for any even integer l1 satisfying d∗ ≤ l1 ≤ 2n−1− 2.
Let T1 be a path of length 2n−1 − 2 between u and v in Qj,1n − F (Qj,1n ). Moreover, let A = {(T1(i), T1(i + 1)) | 1 ≤ i ≤ 2n−1− 1, i ≡ 1 (mod 2)} be a set of disjoint links on T1. First, suppose that |F (Qj,1n )| > 0. Since |A| = ⌈2n−12−2⌉ > 2n − 5 − |F (Qj,1n )| for n ≥ 4, there exists a link (w, b) ∈ A such that (w, (w)j), (b, (b)j), and ((w)j, (b)j) are all fault-free.
Next, suppose that |F (Qj,1n )| = 0 and n ≥ 5. Since |A| = ⌈2n−12−2⌉ > 2n − 5, there still exists a link (w, b) ∈ A such that (w, (w)j), (b, (b)j) and ((w)j, (b)j) are all fault-free. Finally, suppose that |F (Qj,1n )| = 0 and n = 4. If there does not exist any node z of V1(Qj,14 ) such that (z, (z)j) is faulty, there exists a link (w, b) on T1 such that (w, (w)j), (b, (b)j) and ((w)j, (b)j) are all fault-free. If there exists a node z of V1(Qj,14 ) such that (z, (z)j) is faulty, Theorem 4.3 ensures that Qj,14 − {z} has a hamiltonian path, still namely T1, between u and v. Obviously, there also exists a link (w, b) on T1 such that (w, (w)j), (b, (b)j) and ((w)j, (b)j) are all fault-free. In summary, T1 can be written as hu, T1′, w, b, T1′′, vi. Since
|F (Qj,0n )| ≤ 2n − 7, it follows from Theorem 6.4 that Qj,0n − F (Qj,0n ) contains a path T0 of length l0 between (w)j and (b)j for any odd integer l0 from 1 to 2n−1− 1 excluding 3. As a result, hu, T1′, w, (w)j, T0, (b)j, b, T1′′, vi is a path of any even length in the range from 2n−1 to 2n− 2, excluding 2n−1+ 2.
The path of length 2n−1 + 2 is discussed as follows. When n = 4, |F (Qj,0n )| ≤ 1.
Thus, there exists an integer k of {0, 1, 2, 3} − {j, dim((w, b))} such that ((w)j, ((w)j)k), ((b)j, ((b)j)k), and (((w)j)k, ((b)j)k) are all fault-free. Hence, hu, T1′, w, (w)j, ((w)j)k, ((b)j)k, (b)j, b, T1′′, vi is a path of length 10. When n ≥ 5, we have |A| − |F | = ⌈2n−12−2⌉ − (2n − 5) ≥ 2. Thus, there is another link (x, y) of A, other than (w, b), such that (x, (x)j), (y, (y)j), and ((x)j, (y)j) are all fault-free. Without loss of generality, T1 can be written as hu, R1′, w, b, R′′1, x, y, R′′′1, vi. Hence, hu, R′1, w, (w)j, (b)j, b, R′′1, x, (x)j, (y)j, y, R1′′′, vi is a path of length 2n−1+ 2.
Subcase II.1.3: Suppose that u is in Qj,0n and v is in Qj,1n . By Theorem 6.1, there exists a shortest path P∗ between u and v in Qn− F such that P∗ crosses the dimension j exactly once. Thus, P∗ can be written as hu, P0, x, (x)j, P1, vi, where P0 is a shortest path joining u to some node x in Qj,0n − F (Qj,0n ) and P1 is a shortest path joining (x)j to v in Qj,1n − F (Qj,1n ).
Subcase II.1.3.1: Suppose that ℓ(P0) > 0 and ℓ(P1) > 0. By Theorem 6.2, Qj,1n −F (Qj,1n ) has a path T1 of length l1 between (x)j and v for each l1 satisfying ℓ(P1) ≤ l1 ≤ 2n−1− 1 and 2|(l1 − ℓ(P1)). Suppose that ℓ(P0) = 1. By Theorem 6.4, Qj,0n − F (Qj,0n ) has a path T0 of length l0 between u and x for any odd integer l0 from 1 to 2n−1 − 1 excluding 3.
Suppose that ℓ(P0) > 1. By the inductive hypothesis, Qj,0n − F (Qj,0n ) has a path T0 of length l0 between u and x for each l0 satisfying ℓ(P0) ≤ l0 ≤ 2n−1− 1 and 2|(l0 − ℓ(P0)). Hence, hu, T0, x, (x)j, T1, vi is a path of even length l0+ l1+ 1 in the range from d∗ to 2n− 2.
Subcase II.1.3.2: Suppose that ℓ(P0) = 0 or ℓ(P1) = 0. With symmetry, we assume u= x. By the inductive hypothesis, Qj,1n − F (Qj,1n ) contains a path T1 of length l1 between (u)j and v for any odd integer l1 form ℓ(P1) to 2n−1− 1. Then hu, (u)j, T1, vi is a path of even length l1+ 1 in the range from ℓ(P1) + 1 = d∗ to 2n−1.
The paths of lengths greater than 2n−1are constructed as follows. Since |V (Qj,0n )−{u}|−
(2n − 5) > 1 for n ≥ 4, we can choose a node y from V (Qj,0n ) − {u} such that (y, (y)j) is fault-free and (y)j is not v. Let R0 be a path joining u to y in Qj,0n −F (Qj,0n ) and R1 be a path joining (y)j to v in Qj,1n − F (Qj,1n ). Similar to Subcase II.1.3.1, H = hu, R0, y, (y)j, R1, vi is a path of even length in the range from d′ = dQj,0
n −F (Qj,0n )(u, y) + dQj,1
n −F (Qj,1n )((y)j, v) + 1 to 2n− 2. By Theorem 5.3, we have d′ ≤ (n + 1) + (n − 1) + 1 ≤ 2n−1+ 2 for n ≥ 4. Therefore, H is a path of even length in the range from 2n−1+ 2 to 2n− 2.
Subcase II.2: Suppose that |F (Qj,0n )| ≤ 2n − 6 or |F (Qj,1n )| ≤ 2n − 6. Without loss of generality, we assume that |F (Qj,0n )| = 2n − 6. Thus, Qj,1n is fault-free. It is noticed that the faulty links are distributed as shown in Figure 6.1.
Subcase II.2.1: Suppose that both u and v are in Qj,0n . Let (w, b) be a faulty link of Qj,0n such that both (w, (w)j) and (b, (b)j) are fault-free. Let F0 = F (Qj,0n ) − {(w, b)}. By the inductive hypothesis, Qj,0n −F0has a path Plof length l between u and v for any even integer l from dQj,0
n −F0(u, v) to 2n−1−2. If (w, b) is on Pl, we write Plas hu, Pl′, w, b, Pl′′, vi and define Pel = hu, Pl′, w, (w)j, (b)j, b, Pl′′, vi. Otherwise, Pl can be written as hu, Pl′, x, y, Pl′′, vi, where (x, y) is a link on Pl such that both (x, (x)j) and (y, (y)j) are fault-free. Similarly, we define ePl = hu, Pl′, x, (x)j, (y)j, y, Pl′′, vi. Then ePlis a path of length l+2. By Proposition 5.3, we have d∗ = dQn−F(u, v) ≤ h(u, v) + 4 and dQj,0n −F0(u, v) ≤ h(u, v) + 4. If dQj,0n −F0(u, v) = d∗, then path ePl is the desired path. Otherwise, if dQj,0n −F0(u, v) = d∗+ 2, then ePl is a path of even length in the range from d∗+ 4 to 2n−1. It is noticed that a shortest path between u and v in Qn− F can be constructed based on a breadth-first search.
By Theorem 6.2, Qj,1n contains a path T1 of length l1 between (w)j and (b)j or a path R1 of odd length l1 between (x)j and (y)j for any odd integer l1 from 1 to 2n−1− 1. Thus, hu, P2′n−1−2, w, (w)j, T1, (b)j, b, P2′′n−1−2, vi (or hu, P2′n−1−2, x, (x)j, R1, (y)j, y, P2′′n−1−2, vi) is a path of even length in the range from 2n−1 to 2n− 2.
Subcase II.2.2: Suppose that both u and v are in Qj,1n . Let (w, (w)i) be a faulty link of Qj,0n such that both (w, (w)j) and ((w)i, ((w)i)j) are fault-free. Since n ≥ 4, we assume that t ∈ {0, 1, . . . , n − 1} − {j, i}. Moreover, we assume that w ∈ V0(Qj,0n ). Let X = {((x)j, ((x)j)k) | k /∈ {i, j, t}}. Since |X| = n − 3, our inductive hypothesis ensures that Qj,1n − X contains a path T1 of even length l1 between u and v for d∗ ≤ l1 ≤ 2n−1− 2. Let T1 denote the longest path between u and v in Qj,1n − X. It is noted that ((w)j, ((w)j)i) is on T1. Hence, T1 can be represented as hu, T1′, (w)j, ((w)j)i, T1′′, vi. By the inductive hypothesis, Qj,0n − (F (Qj,0n ) − {(w, (w)i)}) contains a path T0 of odd length l0 between w to (w)i for 5 ≤ l0 ≤ 2n−1− 1. As a result, hu, T1′, (w)j, w, T0, (w)i, ((w)j)i, T1′′, vi is a path of even length 2n−1+ l0− 1, in the range from 2n−1+ 4 to 2n− 2.
Let A = {(T1(k), T1(k + 1)) | 1 ≤ k ≤ 2n−1− 1, k ≡ 1 (mod 2)} be a set of disjoint links on T1. Then the paths of lengths 2n−1 and 2n−1+ 2 can be obtained as follows. When n = 4, we suppose that {p, q, j, i} = {0, 1, 2, 3}. Since (w, (w)i) is faulty, we have either {(w, (w)p), ((w)p, ((w)p)i), ((w)p)i, (w)i)} ∩ F = ∅ or {(w, (w)q), ((w)q, ((w)q)i), ((w)q)i,
(w)i, (w)q)i)} ∩ F = ∅. Without loss of generality, we assume {(w, (w)p), ((w)p, ((w)p)i), ((w)i, (w)p)i)} ∩ F = ∅. Obviously, hu, T1′, (w)j, w, (w)p, ((w)p)i, (w)i, ((w)j)i, T1′′, vi is a path of length 2n−1+ 2. Moreover, since |A| − |F | = ⌈2n−12−1⌉ − (2n − 5) = 1 for n = 4, there exists one link (x, y) ∈ A such that (x, (x)j), (y, (y)j), and ((x)j, (y)j) is fault-free.
Hence, T1 can be represented as hu, R1, x, y, R2, vi. Obviously, hu, R1, x, (x)j, (y)j, y, R2, vi is a path of length 2n−1. When n ≥ 5, we have |A| − |F | = ⌈2n−12−2⌉ − (2n − 5) ≥ 2. Thus, there are two links (x1, y1), (x2, y2) ∈ A such that {(xk, (xk)j), (yk, (yk)j), ((xk)j, (yk)j) | k = 1, 2} ∩ F = ∅. Hence, T1 can be represented as hu, R1, x1, y1, R2, x2, y2, R3, vi. Ob-viously, hu, R1, x1, (x1)j, (y1)j, y1, R2, x2, y2, R3, vi and hu, R1, x1, (x1)j, (y1)j, y1, R2, x2, (x2)j, (y2)j, y2, R3, vi are paths of length 2n−1 and of length 2n−1+ 2, respectively.
Subcase II.2.3: Suppose that u is in Qj,0n and v is in Qj,1n . If (u, (u)j) is fault-free, the shortest path between u and v can be of the form hu, (u)j, P1, vi, where P1 is a shortest path joining (u)j to v in Qj,1n . By the inductive hypothesis, Qj,1n contains a path T1 of odd length l1 between (u)j and v for d∗ − 1 ≤ l1 ≤ 2n−1 − 1. Then hu, (u)j, T1, vi is a path of even length in the range from d∗ to 2n−1. If (u, (u)j) is faulty, we choose a neighbor of u in Qj,0n − F (Qj,0n ), namely x, such that (x)j 6= v. Obviously, we have either h((x)j, v) = h(u, v) − 2 or h((x)j, v) = h(u, v). Let R1 be a shortest path joining (x)j to v in Qj,1n . Then hu, x, (x)j, R1, vi is a path of length h(u, v) or h(u, v) + 2. By Theorem 6.2, Qj,1n contains a path T1 of even length l1 between (x)j and v for any even integer l1 from h((x)j, v) to 2n−1 − 2. Then hu, x, (x)j, T1, vi is a path of even length in the range from d∗+ 2 to 2n−1.
The paths of lengths greater than 2n−1 are obtained as follows. Let (w, b) be a faulty link in Qj,0n such that both (w, (w)j) and (b, (b)j) are fault-free. Depending on whether (v, (v)j) is faulty, we distinguish two subcases.
Subcase II.2.3.1: Suppose that (v, (v)j) is fault-free. By the inductive hypothesis, Qj,0n − (F (Qj,0n ) − {(w, b)}) contains a path H0 of length 2n−1− 1 between u to (v)j.
Subcase II.2.3.1.a: Suppose that (w, b) is not located on H0. We choose a link (x, y) on H0 such that (x, (x)j) and (y, (y)j) are fault-free and ((x)j, (y)j) is not incident with v.
Thus, H0 can be represented as hu, H0′, x, y, H0′′, (v)ji. By Lemma 6.2, Qj,1n − {v} contains a path T1 of odd length l1 between (x)j and (y)j for any odd integer l1 from 1 to 2n−1− 3.
Consequently, hu, H0′, x, (x)j, T1, (y)j, y, H0′′, (v)j, vi is a path of even length 2n−1+ l1 + 1, in the range from 2n−1+ 2 to 2n− 2.
Subcase II.2.3.1.b: Suppose that (w, b) is located on H0 and (w, b) is not incident with (v)j. Thus, H0 can be represented as hu, H0′, w, b, H0′′, (v)ji. By Lemma 6.2, Qj,1n − {v}
contains a path T1 of odd length l1 between (w)j and (b)j for any odd integer l1 from 1 to 2n−1− 3. Then hu, H0′, w, (w)j, T1, (b)j, b, H0′′, (v)j, vi is a path of even length 2n−1+ l1+ 1, in the range from 2n−1+ 2 to 2n− 2.
Subcase II.2.3.1.c: Suppose that (w, b) is on H0 and (w, b) is incident with (v)j. Let b= (v)j. Thus, H0 can be written as hu, H0′, w, b = (v)ji. By Theorem 6.2, Qj,1n has a path
T1 of odd length l1 between (w)j and v for 1 ≤ l1 ≤ 2n−1− 1. Thus, hu, H0′, w, (w)j, T1, vi is a path of even length 2n−1+ l1− 1, in the range from 2n−1 to 2n− 2.
Subcase II.2.3.2: Suppose that (v, (v)j) is faulty. According to procedure Partition(Qn, F , u, v), this subcase occurs only when n = 4 and there is a unique node z of V1(Q4) such that both (z, u) and (z, v) are faulty links. In addition, each faulty link corresponds to a unique dimension. By transitivity, we assume that z = 0001, u = 0101, and v = 1001. Then the paths obtained by brute force are listed in Table 6.3.
Table 6.3: The paths of lengths 10, 12, and 14 between u = 0101 and v = 1001 in Q4 − {ef, (0001, 0101), (0001, 1001)}.
ef∈ {(0000, 0010), (0010, 0011)} hu = 0101, 0100, 0110, 0111, 0011, 0001, 0000, 1000, 1100, 1101, 1001 = vi
hu = 0101, 0100, 0110, 0111, 0011, 0001, 0000, 1000, 1100, 1110, 1111, 1101, 1001 = vi
hu = 0101, 0100, 0110, 0111, 0011, 0001, 0000, 1000, 1100, 1110, 1010, 1011, 1111, 1101, 1001 = vi ef= (0100, 0110) hu = 0101, 0111, 0110, 0010, 0011, 0001, 0000, 1000, 1100, 1101, 1001 = vi
hu = 0101, 0111, 0110, 0010, 0011, 0001, 0000, 1000, 1100, 1110, 1111, 1101, 1001 = vi
hu = 0101, 0111, 0110, 0010, 0011, 0001, 0000, 1000, 1100, 1110, 1010, 1011, 1111, 1101, 1001 = vi ef= (0110, 0111) hu = 0101, 0111, 0011, 0010, 0110, 0100, 0000, 1000, 1100, 1101, 1001 = vi
hu = 0101, 0111, 0011, 0010, 0110, 0100, 0000, 1000, 1100, 1110, 1111, 1101, 1001 = vi
hu = 0101, 0111, 0011, 0010, 0110, 0100, 0000, 1000, 1100, 1110, 1010, 1011, 1111, 1101, 1001 = vi