Pan-connectivity of AQ n

In this chapter, we will prove augmented cubes are pan-connected by induction.

Theorem 2 The augmented cubes with dimension n ≥ 1, AQ_n, are pan-connected.

Base on the results of chapter 3, AQ2, AQ3, and AQ4 are pan-connected, we suppose that AQ_nis pan-connected when n ≥ 4. Then we prove that AQ_n+1is also pan-connected.

Let a, b be any two vertices in AQ_n+1, we want to find the path P of length ≤ l join a and b with d(a, b) ≤ l ≤ 2ⁿ⁺¹− 1.

case 1: a, b are in AQ⁰_n or a and b are in AQ¹_n.

Without loss of generality, let a and b are in AQ⁰_n. By induction, there exist the paths of length from d(a, b) to 2ⁿ− 1 joining a to b.

case 1.1: a and b are not adjacent

a

c

b

a’

c’

AQ

Figure 4.1: a and b are not adjacent in the augmented cubeAQn+1.

Let T-set(a)={c, a⁰, c⁰}. Without loss of generality, let (a, a⁰), (c, c⁰) are hypercube edges and a, c are in AQ⁰_n.

By theorem 1, there exist a hamiltonian path P₁ of AQ⁰_n− {a} joining c to b. Hence, l(P₁) = 2ⁿ− 2. And by induction, there exist paths Q_i of lengths from 1 to 2ⁿ− 1 in AQ¹_n joining a⁰ to c⁰. l(Q_i) = 1 ∼ 2ⁿ− 1.

We trace P in the following sequences to get different lengths:

First, to get the path of length 2ⁿ:

Let P = ha, a⁰, c, P₁, bi, and let l(hc,..., bi) = 2ⁿ− 2. Then l(P ) = 2ⁿ.

Second, to get the path of lengths from 2ⁿ+ 1 to 2ⁿ⁺¹− 1:

Let P_i = ha, a⁰, Q_i, c⁰, c, P₁, bi, i = 1 ∼ 2ⁿ− 1. l(P_i) = 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1. So this case is completed.

case 1-2: a and b are adjacent

case 1-2-1: b is in a⁰s T-set

a

c

b

a’ c’

AQn+1

d

b’

d’

Figure 4.2: b is in a⁰s T-set

Lemma 4 For any node a in AQ_n, n ≥ 4, suppose {a⁰, b, b⁰} is T-set(a) and both a and b are in AQ⁰_n−1 or AQ¹_n−1. There exist two pairs of distinct nodes, c, d in AQ⁰_n−1 and c⁰, d⁰ in AQ¹_n−1, that have the following relations: (1). c and d are adjacent and c is adjacent to a and d is adjacent to b. (2). c⁰ and d⁰ are adjacent and c⁰ is adjacent to a⁰ and d⁰ is adjacent to b⁰. (3). c is adjacent to c⁰ and d is adjacent to d⁰.

Proof. Without loss of generality, let a and b are in AQ⁰_n−1and a = (0, v₁, v₂,...v_i...v_n−2), b = (0, v1, v2,...vi..., vn−2). Let c = (0, v1, v2,...vi...vn−2) , d = (0, v1, v2,...vi..., vn−2), c⁰ = (1, v₁, v₂,...v_i...v_n−2), and d0 = (1, v₁, v₂,...v_i..., v_n−2) for some i, 1 ≤ i ≤ 2ⁿ − 2. Such c, d, c⁰, d⁰ conform to the conditions (1), (2), (3) of lemma 4.

This case can be constructed as Figure 4.2. Let T-set(a)={b, a⁰, b⁰}. Without loss of generality, let (a, a⁰), (b, b⁰) are hypercube edges and a, b are in AQ⁰_n.

First, to get the path of length 2ⁿ:

By induction, there exist the path P₁ of length l = 2ⁿ− 2 in AQ¹_n joining a⁰ to b⁰. Let P = ha, a⁰, P₁, b⁰, bi, then l(P ) = 2ⁿ.

Second, to get the path of length 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1:

By theorem 1, there exist the hamiltonian path P_a in AQ⁰_n joining a to c and not pass through b. Thus, l(Pa) = 2ⁿ− 2. By induction, there exist paths Qi of lengths from 1 to 2ⁿ− 1 in AQ¹_n joining c⁰ to a⁰.

Let P_i = ha, P_a, c, c⁰, Q_i, a⁰, bi, 1 ≤ i ≤ 2ⁿ− 1. l(P_i) = 2ⁿ+ 1 ∼ 2ⁿ⁺¹ − 1. Which completes this case.

case 1-2-2: b is not in a⁰s T-set

a

In AQ_n, since AQ⁰_n−1 and AQ¹_n−1 are isomorphic, we have the following lemma.

Lemma 5 If a and b are adjacent in AQ⁰_n−1, let a⁰ = (a^h) and b⁰ = (b^h), then a⁰ and b⁰ are also adjacent in AQ¹_n−1.

By lemma 5, we can trade a and b as in figure 4.3.a in this case. But it is not enough to explain all subcases. Thus, we need a more detail descriptions about a and b. We see that b is not in a’s T-set, so there exists another node c 6= b and c is in a’s T-set. Watch that c and b are not always adjacent, so we dotted it in figure 4.3.b. Of course, there exist another two nodes c⁰ and a⁰ which are both in T-set(a). Without loss of generality, let a and c are both in AQ⁰_n.

We begin our subcases that to find the paths of lengths from 2ⁿ to 2ⁿ⁺¹− 1.

First, to get the path of length 2ⁿ:

See figure 4.3.a. By induction, there exist a path P₁ of length 2ⁿ− 2 in AQ¹_n joining c⁰ to a⁰. Let P = ha, a⁰, P1, b⁰, bi. As a result, l(P ) = 2ⁿ.

Second, to get the path of lengths 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1:

See figure 4.3.b. By induction, there exist paths Q_i of lengths from 1 to 2ⁿ− 1 in AQ¹_n joining a⁰ to c⁰. And by theorem 1, there exist the hamiltonian path P1 that does not go through a joining c to b. l(P₁) = 2ⁿ− 2. Let P_i = ha, a⁰, Q_i, c⁰, c, P₁, bi, i = 1 ∼ 2ⁿ− 1.

Thus, l(Pi) = 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1. And it completes this case.

case 2: a is in AQ⁰_n and b is in AQ¹_n, or a is in AQ¹_n and b is in AQ⁰_n

Without loss of generality, let a is in AQ⁰_n and b is in AQ¹_n.

case 2-1: a, b are adjacent

In this case, it is obvious to see that b is in set(a). And d(a, b) = 1. Let T-set(a)={a⁰, b, b⁰}. It is shown in figure 4.4.

By induction, there exist paths Qi of lengths from 1 to 2ⁿ− 1 in AQ⁰_n joining a to a⁰, 1 ≤ i ≤ 2ⁿ− 1. l(Q_i) = 1 ∼ 2ⁿ− 1.

First, to get the path of lengths 2 ∼ 2ⁿ:

Let Pi = ha, Qi, a⁰, bi, 1 ≤ i ≤ 2ⁿ− 1. l(Pi) = 2 ∼ 2ⁿ.

a b

a’

AQ

b’

Figure 4.4: a, b are adjacent Second, to get the path of lengths 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1:

By induction, there exist the hamiltonian path P1 in AQ⁰_n joining a to a⁰. Also, there exist paths Q_i of lengths from 1 to 2ⁿ− 1 in AQ¹_n joining b to b⁰.

Let P_i = ha, P₁, a⁰, b, Q_i, b⁰i, 1 ≤ i ≤ 2ⁿ− 1. l(P_i) = 2ⁿ+ 1 ∼ 2ⁿ⁺¹− 1.

case 2-2: a, b are not adjacent

Without loss of generality, let a be in AQ⁰_n, and let d(a, b) = k ≥ 2. Since a is not adjacent to b, there exist another node c 6= a in AQ⁰_n such that c is in T-set(b), and b⁰ in AQ¹_n is also in T-set(b). So d(a, c) = k − 1.

Without loss of generality, let (c, b) be the hypercube edge. When tracing a path joining a to b, if we take each step when going from node u to v as a bit-operation,

a

b c

AQ

b’

Figure 4.5: a, b are not adjacent

where u and v are adjacent on the path. Then tracing a path from a to b is as many bit-operations. And any combination of those bit-operations have the same result. For example, if a = (000) and b = (110), then we can go from a to b by the following steps : (1). Change the most right bit to 1, which acts as go from a=(000) to (001).(2). Inverse all bits, which acts as go from (001) to (110)=b. Or we can do the (2)-step first, and then do the (1)-step. It does not matter to have any combinations. See figure 4.5. As a result, we can say that if the path joining a to b exist, then the path going from a to c and from c to b⁰ and from b0 to b exist, and the passing nodes in the segment from a to c are all in AQ⁰_n, and the passing nodes in the segment from b⁰ to b are all in AQ¹_n.

By induction, there exist paths Q_i of lengths from k − 1 to 2ⁿ− 1 in AQ⁰_n joining a to c. Also, there exist paths R_j of lengths from 1 to 2ⁿ− 1 in 2ⁿ− 1 in AQ¹_n joining b⁰ to b.

To get the paths of lengths k + 1 ∼ 2ⁿ⁺¹− 1:

Let P_ij = ha, Q_i, c, b⁰, R_j, bi, k−1 ≤ i ≤ 2ⁿ−1, 1 ≤ j ≤ 2ⁿ−1. l(P_ij) = k+1 ∼ 2ⁿ⁺¹−1.

With all above, it is completed of proof of theorem 2.