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Mutually Independent Hamiltonian Cycles in k-ary n-cubes when k is odd Shin-Shin Kao Chung-Yuan Christian University Department of Applied Mathematics No. 200, Jong-Bei Rd., Chung-Li City Taiwan, R.O.C.
[email protected]
Pi-Hsiang Wang Chung-Yuan Christian University Department of Applied Mathematics No. 200, Jong-Bei Rd., Chung-Li City Taiwan, R.O.C.
[email protected]
Abstract: The k-ary n-cubes, Qkn , is one of the most well-known interconnection networks in parallel computers. Let n ≥ 1 be an integer and k ≥ 3 be an odd integer. It has been shown that any Qkn is a 2n-regular, vertex symmetric and edge symmetric graph with a hamiltonian cycle. In this article, we prove that any k-ary n-cube contains 2n mutually independent hamiltonian cycles. More specifically, let vi ∈ V (Qkn ) for 0 ≤ i ≤ |Qkn | − 1 and let v0 , v1 , . . . , v|Qkn |−1 , v0 be a hamiltonian cycle of Qkn . We prove that Qkn contains 2n hamiltonian cycles l l l of the form v0 , v1l , . . . , v|Q k |−1 , v0 for 0 ≤ l ≤ 2n − 1, where vi = vi whenever l = l . The result is optimal n
since each vertex of Qkn has exactly 2n neighbors.
Key–Words: k-ary n-cubes, hamiltonian cycle, mutually independent, hypercubes.
1 Introduction
(0,1)
The hypercube family, denoted by Qn , is one of the most well-known and popular interconnection networks due to its excellent properties such as the recursive structure, symmetry, small diameter, low degree, easy routing and so on. See [3, 7, 11] and their references. Therefore, numerous studies have been devoted to the hypercube family. See [2, 4, 6, 13, 14] for example.
0
0
(0,0)
1
1
1
(1,0)
(2,0)
2
2
2
Q 31
Q 31
Q 31
(0,2)
(1,1)
(1,2)
(2,2) (2,1)
Q 32
Figure 1: Q32 consists of three Q31 ’s. recursive structure of k-ary n-cubes, where k ≥ 3, is as follows:
The k-ary n-cube, denoted by Qkn , is an enlarged family of Qn that preserves many pleasing properties of the hypercubes. More precisely, Qkn is a graph with kn vertices, and each vertex u can be labeled by an nbit string, u = (u(n − 1), u(n − 2), . . . , u(1), u(0)), where 0 ≤ u(i) ≤ k − 1 for all 0 ≤ i ≤ n − 1. There is an edge between two vertices u = (u(n − 1), u(n−2), . . . , u(1), u(0)) and v = (v(n−1), v(n− 2), . . . , v(1), v(0)) in Qkn if and only if there exists some i, 0 ≤ i ≤ n−1, such that u(i) = v(i)±1 (mod k) and u(j) = v(j) for all j = i. Obviously, the hypercube Qn is the same as Q2n and hence is a subclass of the k-ary n-cube family. The following recursive interpretation of Qkn can be found in [10]. First, we define a cycle of k vertices labeled 0, 1, ..., k − 1 to be a graph with edges between i and i + 1(mod k) for i = 0, 1, .., k − 1. When k = 1, the “cycle” is a vertex. When k = 2, it consists of two vertices sharing an edge. When k ≥ 3, it is a conventional cycle. The ISSN: 1790-2769
0
• A k-ary 1-cube is a cycle of k vertices. Without loss of generality, we place the k vertices on a line and call the leftmost vertex the 0th position vertex and the rightmost vertex the (k − 1)th position vertex. • A k-ary n-cube contains k composite subcubes, each of which is a k-ary (n−1)-cube placed from left to right. For each position i = {0, ..., kn−1 − 1}, edges between composite subcubes are define by connecting all k ith position vertices into a cycle. See Figure 1 for an illustration. It is shown that Qkn is a bipartite graph when k is an even integer [5]. The concept of mutually independent hamiltonian cycles arises from the following application [8]. If k
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with a hamiltonian cycle. The length of a cycle C is the number of edges/vertices in C. Two cycles C1 = u0 , u1 , · · · , uk , u0 and C2 = v0 , v1 , · · · , vk , v0 beginning at s are independent if u0 = v0 = s and ui = vi for 1 ≤ i ≤ k [9]. Cycles beginning at s with the same length are mutually independent if every two different cycles are independent. A graph G is said to contain n mutually independent hamiltonian cycles if there exist n hamiltonian cycles in G beginning at any vertex s such that the n cycles are mutually independent. Recently, there are many studies in mutually independent hamiltonian cycles. Readers can refer to [4, 8, 9, 12, 13].
pieces of data must be sent from a message center u, and the data has to be processed at each intermediate receiver(and the process takes time), then the existence of mutually independent cycles from u guarantees that there will be no waiting time at a processor. Recently, many studies about mutually independent hamiltonian cycles on hypercubes and other interconnection networks are published [8, 9, 12, 13]. In this article, we prove that any k-ary n-dimensional cube Qkn contains 2n mutually independent hamiltonian cycles for any integer n ≥ 1 and any odd integer k ≥ 3. The result is optimal since each vertex of Qkn has exactly 2n neighbors. The article is organized as follows. In Section 2, we introduce the graph terminologies and notations. In Section 3, we present the main results. Our conclusion is given in the last section.
3 Main Results Theorem 1 There exist two mutually independent hamiltonian cycles in Qk1 , where k ≥ 3 is an odd integer.
2 Preliminaries
Proof. By definition, Qk1 is a cycle with k vertices. Let the graph of Qk1 be (V, E), where V = {i | 0 ≤ i ≤ k − 1} and E = {(i, i + 1(mod k)) | 0 ≤ i ≤ k − 1}. Let C1 = 0, 1, 2, . . . , k − 1, 0 and C2 = 0, k − 1, k − 2, . . . , 2, 1, 0. Obviously, C1 and C2 are two mutually independent hamiltonian cycles of Qk1 .
For the graph definitions and notations we follow [1]. G = (V, E) is a graph if V is a finite set and E is a subset of {(u, v) | (u, v) is an unorder pair of V }. We say that V is the vertex set and E is the edge set of G. Two vertices u and v are adjacent if (u, v) ∈ E. The total number of vertices of G is denoted by |V (G)| or |G|. For a vertex u of G, we denote the degree of u by deg(u) = |{v | (u, v) ∈ E}|. A graph G is k-regular if, for every vertex u ∈ G, deg(u) = k. A path is represented by v0 , v1 , v2 , · · · , vk , where every two consecutive vertices are connected by an edge. We also write the path v0 , v1 , v2 , · · · , vk as v0 , P1 , vi , vi+1 · · · , vj , P2 , vt , · · · , vk , where P1 is the path v0 , v1 , · · · , vi−1 , vi and P2 is the If a path P = path vj , vj+1 , · · · , vt−1 , vt . v0 , v1 , v2 , · · · , vk−1 , vk , then P −1 denotes the path vk , vk−1 , · · · , v2 , v1 , v0 . A hamiltonian path between u and v, where u and v are two distinct vertices of G, is a path joining u to v that visits every vertex of G exactly once. Two paths P1 = u0 , u1 , · · · , um and P2 = v0 , v1 , · · · , vm from a to b are independent if u0 = v0 = a, um = vm = b, and ui = vi for 1 ≤ i ≤ m−1 [8]. Paths with the same number of vertices from a to b are mutually independent if every two different paths are independent [8]. In [4], two paths P1 = u0 , u1 , · · · , um and P2 = v0 , v1 , · · · , vm are full-independent if ui = vi for all 0 ≤ i ≤ m. Similarly, paths with the same number of vertices are mutually full-independent if every two different paths are full-independent. A cycle is a path of at least three vertices such that the first vertex is the same as the last vertex. A hamiltonian cycle of G is a cycle that traverses every vertex of G exactly once. A hamiltonian graph is a graph
ISSN: 1790-2769
Lemma 2 There exist four mutually independent hamiltonian cycles in Q32 , denoted by Ci3 for 0 ≤ i ≤ 3, such that ((2, 2), (2, 0)) ∈ C03 , ((2, 0), (0, 0)) ∈ C13 , ((0, 0), (0, 2)) ∈ C23 and ((0, 2), (2, 2)) ∈ C33 . Proof. We construct the four mutually independent hamiltonian cycles below, where the edges required in the lemma are underlined. C03
=
(1, 1), (2, 1), (2, 2), (2, 0), (1, 0), (0, 0), (0, 1),
C13
=
(0, 2), (1, 2), (1, 1); (1, 1), (1, 0), (2, 0), (0, 0), (0, 1), (0, 2), (1, 2),
C23
=
(2, 2), (2, 1), (1, 1); (1, 1), (0, 1), (0, 0), (0, 2), (1, 2), (2, 2), (2, 1),
C33
=
(2, 0), (1, 0), (1, 1); (1, 1), (1, 2), (0, 2), (2, 2), (2, 1), (2, 0), (1, 0), (0, 0), (0, 1), (1, 1).
The lemma is proved.
We shall rewrite the four mutually independent hamiltonian cycles in Lemma 2 to simplify the proof in Lemma 3 as follows: 3
C03 = (1, 1), P03 , (2, 2), (2, 0), P 0 , (1, 1); 3
C13 = (1, 1), P13 , (2, 0), (0, 0), P 1 , (1, 1); 3
C23 = (1, 1), P23 , (0, 0), (0, 2), P 2 , (1, 1); 3
C33 = (1, 1), P33 , (0, 2), (2, 2), P 3 , (1, 1). 117
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(0,0)
(0,0) (0,0)
(0,1)
(0,4)
N -1(4,1)
N(0,3) (0,2)
(0,3)
(0,4)
(0,0)
(1,4)
(1,0)
(0,1)
(0,2)
(0,3)
f (0,0)=(1,1)
(0,2)
f (0,2)=(1,3)
(0,4)
f (1,1)=(2,2) (1,0)
(1,4) -1
E(0,3) W -1(3,0)
W(0,3)
E (3,0)
(2,0)
(2,4)
(2,0)
(2,4)
(3,0)
(3,4)
(3,0)
(3,4)
(4,0)
(4,1)
(4,2)
(4,3)
(4,1)
(4,2)
(4,3)
(2,0)
(a)
C 03
(2,2)
f (2,0)=(3,1) (4,0)
(4,4)
S-1(4,1)
S(0,3)
(b)
f (2,2) =(3,3)
C 05
(4,3)
(4,4)
Figure 3: The hamiltonian cycle C05 of Q52 extended by C03 of Q32 .
Figure 2: In Q52 , (a) N (0, 3), S(0, 3), W (0, 3), and E(0, 3); (b) N −1 (4, 1), S −1 (4, 1), W −1 (3, 0), and E −1 (3, 0).
1. For 0 ≤ i ≤ 3, Ci5 is a cycle extended by the hamiltonian cycle Ci3 of Q32 in Lemma 2. More specifically, f (e) is an edge on Ci5 for all e ∈ E(Q32 ) ∩ Ci3 \{((2, 2), (2, 0)), ((2, 0), (0, 0)), ((0, 0), (0, 2)), ((0, 2), (2, 2))}.
Note that P03 = (1, 1), (2, 1), (2, 2) is a path of 3 length 2 between (1, 1) and (2, 2) on C03 , and P 0 = (2, 0), (1, 0), (0, 0), (0, 1), (0, 2), (1, 2), (1, 1) is a path of length 6 connecting (2, 0) and (1, 1) on C03 . 3 Similarly for Pi3 and P i , where 1 ≤ i ≤ 3.
2. C05 contains ((4, 0), (4, 4)), C15 contains ((0, 0), (4, 0)), C25 contains ((0, 4), (0, 0)), and C35 contains ((4, 4), (0, 4)).
Let 0 ≤ i ≤ j ≤ k − 1. We define eight path patterns to facilitate our derivation in the following:
Proof. By brute force, we construct the four mutually independent hamiltonian cycles in Q52 extended by Q32 as follows.
N (i, j) =
(0, i), (0, i + 1), (0, i + 2), ..., (0, j − 1), (0, j);
S(i, j) =
(k − 1, i), (k − 1, i + 1), (k − 1, i + 2)..., (k − 1, j − 1), (k − 1, j);
W (i, j) =
(i, 0), (i + 1, 0), (i + 2, 0), ..., (j − 1, 0), (j, 0);
E(i, j) =
(1,1)
C05
C15
(i, k − 1), (i + 1, k − 1), (i + 2, k − 1), ..., (j − 1, k − 1), (j, k − 1);
N −1 (j, i) =
(0, j), (0, j − 1), (0, j − 2), ..., (0, i + 1), (0, i);
S −1 (j, i) =
(k − 1, j), (k − 1, j − 1), (k − 1, j − 2), ...,
W −1 (j, i) =
(k − 1, i + 1), (k − 1, i); (j, 0), (j − 1, 0), (j − 2, 0), ..., (i + 1, 0),
E −1 (j, i) =
(i, 0); (j, k − 1), (j − 1, k − 1), (j − 2, k − 1), ..., (i + 1, k − 1), (i, k − 1).
C25
= f ((1, 1)), f (P03 ), f ((2, 2)), (4, 3), S −1 (3, 0),
(4, 0), (4, 4), E −1 (4, 0), (0, 4), N −1 (4, 0), (0, 0),
W (0, 3), (3, 0), f ((2, 0)), f (P03), f ((1, 1));
= f ((1, 1)), f (P13 ), f ((2, 0)), (3, 0), W −1 (3, 0), (0, 0), (4, 0), S(0, 4), (4, 4), E −1 (4, 0), (0, 4), N −1 (4, 1), (0, 1), f ((0, 0)), f (P13), f ((1, 1));
= f ((1, 1)), f (P23 ), f ((0, 0)), (0, 1), N (1, 4), (0, 4), (0, 0), W (0, 4), (4, 0), S(0, 4), (4, 4),
C35
E −1 (4, 1), (1, 4), f ((0, 2)), f (P23 ), f ((1, 1)); = f ((1, 1)), f (P33 ), f ((0, 2)), (1, 4), E(1, 4),
(4, 4), (0, 4), N −1 (4, 0), (0, 0), W (0, 4), (4, 0),
S(0, 3), (4, 3), f ((2, 2)), f (P33), f ((1, 1)).
Note that the underlined edges are required by the lemma. Please see Figure 3 for an illustration. Obviously, Ci5 ’s satisfy the two requirements in the lemma, where 0 ≤ i ≤ 3.
Please see Figure 2 for an illustration in Q52 . Deas follows. If x = fine a function f : Qk2 → Qk+2 2 k (i, j) ∈ V (Q2 ), then f (x)=(i + 1, j + 1) and f (x) ∈ V (Qk+2 2 ); if m ≥ 2 and P = x1 , x2 , . . . , xm is a path of Qk2 , then f (P ) = f (x1 ), f (x2 ), ..., f (xm ) is a path of Qk+2 2 .
Theorem 4 Let k ≥ 5 be an odd integer. There exist four mutually independent hamiltonian cycles in Qk2 , {C0k , C1k , C2k , C3k }, that satisfy the following requirements:
Lemma 3 There exist four mutually independent hamiltonian cycles in Q52 , {C05 , C15 , C25 , C35 }, that satisfy the following requirements:
1. For 0 ≤ i ≤ 3, Cik s are cycles extended by hamiltonian cycles Cik−2 ’s of Qk−2 2 . More specifically, f (e) is an edge on Cik for
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k−2 all e ∈ E(Qk−2 \{((k − 3, k − 2 ) ∩ Ci 3), (k − 3, 0)), ((k − 3, 0), (0, 0)), ((0, 0), (0, k − 3)), ((0, k − 3), (k − 3, k − 3))}.
∗ N −1 (k ∗ − 1, 1), (0, 1), f ((0, 0)), f (P¯1k −2 ), f ((t, t));
C2k
2. C0k contains ((k − 1, 0), (k − 1, k − 1)), C1k contains ((0, 0), (k − 1, 0)), C2k contains ((0, k − 1), (0, 0)), and C3k contains ((k − 1, k − 1), (0, k − 1)).
∗
∗
= f ((t, t)), f (P2k −2 ), f ((0, 0)), (0, 1), N (1, k ∗ − 1), (0, k ∗ − 1), (0, 0), W (0, k ∗ − 1), (k ∗ − 1, 0), S(0, k ∗ − 1), (k ∗ − 1, k ∗ − 1), E −1 (k ∗ − 1, 1), (1, k ∗ − 1), f ((0, 2t)), ∗ f (P¯ k −2 ), f ((t, t));
∗
2
∗
= f ((t, t)), f (P3k −2 ), f ((0, 2t)), (1, k ∗ − 1), E(1, k ∗ − 1), (k ∗ − 1, k ∗ − 1), (0, k ∗ − 1),
Proof. The theorem will be proved using mathematical induction. With Lemma 3, we know that the theorem holds for Qk2 for k = 5. Using the induction hypothesis, we assume the theorem holds when k = k∗ −2, where k∗ is an odd integer and k∗ ≥ 7. We want to prove that the statement is true when k = k∗ .
C3k
Suppose the four mutually independent hamilto∗ nian cycles of Qk2 −2 that satisfy the requirements are ∗ ∗ Cik −2 , 0 ≤ i ≤ 3. Namely, Cik −2 ’s satisfy the following:
It is easy to check that Cik , 0 ≤ i ≤ 3, satisfy the requirements in the theorem. Therefore, Theorem 4 is proved by the mathematical induction.
N −1 (k ∗ − 1, 0), (0, 0), W (0, k ∗ − 1), (k ∗ − 1, 0), S(0, k ∗ − 2), (k ∗ − 1, k ∗ − 2), ∗ f ((2t, 2t)), f (P¯ k −2 ), f ((t, t)). 3
∗
Lemma 5 Let k ∈ {3, 5, 7}. There exist four mutually independent hamiltonian cycles in Qk2 of the form Cik = xi (0), xi (1), xi (2), xi (3), . . . , xi (k2 − 1), xi (0), where 0 ≤ i ≤ 3, such that we can find four mutually full-independent hamiltonian paths between xi (2) and xi (3).
∗ Cik −2
is a cycle extended by (i) For 0 ≤ i ≤ 3, ∗ k ∗ −4 of Qk2 −4 . More the hamiltonian cycle Ci ∗ specifically, f (e) is an edge on Cik −4 for all ∗ ∗ e ∈ E(Qki −4 )∩Cik −4 \{((k∗ −5, k∗ −5), (k∗ − 5, 0)), ((k∗ − 5, 0), (0, 0)), ((0, 0), (0, k∗ − 5)), ((0, k∗ − 5), (k∗ − 5, k∗ − 5))}.
Proof. There are three cases. Case 1. k = 3. Let Ci3 , where 0 ≤ i ≤ 3, be the same as in Lemma 2. With the four edges ((2, 2), (2, 0)), ((2, 0), (0, 0)), ((0, 0), (0, 2)) and ((0, 2), (2, 2)), we can construct four mutually full-independent hamiltonian paths as follows:
∗
(ii) C0k −2 contains ((k∗ − 3, 0), (k∗ − 3, k∗ − 3)), ∗ ∗ C1k −2 contains ((0, 0), (k∗ − 3, 0)), C2k −2 con∗ tains ((0, k∗ − 3), (0, 0)), and C3k −2 contains ((k∗ − 3, k∗ − 3), (0, k∗ − 3)). Define t = C0k
∗
C1k
∗
C2k
∗
C3k
∗
−2 −2 −2 −2
(k ∗ −3) . 2
= (t, t), P0k
∗
= (t, t), P1k
∗
= (t, t), P2k
∗
= (t, t), P3k
∗
By (ii) above, let
−2
, (2t, 2t), (2t, 0), P¯0k , (2t, 0), (0, 0), P¯1k
∗
−2
, (0, 0), (0, 2t), P¯2k
∗
−2
, (0, 2t), (2t, 2t), P¯3k
−2
∗
−2
−2 −2 ∗
, (t, t);
, (t, t); , (t, t);
−2
, (t, t).
HP03
= (2, 2), (0, 2), (1, 2), (1, 1), (2, 1), (0, 1), (0, 0), (1, 0), (2, 0);
HP13
= (2, 0), (2, 1), (2, 2), (1, 2), (1, 0), (1, 1), (0, 1),
HP23
(0, 2), (0, 0); = (0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (1, 2), (1, 1),
HP33
(0, 1), (0, 2); = (0, 2), (1, 2), (1, 1), (0, 1), (0, 0), (1, 0), (2, 0), (2, 1), (2, 2).
When k = k∗ , we construct∗ the mutually independent hamiltonian cycles of Qk2 as follows: C0k
∗
∗
= f ((t, t)), f (P0k −2 ), f ((2t, 2t)), (k ∗ − 1, k ∗ − 2), S −1 (k ∗ − 2, 0), (k ∗ − 1, 0),(k ∗ − 1, k ∗ − 1), E −1 (k ∗ − 1, 0), (0, k ∗ − 1), N −1 (k ∗ − 1, 0), (0, 0), W (0, k ∗ − 2), (k ∗ − 2, 0), f ((2t, 0)), ∗ f (P¯ k −2 ), f ((t, t));
C1k
∗
0
= f ((t, t)), f (P1k
∗
−2
HP05
(k ∗ − 1, k ∗ − 1), E −1 (k ∗ − 1, 0), (0, k ∗ − 1),
119
= (3, 3), (3, 4), (2, 4), (2, 3), (1, 3), (1, 4), (0, 4), (0, 3), (0, 2), (0, 1), (0, 0), (1, 0), (2, 0), (3, 0), (3, 1), (2, 1), (1, 1), (1, 2), (2, 2), (3, 2), (4, 2), (4, 1), (4, 0), (4, 4), (4, 3);
), f ((2t, 0)), (k ∗ − 2, 0),
W −1 (k ∗ − 2, 0), (0, 0), (k ∗ − 1, 0), S(0, k ∗ − 1),
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Case 2. k = 5. Let Ci5 , where 0 ≤ i ≤ 3, be the same as in Lemma 3. With the four edges ((3, 3), (4, 3)), ((3, 1), (3, 0)), ((1, 1), (0, 1)) and ((1, 3), (1, 4)), we can construct four mutually full-independent hamiltonian paths as follows:
HP15
= (3, 1), (4, 1), (0, 1), (1, 1), (2, 1), (2, 0), (1, 0), (0, 0), (0, 4), (0, 3), (0, 2), (1, 2), (1, 3), (1, 4),
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(2, 4), (2, 3), (2, 2), (3, 2), (4, 2), (4, 3), (3, 3), (3, 4), (4, 4), (4, 0), (3, 0); HP25
u0 yi0(N -2)
= (1, 1), (1, 2), (1, 3), (1, 4), (2, 4), (2, 3), (2, 2), (2, 1), (3, 1), (3, 2), (3, 3), (3, 4), (4, 4), (4, 3), (4, 2), (4, 1), (4, 0), (3, 0), (2, 0), (1, 0), (0, 0), (0, 4), (0, 3), (0, 2), (0, 1);
HP35
yi1 (1) = xi1 (0)
xi2 (1)
x ik-2(1) k-2
yi0 (2)
k,0 Q n-1
(2, 2), (1, 2), (0, 2), (4, 2), (4, 3), (4, 4), (3, 4), (3, 3), (2, 3), (2, 4), (1, 4).
x ik-1(0)
yi1 (2) = xi1 (1) k-1
Ri
2 -1
(Ri )
1
Pi
Ri
xi2(N -1)
xi1(N -1)
= (1, 3), (0, 3), (0, 4), (0, 0), (1, 0), (1, 1), (0, 1), (4, 1), (4, 0), (3, 0), (2, 0), (2, 1), (3, 1), (3, 2),
k,1 Q n-1
k,2 Q n-1
xik-2(N -1)
xik-1(N -1)
k,k-2 Q n-1
k,k-1 Q n-1
Figure 4: {Ci |0 ≤ i ≤ 2n − 3} of Theorem 7, where the 2n − 2 mutually independent hamiltonian cycles for Qkn are constructed. With the induction hypothesis, we assume that the the orem holds for each Qk,k n−1 , where 0 ≤ k ≤ k − 1. k,k More precisely, let N = |V (Qn−1 )|, which equals kn−1 . Then there exist 2n − 2 mutually indepen k dent hamiltonian cycles in Qk,k n−1 . Let them be Ci = uk , yik (0), yik (1), . . . , yik (N −2), uk in Qk,k n−1 for 0 ≤ i ≤ 2n − 3. Moreover, there exist 2n − 2 mutually full-independent hamiltonian paths of the form yik (1) = xki (0), xki (1), . . . , xki (N − 1) = yik (2) in Qk,k n−1 .
Case 3. k = 7. Let Ci7 , where 0 ≤ i ≤ 3, be the same as in Theorem 4. With the four edges ((4, 4), (5, 4)), ((4, 2), (4, 1)), ((2, 2), (1, 2)) and ((2, 4), (2, 5)), we can construct four mutually full-independent hamiltonian paths by brute force. These four paths are much more lengthy and hence are skipped here. Lemma 6 Let k ≥ 7 be an odd integer. For 0 ≤ i ≤ 3, there exist four mutually independent hamiltonian cycles in Qk2 of the form Cik = xki (0), xki (1), xki (2), . . . , xki (k2 −1), xki (0) such that one can construct four mutually full-independent hamiltonian paths HPik between the four edges {(xi (2), xi (3)) | 0 ≤ i ≤ 3}.
Let R = {Rik = xki (0) = yik (1), xki (N − 1), xki (N −2), . . . , xki (1) | 0 ≤ i ≤ 2n−3}. Obviously, R is a set of mutually full-independent hamilto k k nian paths of Qk,k n−1 . By deleting xi (0) from Ri , we ¯ k = xk (N − 1), xk (N − 2), . . . , xk (1). define R i i i i ¯ k | 0 ≤ i ≤ 2n − 3} is a Then {R i set of mutually full-independent paths of Thus Ci = u0 , yi0 (0), yi0 (1), yi1 (1) = Qk,k n−1 . (0), Pik−1 , xk−1 (N − 1), x1i (0), x2i (0), . . . , xk−1 i i k−2 k−2 k−2 k−3 ¯ ¯k−3 )−1 , xi (N − 1), Ri ,xi (1), xi (1), (R i k−3 2l+1 ¯ 2l+1 , xi (N − 1), . . . , xi (N − 1), R i 2l (1), (R ¯ 2l )−1 , x2l (N − 1), . . . , x1 (N − (1),x x2l+1 i i i i i ¯ 1 , x1 (1), x0 (1) = y 1 (2), . . . , y 0 (N − 2), u0 , 1), R i i i i i where 1 ≤ l ≤ k−3 2 is an integer, is a hamiltonian cycle of Qkn . In addition, {Ci | 0 ≤ i ≤ 2n − 3} is a set of mutually independent hamiltonian cycles of Qkn . Please see Figure 4 for {Ci |0 ≤ i ≤ 2n − 3}, where the 2n − 2 mutually independent hamiltonian cycles for Qkn are constructed.
Proof. The lemma will be proved using mathematical induction. Obviously, the base case for k = 7 is proved by Lemma 5. With Theorem 4, we know that the four mutually independent hamiltonian cycles of Qk2 can be obtained using the four mutually independent hamiltonian cycles of Qk−2 2 . Therefore, the corresponding mutually full-independent hamiltonian paths between (xi (2), xi (3)) for 0 ≤ i ≤ 3 are constructed in a very similar way. The proof is skipped here. Theorem 7 Let n ≥ 2 be an integer and k ≥ 3 be an odd integer. Let i be an integer with 0 ≤ i ≤ 2n − 1. There exist 2n mutually independent hamiltonian cycles in Qkn of the form Cik = x(0), xi (1), xi (2), . . . , xi (kn − 1), x(0) such that one can find 2n mutually full-independent hamiltonian paths between xi (j) and xi (j + 1) for some 1 ≤ j ≤ kn − 2.
In every Qk,k n−1 , there is a hamiltonian cy k = uk , yik (0), yik (1), · · · , yik (N − cle C Let Lk = 3), yik (N − 2), uk . k k k k k yi (0), yi (1), yi (2), · · · , yi (N − 3), yi (N − 2) k and let L = yik (1), yik (2), · · · , yik (N − 3). Then let C2n−2 be u0 , uk−1 , yik−1 (N − 2), yik−2 (N −
Proof. The theorem will be proved using mathematical induction. With Lemma 5 and 6, we know that the theorem holds for Qk2 for any odd integer k ≥ 3 and j = 2. Note that Qkn consists of k copies of Qkn−1 , k,1 k,k−1 where n ≥ 3. Let them be Qk,0 n−1 , Qn−1 , . . . , Qn−1 . ISSN: 1790-2769
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[1] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, North-Holland, New York, u u u 1980. y (0) y (0) y (0) y (0) y (0) [2] J. Fink, Perfect Matchings Extend to Hamilton y (1) y (1) y (1) y (1) y (1) Cycles in Hypercubes, Journal of Combinatorial Theory, Series B, Vol. 97, 2007, pp. 1074–1076. k,0 k,1 k,k-2 k,k-1 k,k-3 Q n-1 Q n-1 Q n-1 Q n-1 Q n-1 [3] F. Harary, J. P. Hayes, and H.-J. Wu, A Survey of the Theory of the Hypercube Graphs, Computer k Figure 5: The hamiltonian cycle C2n−2 of Qn in TheMathematics with Applications, Vol. 15, 1988, orem 7. pp. 277–289. [4] S.-Y. Hsieh and P.-Y. Yu, Fault-free Mutually Independent Hamiltonian Cycles in Hypercubes 2), uk−1 , uk−2 , yik−2 (0), yik−3 (0), yik−4 (0), Lk−4 , yik−4 (N − with Faulty Edges, Journal of Combinatorial 2), yik−5 (N − 2), (Lk−5 )−1 , yik−5 (0), Optimization, Vol. 13, 2007, pp. 153–162. − 2), yi2l (N − yi2l+1 (0), L2l+1 , yi2l+1 (N [5] C.-H. Huang, Strongly Hamiltonian laceability − 2), (L2l )−1 , yi2l (0), · · · , yi1 (0), L1 , yi1 (N of the even k-ary n-cube, Comput. Electr. Eng., k−1 doi:10.1016/j.compeleceng. 2009.01.002. 2), yi0 (N −2), (L0 )−1 , yi0 (0), yik−1 (0), yik−1 (1), L , k−1 −1 k−2 k−1 k−2 [6] M. Kobeissi and M. Mollard, Disjoint Cycles yi (N − 3), yi (N − 3), (L ) , yi (1), k−3 k−3 and Spanning Graphs of Hypercubes, Discrete k−3 k−3 − 3), yi (N − yi (1), L , yi (N Mathematics, Vol. 288, 2004, pp. 73–87. 2), uk−3 , uk−3 , · · · , u1 , u0 where 0 ≤ l ≤ k−5 2 [7] F. T. Leighton, Introduction to Parallel Algois an integer. Please see Figure 5 for an illustration. rithms and Architecture: Arrays, Trees, Hypercubes, Morgan Kaufmann, San Mateo, CA, Finally, let C2n−1 be u0 , u1 , yi1 (0), L1 , yi0 (N − 1992. − 2), (L1 )−1 , yi0 (0), yik−1 (0), Lk−1 , yik−1 (N [8] C.-K. Lin, H.-M. Huang, L.-H. Hsu and S. Bau, 2), yik−2 (N −2), (Lk−2 )−1 , yik−2 (0), yi2l (0), L2l , yi2l (N − Mutually Independent Hamiltonian Paths in Star 2), yi2l−1 (N − 2), (L2l−1 )−1 , · · · , yi2 (0), Networks, Networks, Vol. 46, 2005, pp. 110– where L2 , yi2 (N − 2), u2 , u3 , · · · , uk−1 , u0 117. is an integer. 2 ≤ l ≤ k−3 [9] C.-K. Lin, H.-M. Huang, J. J.-M. Tan, and L.-H. 2 Hsu, Mutually Hamiltonian Cycles of the Pancake Graphs and the Star Graphs, Discrete MathWith {Ci | 0 ≤ i ≤ 2n − 1}, we obtain the 2n ematics, Vol. 309, 2009, pp. 5474–5483. mutually independent hamiltonian cycles of Qkn . The theorem is proved. [10] W.-Z. Mao, S.-Y. Hsieh and David M. Nicol, On k-ary n-cubes: theory and applications, Discrete Applied Mathematics, Vol.129, 2003, pp. 171– 193. [11] Y. Saad and M. H. Schultz, Topological Proper4 Conclusion ties of Hypercubes, IEEE Transactions on Computers, Vol. 37, 1988, pp. 867–872. Let n ≥ 1 be an integer and k ≥ 3 be an [12] Y.-K. Shih, C.-K. Lin, D. FrankHsu, Jimmy J. odd integer. There exist 2n mutually independent k M. Tan, and L.-H. Hsu, The Construction of hamiltonian cycles in Qn . Moreover, let the 2n MIH Cycles in Bubble-Sort Graphs, Internamutually independent hamiltonian cycles be C1 = n tional Journal of Computer Mathematics, acxi (0), xi (1), xi (2), . . . , xi (k − 1), xi (0), where cepted. 0 ≤ i ≤ 2n − 1. We show that there exists 2n mutu[13] C.-M. Sun, C.-K. Lin, H.-M. Huang and L.-H. ally full-independent hamiltonian paths between xi (j) Hsu, Mutually Independent Hamiltonian Paths and xi (j + 1) for some j. The result is optimal since and Cycles in Hypercubes, Journal of Interconeach vertex in Qkn has exactly 2n neighbors. nection Networks, Vol. 7, 2006, pp. 235–255. [14] C.-H. Tsai, Cycles Embedding in Hypercubes Acknowledgements: The research was supported with Node Failures, Information Processing Letby the the National Science Council of the Repubters, Vol. 102, 2007, pp. 242–246. lic of China under contract NSC 98-2115-M-033-003MY2. y i0(N -2) 0
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