Ring embedding in faulty honeycomb rectangular torus
✩Hsun-Jung Cho
∗, Li-Yen Hsu
Department of Transportation Technology and Management, National Chiao Tung University, Hsinchu, Taiwan 30050, R.O.C. Received 10 September 2001; received in revised form 31 March 2002
Communicated by M. Yamashita
Abstract
Assume that m and n are positive even integers with n 4. The honeycomb rectangular torus HReT(m, n) is recognized as another attractive alternative to existing torus interconnection networks in parallel and distributed applications. It is known that any HReT(m, n) is a 3-regular bipartite graph. We prove that any HReT(m, n)− e is hamiltonian for any edge e∈ E(HReT(m, n)). Moreover, any HReT(m, n) − F is hamiltonian for any F = {a, b} with a ∈ A and b ∈ B where A and B are the bipartition of HReT(m, n), if n 6 or m = 2.
2002 Elsevier Science B.V. All rights reserved.
Keywords: Interconnection networks; Honeycomb torus; Hamiltonian cycle; Ring embedding
1. Introduction
Network topology is a crucial factor for interconnection networks since it determines the performance of the network. Many interconnection network topologies have been proposed in the literature for the purpose of connecting a large number of processing elements [6]. Network topology is always represented by a graph where nodes represent processors and edges represent links between processors. One of the most popular architectures is the mesh connected computers [6]. Each processor is placed in a square or rectangular grid and is connected by a communication link to its neighbors up to four directions.
It is well known that there are three possible tessellations of a plane with regular polygons of the same kind: square, triangular, and hexagonal, corresponding to the division of a plane into regular squares, triangles, and hexagons, respectively. Based on this observation, some computer and communication networks have been built. The square tessellation is the basis for mesh-connected computers. The triangle tessellation is the basis for defining hexagonal mesh multiprocessors [3,11]. The hexagonal tessellation is the basis for defining the honeycomb meshes [2,10].
Tori are meshes with wraparound connections to achieve vertex and edge symmetry. Meshes and tori are among the most frequent multiprocessor networks available on the market. Stojmenovic [10] also introduced honeycomb
✩
This research was partially supported by the National Science Council of the Republic of China under contract NSC 90-2415-H-009-001.
* Corresponding author.
E-mail address: [email protected] (H.-J. Cho).
0020-0190/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved. PII: S 0 0 2 0 - 0 1 9 0 ( 0 2 ) 0 0 3 1 0 - 1
torus into generalized honeycomb torus, and some generalized honeycomb tori are proved to be hamiltonian. Throughout this paper, we assume that m and n are positive even integers with n 4. The honeycomb rectangular torus HReT(m, n) is another attractive alternative to existing torus, which is also introduced by Stojmenovic [10]. Its topological properties and its generalization are studied by Parhami and Kwai [9]. Some applications of the honeycomb rectangular torus are studied in [4]. In this paper, we study the hamiltonian properties of honeycomb rectangular torus. We will prove that the honeycomb rectangular torus HReT(m, n) is hamiltonian. Moreover, any HReT(m, n) remains hamiltonian when any edge is faulty. The honeycomb rectangular torus we proposed is a bipartite graph with bipartition A and B. Thus, any cycle of it contains the same number of vertices in each part. For this observation, we will prove that any HReT(m, n)− F , with n > 4 or m = 2, remains hamiltonian for any F= {a, b} with a ∈ A and b ∈ B.
In the following section, we give some graph terms that are used in this paper and a formal definition of honeycomb rectangular torus. In Section 3, we present a recursive property of the ring embeddings in HReT(m, n). In Section 4, we discuss the ring embedding properties of HReT(2, n). With the recursive property presented in Section 3, we can prove that any HReT(m, n) remains hamiltonian when any edge is faulty. In Section 5, we discuss the ring embedding property of HReT(4, n)− F for any F = {a, b} with a ∈ A and b ∈ B. In the final section, we discuss the ring embedding properties of any HReT(m, n)− F where F = {a, b} with a ∈ A and b ∈ B.
2. Honeycomb rectangular torus
Usually, computer and communication networks are represented by graphs where nodes represent processors and edges represent links between processors. In this paper, a network is represented as an undirected graph. For the graph definition and notation we follow [1]. G= (V, E) is a graph if V is a finite set and E is a subset of
{(a, b) | (a, b) is an unordered pair of V }. We say that V is the node set and E is the edge set of G. Two nodes
a and b are adjacent if (a, b)∈ E. A path is a sequence of nodes such that two consecutive nodes are adjacent.
A path is delimited byx0, x1, x2, . . . , xn−1 . We use P−1to denote the pathxn−1, . . . , x2, x1, x0 if P is the path
x0, x1, x2, . . . , xn−1 . A path is called a hamiltonian path if its nodes are distinct and span V . A cycle is a path
of at least three nodes such that the first node is the same as the last node. A cycle is called a hamiltonian cycle if its nodes are distinct except for the first node and the last node and if they span V . A graph is called hamiltonian if it has a hamiltonian cycle. A graph G= (V, E) is 1-edge hamiltonian if G − e is hamiltonian for any e ∈ E. A hamiltonian bipartite graph G is 1p-hamiltonian if G− F remains hamiltonian for any F = {a, b} with a ∈ A
and b∈ B where A and B are the bipartition of G.
For any two positive integers r and s, we use[r]s to denote r (mod s). We use the brick drawing, proposed
in [10], to define the honeycomb rectangular torus. The honeycomb rectangular torus HReT(m, n) is the graph with the vertex set{(i, j) | 0 i < m, 0 j < n} such that (i, j) and (k, l) are adjacent if they satisfy one of the following conditions:
(1) i= k and j = [l ± 1]n;
(2) j = l and k = [i − 1]mif i+ j is even; and
Fig. 1. The graph HReT(6, 8).
For example, the graph HReT(6, 8) is shown in Fig. 1. From the illustration, it is easy to see that HReT(m, n) is a subgraph of the torus T (m, n) [6]. Obviously, any honeycomb rectangular torus is a 3-regular bipartite graph. We set A as{(i, j) | (i, j) ∈ V (HReT(m, n)), and i +j is even} and set B as {(i, j) | (i, j) ∈ V (HReT(m, n)), and i +j is odd}. Moreover, any honeycomb rectangular torus is vertex transitive. The recursive structure of HReT(m, n) can easily be observed by inserting a pair of rows and/or a pair of columns.
Since the honeycomb rectangular torus is a bipartite graph, any spanning cycle of it contains the same number of vertices in each part. We will prove that any HReT(m, n) is 1-edge hamiltonian. Moreover, HReT(m, n) is 1p-hamiltonian if and only if n > 4 or m= 2.
To discuss the 1p-hamiltonian property of HReT(m, n), let F = {a, b} with a ∈ A and b ∈ B. We may
assume that (0, 0)∈ F because HReT(m, n) is vertex transitive. For this reason, we use F(m, n) to denote
{F |F = {(0, 0), (x, y)} | (x, y) ∈ B}. We use (x, y) to denote the unique element in F −{(0, 0)}. By the assumption,
x+ y is odd. We use P (i, j, k) to denote the path (i, j), (i, [j + 1]n)), (i,[j + 2]n)), . . . , (i, k) and use Q(i, k, j)
to denote the path P−1(i, j, k).
3. A recursive property
In this section, we use Fto denote a subset of V (HReT(m, n))∪ E(HReT(m, n)). We will present a recursive algorithm to obtain hamiltonian cycle of HReT(m, n)− F.
Assume that 0 i < m. We define a function from the vertex set of HReT(m, n) into the vertex set of HReT(m+ 2, n) by assigning fi((k, l))= (k, l) if k i and fi((k, l))= (k + 2, l) otherwise. We define fi(F) to
be the set fi(k, l)| (k, l) ∈ V HReT(m, n)∩ F ∪fi(k, l), fi(k, l) |(k, l), (k, l)∈ EHReT(m, n)∩ Fwith{k, k} =i,[i + 1]m ∪(i, l), (i+ 1, l)|(i, l), ([i + 1]m, l) ∈ EHReT(m, n)∩ F.
Let H be a hamiltonian cycle of HReT(m, n)− F such that there are some edges of H joining vertices of column i to vertices of column[i + 1]m); i.e., ((i, j ), ([i + 1]m, j )∈ E(H ) for some j. Now, we construct a
hamiltonian cycle fi(H ) of HReT(m+ 2, n) − fi(F) as follows:
Let 0 k0< k1<· · · < kt−1 n − 1 be the indices such that ((i, kj), ([i + 1]m, kj)) is an edge of H . Let Hi
be the image of H− {(i, j), ([i + 1]m, j ))| 0 j < n} under fi. For 0 j < t, we set Qj as the path (i, kj), [i + 1]m+2, kj P ([i+1]m+2,kj,[k[j+1]t−1]n) −−−−−−−−−−−−−−−−→[i + 1]m+2,[k[j+1]t − 1]n , [i + 2]m+2,[k[j+1]t− 1]n Q([i+2]m+2,[k[j+1]t−1]n,kj) −−−−−−−−−−−−−−−−→[i + 2]m+2, kj ,[i + 3]m+2, kj .
Fig. 2. (a) A hamiltonian cycle H in HReT(6, 8)− {(0, 0), (5, 0)}, (b) f5(H ), and (c) f1(H ).
Obviously, Qj is a path joining (i, kj) and ([i + 3]m+2, kj) for 0 j < t. It is easy to see that edges of Hi
together with edges of Qj, with 0 j < t, form a hamiltonian cycle of HReT(m + 2, n) − fi(F). We denote this
cycle as fi(H ). For example, a hamiltonian cycle H of HReT(6, 8)− {(0, 0), (5, 0)} is shown in Fig. 2(a). The
corresponding f5(H ) and f1(H ) are shown in Figs. 2(b) and 2(c). We have following lemmas.
Lemma 1. Assume that 0 i < m. Let H be a hamiltonian cycle of HReT(m, n) − Fsuch that there are some edges of H joining vertices of column i to vertices of column[i + 1]m. Then, fi(H ) is a hamiltonian cycle of
HReT(m+ 2, n) − fi(F). Moreover, fi(H ) contains some edges joining column t to column[t + 1]m+ 2 for any t in{i, [i + 1]m,[i + 2]m}.
Lemma 2.
(1) Suppose that H is a hamiltonian cycle of HReT(2, n)−Fsuch that H contains some edges in{((0, j), (1, j)) |
j is odd}. Then f0(H ) is a hamiltonian cycle of HReT(4, n)− f0(F). Moreover, f0(H ) contains some edges
joining column t to column t+ 1 for any t in {0, 1, 2}.
(2) Suppose that H is a hamiltonian cycle of HReT(2, n)−Fsuch that H contains some edges in{((0, j), (1, j)) |
j is even}. Then f1(H ) is a hamiltonian cycle of HReT(4, n)− f1(F). Moreover, f1(H ) contains some edges joining column t to column t+ 1 for any t in {1, 2, 3}.
We say a hamiltonian cycle of HReT(2, n)−Fis regular if H contains some edges in{((0, j), (1, j)) | j is odd} and some edges in{((0, j), (1, j)) | j is even}. Assume that m 4. A hamiltonian cycle H of HReT(m, n) − F is regular if H contains some edges joining column i to column[i + 1]mfor 0 i < m. The following lemma is
derived from the above two lemmas.
Lemma 3. Suppose that H is a regular hamiltonian cycle for HReT(m, n)− F. Then fi(H ) is a regular
hamiltonian cycle of HReT(m+ 2, n) − fi(F) for every 0 i < m.
4. Hamiltonian properties of HReT(2, n)
Obviously, (0, 0), (1, 0), (1, 1), (0, 1), (0, 2), . . ., (0, n− 2), (1, n − 2), (1, n − 1), (0, n − 1), (0, 0) and (0, 0)−−−−−−→ (0, n − 1), (1, n − 1)P (0,0,n−1) −−−−−−→ (1, 0), (0, 0)Q(1,n−1,0)
are regular hamiltonian cycles of HReT(2, n). With these two hamiltonian cycles and the symmetric property of HReT(2, n), HReT(2, n) is 1-edge hamiltonian.
Now, we discuss the 1p-hamiltonian property of HReT(2, n). Assume that F ∈ F(2, n) and that (x, y) is the
unique element in F− {(0, 0)}. Suppose that x= 0. Then
(0, 1), (0, 2), (1, 2), (1, 3), . . ., (0, y− 1), (1, y − 1), (1, y), (1, y + 1), (0, y + 1), . . . , (0, n− 3), (1, n − 3), (1, n − 2), (0, n − 2), (0, n − 1), (1, n − 1), (1, 0), (1, 1), (0, 1)
forms a hamiltonian cycle of HReT(2, n)− F . Suppose that x= 1. Then
(0, 1), (0, 2), (1, 2), (1, 3), . . ., (1, y− 1), (0, y − 1), (0, y), (0, y + 1), (1, y + 1), . . . , (0, n− 3), (1, n − 3), (1, n − 2), (0, n − 2), (0, n − 1), (1, n − 1), (1, 0), (1, 1), (0, 1)
forms a hamiltonian cycle of HReT(2, n)− F .
Lemma 4. HReT(2, n) is 1-edge hamiltonian and 1p-hamiltonian. Moreover, there exists a regular hamiltonian
cycle in HReT(2, n)− e for any e ∈ E(HReT(2, n)). Furthermore, there exists a regular hamiltonian cycle in
HReT(2, n)− F for any F ∈ F(2, n) with F = {(0, 0), (1, 0)}.
5. 1p-hamiltonian property of HReT(4, n)
We first consider the case HReT(4, 4). Suppose that F = {(0, 0), (1, 0)}. Obviously, degG−F(v)= 2 if v ∈
{(0, 1), (0, 3), (1, 1), (1, 3)}. For this reason, any hamiltonian cycle of HReT(4, 4) − F must include the following
edge set:
(0, 1), (0, 2),(0, 2), (0, 3),(0, 3), (1, 3),(1, 3), (1, 2),(1, 2), (1, 1),(1, 1), (0, 1).
However, this edge set induces a cycle of length 6. Thus, HReT(4, 4)− F is not hamiltonian.
In the following, we will prove that every HReT(4, n) with n 6 is 1p-hamiltonian. Assume t is an integer with
0 t < (14n− 1). For 0 i 2t, let Di denote the path(3, 2i), (3, 2i + 1), (2, 2i + 1), (2, 2i + 2), (1, 2i + 2), (1, 2i+ 3), (0, 2i + 3), (0, 2i + 4) .
We set Rt as the path
(3, 0)−→ (0, 4), (3, 4)D0 −→ (0, 8), (3, 8) · · ·D2 −−−→ (0, 4t)D2t−2
and set St as the path
(3, 2)−→ (0, 6), (3, 6)D1 −→ (0, 10), (3, 10) · · ·D3 −−−→ (0, 4t + 2)D2t−1 .
Let F∈ F(4, n) and let (x, y) be the unique element in F − {(0, 0)}.
Case 1: x= 0. By Lemma 4, there exists a hamiltonian cycle H of HReT(2, n) − F . By Lemma 2, f0(H ) is a
regular hamiltonian cycle of HReT(4, n)− F .
Case 2: x= 1. Assume that (x, y) = (1, 0). By Lemma 4, there exists a regular hamiltonian cycle H of
HReT(2, n)− F . By Lemma 2, f1(H ) is a regular hamiltonian cycle of HReT(4, n)− F . Suppose that (x, y) =
(1, 0). It can be checked that
(0, 1), (0, 2), (0, 3), (1, 3)−−−−−−→ (1, n − 1), (0, n − 1)P (1,3,n−1) −−−−−−→ (0, 4),Q(0,n−1,4) (3, 4)−−−−→ (3, 3), (2, 3)P (3,4,3) −−−−→ (2, 2), (1, 2), (1, 1), (0, 1)P (2,3,2)
Subcase 3.2: y= 4t + 1 for some positive integer t. Then the path (3, 0)−→ (0, 4t), (3, 4t)(3, 4t + 1), (3, 4t + 2), (0, 4t + 2)Rt S −1 t −→ (3, 2), (0, 2), (0, 1), (1, 1), (1, 0), (2, 0)−−−−−−→ (2, 4t + 4), (1, 4t + 4)Q(2,0,4t+4) −−−−−−−−→ (1, n − 1),P (1,4t+4,n−1) (0, n− 1)Q(0,n−−−−−−−−→ (0, 4t + 3), (1, 4t + 3), (1, 4t + 2), (2, 4t + 2),−1,4t+3) (2, 4t+ 3), (3, 4t + 3)−−−−−−→ (3, 0)P (3,4t+3,0)
forms a hamiltonian cycle of HReT(4, n)− F . See Fig. 3(c) for illustration.
Subcase 3.3: y= 4t + 3 for some nonnegative integer t. Then the path
(3, 0)−→ (0, 4t)Rt −→ (0, 4t + 4)D2t P (0,4t−−−−−−−−→ (0, n − 1),+4,n−1)
(1, n− 1)Q(1,n−−−−−−−−→ (1, 4t + 4), (2, 4t + 4)−1,4t+4) −−−−−−→ (2, 0), (1, 0), (1, 1),P (2,4t+4,0) (0, 1), (0, 2), (3, 2)−→ (0, 4t + 2), (3, 4t + 2)St −−−−−−→ (3, 0)P (3,4t+2,0)
forms a hamiltonian cycle of HReT(4, n)− F . See Fig. 3(d) for illustration.
Fig. 3. (a) A hamiltonian cycle H in HReT(4, 20)−{(0, 0), (1, 0)}, (b) a hamiltonian cycle H in HReT(4, 20)−{(0, 0), (2, 1)}, (c) a hamiltonian cycle H in HReT(4, 20)− {(0, 0), (2, 9)}, (d) a hamiltonian cycle H in HReT(4, 20) − {(0, 0), (2, 7)}, and (e) a hamiltonian cycle H in HReT(4, 20)− {(0, 0), (3, 0)}.
Case 4: x= 3. Assume that (x, y) = (3, 0). By Lemma 4, there exists a hamiltonian cycle H of HReT(2, n) − {(0, 0), (1, y)}. By Lemma 2, f0(H ) is a regular hamiltonian cycle of HReT(4, n)−F . Assume that (x, y) = (3, 0).
Suppose that n 8. It can be checked that
(0, 1), (0, 2), (0, 3), (1, 3), (1, 2), (2, 2), (2, 3), (2, 4), (1, 4)−−−−−−→ (1, n − 2),P (1,4,n−2) (2, n− 2)−−−−−−→ (2, 5), (3, 5)Q(2,n−2,5) −−−−−−→ (3, n − 1), (2, n − 1), (2, 0), (2, 1),P (3,5,n−1) (3, 1)−−−−→ (3, 4), (0, 4)P (3,1,4) −−−−−−→ (0, n − 1), (1, n − 1), (1, 0), (1, 1)(0, 1)P (0,4,n−1)
forms a regular hamiltonian cycle of HReT(4, n)− F . See Fig. 3(e) for illustration. Hence, we have the following lemma.
Lemma 5.
(1) HReT(4, n) is 1p-hamiltonian if and only if n 6.
(2) Suppose that n 6. There exists a regular hamiltonian cycle in HReT(m, n) − F for any F ∈ F(4, n) except
the case that F= {(0, 0), (3, 0)} and n = 6.
6. Hamiltonian properties of HReT(m, n)
Theorem 1.
(1) Any rectangular honeycomb torus HReT(m, n) is 1-edge hamiltonian. (2) HReT(m, n) is 1p-hamiltonian if and only if either n 6 or m = 2.
(3) Assume that m 4, n 6. There exists a regular hamiltonian cycle in HReT(m, n) − F for any F ∈ F(m, n)
except the case that F = {(0, 0), (m − 1, 0)} and n = 6.
Proof. With Lemma 4, there exists a regular hamiltonian cycle in HReT(2, n)− e for any e ∈ E(HReT(2, n)).
Recursively applying Lemma 3, any rectangular honeycomb torus HReT(m, n) is 1-edge hamiltonian.
Now, we discuss the 1p-hamiltonian property of HReT(m, n). Let F∈ F(m, n) and (x, y) be the unique element
in F− {(0, 0)}. By Lemma 4, HReT(2, n) is 1p-hamiltonian.
Now, we consider the case n= 4. Suppose that F = {(0, 0), (1, 0)}. Obviously, degG−F(v)= 2 if v ∈
{(0, 1), (0, 3), (1, 1), (1, 3)}. Therefore, any hamiltonian cycle of HReT(m, n) − F must include the following
edge set:
(0, 1), (0, 2),(0, 2), (0, 3),(0, 3), (1, 3),(1, 3), (1, 2),(1, 2), (1, 1),(1, 1), (0, 1).
However, this edge set induces a cycle of length 6. Thus, HReT(m, n)− F is not hamiltonian. Hence, HReT(m, n) is not 1p-hamiltonian if m 4 and n = 4.
Now, we prove that HReT(m, n) is 1p-hamiltonian if n 6. We prove the statement by induction on m. With
Lemma 5, our theorem holds for m= 4. Hence, we assume that the theorem holds for HReT(m, n) when mis any even integer with 4 m< m. Now, we consider the case that m 6.
We first consider the case that n 8. Suppose that x < m − 2. By induction, there exists a regular hamiltonian cycle H of HReT(m− 2, n) − F . By Lemma 2, fm−1(H ) is a regular hamiltonian cycle of HReT(m, n)− F .
Suppose that x m−2. By induction, there exists a regular hamiltonian cycle H of HReT(m−2, n)−{(0, 0), (x − 2, y)}. By Lemma 2, f0(H ) is a regular hamiltonian cycle of HReT(m, n)−F . Hence, the theorem holds for n 8.
Now, we consider the case that n= 6. Suppose that (x, y) is neither (m − 3, 0) nor (m − 1, 0). By induction, there exists a regular hamiltonian cycle H of HReT(m− 2, n) − F . By Lemma 2, fm−1(H ) is a regular
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