Spider web networks: a family of optimal,
fault tolerant, hamiltonian bipartite graphs
Shin-Shin Kao
a,*, Lih-Hsing Hsu
baDepartment of Applied Mathematics, Chung-Yuan Christian University, Chong-Li City 320,
Taiwan, ROC
bDepartment of Computer and Information Science, National Chiao Tung University, Taiwan, ROC
Abstract
In this paper, we propose a honeycomb mesh variation, called a spider web network. Assume that m and n are positive even integers with m P 4.A spider web network SWðm; nÞ is a 3-regular bipartite planar graph with bipartition C and D.We prove that the honeycomb rectangular mesh HREMðm; nÞ is a spanning subgraph of SWðm; nÞ.We also prove that SWðm; nÞ e is hamiltonian for any e 2 E and SWðm; nÞ fc; dg re-mains hamiltonian for any c2 C and d 2 D.These hamiltonian properties are optimal. Ó 2003 Elsevier Inc.All rights reserved.
Keywords: Bipartite; 1-edge hamiltonian; 1p-hamiltonian; Optimal
1. Introduction
Throughout this paper, we assume that m; n are positive even integers with mP4.We use ½rs to denote rðmod sÞ.
Network topology is a crucial factor for an interconnection network 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.Network topology is always represented by a graph where the nodes represent processors and the edges represent the links between processors.One of the most popular architectures is mesh-con-nected computers [1].Each processor is placed into a square or rectangular grid
*
Corresponding author.
E-mail address:[email protected](S.-S. Kao).
0096-3003/$ - see front matter Ó 2003 Elsevier Inc.All rights reserved. doi:10.1016/j.amc.2003.06.005
and connected by a communication link to its neighbors in up to four direc-tions.
It is well known that there are three possible tessellations of a plane with regular polygons of the same kind: square, triangular and hexagonal, corre-sponding to dividing a plane into regular squares, triangles and hexagons, respectively.Some computer and communication networks have been built based on this observation.The square tessellation is the basis for mesh-con-nected computers.The triangular tessellation is the basis for defining hexag-onal meshed multiprocessors [2,3].The hexaghexag-onal tessellation is the basis for defining honeycombed meshes [4,5].
Stojmenovic [5] introduced three different honeycomb meshes, the honey-comb rectangular mesh, honeyhoney-comb rhombic mesh and honeyhoney-comb hexagonal mesh.Most of these meshes are not regular.Moreover, any honeycomb mesh is not hamiltonian unless it is small in size [6].To remedy these drawbacks, the honeycomb rectangular torus, honeycomb rhombic torus and honeycomb hexagonal torus are proposed [5].Any such torus is 3-regular.However, all honeycomb tori are not planar.In this paper, we propose a variation of honeycomb meshes, called a spider web network.
In the following section, we give some graph terms that are used in this pa-per and a formal definition of spider web networks.The spider graph SWðm; nÞ is a bipartite graph with bipartition C and D.Moreover, the honeycomb mesh HREMðm; nÞ forms a spanning subgraph of SWðm; nÞ.In Section 3, we prove that SWðm; nÞ e is hamiltonian for any e 2 E.In Section 4, we prove that SWðm; nÞ fc; dg remains hamiltonian for any c 2 C and d 2 D. These hamiltonian properties are optimal.A conclusion is given in the final section.
2. Spider web networks
Usually, computer networks are represented by graphs where nodes repre-sent processors and edges reprerepre-sent the links between processors.In this paper, a network is represented as an undirected graph.For the graph definition and notation, we follow [7]. G¼ ðV ; EÞ is a graph if V is a finite set and E is a subset offða; bÞjða; bÞ is an unordered pair of V g.We say that V is the node set and E is the edge set of G.Two nodes a and b are adjacent ifða; bÞ 2 E.
The honeycomb rectangular mesh HREMðm; nÞ is the graph with the node set fði; jÞj0 6 i < m; 0 6 j < ng such that ði; jÞ and ðk; lÞ are adjacent if they satisfy one of the following conditions:
1. i¼ k and j ¼ l 1;
2. j¼ l and k ¼ i þ 1 if i þ j is odd; and 3. j¼ l and k ¼ i 1 if i þ j is even.
For example, a honeycomb rectangular mesh HREMð8; 6Þ is shown in Fig.1. A spider web network SWðm; nÞ, where m, n are even integers with m P 4, nP2, is the graph with the vertex setfði; jÞj0 6 i < m; 0 6 j < ng such that ði; jÞ andðk; lÞ are adjacent if they satisfy one of the following conditions:
1. i¼ k and j ¼ l 1;
2. j¼ l and k ¼ ½i þ 1m if iþ j is odd or j ¼ n 1; and 3. j¼ l and k ¼ ½i 1m if iþ j is even or j ¼ 0.
For example, a spider graph SWð8; 6Þ is shown in Fig.2(a).Another layout of SWð8; 6Þ is shown in Fig.2(b) with the dashed lines indicating the edges of SWðm; nÞ that are not in HREMðm; nÞ.Obviously, HREMðm; nÞ is a spanning subgraph of SWðm; nÞ.The inner cycle of SWðm; nÞ is hð0; 0Þ; ð1; 0Þ; . . . ; ðm 1; 0Þ; ð0; 0Þi whereas the outer cycle of SWðm; nÞ is hð0; n 1Þ; ð1; n 1Þ; . . . ;ðm 1; n 1Þ; ð0; n 1Þi.It is obvious that any spider web network is a planar 3-regular bipartite graph.A vertex ði; jÞ is labeled black when i þ j is even and white if otherwise.
One of the major requirements of designing the network topology is a networkÕs hamiltonian properties.For example, the ‘‘token ring’’ approach is
Fig.1.HREM(8,6).
used in distributed operating systems.Fault tolerance is also desirable in massive parallel systems that have a relatively high probability of failure.
A path is a sequence of consecutive adjacent nodes.A path is usually de-limited byhx0; x1; x2; . . . ; xn1i.We use P1 to denote the path hxn1; xn2; . . . ;
x1; x0i if P is the path hx0; x1; x2; . . . ; xn1i.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 hamiltonian graph is a graph with a hamiltonian cycle.The honeycomb rectangular mesh HREMð8; 6Þ is not hamiltonian because degHREMð8;6Þð0; 0Þ ¼ 1.
A graph G¼ ðV ; EÞ is 1-edge hamiltonian if G e is hamiltonian for any e2 E.Obviously, any 1-edge hamiltonian graph is hamiltonian.A 1-edge hamiltonian graph G is optimal if it contains the least number of edges among all 1-edge hamiltonian graphs with the same number of vertices as G.A graph G¼ ðV ; EÞ is node hamiltonian if G v is hamiltonian for any v 2 V . A 1-node hamiltonian graph G is optimal if it contains the least number of edges among all 1-node hamiltonian graphs with the same number of vertices as G. A graph G¼ ðV ; EÞ is 1-hamiltonian if it is 1-edge hamiltonian and 1-node hamiltonian.A 1-hamiltonian graph G is optimal if it contains the least number of edges among all 1-hamiltonian graphs with the same number of vertices as G.The study of optimal 1-hamiltonian graphs is motivated by the design of optimal fault-tolerant token rings in computer networks.Numbers of optimal 1-hamiltonian graphs have been proposed [8–10].Obviously, degGðxÞ P 3 for any vertex x in a 1-edge hamiltonian, 1-node hamiltonian, or 1-hamiltonian graph G.
However, any bipartite graph is not 1-hamiltonian.Any cycle of a bipartite graph contains the same number of vertices in each partite set.Thus, the de-letion of a vertex from a hamiltonian bipartite graph results in a non-hamil-tonian graph.Let G be a bipartite graph with bipartition C and D.We use FðGÞ to denote ffc; dgjc 2 C; d 2 Dg.A hamiltonian bipartite graph is 1p
-hamiltonian if G F remains hamiltonian for any F 2 FðGÞ.Obviously, degGðxÞ P 3 for any vertex x in a 1p-hamiltonian graph G. A 1p-hamiltonian
graph G is optimal if it contains the least number of edges among all 1p
-hamiltonian graphs with the same number of vertices as G.
3. A recursive property of SW(m; n)
Using the definition of a spider web network, SWðm; n þ 2Þ can be constructed from SWðm; nÞ as follows: Let S denote the edge subset fðði; n 1Þ; ð½i 1m; n 1ÞÞji ¼ 0; 2; 4; . . . ; m 2g of SWðm; nÞ.Let SWðm; nÞ denote the spanning subgraph of SWðm; nÞ with edge set EðSWðm; nÞÞ S.Let
Vn¼ fði; kÞj0 6 i < m; k ¼ n; n þ 1g, and En¼ fðði; kÞ; ði; k þ 1ÞÞj0 6 i < m; k ¼
n 1; ng [ fðði; nÞ; ð½i 1m; nÞÞji ¼ 0; 2; 4; . . . ; m 2g [ fðði; n þ 1Þ; ð½i þ 1m; nþ 1ÞÞj0 6 i < mg.Then V ðSWðm; n þ 2ÞÞ ¼ V ðSWðm; nÞÞ [ Vn, EðSWðm; n þ
2ÞÞ ¼ ðEðSWðm; nÞÞ SÞ [ En.For this reason, we can view SWðm; nÞ as a
substructure of SWðm; n þ 2Þ if there is no confusion.
Let F0 V ðSWðm; nÞÞ [ EðSWðm; nÞÞ be a faulty set with jF0j 6 2, such
that F0contains an edge in EðSWðm; nÞÞ if jF0j ¼ 1 and F0 FðSWðm; nÞÞ if
jF0j ¼ 2.Suppose that C is a hamiltonian cycle of SWðm; nÞ F0, in which
ði; n 1Þ is fault free for some 0 6 i < m.Now, we are going to construct a hamiltonian cycle of SWðm; n þ 2Þ as follows:
Case 1: there is some edge in S\ EðCÞ.We can pick an edge ððr; n 1Þ; ð½r 1m; n 1ÞÞ 2 C for some even integer 0 6 r < m 1.For 0 6 i 6 m 2, we define e¼ ðð½r þ i m; n 1Þ; ð½r þ i þ 1m; n 1ÞÞ, and Qi as Qi¼ hð½r þ im; nþ 1Þ; ð½r þ i þ 1m; nþ 1Þi if ½r þ i2¼ 0; Qi¼ hð½r þ im; nþ 1Þ; ð½r þ i þ 1m; nþ 1Þi if ½r þ i2¼ 1 and e2 C; Qi¼ hð½r þ im; nþ 1Þ; ð½r þ im; nÞ; ð½r þ i þ 1m; nÞ; ð½r þ i þ 1m; nþ 1Þi if otherwise:
Then set the path Q as hðr; n þ 1Þ; Q0;ð½r þ 1m ; nþ 1Þ; Q1;ð½r þ 2m; nþ 1Þ . . .
ð½r 2m; nþ 1Þ; Qm2;ð½r 1m; nþ 1Þi.
Now we perform the following algorithm on C: Algorithm 1 (ExtendðCÞ)
1.Replace those edges ðði; n 1Þ; ð½i 1m; n 1ÞÞ 2 C, where i 6¼ r and i is
even, with the pathhði; n 1Þ; ði; nÞ; ð½i 1m; nÞ; ð½i 1m; n 1Þi.
2.Replace the edge ððr; n 1Þ; ðr 1; n 1ÞÞ with the path hðr; n 1Þ; ðr; nÞ; ðr; n þ 1Þ; Q; ðr 1; n þ 1Þ; ðr 1; nÞ; ðr 1; n 1Þi.
Obviously, the resultant of Algorithm 1 is a hamiltonian cycle of SWðm; n þ 2Þ F0.
Case 2: there is no edge in S\ EðCÞ.Obviously, ðði; n 1Þ; ði 1; n 1ÞÞ 2 C for every odd i with 1 6 i < m.The hamiltonian cycle of SWðm; n þ 2Þ F0can
be easily constructed by replacing every ðði; n 1Þ; ði 1; n 1ÞÞ, where i is odd and 1 6 i < m, with the path hði; n 1Þ; ði; nÞ; ði; n þ 1Þ; ði 1; n þ 1Þ; ði 1; nÞ; ði 1; n 1Þi.
Thus, we have the following theorem.
Theorem 3.1. Let F0 V ðSWðm; nÞÞ [ EðSWðm; nÞÞ be a faulty set with jF0j 6 2, such that F0 contains an edge in EðSWðm; nÞÞ if jF0j ¼ 1 and F0
FðSWðm; nÞÞ if jF0j ¼ 2. Suppose that ði; n 1Þ with 0 6 i < m is faulty free,
then SWðm; n þ 2Þ F0is hamiltonian if SWðm; nÞ F0is hamiltonian.
4. SWðm; nÞ is 1-edge hamiltonian
For j¼ 0 or n 1, Ijði; kÞ denotes hði; jÞ; ð½i þ 1m; jÞ; ð½i þ 2m; jÞ; . . . ; ðk; jÞi,
and I1
j ði; kÞ denotes hðk; jÞ; ð½k 1m; jÞ; ð½k 2m; jÞ; . . . ; ði; jÞi.In addition, let
Hiðj; kÞ denote the path hði; jÞ; ði; j þ 1Þ; ði; j þ 2Þ; . . . ; ði; kÞi, and Hi1ðj; kÞ ¼
hði; kÞ; ði; k 1Þ; . . . ; ði; jÞi for 0 6 i < m, 0 6 j; k < n.
Theorem 4.1. SWðm; nÞ is 1-edge hamiltonian for any even integers m, n with mP4, n P 2.
Proof. We prove this theorem by induction.We first prove SWðm; 2Þ is 1-edge hamiltonian.Let e be an edge of SWðm; 2Þ.By the symmetric property of SWðm; 2Þ, we may assume that e is either ðð0; 0Þ; ðm 1; 0ÞÞ or ðði; 0Þ; ði; 1ÞÞ with i6¼ 0; m 1.Obviously, hð0; 0Þ; I0ð0; m 1Þ; ðm 1; 0Þ; ðm 1; 1Þ; I11ð0;
m 1Þ; ð0; 1Þ; ð0; 0Þi forms a hamiltonian cycle of SWðm; 2Þ e.
Consider SWðm; 4Þ.Let e 2 EðSWðm; 4ÞÞ.There are three cases: (1) e¼ ðði; jÞ; ð½i þ 1m; jÞÞ for 0 6 i < m if j ¼ 0; 3, or i ¼ 0; 2; 4; . . . ; m 2 if j ¼ 1,
or i¼ 1; 3; . . . ; m 1 if j ¼ 2; (2) e ¼ ðði; jÞ; ði; j þ 1ÞÞ for 0 6 i < m, j ¼ 0; 2; (3) e¼ ðði; 1Þ; ði; 2ÞÞ for 0 6 i < m.In Case 1 and Case 2, we may assume that e2 EðSWðm; 2ÞÞ since the inner cycle and the outer cycle are
symmetric.Be-cause SWðm; 2Þ is 1-edge hamiltonian, there exists a hamiltonian cycle of SWðm; 4Þ e using Theorem 3.1. For Case 3, suppose e ¼ ðð0; 1Þ; ð0; 2ÞÞ using the symmetric property of SWðm; 4Þ.Let Pi¼ hði þ 1; 0Þ; ði; 0Þ; Hið0; n 1Þ;
ði; n 1Þ; ði 1; n 1Þ; H1
i1ð0; n 1Þ; ði 1; 0Þi.Obviously, hð0; 0Þ; ð0; 1Þ; ð1;
1Þ; H1ð1; 3Þ; ð1; 3Þ; ð0; 3Þ; ð0; 2Þ; ðm 1; 2Þ; ðm 1; 3Þ; ðm 2; 3Þ; Hm21 ð1; 3Þ; ðm
2; 1Þ; ðm 1; 1Þ; ðm 1; 0Þ; ðm 2; 0Þ; Pm3;ðm 4; 0Þ; Pm5;ðm 6; 0Þ; . . . ; P3;
ð2; 0Þ; ð1; 0Þ; ð0; 0Þi forms a hamiltonian cycle of SWðm; 4Þ e.Thus, SWðm; 4Þ is 1-edge hamiltonian.
By inductive hypothesis, assume SWðm; kÞ is 1-edge hamiltonian for some even integer k with k P 4.Let e be an edge of SWðm; k þ 2Þ.Since the inner
cycle and the outer cycle of SWðm; k þ 2Þ are symmetrical, we may assume that eis in SWðm; kÞ.Then there exists a hamiltonian cycle of SWðm; kÞ e.Ap-plying Theorem 3.1, SWðm; k þ 2Þ e is hamiltonian.
Hence any spider web network SWðm; nÞ is 1-edge hamiltonian.Fig.3 gives an illustration. h
5. SW(m; n) is 1p-hamiltonian
Lemma 5.1. SWðm; 2Þ is 1p-hamiltonian for m P 4.
Proof. Let F 2 FðSWðm; 2ÞÞ.By the symmetric property of SWðm; 2Þ, we may assume thatð0; 0Þ 2 F .So, the other vertex in F is ðx; yÞ, where x þ y is odd. Define two paths:
piðk; k þ 1Þ ¼ hði 1; kÞ; ði 1; k þ 1Þ; ði; k þ 1Þ; ði; kÞ; ði þ 1; kÞi;
qiðk þ 1; kÞ ¼ hði 1; k þ 1Þ; ði 1; kÞ; ði; kÞ; ði; k þ 1Þ; ði þ 1; k þ 1Þi:
To simplify the notation, pi¼ pið0; 1Þ and qi¼ qið1; 0Þ.
Suppose that y¼ 1.Then we have a hamiltonian cycle of SWðm; 2Þ F : hð1; 0Þ; ð2; 0Þ; p3;ð4; 0Þ; p5;ð6; 0Þ; . . . ; ðx; 0Þ; ðx þ 1; 0Þ; pxþ2;
ðx þ 3; 0Þ; pxþ4; . . . ;ðm 1; 0Þ; ðm 1; 1Þ; ð0; 1Þ; ð1; 1Þ; ð1; 0Þi:
Suppose that y¼ 0.There exists a hamiltonian cycle of SWðm; 2Þ F : hð0; 1Þ; ð1; 1Þ; q2;ð3; 1Þ; q4;ð5; 1Þ; . . . ; ðx; 1Þ; ðx þ 1; 1Þ; qxþ2;ðx þ 3; 1Þ; . . . ;
qm3;ðm 2; 1Þ; ðm 2; 0Þ; ðm 1; 0Þ; ðm 1; 1Þ; ð0; 1Þi:
Hence SWðm; 2Þ is 1p-hamiltonian. h
Lemma 5.2. There exist m
2 1 disjoint paths, P n 1; P2n; . . . ; Pmn 21, that span SWðm; nÞ fð0; 0Þg such that Pn l joins ð2l; n 1Þ to ð2l þ 1; n 1Þ for 1 6 l <m 2 1, and P n m 21 joinsð0; n 1Þ to ðm 2; n 1Þ.
Proof. We prove this lemma by induction.For n¼ 2, we set P2 l as hð2l; 1Þ; ð2l þ 1; 1Þi for 1 6 l <m 2 1, and set P 2 m 21 as hð0; 1Þ; ð1; 1Þ; ð1; 0Þ; I0ð1; m 1Þ; ðm 1; 0Þ; ðm 1; 1Þ; ðm 2; 1Þ.Obviously, P2
lÕs satisfy the requirement
of the lemma for 0 6 l 6m
2 1.Now assume that the lemma holds for n ¼ k,
where k is even.Then, there existm
2 1 disjoint paths, P k 1; P2k; . . . ; Pmk 21, that span SWðm; kÞ fð0; 0Þg such that Pk l joins ð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l <m 2 1, and P k m 21 joins ð0; k 1Þ to ðm 2; k 1Þ. Now, we set Pkþ2
l as hð2l; k þ 1Þ; ð2l þ 1; k þ 1Þi for 1 6 l <m2 1.Define
fi¼ hði; k 1Þ; ði; kÞ; ði þ 1; kÞ; ði þ 1; k 1Þ; Pðiþ1Þ=2k ;ði þ 2; k 1Þi and set Pmkþ2 21
hð0; k þ 1Þ; ð1; k þ 1Þ; ð1; kÞ; ð2; kÞ; ð2; k 1Þ; Pk 1;ð3; k 1Þ; f3;ð5; k 1Þ; f5;ð7; k 1Þ; . . . ; fm5;ðm 3; k 1Þ; ðm 3; kÞ; ðm 2; kÞ; ðm 2; k 2Þ; ðPk m 21Þ 1; ð0; k 1Þ; ð0; kÞ; ðm 1; kÞ; ðm 1; k þ 1Þ; ðm 2; k þ 1Þi: Pkþ2
l , 1 6 l 6m2 1, satisfies the requirement of lemma.Hence the lemma is
proved.See Fig.4(a) for an illustration. h
Lemma 5.3. Assume that r is an even integer, 0 < r 6 m 2. There exist r 2 dis-joint paths, Qn 1; Q n 2; . . . ; Q n r 2, that span SW ðm; nÞ fðr; 0Þg, such that Qn l joins ð2l; n 1Þ to ð2l þ 1; n 1Þ for 1 6 l 6r 2 1, and Q n r 2 joins ð0; n 1Þ to ðr; n 1Þ.
Proof. We prove this lemma by induction.For n¼ 2, we set Q2 l as hð2l; 1Þ; ð2l; 0Þ; ð2l þ 1; 0Þ; ð2l þ 1; 1Þi for 1 6 l 6r 2 1, and set Q 2 r 2 as hð0; 1Þ; ð1; 1Þ; ð1; 0Þ; ð0; 0Þ; ðm 1; 0Þ; ðm 1; 1Þ; q1 m2;ðm 3; 1Þ; q1m4;ðm 5; 1Þ; . . . ; q1rþ2;ðr þ 1; 1Þ; ðr; 1Þi.Obviously, Q2
lÕs satisfy the requirement of the lemma for
1 6 l 6r
2.We assume that the lemma holds for n¼ k where k is even.Then,
there exist r 2 disjoint paths, Q k l, 1 6 l 6r2, that span SW ðm; kÞ fðr; 0Þg such that Qk
ljoinsð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l <r2, and Q k r
2joinsð0; k 1Þ to
ðr; k 1Þ.
Now, we set Qkþ2l as hð2l; k þ 1Þ; ð2l þ 1; k þ 1Þi for 1 6 l <r
2.Define gi¼
hði; k 1Þ; Qk
i=2;ði þ 1; k 1Þ; ði þ 1; kÞ; ði þ 2; kÞ; ði þ 2; k 1Þi, and set Q kþ2 r 2 as:
hð0; k þ 1Þ; ð1; k þ 1Þ; ð1; kÞ; ð2; kÞ; ð2; k 1Þ; g2;ð4; k 1Þ; g4; ð6; k 1Þ; . . . ; gr2;ðr; k 1Þ; ðQkr 2Þ 1 ;ð0; k 1Þ; ð0; kÞ; ðm 1; kÞ; ðm 1; k þ 1Þ; q1 m2ðk þ 1; kÞ; ðm 3; k þ 1Þ; . . . ; q1rþ2ðk þ 1; kÞ; ðr þ 1; k þ 1Þ; ðr; k þ 1Þi: Obviously, Qkþ2
l , for 1 6 l 6r2, satisfies the requirement of lemma.Hence the
lemma is proved.See Fig.4(b) for an illustration, where r¼ 4. h Lemma 5.4. Assume that s is a positive odd integer. There existm
2 1 disjoint paths, Rn l, where 1 6 l < m 2 that span SW ðm; nÞ fðs; 1Þg such that Rn l joins ð2ðl 1Þ; n 1Þ to ð2l 1; n 1Þ for l 6¼sþ1 2 , and Rsþ12 joins ðs 1; n 1Þ to ðm 2; n 1Þ.
Proof. We prove this lemma by induction.For n¼ 2, we set Rl¼ hð2ðl 1Þ; 1Þ; ð2ðl 1Þ; 0Þ; ð2l 1; 0Þ; ð2l 1; 1Þi for 1 6 l 6s 1 2 ; Rl¼ hð2ðl 1Þ; 1Þ; ð2l 1; 1Þi for sþ 3 2 6l 6 m 2 1: Besides, R2 sþ1 2 ashðs 1; 1Þ; ðs 1; 0Þ; I0ðs 1; m 1Þ; ðm 1; 0Þ; ðm 1; 1Þ; ðm 2; 1Þi.Obviously, R2
l satisfies the requirement of the lemma for 1 6 l 6m2 1.
Now assume that the lemma holds for n¼ k where k is even.Then, there exist
m 2 1 disjoint paths, R k lÕs, that span SW ðm; kÞ fðs; 1Þg such that Rk l joins ð2ðl 1Þ; k 1Þ to ð2l 1; k 1Þ for 1 6 l <m 2; l6¼ sþ1 2, and R k sþ1 2 joins ðs 1; k 1Þ to ðm 2; k 1Þ.
Now, we set Rkþ2l ashð2ðl 1Þ; k þ 1Þ; ð2l 1; k þ 1Þi for 1 6 l <m2; l6¼sþ12 .
Define gi¼ hði; k 1Þ; Rki=2;ði þ 1; k 1Þ; ði þ 1; kÞ; ði þ 2; kÞ; ði þ 2; k 1Þi, and
set Rkþ2sþ1 2 as: hðs 1; k þ 1Þ; ðs; k þ 1Þ; ðs; kÞ; ðs þ 1; kÞ; ðs þ 1; k 1Þ; gsþ1;ðs þ 3; k 1Þ; . . . ; gm4;ðm 2; k 1Þ; ðRkðsþ1Þ=2Þ 1;ðs 1; k 1Þ; g1s3;ðs 3; k 1Þ; . . . ; g10 ;ð0; k 1Þ; ð0; kÞ; ðm 1; kÞ; ðm 1; k þ 1Þ; ðm 2; k þ 1Þi: Since Rkþ2
l , for 1 6 l 6s2, satisfies the requirement of lemma, the lemma is
Lemma 5.5. There exist m 2 1 disjoint paths, S n l, where 1 6 l < m 2 that span SWðm; nÞ fð0; 1Þg such that Sn l joins ð2l þ 2; n 1Þ to ð2l þ 3; n 1Þ for 1 6 l 6m 2 2 and S n m 21 joinsð1; n 1Þ to ð3; n 1Þ.
Proof. We prove this lemma by induction.For n¼ 2, we set S2
l ¼ hð2l þ 2; 1Þ; ð2l þ 3; 1Þi for 1 6 l 6 m 2 2, and S 2 m 21 as hð1; 1Þ; ð1; 0Þ; ð0; 0Þ; ðm 1; 0Þ; I1 0 ð2; m 1Þ; ð2; 0Þ; ð2; 1Þ; ð3; 1Þi.Obviously, S 2
lÕs satisfy the requirement of the
lemma for 1 6 l 6m
2 1.Now assume that the lemma holds for n ¼ k where k is
even.Then, there existm
2 1 disjoint paths, S k lÕs, that span SW ðm; kÞ fð0; 1Þg such that Sk l joins ð2l þ 2; k 1Þ to ð2l þ 3; k 1Þ for 1 6 l 6 m 2 2, and S k m 21 joinsð1; k 1Þ to ð3; k 1Þ.
Now, we set Slkþ2ashð2l þ 2; k þ 1Þ; ð2l þ 3; k þ 1Þi for 1 6 l 6m2 2.Define
hi¼ hði; k 1Þ; ði; kÞ; ði þ 1; kÞ; ði þ 1; k 1Þ; Si1k 2
;ði þ 2; k 1Þi, and set Skþ2 m 21 as: hð1; k þ 1Þ; ð0; k þ 1Þ; ð0; kÞ; ðm 1; kÞ; ðm 1; k 1Þ; h1 m3; ðm 3; k 1Þ; h1m5;ðm 5; k 1Þ; . . . ; h13 ;ð3; k 1Þ; ðS k m 21Þ 1; ð1; k 1Þ; ð1; kÞ; ð2; kÞ; ð2; k þ 1Þ; ð3; k þ 1Þi: Skþ2
l , 1 6 l 6m2 1, satisfies the requirement of lemma, so the lemma is proved.
See Fig.4(d) for an illustration. h
Lemma 5.6. Assume that t is an even integer, 0 < t 6 m 2. There exist m 2 1
disjoint paths, Tn
l, where 1 6 l <m2 that span SW
ðm; nÞ fðt; 1Þg such that Tn l joins ð2l; n 1Þ to ð2l þ 1; n 1Þ for 1 6 l 6m 2 1 and l 6¼ t 2, and T n t 2 joins ð1; n 1Þ to ðt þ 1; n 1Þ.
Proof. We prove this lemma by induction.For n¼ 2, we set T2 t 2 ¼ hð1; 1Þ; ð0; 1Þ; ð0; 0Þ; I0ð0; t þ 1Þ; ðt þ 1; 0Þ; ðt þ 1; 1Þi. Tl¼ hð2l; 1Þ; ð2l þ 1; 1Þi for 1 6 l 6 t 2 2 ; Tl¼ hð2l; 1Þ; ð2l; 0Þ; ð2l þ 1; 0Þ; ð2l þ 1; 1Þi for tþ 2 2 6l 6 m 2 2 : Obviously, T2
lÕs satisfy the requirement of the lemma for 1 6 l 6 m
2 1.Now
assume that the lemma holds for n¼ k where k is even.Then, there existm 2 1 disjoint paths, Tk lÕs, that span SW ðm; kÞ fðt; 1Þg such that Tk l joinsð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l 6m 2 1 and l 6¼ t 2, and T k t 2 joins ð1; k 1Þ to ðt þ 1; k 1Þ. Now, we set Tkþ2
l ashð2l; k þ 1Þ; ð2l þ 1; k þ 1Þi for 1 6 l 6m2 1, and l 6¼ t 2.
Define hi¼ hði; k 1Þ; ði; kÞ; ði þ 1; kÞ; ði þ 1; k 1Þ; Tiþ1k 2
;ði þ 2; k 1Þi, and set Ttkþ2
hð1; k þ 1Þ; ð0; k þ 1Þ; ð0; kÞ; ðm 1; kÞ; ðm 1; k 1Þ; h1 m3; ðm 3; k 1Þ; h1m5;ðm 5; k 1Þ; . . . ; h1tþ1;ðt þ 1; k 1Þ; ðTk t 2Þ 1 ; ð1; k 1Þ; h1;ð3; k 1Þ; . . . ; ht3;ðt 1; k 1Þ; ðt 1; kÞ; ðt; kÞ; ðt; k þ 1Þ; ðt þ 1; k þ 1Þi: Tlkþ2, 1 6 l 6m
2 1, satisfies the requirement of lemma, so the lemma is proved.
See Fig.4(e), where t¼ 4. h
Theorem 5.1. SWðm; nÞ is 1p-hamiltonian for any even integers m, n with
mP4, n P 2.
Proof. This theorem is proved by induction.Using Lemma 5.1, SWðm; 2Þ is 1p
-hamiltonian.Assume that SWðm; kÞ is 1p-hamiltonian for some even integer k
with k P 2.
Now, we want to prove SWðm; k þ 2Þ is 1p-hamiltonian.Let
F 2 FðSWðm; k þ 2ÞÞ.Obviously, one of the following cases holds: (1) fði; jÞj0 6 i < m; j ¼ k; k þ 1g \ F ¼ ;, (2) fði; jÞj0 6 i < m; j ¼ 0; 1g \ F ¼ ;, and (3) jfði; jÞj0 6 i < m; j ¼ k; k þ 1g \ F j ¼ 1 and jfði; jÞj0 6 i < m; j ¼ 0; 1g \ F j ¼ 1.
Case 1: fði; jÞj0 6 i < m; j ¼ k; k þ 1g \ F ¼ ;.Then F 2 FðSWðm; kÞÞ.By induction, SWðm; kÞ F is hamiltonian.Applying Theorem 3.1, SWðm; kþ 2Þ F is hamiltonian.
Case 2: fði; jÞj0 6 i < m; j ¼ 0; 1g \ F ¼ ;.Since the inner cycle and the outer cycle are symmetrical in any spider web network, SWðm; k þ 2Þ F is hamiltonian as in Case 1.
Case 3: jfði; jÞj0 6 i < m; j ¼ k; k þ 1g \ F j ¼ 1 and jfði; jÞj0 6 i < m; j ¼ 0; 1g \ F j ¼ 1.By the symmetric property of the spider web networks, we have the following five cases: (3.1) F ¼ fð0; 0Þ; ð0; k þ 1Þg, (3.2) F ¼ fðr; 0Þ; ð0; k þ 1Þg with r an non-zero even integer, (3.3) F ¼ fðs; 1Þ; ð0; k þ 1Þg with s an odd integer, (3.4) F ¼ fð0; 1Þ; ð0; kÞg, and (3.5) F ¼ fðt; 1Þ; ð0; kÞg with t an non-zero even integer.
Case (3.1): F ¼ fð0; 0Þ; ð0; k þ 1Þg.Using Lemma 5.2, there existm
2 1 dis-joint paths, Pk 1; P2k; . . . ; Pmk 21, that span SW ðm; kÞ fð0; 0Þg such that Pk l joins ð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l <m 2 1, and P k m 21 joins ð0; k 1Þ to ðm 2; k 1Þ.
Define C1ðiÞ ¼ hði; k 1Þ; ði; kÞ; ði 1; kÞ; ði 1; k 1Þ; ðPi2k 2Þ 1 ;ði 2; k 1Þi. Obviously, hð0; k 1Þ; Pk m 21;ðm 2; k 1Þ; C1ðm 2Þ; ðm 4; k 1Þ; . . . ;C1ð4Þ; ð2; k 1Þ; ð2; kÞ; ð1; kÞ; ð1; k þ 1Þ; Ikþ1ð1; m 1Þ; ðm 1; k þ 1Þ; ðm 1; kÞ; ð0; kÞ;
ð0; k 1Þi forms a hamiltonian cycle of SWðm; k þ 2Þ F .See Fig.5(a). Case (3.2): F ¼ fðr; 0Þ; ð0; k þ 1Þg.By Lemma 5.3, there exist r
2 disjoint paths, Qk 1; Qk2; . . . ; Qkr 2, that span SW ðm; kÞ fðr; 0Þg such that Qk l joins ð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l <r 2, and Q k r 2 joins ð0; k 1Þ to ðr; k 1Þ.
Define C2ðiÞ ¼ hði; k 1Þ; ði; kÞ; ði 1; kÞ; ði 1; k 1Þ; ðQki2 2
Þ1;ði 2; k 1Þi, BðiÞ hði;k þ 1Þ; ði; kÞ; ði þ 1;kÞ;ði þ 1;k þ 1Þ;ði þ 2;k þ 1Þi.Obviously, hð0;k 1Þ; Qk
r
2;ðr; k 1Þ; C2ðrÞ; ðr 2; k 1Þ; . . . ; C2ð4Þ; ð2; k 1Þ; ð2; kÞ; ð1; kÞ; ð1; k þ 1Þ;
Ikþ1ð1; r þ 1Þ; ðr þ 1; k þ 1Þ; Bðr þ 1Þ; ðr þ 3; k þ 1Þ; . . . ; Bðm 3Þ; ðm 1; k þ 1Þ; ðm 1; kÞ; ð0; kÞ; ð0; k 1Þi forms a hamiltonian cycle of SWðm; k þ 2Þ F . See Fig.5(b), where r¼ 4.
Case (3.3): F ¼ fðs; 1Þ; ð0; k þ 1Þg.By Lemma 5.4, there existm
2 1 disjoint paths, Rk 1; R k 2; . . . ; R k m 21, that span SW ðm; kÞ fðs; 1Þg such that Rk l joins ð2l 2; k 1Þ to ð2l 1; k 1Þ for 1 6 l <m 2and l6¼ sþ1 2, and R k sþ1 2 joinsðs 1; k 1Þ to ðm 2; k 1Þ.
Define C3ðiÞ ¼ hði; k 1Þ; ði; kÞ; ði 1; kÞ; ði 1; k 1Þ; ðRki 2Þ
1;ði 2;k 1Þi,
C03ðiÞ hði; k 1Þ; Rk iþ2
2
;ði þ 1; k 1Þ; ði þ 1; kÞ; ði þ 1; k þ 1Þ; ði þ 2; k þ 1Þ; ði þ 2; kÞ; ði þ 2; k 1Þi.Obviously, hðs 1; k 1Þ; Rk sþ1 2 ;ðm 2; k 1Þ; C3ðm 2Þ; ðm 4; k 1Þ; C3ðm 4Þ; ðm 6; k 1Þ . . . C3ðs þ 3Þ; ðs þ 1; k 1Þ; ðs þ 1; kÞ; ðs; kÞ; ðs; k þ 1Þ; Ikþ1ðs; m 1Þ; ðm 1; k þ 1Þ; ðm 1; kÞ; ð0; kÞ; ð0; k 1Þ; C30ð0Þ; ð2; k 1Þ; C0
3ð2Þ; ð4; k 1Þ . . . ; C30ðs 3Þ; ðs 1; k 1Þi forms a hamiltonian
cycle of SWðm; k þ 2Þ F .See Fig.5(c), where s ¼ 3.
Case (3.4): F ¼ fð0; 1Þ; ð0; kÞg.By Lemma 5.5, there exist m 2 1 disjoint paths, Sk 1; S2k; . . . ; Smk 21, that span SW ðm; kÞ fð0; 1Þg such that Sk l joins ð2l þ 2; k 1Þ to ð2l þ 3; k 1Þ for 1 6 l 6m 2 2, and S k m 21 joins ð1; k 1Þ to ð3; k 1Þ.
Define C4ðiÞ ¼ hði; k 1Þ; ði; kÞ; ði; k þ 1Þ; ði þ 1; k þ 1Þ; ði þ 1; kÞ; ði þ 1; k
1Þ; Sk i1 2 ;ði þ 2; k 1Þi.Obviously, hð1; k 1Þ; Sk m 21;ð3; k 1Þ; C4ð3Þ; ð5; k 1Þ; C4ð5Þ; ð7; k 1Þ; . . . ; C4ðm 3Þ; ðm 1; k 1Þ; ðm 1; kÞ; ðm 1; k þ 1Þ; ð0; kþ
1Þ; ð1; k þ 1Þ; ð2; k þ 1Þ; ð2; kÞ; ð1; kÞ; ð1; k 1Þi forms a hamiltonian cycle of SWðm; k þ 2Þ F .See Fig.5(d).
Case (3.5): F ¼ fðt; 1Þ; ð0; kÞg.By Lemma 5.6, there exist m
2 1 disjoint paths, Tk 1; T2k; . . . ; Tmk 21, that span SW ðm; kÞ fðt; 1Þg such that Tk l joins ð2l; k 1Þ to ð2l þ 1; k 1Þ for 1 6 l 6m 2 1 and l 6¼ t 2, and T k t 2 joins ð1; k 1Þ to ðt þ 1; k 1Þ.
Define C5ðiÞ ¼ hði; k 1Þ; ði; kÞ; ði; k þ 1Þ; ði þ 1; k þ 1Þ; ði þ 1; kÞ; ði þ 1; k 1Þ;
Tk iþ1
2
;ði þ 2; k 1Þi, and C0
5ðiÞ ¼ hði;k 1Þ;ðTi1k 2Þ
1
;ði 1;k 1Þ;ði 1;kÞ;ði 2;kÞ; ði 2; k 1Þi.Obviously, hð1; k 1Þ; Tt
2;ðt þ 1; k 1Þ; C5ðt þ 1Þ; ðt þ 3; k 1Þ;
C5ðt þ 3Þ; ðt þ 5; k 1Þ; . . . ; C5ðm 3Þ; ðm 1; k 1Þ; ðm 1; kÞ; ðm 1; k þ 1Þ;
ð0; k þ 1Þ; Ikþ1ð0; tÞ; ðt; k þ 1Þ; ðt; kÞ; ðt 1; kÞ; ðt 1; k 1Þ; C50ðt 1Þ; ðt 3; k 1Þ;
C0
5ðt 3Þ; ðt 5; k 1Þ; . . . ; C50ð3Þ; ð1; k 1Þi forms a hamiltonian cycle of
SWðm; k þ 2Þ F .See Fig.5(e), where t ¼ 4. Thus we have proved the theorem. h 6. Concluding remarks
Since the honeycomb rectangular mesh HREMðm; nÞ is a spanning subgraph of SWðm; nÞ, the spider web network can be viewed as a variation of the honeycomb mesh.The spider web networks we proposed are 3-regular planar graphs.Moreover, they are 1-edge hamiltonian and 1p-hamiltonian.Since the
spider web network is 3-regular, it is optimal.
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