Embeddings of circulant networks

Embeddings of Circulant NetworksIndra Rajasingh1, Paul Manuel2, M. Arockiaraj1, Bharati Rajan1

1 Department of Mathematics, Loyola College, Chennai 600 034, India.2 Department of Information Science, Kuwait University, Safat, Kuwait.

Abstract. In this paper we solve the edge isoperimetric problem for circulant networks and considerthe problem of embedding circulant networks into various graphs such as arbitrary trees, cycles, certainmulticyclic graphs and ladders to yield the minimum wirelength.

Keywords: Circulant networks; multicyclic graphs; embedding; congestion; wirelength.

1 Introduction

Graph embedding has been known as a powerful tool for implementation of parallel algorithms or simu-lation of di¤erent interconnection networks. A parallel algorithm can be modeled by a task interactiongraph, where nodes and edges represent tasks and direct communications between tasks, respectively.Thus, the problem of e¢ciently executing a parallel algorithm on a parallel computer can be oftenreduced to the problem of mapping the graph , representing , on the graph , representing , sothat the communication overhead is minimized. This is called graph embedding [18], which is de…nedmore precisely as follows:Let () and () be …nite graphs with n vertices. An embedding of into is de…ned

[3] as follows:

1. is a bijective map from ()! ()2. is a one-to-one map from () to f (() ()) : (() ()) is a path in between ()and () for ( ) 2 ()g

Here is called the guest graph and , the host graph. The edge congestion of an embedding of into is the maximum number of edges of the graph that are embedded on any single edge of. Let (()) denote the number of edges ( ) of such that is in the path (() ())between () and () in . In other words,

(()) = jf( ) 2 () : 2 (() ())gjwhere (() ()) denotes the path between () and () in with respect to . For convenienceof notation we write () instead of (()) in the sequel.

0 1

2 3

4 5

6 7

f

G : H :

Figure 1: Wiring diagram of a hypercube into a cycle with () = 20. The edge congestionsare marked on the edges of .

1

If we think of as representing the wiring diagram of an electronic circuit, with the vertices represent-ing components and the edges representing wires connecting them, then the edge congestion ()is the minimum, over all embeddings : () ! (), of the maximum number of wires that crossany edge of [4].

The Wirelength Problem : The wirelength [17] of an embedding of into is given by

() =P

2() () =

P()2()

(() ())

where (() ()) denotes the length of the shortest path (() ()) in . See Figure 1. Then,the minimum wirelength of into is de…ned as

() = min ()

where the minimum is taken over all embeddings of into. The wirelength problem [3, 4, 6, 17, 18, 20]of a graph into is to …nd an embedding of into that induces the minimum wirelength().The wirelength of a graph embedding arises from VLSI designs, data structures and data representa-

tions, networks for parallel computer systems, biological models that deal with cloning and visual stimuli,parallel architecture, structural engineering and so on [16, 23].Grid embedding plays an important role in computer architecture. VLSI Layout Problem [1], Crossing

Number Problem [8], Edge Embedding Problem [12] are all a part of grid embedding. Embeddingproblems have been considered for binary trees into paths [16], complete binary trees into hypercubes[2], tori and grids into twisted cubes [15], meshes into locally twisted cubes [13], paths into twisted cubes[10], cycles into twisted cubes [11], meshes into faulty crossed cubes [25], star graph into path [24], snarksinto torus [22], generalized ladders into hypercubes [5], grids into grids [21], binary trees into grids [18],hypercubes into cycles [6], generalized wheels into arbitrary trees [20], and hypercubes into grids [17].Even though there are numerous results and discussions on the wirelength problem, most of them dealwith only approximate results and the estimation of lower bounds [3, 6]. All the embeddings discussedin this paper produce exact wirelengths.Another interesting -complete problem [12] is the edge isoperimetric problem [14] which will be

used to solve the wirelength problem. We consider the following version of the edge isoperimetric problemof a graph ().Find a subset of vertices of a given graph, such that the number of edges in the subgraph induced by

this subset is maximal among all induced subgraphs with the same number of vertices. Mathematically,for a given , if () = max

µ , jj=j()j where () = f( ) 2 : 2 g, then the problem

is to …nd µ such that jj = and () = j()j. We call such a set optimal. Further if asubset of vertices is optimal, then its complement is also an optimal set. In the literature, this problemis also de…ned as the maximum subgraph problem.

2 Background

Lemma 1 (Congestion Lemma) [17] Let be an -regular graph and be an embedding of into. Let be an edge cut of such that the removal of edges of leaves into 2 components 1 and2 and let 1 = ¡1(1) and 2 = ¡1(2) Also satis…es the following conditions:

(i) For every edge ( ) 2 , = 1 2 (() ()) has no edges in .(ii) For every edge ( ) in with 2 1 and 2 2, (() ()) has exactly one edge in .(iii) 1 is a maximum subgraph on vertices where = j (1)j.

Then () is minimum and () = ¡ 2 j(1)j. ¤

2

Lemma 2 (Partition Lemma) [17] Let : ! be an embedding. Let f1 2 g be a partitionof () such that each is an edge cut of . Then

() =

X=1

(). ¤

Lemma 3 (Generalized Partition Lemma) Let : ! be an embedding. For 1 · · ,suppose = f1 2 g partitions () n for mutually disjoint ’s such that , 1 · · ,

1 · · and =[=1

are all edge cuts of . Then

() =1

24 X=1

X=1

() + ()

35 .Proof. For 1 · · , we have () =

P=1 () + (). By summing up for all , we get

() =1P=1 (1 )+

2P=1 (2 )+ ¢ ¢ ¢+

P=1 ( )+ (1)+ (2)+ ¢ ¢ ¢+ ().

Therefore () = 1

"P=1

P=1 (

) + ()

#. ¤

Remark 1 When = 1, we obtain the Partition Lemma.

3 Edge Isoperimetric Problem for Circulant Networks

In 2004, L. H. Harper [14] quotes "In analyzing Harary graph he wished to solve its edge isoperimetricproblem". In this section we solve the maximum subgraph problem for circulant networks which is aparticular class of Harary graphs.

De…nition 1 [23] A circulant undirected graph, denoted by (;§) where µ f1 2 b2cg, ¸ 3is de…ned as a graph consisting of the vertex set = f0 1 ¡1g and the edge set = f( ) : j ¡ j ´(mod) 2 g.

The circulant graph shown in Figure ?? is (10;§f1 2 3g). It is clear that (;§1) is theundirected cycle and (;§f1 2 b2cg) is the complete graph . Further (;§f1 2 g),1 · b2c, ¸ 3 is a 2-regular graph.

Lemma 4 Let denote the cycle (;§1) in (;§f1 2 g), 1 · b2c, ¸ 3. Let 1 and 2be disjoint segments induced by 1 and 2 consecutive vertices on respectively such that 1+2 · b2c.Then [1 [2], the subgraph induced by 1 [2, contains a vertex of degree at most .

Proof. Let the complement of 1 [2 in consist of two disjoint segments 1 and 2 induced by1 and 2 consecutive vertices on respectively. Then 1 + 2 ¸ d2e. Without loss of generality, let2 · 1

2 b2c and 2 ¸ 12 d2e. This implies 2 · 2. Suppose 1 = 1 and 1 = fg. Then

deg[1[2] =

8>><>>:2 ¡ (1 + 2) : 1 2 · 0 : 1 2 ¸ ¡ 1 : 2 ¸ 1 · ¡ 2 : 1 ¸ 2 ·

3

K1 K2

D1

D2(a)

K1 K2

(b)

K1 K2

(c)

Figure 2: () 121 and 2 are disjoint segments on () 2 ¸ () 2 1, 2 1

But 2 ¡ (1 + 2) · 2 ¡ d2e · 2 ¡ = . Therefore deg[1[2] · . The same argument holdswhen 2 = 1. We now assume that 1 ¸ 2 and 2 ¸ 2. Let be the end vertices of 1 and be theend vertices of 2 taken in the clockwise sense. See Figure 2(). We claim that deg[1[2] · .

Case 1 (2 ¸ ) : In this case deg[1[2] · . See Figure 2().

Case 2 (2 ) : Let be a vertex on such that ( ) = measured in the clockwise directionand let the shortest path of length on with origin and passing through have its other end at .

Subcase 2.1 ( 2 1) : In this case 2 1. Therefore deg[1[2] · 2 ¡ (1 + 2) · .

Subcase 2.2 ( 2 1) : If lies on 1 then deg[1[2] · (2¡1)+ ¡2. But (2¡1)+ ¡2 =(2 ¡ 2) + ¡ 1 · (2 ¡ 2) + ¡ 1 . Therefore deg[1[2] . See Figure 2(). If lies on 1then deg[1[2] · (2 ¡ 1) + 1. In this case 1 ¡ 2. Therefore deg[1[2] .

Subcase 2.3 ( 2 2) : Since 2 · 2 and 2 , we have 2 . Hence this case does not occur. ¤

Lemma 5 Let denote the cycle (;§1) in (;§f1 2 g), 1 · b2c, ¸ 3. Let be asegment on induced by consecutive vertices on where · b2c. If1 and2 are disjoint segmentsinduced by 1 and 2 consecutive vertices on respectively such that 1+2 = then j([1 [2])j ·j([])j.

Proof. The proof is by induction on . Suppose = 2. Then j([])j = 1. If and are nonadjacentvertices on such that ( ) · then j([f g])j = 1 and if ( ) , then j([f g])j = 0.Thus j([f g])j · j([])j. Assume the result to be true for ¡ 1 consecutive vertices. Consider consecutive vertices on , · b2c. If · + 1 then [] is the complete graph on verticesand hence it contains the maximum number of edges. Suppose + 1. Let 1 and 2 be disjointsegments induced by 1 and 2 consecutive vertices on respectively such that 1+ 2 = . By Lemma4, [1 [2] contains an end vertex in 1 or 2 of degree at most . Without loss of generality, let be an end vertex of 1 with deg[1[2] · . Delete the vertex from 1 [2 to obtain 01 [2with ¡1 vertices. By induction hypothesis j([ 01 [2])j · j([ 0])j where 0 is induced by ¡1consecutive vertices on . Thus, j([1 [2])j = j([01 [ fg [2])j · j([ 01 [2])j + ·j([0])j+ = j([])j. ¤

Lemma 6 A set of consecutive vertices of (;§1) induces a maximum subgraph of (;§) on vertices, · b2c, = f1 2 g, 1 · b2c, ¸ 3.

4

X1

X2

XpXp-1

Xi

Xp

X A

Figure 3: jj = ¡ jj and jj =

Proof. Let the cycle (;§1) be denoted by and let be a set of consecutive vertices on . Let be a set of non-consecutive vertices on . Then =

[=1

where ¸ 2, ’s are mutually disjoint and

each is a set of consecutive vertices on such thatX=1

jj = . We claim that j([])j · j([])j.We prove this claim by induction on . When = 2, by Lemma 5, we get j([])j · j([])j.

Assume that the claim is true for ¡ 1. Then¯̄̄̄¯(

"¡1[=1

#)

¯̄̄̄¯ · j([])j where is induced by

¡ jj consecutive vertices on . Now,¯̄̄̄¯(

"[=1

#)

¯̄̄̄¯ =¯̄̄̄¯(

"¡1[=1

[#)

¯̄̄̄¯ · j([ [])j ·

j([])j. See Figure 3. ¤The following result shows that Lemma 6 is true even if the restriction on the upper bound of is

relaxed.

Theorem 1 A set of consecutive vertices of (;§1), 1 · · induces a maximum subgraph of(;§), where = f1 2 g, 1 · b2c, ¸ 3.

Proof. By Lemma 6, a set of · b2c consecutive vertices of (;§1) in (;§f1 2 g) inducesa maximum subgraph of (;§f1 2 g). Since (;§f1 2 g), 1 · b2c is a regular graph,the remaining ¡ vertices also induce a maximum subgraph of . ¤

Theorem 2 The number of edges in a maximum subgraph on vertices of (;§), = f1 2 g,1 · b2c, 1 · · , ¸ 3 is given by

=

8<: ( ¡ 1)2 ; · + 1 ¡ ( + 1)2 ; + 1 · ¡ 12f(¡ )2 + (4 + 1) ¡ (2 + 1)g ; ¡ · .

Proof. Let = f1 2 g be a set of consecutive vertices of (;§1) in (;§). By Theorem1, [] is a maximum subgraph of (;§) on vertices.

Case 1 ( · ¡ ) : The number of edges induced by in (;§), 1 · · is given by

=

½ ¡ ; 0 ; otherwise

5

0

1

2

3

4

5

6

7

0

1

2

3

4

5

6

7

0

1

2

3

4

5

6

7

(a) (b) (c)

Figure 4: Circulant graph () (8;§1) () (8;§2) () (8;§3)

See Figure 4. Therefore, j([])j =X=1

=

minf¡1gX=1

( ¡ ). This implies that

=

½( ¡ 1)2 ; · + 1 ¡ ( + 1)2 ; + 1.

Case 2 ( ¡ ) :

j([])j = j()j ¡ f2(¡ )¡ number of edges induced by (¡ ) consecutive vertices of g= ¡ 2(¡ ) + (¡ )(¡ ¡ 1)2=1

2f(¡ )2 + (4 + 1) ¡ (2 + 1)g. ¤

4 Wirelength of Circulant Networks into Arbitrary Trees

A tree is a connected graph having no cycles. Any two vertices of a tree are joined by a unique path.The most common type of tree is the binary tree. It is so named because each node can have at mosttwo descendents. A binary tree is said to be a complete binary tree if each internal node has exactly twodescendents. These descendents are described as left and right children of the parent node. The binarysearch tree property [7] of a binary tree states that all labels in the left subtree of any vertex are allless than , and all labels in the right subtree of are greater than . The consecutive label property ismotivated by the binary search tree property [20].

De…nition 2 Let be an ordered rooted tree with vertex labels 1 2 . A subtree 0 of the tree isconsecutively labelled if the labels of 0 are consecutive numbers +1 +2 + where denotes thenumber of vertices of 0.

De…nition 3 Let be an ordered rooted tree with vertex labels 1 2 . A labeling of satis…es theconsecutive label property if for every vertex of , the subtrees 1 2 rooted at are consecutivelylabeled.

Remark 2 Preorder, Inorder and Postorder labeling [19] of the vertices of trees induce consecutive labelproperty in trees.

Embedding Algorithm A

6

Input : A circulant network (;§f1 2 g), 1 · b2c and an arbitrary rooted tree on vertices.

Algorithm : Label the consecutive vertices of (;§1) in (;§f1 2 g) as 0 1 ¡ 1 in theclockwise sense. Label the vertices of the tree as 0 1 ¡ 1 using inorder labeling.

Output : An embedding of (;§f1 2 g) into given by () = with minimum wirelength.

Proof of correctness : Let be an edge of . Then ¡ yields a component 1 which is consecu-tively labeled [20]. By Theorem 1, the subgraph of (;§f1 2 g) induced by ©¡1() : 2 1ª ismaximum. By Congestion Lemma, the congestion on is minimum. This is true for every edge of .Partition Lemma implies that the wirelength is minimum.The proof of the following result is an easy consequence of Lemma 2 and of the discussion in the

proof of Theorem 2.

Theorem 3 The wirelength of (;§f1 2 g) 1 · b2c into is given by

((;§f1 2 g) ) = 2X2( )

8<:()¡minf()¡1gX

=1

f()¡ g9=;

where () is the number of vertices in the component 1 of ¡ with () · b2c. ¤

As (;§f1 2 b2cg) ' , we have the following result.

Theorem 4 The wirelength of the complete graph into is given by

( ) =X2( )

()f¡ ()g

where () is the number of vertices in the component 1 of ¡ with () · b2c. ¤

5 Wirelength of Circulant Networks into Cycle Related Graphs

In this section we consider the embedding of circulant network into cycles and certain multicyclic graphs.Here denotes a cycle on vertices.

5.1 Even and Odd Cycles

Embedding Algorithm B

Input : A circulant network (;§f1 2 g), 1 · b2c and .

Algorithm : Label the consecutive vertices of (;§1) in (;§f1 2 g) and as 0 1 ¡ 1in the clockwise sense.

Output : An embedding of (;§f1 2 g) into given by () = with minimum wirelength.

Proof of correctness :

7

01

n-1

n/2

i-1i

n/2+i-1n/2+i Si

0

14

3 2

S1

S2S2

S

S

S1

S1S2

S2

(a) (b)

1

S11

11

2

2 2

2

Figure 5: () contains two diametrically opposite edges of , even () edge cuts of 5

Case 1 ( even) : Let the edge set of be partitioned into f1 2 2g where each containstwo diametrically opposite edges of . In other words, = f(¡1 ) (2+¡1 2+)g, 1 · · 2where the labels are taken . See Figure 5(). For each , () n has two components 1and 2. Let 1 = ¡1(1) and 2 = ¡1(2). Then each , = 1 2 is on 2 consecutivevertices of (;§1). By Theorem 1, these vertices induce a maximum subgraph of (;§f1 2 g),1 · b2c. Thus each satis…es conditions (i), (ii) and (iii) of the Congestion Lemma. Therefore () is minimum. Partition Lemma implies that the wirelength is minimum.

Case 2 ( odd) : For 1 · · 2, let = f1 2 (¡1)2g where 1 = f( ¡ 1 ) (¡32 + ¡12 +)g and 2 = f( ¡ 1 ) (¡12 + +12 + )g, 1 · · ( ¡ 1)2, the labels taken . Let1 = f( ¡ 1 0)g and 2 = f(¡12 +12 )g. Then partitions () n , = 1 2. The sets 1 2 aremutually disjoint and = 1 [2 is an edge cut of . See Figure 5(). For each , () n has twocomponents 1 and

2 induced by consecutive vertices on with

¯̄1¯̄= b2c and ¯̄2 ¯̄ = d2e.

Let 1 = ¡1(1) and

2 =

¡1(2). Then 1 is on b2c consecutive vertices of (;§1). By

Theorem 1, these vertices induce a maximum subgraph of (;§f1 2 g), 1 · b2c. Thuseach satis…es conditions (i), (ii) and (iii) of the Congestion Lemma. Therefore (

) is minimum.

Similarly () is minimum. The 2-Partition Lemma implies that the wirelength is minimum.

Theorem 5 ((;§f1 2 g); 1 · b2c ) = (+1)2 .

Proof. We have already proved that the embedding de…ned in Embedding Algorithm B inducesminimum wirelength of (;§f1 2 g) onto .

Case 1 ( even) : Following the notation used in Case 1 of the algorithm, we have by Lemma 1,

(1) = 2 b2c ¡ 2X=1

(b2c ¡ ) = ( + 1). But n is isomorphic to n for 6= .

Therefore, ((;§f1 2 g) ) = 2 (1) =

(+1)2 .

Case 2 ( odd) : Following the notation used in Case 2 of the algorithm, we have by Lemma 1, () = ( + 1). But n is isomorphic to n for 1 · · 2, 1 · · ( ¡ 1)2. Therefore,((;§f1 2 g) ) = 1

2f2P=1

(¡1)2P=1

() + ()g = 1

2f( ¡ 1) () + ()g =(+1)2 . ¤

Theorem 6 ( ) = 2 b2c d2e. ¤

8

5.2 Unicyclic Graphs

A connected unicyclic graph arises from a tree by adding an extra edge. In other words, a graphwhich contains exactly one cycle : 121 is said to be a unicyclic graph, denoted by . Let1 2 be trees with roots at 1 2 respectively. Then contains vertices where = j (1)j+ j (2)j+ + j ()j.

Embedding Algorithm C

Input : A circulant network (;§f1 2 g), 1 · b2c and .

Algorithm : Label the consecutive vertices of (;§1) in (;§f1 2 g) as 0 1 ¡ 1 in theclockwise sense. Label the vertices of the unicyclic graph as follows: Label the vertex 1 as 0 and the

vertices +1, 1 · · ¡ 1 asX=1

j ()j. Label the vertices of (1)Â1 from 1 to j (1)j ¡ 1 and

the vertices of (+1)Â+1, 1 · · ¡ 1 fromX=1

j ()j + 1 toX=1

j ()j + j (+1)j ¡ 1 usinginorder labeling.

Output : An embedding of (;§f1 2 g) into given by () = with minimum wire-length.

Proof of correctness : We partition the edges of each tree as in Section 4 and the edges of the cycleas in Section 5.1. A straightforward computation yields minimum wirelength.

5.3 Multicyclic Graphs

A multicyclic graph is obtained from a tree by replacing at least two of the edges by two parallel edges andsubdividing the parallel edges to obtain paths. The multicyclic graph can also be viewed as a particularclass of series-parallel graphs. Series-parallel graphs are an important class of recursively de…ned graphsthat can be characterized in many ways. The oldest and the most popular characterization due to Du¢n[9] provides a Kuratowski-like condition which states that a graph is series-parallel if and only if itcontains no subgraph homeomorphic to 4, the complete graph on four vertices.A series–parallel graph is usually de…ned recursively by using parallel and series compositions. This

classical de…nition justi…es another name of these graphs, namely, 2-terminal series–parallel graphs, sincewe assume that every such graph has two nodes distinguished as poles and denoted by (for South) and (for North). A series–parallel graph with poles and is de…ned as either

(i) an edge ()

or can be constructed as in (ii) or (iii):

(ii) is a parallel composition of at least two series-parallel graphs 1 2 ( ¸ 2), denoted by = 1 k 2 k k . This operation identi…es the South Poles of the component graphs intothe South Pole of , and similarly the North Poles become of .

(iii) is a series composition of at least two series-parallel graphs 1 2 ( ¸ 2), denoted by = 1 ±2 ± ±. This operation identi…es and +1 for = 1 2 ¡ 1, and assigns 1to and to .

Embedding Algorithm D

9

G1 G i G l

S1 Sn/2S1 Sn/2 S1 Sn/2

S=S1 S2 Si Si+1 Sl N=Sl+1

1 2

0 3

4 5

6

7 8

9

10 11

12

13 14

15

16171819202122232425

1 1 i i l l

Figure 6: 1 and 2 contain two diametrically opposite edges of ' 6, = 5, = 6

Input : A circulant graph ((¡1)+1;§f1 2 g), 1 · j(¡1)+1

2

kand a series-parallel graph

= 1 ±2 ± ± where ' 1 · · , even and the south pole and the north pole of each are diametrically opposite vertices of .

Algorithm : Label the consecutive vertices of ((¡ 1) + 1;§1) in ((¡ 1) + 1;§f1 2 g) as0 1 (¡ 1) in the clockwise sense. Label the vertices of the series-parallel graph as follows: Withoutloss of generality, we name north pole as +1. Label south poles as

(¡1)2 , 1 · · + 1. For

each , there are two edge disjoint paths between the south pole and the north pole. For 1 · · , theinternal vertices of a path from to +1 are labeled consecutively from

(¡1)2 + 1 to 2 ¡ 1 and the

internal vertices of the other path from to +1 are labeled consecutively from(2¡+1)

2 ¡ + ¡ 1 to(2¡)2 ¡ + + 1.

Output : An embedding of ((¡1)+1;§f1 2 g), 1 · j(¡1)+1

2

kinto = 1±2±±

where ' 1 · · , even, given by () = with minimum wirelength.

Proof of correctness : As in the case of embedding circulant graph into an even cycle, let edge set of the series-parallel graph be partitioned into f : 1 · · , 1 · · 2g where each contains two diametrically opposite edges of , 1 · · , in 1 ± 2 ± ± . See Figure 6. For1 · · , 1 · · 2, () n has two components 1 and 2 induced by consecutive verticeson . Let 1 =

¡1(1) and 2 = ¡1(2). Then 1 is induced by consecutive vertices of((¡ 1) + 1;§f1 2 g). Again by Lemmas 1 and 2, induces minimum wirelength.

Theorem 7 The wirelength of ((¡1)+1;§f1 2 g), 1 · j(¡1)+1

2

kinto = 1±2±±

where ' 1 · · , even, is given by

() =

8>>>>>><>>>>>>:2

2X=1

8<:¡minf¡1gX=1

(¡ )9=; ; even

2

(¡1)2X=1

8<:¡minf¡1gX=1

(¡ )9=;+

8<: ¡minf¡1gX=1

( ¡ )9=; ; odd

where = [2(¡ 1)(¡ 1) + ] 2 and = [(¡ 1)(¡ 1) + ] 2.

Proof. Clearly, n is isomorphic to n (¡+1) for 1 · · b2c, 1 · · 2. Therefore, (

) = (

(¡+1) ).

10

Case 1 ( even) :

() =X=1

2X=1

()

= 2

2X=1

h (

1) + (

2) + ¢ ¢ ¢+ ((2))

i

= 2

2X=1

2

8<:2¡ 2minf¡1gX=1

(¡ )9=; where = [2(¡ 1)(¡ 1) + ] 2

= 2

2X=1

8<:¡minf¡1gX=1

(¡ )9=;

Case 2 ( odd) :

() =X=1

2X=1

()

= 2

(¡1)2X=1

2X=1

() +

2X=1

(( +12 ) )

= 2

(¡1)2X=1

2

8<:2¡ 2minf¡1gX=1

(¡ )9=;+ 2

8<:2 ¡ 2minf¡1gX=1

( ¡ )9=;

= 2

(¡1)2X=1

8<:¡minf¡1gX=1

(¡ )9=;+

8<: ¡minf¡1gX=1

( ¡ )9=;

where = [2(¡ 1)(¡ 1) + ] 2 and = [( ¡ 1)(¡ 1) + ] 2. ¤

Remark 3 The techniques adopted in this Section allow us to compute the exact wirelength of circulantnetworks into all classes of multicyclic graphs.

6 Wirelength of Circulant Networks into Ladders

In this section we consider the embedding of circulant network (2;§f1 2 g), 1 · into theladder 2 £ .

Embedding Algorithm E

Input : A circulant network (2;§f1 2 g), 1 · and the ladder 2 £ .

Algorithm : Label the consecutive vertices of (2;§1) in (2;§f1 2 g) as 0 1 2 ¡ 1 inthe clockwise sense. Label the vertices of the ladder as follows: The …rst row is labeled 0 to ¡ 1 fromleft to right and the second row is labeled to 2¡ 1 from right to left.

Output : An embedding of (2;§f1 2 g) into 2 £ given by () = with minimumwirelength.

11

01

n-1n

2n-1

0 n-1

n2n-1

X

Yii-1 i

2n-i

Figure 7: is a horizontal edge cut and is a vertical edge cut

Proof of correctness : Let be a horizontal edge cut of the ladder such that disconnects theladder into two components 1 and 2 where (1) = f0 1 ¡1g and (2) = f +1 2¡1g.Let be a vertical edge cut of the ladder such that disconnects the ladder into two components and 0 where () = f0 1 ¡ 1g [ f2¡ 2¡ +1 2¡ 1g and (0) = (2 £) n ().See Figure 7. Let 1 and 2 be the inverse images of 1 and 2 under this labeling. The edge cut satis…es the conditions (i) and (ii) of the Congestion Lemma. Since 1 is on consecutive vertices of(2;§1) in (2;§f1 2 g), by Theorem 1, j(1)j is maximum satisfying the condition (iii) of theCongestion Lemma. Thus by the Congestion Lemma, () is minimum. Again let and 0 be theinverse images of and 0 under this labeling. The edge cut satis…es the conditions (i) and (ii) of theCongestion Lemma. Also is induced by 2 consecutive vertices of (2;§1) in (2;§f1 2 g).Thus () is minimum for = 1 2 ¡ 1. Hence by Partition Lemma the wirelength is minimum.Theorem 8 The wirelength of (2;§f1 2 g) 1 · into 2 £ is given by

((2;§f1 2 g) 2£) =

8>>>><>>>>:( + 1) + (2 ¡ 1)¡ 4

(¡1)2X=1

minf2¡1gX=1

(2¡ ) ; odd

(+ 2)¡ 42X=1

minf2¡1gX=1

(2¡ ) ; even.

Proof. We have () = 2¡ 2X=1

(¡ ) = ( + 1).

Case 1 ( odd) : Clearly, () = (¡) for 1 · · (¡ 1)2.

Therefore,¡1X=1

() = 2

(¡1)2X=1

()

= 2

8<:4 ¡ 2minf1gX=1

(2¡ ) + 8 ¡ 2minf3gX=1

(4¡ ) + ¢ ¢ ¢+ 2(¡ 1)¡ 2minf¡2gX

=1

(¡ 1¡ )9=;

= (2 ¡ 1)¡ 4(¡1)2X=1

minf2¡1gX=1

(2¡ ).

Thus, ((2;§f1 2 g) 1 · 2£) = (+1)+(2¡1)¡4(¡1)2X=1

minf2¡1gX=1

(2¡ ).

Case 2 ( even) : Clearly, () = (¡) for 1 · · 2¡ 1.

12

Therefore,¡1X=1

() =

2¡1X=1

() + (2) +¡1X

=2+1

()

= 2

8<:4 ¡ 2minf1gX=1

(2¡ ) + 8 ¡ 2minf3gX=1

(4¡ ) + ¢ ¢ ¢+ 2(¡ 2)¡ 2minf¡3gX

=1

(¡ 2¡ )9=;

+2¡ 2X=1

(¡ )

= 2

8<:4 ¡ 2minf1gX=1

(2¡ ) + 8 ¡ 2minf3gX=1

(4¡ ) + ¢ ¢ ¢+ 2¡ 2minf¡1gX

=1

(¡ )9=;

¡2+ 2X=1

(¡ )

= (+ 2)¡ 42X=1

minf2¡1gX=1

(2¡ )¡ ( + 1).

Thus, ((2;§f1 2 g) 1 · 2 £ ) = (+ 2)¡ 42X=1

minf2¡1gX=1

(2¡ ). ¤

7 Conclusion

We obtain the exact wirelength of circulant networks into arbitrary trees, certain multicyclic graphs andladders. All the embeddings constructed in this paper are simple, elegant and produce exact wirelengths.It would be an interesting line of research to solve the following problem.

Open Problem : To …nd the exact wirelength of circulant networks into grids.

References

[1] S. N. Bhatt and F. T. Leighton, A framework for solving VLSI graph layout problems, J. Computerand System Sciences, 28 (1984), 300 - 343.

[2] S. L. Bezrukov, Embedding complete trees into the hypercube, Discrete Applied Mathematics, 110(2001), 101 - 119.

[3] S. L. Bezrukov, J. D. Chavez, L. H. Harper, M. Röttger and U. P. Schroeder, Embedding of hyper-cubes into grids, MFCS, 1998, 693 - 701.

[4] S. L. Bezrukov, J. D. Chavez, L. H. Harper, M. Röttger and U. P. Schroeder, The congestion of-cube layout on a rectangular grid, Discrete Mathematics, 213 (2000), 13 - 19.

[5] R. Caha and V. Koubek, Optimal embeddings of generalized ladders into hypercubes, DiscreteMathematics, 233 (2001), 65 - 83.

[6] J. D. Chavez and R. Trapp, The cyclic cutwidth of trees, Discrete Applied Mathematics, 87 (1998),25 - 32.

[7] T. H. Cormen, C. E. Leiserson, R. L. Rivest and C. Stein, Introduction to algorithms, MIT Pressand McGraw-Hill, New York, 2001.

13

[8] H. N. Djidjev and I. Vrto, Crossing numbers and cutwidths, Journal of Graph Algorithms andApplications, 7 (2003), 245 - 251.

[9] R. J. Du¢n, Topology of series - parallel networks, J. Math. Analysis Appl., 10 (1965), 303 - 318.

[10] J. Fan, X. Jia and X. Lin, Optimal embeddings of paths with various lengths in twisted cubes, IEEETrans. Parallel Distrib. Syst, 18 (2007), no.4, 511 - 521.

[11] J. Fan, X. Jia and X. Lin, Embedding of cycles in twisted cubes with edge-pancyclic, Algorithmica,51 (2008), no.3, 264 - 282.

[12] M. R. Garey and D. S. Johnson, Computers and Intractability, A Guide to the Theory of NP-Completeness, Freeman, San Francisco 1979.

[13] Y. Han, J. Fan, S. Zhang, J. Yang and P. Qian, Embedding meshes into locally twisted cubes,Information Sciences, 180 (2010), no.19, 3794 - 3805.

[14] L. H. Harper, Global Methods for Combinatorial isoperimetric Problems, Cambridge UniversityPress, 2004.

[15] P.-L. Lai and C.-H. Tsai, Embedding of Tori and Grids into Twisted Cubes, Theoretical ComputerScience, 411 (2010), no.40-42, 3763 - 3773.

[16] Y. L. Lai and K. Williams, A survey of solved problems and applications on bandwidth, edgesum,and pro…le of graphs, J. Graph Theory, 31 (1999), 75 - 94.

[17] P. Manuel, I. Rajasingh, B. Rajan and H. Mercy, Exact wirelength of hypercube on a grid, DiscreteApplied Mathematics, 157 (2009), no. 7, 1486 - 1495.

[18] J. Opatrny and D. Sotteau, Embeddings of complete binary trees into grids and extended grids withtotal vertex-congestion 1, Discrete Applied Mathematics, 98 (2000), 237 - 254.

[19] J. Quadras, Embeddings and interconnection networks, Ph.D. dissertation, University of Madras,India, 2005.

[20] I. Rajasingh, J. Quadras, P. Manuel and A. William, Embedding of cycles and wheels into arbitrarytrees, Networks, 44 (2004), 173 - 178.

[21] M. Rottger and U. P. Schroeder, E¢cient embeddings of grids into grids, Discrete Applied Mathe-matics, 108 (2001), no. 1-2, 143 - 173.

[22] A. Vodopivec, On Embeddings of snarks in the torus, Discrete Mathematics, 308 (2008), no.10, 1847- 1849.

[23] J. -M. Xu, Topological structure and analysis of interconnection networks, Kluwer Academic Pub-lishers, 2001.

[24] M.-C. Yang, Path embedding in star graphs, Applied Mathematics and Computation, 207 (2009),no.2, 283 - 291.

[25] X. Yang, Q. Dong and Y. Y. Tang, Embedding meshes/tori in faulty crossed cubes, InformationProcessing Letters, 110 (2010), no.14-15, 559 - 564.

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