embedding meshes into locally twisted cubes

Information Sciences 180 (2010) 3794–3805

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Information Sciences

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Embedding meshes into locally twisted cubes

Yuejuan Han a,b, Jianxi Fan a,*, Shukui Zhang a, Jiwen Yang a,b, Peide Qian a

a School of Computer Science and Technology, Soochow University, Chinab Network Center, Soochow University, Suzhou 215006, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 22 March 2009Received in revised form 6 May 2010Accepted 3 June 2010

Keywords:Locally twisted cubeMeshEmbeddingDilationExpansionParallel computing system

0020-0255/$ - see front matter � 2010 Elsevier Incdoi:10.1016/j.ins.2010.06.001

* Corresponding author.E-mail address: [email protected] (J. Fan).

As a newly introduced interconnection network for parallel computing, the locally twistedcube possesses many desirable properties. In this paper, mesh embeddings in locallytwisted cubes are studied. Let LTQn(V,E) denote the n-dimensional locally twisted cube.We present three major results in this paper: (1) For any integer n P 1, a 2 � 2n�1 meshcan be embedded in LTQn with dilation 1 and expansion 1. (2) For any integer n P 4, twonode-disjoint 4 � 2n�3 meshes can be embedded in LTQn with dilation 1 and expansion2. (3) For any integer n P 3, a 4 � (2n�2 � 1) mesh can be embedded in LTQn with dilation2. The first two results are optimal in the sense that the dilations of all embeddings are 1.The embedding of the 2 � 2n�1 mesh is also optimal in terms of expansion. We also presentthe analysis of 2p � 2q mesh embedding in locally twisted cubes.

� 2010 Elsevier Inc. All rights reserved.

1. Introduction

Interconnection networks play an important role in parallel computing systems. An interconnection network can be rep-resented by a graph G = (V,E), where V represents the node set and E represents the edge set.

One of the critical performance factors of interconnection network is how well other existing networks can be embeddedinto this one. This can be modeled by the following graph embedding problem: given a host graph G2 = (V2,E2) and a guestgraph G1 = (V1,E1), which represent the network where other networks are to be embedded and the network to be embedded,respectively. The problem becomes finding an injective mapping from each node of G1 to a node of G2, and a mapping fromeach edge of G1 to a path in G2. Two common measures of how effective an embedding is are dilation and expansion. Thedilation of an embedding w is defined as

dilðG1;G2;wÞ ¼ maxfdistðG2;wðuÞ;wðvÞÞjðu;vÞ 2 E1g;

where dist(G2,w(u),w(v)) denotes the distance between the two nodes w(u) and w(v) in G2. The smaller the dilation of anembedding is, the shorter the communication delay is when G2 simulates G1 [1]. The expansion of an embedding is definedas

expðG1;G2;wÞ ¼ jVðG2Þj=jVðG1Þj;

which measures the processor utilization. The smaller the expansion of an embedding is, the more efficient processor utili-zation is when G2 simulates G1. Obviously, exp(G1,G2,w) P 1. Graph embedding has important applications in transplantingparallel algorithms developed for one network to a different one, and allocating concurrent processes to processors in the

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Y. Han et al. / Information Sciences 180 (2010) 3794–3805 3795

network. As the common interconnection structures used in parallel computing, meshes [8,10,11,13,16], paths [5–7] and cy-cles [2,4,9] are the fundamental guest graphs.

The hypercube network Qn has been proved to be one of the most popular interconnection networks. The locally twistedcube LTQn is a variant of Qn, proposed by Yang et al. [14]. It has many attractive features as compared to the hypercube, forinstance, the diameter is only about half of that of Qn. In particular, Yang et al. [15] showed that LTQn is Hamiltonian con-nected and contains cycles of lengths from 4 to 2n for any integer n P 3. Furthermore, for any integer n P 3, LTQn was provedto be (n � 2)-hamiltonian, (n � 3)-hamiltonian connected [12] and (n � 2)-pancyclic [3].

In this paper, mesh embeddings in locally twisted cubes are studied. We present three major results in this paper:

(1) For any integer n P 1, a 2 � 2n�1 mesh can be embedded in LTQn with dilation 1 and expansion 1.(2) For any integer n P 4, two node-disjoint 4 � 2n�3 meshes can be embedded in LTQn with dilation 1 and expansion 2.(3) For any integer n P 3, a 4 � (2n�2 � 1) mesh can be embedded in LTQn with dilation 2.

The rest of this paper is organized as follows. Section 2 provides the definitions and notations used in the paper. We pro-vide the proofs of the results in Section 3, Sections 4 and 5. Section 6 is the discussion on 2p � 2q mesh embedding. In theend, we summarize the paper in Section 7.

2. Preliminaries

Notation 1. Given two integers m and n, an m � n mesh M can be denoted by an m � n matrix

a11 a12 . . . a1n

a21 a22 . . . a2n

. . . . . . . . . . . .

am1 am2 . . . amn

0BBB@

1CCCA;

where V(M) = {aijj1 6 i 6m, and 1 6 j 6 n}, (aij,ai, j+1) 2 E(M) for any integers i, j with 1 6 i 6m and 1 6 j 6 n � 1, and(akl,ak+1,l) 2 E(M) for any integers k, l with 1 6 k 6m � 1 and 1 6 l 6 n.h a11,a12, . . .,a1ni and h am1,am2, . . .,amni are calledthe row-borders; ha11,a21, . . .,am1 i and ha1n,a2n, . . .,amni are called the column-borders.

A binary string x of length n is denoted by x1x2 . . .xn�1xn, where x1 is the most significant bit and xn is the least significantbit. The ith bit xi of x can also be written as bit (x, i). Suppose that z is a binary string of length k. zi denotes the new binarystring by repeating z string i times. If i = 0,zi denotes the empty string.

Similar to the n-dimensional hypercube, the n-dimensional locally twisted cube LTQn is an n-regular graph of 2n nodes.Every node of LTQn is identified by a unique binary string of length n. LTQn can be recursively defined as follows.

Definition 1 [14]. For n P 2, an n-dimensional locally twisted cube, LTQn, is defined recursively as follows:

(1) LTQ2 is a graph consisting of four nodes labeled with 00, 01, 10, and 11, respectively, connected by four edges (00, 01),(00, 10), (01, 11), and (10, 11).

(2) For n P 3, LTQn is built from two disjoint copies of LTQn�1 with the following steps. Let LTQ0n�1 denote the graph

obtained by prefixing the label of each node of one copy of LTQn�1 with 0, and LTQ 1n�1 denote the graph obtained

by prefixing the label of each node of the other copy of LTQn�1 with 1. Connect each node x = 0x2x3. . .xn of LTQ 0n�1

to the node 1(x2 + xn)x3. . .xn of LTQ 1n�1 with an edge, where 0 + 0 represents the modulo 2 addition.

Figs. 1 and 2 demonstrate LTQ3, LTQ4 and LTQ5.

Fig. 1. (a) The 3-dimensional locally twisted cube LTQ3; (b) the 4-dimensional locally twisted cube LTQ4.

Fig. 2. The 5-dimensional locally twisted cube LTQ5.

3796 Y. Han et al. / Information Sciences 180 (2010) 3794–3805

3. Embedding a 2 � 2n�1 mesh into the n-dimensional locally twisted cube

In this section, we discuss the 2 � 2n�1 mesh embedding in the n-dimensional locally twisted cube by induction on itsdimension n. For any integer n 2 {3, 4, 5}, according to Definition 1, we can easily verify lemmas 1, 2 and 3 as follows.

Lemma 1. There is a 2 � 4 mesh in LTQ3.

Proof. Obviously,

000 001 111 110010 011 101 100

� �;

is a 2 � 4 mesh in LTQ3. h

Lemma 2. There is a 2 � 8 mesh in LTQ4, whose column-borders are h0000,0010i and h1000,1010i.

Proof. Obviously,

0000 0001 0111 0110 1110 1111 1001 10000010 0011 0101 0100 1100 1101 1011 1010

� �;

is a 2 � 8 mesh in LTQ4, whose column-borders are h0000,0010i and h1000,1010i. h

Lemma 3. There is a 2 � 16 mesh in LTQ5, whose column-borders are h00000,00010i and h10000,10010i.

Proof. Let

a1 a2 � � � a8

b1 b2 � � � b8

� �;

be a 2 � 8 mesh in LTQ4, where a1 = 0000, b1 = 0010, a8 = 1000, b8 = 1010. Then,


M0 ¼0a1 0a2 � � � 0a8

0b1 0b2 � � � 0b8

� �and M1 ¼

1a1 1a2 � � � 1a8

1b1 1b2 � � � 1b8

� �;

are 2 � 8 meshes in LTQ04 and LTQ 1

4, respectively. Clearly, V(M0)T

V(M1) = ;.According to Definition 1, we have h0a8,1a8 i = h01000,11000i 2 E(LTQ5) and h 0b8,1b8i = h01010,11010i 2 E(LTQ5).

Therefore,

0a1 0a2 � � � 0a8 1a8 � � � 1a2 1a1

0b1 0b2 � � � 0b8 1b8 � � � 1b2 1b1

� �

is a 2 � 16 mesh in LTQ5, whose column-borders are h0a1,0b1i = h00000,00010i and h1a1,1b1i = h10000,10010 i. h

We will prove the main lemma as follows.

Lemma 4. For any integer n P 4, there is a 2 � 2n�1 mesh in LTQn, whose column-borders are h00n�300,00n�310 i andh10n�300,10n�310i.

Proof. We prove this lemma by induction on the dimension n of LTQn. According to Lemmas 2 and 3, this lemma holds whenn = 4,5. Supposing that the lemma holds for n = s(s P 4), we will prove that the lemma holds for n = s + 1.

According to the induction hypothesis, for any integer s P 4,

M1 ¼a1 a2 � � � a2s�1

b1 b2 � � � b2s�1

� �and M2 ¼

a2s�1 � � � a2 a1

b2s�1 � � � b2 b1

� �;

are the same 2 � 2s�1 meshes in LTQs with different notations, where a1 ¼ 00s�300; b1 ¼ 00s�310;a2s�1 ¼ 10s�300 andb2s�1 ¼ 10s�310. Then,

M01 ¼

0a1 0a2 � � � 0a2s�1

0b1 0b2 � � � 0b2s�1

� �and M1

2 ¼1a2s�1 � � � 1a2 1a1

1b2s�1 � � � 1b2 1b1

� �;

are two 2 � 2s�1 meshes in LTQ0s and LTQ 1

s , respectively. Obviously, VðM01ÞT

VðM12Þ ¼ ;.

According to Definition 1, we have ð0a2s�1 ;1a2s�1 Þ ¼ ð010s�300;110s�300Þ 2 EðLTQsþ1Þ and ð0b2s�1 ;1b2s�1 Þ ¼ ð010s�310;110s�310Þ 2 EðLTQsþ1Þ. Therefore,

0a1 0a2 � � � 0a2s�1 1a2s�1 � � � 1a2 1a1

0b1 0b2 � � � 0b2s�1 1b2s�1 � � � 1b2 1b1

� �;

is a 2 � 2(s + 1)�1 mesh in LTQs + 1, whose column-borders are h0a1,0b1i = h 00(s + 1)�300,00(s + 1)�310i and h 1a1,1b1i =h 10(s + 1)�300,10(s + 1)�310i. Hence, n = s + 1 holds. h

Based on these lemmas, we prove the first result in this paper as follows.

Theorem 1. For any integer n P 1, there is a 2 � 2n�1 mesh in LTQn.

Proof. The theorem holds for n=1 and 2, clearly. According to Lemma 1, it holds for n = 3. According to Lemma 4, it holds forn P 4. Therefore, the theorem holds. h

According to Theorem 1, we have the following corollary.

Corollary 1. For any integer n P 1, a 2 � 2n�1 mesh can be embedded with dilation 1 and expansion 1 in LTQn.

4. Embedding two 4 � 2n�3 meshes into the n-dimensional locally twisted cube

An m � n mesh denoted by an m � n matrix in Notation 1 can also be called an n �m mesh. a b c de f g h

� �is a 2 � 4

mesh in Section 3. In this Section, to express 4 � 2n�3(n P 4) meshes more clearly, we also use a b c de f g h

� �to represent

a 4 � 2 mesh.In this section, the 4 � 2n�3 mesh embedding in the n-dimensional locally twisted cube is discussed. Similar to the 2

� 2n�1 mesh embedding as discussed in the last section, the 4 � 2n�3 mesh embedding is also achieved by induction on n.According to Definition 1, we can easily verify the following three lemmas.

Lemma 5

ð1Þ M1 ¼0001 0011 1111 11010111 0101 1001 1011

� �and M2 ¼

1111 1101 0001 00111001 1011 0111 0101

� �;

are two 4 � 2 meshes in LTQ4 with the same eight nodes labeled by strings ended with 1.


ð2Þ M3 ¼0000 0010 1010 10000100 0110 1110 1100

� �;

is a 4 � 2 mesh in LTQ4 with the nodes labeled by strings ended with 0.

Lemma 6

ð1Þ M1 ¼

00001 00011 01111 0110100111 00101 01001 0101111111 11101 10001 1001111001 11011 10111 10101

0BBB@

1CCCA and M2 ¼

01001 01011 00111 0010101111 01101 00001 0001110111 10101 11001 1101110001 10011 11111 11101

0BBB@

1CCCA;

are two 4 � 4 meshes in LTQ5 with the same sixteen nodes labeled by strings ended with 1.

ð2Þ M3 ¼

00000 00010 01010 0100000100 00110 01110 0110010100 10110 11110 1110010000 10010 11010 11000

0BBB@

1CCCA;

is a 4 � 4 mesh in LTQ5 with the nodes labeled by strings ended with 0.

Lemma 7

ð1Þ M1 ¼

000001 000011 001111 001101000111 000101 001001 001011011111 011101 010001 010011011001 011011 010111 010101101001 101011 100111 100101101111 101101 100001 100011110111 110101 111001 111011110001 110011 111111 111101

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

and M2 ¼

010001 010011 011111 011101010111 010101 011001 011011001111 001101 000001 000011001001 001011 000111 000101111001 111011 110111 110101111111 111101 110001 110011100111 100101 101001 101011100001 100011 101111 101101

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

;

are two 4 � 8 meshes in LTQ6 with the same thirty-two nodes labeled by strings ended with 1.

ð2Þ M3 ¼

000000 000010 001010 001000000100 000110 001110 001100010100 010110 011110 011100010000 010010 011010 011000110000 110010 111010 111000110100 110110 111110 111100100100 100110 101110 101100100000 100010 101010 101000

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

;

is a 4 � 8 mesh in LTQ6 with the nodes labeled by strings ended with 0.With these lemmas we have the following two theorems.

Theorem 2. For any integer n P 6, there are two 4 � 2n�3 meshes with the same nodes in LTQn:

M1 ¼

a1 b1 c1 d1

a2 b2 c2 d2

. . . . . . . . . . . .

a2n�3 b2n�3 c2n�3 d2n�3

0BBB@

1CCCA and M2 ¼

a01 b01 c01 d01a02 b02 c02 d02. . . . . . . . . . . .

a02n�3 b02n�3 c0

2n�3 d02n�3

0BBBB@

1CCCCA;

where the row-borders of M1 are ha1,b1,c1,d1i = h000n�60001,000n�60011, 000n�61111,000n�61101i and ha2n�3 ; b2n�3 ;

c2n�3 ; d2n�3 i ¼ h110n�60001;110n�60011;110n�61111;110n�61101i in LTQn, such that bit (x,n)=1 for all x 2 V(M1); and the row-borders of M2 are ha01; b

01; c01; d

01i ¼ h010n�60001;010n�60011;010n�61111;010n�61101i and ha0

2n�3 ; b02n�3 ; c02n�3 ; d

02n�3 i ¼

h100n�60001;100n�60011;100n�61111;100n�61101i in LTQn, such that bit (x,n)=1 for all x 2 V(M2).

Proof. The proof is by induction on n. First, we prove the theorem holds for n=7. We use g1,h1,l1,m1,g2,h2,l2, m2 to denote0001,0011,1111,1101,0111,0101,1001,1011, respectively. And, according to Lemma 7, we can easily obtain:


M1 ¼

00g1 00h1 00l1 00m1

00g2 00h2 00l2 00m2

01l1 01m1 01g1 01h1

01l2 01m2 01g2 01h2

10l2 10m2 10g2 10h2

10l1 10m1 10g1 10h1

11g2 11h2 11l2 11m2

11g1 11h1 11l1 11m1

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

and M2 ¼

01g1 01h1 01l1 01m1

01g2 01h2 01l2 01m2

00l1 00m1 00g1 00h1

00l2 00m2 00g2 00h2

11l2 11m2 11g2 11h2

11l1 11m1 11g1 11h1

10g2 10h2 10l2 10m2

10g1 10h1 10l1 10m1

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

;

which are two 4 � 8 meshes in LTQ6. Then,

M01 ¼

000g1 000h1 000l1 000m1

000g2 000h2 000l2 000m2

001l1 001m1 001g1 001h1

001l2 001m2 001g2 001h2

010l2 010m2 010g2 010h2

010l1 010m1 010g1 010h1

011g2 011h2 011l2 011m2

011g1 011h1 011l1 011m1

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

and M12 ¼

101g1 101h1 101l1 101m1

101g2 101h2 101l2 101m2

100l1 100m1 100g1 100h1

100l2 100m2 100g2 100h2

111l2 111m2 111g2 111h2

111l1 111m1 111g1 111h1

110g2 110h2 110l2 110m2

110g1 110h1 110l1 110m1

0BBBBBBBBBBBBB@

1CCCCCCCCCCCCCA

;

are two 4 � 8 meshes in LTQ 06 and LTQ1

6, respectively. Since the most significant bit of every node of M01 is 0 and the most

significant bit of every node of M12 is 1, VðM0

1ÞT

VðM12Þ=;. According to Definition 1, we can easily verify that

(011g1,101g1), (011h1,101h1), (011l1,101l1), (011m1, 101m1) 2 E(LTQ7). Thus, we can get M01 by connecting M0

1 and M12:

M01 ¼

000g1 000h1 000l1 000m1

000g2 000h2 000l2 000m2

001l1 001m1 001g1 001h1

001l2 001m2 001g2 001h2

010l2 010m2 010g2 010h2

010l1 010m1 010g1 010h1

011g2 011h2 011l2 011m2

011g1 011h1 011l1 011m1

101g1 101h1 101l1 101m1

101g2 101h2 101l2 101m2

100l1 100m1 100g1 100h1

100l2 100m2 100g2 100h2

111l2 111m2 111g2 111h2

111l1 111m1 111g1 111h1

110g2 110h2 110l2 110m2

110g1 110h1 110l1 110m1

0BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCA

;

which is a 4 � 16 mesh in LTQ7. We can easily verify that M01 satisfies the conditions in the theorem. Similarly, we can obtain

the 4 � 16 mesh M02 by connecting M0

2 and M11, such that M0

2 satisfies the conditions in the theorem. Thus, we prove that thetheorem holds for n = 7.

Supposing that the theorem holds for n = s � 1, we will prove that the theorem still holds for n = s. For n = s � 1, accordingto the induction hypothesis, let

M1 ¼

a1 b1 c1 d1

a2 b2 c2 d2

. . . . . . . . . . . .

a2s�4 b2s�4 c2s�4 d2s�4

0BBB@

1CCCA and M2 ¼

a01 b01 c01 d01a02 b02 c02 d02. . . . . . . . . . . .

a02s�4 b02s�4 c0

2s�4 d02s�4

0BBBB@

1CCCCA;

be two 4 � 2s�4 meshes in LTQs�1. The row-borders of M1 are ha1,b1,c1, d1i = h000s�7g1, 000s�7h1,000s�7l1,000s�7m1 i =h000s�70001,000s�70011,000s�71111, 000s�71101i and ha2s�4 ;b2s�4 ;c2s�4 ;d2s�4 i ¼ h110s�7g1;110s�7h1;110s�7l1;110s�7m1i ¼h110s�70001;110s�70011;110s�71111;110s�71101i in LTQs�1. And the row-borders of M2 are ha01;b

01;c01;d

01i ¼

h010s�7g1;010s�7h1;010s�7l1;010s�7m1i ¼ h010s�70001;010s�70011;010s�71111;010s�71101i and ha02s�4 ;b

02s�4 ;c02s�4 ;d

02s�4 i ¼

h100s�7g1;100s�7h1;100s�7l1;100s�7m1i ¼ h100s�70001;100s�70011;100s�71111;100s�71101i in LTQs�1.

Then,


M01 ¼

0a1 0b1 0c1 0d1

0a2 0b2 0c2 0d2

. . . . . . . . . . . .

0a2s�4 0b2s�4 0c2s�4 0d2s�4

0BBB@

1CCCA ¼

0000s�7g1 0000s�7h1 0000s�7l1 0000s�7m1

0a2 0b2 0c2 0d2

. . . . . . . . . . . .

0110s�7g1 0110s�7h1 0110s�7l1 0110s�7m1

0BBBB@

1CCCCA

and

M12 ¼

1a01 1b01 1c01 1d011a02 1b02 1c02 1d02. . . . . . . . . . . .

1a02s�4 1b02s�4 1c0

2s�4 1d02s�4

0BBBB@

1CCCCA ¼

1010s�7g1 1010s�7h1 1010s�7l1 1010s�7m1

1a02 1b02 1c02 1d02. . . . . . . . . . . .

1100s�7g1 1100s�7h1 1100s�7l1 1100s�7m1

0BBBB@

1CCCCA;

are two 4 � 2s�4 meshes in LTQ0s�1 and LTQ1

s�1, respectively. According to Definition 1, we can easily verify thatð0a2s�4 ;1a01Þ; ð0b2s�4 ;1b01Þ; ð0c2s�4 ;1c01Þ; ð0d2s�4 ;1d01Þ 2 EðLTQsÞ. As a consequence, we can get M0

1 by connecting M01 and M1

2:

M01 ¼

0a1 0b1 0c1 0d1

0a2 0b2 0c2 0d2

. . . . . . . . . . . .0a2s�4 0b2s�4 0c2s�4 0d2s�4

1a01 1b01 1c01 1d011a02 1b02 1c02 1d02. . . . . . . . . . . .

1a02s�4 1b02s�4 1c0

2s�4 1d02s�4

0BBBBBBBBBB@

1CCCCCCCCCCA¼

0000s�7g1 0000s�7h1 0000s�7l1 0000s�7m1

0a2 0b2 0c2 0d2

. . . . . . . . . . . .1100s�7g1 1100s�7h1 1100s�7l1 1100s�7m1

0BB@

1CCA

¼000s�6g1 000s�6h1 000s�6l1 000s�6m1

0a2 0b2 0c2 0d2

. . . . . . . . . . . .110s�6g1 110s�6h1 110s�6l1 110s�6m1

0BB@

1CCA;

which is a 4 � 2s�3 mesh in LTQs. We can easily verify that M01 satisfies the conditions in the theorem. Similarly, we can get

the 4 � 2s�3 mesh M02 by connecting M0

2 and M11, such that M0

2 satisfies the conditions in the theorem. Hence, we prove thatthe theorem holds for n = s. h

Theorem 3. For any integer n P 5, there is a 4 � 2n�3 mesh

M ¼

a1 b1 c1 d1

a2 b2 c2 d2

. . . . . . . . . . . .

a2n�3 b2n�3 c2n�3 d2n�3

0BBB@

1CCCA;

such that the row-borders of M are ha1,b1,c1,d1i = h00n�50000,00n�50010,00n�51010,00n�51000i and ha2n�3 ; b2n�3 ; c2n�3 ; d2n�3 i ¼h10n�50000;10n�50010;10n�51010;10n�51000i in LTQn and bit (x,n)=0 for all x 2 V(M).

Proof. Beginning with the 4 � 4 mesh M3 in LTQ5 in Lemma 6, the proof method is similar to that in Theorem 2. We skip thediscussion here. h

Theorem 4. For any integer n P 4, there are two 4 � 2n�3 meshes M and M0

in LTQn, such that V(M)T

V(M0) = ;.

Proof. According to Lemmas 5–7 and Theorems 2, 3, there are two 4 � 2n�3 meshes M and M0in LTQn, such that bit (x,n)=0

for all x 2 V(M) and bit (y,n)=1 for all y 2 V(M0). As a result, V(M)

TV(M

0) = ;. h

According to the Theorem 4, we obtain the following corollary.

Corollary 2. For any n P 4, two 4 � 2n�3 meshes can be embedded with dilation 1 and expansion 2 in LTQn, such that there is nonode-overlapping between the two embedded meshes and these two 4 � 2n�3 meshes cover all nodes of LTQn.

Corollary 2 shows that two embedded meshes can work independently in a n-dimensional locally twisted cube withoutinterfering with each other, which is a desirable feature in parallel computing.


5. Embedding a 4 � (2n�2 � 1) mesh into the n-dimensional locally twisted cube

In the last two sections, we prove that a 2 � 2n�1 mesh can be embedded with dilation 1 and expansion 1 in LTQn for anyinteger n P 1, and two node-disjoint 4 � 2n�3 meshes can be embedded with dilation 1 and expansion 2 in LTQn for any inte-ger n P 4. In this section, the 4 � (2n�2 � 1) mesh embedding in the n-dimensional locally twisted cube is discussed.

Lemma 8. A 4 � 1 mesh can be embedded into LTQ3 with dilation 1.

Proof. Obviously, a 4 � 1 mesh

M ¼ 000 010 011 001ð Þ;

can be embedded into LTQ3 with dilation 1. h


Proof. According to Lemma 5, 0000 0010 1010 10000100 0110 1110 1100

� �and 0001 0011 1111 1101

0111 0101 1001 1011

� �are two node-disjoint

4 � 2 meshes in LTQ4. According to Definition 1, (0000,0001), (0010,0011) 2 E(LTQ4). Furthermore, considering that

(1010,1110), (1110,1111), (1000,1100), (1100,1101) 2 E(LTQ4), a 4 � 3 mesh0000 0010 1010 10000001 0011 1111 11010111 0101 1001 1011

0@

1A can be embed-

ded into LTQ4 with dilation 2. h


Proof. By applying the similar proof method used in Lemma 9, considering that (10100,10000), (10000,10001),(10110,10010), (10010,10011) 2 E(LTQ5), a 4 � 7 mesh

M ¼

00000 00010 01010 01000

00100 00110 01110 01100

10100 10110 11110 11100

10001 10011 11111 11101

10111 10101 11001 11011

01111 01101 00001 00011

01001 01011 00111 00101

0BBBBBBBBBBBB@

1CCCCCCCCCCCCA

;



Proof. By applying the similar proof method used in Lemma 9, considering that (100100,100000), (100000,100001),(100110,100010), (100010,100011) 2 E(LTQ6), a 4 � 15 mesh

M ¼

000000 000010 001010 001000000100 000110 001110 001100010100 010110 011110 011100010000 010010 011010 011000110000 110010 111010 111000110100 110110 111110 111100100100 100110 101110 101100100001 100011 101111 101101100111 100101 101001 101011111111 111101 110001 110011111001 111011 110111 110101001001 001011 000111 000101001111 001101 000001 000011010111 010101 011001 011011010001 010011 011111 011101

0BBBBBBBBBBBBBBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCCCCCCCCCCCCCA

;



Lemma 12. For any integer n P 6, a 4 � (2n�2 � 1) mesh

M ¼

a1 b1 c1 d1

. . . . . . . . . . . .

a2n�3�1 b2n�3�1 c2n�3�1 d2n�3�1

a2n�3 b2n�3 c2n�3 d2n�3

. . . . . . . . . . . .

a2n�2�1 b2n�2�1 c2n�2�1 d2n�2�1

0BBBBBBBB@

1CCCCCCCCA¼

000n�60000 000n�60010 000n�61010 000n�61000. . . . . . . . . . . .

100n�60100 100n�60110 100n�61110 100n�61100100n�60001 100n�60011 100n�61111 100n�61101

. . . . . . . . . . . .

010n�60001 010n�60011 010n�61111 010n�61101

0BBBBBBBBB@

1CCCCCCCCCA;

can be embedded into LTQn with dilation 2.

Proof. According to Theorem 2, we know that for any integer n P 6,

010n�60001 010n�60011 010n�61111 010n�61101. . . . . . . . . . . .

100n�60001 100n�60011 100n�61111 100n�61101

0B@

1CA;

is a 4 � 2n�3 mesh which can be embedded into LTQn.According to Theorem 3, we know that for any integer n P 6,

000n�60000 000n�60010 000n�61010 000n�61000. . . . . . . . . . . .

100n�60100 100n�60110 100n�61110 100n�61100100n�60000 100n�60010 100n�61010 100n�61000

0BBB@

1CCCA;

is a 4 � 2n�3 mesh which can be embedded into LTQn.According to Definition 1, (100n�61110,100n�61111),(100n�61100,100n�61101) 2 E(LTQn). Furthermore, considering that

(100n�60100,100n�60000), (100n�60000,100n�60001), (100n�60110,100n�60010), (100n�60010,100n�60011) 2 E(LTQn), a4 � (2n�2 � 1) mesh

000n�60000 000n�60010 000n�61010 000n�61000. . . . . . . . . . . .

100n�60100 100n�60110 100n�61110 100n�61100100n�60001 100n�60011 100n�61111 100n�61101

. . . . . . . . . . . .

010n�60001 010n�60011 010n�61111 010n�61101

0BBBBBBBBB@

1CCCCCCCCCA;

can be embedded into LTQn with dilation 2. h

With these lemmas, we obtain the following theorem.

Theorem 5. For any integer n P 3, there is a 4 � (2 n�2 � 1) mesh, which can be embedded into LTQn with dilation 2.

Proof. According to Lemmas 8–12, clearly, the theorem holds. h

6. Discussion on 2p � 2q mesh embedding in locally twisted cubes

For any integers p P 1 and q P 1, we consider embedding 2p � 2q meshes into locally twisted cubes in this section. SinceLTQn has 2n nodes, meshes of 2p � 2q may be embedded into LTQn if 2p � 2q 6 2n, in other words, p � q 6 2n�2. Therefore, forany integers p P 1 and q P 1 with p � q > 2n�2, meshes of size 2p � 2q cannot be embedded into LTQn.

Lemma 13. For any integers n,p and q with n = 2,3 and p � q 6 2n�2, all the meshes of size 2p � 2q can be embedded into LTQn

with dilation 1.

Proof. According to Theorem 1, there is a 2 � 2 mesh which can be embedded into LTQ2 with dilation 1, for p = 1 and q = 1.Also according to Theorem 1, for any integers p and q with p � q 6 2, a 2 � 4 mesh in LTQ3 can be embedded into LTQ3 withdilation 1. A 2 � 4 mesh can also be considered as a 4 � 2 mesh. Therefore, the theorem holds (see Fig. 3). h

Fig. 3. 2p � 2q meshes which can be embedded into LTQ2 and LTQ3 with dilation 1.


Lemma 14. For any integers p and q with p � q 6 4, all the meshes of size 2p � 2q can be embedded into LTQ4 with dilation 1,except the 4 � 4 mesh.

Proof. According to Theorem 1, meshes of size 2 � 2, 2 � 4, 2 � 8 can be embedded into LTQ4 with dilation 1. For the 2 � 8mesh can be embedded into LTQ4 with dilation 1, clearly, 2 � 6 mesh can also be embedded into LTQ4 with dilation 1. There-fore, meshes of size 4 � 2, 6 � 2 and 8 � 2 can be embedded into LTQ4 with dilation 1 too.

All the 2p � 2q meshes which can be embedded into LTQ4 with dilation 1 are listed in Fig. 4 except the 4 � 4 mesh.We prove that a 4 � 4 mesh cannot be embedded into LTQ4 with dilation 1 as follows.A 4 � 4 mesh (see Fig. 5(a)) has 9 cycles of length 4, and 12 edges are used twice in these 9 cycles. The 12 edges intersect

at 4 nodes labeled with A,B,C,D, whose degrees are 4.We present all the 16 cycles of length 4 in LTQ4 (see Fig. 1(b)) as follows:

1. (0000,0010), (0010,0110), (0110,0100), (0100,0000);2. (0100,0110), (0110,0111), (0111,0101), (0101,0100);3. (0111,0101), (0101,0011), (0011,0001), (0001,0111);4. (0000,0010), (0010,0011), (0011,0001), (0001,0000);5. (1010,1000), (1000,1100), (1100,1110), (1110,1010);6. (1110,1100), (1100,1101), (1101,1111), (1111,1110);7. (1101,1111), (1111,1001), (1001,1011), (1011,1101);8. (1010,1000), (1000,1001), (1001,1011), (1011,1010);9. (0000,1000), (1000,1100), (1100,0100), (0100,0000);

10. (0000,0010), (0010,1010), (1010,1000), (1000,0000);11. (0100,0110), (0110,1110), (1110,1100), (1100,0100);12. (0010,1010), (1010,1110), (1110,0110), (0110,0010);13. (0111,0101), (0101,1001), (1001,1011), (1011,0111);14. (1101,1111), (1111,0011), (0011,0001), (0001,1101);15. (0101,0011), (0011,1111), (1111,1001), (1001,0101);16. (0111,0001), (0001,1101), (1101,1011), (1011,0111).

Among the 64 edges in those 16 cycles, we find that 24 edges are used at least twice: (0000,0010), (0010,0110),(0110,0100), (0100,0000), (1010,1000), (1000,1100), (1100,1110), (1110,1010), (0000,1000), (0010,1010), (0100,1100),(0110,1110), (0111,0101), (0101,0011), (0011,0001), (0001,0111), (1101,1111), (1111,1001), (1001,1011), (1011,1101),(0111,1011), (0101,1001), (0001,1101), (0011,1111).

These 24 edges can be represented as Fig. 5(b). According to this Figure, we can verify that all the node degrees are lessthan 4. Therefore, we prove that a 4 � 4 mesh cannot be embedded into LTQ4 with dilation 1. h

Fig. 4. 2p � 2q meshes which can be embedded into LTQ4 with dilation 1.

Fig. 5. (a) a 4 � 4 mesh; (b) the 24 edges used at least twice in LTQ4.

Fig. 6. 2p � 2q meshes which can be embedded into LTQ5 with dilations no more than 2.


Lemma 15

(1) For any integers p and q with p � q 6 8, all the meshes of size 2p � 2q can be embedded into LTQ5 with dilations no morethan 2, except the 4 � 8(or 8 � 4) mesh. (2) A 4 � 8(or 8 � 4) mesh cannot be embedded into LTQ5 with dilation 1.

Proof. According to Theorem 1, meshes of size 2 � 2, 2 � 4, 2 � 8 and 2 � 16 can be embedded into LTQ5 with dilation 1. Thisis the same for meshes of size 2 � 6, 2 � 10, 2 � 12, 2 � 14, 4 � 2, 6 � 2, 8 � 2, 10 � 2, 12 � 2, 14 � 2 and 16 � 2 (see Fig. 6).

According to Theorem 2, a 4 � 4 mesh can be embedded into LTQ5 with dilation 1.According to Theorem 5, a 4 � 7 mesh can be embedded into LTQ5 with dilation 2. Therefore, a mesh of size 4 � 6(or

6 � 4) can be embedded into LTQ5 with dilation 2.With the similar proof method used in Lemma 14, we can prove that a 4 � 8(or 8 � 4) mesh cannot be embedded into

LTQ5 with dilation 1. h

We show all the 2p � 2q meshes which can be embedded into LTQ2,LTQ3,LTQ4 and LTQ5 in Figs. 3, 4 and 6, respectively,where the meshes with double strikethroughs cannot be embedded into LTQn with dilation 1. We also consider the 4 � 16mesh and the 8 � 8 mesh, which cannot be embedded into LTQ6 with dilation 1 either. We conjecture that 2p � 2q mesheswhich have the same node numbers of LTQn cannot be embedded into LTQn with dilation 1 for n P 4.

7. Conclusions

This paper provides the study of embedding three different types of special meshes into locally twisted cubes. Three ma-jor results are as follows:

(1) A 2 � 2n�1 mesh can be embedded with dilation 1 and expansion 1 in the n-dimensional locally twisted cube for anyinteger n P 1.

(2) Two node-disjoint 4 � 2n�3 meshes can be embedded with dilation 1 and expansion 2 in the n-dimensional locallytwisted cube for any integer n P 4.


(3) A 4 � (2n�2 � 1) mesh can be embedded with dilation 2 in the n-dimensional locally twisted cube for any integern P 3.

The first two results are optimal in the sense that the dilations of all embeddings are 1. The embedding of the 2 � 2n�1

mesh is also optimal in terms of expansion because it has the smallest expansion 1. We also present the analysis of2p � 2q mesh embedding in locally twisted cubes.

Acknowledgment

This work is supported by Natural Science Foundation of China under Grant No. 60873047 and Natural Science Founda-tion of Jiangsu Province under Grant No. BK2008154.

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