9
Study of Link Utilization of Perfect Difference Network and Hypercube Rakesh Kumar Katare Department of Computer Science A.P.S. University, Rewa (M.P.), India - 486003 katare [email protected] N.S. Chaudhari Department of CSE, lIT, Indore, (M.P.), India - 452017 [email protected] Shazad Ahmed Mughal Department of Computer Science A.P.S. University, Rewa (M.P.), India - 486003 [email protected] Shashi Kant Verma Department of MCA ITM Universe, Gwalior (M.P.), India-474001 [email protected] Shah Imran Department of Computer Science A.P.S. University, Rewa (M.P.), India - 486003 [email protected] Rajesh Roshan Raina Department of Computer Science A.P.S. University, Rewa (M.P.), India - 486003 [email protected] Abstract- In this paper, the comparision of perfect difference network and hypercube shows the performance and robustness of the two architectures. The topological properties of the two architectures are presented in the form of lemmas. We have made attempts to study the circuits to show the robustness of these architectures. In these architectures, the diameter remains same during normal course and during link failure. The comparison between these two architectures and their diameters can be shown on their adjacency matrices as well. In the lemmas, we have shown that the degree of a node of these architectures can be changed while changing in the architectural design. Keywords - Perfect Difference Set (PDS), Perfect Difference Network (PDN), Hypercube, Circuits, Adjacency Matrix. I. INTRODUCTION This paper presents study of link utilization of the Perfect Difference Network (PDN) architecture. When data is distributed in an interconnected network it passes through certain nodes and takes certain paths. The PDN architecture is presented in the form of circuits to study the utilization of nodes, its properties and performance. To prove all these properties and characteristics the lemmas are presented. It is shown that the diameter of this architecture remains the same during removing one node and during the link failure; the degree of a node may be changed while a change is brought in the architecture design A. Perfect Difference Set J) Formulation of Perfect Difference Set: As we know from remainder theorem that Numerator=Remainder+ Denominator * Quotient. The Perfect Difference Set (PDS) can be formulated from remainder theorem as [I] an integer (numerator) is equal to the addition of remainder and denominator where quotient is one. Now integer is a member of set of (I, 2, ...,0 2 + 0), where o is a prime or power of prime and the remainder is the difference Sj- Sj, where i :/:j, O:S i,j:S o. So we can write that or 2) Definition of Perfect Difference Set: A set {sO, s 1, , so} of 0 + 1 integers having the property that their 0 2 +0 differences, 0 ::; i :/:j :s 0, are congruent modulo 0 2 + 0 + 1, to the integers 1, 2, ,0 2 + 0 in some order is a perfect difference set of order O. Perfect Difference sets [13] are sometimes called simple difference sets, given that they correspond to the special I) = I case of difference sets for which each of the possible differences is formed in exactly 0 ways, where 0 is a prime or power of prime and n=0 2 + 0 + I and (S, - Sj) = (0 2 + 0) mod n 3) Evaluation of PDS: The Perfect Difference Set of each node of the PDN can be evaluated by the remainder theorem i.e., N=R+D * Q, Where N = Numerator, R = Remainder, D and Q = Quotient The above equation can be written as Integer = (S, - Sj) + (0 2 + 0 + 1) * 1 Where integer is a member of the set (l, 2, ,0 2 + 0) and S, - Sj is numerator or the difference set. Denominator So we can write as - (S, - Sj) = (Integer) mod (0 2 + 0 + 1) (1)

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Page 1: Study of Link Utilization of Perfect Difference Network ...worldcomp-proceedings.com/proc/p2013/PDP3426.pdfStudy of Link Utilization of Perfect Difference Network and Hypercube Rakesh

Study of Link Utilization of Perfect Difference Network and Hypercube

Rakesh Kumar KatareDepartment of Computer Science

A.P.S. University,Rewa (M.P.), India - 486003katare [email protected]

N.S. ChaudhariDepartment of CSE,

lIT,Indore, (M.P.), India - 452017

[email protected]

Shazad Ahmed MughalDepartment of Computer Science

A.P.S. University,Rewa (M.P.), India - [email protected]

Shashi Kant VermaDepartment of MCA

ITM Universe,Gwalior (M.P.), India-474001

[email protected]

Shah ImranDepartment of Computer Science

A.P.S. University,Rewa (M.P.), India - 486003

[email protected]

Rajesh Roshan RainaDepartment of Computer Science

A.P.S. University,Rewa (M.P.), India - 486003

[email protected]

Abstract- In this paper, the comparision of perfect differencenetwork and hypercube shows the performance and robustness ofthe two architectures. The topological properties of the twoarchitectures are presented in the form of lemmas. We have madeattempts to study the circuits to show the robustness of thesearchitectures. In these architectures, the diameter remains sameduring normal course and during link failure.

The comparison between these two architectures and theirdiameters can be shown on their adjacency matrices as well. Inthe lemmas, we have shown that the degree of a node of thesearchitectures can be changed while changing in the architecturaldesign.

Keywords - Perfect Difference Set (PDS), Perfect DifferenceNetwork (PDN), Hypercube, Circuits, Adjacency Matrix.

I. INTRODUCTION

This paper presents study of link utilization of the PerfectDifference Network (PDN) architecture. When data isdistributed in an interconnected network it passes throughcertain nodes and takes certain paths. The PDN architecture ispresented in the form of circuits to study the utilization ofnodes, its properties and performance. To prove all theseproperties and characteristics the lemmas are presented. It isshown that the diameter of this architecture remains the sameduring removing one node and during the link failure; thedegree of a node may be changed while a change is brought inthe architecture design

A. Perfect Difference Set

J) Formulation of Perfect Difference Set: As we know fromremainder theorem that

Numerator=Remainder+ Denominator * Quotient.

The Perfect Difference Set (PDS) can be formulatedfrom remainder theorem as [I] an integer (numerator) is equalto the addition of remainder and denominator where quotient isone. Now integer is a member of set of (I, 2, ...,02 + 0), whereo is a prime or power of prime and the remainder is thedifference Sj - Sj,

where i :/:j, O:S i,j:S o.So we can write that

or

2) Definition of Perfect Difference Set: A set {sO, s 1, ,so} of 0 + 1 integers having the property that their 02 + 0differences, 0 ::; i :/:j :s 0, are congruent modulo 02 + 0 + 1, tothe integers 1, 2, , 02 + 0 in some order is a perfectdifference set of order O. Perfect Difference sets [13] aresometimes called simple difference sets, given that theycorrespond to the special I) = I case of difference sets forwhich each of the possible differences is formed in exactly 0ways, where 0 is a prime or power of prime and

n = 02 + 0 + I and (S, - Sj) = (02 + 0) mod n

3) Evaluation of PDS: The Perfect Difference Set of each nodeof the PDN can be evaluated by the remainder theorem i.e.,

N=R+D * Q,

Where N = Numerator, R = Remainder, Dand Q = Quotient

The above equation can be written as

Integer = (S, - Sj) + (02 + 0 + 1) * 1

Where integer is a member of the set (l, 2, , 02 + 0) andS, - Sj is numerator or the difference set.

Denominator

So we can write as -

(S, - Sj) = (Integer) mod (02 + 0 + 1) (1)

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(0- 0)

I~·ll

Fig. 1: PDN with n = 7, () = 2 and PDS = {O, 1,3}

Example 1 : Let Sj - Sj = 1 - 0, 0 = 2, n = 02 + 0 + 1 = 7 andinteger = I as shown in the Fig. 1.1

Therefore, eq( I) can be written as:

(1 - 0) = 1 mod 7

=> 1= (1- 0) + 7

1 = 1 + 7

1 = 8

=>

=>But 8 mod 7 = I

Hence (1 - 0) is the Perfect Difference Set for node 1.

Example 2 : Let S, - Sj = 3 - 1, 0 = 2, n = 02 + b + 1 = 7 andinteger = 2 as shown in the Fig. 1.1

Therefore, eq( 1) can be written as:

(3 - 1) = 2 mod 7

=> 2 = (3 - 1) + 7

=> 2=2+7

=> 2=9

But 9 mod 7 =2

Hence (3 - 1) is the Perfect Difference Set for node 2.

Example 3 : Let S, - Sj = 3 - 0, b = 2, n = 02 + 0 + 1 = 7 andinteger = 3 as shown in the Fig. 1.1

Therefore, eq(l) can be written as:

(3 - 0) = 3 mod 7

=> 3 = (3 - 0) + 7

=> 3=3+7

=> 3 = 10

But 10 mod 7 = 3

Hence (3 - 0) is the Perfect Difference Set for node 3.

Example 4 : Let S, - Sj = 0 - 3, 0 = 2, n = 02 + 0 + 1 = 7 andinteger = 4 as shown in the Fig. 1.1

Therefore, eq( 1) can be written as:

(0-3)=4mod 7

=> 4 = (0 - 3) + 7

=> 4 = -3 + 7

=> 4=4

Hence (0 - 3) is the Perfect Difference Set for node 4.

Example 5 : Let S, - Sj = 1 - 3, 0 = 2, n = 02 + 0 + 1 = 7 andinteger = 5 as shown in the Fig. 1.1

Therefore, eq(l) can be written as:

(1- 3) = 5 mod 7

=> 5 = (1- 3) + 7

=> 5 = -2 + 7

=> 5=5

Hence (l - 3) is the Perfect Difference Set for node 5.

Example 6: Let Sj - Sj = 0 - 1, 0 = 2, n = 02 + 0 + 1 = 7 andinteger = 6 as shown in the Fig. 1.1

Therefore, eq(1) can be written as:

(0 - 1) = 6 mod 7

=> 6 = (0 - I) + 7

=> 6 = -1 + 7

=> 6=6

Hence (0 - 1) is the Perfect Difference Set for node 6.

Example 7 : Let S, - Sj = 0 - 0, 0 = 2, n = 02 + 0 + 1 = 7 andinteger = 0 as shown in the Fig. 1.1

Therefore, eq(1) can be written as:

(0 - 0) = 0 mod 7

=> 0=0

Hence (0 - 0) is the Perfect Difference Set for node O.

This way we can find out any PDS appropriate to whichnode number in the Perfect Difference Network architecture.

4)Perfect Difference Network: Perfect Difference Network[1][8][9][11] is the network architecture, in which the diameteris always 2, i.e., every node ith needs to visit only two links tocommunicate with other nodes i ± 1 & i ± Sj (mod n), for 2 S jS O. In a Perfect Difference Network, the total number ofnodes is 02 + 0 + 1, i.e., if b = 2 then the total number of nodesin PDN is 7 and if 0 = 3, then number of nodes in PDN is 13.Also the degree of every node in a PDN is 20 i.e., if 0 = 2 thendegree of every node in a PDN is 4 and similarly for otherprime or power of prime numbers.

The design of Perfect difference network is done in such away where each node is connected via directed links to everyother node. The links in PDN architecture are bidirectional in

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nature and the connectivity leads to a chordal ring of degree 28i.e., 0 in-degree and 0 out-degree and diameter D = 2 [3][9].

n

7

1>~ , 'j

~i

~?

::.j~n

ill. 1 ~,

Fig. 2: PDS of order Ii in normal form

5) Topological properties of PDN: Some of the topologicalproperties ofPDN [2][5][6] are as under:

• Average inter-node distance: - Each node has distance of 0to itself, 1 to its 20 neighbors and 2 to the other 02

- 0nodes. Hence, D = [20 + 2(02

- o)]/n = 202/n. Ifwe did notcount the distance of a node to itself, the average inter-node distance would become 202/(n - 1) = 20 1(0 + 1).Hence the average inter-node distance of PDN of order 0is D = 252/n.

• The upper bound of the PDN is min(2Sall, nModd- SOdd+Seven),where Moddis the number of odd elements in thePDS, Sevenand SOddrepresent the sum of all PDS elementsthat are even and odd respectively, and Sall is the sum ofall s, values for the PDS. For an element s, of a specificPDS of order define s, as s, if s, < nl2 and as n - s, > nl2.

• The total number of links going between odd and evennodes is:

I (n - s) + I Sj = nModd- SOdd+ Seven

• Oddskips and Evenskips

• The lower bound on the bisection width of PDN is«0+ l)(n+ 1) 14) [7].

• The calculation of bisection width for an arbitrary graph isan NP-Complete problem.

6) Hypercube: Hypercube are loosely coupled parallelprocessors based on binary n. The hypercube network n-cubeparallel processor consists of 2n identical processors. Inhypercube architecture the degree and diameter of the graph issame i.e. 3, because of this equality they achieve a goodbalance between the communication speed and complexity ofthe topologic network [1].

The hypercube architecture has many other limitations.Primarily k-dimensional hypercube have N=2n vertices, sotheir structures are restricted to having exactly 2k nodes.Because structures size must be 2, there are large gaps in thesize of systems that can be built with the hypercube. This

restricts the numher of possihle nodes The perfect differenceset establishes the structure that can be constructed for everyprime power n=pr. This provides a large advantage over thehypercube architecture, where structures exist only for thepowers of2 [6][12].

The hypercube is well known as one of the most efficientnetwork topologies, especially for interconnection in parallelcomputers. The configuration of a three-dimensional (3D)hypercube (the familiar cube) network is shown in Figure-3 asan example. A 3D hypercube has 8 vertices and 12 edges,which correspond to network nodes and links, respectively.We defme an n-dimensional hypercube network as follows. Itconsists of N=2n nodes, each of which is labeled by a uniquebinary node number. Nodes whose nodes numbers differ byonly one bit from each other are interconnected by abidirectional link. As shown in Figure-3, Node (000) is link-connected to Node (001), (010), and (l00) since their numbersdiffer by only one bit from Node (000), in accordance with thedefinition. The total number of bidirectional links for an n-dimensional hypercube is (N/2) Log2N. This characteristicmakes the hypercube scalable since the number of links in thisnetwork is proportional to 0 (NLog2N), which for large-scalenetworks is much smaller than 0 (N2), the number of links ina full mesh network, which is the richest network topology.The hypercube network subsumes lower-order networktopologies such as mesh, tree, and ring, and thus exhibits thefeatures of these network topologies [6][10].

A unique feature of the hypercube with three or moredimensions is that it can form at least three disjoint pathsbetween any arbitrary pair of nodes, which makes thehypercube robust and reliable enough to secure the networkagainst multiple failures, these features make the hypercubenetwork attractive.

7) Topological properties of Hypercube: A hypercube is amultidimensional mesh of nodes with exactly two nodes ineach dimension. A d-dimensional hypercube consists of knodes, where k=2n [13].

• A hypercube has n dimensions; where n can be anypositive integer (including zero).

• The n cube is a connected graph of diameter n.

Fig. 3: Configuration of a 3D hypercube network.

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• A hypercube has 2n vertices.

• There are n connections (line) that meet at each vertex of ahypercube.

• All connections at a hypercube vertex meet at right angleswith respect to each other.

• The hypercube can be constructed recursively from lowerdimensional cubes.

• An architecture where the degree and diameter of a graphis the same then they will achieve a good balance between,the communication speed and complexity of the topologynetwork [7].

• An n-cube graph is an undirected graph of k=211verticeslabeled from 0 to 211-1 and such that there is an edgebetween any two vertices if and only if the binaryrepresentations of their labels differ by one and only onebit [8].

• Any two adjacent nodes A and B of an N cube are suchthat the nodes adjacent to A and those adjacent to Bareconnected in one-to-one fashion.

• There are n different ways of tearing an n-cube, i.e., ofsplitting it into two (n - 1) sub cubes so that theirrespective vertices are connected in a one-to-one way [4].

• There are n! 211different ways in which the 211nodes of ann-cube can be numbered.

• A graph G=(V,E) is an n-cube if and only if

• V has 211vertices;

• Every vertex has degree n;

• G is connected;

• Any two adjacent nodes A and B are such that thenodes adjacent to A and those adjacent to B are linkedone-to-one fashion.

II LINK UTILIZATION OF PDN AND HYPERCUBE

Network topologies offer additional design points toaccommodate the needs of new and emerging technologies.Therefore, further study is needed to resolve some openquestions and to derive cost/performance, robustness in termsof lemmas to study comparisons between PDN and Hypercubearchitectures.

Lemma 1: The total number of chordal diagonals in a PDN isalways even.

Proof - Consider a PDS of order n, (where n = &2+ &+ 1 ando is a prime or power of prime) then it gives the followingconclusions.

In PDN, we have total degree of a node = 25 vertices andsince each node is connected to two ring links, therefore, wehave total number of chordal diagonals = 25 - 2.

Case 1: if 5 is even then 2 * 5 result an even integer andtherefore (2*5 - 2) also result an even integer.

Case 2: if 5 is odd then 2 * 5 again result an even integerand therefore (2*& - 2) also result an even integer.

As, we know that subtracting 2 in even integer alwaysturns an even integer.

So, the total number of chordal diagonals in PDN is alwayseven.

Lemma 2: The 'diameter of the PDN' remains always samewhile removing one node of the PDN

Proof: We know that 5 = 2, n = 52 + 5 + 1, then on removingone node of the PDN will loses its properties and n becomes n- 1 nodes i.e., n = 02 + 5.

The other properties becomes as:

• Total number of edges 'n.S' becomes 'n.S - 25'.

• Total degree 2n.5 of PDN architecture becomes 25(n -I). These changes can be reflected in chordal diagonalsalso.

But the only property of the PDN i.e., diameter of the PDNremains always same as shown in the Fig. 4. Hence theremaining network works as a normal network with maximumdiameter 2.

From the Fig. 4 if we have to visit node 6 from node 4, wehave to visit only two edges to reach to the destination nodei.e., 4-5-6 or 4-3-6.similarly, if the source node is 4 anddestination node is 2 then paths can be 4-3-2, 4-5-2 and 4-1-2.It shows that paths can be more than 2 but the maximumdiameter is always 2. In other words, PDN is robust.

Lemma 3: The 'diameter of the PDN' remains always samewhile removing one edge of the PDN.

Proof: As we know 0 = 2, n = 02 + 0 + 1, then on removing oneedge from the PDN, its properties becomes as:

• Total number of edges 'n.S' becomes 'n.S - I'.

• Total degree of PDN is equal to 2no - 2 as one edge isremoved.

• The chordal ring property is also lost.

Fig. 4: PDN without one node

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Fig. 5: PDN without one edge

• Lesser number of circuits are formed.

The only property of the PDN i.e., diameter of the PDNremains always same as shown in the Fig. 5. Hence theremaining network works as a normal network with maximumdiameter 2.

From the figure if we have to visit node I from node 6, wehave to visit only two edges to reach to the destination nodei.e., 6-2-1 or 6-5-1. Similarly, if the source node is 0 anddestination node is I then path can be 0-4-1. It shows that thereis change is path formation circuits but the maximum diameteris always 2. So it is clear from the figure that PDN loses itsproperties except the 'diameter' and the remaining networkwill be used as a normal network with maximum diameter 2. Italso shows that PDN is robust in nature.

Lemma 4: The 'chordal ring property of the PDN' remainssame while removing more than one chordal diagonals of thePDN. The remaining network will work as ring topology.

Proof: As we know 8 = 2, n = 82 + 8 + 1, then on removingmore than one chordal diagonal from the PDN, its propertiesbecomes as:

• Total number of edges 'n.S' becomes 'n.8 - number ofedges to be removed'.

• Total degree of PDN is not equal to 2n& as more thanone diagonal edge is removed.

• The diameter property in this situation is also lost.

The only chordal ring property of the PDN remains sameas shown in the Fig. 6. Hence the remaining network works asa normal ring topology network.

In the Fig. 6, given above, it is clear that the chordal ringproperty of the PDN is not lost by removing the chordaldiagonals. It is also clear from the figure that the diameter

Fig. 6: PDN without More than One Chordal Diagonal

property is not no more exist. For example, if we have tomove from the node 0 to node 4, there are different ways toreach to the destination node, like 0 - 6 - 5 - 4 or 0 - 6 - 2 - 3- 4 or 0 - 6 - 3 - 4 or 0 - I - 2 - 3 - 4. So it is clear that thediameter of the PDN is lost by removing the chordal diagonals(more than one).

Lemma 5: Diameter of Perfect Difference Network (PDN) isequal to total degree of node/Prime number or power of prime.

Proof: To prove that diameter of a PDN = 2

Let, Prime or power of prime = 8

Then, Total degree of a node = 28

Therefore,

Diameter = 28/8

= 2 Hence it is proved.

Lemma 6: Total number of vertices in a PDN is always odd.Each vertex of PDN has always even number of edges.

Proof: Consider a PDS is of order n, (where n = 82 + 8 +1 and8 is a prime or power of prime) then it gives the followingconclusions.

Since, a PDN has n = 62 + 6 + I vertices.

Case 1: if 6 is even then 62 + 6 result in an even integer

Case 2: if 8 is odd then 82 + 8 again result an even integer

As, we know that adding I in even integer always turns anodd integer.

So, the total number of vertices in PDN is always odd.

Again, since each ilh node of PDN is connected to (i ± 1)mod nand (i ± Sj ) mod n node, where 2 ::;j ::S6 and any twovertex having one edge between them.

Therefore d (v) = 2.8, where 6 is also equal to total numberof non-zero elements in a PDS.

So each node ofPDN has even number of edges.

Lemma 7: The degree of a node is increased by 2 in case ofPDN of PDNs. In this case the diameter also changes whiletraversingfrom one PDN to another PDN.

Proof: Since each ilh node of PDN is connected to (i±l) nodeof other PDN, and any two vertexes having one edge betweenthem. Therefore, each node requires 2 more edges to connectto the 2 nodes of the PDN in a PDN ofPDN's.

The given below Figure-7 depicts the above assumption.Similarly if we want to traverse from one node of a PDN toanother node of another PDN in a PDN of PDN s, the diametercan be changed as shown in the figure given below:

In this case the diameter changes from one PDN to anotherPDN and can be equal to 2 + shortest distance between twoconcerned PDN's.

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PUN I PUN 2

Fig. 7: PDN ofPDN's

Lemma 8: The circuits formed in a PDN are a combination ofodd and even length. And the smallest circuit in a PDN is oflength 3.

Proof: The circuits formed from the PDN with 7 nodes areshown below:0---+4---+3---+00---+4---+1---+00-..3-..6---+00---+6-..5-..1-..00---+6---+5---+2---+1---+00---+6---+5---+4---+1---+00---+6-..5---+4---+00---+6---+5---+4---+3---+00---+6---+5---+4---+3---+2---+1---+00---+1---+2---+6---+00---+1---+2---+3---+00---+1---+2-..3---+6---+00-..1---+2---+3-..4---+01---+5---+4-"11---+5---+2---+11---+5---+4---+3---+2---+11---+4---+3---+2---+12---+6---+5---+22---+3---+6---+22---+5---+4---+3---+22-..6---+5---+4---+3-23-..6-..5-..4-..3

(Odd Length)(Odd Length)(Odd Length)(Even Length)(Odd Length)(Odd Length)(Even Length)(Odd Length)(Odd Length)(Even Length)(Even Length)(Odd Length)(Odd Length)(Odd Length)(Odd Length)(Odd Length)(Even Length)(Odd Length)(Odd Length)(Even Length)(Odd Length)(Even Length)

From the circuit calculations given above, it is clear thatthe circuits formed in PDN are a combination of odd and evenlengths.

Lemma 9: The matrix of PDN is always n <n. In a PDNmatrix, the row i is formed from row (i±l) by shifting i numberof bits.

Proof: The adjacency matrix of the PDN is shown below:

From the adjacency matrix of the Perfect DifferenceNetwork, the second row of the PDN is obtained by shiftingthe bits of first row towards right as given below:

First row:

0101101

Now shift the bits of first row by one towards right, thenthe row will be:

1010110The row that we obtained after the shifting of bits of first

row towards right is same as the second row:

1010110Now we will do this for the third row of the adjacency

matrix. For this we have to shift the bits of second row by onetowards right, already obtained from first row by shifting ofbits by one, as given below:

Second row:

010 1 0

Now shift the bits towards right by one, the row will be:

o I 0 I 01The row that is obtained from the second row is same as

third row of the adjacency matrix of the Perfect DifferenceNetwork. Ifwe do it for the remaining rows, we can easily getthe next rows of adjacency matrix of the Perfect DifferenceNetwork.

Lemma 10: Diameter of a Perfect Difference Network andHypercube can be calculatedfrom their adjacency matrices.

Proof: The adjacency matrix of the PDN having 7 nodes isshown in Fig. 9. It shows dotted lines between rows/columnsfor Diameter ofPDN when n=7.

{I 1 .2 3 ••(o

J

0 1 I} 1 11 0 0

2 'it 1/ '0

.'1

lo 0

4- 0 1 .'0

~ n 0 1

6 1 0 '0\.

5o

Fig. 8: Adjacency Matrix ofPDN Having 7 Nodes

n Il} n 1

0

2 It 1

2 3 •• 5

+-+ (}o -1--41

o n -l ->-: l-~--; c~ a -+--+ (}6'1..10+-+

(} ()

o 0

Fig. 9: Diameter ofPDN when n=7

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~---~

--1-- --";;.

lJJ

-1- _.__.•..

11

Fig. 10: Diameter ofPDN when n=13

w .U .U

The adjacency matrix of the PDN having 13 nodes isshown in Fig. 10.

From the matrices Fig. 10 & Fig.Ll , the consecutive two I' srepresented as dotted lines in each matrices shows the diameterof these two architecture i.e., the maximum length from anysource node to the destination node in Perfect DifferenceNetwork is 2 while as it is 3 in case of Hypercube. It showsdotted lines between rows/columns for Diameter of PDN whenn=13.

Similarly the diameter of Hypercube can be calculated asshown below in the Fig.l1. It shows dotted lines between tworows/columns as diameter of Hypercube.

Lem mall: The number of circuits formed in hypercube is ofeven length.

Proof: 3D hypercube and its circuits whose source anddestination node is '0' is shown below as:

Circuits Length0----> 1 ---->2 ---->3---->00----> 1 ---->2 ---->3---->4---->7---->00----> 1 ---->2 ---->3---->4---->5---->6---->7---->00----> 1 ---->2 ---->5 ---->4---->3---->00---->1---->2---->5->4---->7->0o ---->1 -> 2 ---->5 -> 6 -> 7---->00----> 1 -> 2 -> 5 -> 6---->7->4---->3->0o ---->I -> 6 ---->5 ---->4 -> 7->0o ---->1 ->6 ---->7 -> 0o ---->1 ---->6 -> 7 ---->4---->3-> 0o -> 1 -> 6 -> 5 -> 2-> 3-> 0o ---->1 ---->6 ---->5 -> 2 -> 3 ->4----> 7 ->00----> 1 -> 6 -> 7 -> 4 -> 5-> 2---->3---->00---->3->2->5---->4->7---->00---->3->2---->5->4->7---->6---->1->00->3->2->5---->6->7->00---->3->2->5->6->1->00---->3->2->1---->6->7---->00->3->2---->1---->6->5->4---->7---->00->3->4---->7->0

(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)

0---->3---->4->7->6---->00---->3->4->5---->6---->7---->00---->3---->4---->7->6---->5---->2---->1---->00---->3---->4->5->6->1->00---->3---->4---->5->2---->1->00---->3---->4---->5---->2->1->6->7---->00->7->4->3---->2---->5---->6---->1---->00---->7->4---->5---->2->1---->00->7---->4---->5---->2---->3---->00->7---->4---->5---->6->1->2->3---->00->7->4->5->6---->1---->00->7->6---->5---->4---->3---->00->7->6---->5---->2---->3---->00->7->6->5---->2---->1---->00->7->6->5---->4---->3---->2---->1---->00->7->6---->1---->2---->3->00->7->6---->1---->2---->5---->4---->3---->0

(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)(Even Length)

From the above circuit calculations it is shown that thelength of each circuit is even. Therefore, number of nodes inhypercube is Z", if the nodes are even then the associatedadjacent edges are also even, hence proved.

a 1 :2 s 4- 5 (> '7

0 o ,f 0 _'1. e 0 i) •.,,1" "

,:t ~,- {} ~: u 0 0

~~..• oZ 0: .A' n /1 0 ,"1 0 o

" ,.."~ ;'" J

0 ;£' ~'0 Q il Q

" 0 0 0 ,{( 0 /t ;0 ;1.-,- ". .,

:$ (} 0 ~- 0 .;r o ;.,,' (J

6 0 "l 0 (I o A a ,.3,.J . -."7 J-' a 0 a ~ 0 }o' 0

Fig. 11: Diameter of Hypercube.

o

Fig. 12: 3D Hypercube

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Lemma 12: The minimum cycle length of the circuits formedinHypercube is 4 and maximum is 8.

Proof: As shown in the Lemma 11, it is clear that theminimum length of the circuit is 4 and its maximum length is8, hence proved.

Lemma 13: All the rows and columns of sub-graph of matrixof hypercube is a circuit if the sub-graph matrix of matrix ofhypercube is minimum 4 x4 matrix.

Proof: From the Lemma 11, it is clear that each face of thehypercube is a combination of 4 vertices; therefore each faceof the hypercube is a sub graph of the 3D hypercube graph. Itis also shown in the above matrix (Fig. 13) that sub-graphmatrix of matrix of hypercube is minimum of 4x4 matrix,hence proved.

Lemma 14: When two Hypercubes are connected then thedegree ofmiddle vertices are increased by one and the degreeor corner vertices remains same.

Proof: As we know that the degree of the hypercube is n i.e., 3, but it can be seen from the Fig. 14 that node' 1', '4', '7', and, 10' has degree n+ 1 i.e. 4 and the remaining nodes have thesame degree. Hence when two hypercubes are connected thenthe degree of middle vertices are increased by one and thedegree or corner vertices remains same.

0 2 2, ::I •• ~ G ,.0 0 1. I) .1 ! 0- 0 0 i

'1 :t Q 2 0 ! [I () :i. o

:2 Q Ji Q Ol 0 Cl 0 o

s ; 1, (I ::I.• o :1 o " {;I.- -.,---_. -..•..-~---.-- _ .._.'_ ..- .--,.-~.. -,----" _._.4 Q Q 0 Ji {;I '1 0 '1

5 0: () () ,I Q ."IS 0 1 j) () n- o7' 1 o 0 00 1. 0 .1 "

Fig. 13: Matrix of hypercube is 4 sub-matrices of 4x4

1

Fig. 14: Two connected hypercubes.

III CONCLUSION

A comparative study of hypercube and perfect differencenetwork is done on the basis of topological properties.

Hypercubes are loosely coupled parallel processors basedon the binary n-cube network; n-cube parallel processorconsists of 2" identical processors. In hypercube architecture,the degree and diameter of the graph is same i.e. n. So becauseof this equality they achieve a good balance between thecommunication speed and the complexity of the topologynetwork Perfect difference networks (PDNs) that are based onthe mathematical notion of perfect difference sets to comprisean asymptotically optimal method for connecting a number ofnodes into a network with diameter 2 and its performance liesbetween hypercube and complete graph.

In this paper, we have derived the properties of hypercubeand perfect difference network in the form of lemmas forcomparing these two architectures. The comparison betweenthese architectures has been shown with the help of circuits.The circuits formed in hypercube are of even length while asin PDN, it is a combination of odd and even lengths. We havederived that the minimum length of circuits formed inhypercube is 4 and maximum 2" while as in PDN, minimumlength is 3 and maximum length is the total number of nodesformed in PDN. We have studied that the adjacency matrix ofhypercube is a combination of 4 sub-matrices of order 4x4,while as in PDN, there is only nxn matrix. We have exploredthat the degree of middle vertices of two connected hypercubesis increased by one while as in PDN of PDN s, the degree of anode is increased by 2. We have studied that the number ofvertices in PDN is odd and has even number of edges while as,it is not the case in Hypercube. We have made attempts tostudy the circuits to show the robustness of these architectures.In these architectures, the diameter remains same duringnormal course and during link failure.

The comparison between these two architectures and theirdiameters can be shown on their adjacency matrices as well.

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