cut-set

Fundamental Circuits And Fundamental Cut-Sets Yashaswini Hegde

 Abstract—The

objective of this paper is to discuss the relationshi relationship p between between fundamental fundamental circuits circuits and fundamenfundamental cut-sets such as with with resp respec ectt to a span spannin ning g tree tree a chor chord d that that dedetermin termines es a fun fundam dament ental al circui circuitt τ  occurs occurs in every every fundamental cut set associated with the branches in τ  and not in any other. every circuit has even number of egdes in common with any cut-set fundamental cut-set matrix and transpose of fundamental circuit matrix are orthogonal. •

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Figu Figure re 1.

A Gra Graph ph,G ,G

Figure Figure 2.

A Spanning Spanning Tree Tree Of The The Graph,T Graph,T

I. I NTRODUCTION •

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spannin ing g tree tree of  Fundamental Circuits:Circuits:- A spann a graph is a sub graph, which is a tree and connect connectss all the vertic vertices es togeth together er.. Suppos Supposee if  an edge is added between any two vertices of  a tree a circuit is created, because there already exists one path between any two vertices of a tree.Suppose if a spanning tree T is considered in connected graph G, adding any one chord(an edge which is absent in T but present in G)to T, will create exactly one circuit. Such a circuit, formed by adding a chord to a spanning tree, is called fundamental circuit. A graph will have fundamental circuits as many as its chords with respect respect to its spannin spanning g tree, tree, under under conside consideraration.Along with this, – Circuit is a fundamental circuit only with respect to a given spanning tree – A given circuit may be fundamental with resp respec ectt to one one give given n tree tree,, but but not not with with respect to a different spanning tree of the same graph. ough the the numb number er of the the funda fundame ment ntal al – Tough circu circuit itss in a grap graph h is fixed fixed,t ,the he circ circui uits ts that become fundamental, change with the spanning trees. Here is a graph and its spanning tree Fundamenta Fundamentall Cut-set Cut-set A cut-set is a set of edges edges whose whose remov removal al would would discon disconnect nect a

Department of Computer Science, University of Mysore, Mysore – 570006, Karnataka, India

graph. And the concept of the cut-set is closely rela relate ted d to circ circui uit. t. If the the verti ertice cess of a non non direct directed ed graph graph G=(X,A G=(X,A)) are partit partition ioned ed into into two sets X1 ad X2 where X1 belongs to X and X2 is the complement of X1 in X.Then the set of links of G whose terminal vertices lie one in X1 and other in X2 is called cut-set of G. To underst understand and the relati relation on betwee between n the cutcutset set and and fund fundam amen enta tall circ circui uitt it is requi required red to unde underst rstan and d the the relat relatio ion n betw betwee een n a span spanni ning ng tree tree and and the the cutcut-set set.. Sinc Sincee a spann spannin ing g Tree is mini minima mall set set of edge edgess that that conne connect ctss all all the the vertices and the cut-set is minimal set of edges that disconnects some vertices from the other, both should have at least an edge in common. The The fund fundam amen enta tall cutcut-se sets ts with with resp respect ect to a spanning tree T is defined as the n-1 cut-sets each one of which contains one and only edge

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which is in the spanning tree T. And thus if T is a spanning tree of a non-directed graph G, the fundamental cut-set determined by an edge ei of T is composed of  ei and those edges of  G not in T, which when added to T lead to fundamental circuits containing ei . Fundamental cut-set and Fundamental circuit Matrices The fundamental circuit is of the form Bf =[IB] where I is the unit matrix. Similarly Fundamental cut-set matrix is defined as C f =[CI]

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The transpose of fundamental circuit matrix Bf T  and C f  the fundamental cut-set matrix C are orthogonal. ie Bf T .C f  = 0. This information tells that- Each circuit cut by a cut- set has an even edges in common with the cut-set. Bf T  + C f  = 0. II . P ROBLEM S TATEMENT- R ELATION B ETWEEN F UNDAMENTAL C UT-S ET AND F UNDAMENTAL C IRCUITS

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Generation of a graph, which can be generated randomly. Creation of a spanning tree for the given graph. This is generated using DFS technique and in the resulting adjacency matrix only branches that in the array resulting from the algorithm is retained as 1 rest is set to 0 Generating the the fundamental circuit matrix. This is generated as a matrix of chord X branch with respect to the spanning tree. Introducing a chord a cycle is detected as the corresponding row is updated. Generating the fundamental cut-set matrix. Checking for the defined relation. Checking both matrices are orthogonal.

Generating the fundamental cut-set matrix and checking for the relation(even number of edges) and also checking both matrices are orthogonal are explained below. The following figures shows all fundamental circuits and respective fundamental cut-sets.

Prove the relation between Fundamental Circuit and all fundamental cut-sets. Such as •

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The fundamental cut-sets with respect to a spanning tree T is defined as the n-1 cut-sets each one of which contains one and only edge which is in the spanning tree T. Each circuit cut by a cut- set has an even edges in common with the cut-set. The transpose of fundamental circuit matrix Bf T  and C f  the fundamental cut-set matrix C are orthogonal. ie Bf T .C f  = 0.

Figure 3.

Fundamental Circuit, with edge e3

Figure 4.

Fundamental Circuit, with edge e4

 A. Data Structures Used  •

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Double dimensional array to hold the fundamental cut-set. Double dimensional array to hold intermediate transposed matrix Double dimensional array to hold the result of  the multiplication.

 B. Development Of Algorithm

The development of the algorithm contains 6 stages.

The fundamental circuit matrix is -

Figure 5.

Fundamental Circuit, with edge e7

Figure 8.

Fund cut-set,with edge e2,e3,e4

Figure 6.

Fundamental Circuit, with edge e9

Figure 9.

Fund cut-set,with edge e4,e5,e7,e9

c c1 Bf  = c2 c3 c4

e3 e4 e7 e9 e1 e2 e5 e6 e8

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 0 0

1 1 0 0

0 1 1 1

0 0 1 1

0 0 0 1

let us consider cut-set, e4,e5,e7,e9 due to the removal of a edge e5 in a spanning tree. From the fundamental circuit matrix get the rows which contain e5=1. Such rows aree1,e2,e4,e5 -considering only ones in row c2 0 1001110 0 e5,e6,e7 - in row c3, 0 0 1 0 0 0 1 1 0 e5,e6,e8,e9- in row c4 0 0 0 1 0 0 1 1 1 To prove the relation-1, it can be checked that e5=1 in all the three above mentioned fundamental circuits. and to prove the relation 2- for a row in cut set with branch e5=1 all the fundamental circuits with e5=1 ie e5 on are checked by doing  ringsum operation and see any two bits are reset. If so, that means there are two branches are common. •

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Figure 7.

Fund cut-set,with edge e1,e3,e4

The fundamental cut-set matrix with respect to the same spanning tree.

k k1 k2 C f  = k3 k4 k5

e3 e4 e7 e9 e1 e2 e5 e6 e8

1 1 0 0 0

1 1 1 0 0

0 0 1 1 0

0 0 1 1 1

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1



0111001 00 0100111 00  0111001 00 0010001 10  0111001 00 0001001 11 It can be observed exactly two fields have been  reset after the ringsum operation. From the above described fundamental circuit it can be observed that its of the form

Figure 10.

Figure 11.

Algorithm 1 create sptree(int vertex,int numVertices) 1: setting the visiting vertex as 1 to avoid cycles 2: g visited[vertex]=1 3: for (i = 0; i <= numV ertices; i + +) do 4: if  (g adj list[vertex][i] == 1) then g dst lbl[i].dist = g adj list[vertex][i] 5: g dst lbl[i].nei node = vertex 6: iff neighbour exists and not yet visited 7: 8: if  (!g visited[i]) then go visit next 9: create sptree(i,numVertices) 10: 11: end if  12: end if  13: end for

Fund cut-set,with edge e6,e7,e9

Algorithm 2 create fundckt(int vertex,int ptr,int vstart) 1: g visited[vertex]=1 2: for (i = 0; i <= numV ertices; i + +) do 3: if  (g adj list[vertex][i] == 1) then 4: if  (!g visited[i]) then path[vertex][i]=i 5: ptr++ 6: create fundckt(i,ptr,vstart) 7: 8: end if  9: else if  (vstart==i) then brek  10: 11: end if  12: end for

Fund cut-set, with edge e8,e9

Bf =[IB] where I is the unit matrix. Similarly from the Fundamental cut-set matrix, it can be observed that,it is of form

C. Apriory Analysis

C f =[CI]

The algorithm contain the following modules

and that could be generated as an orthogonal matrix to the fundamental circuit matrix. M = [B T I ]

k k1 k2 C f  = k3 k4 k5

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e3 e4 e7 e9 e1 e2 e5 e6 e8

1 1 0 0 0

1 1 1 0 0

0 0 1 1 0

0 0 1 1 1

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

If a fundamental circuit has rows of number of  chords the fundamental cut-set will have number of  branches as number of rows.

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Generation of spanning tree This takes the time complexity of  O(n2 ),where n is the number of vertices. Creation Of Fundamental Circuits This also takes of  O(n) since it is checking the spanning tree with a cycle. Creation of Fundamental Cut-set. This involves majorly two loops with number of branch and number of edges.Hence the total time complexity is O(rank X edges) Checking, if the fundamental circuit and fundamental cut-set is having even number of edges. This takes the time complexity f O(n) since it is just checking which bits are reset after ring-

Algorithm 3 create fundcutset by fundckt(int Algorithm 5 check orthogonal(int **BFT,int **BF,int brch,int edg) **CF,int m1,int n1,int m2,int n2) if  (n1==m2) then This loop is to initialize the matrix for (i=1;i <= m1;i++) do for (i=1;i<=brach;i++) do for (j=1; j <= n2;j++) do 3: for (j=1;j<=edg;j++) do g resmatrix[i][j]=0 g csmatrix[i][j]=0 5: for (k=1;k <= n1;k++) do end for g resmatrix[i][j]= 6: end for g resmatrix[i][j]+BFT[i][k] * this loop is to hold the transposed ’B’ part of  CF[k][j] BF matrix. if  (g resmatrix[i][j] mod 2) then for (i=1;i<=(edg-brach);i++) do error,break  k=1 9: end if  for (j=(edg-brch+1);j<=edg;j++) do 10: end for g csmatrix[k][l] = g Bf[i][j] end for k = k +1 12: end for end for end if  l = l+1 return 15: end for this loop is to fill the identity part of the CF matrix. i=1 O(edgesXrankXnullity ) since the usual 18: for (j=(edg-brch+1);j<=edg;j++) do matrix multiplication involves three loops. g csmatrix[i][j] = 1 i = i+1  D. Experimental Analysis 21: end for The experiment is done by generating Fundamental Circuit matrix and then deducing the FunAlgorithm 4 check evennum common edjes(int damental Cut-set matrix out of it. Then a specific *cktrow,int *cutsetrow,int br) row from the Fundamental Cut-set matrix with ctr=0 respect to a branch is chosen and from the Funfor (i=1;i <= br;i++) do damental Circuit matrix those in which the same g reseven[i]=cktrow[i]ˆcutsetrow[i] branch is set to 1 is passed on to the routine 4: if  (cutsetrow[i]==1) && g reseven[i]==0) (check evennum common edjes) one by one for then testing. The counter will be 0 under mod 2 if even ctr++ number of branches are present. To check, if both end if  transpose of Fundamental Circuit matrix and Fundaend for mental cut-set matrix are orthogonal, the transpose 8: if  (ctr mod 2) then of Fundamental Circuit matrix is generated and it relation proved is multiplied with the Fundamental cut-set matrix end if  and each element is checked if 0 under mod 2. If  so they are orthogonal. The out put of the program sum operation. The Fundamental Circuit matrix generating the transpose of the Fundamental 1001101 Circuit matrix. 0100101 The time complexity is O(nullity X edges) Checking, if the transpose of fundamental 0 0 1 0 0 1 1 circuit matrix and fundamental cut-set matrix are orthogonal. The Fundamental Cut-set Matrix This takes the time complexity of   1 0 0 1 0 0 0 •

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1100100 0010010 1110001 The reduced matrix 1001000 1100100 0010010 The resulting matrix multiplication under mod 2 which is reduced 1001000 1100100 0010010 1001000 0101100 0010010 0111110

1 0 1 0 0 0 1

The corrected matrix for mod 2. 1 0 0 1 0 0 0 100100 010010 001000 101100 010010 111110 110110

The ringsum of 1100100 from the cutset matrix and 0100101 ctr=2.

III. C ONCLUSION The test for finding the relation between fundamental circuit and fundamental cut-set mainly involves •

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finding the ringsum of few rows of each with the time complexity of O(edges) and an additional array to hold the result finding the multiplication of transpose of  Bf  and C f  is 0 under mod 2 takes the time complexity of O(edges X rank X nullity) but few more two dimensional arrays to hold the resulting matrix and reduced matrices. R EFERENCES

[1] N.Christofides Graph Theory An Algorithmic Approach. demic Press Inc(London)ltd,1975 [2] N.Deo. Graph Theory. PHI,1974

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