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ScienceDirect Procedia Computer Science 47 (2015) 360 – 367
soEnergy of some standard graphs S.P.Jeyakokila a, P.Sumathi b a Department of Mathematics, Lady Doak College, Madurai.E-mail:
[email protected] b Department of Mathematics, C.Kandasamy Naidu College, Chennai. E-mail:
[email protected]
Abstract Let G be finite non trivial graph. In the earlier papers idegree, odegree ,iodegree, oEnergy and soEnergy were discussed and they are calculated for some standard graphs. In this paper energy graph is obtained for path and cycle and soEnergy is also found out. 2015The TheAuthors. Authors. Published Elsevier ©©2015 Published by by Elsevier B.V.B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Peer-review under responsibility of organizing committee of the Graph Algorithms, High Performance Implementations Peer-review under responsibility of organizing committee of the Graph Algorithms, High Performance Implementations and Applications (ICGHIA2014). and Applications (ICGHIA2014) Keywords: idegree; odegree; oidegree; oEnergy; Energy curve; soEnergy;
Section 1 1.1 Introduction: Given a finite graph G=(V,E) and any proper subset D of the vertex set V=V(G) of G, we associate a non negative integral matrix ܣ ሺܩሻ ൌ ሺܽ ሻ of order ȁܦȁܺȁܦȁ with D so that the ith diagonal entry in the matrix counts precisely the number of edges that join the i th vertex of G with vertices in V-D so that these partial degrees of the vertices in D are precisely the eigen values of ܣ ሺܩሻ whence their sum may conceived as energy ߝீ ሺܦሻ of the given set D. This is the classical way of defining energy of a set. In the earlier papers Energy graph for some standard graphs like complete graph, bipartite graph , star graph and caterpillar graphs are calculated and soEnergy was also calculated.In this paper soEnergy of Path and cycle are calculated . The basic definitions are as follows
x
S.P.Jeyakokila Email:
[email protected]
1877-0509 © 2015 The Authors. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Peer-review under responsibility of organizing committee of the Graph Algorithms, High Performance Implementations and Applications (ICGHIA2014) doi:10.1016/j.procs.2015.03.217
S.P. Jeyakokila and P. Sumathi / Procedia Computer Science 47 (2015) 360 – 367
Definition 1.1: idegree Let G be a graph and D be a minimal dominating set, the number of edges which join the vertices of -D into itself is called idegree with respect to D and it is denoted by id D Definition 1.2: odegree Let G be a graph and D be a minimal dominating set, the number of edges which join the vertices of V-D onto the vertices of D is called odegree with respect to D and is denoted as od D. Definition 1.3: oidegree Let G be graph and D be a minimal dominating set, oidegree is the difference between od and id if od > id and it is denoted by oiD . Definition 1.4: iodegree Let G be a graph and D be a minimal dominating set, iodegree is the difference between od and id if id > od and it is denoted by ioD, Definition 1.5: oEnergy and D be a minimal dominating set, oEnergy of a graph with respect to D oLet H GG(Dbe) ais graph the summation of all oid’s if od greater than id . denoted by Definition 1.6: Energy graph Let G be a graph and the D be a minmal dominating set ,then energy graph is the curve obtained by joining all the oEnergy, taking the no of vertices of the set D, plot the number of points of V-D as the x axis and the oEnergy as the y axis Definition 1.7: soEnergy Let G be a graph , soEnergy of a graph with respect to a minimal dominating set D is n
¦
Di V G
oH G ( Di ) where Di
Di 1 V j ,V j ,is the vertex with minimum oi degree, Di is the cardinality
of minimal dominating set D, is denoted by so G (D) .
Algorithm to draw an Energy Curve: Step1: Find a minimal dominating set D . Step 2: Find the idegree, odegree and energy for the vertices in the graph induced by . Step 3: Shift a vertex with minimum positive oEnergy to the set D. If no such positive oEnergy exists, shift a vertex with oEnergy zero to the set D otherwise shift a vertex of maximum negative oEnergy to the set D. Step 4: Find the idegree, odegree and energy for the vertices in the new with respect to new D set. Step 5: Repeat steps 2, 3 and 4 until V D 0 Step 6: Plot the x axis with the number of vertices of the set V-D and y axis with the oEnergies. Plot the points (|V-D|, oH G ) , and draw the corresponding graph. Section 2 2.1 Energy Graph of P7 for different dominating sets . Energy graph of P7 for four different sets. The sets are as follows:
Energy Curve of P7 with `D={1,4,7}
Energy curve of P7 with D={2,4,6}
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energygraph pf p7 with D=1,4,7 energygraph of p7 with D=2,4,6
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energy graph of p7 with D={2,5,7)
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Energy graph of p7 with D=2,5,6
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Fig. 1. (a) Energy Curve of P7 with `D={1,4,7}; (b) Energy curve of P7 with D={2,4,6} (c) Energy curve of P7 with D={2,5,7}; (d) Energy curve of P7 with D={2,5,6)
Theorem 2.1 Let G be a path graph Pn, (i) Let D be the dominating set with cardinality n=3m,m t 1,then
n m 1 if d (v) 2 except for one vertex in D soH G ( D) ® if d (v) 2 v D 2m ¯ (ii)
Let D be the dominating set with cardinality n=3m+1,
m t 1 then,
6.5
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so G ( D)
3 2 0 2(m 1) if d (vi ) 1 for exactly one vertex ° 2 1 2(m 1) if d (vi ) 2 vi with one adjacent vertex ° ® °6 5 4 2 0 2(m 2) if d (vi ) 2i with no adjacent vertices, m 2 ! 0 °¯ 2m if d (vi ) 1 for exactly two vertices
(iii) Let D be the dominating set with cardinality n=3m+2,
m t1
1 2m if d (vi ) 1 for exactly one vertex ° 4 3 2 0 {m 1)2 if d (vi ) 2vi so G ( D) ® °4 2 0 2(m 1) if d (v ) 2 for exactly 2 vertices i ¯ proof: Let G be a path with n number of vertices. Case i: Let D be the dominating set with cardinality 3m. Sub case (i) Let D={ v2 , v5 , v8 ,...., vn1 }} then V-D={ v1 , v3 , v 4 ,...., v n 2 , v n } with cardinality n-m. Initially vertices vi and vn have oidegree 1 and hence oEnergy=2 and all other vertices have idegree and odegree 1 in the consecutive stage oenergy will 1 and 0 respectively. At this stage all other vertices have iodegree 0, so shifting any one of the vertices to the dominating set oEnergy is 2 in the consecutive stage oEnergy is 2 and 0 respectively and this happens for Therefore soEnergy = 2+1+(
nm 1 times. 2
nm 1 )2= n-m+1 2
Subcase (ii) Let D={ v1 , v4 , v7 ,...., vn2 , vn } with d(vi)=2 for exactly 2 vertices then V-D={ v2 , v3 , v5 , v6 ,...., vn3 , vn1 } Initially vn have idegree 0 and odegree 2 and hence oidegree of vn is 2 and all other vertices have idegree and odegree 1 ,oidegree 0, oEnergy at this stage is 2. Since idegree of the shifted vertices is 0 there is no change in the idegree ,odegree and iodegree of the remaining vertices and hence oEnergy is 0. Shifting any one of the vertices to the dominating set the vertex adjacent to the shifted vertex will alone have idegree 0 and odegree as 2 and hence oEnergy is 2. This procedure happens for m number of times till all the vertices are shifted to D, Therefore soEnergy=2m, Case(ii) Let D be the dominating set with cardinality n=3m+1. SubCase (i) Let D= { v2 , v5 , v8 ,...., vn2 , vn } with d(vi)=1 for exactly one vertex then V-D={ v1 , v3 , v4 ,...., vn4 , vn3 , vn1 }. Initially vertex v1 and vn-1 have iodegree 1 and 2 respectively hence oEnergy is 3. Shifting vertices of minimum oidegree to the dominating set, in the consecutive steps oEnergies are 2 and zero. All other vertices have idegree and odegree 1 and 1 till this stage and oEnergy is 0. Shifting any one of the vertices to the dominating set oEnergy is 2 and 0 respectively. oEnergy happens to be 2 for m-1 number of times. Therefore soEnergy=3+2+0+2(m-1) Subcase(ii) Let D={ v2 , v5 , v8 ,...., vn2 , vn1 } with d(vi)=2 for exactly 2 vertices then V-D={ v1 , v3 , v4 ,...., vn4 , vn3 , vn }. Initially vertices v1 and vn have iodegree 1 and all other vertices have 0 and hence oEnergy=2.
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In the consecutive stages oEnergies are 1 and zero as in the remaining cases 2 and 0 happens alternatively for m-1 times. Therefore soEnergy = 2+1+0+2(m-1). Subcase(iii) Let D={ v2 , v5 ,...., vn3 , vn1 } with d(vi)=2 for every i with no adjacent vertices then V-D={ v1 , v3 , v4 ,...., vn2 , vn }. Always the idegree and odegree of v1& vn are zero and one, hence iodegree of v1 & vn are 1. The vertices that lies in between two vertices have iodegree 2 , according to our choice this happens for two vertices and all other vertices have iodegree 0. Hence oEnergy is 6. Shifting the vertices of minimum degree to the dominating set the oEnergy in the consecutive steps are 5 and 4. After this stage the vertex of minimum degree are two.hence oEnergy are respectively 2 and zero. Till this stage all other vertices have iodegree 0, shifting the vertices oEnergy is 2 and this happens for m-2 times. Therefore soEnergy=6+5+4+2+0+2(m-2) when (m-2)>0. Sub case (iv) Let D={ v1 , v4 , v7 ,...., vn2 , vn } with d(vi)=2 for exactly 2 vertices then V-D={ v2 , v3 , v5 , v6 ,...., vn3 , vn1 } The case is similar to subcase ii of case i. soEnergy=2m. Case (iii) Let D be the dominating set with cardinality n=3m+2 Subcase(i) Let D={ v2 , v5 ,...., vn3 , vn1 } with d(vi)=1 for exactly one vertex and V-D={ v1 , v3 , v4 ,...., vn4 , vn2 , vn1 }. At the initial stage vertex v1 alone have iodegree 1 and all other vertices have iodegree 0 and hence oEnergy is 1 and zero in the next stage. As discussed in other cases oEnergy is 2 and it appears for m number of times. Therefore soEnergy=1+2m. Subcase(ii) Let D={ v2 , v5 ,...., vn3 , vn1 } with d(vi)=2 for all vertices then V-D={ v1 , v3 , v4 ,...., vn4 , vn2 , vn1 }. In the initial stage vertices v1 and vn have iodegree 1,vn-2 lies between vn-3 and vn-1 hence iodegree of vn2 is 2. Hence oEnergy=4, In the consecutive stages oEnergies are 3,2 and zeroes. In the remaining stages oEnergy appears to be 2 for (m-1) stages as in the other cases. Therefore soEnergy=4+3+2+0+2(m-1). Subcase(iii) Let D={ v1 , v3 , v6 ,...., vn2 , vn } with d(vi)=1 for exactly 2 vertices then V-D={ v2 , v4 , v5 ,....vn1 } At the first stage vertices v2 and vn-1 have idegree 0 and odegree 2 and hence oidegree 4, Shifting any one of vertices with oidegree 2 and in the consecutive stages oEnergies are 0 and 2 respectively. As in the other cases 2 and zero appears for m-1 times. Therefore soEnergy=4+2+0+(m-1)2 Section 4 4.1 Energy Graphs of Cycle Cn Energy Graphs of Cycle Cn for different dominating sets are given below using the above said algorithm.
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energ graph of c 9 with respect to dominating set D=2,5,8 6
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energ graph of c 8 with respect to dominating set D=2,5,8
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energ graph of c 8 with respect to dominating set D=2,4,6,8
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Fig. 1. (a) Energy graph of C9 with respect to dominating set D=2,5,8; (b) ) Energy graph of C9 with respect to dominating set D=2,4,7,9; (c) ) Energy graph of C8 with respect to dominating set D=2,5,8; (d) ) Energy graph of C8 with respect to dominating set D=2,4,6,8;
Theorem 4.1: soEnergy of cycle (i) When n is odd D={ v1 , v3 , v5 ,...., vn2 } where n is odd and n=2m+1 then
soH =(m-1)2+(m-2)2+----+2+0+2+0
n(n 2) (ii) When n is even D is minimal but not minimum D={ v2 , v5 , v7 ,...., vn }then soH = D is minimum and cardinality of D is 2m then soH =2(m-1) Proof: Let G be cyclic graph Cn with n number of vertices.
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Case (i) When n is odd and n=2m+1. Let D be the dominating set { v1 , v3 , v5 ,...., vn2 } , V-D={ v2 , v4 , v6 ...., vn3 , vn1 , v1 } At the initial stage except for vn-1 and vn all other vertices have idegree 0 and odegree 2, iodegree 2. As vn-1 and vn are adjacent vertices they have idegree and odegree 1 and oidegree 0. And hence oEnergy=(m-1)2. As idegree is zero for these m-1 vertices consecutive oEnergies are (m-2)2, (m-3)2,...,0. Now for the remaining two vertices vn-1 and vn, shifting any one of the vertices to the dominating set. oEnergy=2+0 Therefore soEnergy= (m-1)2+(m-2)2+----+2+0+2+0. Case(ii) Let cardinality of V be n where n is even and n=2n1. Subcase(i) Let D be the minimum dominating set { v1 , v3 , v5 , v7 ,...., vn1 } and V-D={ v2 , v4 , v6 ,...., vn } . By the choice of D idegee of each vertices vi would be 0 and odegree of each vertex is 2 and cardinality of D is n/2 and hence oEnergy at the first stage is (n/2)2. As idegree of all the vertices are zero there is no change in the idegree or odegree of each vertex except the oEnergy decreases by 2. Therefore soEnergy=n+(n-2)+...+2+0 As n is even soEnergy=2(1+2+...+ =
n2 n +) 2 2
n(n 2) 4
Subcase (ii) Let cardinality of v be 2m. Let d be the minimal dominating set idegree of each vertex would be 1 or zero and odegree be 1 or two, oidegree of the vertices would be 0 or 2. oEnergy at this stage is 0 or 2. Shifting the vertices of minimum oidegree to the dominating set at the second stage the vertex corresponding to the oEnergy 0 or 2 is shifted to D. If vertex of oEnergy 2 is shifted then there is no change in the idegee and odegree and oidegree of other vertices since idegree of the vertex is zero. Therefore oEnergy=0 If the vertex of oEnergy 0 is shifted then idegree of any one of the vertices will be zero and odegree 2 oidegree of that vertex is 2. Therefore oEnerg=2 Either one of the cases happens alternatively or it happens for m-1 tmes. Therefore soH =0+2+0+2+...+2+0 where 2 appears for m times. Therefore soH =2(m-1) References: [1] Acharya B.D and Mukti Acharya, New algebraic models of a social system Indian J.Pure and Appl.Math., 17(2)(1986),150168. [2] Acharya B.D., set-valuations of graphs and their applications, In:Proc.symp. optimization, Design of Experiments in Graph Theory, (Dec. 15-18, 1986); (Eds:H.Narayanan and G.A. Patwardhan, IIT Bombay, 1987), 218-249. [3] Everett M. and Nieminen J., Partitions and homomorphisms in directed and undirected Graphs, J.Math. Sociology, 7(1980), 91-111. [4] Cartwright D. and Harary F., A graph-theoretic approach to investigation of system enviromnet relationship, J.Math. Sociology, 5(1977), 87-111. [5] Peay E.R., Matrix operations and the properties of networks and directed graphs, J.Math.Psychol.,15(1)(1977), 89-101. [6] Peay E.R., structural models with qualitative values, J.Math.Psychol., 8(2)(1982), 161- 1992. [7] Reis H.T., Collins W.A. and Berscheid E., The relationship context of human behavior and development, Psychol.. Bull., 126(3)(2003, 844-872. 0[8] Seidman S.B., LS sets as cohesive subsets of graphs and hypergraphs, In: Proc. SIAM Conf. on Appl. of Discrete Mathematics, Rensselare Polytechnic Institute, 1981. [9] Seidman S.B., Structures included by collections of subsets: A hypergraph approach, Math. Soc.Sci.,1(1981), 381-396. [10] Acharya B.D., Rao S.B., Sumathi P., Swaminathan V., Energy of a set of vertices in a graph, AKCE J.Graphs. Combin., 4
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