Negation switching invariant 3-Path signed graphs

Journal of Discrete Mathematical Sciences & Cryptography Vol. ? (?), No. ?, pp. 1–16 DOI : 10.1080/09720529.2016.1187959

Negation switching invariant 3-Path signed graphs Deepa Sinha * South Asian University Akbar Bhawan, Chanakyapuri New Delhi-110 021 India Ayushi Dhama † Centre for Mathematical Sciences Banasthali University Banasthali-304 022 Rajasthan India Abstract Formally, a signed graph (sigraph) S is a pair (G, s) that consists of a graph G = (V, E) and a sign mapping or signature s from E to the sign group +, -. Given a sigraph S and a positive integer t, the t-path sigraph (S)t of S is a sigraph whose vertex set is V(S) and two vertices are adjacent if and only if there exist a path of length t between these vertices and then by defining its sign st(e) to be ‘-’ if and only if in every such path of length t in S all the edges are negative. The negation h(S) of a sigraph S is a sigraph obtained from S by reversing the sign of every edge of S. Two sigraphs S1 and S2 on the same underlying graph are switching equivalent if it is possible to assign signs ‘+’ (‘plus’) or ‘-’ (‘minus’) to the vertices of S1 such that by reversing the sign of each of its edges that has received opposite signs at its ends one obtains S2. In this paper, we characterize sigraphs whose negations are switching equivalent to their t-path sigraphs for t = 3. Keywords: Balanced sigraph, Marked sigraph, Signed isomorphism, Switching equivalence, t-Path sigraph. 2010 Mathematics Subject Classification: Primary 05C 22, Secondary 05C 75.  esearch is supported by the Department of Science and Technology (Govt. of India), New R Delhi, India under the Project SR/S4/MS: 409/06. *E-mail:  [email protected] † E-mail:  [email protected]

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1. Introduction We use standard terminology of graph theory Harary [7] and West [12], where a graph is considered to be simple, finite and loopless. For the signed graphs Zaslavsky [14, 15] is followed. Formally, a sigraph is an ordered pair S = (Su , σ ), where Su = (V , E) is a graph called the underlying graph of S and σ : E → {+ , −} is a function from the edge set E of Su into the set {+, –}, called the signature (or sign in short) of S. An independent positive (negative ) edge of S is a positive (negative) edge of S such that there is no positive (negative) edge incident at each end of this edge. The positive (negative) degree of a vertex v ∈ V (S) denoted by d + (v )(d − (v )) is the number of positive (negative) edges incident to the vertex v. A sigraph is all-positive (respectively, all-negative) if all its edges are positive (negative); further, it is said to be homogeneous if it is either allpositive or all-negative and heterogeneous otherwise. A graph (sigraph) is said to be totally disconnected if there is no edge in the graph (sigraph) i.e., it is a null graph (sigraph). The negation of a sigraph S is a sigraph obtained from S by reversing the sign of every edge of S. 〈Vi 〉 determines the induced subsigraph of S on the vertex subset Vi of V(S) whereas 〈Vi 〉 u determines the underlying subgraph of Su which is induced on the vertex subset Vi of V(Su). The sign of a cycle (this is the edge set of a simple cycle) is defined to be the product of the signs of its edges; in other words, a cycle is positive if it contains an even number of negative edges and negative if it contains an odd number of negative edges. A sigraph is said to be balanced if every cycle in it is positive. A sigraph S is called totally unbalanced if every cycle in S is negative. A chord is an edge joining two non adjacent vertices in a cycle. The following characterization of balanced sigraph is well known. Theorem 1. [6] A sigraph S is balanced if and only if its vertex set V(S) can be partitioned into two subsets V1 and V2 (one of them possibly empty) such that every negative edge of S joins a vertex of V1 with one of V2 while no positive edge does so. Lemma 2. [16] A signed graph in which every chordless cycle is positive, is balanced. A marked sigraph is an ordered pair Sµ = (S, µ ), where S = (Su , σ ) is a sigraph and µ : V (Su ) → {+ , −} is a function from the vertex set V(Su) of Su into the set {+, –}, called a marking of S. Switching S with respect to a marking m is the operation of changing the sign of every edge of S to its opposite whenever its end vertices are of opposite signs. The resulting

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sigraph Sm(S) is said switched sigraph. A sigraph S is called to switch to another sigraph S¢ written S ~ S¢, whenever their exists a marking m such that S′ ≅ Sµ (S), where ‘@’ denotes the usual equivalence relation of isomorphism in the class of sigraphs. Hence, if S ~ S¢ we shall say that S and S¢ are switching equivalent. Two sigraphs S1 and S2 are signed isomorphic (written S1 @ S2 or sometimes S1 = S2) if there is a one-to-one correspondence between their vertex sets which preserve adjacency as well as sign. Two sigraphs S1 and S2 are said to be weakly isomorphic (e.g., see Sozánski [11]) or cycle isomorphic (e.g., see Zaslavsky [13]) if there exists an isomorphism f : (S1 )u → (S2 )u such that the sign of every cycle Z in S1 equals the sign of f(Z) in S2 (i.e., f preserves both vertex adjacencies and the signs of the cycles of S1 in S2), where the sign of any subsigraph of a sigraph is defined as the product of the signs of its edges. When y(G) denotes the set of all the sigraphs possible for a graph G, following theorem will be also useful in our further investigation Theorem 3. [11] Given a graph G, any two sigraphs in y(G) are switching equivalent if and only if they are cycle isomorphic. Let Z be a heterogeneous cycle in a sigraph S. By a negative section of Z we mean a maximal set P of vertices of Z such that the subsigraph consisting of the edges of Z joining vertices in P is all-negative and connected. 2.  t-Path Invariant Graphs Escalante et al. (see [3]) introduced t-path graphs in the theory of graphs. t-path graphs are the generalization of open neighborhood graphs introduced by Acharya [1]. t-path invariant graph by Escalante and Montejano [3] was the graphs which satisfied the following equation (G)t @ G (1) The solution for (1) was discovered only for t = 2, 3. For the higher value of t, it is an open problem to find out. Now we give a characterization to define the structure of 3-path invariant graphs. Theorem 4. [3] A graph G of order p is a 2-path invariant graph if and only if G = K p or Kp with p ≥ 3 or the odd p-cycle Cp, p = 2m + 1, m ≥ 2. The structure of 3-path invariant graph is given by the following theorem:

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Figure 1 Double Star K tm,2, n

Theorem 5. [4] A graph G is a 3-path invariant graph if and only if it is isomorphic to any one of the following graphs: (a) Cm, 4 £ m π 3k, k is a positive integer (b) Kn, n ≥ 4 (c) Km, n, 2 £ m £ n (d) The double star Ktm,2, n as shown in Figure-1 (e) E1 = K4 – x, i.e., E1 is obtained from K4 by removing an edge. (f) E2 = the subdivision graph of E1 (g) E3 (h) E4

where E2, E3, E4 are shown in Figure-2.

3.  A System of Sigraph Equivalence Mishra [8] extended the notion of t-path graphs in the theory of sigraphs as follows: a t-path sigraph of a sigraph S is a sigraph whose underlying structure is ((S)t)u and for any edge e = uv, s(uv) = – if and only if every u – v path of length t in S all edges are negative. By the same pattern as discussed above, switching invariant t-path sigraphs are the sigraphs which satisfy the following equation (S)t ~ S, t = 2, 3, ...

(2)

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Figure 2 Showing E2, E3 and E4

The solution of equation (2) was given by Gill and Patwardhan [5], Acharya [2], for t = 2 and 3 respectively. Here, in this paper, we describe all the solutions to

η (S)  (S)t , t ∈ {2, 3} (3)



which we will call negation-switching invariant t-path sigraph. For t = 1 the solution already exists [9]. The following theorem determines the solution to S ~ h(S). Theorem 6: [9] For a connected sigraph S = (Su, s), S ~ h(S) if and only if either (i) Su is bipartite or (ii) there exist subsets V1 and V2 of V(S) such that (a) S = 〈V1 〉 ∪ 〈V2 〉 and 〈V1 ∩ V2 〉 is bipartite, (b)  〈V1 〉 u ≅ 〈V2 〉 u such that degrees of corresponding vertices are preserved in S also and

(c) each odd (even) cycle in 〈V1 〉 is of opposite (same) sign to the corresponding cycle in 〈V2 〉.

Theorem 7: [9] For a given sigraph S = (Su , σ ), S ≅ η (S) if and only if the edge set E(S) can be partitioned into two subsets E1 and E2 such that 〈E2 〉 ≅ 〈η (E1 )〉 and degrees of corresponding vertices are preserved in S.

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4.  Negation-Switching Invariant 2-Path Sigraphs In view of Theorem 4, it has been proved by the authors that the solutions of (3) for t = 2 lies in the sets ψ (C2 m+1 ), m ≥ 2 and ψ (K p ), p ≥ 3. The following theorem determines the solution for (S)2  η (S) that may be found in ψ (C2 m+1 ), m ≥ 2. Theorem 8. [10] For any integer m ≥ 2, a heterogeneous sigraph S ∈ψ (C2m+1 ), is a solution to (S)2 ~ h(S) if and only if it has an odd number of negative sections. The following lemma has been used by the authors to derive the theorem that gives solution to (S)2 ~ h(S) for S ∈ψ (K n ), n ≥ 3. Lemma 9. [10] Let S ∈ψ (K n ) and uv ∈ E(S). If all the other edges of S incident at u and v are negative then uv is a negative edge in (S)2. Theorem 10: [10] S ∈ψ (K n ), n ≥ 3, is a solution to (S)2 ~ h(S) if and only if any one of the following conditions is satisfied: (i) E+(S) consists of

n 2 ,  

positive independent edges or

(ii) V(S) can be partitioned into two subsets V1 and V2 such that the induced subsigraphs ·V1Ò and ·V2Ò are all-negative and the edges across ·V1Ò and ·V2Ò are all-positive or (iii) n is even, n ≥ 6 and the induced subsigraphs ·V1Ò and ·V2Ò on

n 2

vertices are homogeneous of different parities and the edges across ·V1Ò and ·V1Ò are all-negative. 5.  Negation-Switching Invariant 3-Path Sigraphs The following Theorem gives solution to (S)3 ~ h(S) that may be found in ψ (Cm ), 4 ≤ m ≠ 3k for any integer k ≥ 2. Theorem 11. If m is an integer such that 4 £ m π 3k for any integer k ≥ 2 then S Œy(Cm) is a solution to (S)3 ~ h(S) if and only if the number of negative sections of length one in Cm is of the same parity as that of m. Proof: Suppose p and q are the number of negative sections whose length is one and two respectively and the other negative sections are of length k1, k2, ..., kr in some order. By definition of (S)3, it is easy to see that it has a negative edge for each consecutive triple of negative edges in S. Hence (S)3 has two less negative edges for each negative section of length ≥ 3 in S. Thus,

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|E− (S)3 |= ∑(ki − 2).



i =1

On the other hand, r

|E− (S)|= p + 2q + ∑(ki )



i =1

and r   |E− (η (S))|= m −  p + 2q + ∑(ki )  . i =1  



Hence, (S)3 ~ h(S) if and only if they are cycle isomorphic. Therefore, due to Theorem 3 following implications complete the proof:

h(S) and (S)3 are switching equivalent

¤ h(S) and (S)3 are cycle isomorphic ¤

∑

r

(ki − 2) + m − p − 2q − ∑ i=1(ki ) ≡ 0(mod 2) r

i =1

¤ −2r + m − p ≡ 0(mod 2) ¤ m − p ≡ 0(mod 2). This completes the proof. Below, we state a lemma which will be useful in our further investigation. Lemma 12: For S ∈ψ (K n ), n ≥ 4 , and S ∈ψ (K m,n ), 2 ≤ m ≤ n, (S)3 heterogeneous if and only if S has exactly one positive edge.

is

Proof: Suppose S ∈ψ (K n ) n ≥ 4 or S ∈ψ (K m , n ) 2 ≤ m ≤ n, is heterogeneous. If possible, let S has at least two positive edges, say ij and kl (may be j = k). Now, for both the cases, for any two vertices m and n in S, except i and j, there exists a heterogeneous or all-positive 3-path mi, ij, jn in S. Thus, by the definition of (S)3, there is a positive edge mn in (S)3. Also, for the vertices i and j there exists a heterogeneous or all-positive 3-path ik, kl, lj and if j = k, then we have a heterogeneous or all-positive 3-path is, sl, lj, where s is any other vertex in S. Again, by the definition of (S)3, we have a positive edge ij in (S)3. Thus, (S)3 is all-positive, a contradiction to the hypothesis. Hence, S has exactly one positive edge. Conversely, suppose S has exactly one positive edge, say ij. Then, for any two vertices m and n in S, except i and j, there exists a heterogeneous 3-path mi, ij, jn in S. Thus, by the definition of (S)3, there is a positive edge

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mn in (S)3. But, by the definition of (S)3, ij is a negative edge in (S)3. Hence, S is heterogeneous. The following theorem describes the solution in (S)3  η (S) that may be found in ψ (K n ), n ≥ 4. Theorem 13: For any integer n ≥ 4, y(Kn) is a solution to (S)3 ~ h(S) if and only if either S has exactly one positive edge or V(S) can be partitioned into two subsets V1 and V2 such that induced subsigraphs ·V1Ò and ·V2Ò are all-negative and the edges across ·V1Ò and ·V2Ò are all-positive. Proof: Let S be a solution to (S)3 ~ h(S), such that S ∈ψ (K n ), n ≥ 4. If possible, suppose S has more than one positive edge and V(S) can not be partitioned into two subsets V1 and V2 such that induced subsigraphs ·V1Ò and ·V2Ò are all-negative and the edges across ·V1Ò and ·V2Ò are all-positive. Now, by Lemma 12, (S)3 is all-positive and hence balanced. On the other hand h(S) is not balanced. Thus, by Theorem 3, S  η (S), a contradiction to the hypothesis. Hence conditions are necessary. Conversely, let S has exactly one positive edge. Then, (S)3 has exactly one negative edge and h(S) also has exactly one negative edge and hence η (S) ≅ (S)3 which also implies that η (S)  (S)3 . Next, suppose V(S) can be partitioned into two subsets V1 and V2 such that induced subsigraphs ·V1Ò and ·V2Ò are all-negative and the edges across ·V1Ò and ·V1Ò are all-positive. In this case it is clear that S has more than one positive edge then, by Lemma 12, (S)3 is all-positive and hence trivially balanced. It is clear that V (S) = V (η (S)). Now, by the definition of h(S), ·V1Ò and ·V2Ò are all-positive and all the edges across ·V1Ò and ·V2Ò are all-negative in h(S). Thus, by Theorem 1, h(S) is balanced and by Theorem 3, (S)3 ~ h(S). Hence the result. In the next result, we find the solution to (S)3 ~ h(S) for S Œy(Km,n), m, n≥2 Theorem 14: S Œy (Km,n), m, n ≥ 2 is a solution to (S)3 ~ h(S) if and only if either S has exactly one positive edge or S is balanced. Proof: Suppose S Œy(Km,n), m, n ≥ 2 is a solution to (S)3 ~ h(S). If possible, suppose S is unbalanced and has more than one positive edge. Now, by Lemma 12, (S)3 is all-positive and hence trivially balanced. Km,n is a bipartite graph so S and h(S) are also bipartite sigraph. Since, S is balanced, h(S) is also balanced sigraph. Thus, (S)3 is balanced and h(S) is unbalanced. Hence, by Theorem 3, (S)3  η (S), a contradiction.

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Conversely, suppose S has exactly one positive edge. Then, by the definition of (S)3, (S)3 has exactly one negative edge and h(S) also has exactly one negative edge and hence η (S) ≅ (S)3 which also implies that (S)3 ~ h(S). Next, suppose S is balanced. If S is homogeneous then result is trivial as S is a bipartite sigraph. Now, suppose S is heterogeneous. It is clear that in this case S has more than one positive edge. Thus, by Lemma 12, (S)3 is all-positive and hence trivially balanced. Since, S is a balnced bipartite sigraph, h(S) is also balanced. Thus, S and h(S) both are balanced sigraphs and hence by Theorem 3, (S)3 ~ h(S). Hence the result. To find the solution to (S)3 ~ h(S) for S ∈ψ (Ktm,2, n ) we state the following Lemma. m,n m,n Lemma 15: For any heterogeneous signed graph S ∈ψ (Kt,2 ), where Kt,2 is shown in Figure 1, m π 0 or n π 0 and t ≥ 2, (S)3 is heterogeneous if and only if core 〈 x, u1 , u2 ,..., ut , y 〉 of S is either all-negative or has exactly one positive edge.

Proof: Suppose for a heterogeneous signed graph S ∈ψ (Ktm,2, n ) , where Ktm,2, n is shown in Figure 1, m π 0 or n π 0 and t ≥ 2, (S)3 is heterogeneous. If possible, suppose core has at least two positive edge. In core any positive edge will be of the type either xun or uny. Without loss of generality, suppose one positive edge is xun. Now, by the definition of (G)3, the edges in (Ktm,2, n )2 is of the four type yix, xuk, uky and xjy for i = 1, 2, ..., n, j = 1, 2, ..., m and k = 1, 2, ..., t. For any two vertices of type x and yi there is a heterogeneous or all-positive 3-path y i y , yun , un x in S. Then, by the definition of (S)3, we have a positive edge xyi in (S)3. For any two vertices of type xj and y there is a heterogeneous or all-positive 3-path x j x , xun , un y in S. Again we have a positive edge xjy in (S)3. For any two vertices of type uk and y there is a heterogeneous or all-positive 3-path which passes through unx. Then, by the definition of (S)3, we have a positive edge uky in (S)3. Also, for any two vertices of type uk and x, except uk = un, there is a heterogeneous or allpositive 3-path which passes through unx. When, uk = un, then there exists a heterogeneous or all-positive 3-path which passes through the other positive edge. Then, by the definition of (S)3, we have a positive edge ukx in (S)3. Thus, (S)3 is all-positive, a contradiction to the hypothesis. Hence core of S is either all-negative or has exactly one positive edge. Conversely, suppose core of S is all-negative. Then, by the definition of (S)3, core in (S)3 is also all-negative. Since S is heterogeneous, there is at least one positive edge in S and this positive edge is clearly a pendant

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edge. Then, there will be atleast one positive edge in (S)3 also. Hence, (S)3 is heterogeneous. Next, suppose core of S has exactly one positive edge. Then, by the definition of (S)3, this positive edge is negative in the core of (S)3. While other edges in core of (S)3 are positive as there exists a heterogeneous 3-path between these vertices in S which passes through that positive edge. m,n Theorem 16. S ∈ψ (Kt,2 ), m π 0 or n π 0 and t ≥ 2 is a solution to (S)3 ~ h(S) if and only if S is balanced or the CORE of S has exactly one positive edge.

Proof: Suppose S ∈ψ (Ktm,2, n ), m π 0 or n π 0 and t ≥ 2 is a solution to (S)3 ~ h(S). If possible, let S is unbalanced and S has more than one positive edge. It is clear that S is not homogeneous otherwise S will be balanced as S is a bipartite. Now, by Lemma 15, (S)3 is all-positive and hence trivially balanced. According to our assumption S is unbalanced and bipartite so h(S) is also unbalanced. Thus, by Theorem 3, S  η (S), a contradiction to the hypothesis. Hence conditions are necessary. Conversely, suppose core of S is all-negative. Since, S is bipartite, S is balanced and h(S) is also balanced. Now, by the definition of (S)3, core of (S)3 is also all-negative and hence (S)3 is also balanced. Thus, by Theorem 3, (S)3 ~ h(S). Next, suppose core of S has exactly one positive edge. Then, by the definition of (S)3, core of (S)3 has exactly one negative edge. Also in h(S), core of h(S) has exactly one negative edge. Thus, core of (S)3 and h(S) are isomorphic and cycles in h(S) and (S)3 are only due to edges of cores, hence (S)3 ~ h(S). Hence the result.

Figure 3 Solutions to (S)3 ~ h(S) on E1

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Figure 4 Showing E2 and (E2)3

Now we proceed to give solutions to (S)3 ~ h(S) for S Œy(Ei), 1 £ i £ 4. Figure 3 exhibits the solutions to (S)3 ~ h(S) for S Œy(E1), where E1 = K4 – x. Next, we observe that E2 is the subdivision graph of E1 which is shown in Figure 2. By 〈〈l , m〉〉 we denote a three length path between any two points l, m of E2. The following Theorem gives us the solutions to (S)3 ~ h(S) for S Œy (K2). Theorem 17: A sigraph S Œy (K2) is a solution to (S)3 ~ h(S) if and only if (i) if S is balanced or totally unbalanced then number of all-negative 3-paths ··1, dÒÒ and ··4, aÒÒ are of same parity as that of all-negative 3-paths ··1, cÒÒ and ··4, bÒÒ and (ii) if S is unbalanced then all-negative 3-paths ··1, dÒÒ and ··4, aÒÒ are of opposite parity as that of all-negative 3-paths ··1, cÒÒ and ··4, bÒÒ. The result immediately comes from Figure 4 and Theorem 3. The following theorems determine the solutions to (S)3 ~ h(S) for S Œy(E3) and S Œy(E4). Theorem 18: S Œy (E3) is a solution to (S)3 ~ h(S) if and only if the number of all-negative 3-paths containing cut points of E3 is of same parity as that of the number of positive edges in the unique triangle of S. Proof: Suppose S Œy(E4) is a solution to (S)3 ~ h(S). It is clear that in (S)3 and h(S) there is only one triangle. By Figure 5, it is clear that vertices of the triangle in (S)3 are 1, 4 and 5 while vertices of triangle in h(S) are 1, 2 and 3. 2 and 3 are the cut vertices of S. If possible, let conclusion is false. Suppose all-negative 3-paths containing cut points of S is of different

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Figure 5 Showing E3 and (E3)3

parity as that of positive edges in the unique triangle of S. Then, allnegative 3-paths containing cut points of S is of different parity as that of negative edges in the unique triangle of h(S). This shows that (S)3 is balanced while h(S) is unbalanced or vice-versa. Thus, by Theorem 3, (S)3  η (S), a contradiction. Thus condition is necessary. Conversely, suppose condition is true. Then, result is trivial by Figure 5 and Theorem 3. Theorem 19: S Œy (E4) is a solution to (S)3 ~ h(S) if and only if the number of all-negative 3-path each of which contains two lines of the unique hexagon of S is of the same parity as that of the number of positive edges in the hexagon. By the same argument as in the above Theorem, the result follows from Figure 6 and Theorem 3.

Figure 6 Showing E4 and (E4)3

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6. Acknowledgement The authors wishes to thank all the experts who have helped the paper to bring in the present form. References [1] B.D Acharya, Open-neighborhood graphs, Technical Report Number DAE: 22/8/72-G, Indian Institute of Technology, Bombay, May 1973. Also, see: Graph Theory Newsletter, 2(4)(1973). [2] M. Acharya, Switching-invariant three-path signed graphs, Proceeding of Symposium On Optimization, Design of Experiments and Graph Theory (Eds. M.N. Gopalan and G.A. Patwardhan), IIT, Bombay, December 15-17, 1986, 342-345. [3] F. Escalante, L. Montejano and T. Rojano, A characterization of n-path graphs and of graphs having n-th root, J. Combin. Th., Ser. B, 16(3) 282289, 1974(a). [4] F. Escalante and L. Montejano, Trees and n-path invariant graphs, Graph Theory Newsletter, 3(3), 1974(b). [5] M.K. Gill and G.A. Patwardhan, Switching-invariant two-path sigraphs, Discrete Mathematics, 61(1986), 189-196. [6] F. Harary, On the notion of balance of a signed graph, Michigan Math. J., 2(1953), 143-146. [7] F. Harary, Graph Theory, Addison-Wesley Publ. Comp., Reading, Massachusetts, 1969. [8] V. Mishra, Graphs associated with (0, 1) and (0,1,-1) matrices, Ph.D. Thesis, IIT Bombay, 1974. [9]  D. Sinha and A. Dhama, Negation Switching Invariant Signed Graphs, Electronic Journal of Graph Theory and Applications, 2(1)(2014), 32-41. [10] D. Sinha and A. Dhama, Negation Switching Invariant 2-Path SignedGraphs, submitted. [11] T. Sozánsky, Enumeration of weak isomorphism classes of signed graphs, J. Graph Theory, 4(2)(1980), 127-144. [12] D.B. West, Introduction to Graph Theory, Prentice-Hall of India Pvt. Ltd., 1996. [13] T. Zaslavsky, Signed graphs, Discrete Appl. Math, 4 (1)(1982), 47-74.

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[14] T. Zaslavsky, A mathematical bibliography of signed and gain graphs and allied areas, VII Edition, Electron. J. Combin., #DS8(1998). [15] T. Zaslavsky, Glossary of signed and gain graphs and allied areas, II Edition, Electron. J. Combin., #DS9(1998). [16] T. Zaslavsky, Signed analogs of bipartite graphs, Discrete Mathematics, 179(1998), 205-216.

Received July, 2015