On derived t-path, t=2,3 signed graph and t-distance signed graph

Abstract

A signed graph $\Sigma$ is a pair $\Sigma =\left({\Sigma }^u,\sigma \right)$that consists of a graph $\left({\Sigma }^u,E\right)$ and a sign mapping called signature $\sigma$ from E to the sign group $\left\{+,-\right\}$. In this paper, we discuss the t-path product signed graph ${\widehat{\left(\Sigma \right)}}_t$where vertex set of ${\widehat{\left(\Sigma \right)}}_t$ is the same as that of $\Sigma$ and two vertices are adjacent if there is a path of length t, between them in the signed graph $\Sigma$. The sign of an edge in the t-path product signed graph is determined by the product of marks of the vertices in the signed graph $\Sigma$, where the mark of a vertex is the product of signs of all edges incident to it. In this paper, we provide a characterization of $\Sigma$ which are switching equivalent to t-path product signed graphs ${\widehat{\left(\Sigma \right)}}_t$ for $t=2,3$ which are switching equivalent to $\Sigma$ and also the negation of the signed graph ŋ$\left(\Sigma \right)$ that are switching equivalent to ${\widehat{\left(\Sigma \right)}}_t$ for $t=2,3$. We also characterize signed graphs that are switching equivalent to $t$-distance signed graph ${\left(\overline{\Sigma }\right)}_t$ for $t=2$ where 2-distance signed graph ${\left(\overline{\Sigma }\right)}_2=\left(V^{\prime },E^{\prime },{\sigma }^{\prime }\right)$ defined as follows: the vertex set is same as the original signed graph $\Sigma$ and two vertices $u,v\in {\left(\overline{\Sigma }\right)}_2$, are adjacent if and only if there exists a distance of length two in $\Sigma$. The edge $uv\in {\left(\overline{\Sigma }\right)}_2$ is negative if and only if all the edges, in all the distances of length two in $\Sigma$ are negative otherwise the edge is positive. The t-path network along with these characterizations can be used to develop model for the study of various real life problems communication networks.

• •t-path product signed graph.
• •t-distance signed graph.

Graphical abstract

Specifications table

Subject area:	Mathematics
More specific subject area:	Graph Theory
Name of your method:	t-path product and t-distance signed graph
Name and reference of original method:	N/A
Resource availability:	Provided in the Article

Method details

Introduction

A signed graph is a dynamic and vibrant area of research in graph theory, with strong connections to behavioural and social sciences. Its origin can be traced back to Heider [1] and Cartwright [2]. Since then, signed graph theory has rapidly evolved, with signed graphs being associated with social networks, various models, and graph spectra. In graph theory, signed graphs have been instrumental in defining various properties and introducing new concepts.

Harary formulated and restructured the signed networks by introducing structural balance theory [3] for social balance, which was used for portfolio management [4], where they used signed graphs to analyse the extent of hedging in a portfolio. These networks are widely used in data clustering [[5], [6], [7]]. Signed networks of some intersection networks have been already studied [8,9,10,11]. Also, path graphs of signed graphs are discussed in [[12], [13], [14], [15], [16], [17],18].

The study of networks has come a long way with its applicability in many fields [19,[20], [21], [22]]. One such application is in wireless networks and frequency allocation problems. The frequency allocation [23] in a wireless network [24] or radiofrequency allocation [25] is one of the typical problems that we still face. The interference occurs place in a network when the transmission from one station interacts with the transmission from another. There are three dimensions to the frequency spectrum first being the space in which the emission is radiated, second is the frequency bandwidth, and third is the time. If these three dimensions co-occur for a channel receiving transmission, then interference occurs. Many methods and models, such as dipole-moment models [26], the Kurtosis detection algorithm and its versions [27], and the decomposition method based on reciprocity [28], have been studied for this problem. A model with a very different approach using the theoretic aspects of signed networks and t-path networks is studied by Sinha and Sharma in [29], which aims to detecting and reduce interference by assuming the stations/channels as vertices and transmitting between them as edges. Furthermore, in [29], authors assume that two channels (vertices) are joined by a positive edge if and only if they are in different time or frequency bandwidth (assuming that space is always the same) and by a negative edge if both are the same in time and frequency bandwidth. If the given signed network $\Sigma$ represents a channel network, then its 2-path signed network ($\Sigma$)₂ provides the interference pattern of different channels at a specific channel. In the 2-path signed network, the edge $uw$ is given a negative sign if and only if there exist all-negative two paths in $\Sigma$, say $uv$ and $vw$ between them; that is, they are the same in time as well as frequency, and thus, the interference takes place at $v$ because of the vertices $u$ and $w$.

The graph-based approach discussed in this paper can enhance Non-Terrestrial Network (NTN) routing by using two-hop neighbor relationships to create a resultant graph that optimizes connectivity and resource utilization. Initially, all vertices (e.g., satellites, UAVs, ground stations) are connected by edges, representing direct communication links. In the resultant t-path product signed graph, new edges are added based on t-hop reachability, with edges receiving a positive sign denoting strong, reliable connections (e.g., high-bandwidth or low-latency links) and negative edges representing weaker, constrained links (e.g., backup or inter-layer paths). This model supports a depth-based or layer-based routing strategy: intra-layer routing (e.g., within LEO satellites) uses positive edges for efficiency, while inter-layer communication (e.g., LEO to ground stations) relies on negative edges as transitional links. By dynamically adapting to the mobility and resource constraints of NTN environments, this approach may ensure minimal latency for critical data, robust fault tolerance, and effective load balancing across layers, ultimately improving the network's performance and resilience.

Also, the proposed graph model has broad applications in many fields, including Community Detection in Multilayer networks where the links can be used to model complex real-world interaction. In social networks this can be used to identify positive and negative interactions. In the field of market analysis, this model could reveal patterns in consumer behaviour. Recommendation Systems is another application of the model where users (nodes) can be linked to items (products, movies, etc.), and these links can be positive (likes, purchases) or negative (dislikes, dislikes), and the collaborative filtering approach of the model can be put to use to improve the existing models. This approach can be applied across various domains, from social networks and market segmentation to recommendation systems and community detection, to discover hidden structures and latent interactions that may not be apparent through direct connections alone. The clustering behaviour that divides the graph into two groups with positive intra-cluster relationships and negative inter-cluster relationships is particularly useful for detecting group formation and hidden dynamics within networks.

In this paper, we try to establish the results for signed networks defined on t-path networks, t=2,3.

For standard terminology and notation in graph theory, one can refer to West [30], and for signed graph literature, one can read Zaslavsky [31,32].

In a signed graph, denoted as $\Sigma =\left({\Sigma }^u,\sigma \right)$, it is an ordered pair where${\Sigma }^u$ represents the underlying graph (V, E) and $\sigma :E\to \left\{+,-\right\}$ denotes the signature function, assigning each edge in $E$ a sign from the set$\left\{+,-\right\}$. The edge receiving ‘${-}^{\prime }$ sign is called negative edge and the edge receiving ‘${+}^{\prime }$ sign is called positive edge. Pictorially, negative edges are denoted by dotted lines and positive edges by the bold lines in the signed graph. $d_{\Sigma }^{+}\left(v_i\right)\left(d_{\Sigma }^{-}\left(v_i\right)\right)$ denotes the number of positive edges(negative edges) incident with vertex $v_i$. If $d_{\Sigma }^{+}\left(v_i\right)=0$or $d_{\Sigma }^{-}\left(v_i\right)=0$ then degree of $v_i$ is homogeneous. The net-degree of vertex $v_i$ and is given by |$d_{\Sigma }^{+}\left(v_i\right)-d_{\Sigma }^{-}\left(v_i\right)$|.

A signed graph is all-positive (resp., all-negative) if all its edges are positive (negative); further, it is said to be homogeneous if it is either all-positive or all-negative and heterogeneous otherwise. The negation of a signed graph ŋ$\left(\Sigma \right)$ is obtained by reversing the sign of edges of $\Sigma$. A cycle is positive(negative) if it has even(odd) number of negative edges. A signed graph is balanced if all its cycles are positive.

A function $\mu :V\left(\Sigma \right)\to \left\{-1,+1\right\}$ is called a marking of the signed graph $\Sigma$. A signed graph $\Sigma$ together with one of its making $\mu$ is denoted by ${\Sigma }_{\mu }$.

Abelson and Rosenberg [33] introduced the concept of switching in signed graphs. The switching of a signed graph $\Sigma$ with respect to a marking $\mu$ involves changing the sign of each edge in $\Sigma$ to its opposite whenever its end vertices have opposite marks in ${\Sigma }_{\mu }$. For any vertex $v\in \Sigma$, if $\mu \left(v\right)={\Pi }_{u\in N\left(v\right)}\sigma \left(vu\right)\mu$ is called canonical-marking or C-marking [34].

$\Psi \left(\Sigma \right)$ is the class of all signed graphs such that they are isomorphic to underlying graphs ${\Sigma }^u$ .

A negative section [35] of a signed graph$\Sigma$, we mean a connected component (that is not an isolated vetex) of the subgraph induced by negative edges.

Two signed graphs ${\Sigma }_1$ and ${\Sigma }_2$ are said to be cycle-isomorphic if there exists an isomorphism f from ${\Sigma }_1^u$ onto${\Sigma }_2^u$ such that the sign of every cycle C in ${\Sigma }_1$ equals the sign of f(C) in ${\Sigma }_2$ where the sign of any subgraph of a signed graph is defined as the product of the sign of its edges.

Theorem 1

[34] A signed graph is balanced if and only if it is possible to assign marks to its vertices such that the sign of each edge uv is equal to the product of the marks of u and v.

The partition criterion to characterize the balance property of a signed graph is given by Harary.

Theorem 2

[35] A signed graph $\Sigma$ is balanced if and only if its vertex set $V\left(\Sigma \right)$ can be partitioned into two subsets $V_1$ and$V_2$ one of them possibly empty, such that every positive edge joins two vertices in the same subset and every negative edge joins two vertices from different subsets.

Let $G=\left(V,E\right)$ be a finite connected graph then n-path graph $G_n$ is the graph having $V\left(G\right)$ as a vertex set and for which two vertices $u,v$ are adjacent if and only if there exist a $u-v$ path of length exactly $n$ in $G.$

Let $G=\left(V,E\right)$ be a finite connected graph then n-distance graph $\overline{G_n}$ is the graph having $V\left(G\right)$ as a vertex set and for which two vertices $u,v$ are adjacent if and only if there exist a $u-v$ distance of length exactly $n$ in $G.$ The study of n-distance graph $\overline{G_n}$ was initiated by Simic [36] while solving the graph equation $\overline{G_n}\cong L\left(G\right)$, where $L(G$) is line graph of $G$. A graph is said to be self n-distance graph if it is isomorphic to its n-distance graph.

Theorem 3

[37] A graph G of order p is a 2-path invariant graph if and only if $G=K_p$ or ${\overline{K}}_p$with $p\ge 3$, or the odd p-cycle $C_p$, p =2m + 1, $m\ge 2$.

Theorem 4

[37] A graph G is a 3-path invariant graph if and only if it is isomorphic to any one of the following graphs

• (a)$C_m$,$4\le m\ne 3k$, k is a positive integer.
• (b)$K_n$,$n\ge 4$
• (c)$K_{m,n},2\le m\le n$
• (d)The double star$K_{m,n}^{t,2}$as shown inFig. 4
• (e)$E_1=K_4-x$
• (f)$E_2=$the subdivision graph of$E_1$
• (g)$E_3$
• (h)$E_4$

where $E_2$, $E_3$, $E_4$ are shown Fig. 1,2,3.

Fig. 4.

Double star $K_{m,n}^{t,2}.$

Fig. 1.

Graph$E_2.$

Fig. 2.

Graph $E_3.$

Fig. 3.

Graph$E_4.$

n-path product signed graph

The n-path product signed graph ${\widehat{\left(\Sigma \right)}}_n=\left({\widehat{\left(\Sigma \right)}}_n,{\sigma }^{\prime }\right)$ is defined as follows: The vertex set is same as that of the original signed graph $\Sigma$ and two vertices $u,v\in {\widehat{\left(\Sigma \right)}}_n$, are adjacent if and only if there exists a path of length n in $\Sigma$. The sign of an edge$,{\sigma }^{\prime }\left(uv\right)=\mu \left(u\right)\mu \left(v\right)$, where $\mu$ is C-marking on the $\Sigma$.

Lemma 5

[34] The n-path product signed graph ${\widehat{\left(\Sigma \right)}}_n$ of a signed graph $\Sigma$ is always balanced.

Theorem 6

[32] Given a graph G, any two signed graphs in $\Psi \left(G\right)$ are switching equivalent if and only if they are cycle-isomorphic.

2-path product signed graph

The 2-path product signed graph ${\widehat{\left(\Sigma \right)}}_2$ is defined as follows: The vertex set is same as the original signed graph $\Sigma$ and two vertices$u,v\in {\widehat{\left(\Sigma \right)}}_2$, are adjacent if and only if there exists a path of length two in $\Sigma$. The sign of an edge $uv,{\sigma }^{\prime }\left(uv\right)=\mu \left(u\right)\mu \left(v\right)$, where $\mu$ is C-marking on $\Sigma$.

Algorithm to obtain a 2-path product signed graph for a given signed graph

Input: Adjacency matrix A and dimension n.

Output: Adjacency matrix of 2-path product signed graph.

Process:

1	Enter the order n and adjacency matrix A of for a given signed graph $\Sigma$.
2	Collect all the q P pairs for given signed graph $\Sigma$.
3	$\text{for}\mathrm{i}=1\text{to}n\text{do}$
4	$\mathrm{d}\left[\mathrm{i}\right]\text{for}j=1\text{to}n\text{do}$
5	$\mathrm{b}\left[\mathrm{i}\right]\left[\mathrm{j}\right]=0\text{if}\mathrm{a}\left[\mathrm{i}\right]\left[\mathrm{j}\right]\ne 0\text{then}$
6	$\mathrm{d}\left[\mathrm{i}\right]=\mathrm{d}\left[\mathrm{i}\right]\times \mathrm{a}\left[\mathrm{i}\right]\left[\mathrm{j}\right];$
7	$fori=1tondo$
8	$\text{for}\mathrm{j}=1\text{to}\mathrm{n}\text{do}$
9	$\text{for}\mathrm{k}=1\text{to}\mathrm{n}\text{do}$
10	$if\left(\mathrm{a}\left[\mathrm{i}\right]\left[\mathrm{j}\right]\ne 0\right)\&\&\left(\mathrm{a}\left[\mathrm{i}\right]\left[\mathrm{k}\right]\ne 0\right)\text{then}$
11	$if\left(\mathrm{d}\left[\mathrm{k}\right]\times \mathrm{d}\left[\mathrm{j}\right]=-1\right)\text{then}$
12	$\mathrm{b}\left[\mathrm{k}\right]\left[\mathrm{j}\right]=-1$
13	$\mathrm{b}\left[\mathrm{j}\right]\left[\mathrm{k}\right]=-1$
14	else
15	$\mathrm{b}\left[\mathrm{k}\right]\left[\mathrm{j}\right]=1$
16	$\mathrm{b}\left[\mathrm{j}\right]\left[\mathrm{k}\right]=1$

Theorem 7

A signed graph$\Sigma \in \Psi \left(C_{2m+1}\right)$,$m\ge 2$is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$if and only if$\Sigma$is all-positive or has even number of odd negative sections.

Proof. Given that a signed graph $\Sigma \in \Psi \left(C_{2m+1}\right)$, $m\ge 2$ is a solution to $\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$. By the definition of ${\widehat{\left(\Sigma \right)}}_2$, ${\widehat{\left(\Sigma \right)}}_2$ is balanced, and by the definition of switching equivalence, ${\widehat{\left(\Sigma \right)}}_2$ and $\Sigma$ are cycle-isomorphic. This implies $\Sigma$ is also balanced, then the possibilities of $\Sigma$ is either all-positive or contains even number of odd negative sections.

Conversely, since $\Sigma$ is all-positive or has even number of odd negative sections, and ${\widehat{\Sigma }}_2^u$ and${\Sigma }^u$ both have same underlying graph, ${\widehat{\left(\Sigma \right)}}_2$ and $\Sigma$ are cycle-isomorphic being balanced. Hence $\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$.

Theorem 8

A signed graph$\Sigma \in \Psi \left(K_n\right)$, for any integer$n\ge 3$is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$if and only if the vertex set of$\Sigma$can be partitioned into two subsets$V_1$and$V_2$one of them possibly empty such that <$V_1$> and <$V_2$> are all-positive and all the other edges are negative.

Proof. Given that a signed graph $\Sigma \in \Psi \left(K_n\right)$, for any integer $n\ge 3$ is a solution to $\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$. Since ${\widehat{\left(\Sigma \right)}}_2$ is always balanced implies $\Sigma$ is also balanced as both ${\widehat{\left(\Sigma \right)}}_2$and $\Sigma$ are cycle-isomorphic. Hence, vertex set can be partitoned into two subsets $V_1$ and $V_2$ such that <$V_1$> and <$V_2$> are all-positive and all the other edges are negative.

Conversely, let $\Sigma$ and ${\widehat{\left(\Sigma \right)}}_2$are balanced and ${\Sigma }^u\cong {\widehat{\Sigma }}_2^u$, for $\Sigma \in \Psi \left(K_n\right)$, they will be cycle-isomorphic and hence, $\Sigma \sim {\widehat{\left(\Sigma \right)}}_2$.

Theorem 9

A signed graph$\Sigma \in \Psi \left(C_{2m+1}\right)$,$m\ge 2$is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_2$ if and only if $\Sigma$ has an even number of odd positive sections.

Proof. Given that a signed graph $\Sigma \in \Psi \left(C_{2m+1}\right)$; $m\ge 2$ is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_2$. Since ${\widehat{\left(\Sigma \right)}}_2$ is balanced, ŋ$\left(\Sigma \right)$ is also balanced because they are switching equivalence to each other. Thus, ŋ$\left(\Sigma \right)$ has an even number of odd negative sections which implies $\Sigma$ has an even number of odd positive sections.

Conversely, let $\Sigma \in \Psi \left(C_{2m+1}\right)$ and $\Sigma$ has an even number of odd positive sections which implies ŋ$\left(\Sigma \right)$ has even number of odd negative sections. Thus ŋ$\left(\Sigma \right)$ is balanced and also (ŋ${\left(\Sigma \right))}^u\cong {\widehat{\Sigma }}_2^u$ for $\Sigma \in \Psi \left(C_{2m+1}\right)$ and ŋ$\left(\Sigma \right)$and ${\widehat{\left(\Sigma \right)}}_2$ both being balanced and cycle-isomorphic, implies ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_2$.

Theorem 10

A signed graph$\Sigma \in \Psi \left(K_n\right),$for any integer$n\ge 3$is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_2$ if and only if the vertex set can be partitioned into two subsets $V_1$ and $V_2$ one of them possibly empty such that <$V_1$> and <$V_2$> are all-negative and all the other edges are positive.

Proof. Let $\Sigma \in \Psi \left(K_n\right)$and it is a solution of ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_2$. Now, ${\widehat{\left(\Sigma \right)}}_2$being balanced, ŋ$\left(\Sigma \right)$ is also balanced by switching equivalence. ŋ$\left(\Sigma \right)$ is balanced then the vertex set of ŋ$\left(\Sigma \right)$can be partitoned into two sets $V_1$ and $V_2$ such as the induced subgraph <$V_1$> and induced subgraph <$V_2$> are both positive by Harary partition in ŋ$\left(\Sigma \right)$ and all other edges are negative. Hence, it follows that vertices of $\Sigma$ can be partitoned into two subsets $V_1$ and $V_2$such that <$V_1$> and <$V_2$> are all-negative and all other edges are positive.

Converse is trivial to see.

3 - path product signed graph

The 3-path product signed graph ${\widehat{\left(\Sigma \right)}}_3$ is defined as follows: The vertex set is same as the original signed graph $\Sigma$ and two vertices $u,v\in {\widehat{\left(\Sigma \right)}}_3$, are adjacent if and only if there exists a path of length three in $\Sigma$. The sign of an edge $uv,{\sigma }^{\prime }\left(uv\right)=\mu \left(u\right)\mu \left(v\right)$, where$\mu$ is C-marking on $\Sigma$.

Theorem 11

For a signed graph$\Sigma \in \Psi \left(C_m\right)$,$4\le m\ne 3k$, k is a positive integer, is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if negative sections of odd length are even in number.

Proof. Given that $\Sigma \in \Psi \left(C_m\right)$, $4\le m\le 3k$, k a positive integer, is a signed graph which is solution of $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$. Since ${\widehat{\left(\Sigma \right)}}_3$ is balanced by the definition and $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$ this implies $\Sigma$ is also balanced by the definition and since $\Sigma$ is balanced it has an even number of negative edges, and therefore, the number of negative sections with odd lengths must be even .

Conversely, the numbers of negative sections of odd length are even in any $\Sigma \in \Psi \left(C_m\right)$, $4\le m\ne 3k.$ This implies number of negative edges in $\Sigma$ are even. Thus, $\Sigma$ being balanced and ${\widehat{\left(\Sigma \right)}}_3$ being balanced by definition implies that $\Sigma$ is cycle-isomorphic to ${\widehat{\left(\Sigma \right)}}_3$ . Hence, $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$.

Theorem 12

A signed graph$\Sigma \in \Psi \left(K_n\right)$, for any integer$n\ge 4$is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if the vertex set of$\Sigma$can be partitioned into two subsets$V_1$and$V_2$one of them possibly empty such that <$V_1$> and <$V_2$> are all-positive and all the other edges are negative.

Proof. Proof follows by Theorem 2.

Theorem 13

A signed graph$\Sigma \in \Psi \left(K_{m,n}\right)$;$2\le m\le n$is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if for every vertex in one of the partition, the degree of the every vertex is homogeneous, or the vertex set can be partitioned into two subsets such that every negative edge joins two vertices in the same subset and every positive edge joins two vertices in different subsets.

Proof. Given that $\Sigma \in \Psi \left(K_{m,n}\right)$; $2\le m\le n$. Suppose, for every vertex in one of the partition say $V_1$ of $\Sigma$, the degree of the every vertex is homogeneous. Putting the collection of all such vertices having all negative degree in $V_1$ in the set $V_1^{\prime }$ and rest of the vertices of $\Sigma$ in $V_2^{\prime }$ gives a Harary's partition of the balanced signed graph. Now, $\Sigma$ is balanced implies $\Sigma$ and ${\widehat{\left(\Sigma \right)}}_3$are cycle-isomorphic and hence to $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$.

Further, we partition the vertices of $\Sigma$ into two subsets say $U_1$ and $U_2$ in such a way that every negative edge joins vertices in the same set and all positive edges has one end vertex in $U_1$ and other end vertex in $U_2$. Now, we consider a cycle $v_1{v}_2$ $v_3\dots \dots v_{2r}{v}_1$and suppose the vertex $v_1$lies in the first set $U_1$ then there are two possibilities. $v_2$ may lie on $U_1$ or may lie in $U_2$. In either cases the number of the positive edges and negative edges will always be even in the cycle $v_1{v}_2$ $v_3\dots \dots v_{2r}{v}_1$ due to the partition. Hence the cycle is positive and similarly all the cycles will be positive and therefore $\Sigma$ is balanced. Now, $\Sigma$ and ${\widehat{\left(\Sigma \right)}}_3$ being cycle isomorphic, $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$.

Conversely, suppose $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$. This implies $\Sigma$ is balanced. Thus every cycle in $\Sigma$ has even number negative and positive edges. Thus, it is easy to partition the vertex set into two sets such that all positive edges have end vertices in different partitions and the negative edges are within the sets.

Also if $\Sigma$ is balanced, then its vertices set can be partitioned into two subsets such that all negative edges have end vertices in different partitions and the positive edges are within the sets. So all positive edges have end vertices in different partitions and the negative edges are within the sets. If in this case the homogeneous vertices of all-positive degree are kept in one partition, then this becomes the special case of the arguments given in above case. Hence the proof.

Theorem 14

A signed graph$\Sigma \in \Psi \left(E_1\right)$, is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if the cycle subsigned graph of$\Sigma$is either all-positive or have even number of negative edges.

Proof. Proof follows by Theorem 2. The possible structures are given in Fig. 5 for which $\Sigma \in \Psi \left(E_1\right)$ and$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$.

Fig. 5.

Showing $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$; $\Sigma \in \Psi \left(E_1\right).$

Theorem 15

A signed graph$\Sigma \in \Psi \left(E_2\right)$, is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if$\Sigma$is homogeneous or both the cycles have even number of negative edges.

Proof. The proof follows by Theorem 2.

Theorem 16

A signed graph$\Sigma \in \Psi \left(E_3\right)$, is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if the cycle subsigned graph of$\Sigma$is either all-positive or have even number of negative edges.

Proof. Proof follows by Theorem 2. The possible structures are all balanced cycles on $\Sigma \in \Psi \left(E_3\right)$ with pendent edges having any signature.

Theorem 17

A signed graph$\Sigma \in \Psi \left(E_4\right)$, is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if the cycle subsigned graph of$\Sigma$have even number of negative edges.

Proof. Proof follows by Theorem 2. The possible structures are all balanced cycles on $\Sigma \in \Psi \left(E_4\right)$ with pendent edges having any signature.

Theorem 18

A signed graph$\Sigma \in \Psi \left(K_{m,n}^{t,2}\right)$where$K_{m,n}^{t,2}$is shown inFig. 4,$m\ge 0,n\ge 0$and$t\ge 2$is a solution to$\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$if and only if in each cycle of core, the net-degree of the vertices$u_1,u_2,u_3\dots u_t$in the core <$x,u_1,u_2,u_3\dots u_t,y$> is either$2\forall t$or$1\forall t$.

Proof. Given that $\Sigma \in \Psi \left(K_{m,n}^{t,2}\right)$ is a solution of $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$. Now by the definition, ${\widehat{\left(\Sigma \right)}}_3$ is always balanced. This implies $\Sigma$ is balanced. Now, the core contains each cycle of 4-length and $\Sigma$ being a balanced then the all possible cycles are homogenous or contains even number negative edges. Hence, the each cycle of core <$x,u_1,u_2,u_3\dots u_t,y$>, the net-degree of the vertices $u_1,u_2,u_3\dots u_t$are either $2\forall t$ or $1\forall t$.

Conversely, in each cycle of core < $x,u_1,u_2,u_3\dots u_t,y$>, the net-degree of the vertices $u_1,u_2,u_3\dots u_t$ of cycle is $2\forall t$ or net-degree of every vertices $u_1,u_2,u_3\dots u_t$ is $1\forall t$. This will make every cycle positive in the core and hence $\Sigma$ will be balanced which implies $\Sigma \sim {\widehat{\left(\Sigma \right)}}_3$.

Theorem 19

For a signed graph$\Sigma \in \Psi \left(C_m\right)$;$4\le m\ne 3k$, k a positive integer is a solution ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if and only if positive-sections of odd length are even in number.

Proof. For any $\Sigma \in \Psi \left(C_m\right)$, let ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$. Since ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ and ${\widehat{\left(\Sigma \right)}}_3$ is always balanced, ŋ$\left(\Sigma \right)$ is balanced which further implies that ŋ$\left(\Sigma \right)$has an even number of odd negative sections with no condition on positive section of odd(even) length. This implies positive section of odd length be even in number .

Conversely, let the numbers of positive sections of odd length are even in number in any $\Sigma \in \Psi \left(C_m\right)$;$4\le m\ne 3k$. This implies number of negative edges in ŋ$\left(\Sigma \right)$ is even. Thus, ŋ$\left(\Sigma \right)$being balanced is cycle-isomorphic to ${\widehat{\left(\Sigma \right)}}_3$by the definition of ${\widehat{\left(\Sigma \right)}}_3$. Hence, ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$.

Theorem 20

A signed graph$\Sigma \in \Psi \left(K_n\right)$, for any integer$\mathrm{n}\ge 4$is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$if and only if $\Sigma$ is all-negative or the vertex set of $\Sigma$ can be partitioned into two subsets $V_1$ and $V_2$one of them possibly empty such that <$V_1$> and <$V_2$> are all-negative and all the other edges are positive.

Proof. Given that a signed graph $\Sigma \in \Psi \left(K_n\right)$, for any integer $n\ge 4$ is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ and ${\widehat{\left(\Sigma \right)}}_3$ is always balanced by definition, ŋ$\left(\Sigma \right)$ is also balanced as both ${\widehat{\left(\Sigma \right)}}_3$ and ŋ$\left(\Sigma \right)$are cycle-isomorphic. Hence, ŋ$\left(\Sigma \right)$ is all-positive or by Harary's partition vertex set of ŋ$\left(\Sigma \right)$ can be partitioned into two subsets $V_1$and $V_2$such that $<V_1>and\langle V_2\rangle$are all-positive and all the other edges are negative. This implies $\Sigma$ is all-negative, or the vertex set of $\Sigma$ can be partitioned into two subsets$V_1$and $V_2$such that $<V_1>and\langle V_2\rangle$are all-negative and all the other edges are positive.

Conversely, since ŋ$\left(\Sigma \right)$will be balanced by Harary's partition criterion, ${\widehat{\left(\Sigma \right)}}_3$ is balanced by the definition, and (ŋ${\left(\Sigma \right))}^u\cong {\left({\widehat{\left(\Sigma \right)}}_3\right)}^u$ implies ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$.

Theorem 21

A signed graph$\Sigma \in \Psi \left(K_{m,n}\right)$;$2\le m\le n$is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if for every vertex in one of the partition, the degree of the every vertex is homogeneous, or the vertex set can be partitioned into two subsets such that every negative edge joins two vertices in the same subset and every positive edge joins two vertices in different subsets.

Proof. The proof is easy to see by Theorem 13 and by the fact that every cycle has even number of negative and positive edges.

Theorem 22

A signed graph$\Sigma \in \Psi \left(E_1\right)$, is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if and only if $\Sigma$ is all-negative or the structures shown in Fig. 6.

Proof. Proof follows by Theorem 2.

Fig. 6.

Showing ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$; $\Sigma \in \Psi \left(E_1\right).$

Theorem 23

A signed graph$\Sigma \in \Psi \left(E_2\right)$, is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if and only if $\Sigma$ is homogeneous or the vertex set can be partitioned into two subsets such that every negative edge joins two vertices in the same subset and every positive edge joins two vertices in different subsets.

Proof. The proof follows by Theorem 21 as the underlying graph of the signed graph $\Sigma$ is a bipartite graph.

Theorem 24

A signed graph$\Sigma \in \Psi \left(E_3\right)$, is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$, if and only if for $\Sigma \in \Psi \left(E_3\right)$, the triangle is negative in $\Sigma$.

Proof. Given that $\Sigma \in \Psi \left(E_3\right)$, is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$. Now by the definition, ${\widehat{\left(\Sigma \right)}}_3$ is always balanced. This implies ŋ$\left(\Sigma \right)$ is balanced. Now $\Sigma$ has only one cycle of length 3 and ŋ$\left(\Sigma \right)$ being a balanced signed graph, implies that $\Sigma$ has a negative triangle.

Conversely, let $\Sigma \in \Psi \left(E_3\right)$ contains a negative triangle. Then ŋ$\left(\Sigma \right)$ will have even number of negative edges making ŋ$\left(\Sigma \right)$balanced. Since ${\widehat{\left(\Sigma \right)}}_3$ is balanced, and ŋ$\left(\Sigma \right)$ and ${\widehat{\left(\Sigma \right)}}_3$ both have same underlying graph, they will be cycle-isomorphic to each other. Thus, ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$.

Theorem 25

A signed graph$\Sigma \in \Psi \left(E_4\right)$, is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if and only if the vertex set of $\Sigma$ can be partitioned into two subsets such that the degree of every vertex in one of the partition is homogeneous, or the vertex set can be partitioned into two subsets such that every negative edge joins two vertices in the same subset and every positive edge joins two vertices in different subsets.

Proof. Proof follows by Theorem 21. The possible structures are negative cycle in $\Sigma \in \Psi \left(E_4\right)$with pendent edges taking any signature.

Theorem 26

A signed graph$\Sigma \in \Psi \left(K_{m,n}^{t,2}\right)$where$K_{m,n}^{t,2}$is shown inFig. 4,$m\ge 0,n\ge 0$and$t\ge 2$is a solution to ŋ$\left(\Sigma \right)\sim {\widehat{\left(\Sigma \right)}}_3$ if and only if in each cycle of core, the net-degree of the vertices $u_1,u_2,u_3\dots u_t$ in the core < $x,u_1,u_2,u_3\dots u_t,y$> is either $2\forall t$ or $1\forall t$.

Proof. Proof follows by Theorem 2.

A constructive method to compute balanced signed from an unbalanced signed graph

Now we observe that in all the above cases the signed graph $\Sigma$ will be switching equivalent to $t-$path product signed graph ${\widehat{\left(\Sigma \right)}}_t$ if $\Sigma$ is balanced. Here we give the constructive method to compute balanced signed graph from an unbalanced signed graphs for planar signed graphs and nonplanar signed graphs.

By planar signed graph, we mean a signed graph that can be embedded in the plane. The bounded regions formed by a planar signed graph are termed as interior faces, and the unbounded region is known as the exterior face.

The dual graph D(G), of a planar graph $G$ has a vertex for each face of the graph $G$. It has an edge for each pair of faces in $G$ corresponding to each common edge between these faces and a self loop when same face appears on both sides of an edge. The dual signed graph $D\left(\Sigma \right)$, of a planar signed graph $\Sigma$ has a vertex, for each face of the signed graph $\Sigma$, with a marking same as the sign of the cycle encircling the corresponding face in $\Sigma$, for the exterior face, it is the sign of the circumferential cycle. It has an positive edge for each pair of faces in $\Sigma$ corresponding to each common edge between these faces and a positive self-loop when same face appears on both sides of an edge.

A balancing set of an unbalanced signed graph $\Sigma$ is the set of edges such that changing the signs of these edges results in a balanced signed graph. A balancing set is called a minimal balancing set if and only if any proper subset of it is no longer a balancing set and it is denoted by $L_b$.

• Case 1: When the signed graph$\mathbf{\Sigma }$is planar

• step 1 Find the signs of cycles corresponding to the faces of signed graph $\Sigma$.

• step 2 Make the dual graph D($\Sigma$) of $\Sigma$.

• step 3 Choose the negative vertices and calculate the distance between all the pairs of negative vertices where the distance between two vertices is a shortest path.

• step 4 Make the complete graph whose edges are attached by the calculated distance in step 3.

• step 5 Now, find the optimal pairs $V_1^{*},V_2^{*},V_3^{*}\dots V_m^{*}$ such that $V_u=V_1\cup V_2\cup V_3\dots \cup V_{2m}$ is set of negative vertices and $c\left(V_1^{*},V_2^{*},V_3^{*}\dots V_m^{*}\right)=minc{\left(V_1^{*},V_2^{*},V_3^{*}\dots V_m^{*}\right)}_{V_1^{*},V_2^{*},V_3^{*}\dots V_m^{*}}$where $c\left(V_i\right)$ is the length of shortest path joining vertices $v_{p_i}$ and ${v^{\prime }}_{p_i}$.

• step 6 Find the shortest path $L_i^{*}$ joining the vertices of $v_i^{*}$ for $i=1,2,3\dots m$

• step 7 Derive the minimum balancing set $L_i^{*}\left(b\right)=d^{-1}\left(L_1^{*}\cup L_2^{*}\cup L_3^{*}\dots \cup L_m^{*}\right)$

A sign vector $s=\left(s_1,s_2,\dots s_n\right)$ is a vector whose element $s_i=+1or-1$for $i=1,2,3\dots n$.

A edge $v_i{v}_j$in $L_b$ an inherent edge if and only if both $v_i$and $v_j$ are contained in same equivalence class otherwise noninherent.

Example: Consider the Signed Graph $\Sigma$ in Fig. 7

• Step1 Signs of each cycles corresponding to faces of $\Sigma$ in Fig. 8 are:

Fig. 7.

Signed graph $\Sigma .$

Fig. 8.

Showing the signature of the faces of the $\Sigma .$

• Step 2 Make the dual graph $D\left(\Sigma \right)$ as shown in Fig. 9.

• Step 3 Set of negative vertices in $D\left(\Sigma \right)$ is $\{v_1^{\prime },v_2^{\prime },v_3^{\prime },v_4^{\prime }\}$. Now possible pairs of shortest path between any two vertices

  $$L_1=\left\{v_1^{\prime },v_2^{\prime }\right\},L_2=\left\{v_3^{\prime },v_4^{\prime }\right\},L_3=\left\{v_1^{\prime },v_3^{\prime }\right\},L_4=\left\{v_2^{\prime },v_4^{\prime }\right\},L_5=\left\{v_1^{\prime },v_4^{\prime }\right\},L_6=\left\{v_3^{\prime },v_2^{\prime }\right\}$$

• Step 4 shortest path between vertices lies in $L_1$, $L_2$, $L_5$, $L_6$.

• Step 5 optimal pairs ${L^{*}}_1=\left\{v_1^{\prime }{v}_2^{\prime },v_3^{\prime }{v}_4^{\prime }\right\}$ or ${L^{*}}_2=\left\{v_1^{\prime }{v}_4^{\prime },v_3^{\prime }{v}_2^{\prime }\right\}$ .

• Step 6 shortest path between two vertices in ${L^{*}}_1$ or ${L^{*}}_2$ is length of 1.

• Step 7 ${L^{*}}_b=d^{-1}\left({L^{*}}_1\cup {L^{*}}_2\right)$. Hence sign graph will become balanced signed graph as shown in Fig. 10.

• Case 2: When the signed graph $\mathbf{\Sigma }$ is non- planar

• step 1 First make any minimal balancing set $L_b$ from the signed graph $\Sigma$.

• step 2 Find the sign vector corresponding to the balancing set.

• step 3 Partition the vertex set V of $\Sigma$ into the equivalence classes which is induced by $L_b$.

• step 4 Make the condensed graph $G_c$as each vertex $e_i$ correspond to an equivalence class $E_i$ induced by $L_b$ and two vertices are joined by a line when there exist a non-inherent line of $L_b$joining corresponding equivalence classes.

• step 5 Find all the spanning tree of $G_c$ (Condensed graph).

• step 6 Calculate the assignment vector t for the refinement of minimal balancing set $L_b$.

• step 7 Delete the duplicate assignment vector that is need to find the distinct assignment vector form step 6.

• step 8 From each assignment vector t, delete all the non-inherent classes with different assignment value of t.

Fig. 9.

Showing the dual of the $\Sigma .$

Fig. 10.

Showing balanced signed graph on $\Sigma .$

The final step 8 gives all the minimal balancing sets contained in $L_b$.

If the signed graph is unbalanced then the constructive method can be used so as to make it balanced and becomes switching equivalent to t-path product signed graph when ${\Sigma }^u\cong {\widehat{\Sigma }}_t^u$.

2-distance signed graph

Assume that $\Sigma =\left(V,E,\sigma \right)$ is a signed graph. We associate with $\Sigma$ the 2-distance signed graph ${\left(\overline{\Sigma }\right)}_2=\left(V^{\prime },E^{\prime },{\sigma }^{\prime }\right)$ defined as follows: the vertex set is same as the original signed graph $\Sigma$ and two vertices $u,v\in {\left(\overline{\Sigma }\right)}_2$, are adjacent if and only if there exists a distance of length two between $u$ and $v$ in $\Sigma$. The edge $uv\in {\left(\overline{\Sigma }\right)}_2$ is negative if and only if all the edges, in all the paths of distance two between $u$ and $v$ in $\Sigma$ are negative otherwise the edge is positive.

$C_5|C_3$ is a graph in Fig. 11 which have $v_1{v}_2{v}_3{v}_4{v}_5{v}_6{v}_1$ forms a cycle $C_6$such that $v_2{v}_6$ forms a chord. Here $C_5=v_2{v}_3{v}_4{v}_5{v}_6{v}_2$ and $C_3=v_1{v}_2{v}_6{v}_1$

Fig. 11.

$C_5|C_3$and $\overline{{\left(C_5\right|C_3)}_2}.$

Theorem 27

[38]

$\Gamma$ has a $C_5|C_3$ subgraph, then $\Gamma$ is isomorphic to $C_5|C_3$.

Theorem 28

Signed graph $\Sigma \in \Psi \left(C_5|C_3\right)$ , is a solution to $\Sigma \sim {\left(\overline{\Sigma }\right)}_2$ if and only if $\Sigma$ is homogeneous or

• (i) Number of negative edges in cycle $C_3=v_1{v}_2{v}_6{v}_1$ is of same parity as number of all negative paths in T where $T=\{v_1-v_3,v_3-v_5,v_5-v_1$} and

• (ii) Number of negative sections in $C_5=v_2{v}_3{v}_4{v}_5{v}_6{v}_2$ must be even or $C_5$ is all-negative.

Proof. Given that $\Sigma \in \Psi \left(C_5|C_3\right)$ is a solution to $\Sigma \sim {\left(\overline{\Sigma }\right)}_2$and $T=\{v_1-v_3,v_3-v_5,v_5-v_1$}. We have every cycle of $\Sigma$ is cycle-isomorphic to corresponding cycle of ${\left(\overline{\Sigma }\right)}_2$. Now $\Sigma \sim {\left(\overline{\Sigma }\right)}_2$ trivially follows when $\Sigma$ is homogeneous. Also, for a signed cycle ${\mathrm{C}}_3\in \Sigma$, there exist a corresponding signed cycle $C_3^{\prime }\in {\left(\overline{\Sigma }\right)}_2$ and both are cycle-isomorphic.

Now, the edge between $x$ and $y$ in 2-distance signed graph ${\left(\overline{\Sigma }\right)}_2$ is due to 2-distance between in $x$ and $y$ in $\Sigma$ and

$$\sigma \left(\mathit{xy}\right)=\{\begin{array}{c}-\text{if}\mathrm{x}-\mathrm{y}\text{path}\text{is}\text{all}-\text{negative}\\ +\text{Otherwise}\end{array}$$

Hence, Number of negative edges in signed cycle $C_3\in \Sigma$ is of same parity as number of all-negative paths in T.

(ii) Now, let $q$ be the number of negative sections of length 1 and other negative sections $k_1,k_2,k_3$ of some lengths 2, 3, 4 respectively in same order. Then it is clear from the definition of ${\left(\overline{\Sigma }\right)}_2$ that $C_5^{\prime }$ will have negative edges corresponding to the adjacent pair of negative edges in $C_5$. Hence $C_5^{\prime }$ will have one less negative edge for each negative section in $C_5$ in $\Sigma$. Now, $\Sigma \sim {\left(\overline{\Sigma }\right)}_2$ implies that the number of negative edge in $\Sigma$ must be even which implies ${\sum }_{i=1}^3{k}_i+q+{\sum }_{i=1}^3(k_i-1)\cong 0\left(mod2\right)$ which is possible when $q-3\cong 0\left(mod2\right)$. Hence number of negative section must be even.

Conversely, we have the number of negative edges in signed cycle $v_1{v}_2{v}_6{v}_1$ of same parity as number of all negative paths in $T$where $T=\{v_1-v_3,v_3-v_5,v_5-v_1$} and the number of negative sections in $C_5=v_2{v}_3{v}_4{v}_5{v}_6{v}_2$ must be even or all-negative. This implies $C_3$ is cycle-isomorphic to $C_3^{\prime }$ and $C_5$ is cycle-isomorphic to $C_5^{\prime }$. Hence, $\Sigma \sim {\left(\overline{\Sigma }\right)}_2$.

Theorem 29

Let$C_n:v_1{v}_2{v}_3{v}_4\dots v_n{v}_1$be a cycle of length$n$with$n\ge 6$vetices and${C^c}_n$be complement of$C_n$. For any$Z_a,Z_b\in \Psi \left({C^c}_n\right);$${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$if and only if the number of the number of all-negative induced subgraph in$\left\{v_j,v_{j+1},v_{j+3},v_{j+4},...,v_{j+n-2}withj+n-2\cong mod\left(n\right);j=1,2,3...n\right\}$, in$Z_a$is of same parity as that of$Z_b$.

Proof. Given that $Z_a,Z_b\in \Psi \left({C^c}_n\right)$and ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2.{\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$are cycles of length $n$. Any edge $v_j{v}_{j+1}$ where $j=1,2,3,\dots n$ and $v_{n+1}=v_1$ in ${\overline{\left(Z_a\right)}}_2$ and ${\overline{\left(Z_b\right)}}_2$ corresponds to a path of length two between $v_j$and $v_{j+1}$ in $Z_a,Z_b$. We know that $Z_a$ and $Z_b$ are $n-3$ regular signed graph. Any edge $v_j{v}_{j+1}$ in ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ corresponds to an induced subgraph $<v_j,v_{j+1},v_{j+3},v_{j+4},...,v_{j+n-2}>withj+n-2\cong mod\left(n\right);j=1,2,3...n\}$in $Z_{a}and{Z}_b$ respectively; which is all- negative when corresponding edge is negative. Now ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ implies that number of all-negative induced subgraphs in $\left\{v_j,v_{j+1},v_{j+3},v_{j+4},...,v_{j+n-2}withj+n-2\cong mod\left(n\right);j=1,2,3...n\right\}$ in $Z_a$ will be of the same parity as that of, $Z_b$.

Conversely, if all-negative induced subgraphs in $\left\{v_j,v_{j+1},v_{j+3},v_{j+4},...,v_{j+n-2}withj+n-2\cong mod\left(n\right);j=1,2,3...n\right\}$ in $Z_a,Z_b$ are of the same parity then ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ as the number of negative edges in ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are also of the same parity

Theorem 30

For any $Z_a,Z_b\in \Psi \left(F_1\right)$ , is a solution to ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ if and only if

• (i) The triangle in $Z_a,Z_b$are either all-positive or all-negative, or.

• (ii) The triangle in $Z_a,Z_b$ contains exactly one negative edge, and the number of incident edges on the end vertices of this negative edge in $Z_a$ is of same parity as that of $Z_b$, or.

• (iii) The triangle in $Z_a,Z_b$ contains exactly one positive edge, and the number of incident edges on the end vertices of this positive edge in $Z_a$ is of same parity as that of $Z_b$.

Proof. Given that $Z_a,Z_b\in \Psi \left(F_1\right)$, and ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$. ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ are cycle isomorphic. Every edge in ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$corresponds to a path of length two in $Z_a,Z_b$ respectively.

• Case1: If ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ are all-negative, then each path of length two in $Z_a,Z_b$ is negative. Thus triangle is negative in $Z^{\prime },Z^{″}.$

• Case2: If ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ are all-positive, then each path of length two in $Z_a,Z_b$ is positive. Thus triangle is positive in $Z_a,Z_b$.

Hence, the triangles in $Z_a,Z_b$are either all-positive or all-negative.

• (ii)${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are cycle-isomorphic, and let ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ be balanced, then ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$contain an even number of negative edges corresponds to an even number of negative paths of length two in $Z_a,Z_b$. Now, every edge in ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ corresponds to a path of length two between any pair of vertices passes through the triangle. If ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$have exactly two negative edges then the triangle has exactly one negative edge, creating two negative paths of length two in $Z_a,Z_b$. Since the triangles in ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are balanced which gives the number of incident edges on the end vertices of this negative edge in $Z_a$ is of same parity as that of $Z_b$.
• (iii)${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are cycle-isomorphic, and let ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$be balanced, then ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ contain an even number of negative edges corresponds to an even number of negative paths of length two in $Z_a,Z_b$. Now, every edge in ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ corresponds to a path of length two between any pair of vertices passes through the triangle. If ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$have exactly four negative edges then the triangle has exactly one positive edge, creating four negative paths of length two in $Z_a,Z_b$. Since the cycles ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are balanced which gives the number of incident edges on the end vertices of this positive edge in $Z_a$ is of same parity as that of $Z_b.$

Conversely, by part (i), (ii) and (iii) cycle ${\overline{\left(Z_a\right)}}_2$and ${\overline{\left(Z_b\right)}}_2$ are balanced then ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2.$

$F_2$ is a graph in Fig. 12 which is obtained by identification of $C_4:v_1{v}_2{v}_5{v}_6{v}_1$ and unicycle $v_2{v}_5{v}_3{v}_2$ + ${\mathrm{v}}_3{\mathrm{v}}_4$.

Fig. 12.

Graph $F_2.$

Theorem 31

For any $Z_a,Z_b\in \Psi \left(F_2\right)$ , is a solution to ${\overline{\left(Z_a\right)}}_2\sim {\overline{\left(Z_b\right)}}_2$ if and only

• (i) The triangle in $Z_a,Z_b$are either all-positive or all-negative, or.

• (ii) The triangle in $Z_a,Z_b$ contains exactly one negative edge, and the number of incident edges on the end vertices of this negative edge in $Z_a$ is of same parity as that of $Z_b$, or.

• (iii) The triangle in $Z_a,Z_b$ contains exactly one positive edge $v_2{v}_5$, and the number of incident edges on the end vertices of this positive edge in $Z_a$ is of same parity as that of $Z_b$, or.

Proof. Proof follows by Theorem 30.

Conclusion and Scope

In this paper, the concept of a $t$-path product signed graph for $t$=2, 3 for a given signed graph $\Sigma$ is explored. A $t$-path product signed graph retains the same vertex set as, and vertices $u$ and $v$ are adjacent if there is a path of length $t$in $\Sigma$. The sign of an edge, ${\sigma }^{\prime }\left(uv\right)$, is determined by the canonical marking $\mu$. The paper provides a switching equivalent between $\Sigma$ and $t$-path product signed graph for $t=2,3$ and also for its negation. Additionally, the paper explores the isomorphism between the 2-path signed graph and the 2-path product signed graph. Also some possible directions of applications are discussed which sees the reference to [[39], [40], [41], [42], [43]].

Ethics statements

Nil

CRediT authorship contribution statement

Deepa Sinha: Conceptualization, Supervision, Writing – review & editing. Sachin Somra: Software, Methodology, Writing – original draft.

Declaration of competing interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Data availability

No data were used to support the findings of the study.