Balanced picture fuzzy graph with application

Abstract

Picture fuzzy graphs are an extension of intuitionistic fuzzy graphs. Balanced picture fuzzy graph is a special type of picture fuzzy graph (PFG). In this study, the definition and important properties of PFG like, average PFG, balanced PFG, size, order, density of a PFG, isomorphism, the direct product of two PFG, etc have been studied. The necessary and sufficient conditions for balanced picture fuzzy graphs have also been studied in this article. Beside this, we proposed an algorithm to test whether a PFG is balanced or not. The proof of correctness and an illustration of the proposed algorithm is presented in this article. Lastly, an application of balanced PFG to business alliance is presented.

1 Introduction

A picture fuzzy set is a generalization of intuitionistic fuzzy set (Atanassov 1986). Picture fuzzy models give more precision, flexibility and compatibility to the system as compared to the intuitionistic fuzzy models. The concept of a picture fuzzy set was first introduced by Cuong and Kreinovich (2013). In addition to intuitionistic fuzzy sets, Coung appended new components which determine the degree of neutral membership. The intuitionistic fuzzy set gives the degree of membership and the degree of non-membership of an element, while the picture fuzzy set gives the degree of positive membership, the degree of neutral membership and the degree of negative membership of an element. These memberships are more or less independent from each other, the only requirement is that the sum of these three degrees is less than or equal to 1. Basically, picture fuzzy sets based models may be adequate in situations where we counter several opinions involving more answers of types: yes, no, abstain, refusal. If we take voting as an example , the human voters may be separated into four possible groups with distinct choices of opinions like vote for, abstain, vote against, refusal of the voting. Some properties of the picture fuzzy set and its operators have been studied in Cuong (2014), Dutta and Ganju (2017).

1.1 Review of literature

After the invention of fuzzy graph theory, it increases with its several extensions. In 2012 (Rashmanlou and Pal 2013) have defined balanced interval-valued fuzzy graphs and discuss some important properties, like product of two interval-valued fuzzy graphs, balanced and strictly balanced interval-valued fuzzy graphs, etc. (Samanta and Pal 2011) has introduced the definition of fuzzy threshold graph. Also, they have introduced some important definitions related to fuzzy threshold graphs like, fuzzy alternating four cycle, fuzzy ferrers digraph, threshold dimension of fuzzy graph. Fuzzy planar graphs have been introduced by Samanta et al. (2014),Samanta and Pal (2015). In this article they have discussed the concept of two different edges: effective edge and considerable edge. Also, in 2013, (Pal et al. 2013) have studied about fuzzy k-competition graphs. Interval-valued fuzzy threshold graphs are defined and studied several properties by (Pramanik et al. 2016a, b). They also have considered planarity in bipolar fuzzy graphs and they extended it to bipolar fuzzy planar graphs (Pramanik et al. 2018). In 2017, (Alvi et al. 2017) have studied an adaptive grayscale image de-noising technique by fuzzy inference system. Also, (Pramanik et al. 2016) have extended fuzzy planar graph to interval-valued fuzzy planar graph, interval-valued fuzzy graph. (Voskoglou and Pramanik 2020) have discussed and characterized several fuzzy graph theoretic structures and fuzzy hypergraphs. (Srivastav et al. 2020) have worked on Integration of Multiple Cache Server Scheme for User-Based Fuzzy Logic in Content Delivery Networks.

Sahoo and Pal (2015b) discussed the concept of intuitionistic fuzzy competition graphs. In 2012, (Akram and Davvaz 2012))defined strong intuitionistic fuzzy graphs. They also discuss intuitionistic fuzzy hypergraphs with applications (Akram and Dudek 2013). Also, in 2014, (Akram and Al-Shehrie, 2014) defined intuitionistic fuzzy cycles and intuitionistic fuzzy trees and intuitionistic fuzzy planar graphs (Al-Shehrie and Akram 2014). (Karunambigai et al. 2013) have defined density of intuitionistic fuzzy graph, Balanced intuitionistic fuzzy graph, direct product of intuitionistic fuzzy graph. (Sahoo and Pal 2015b) discussed the concept of intuitionistic fuzzy competition graphs. Also, (Sahoo and Pal 2015a, 2016) discussed intuitionistic fuzzy tolerance graphs with application, different types of products on intuitionistic fuzzy graphs. In 2019, (Zuo et al. 2019) introduced the concept of picture fuzzy set to graph theory and obtained PFG. In this article some types of picture fuzzy graphs such as strong PFG, regular PFG, complete PFG, and complement of PFG are introduced. Also, the isomorphism of PFGs, Cartesian product, composition, join, direct product, lexicographic and strong product on PFGs have been defined. (Ismayil and Bosley 2019) have introduced the domination in PFG and have defined the order and size of a PFG. Also, (Ismayil et al. 2019), have studied edge domination in PFG. (Akram and Habib 2019) have introduced q-rung picture fuzzy line graphs and developed a necessary condition for this graph. Recently, (Mohanta et al. 2020) have introduced the concept of a dombi operator to picture fuzzy graphs and obtained picture Dombi fuzzy graphs. In 2020, (Das and Ghorai 2020) have studied some properties of planer picture fuzzy graph like, strong (weak) edges, strong (weak) picture fuzzy planar graphs, strength of an edge, degree of planarity, picture fuzzy faces, strong (weak) picture fuzzy faces, etc. Many terminologies of fuzzy graphs and their variations and applications are mentioned in the recent published book written by (Pal et al. 2020).

1.2 Motivation

Picture fuzzy graph is an extension of intuitionistic fuzzy graphs. Intuitionistic fuzzy graphs have a lot of applications in the real world, because it has a capability to model several decision making problems in an uncertain environment. A number of generalizations of intuitionistic fuzzy graphs have been introduced to deal with the uncertainty of the complex real life problems. As uncertainties are well expressed using picture fuzzy sets, picture fuzzy graphs would be a prominent research direction for modeling the uncertain optimization problems. We see that one of the important concepts of neutrality degree is lacking in IFS theory. Concept of neutrality degree can be seen in situations when we face human opinions involving more answers of type: yes, abstain, no, refusal. For example, in a democratic election station, the council issues 600 voting papers for a candidate. The voting results are divided into four groups accompanied with the number of papers namely “vote for”(350), “abstain”(70), “vote against” (155) and “refusal of voting”(25). Group “abstain” means that the voting paper is a white paper rejecting both “agree” and “disagree” for the candidate but still takes the vote. Group “refusal of voting” is either invalid voting papers or bypassing the vote.

This situation can not be described by intuitionistic fuzzy graphs. Motivated from this point of view we consider balanced picture fuzzy graphs for this research study and have presented some definitions, properties, theorems, and algorithms to test whether a balanced picture fuzzy graph is balanced or not.

The remaining part of the paper is organized as follows. Some preliminaries are presented in Sect. 1. In Sect. 2, picture fuzzy graphs and some related terms are defined. In Sect. 3, balanced picture fuzzy graphs and its properties have been studied. Also an algorithm to check the balanced property of a PFG and illustration of the algorithm is also presented in this section. In Sect. 4 an application of balanced PFG is presented. Section 5 is for brief discussion and conclusion.

2 Preliminaries

Intuitionistic fuzzy graph is a generalization of fuzzy graphs. The definition of intuitionistic fuzzy graph is given below:

Definition 1

An intuitionistic fuzzy graph is of the form $G=(V,\sigma , \mu )$ where $\sigma =(\sigma _1, \sigma _2)$, $\mu =(\mu _1,\mu _2)$ and

• (i) $V=\{p_{1},p_{2},\ldots , p_{n}\}$ such that $\sigma _{1}:V\rightarrow [0,1]$ and $\sigma _{2}:V \rightarrow [0,1]$, denote the degree of membership and non-membership of the node $p_{i}\in V$ respectively and $0\le \sigma _{1}(p_{i}) + \sigma _{2}(p_{i})\le 1$ for every $p_{i}\in V$ $(i=1,2,\ldots ,n)$.

• (ii) $\mu _{1}:V \times V \rightarrow [0,1]$ and $\mu _{2}:V \times V \rightarrow [0,1]$, where $\mu _{1}(p_{i},p_{j})$ and $\mu _{2}(p_{i},p_{j})$ denote the the degree of membership and non-membership value of the edge $(p_{i},p_{j})$ respectively such that $\mu _{1}(p_{i},p_{j}) \le \min \{\sigma _{1}(p_{i}),\sigma _{1}(p_{j})\}$ and $\mu _{2}(p_{i},p_{j}) \le \max \{\sigma _{2}(p_{i}),\sigma _{2}(p_{j})\}$, $0 \le \mu _{1}(p_{i},p_{j}) + \mu _{2}(p_{i},p_{j})\le 1$ for every $(p_{i},p_{j})\in V\times V$.

Picture fuzzy set is an extension of an intuitionistic fuzzy set. The definition of picture fuzzy set is presented below.

Definition 2

A picture fuzzy set A on a universe X is an object of the form

$A=\{(x,\mu _A(x),\eta _A(x),\nu _A(x)):x\in X\}$, where $\mu _A(x)\in [0,1]$ is called the degree of positive membership of x in A, $\eta _A(x)\in [0,1]$ is called the degree of neutral membership of x in A, $\nu _A(x)\in [0,1]$ is called the degree of negative membership of x in A and they satisfies $0\le \mu _A(x)+\eta _A(x)+\nu _A(x)\le 1$ for all $x\in X$.

Now $1-\{\mu _A(x)+\eta _A(x)+\nu _A(x)\}$ is said to be the degree of refusal of x in A. Basically, picture fuzzy sets based models may be adequate in situations when we face human opinions involving more answers of types: yes, abstain, no , refusal. Voting can be a good example of such a situation as the human voters may be divided into four groups of those who: vote for, abstain, vote against, refusal of the voting.

3 Picture fuzzy graph

In this section, we introduce some basic definitions and properties, theorems related to picture fuzzy graphs.

Definition 3

A picture fuzzy graph (PFG) is of the type $G=(V,E,m_V,m_E)$, where $V=\{p_1,p_2,\ldots ,p_n\}$ be the set of nodes, $m_V$ and $m_E$ denotes the membership function of nodes and edges respectively, where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$ is such that,

(i):

$\mu _V:V \rightarrow [0,1]$, $\eta _V:V \rightarrow [0,1]$ and $\nu _V:V \rightarrow [0,1]$ are respectively positive membership, neutral membership and negative membership function and it satisfy the condition $0\le \mu _V(p_i)+\eta _V(p_i)+\nu _V(p_i)\le 1$ for all $p_i\in V$, for $i=1,2,\ldots ,n$.

(ii):

$\mu _E:V\times V\rightarrow [0,1]$, $\eta _E:V\times V\rightarrow [0,1]$ and $\nu _E:V\times V\rightarrow [0,1]$, are respectively positive membership, neutral membership and negative membership function of edge $(p_i,p_j)$, which satisfies the following condition

$\mu _E(p_i,p_j)\le \min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)\le \min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)\le \max \{\nu _V(p_i), \nu _V(p_j)\}$ and $0\le \mu _E(p_i,p_j)+\eta _E(p_i,p_j)+\nu _E(p_i,p_j)\le 1$ for every $(p_i,p_j)\in V\times V$, for $i=1,2,\ldots ,n$.

Here we notice that G is a picture fuzzy graph with node set V and E be the set of edges, $\mu _V(p_i), \eta _V(p_i), \nu _V(p_i)$ are respectively the positive, neutral, negative membership value of the node $p_i$, and $\mu _E(p_i,p_j), \eta _E(p_i,p_j), \nu _E(p_i,p_j)$ are respectively the positive, neutral, negative membership degree of the edge joining the nodes $p_i, p_j$.

A PFG is shown in Fig. 1. In this figure we see that the degree of positive membership, neutral membership and negative membership value of the node $p_1$ is 0.3, 0.2, 0.4 respectively. Also, the degree of positive membership, neutral membership and negative membership value of the edge joining the nodes $p_1$ and $p_2$ is 0.25, 0.1, 0.35 respectively. Same for the other nodes and edges.

Fig. 1

A picture fuzzy graph

Throughout the paper we denote a PFG $G=(V,E,m_V,m_E)$, where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$.

Definition 4

A PFG $G=(V,E,m_V,m_E)$, where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$ is called complete picture fuzzy graph if $\mu _E(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)= \min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$.

Definition 5

$G=(V,E,m_V,m_E)$ be a PFG then G is said to be a single valued strong picture fuzzy graph if $\mu _E(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}$,

$\eta _E(p_i,p_j)= \min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $(p_i, p_j)\in E$.

Definition 6

A PFG $G=(V,E,m_V,m_E)$ where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$ is called single valued average picture fuzzy graph

if $\mu _E(p_i,p_j)= \frac{1}{2}\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)= \frac{1}{2}\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \frac{1}{2}\max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$.

Definition 7

The complement of a PFG $G=(V,E,m_V,m_E)$ is a PFG $\bar{G}=(\bar{V},\bar{E},\bar{m_V},\bar{m_E})$, where $m_V=(\mu _V,\eta _V,\nu _V)$, $m_E=(\mu _E,\eta _E,\nu _E)$, $\bar{m_V}=(\bar{\mu _V},\bar{\eta _V},\bar{\nu _V})$ and $\bar{m_E}=(\bar{\mu _E},\bar{\eta _E},\bar{\nu _E})$

• if (i) $\bar{V}=V$.

• (ii) $\bar{\mu _V}(p_i)=\mu _V(p_i)$, $\bar{\eta _V}(p_i)=\eta _V(p_i)$, $\bar{\nu _V}(p_i)=\nu _V(p_i)$, for all $p_i \in V$.

• (iii) $\bar{\mu _E}(p_i,p_j)=\min \{\mu _V(p_i), \mu _V(p_j)\}-\mu _E(p_i,p_j)$, $\bar{\eta _E}(p_i,p_j)=\min \{\eta _V(p_i), \eta _V(p_j)\}-\eta _E(p_i,p_j)$ and $\bar{\nu _E}(p_i,p_j)=\max \{\nu _V(p_i), \nu _V(p_j)\}-\nu _E(p_i,p_j)$, for all $(p_i,p_j)\in V\times V$.

Definition 8

Let $G=(V,E,m_V,m_E)$ be a PFG where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$. The order of G is denoted by O(G) and it is defined by $O(G)=(O_{\mu }(G),O_{\eta }(G),O_{\nu }(G))$, where

$O_{\mu }(G)=\frac{1}{n}{\displaystyle \sum\nolimits _{p_i\in V}\mu _V(p_i)}$, $O_{\eta }(G)=\frac{1}{n}{\displaystyle \sum\nolimits _{p_i\in V}\eta _V(p_i)}$, $O_{\nu }(G)=\frac{1}{n}{\displaystyle \sum\nolimits _{p_i\in V}\nu _V(p_i)}$.

The value of $O_{\mu }(G)$, $O_{\eta }(G)$ and $O_{\nu }(G)$ lies between 0 and 1.

Definition 9

Let $G=(V,E,m_V,m_E)$ be a PFG where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$. Then the size of G is denoted by $S_G$ and is defined by $S_G=(S_{\mu }(G),S_{\eta }(G),S_{\nu }(G))$, where

$S_{\mu }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}$, $S_{\eta }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\eta _E(p_i,p_j)}$, $S_{\nu }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\nu _E(p_i,p_j)}$.

4 Balanced picture fuzzy graph

In this section, we have studied the definition and properties of balanced picture fuzzy graphs. Also, we have studied the necessary and sufficient condition for a balanced picture fuzzy graph. An algorithm is also presented to test whether a PFG is balanced or not.

Definition 10

Let $G=(V,E,m_V,m_E)$ be a PFG where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$. Then the weight of G is denoted by w(G) and is defined by

$$\begin{aligned}&w(G)=(w_{\mu }(G),w_{\eta }(G),w_{\nu }(G)), \hbox {where} \\&w_{\mu }(G)= {\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}\\&w_{\eta }(G)= {\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\eta _V(p_i),\eta _V(p_j)\}}\\&w_{\nu }(G)= {\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\max \{\nu _V(p_i),\nu _V(p_j)\}} \end{aligned}$$

Now, one can defined the density of a PFG G based on weight and size of G.

Definition 11

Let $G=(V,E,m_V,m_E)$ be a PFG where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$. Then the density of G is denoted by $\rho (G)$ and is defined by

$\rho (G)=(\rho _{\mu }(G),\rho _{\eta }(G),\rho _{\nu }(G))$, where $\rho _{\mu }(G)= \frac{S_{\mu }(G)}{w_{\mu }(G)}$, $\rho _{\eta }(G)= \frac{S_{\eta }(G)}{w_{\eta }(G)}$, $\rho _{\nu }(G)= \frac{S_{\nu }(G)}{w_{\nu }(G)}$ for all $p_i, p_j\in V$. All the components $\rho _{\mu }(G)$, $\rho _{\eta }(G)$ and $\rho _{\nu }(G)$ lie between 0 and 1.

Definition 12

Let $G=(V,E,m_V,m_E)$ be a PFG. A subgraph S of G is called intense picture fuzzy subgraph if the node set of S is a subset of node set of G and $E(S)\subseteq E(G)$ and $\rho (S)\le \rho (G)$.

Now $\rho (S)\le \rho (G)$ holds if $\rho _{\mu }(S)\le \rho _{\mu }(G)$, $\rho _{\eta }(S)\le \rho _{\eta }(G)$, $\rho _{\nu }(S)\le \rho _{\nu }(G)$.

Definition 13

Let $G=(V,E,m_V,m_E)$ be a PFG. A subgraph S of G is called feeble picture fuzzy subgraph if the node set of S is a subset of node set of G and $V(S)\subseteq V(G)$, $E(S)\subseteq E(G)$ and $\rho (S)> \rho (G)$, i.e.,$\rho _{\mu }(S)> \rho _{\mu }(G)$, $\rho _{\eta }(S)> \rho _{\eta }(G)$, $\rho _{\nu }(S)> \rho _{\nu }(G)$.

Fig. 2

A picture fuzzy graph

Example 1

Here we consider a graph in Fig. 2 and find out the intense picture fuzzy subgraph and feeble picture fuzzy subgraph of G. For this graph $\rho (G)=(0.7,0.3,0.5)$.

Let us consider a subgraph $H_1=(V_1,E_1,m_{V_1},m_{E_1})$ where $V_1=\{p_3,p_4,p_5\}$, $E_1=\{(p_3,p_5),(p_5,p_4),(p_3,p_4) \}$.

Then $\rho (H_1)=(0.7,0.3,0.5)$.

Let us consider a subgraph $H_2=(V_2,E_2,m_{V_2},m_{E_2})$ where $V_2=\{p_4,p_5\}$, $E_2=\{(p_4,p_5) \}$.

Then $\rho (H_2)=(0.8,0.5,0.55)$.

Again consider a subgraph $H_3=(V_3,E_3,m_{V_3},m_{E_3})$ where $V_3=\{p_3,p_4\}$, $E_3=\{(p_3,p_4) \}$.

Then $\rho (H_3)=(0.625,0.167,0.04)$. Also, we consider a subgraph $H_4=(V_4,E_4,m_{V_4},m_{E_4})$ where $V_4=\{p_1,p_2,p_3\}$, $E_4=\{(p_1,p_2),(p_2,p_3),(p_1,p_3), (p_3,p_4) \}$.

Then $\rho (H_4)=(0.675,0.23,0.48)$. Hence, the subgraphs $H_1$ and $H_3$, $H_4$ are intense picture fuzzy subgraphs of G and the subgraphs $H_2$ is a feeble picture fuzzy subgraph of G.

Theorem 1

Let $G=(V,E,m_V,m_E)$ be a PFG. Then G is called self complementary if and only if G is an average picture fuzzy graph.

Proof

Since, $G=(V,E,m_V,m_E)$ be a PFG. Let $\bar{G}=(\bar{V},\bar{E},\bar{m_V},\bar{m_E})$ be its complement, where $m_V=(\mu _V,\eta _V,\nu _V)$, $m_E=(\mu _E,\eta _E,\nu _E)$, $\bar{m_V}=(\bar{\mu _V},\bar{\eta _V},\bar{\nu _V})$ and $\bar{m_E}=(\bar{\mu _E},\bar{\eta _E},\bar{\nu _E})$. Since $\bar{G}$ is the complement of G, then

• (i) $\bar{V}=V$.

• (ii) $\bar{\mu _V}(p_i)=\mu _V(p_i)$, $\bar{\eta _V}(p_i)=\eta _V(p_i)$, $\bar{\nu _V}(p_i)=\nu _V(p_i)$, for all $p_i \in V$.

• (iii) $\bar{\mu _E}(p_i,p_j)=\min \{\mu _V(p_i), \mu _V(p_j)\}-\mu _E(p_i,p_j)$, $\bar{\eta _E}(p_i,p_j)=\min \{\eta _V(p_i), \eta _p(p_j)\}-\eta _E(p_i,p_j)$ and $\bar{\nu _E}(p_i,p_j)=\max \{\nu _V(p_i), \nu _V(p_j)\}-\nu _E(p_i,p_j)$, for every $p_i,p_j\in V$.

Let G be a single valued average PFG then $\mu _E(p_i,p_j)= \frac{1}{2}\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)= \frac{1}{2}\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \frac{1}{2}\max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$. Now,

$$\begin{aligned} \bar{\mu }_E(p_i,p_j)= & {} \min \{\mu _V(p_i), \mu _V(p_j)\}-\frac{1}{2}\min \{\mu _V(p_i), \mu _V(p_j)\}\\= & {} \frac{1}{2}\min \{\mu _V(p_i), \mu _V(p_j)\}\\= & {} \mu _E(p_i,p_j) \text{ for } \text{ all } p_i, p_j\in V. \end{aligned}$$

Similarly, $\bar{\eta }_E(p_i,p_j)=\eta _E(p_i,p_j)$ and $\bar{\nu }_E(p_i,p_j)=\nu _E(p_i,p_j)$. This shows that $\bar{G}$ is similar to G. Hence, G is a self complementary PFG.

Conversely, let G is a self complementary PFG. Then $\bar{\mu }_{E}(p_i,p_j)=\mu _E(p_i,p_j)$, $\bar{\eta }_{E}(p_i,p_j)=\eta _E(p_i,p_j)$, $\bar{\nu }_{E}(p_i,p_j)=\nu _E(p_i,p_j)$ for all $p_i,p_j\in V$. Now, $\mu _E(p_i,p_j)=\bar{\mu }_{E}(p_i,p_j)$ $=\min \{\mu _V(p_i), \mu _V(p_j)\}-\mu _E(p_i,p_j)$. So, $2\mu _E(p_i,p_j)=\min \{\mu _V(p_i), \mu _V(p_j)\}$. Therefore, $\mu _E(p_i,p_j)=\frac{1}{2}\min \{\mu _V(p_i), \mu _V(p_j)\}$. Similarly, $\eta _E(p_i,p_j)=\frac{1}{2}\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\frac{1}{2}\min \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$. Hence, G is single valued average PFG. $\square$

Theorem 2

Let $G=(V,E,m_V,m_E)$ be an average PFG and $\rho (G)$ be the density of G then $\rho (G)=(0.5,0.5,0.5)$.

Proof

Since $\rho (G)$ is the density of the PFG G then $\rho (G)=(\rho _{\mu }(G),\rho _{\eta }(G),\rho _{\nu }(G))$

We know that,

$$\begin{aligned} \rho _{\mu }(G)= & {} \frac{S_{\mu }(G)}{w_{\mu }(G)}\\= & {} \frac{{\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}\\= & {} \frac{\frac{1}{2}{\displaystyle \sum\nolimits _{p_i\ne p_j}\min \{\mu _V(p_i),\mu _V(p_j)\}}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}\\= & {} 0.5. \end{aligned}$$

Similarly, $\rho _{\eta }(G)=0.5$ and $\rho _{\nu }(G)=0.5$. Therefore, $\rho (G)=(0.5,0.5,0.5)$. $\square$

Definition 14

A PFG $G=(V,E,m_V,m_E)$ is said to be balanced if all its subgraphs are intense in G, i.e., $\rho (S)\le \rho (G)$ for any subgraph S of G. $\rho (S)\le \rho (G)$ holds if $\rho _{\mu }(S)\le \rho _{\mu }(G)$, $\rho _{\eta }(S)\le \rho _{\eta }(G)$ and $\rho _{\nu }(S)\le \rho _{\nu }(G)$.

Fig. 3

A PFG

Example 2

We consider a PFG $G=(V,E,m_V,m_E)$ such that $V=\{p_1,p_2,p_3,p_4,p_5\}$, $E=\{(p_1,p_2), (p_1,p_3), (p_1,p_4), (p_2,p_3), (p_2,p_5)\}$ in Fig. 3 and then check whether this PFG is balanced or not. We know that the size of the graph G is $S_G=(S_{\mu }(G),S_{\eta }(G),S_{\nu }(G))$. For this graph, $S_{\mu }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}=1.2$, $S_{\eta }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\eta _E(p_i,p_j)}=0.95$, $S_{\nu }(G)={\displaystyle \sum\nolimits _{p_i\ne p_j}\nu _E(p_i,p_j)}=1.95$. Again, the weight of G is $w(G)=(w_{\mu }(G),w_{\eta }(G),w_{\nu }(G))$. Then $w_{\mu }(G)= {\displaystyle \sum _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}=1.5$.

$$\begin{aligned}&w_{\eta }(G)= {\displaystyle \sum _{(p_i, p_j)\in E}\min \{\eta _V(p_i),\eta _V(p_j)\}}=1.9.\\&w_{\nu }(G)= {\displaystyle \sum _{(p_i, p_j)\in E}\min \{\nu _V(p_i),\nu _V(p_j)\}}=3.0. \end{aligned}$$

Therefore, the density of G is $\rho (G)=(\rho _{\mu }(G),\rho _{\eta }(G),\rho _{\nu }(G))$, where

$\rho _{\mu }(G)= \frac{S_{\mu }(G)}{w_{\mu }(G)}=0.8$, $\rho _{\eta }(G)= \frac{S_{\eta }(G)}{w_{\eta }(G)}=0.5$, $\rho _{\nu }(G)= \frac{S_{\nu }(G)}{w_{\nu }(G)}=0.65$. So, $\rho (G)=(0.8,0.5,0.65)$. From Table 1, we see that the density of the subgroup $S_i$ is (0.8, 0.5, 0.65) for $i=1,2,\ldots ,19,21,23$ and the density of the subgroup $S_j$ is (0, 0, 0) for $j=20,22,24,25,26$. Here, all $\rho (S_r)\le \rho (G)$ for all subgraph $S_r$ of G which is shown in the following table.

Table 1 Density of all subgraph of the PFG in Fig. 3

Hence, G is balanced.

Definition 15

A PFG $G=(V,E,m_V,m_E)$ is said to be strictly balanced if $\rho (G)=\rho (S)$ for all subgraph S of G, i.e., $\rho (S)= \rho (G)$ holds if $\rho _{\mu }(S)= \rho _{\mu }(G)$, $\rho _{\eta }(S)= \rho _{\eta }(G)$ and $\rho _{\nu }(S)= \rho _{\nu }(G)$.

Theorem 3

PFG $G=(V,E,m_V,m_E)$ is strictly balanced iff

$\mu _E(p_i,p_j)=\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\lambda _3\max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i,p_j\in V$, where $\rho (G)=(\lambda _1, \lambda _2, \lambda _3)$.

Proof

Suppose that $G=(V,E,m_V,m_E)$ is strictly balanced PFG with n nodes and $V=\{p_1,p_2,\ldots ,p_n\}$. Then $\rho (G)=\rho (S)$ for all subgraph S of G. Also given that $\rho (G)=(\lambda _1, \lambda _2, \lambda _3)$. Since, G contains n nodes then G has $2^{n}-(n+1)=\Delta$ subgraph. Among the $\Delta$ subgraphs of G, $^nc_2=\Delta _1$ subgraphs are the subgraphs, each containing 2 nodes.

In $\Delta _1$ subgraph, let us consider any arbitrary subgraph, say $S_r$ of G.

Let $S_r=(V_r,E_r,m_{V_r},m_{E_r})$ where $V_r=\{p_{r_i},p_{r_j}\}$.

Now the density of $S_r$, $\rho (S_r)=(\rho _{\mu }(S_r),\rho _{\eta }(S_r),\rho _{\nu }(S_r))$.

Therefore, $\rho _{\mu }(S_r)= \frac{\mu _{E_r(p_{r_i},p_{r_j})}}{\min \{\mu _{V_r}(p_{r_i}),\mu _{V_r}(p_{r_j}) \}}$, $\rho _{\eta }(S_r)= \frac{\eta _{E_r(p_{r_i},p_{r_j})}}{\min \{\eta _{V_r}(p_{r_i}),\eta _{V_r}(p_{r_j}) \}}$,

$\rho _{\nu }(S_r)= \frac{\nu _{E_r(p_{r_i},p_{r_j})}}{\max \{\nu _{V_r}(p_{r_i}),\nu _{V_r}(p_{r_j}) \}}$, for $p_{r_i}, p_{r_j}\in V_r$. Since, $S_r$ is an arbitrary subgraph G with two nodes and G is strictly balanced, so, $\rho (S_r)=\rho (G)$. This implies that $\rho _{\mu }(S_r)=\lambda _1$, $\rho _{\eta }(S_r)=\lambda _2$ and $\rho _{\nu }(S_r)=\lambda _3$.

That is $\mu _{E_r}(p_{r_i},p_{r_j})=\lambda _1\min \{\mu _V(p_{r_i}), \mu _V(p_{r_j})\}$,

$\eta _{E_r}(p_{r_i},p_{r_j})=\lambda _2\min \{\eta _V(p_{r_i}), \eta _V(p_{r_j})\}$, $\nu _{E_r}(p_{r_i},p_{r_j})=\lambda _3\max \{\nu _V(p_{r_i}), \nu _V(p_{r_j})\}$ Since, $S_r$ is arbitrary subgraph with two nodes of G, so the above relation is true for all $p_{r_i}, p_{r_j}\in V$.

Hence, $\mu _E(p_i,p_j)=\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\lambda _3\max \{\nu _V(p_i), \nu _V(p_j)\}$, for all $p_i,p_j\in V$.

Conversely, let $G=(V,E,m_V,m_E)$ be a PFG where $V=\{p_1,p_2,\ldots ,p_n\}$. $\rho (G)$ be the density of G where, $\rho (G)=(\lambda _1, \lambda _2, \lambda _3)$. The membership function of all edges are satisfied the relations $\mu _E(p_i,p_j)=\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\lambda _3\max \{\nu _V(p_i), \nu _V(p_j)\}$, for all $p_i,p_j\in V$. Now we have to prove that G is strictly balanced.

Let $S_t=(V_t,E_t,m_{V_t},m_{E_t})$ be any subgraph of G, where $V_t=\{p_{t_1},p_{t_2},\ldots ,p_{t_m}\}$, $t_1,t_2,\ldots ,t_m \in \{1,2,\ldots ,n\}$ and $t_i\ne t_j$ for all i, j.

Let $\rho (S_t)=(\rho _{\mu }(S_t),\rho _{\eta }(S_t),\rho _{\nu }(S_t))$ be the density of the subgraph $S_t$. Then,

$$\begin{aligned} \rho _{\mu }(S_t)= & {} \frac{{\displaystyle \sum\nolimits _{p_{t_i}\ne p_{t_j}}\mu _{E_t}(p_{t_i},p_{t_j})}}{{\displaystyle \sum\nolimits _{(p_{t_i}, p_{t_j})\in E}\min \{\mu _{V_t}(p_{t_i}),\mu _{V_t}(p_{t_j})\}}}\\= & {} \frac{\lambda _1{\displaystyle \sum\nolimits _{(p_{t_i}, p_{t_j})\in E}\min \{\mu _{V_t}(p_{t_i}),\mu _{V_t}(p_{t_j})\}}}{{\displaystyle \sum\nolimits _{(p_{t_i}, p_{t_j})\in E}\min \{\mu _{V_t}(p_{t_i}),\mu _{V_t}(p_{t_j})\}}}\\= & {} \lambda _1\\ \end{aligned}$$

Similarly, $\rho _{\eta }(S_t)=\lambda _2$ and $\rho _{\nu }(S_t)=\lambda _3$. Therefore, $\rho (S_t)=(\lambda _1, \lambda _2, \lambda _3)$. Since $S_t$ is an arbitrary subgraph of PFG G, so, $\rho (S)=\rho (G)$ for all subgraph S of G. Hence, G is strictly balanced. $\square$

Corollary 1

PFG $G=(V,E,m_V,m_E)$ is balanced iff

$\mu _E(p_i,p_j)=\min \{\mu _V(p_i)\,, \mu _V(p_j)\} \times \lambda _1$, $\eta _E(p_i,p_j)=\min \{\eta _V(p_i)\,, \eta _V(p_j)\} \times \lambda _2$, $\nu _E(p_i,p_j)=\max \{\nu _V(p_i)\,, \nu _V(p_j)\} \times \lambda _3$ for all $(p_i,p_j)\in E$, where $\rho (G)=(\lambda _1, \lambda _2, \lambda _3)$.

Corollary 2

Let $G=(V,E,m_V,m_E)$ be a balanced PFG and S be any subgraph of G then $\rho (S)=\rho (G)$ or $\rho (S)=(0,0,0)$.

Figure 4 shows the inclusion of density of a PFG, balanced PFG and strictly PFG.

Fig. 4

Workflow of definitions

4.1 An Algorithm

In this subsection we proposed an algorithm to test whether a PFG is balanced or not. A proof of correctness of the proposed algorithm is also given in this section.

The proof of correctness of the above algorithm is given by the following theorem.

Theorem 4

Algorithm BPFG correctly tests whether a PFG is balanced or not.

Proof

Let $G=(V,E,m_V,m_E)$ be a PFG and let $\rho (G)$ be the density of the graph G. Then $\rho (G)=(\lambda _1,\lambda _2,\lambda _3)$. According to the definition of density function of G, we have

$$\begin{aligned} \lambda _1= & {} \rho _{\mu }(G)\\= & {} \frac{S_{\mu }(G)}{w_{\mu }(G)}\\= & {} \frac{{\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}\\ \lambda _2= & {} \rho _{\eta }(G)\\= & {} \frac{S_{\eta }(G)}{w_{\eta }(G)}\\= & {} \frac{{\displaystyle \sum\nolimits _{p_i\ne p_j}\eta _E(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\eta _V(p_i),\eta _V(p_j)\}}}\\ \lambda _3= & {} \rho _{\nu }(G)\\= & {} \frac{S_{\nu }(G)}{w_{\nu }(G)}\\= & {} \frac{{\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\nu _V(p_i),\nu _V(p_j)\}}} \end{aligned}$$

So, Algorithm BPFG correctly computes $\rho (G)$, the density of G.

Again from Theorem 3 and Corollary 1 of Theorem 3, we have a PFG

$G=(V,E,m_V,m_E)$ is balanced iff $\mu _E(p_i,p_j)=\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\lambda _3\max \{\nu _V(p_i), \nu _V(p_j)\}$, or $\mu _E(p_i,p_j)=0$, $\eta _E(p_i,p_j)=0$, $\nu _E(p_i,p_j)=0$ for all $p_i,p_j\in V$.

That is G is balanced if $(\mu _E(p_i,p_j),\eta _E(p_i,p_j),\nu _E(p_i,p_j))=(0,0,0)$ or $(\mu _E(p_i,p_j),\eta _E(p_i,p_j),\nu _E(p_i,p_j))=(\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_i), \nu _V(p_j)\})$. Otherwise G is not balanced.

Hence, the above algorithm correctly check the balanced property of PFG. $\square$

4.2 Illustration of Algorithm BPFG

The algorithm BPFG is illustrated for two different PFGs for considering all possible cases.

Illustration 1

Fig. 5

A PFG for illustration 1

Let us consider a PFG $G=(V,E,m_V,m_E)$ where $V=\{p_1,p_2,\ldots ,p_6\}$ and $m_V$, $m_E$ are shown in Fig. 5.

For this graph, $\rho (G)=(0.8,0.75,0.6)=(\lambda _1,\lambda _2,\lambda _3)$. Therefore, $\lambda _1=0.8$, $\lambda _2=0.75$ and $\lambda _3=0.6$.

Iteration 1: For $i=1$.

Let us consider a node $p_1$. Now compute, $(\mu _E(p_1,p_j),\eta _E(p_1,p_j),\nu _E(p_1,p_j))$ for $j=2,3,\ldots ,6$.

When $j=2$.

$$\begin{aligned}&\mu _E(p_1,p_2)=0.32=0.8\min \{0.4,0.6\}=0.8\min \{\mu _V(p_1),\mu _V(p_2)\}.\\&\eta _E(p_1,p_2)=0.075=0.75\min \{0.3,0.1\}=0.7\min \{\eta _V(p_1),\eta _V(p_2)\}.\\&\nu _E(p_1,p_2)=0.18=0.6\min \{0.1,0.3\}=0.6\max \{\nu _V(p_1),\nu _V(p_2)\}. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_2),\eta _E(p_1,p_2),\nu _E(p_1,p_2)) =(\lambda _1\min \{\mu _V(p_1), \mu _V(p_2)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_2)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_2)\})$

When $j=3$.

$$\begin{aligned}&\mu _E(p_1,p_3)=0.32=0.8\min \{0.4,0.5\}=0.8\min \{\mu _V(p_1),\mu _V(p_3)\}.\\&\eta _E(p_1,p_3)=0.15=0.75\min \{0.3,0.2\}=0.7\min \{\eta _V(p_1),\eta _V(p_3)\}.\\&\nu _E(p_1,p_3)=0.12=0.6\min \{0.1,0.2\}=0.6\max \{\nu _V(p_1),\nu _V(p_3)\}. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_3),\eta _E(p_1,p_3),\nu _E(p_1,p_3)) =(\lambda _1\min \{\mu _V(p_1), \mu _V(p_3)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_3)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_3)\})$

When $j=4$.

$$\begin{aligned}&\mu _E(p_1,p_4)=0.24=0.8\min \{0.4,0.3\}=0.8\min \{\mu _V(p_1),\mu _V(p_4)\}.\\&\eta _E(p_1,p_4)=0.075=0.75\min \{0.3,0.1\}=0.7\min \{\eta _V(p_1),\eta _V(p_4)\}.\\&\nu _E(p_1,p_3)=0.12=0.6\min \{0.1,0.2\}=0.6\max \{\nu _V(p_1),\nu _V(p_4)\}. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_4),\eta _E(p_1,p_4),\nu _E(p_1,p_4)) =(\lambda _1\min \{\mu _V(p_1), \mu _V(p_4)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_4)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_4)\})$

When $j=5$.

$$\begin{aligned}&\mu _E(p_1,p_5)=0.24=0.8\min \{0.4,0.3\}=0.8\min \{\mu _V(p_1),\mu _V(p_5)\}.\\&\eta _E(p_1,p_5)=0.225=0.75\min \{0.3,0.3\}=0.7\min \{\eta _V(p_1),\eta _V(p_5)\}.\\&\nu _E(p_1,p_5)=0.18=0.6\min \{0.1,0.3\}=0.6\max \{\nu _V(p_1),\nu _V(p_5)\}. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_5),\eta _E(p_1,p_5),\nu _E(p_1,p_5)) =(\lambda _1\min \{\mu _V(p_1), \mu _V(p_5)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_5)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_5)\})$

When $j=6$.

$$\begin{aligned} \mu _E(p_1,p_6)=0, \eta _E(p_1,p_6)=0, \nu _E(p_1,p_6)=0. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_6),\eta _E(p_1,p_6),\nu _E(p_1,p_6))=(0,0,0)$.

So, we can write,

$$\begin{aligned}&(\mu _E(p_1,p_j),\eta _E(p_1,p_j),\nu _E(p_1,p_j))\\= & {} \left\{ \begin{array} {ll } (\lambda _1\min \{\mu _V(p_1), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_1), \eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_j)\}) &{} \text { for } j=2,3,4,5\\ (0,0,0) &{} \text { for } j=6 \\ \end{array} \right. \end{aligned}$$

Iteration 2: For $i=2$.

Let us consider a node $p_2$. Now compute, $(\mu _E(p_2,p_j),\eta _E(p_2,p_j),\nu _E(p_2,p_j))$ for $j=3,4,5,6$.

When $j=3$.

$$\begin{aligned} \mu _E(p_2,p_3)=0.4=0.8\min \{0.6,0.5\}=0.8\min \{\mu _V(p_2),\mu _V(p_3)\}.\\ \eta _E(p_2,p_3)=0.075=0.75\min \{0.1,0.2\}=0.7\min \{\eta _V(p_2),\eta _V(p_3)\}.\\ \nu _E(p_2,p_3)=0.18=0.6\min \{0.3,0.2\}=0.6\max \{\nu _V(p_2),\nu _V(p_3)\}.\\ \end{aligned}$$

Therefore, $(\mu _E(p_2,p_3),\eta _E(p_2,p_3),\nu _E(p_2,p_3)) =(\lambda _1\min \{\mu _V(p_2), \mu _V(p_3)\}, \lambda _2\min \{\eta _V(p_2)$, $\eta _V(p_3)\}, \lambda _3\min \{\nu _V(p_2), \nu _V(p_3)\})$.

When $j=4$.

$$\begin{aligned} \mu _E(p_2,p_4)=0.24=0.8\min \{0.6,0.3\}=0.8\min \{\mu _V(p_2),\mu _V(p_4)\}.\\ \eta _E(p_2,p_4)=0.075=0.75\min \{0.1,0.1\}=0.7\min \{\eta _V(p_2),\eta _V(p_4)\}.\\ \nu _E(p_2,p_3)=0.18=0.6\min \{0.3,0.2\}=0.6\max \{\nu _V(p_2),\nu _V(p_4)\}.\\ \end{aligned}$$

Therefore, $(\mu _E(p_2,p_4),\eta _E(p_2,p_4),\nu _E(p_2,p_4)) =(\lambda _1\min \{\mu _V(p_2), \mu _V(p_4)\}, \lambda _2\min \{\eta _V(p_2)$, $\eta _V(p_4)\}, \lambda _3\min \{\nu _V(p_2), \nu _V(p_4)\})$.

When $j=5$.

$$\begin{aligned} \mu _E(p_2,p_5)=0, \eta _E(p_2,p_5)=0, \nu _E(p_2,p_5)=0. \end{aligned}$$

Therefore, $(\mu _E(p_2,p_5),\eta _E(p_2,p_5),\nu _E(p_2,p_5))=(0,0,0)$.

When $j=6$.

$$\begin{aligned} \mu _E(p_2,p_6)=0, \eta _E(p_2,p_6)=0, \nu _E(p_2,p_6)=0. \end{aligned}$$

Therefore, $(\mu _E(p_2,p_6),\eta _E(p_2,p_6),\nu _E(p_2,p_6))=(0,0,0)$.

Therefore, we can write,

$$\begin{aligned} (\mu _E(p_2,p_j),\eta _E(p_2,p_j),\nu _E(p_2,p_j))\\ = \left\{ \begin{array} {ll } (\lambda _1\min \{\mu _V(p_2), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_2), \eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_2), \nu _V(p_j)\}) &{} \text { for } j=1,3,4\\ (0,0,0) &{} \text { for } j=5,6 \\ \end{array} \right. \end{aligned}$$

Similarly,

Iteration 3:

$$\begin{aligned} (\mu _E(p_3,p_j),\eta _E(p_3,p_j),\nu _E(p_3,p_j))\\= \left\{ \begin{array} {ll } (\lambda _1\min \{\mu _V(p_3), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_3), \eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_3), \nu _V(p_j)\}) &{} \text { for } j=1,2,4\\ (0,0,0) &{} \text { for } j=5,6 \\ \end{array} \right. \end{aligned}$$

Iteration 4:

$$\begin{aligned}&(\mu _E(p_4,p_j),\eta _E(p_4,p_j),\nu _E(p_4,p_j))\\&\quad = (\lambda _1\min \{\mu _V(p_4), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_4), \\ \eta _V(p_j)\},&\quad \quad \lambda _3\min \{\nu _V(p_4), \nu _V(p_j)\}) \text { for } j=1,2,3,5,6 \end{aligned}$$

Iteration 5:

$$\begin{aligned} (\mu _E(p_5,p_j),\eta _E(p_5,p_j),\nu _E(p_5,p_j))\\= \left\{ \begin{array} {ll } (\lambda _1\min \{\mu _V(p_5), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_5),\,\eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_5), \nu _V(p_j)\}) &{} \text { for} j=1,4,6\\ (0,0,0) &{} \text { for } j=2,3 \\ \end{array} \right. \end{aligned}$$

Therefore, $(\mu _E(p_i,p_j),\eta _E(p_i,p_j),\nu _E(p_i,p_j))=(\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}, \lambda _2\min \{\eta _V(p_i)$, $\eta _V(p_j)\}, \lambda _3\min \{\nu _V(p_i), \nu _V(p_j)\})$ or, $(\mu _E(p_i,p_j),\eta _E(p_i,p_j),\nu _E(p_i,p_j))=(0,0,0)$ for all $i,j=1,2,\ldots ,6$ and $i\ne j$.

Hence, G is a balanced PFG.

Illustration 2

Fig. 6

A PFG for illustration 2

Here we consider another PFG $G=(V,E,m_V,m_E)$ where $V=\{p_1,p_2,\ldots ,p_5\}$ and $m_V$, $m_E$ are shown in Fig. 6.

For this graph, $\rho (G)=(0.7,0.5,0.8)=(\lambda _1,\lambda _2,\lambda _3)$. Therefore, $\lambda _1=0.7$, $\lambda _2=0.5$ and $\lambda _3=0.8$. Now we check whether G is balanced or not using Algorithm BBPFG.

Iteration 1: For $i=1$.

Let us consider a node $p_1$. Now compute, $(\mu _E(p_1,p_j),\eta _E(p_1,p_j),\nu _E(p_1,p_j))$ for $j=2,3,\ldots ,5$.

When $j=2$.

$$\begin{aligned} \mu _E(p_1,p_2)=0.28=0.7\min \{0.5,0.4\}=0.7\min \{\mu _V(p_1),\mu _V(p_2)\}.\\ \eta _E(p_1,p_2)=0.05=0.5\min \{0.1,0.2\}=0.5\min \{\eta _V(p_1),\eta _V(p_2)\}.\\ \nu _E(p_1,p_2)=0.16=0.8\min \{0.2,0.2\}=0.8\max \{\nu _V(p_1),\nu _V(p_2)\}. \end{aligned}$$

Therefore, $(\mu _E(p_1,p_2),\eta _E(p_1,p_2),\nu _E(p_1,p_2)) =(\lambda _1\min \{\mu _V(p_1), \mu _V(p_2)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_2)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_2)\})$.

When $j=3$.

$$\begin{aligned} \mu _E(p_1,p_3)=0.18=0.7\min \{0.5,0.3\}=0.7\min \{\mu _V(p_1),\mu _V(p_3)\}.\\ \eta _E(p_1,p_3)=0.05=0.5\min \{0.1,0.3\}=0.5\min \{\eta _V(p_1),\eta _V(p_3)\}.\\ \nu _E(p_1,p_3)=0.20 \ne 0.8\min \{0.2,0.3\}=0.8\max \{\nu _V(p_1),\nu _V(p_3)\}. \end{aligned}$$

Thus, $(\mu _E(p_1,p_3),\eta _E(p_1,p_3),\nu _E(p_1,p_3)) \ne (\lambda _1\min \{\mu _V(p_1), \mu _V(p_3)\}, \lambda _2\min \{\eta _V(p_1)$, $\eta _V(p_3)\}, \lambda _3\min \{\nu _V(p_1), \nu _V(p_3)\})$.

Hence, G is not a balanced PFG.

Theorem 5

Let $G=(V,E,m_V,m_E)$ be a complete PFG then $\rho (G)=(1,1,1)$.

Proof

Let $G=(V,E,m_V,m_E)$ be a complete PFG, where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$, then $\mu _E(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)= \min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$. Since $\rho (G)$ is the density of G then $\rho (G)=(\rho _{\mu }(G),\rho _{\eta }(G),\rho _{\nu }(G))$

Now,

$$\begin{aligned} \rho _{\mu }(G)= & {} \frac{S_{\mu }(G)}{w_{\mu }(G)}\\= & {} \frac{{\displaystyle \sum\nolimits _{p_i\ne p_j}\mu _E(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}\\= & {} \frac{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in E}\min \{\mu _V(p_i),\mu _V(p_j)\}}}\\= & {} 1 \end{aligned}$$

Similarly, $\rho _{\eta }(G)=1$ and $\rho _{\nu }(G)=1$. Therefore, $\rho (G)=(1,1,1)$. $\square$

Theorem 6

Any single valued complete picture fuzzy graph $G=(V,E,m_V,m_E)$ is strictly balanced.

Proof

Let $G=(V,E,m_V,m_E)$ be a complete PFG, where $m_V=(\mu _V,\eta _V,\nu _V)$ and $m_E=(\mu _E,\eta _E,\nu _E)$. Let $\rho (G)$ be the density of G. Since G is complete PFG then $\rho (G)=(1,1,1)$. So, $\rho _{\mu }(G)=1$, $\rho _{\eta }(G)=1$ and $\rho _{\nu }(G)=1$. Again since G is complete PFG then, $\mu _E(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)= \min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)= \max \{\nu _V(p_i), \nu _V(p_j)\}$ for all $p_i, p_j\in V$. Now the above relations can be written as $\mu _E(p_i,p_j)=\lambda _1\min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\mu _E(p_i,p_j)=\lambda _3\min \{\nu _V(p_i), \nu _V(p_j)\}$, for all $p_i,p_j\in V$, where $\lambda _1=\lambda _2=\lambda _3=1$. Then by Theorem 3 we can say that G is strictly balanced. $\square$

Observation 1

Every average PFG is strictly balanced.

Observation 2

Every strong PFG is balanced.

Theorem 7

Let $G=(V,E,m_V,m_E)$ be a strictly balanced PFG then $\bar{G}$ is also strictly balanced and $\rho (G)+\rho (\bar{G})=(1,1,1)$.

Proof

Since, $G=(V,E,m_V,m_E)$ is a strictly balanced PFG, therefore there exists three real numbers $\lambda _1, \lambda _2, \lambda _3$ in [0, 1] such that $\mu _E(p_i,p_j)=\lambda _1 \min \{\mu _V(p_i), \mu _V(p_j)\}$, $\eta _E(p_i,p_j)=\lambda _2\min \{\eta _V(p_i), \eta _V(p_j)\}$, $\nu _E(p_i,p_j)=\lambda _3\max \{\nu _V(p_i), \nu _V(p_j)\}$, for all $p_i,p_j\in V$ where $\rho (G)=(\lambda _1, \lambda _2, \lambda _3)$. Since, $\bar{G}=(\bar{V},\bar{E},\bar{m_V},\bar{m_E})$ be the complement of G so $\bar{V}=V$ and $\bar{m_V}=(\bar{\mu _V},\bar{\eta _V},\bar{\nu _V})$, $\bar{m_E}=(\bar{\mu _E},\bar{\eta _E},\bar{\nu _E})$, $\bar{\mu _V}(p_i)=\mu _V(p_i)$, $\bar{\eta _V}(p_i)=\eta _V(p_i)$, $\bar{\nu _V}(p_i)=\nu _V(p_i)$, for all $p_i \in V$.

$\bar{\mu _E}(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}-\mu _E(p_i,p_j)$, $\bar{\eta _E}(p_i,p_j)=\min \{\eta _V(p_i), \eta _p(p_j)\}-\eta _E(p_i,p_j)$ and $\bar{\nu _E}(p_i,p_j)= \max \{\nu _V(p_i), \nu _V(p_j)\}-\nu _E(p_i,p_j)$, for all $p_i,p_j\in V$.

Now we have $\bar{\mu _E}(p_i,p_j)= \min \{\mu _V(p_i), \mu _V(p_j)\}-\lambda _1 \min \{\mu _V(p_i), \mu _V(p_j)\} =(1-\lambda _1)\min \{\mu _V(p_i), \mu _V(p_j)\}$, for all $p_i,p_j\in V$.

Similarly, $\bar{\eta _E}(p_i,p_j)=(1-\lambda _1)\min \{\eta _V(p_i), \eta _V(p_j)\}$, for all $p_i,p_j\in V$ and $\bar{\eta _E}(p_i,p_j)=(1-\lambda _1)\min \{\eta _V(p_i), \eta _V(p_j)\}$, for all $p_i,p_j\in V$. Therefore, for the graph $\bar{G}$ there exists three real numbers $1-\lambda _1, 1-\lambda _2, 1-\lambda _3$ which lie on [0, 1] such that the above relation holds. Hence, by Theorem 3, $\bar{G}$ is strictly balanced with $\rho (\bar{G})=(1-\lambda _1, 1-\lambda _2, 1-\lambda _3)$. $\square$

Fig. 7

a A PFG G and b its Complement $\bar{G}$

Example 3

A PFG G and its complement $\bar{G}$ are shown in Fig. 7. Here the graph G is strictly balanced and $\rho (G)=(0.7,0.7,0.7)$ and also $\bar{G}$ is strictly balanced and $\rho (\bar{G})=(0.3,0.3,0.3)$. Therefore, $\rho (G)+\rho (\bar{G})=(1,1,1)$.

Theorem 8

Let $G=(V,E,m_V,m_E)$ be an average PFG then $\rho (\bar{G})=(0.5,0.5,0.5)$.

Proof

Since, $G=(V,E,m_V,m_E)$ is an average PFG then from Theorem 3, $\rho (G)=(0.5,0.5,0.5)$. Since every PFG is strictly balanced then by Theorem 10, $\rho (G)+\rho (\bar{G})=(1,1,1)$. That is $\rho (\bar{G})=(0.5,0.5,0.5)$ $\square$

Definition 16

An isomorphism between two PFG $G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ is a bijection mapping $f:V_1\rightarrow V_2$ which satisfies the following condition:

• (i) $\mu _{V_1}(p_i)=\mu _{V_2}(f(p_i))$, $\eta _{V_1}(p_i)=\eta _{V_2}(f(p_i))$, $\nu _{V_1}(p_i)=\nu _{V_2}(f(p_i))$ for all $p_i\in V$.

• (ii) $\mu _{E_1}(p_i,p_j)=\mu _{E_2}(f(p_i),f(p_j))$, $\eta _{E_1}(p_i,p_j)=\eta _{E_2}(f(p_i),f(p_j))$, $\nu _{E_1}(p_i,p_j)=\nu _{E_2}(f(p_i),f(p_j))$ for all $(p_i,p_j)\in E_1$.

Theorem 9

Let $G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ be two isomorphic PFG. Then if $G_1$ is balanced then $G_2$ is balanced and vice versa.

Proof

An isomorphism between two PFG $G_1$ and $G_2$ is a mapping $f:V_1\rightarrow V_2$ which satisfies the following conditions:

• (i) $\mu _{V_1}(p_i)=\mu _{V_2}(f(p_i))$, $\eta _{V_1}(p_i)=\eta _{V_2}(f(p_i))$, $\nu _{V_1}(p_i)=\nu _{V_2}(f(p_i))$ for all $p_i\in V$.

• (ii) $\mu _{E_1}(p_i,p_j)=\mu _{E_2}(f(p_i),f(p_j))$, $\eta _{E_1}(p_i,p_j)=\eta _{E_2}(f(p_i),f(p_j))$, $\nu _{E_1}(p_i,p_j)=\nu _{E_2}(f(p_i),f(p_j))$ for all $(p_i,p_j)\in E_1$. Then

  $$\begin{aligned}&{\displaystyle \sum _{p_i \in V_1}\mu _{V_1}(p_i)}={\displaystyle \sum _{q_i \in V_2}\mu _{V_2}(q_i)}\\&{\displaystyle \sum _{p_i \in V_1}\eta _{V_1}(p_i)}={\displaystyle \sum _{q_i \in V_2}\eta _{V_2}(q_i)}\\&{\displaystyle \sum _{p_i \in V_1}\nu _{V_1}(p_i)}={\displaystyle \sum _{q_i \in V_2}\nu _{V_2}(q_i)}\\&{\displaystyle \sum _{(p_i,p_j) \in E_1}\mu _{E_1}(p_i,p_j)}={\displaystyle \sum _{(q_i,q_j) \in E_2}\mu _{E_2}(q_i,q_j)}\\&{\displaystyle \sum _{(p_i,p_j) \in E_1}\eta _{E_1}(p_i,p_j)}={\displaystyle \sum _{(q_i,q_j) \in E_2}\eta _{E_2}(q_i,q_j)}\\&{\displaystyle \sum _{(p_i,p_j) \in E_1}\nu _{E_1}(p_i,p_j)}={\displaystyle \sum _{(q_i,q_j) \in E_2}\nu _{E_2}(q_i,q_j)}\\ \end{aligned}$$

Let $H_1$ be any arbitrary subgraph of a PFG $G_1$ and $H_2$ be that of $G_2$. Therefore $H_1$ and $H_2$ are also isomorphic.

Let $H_1=(V_1',E_1',m_{V_1}',m_{E_1}')$, $H_2=(V_2',E_2',m_{V_2}',m_{E_2}')$. Let $G_1$ is balanced. Therefore $\rho (H_1)\le \rho (G_1)$, that is $\rho _{\mu }(H_1)\le \rho _{\mu }(G_1)$, $\rho _{\eta }(H_1)\le \rho _{\eta }(G_1)$, $\rho _{\nu }(H_1)\le \rho _{\nu }(G_1)$.

Now

$$\begin{aligned} \rho _{\mu }(H_1)\le & {} \rho _{\mu }(G_1) \& \text {gives}\\ \frac{{\displaystyle \sum\nolimits _{p_i, p_j\in {V_1'}}\mu _{E_1'}(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in {E_1'}}\min \{\mu _{V_1'}(p_i),\mu _{V_1'}(p_j)\}}}\le & {} \frac{{\displaystyle \sum\nolimits _{p_i, p_j\in {V_1}}\mu _{E_1}(p_i,p_j)}}{{\displaystyle \sum\nolimits _{(p_i, p_j)\in {E_1}}\min \{\mu _{V_1}(p_i),\mu _{V_1}(p_j)\}}}\\ \frac{{\displaystyle \sum\nolimits _{q_i, q_jr\in {V_2'}}\mu _{E_2'}(q_i,q_j)}}{{\displaystyle \sum\nolimits _{(q_i, q_j)\in {E_2'}}\min \{\mu _{V_2'}(q_i),\mu _{V_2'}(q_j)\}}}\le & {} \frac{{\displaystyle \sum\nolimits _{q_i, q_j\in {V_1}}\mu _{E_2}(q_i,q_j)}}{{\displaystyle \sum\nolimits _{(q_i, q_j)\in {E_2}}\min \{\mu _{V_1}(q_i),\mu _{V_1}(q_j)\}}}\\ \end{aligned}$$

Therefore, $\rho _{\mu }(H_2)\le \rho _{\mu }(G_2)$. Similarly, $\rho _{\eta }(H_2)\le \rho _{\eta }(G_2)$ and $\rho _{\nu }(H_2)\le \rho _{\nu }(G_2)$. Since $H_1$ is arbitrary and $H_2$ be corresponding isomorphic subgraph of $G_2$, therefore $H_2$ is balanced. Similarly we can introduced a function $f_1:V_2 \rightarrow V_1$ since $G_1$ and $G_2$ are isomorphic. We can proceed in similar way and prove that $G_1$ is balanced when G is balanced. $\square$

Definition 17

$G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ be two PFGs where,

• (i) $V=V_1\times V_2$ and $m_{V_i}=(\mu _{V_i},\eta _{V_i},\nu _{V_i})$, $m_{E_i}=(\mu _{E_i},\eta _{E_i},\nu _{E_i})$ for $i=1,2$.

• (ii) $E=\{(p_i,q_i)(p_j,q_j): (p_i,p_j)\in E_1, (q_i,q_j)\in E_2\}$.

Then the direct product of $G_1$ and $G_2$ is a PFG and is denoted by $G_1\sqcap G_2=(V,E,m_V,m_E)$ where,

$(\mu _{V_1}\sqcap \mu _{V_2})(p_i,q_i)=\min \{\mu _{V_1}(p_i),\mu _{V_2}(q_i)\}$, $(\eta _{V_1}\sqcap \eta _{V_2})(p_i,q_i)=\min \{\eta _{V_1}(p_i),\eta _{V_2}(q_i)\}$, $(\nu _{V_1}\sqcap \nu _{V_2})(p_i,q_i)=\min \{\nu _{V_1}(p_i),\nu _{V_2}(q_i)\}$ for all $(p_i,q_i)\in V_1 \times V_2$. and $(\mu _{E_1}\sqcap \mu _{E_2})((p_i,q_i)(p_j,q_j))=\min \{\mu _{E_1}(p_i,p_j),\mu _{E_2}(q_i,q_j)\}$, $(\eta _{E_1}\sqcap \eta _{E_2})((p_i,q_i)(p_j,q_j))=\min \{\eta _{E_1}(p_i,p_j),\eta _{E_2}(q_i,q_j)\}$, $(\nu _{E_1}\sqcap \nu _{E_2})((p_i,q_i)(p_j,q_j))=\min \{\nu _{E_1}(p_i,p_j),\nu _{E_2}(q_i,q_j)\}$ for all $(p_i,p_j)\in E_1, (q_i,q_j)\in E_2$

Fig. 8

a $G_1$ and b $G_2$

Fig. 9

The direct product $G_1 \sqcap G_2$

Example 4

Here we consider two PFGs $G_1$ and $G_2$ (See Fig. 8) and their direct product $G_1\sqcap G_2$ (See Fig. 9). The membership function of edges of $G_1\sqcap G_2$ are shown in the Table 2. From this table we see that the positive, neutral, negative membership value of the edge joining the nodes $p_2q_2$ and $p_1q_1$ in the graph $G_1 \sqcap G_2$ are 0.28, 0.08, 0.3. Also from Table 2, we see that there is no edge between the nodes $p_1q_2$ and $p_1q_1$. Same for the other edges of the graph $G_1 \sqcap G_2$.

Table 2 Membership function of edges of $G_1\sqcap G_2$

Theorem 10

$G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ be two complete PFGs. Then the direct product of $G_1$ and $G_2$ are strong PFG.

Proof

Since, $G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ be two PFGs be two complete PFG. Then $\mu _{E_1}(p_i,p_j)= \min \{\mu _{V_1}(p_i), \mu _{V_1}(p_j)\}$, $\eta _{E_1}(p_i,p_j)= \min \{\eta _{V_1}(p_i), \eta _{V_1}(p_j)\}$, $\nu _{E_1}(p_i,p_j)= \max \{\nu _{V_1}(p_i), \nu _{V_1}(p_j)\}$ for all $p_i, p_j\in V$.

and $\mu _{E_2}(p_i,p_j)= \min \{\mu _{V_2}(p_i), \mu _{V_2}(p_j)\}$, $\eta _{E_2}(p_i,p_j)= \min \{\eta _{V_2}(p_i), \eta _{V_2}(p_j)\}$, $\nu _{E_2}(p_i,p_j)= \max \{\nu _{V_2}(p_i), \nu _{V_2}(p_j)\}$ for all $p_i, p_j\in V$. Now, $G_1 \sqcap G_2$ be the direct product of $G_1$ and $G_2$, whose edge set $E=\{((p_i,q_i),(p_j,q_j)): (p_i,p_j)\in E_1, (q_i,q_j)\in E_2 \}$. Then

$$\begin{aligned} (\mu _{E_1}\sqcap \mu _{E_2})((p_i,q_i)(p_j,q_j))= & {} \min \{\mu _{E_1}(p_i,p_j),\mu _{E_2}(q_i,q_j)\}\\= & {} \min \{ \min \{\mu _{V_1}(p_i), \mu _{V_1}(p_j)\}, \min \{\mu _{V_2}(q_i), \mu _{V_2}(q_j)\}\} \\= & {} \min \{ \min \{\mu _{V_1}(p_i), \mu _{V_2}(q_i)\}, \min \{\mu _{V_1}(p_j), \mu _{V_2}(q_j)\}\} \\= & {} \min \{\mu _{V_1}\sqcap \mu _{V_2}(p_i,q_i), \mu _{V_2}\sqcap \mu _{V_2}(p_j,q_j)\} \\&\text {for all} (p_i,p_j)\in E_1, (p_j,q_j)\in E_2.\\= & {} \min \{\mu _{V_1}\sqcap \mu _{V_2}(p_i,q_i), \mu _{V_2}\sqcap \mu _{V_2}(p_j,q_j)\} \\&\text {for all} ((p_i,q_i),(p_j,q_j))\in E. \end{aligned}$$

Similarly, $(\eta _{E_1}\sqcap \eta _{E_2})((p_i,q_i)(p_j,q_j))= \min \{\eta _{V_1}\sqcap \eta _{V_2}(p_i,q_i), \eta _{V_2}\sqcap \eta _{V_2}(p_j,q_j)\}$

for all $((p_i,q_i),(p_j,q_j))\in E$ and $(\nu _{E_1}\sqcap \nu _{E_2})((p_i,q_i)(p_j,q_j)) = \min \{\nu _{V_1}\sqcap \nu _{V_2}(p_i,q_i), \nu _{V_2}\sqcap \nu _{V_2}(p_j,q_j)\}$ for all $((p_i,q_i),(p_j,q_j))\in E$. This shows that $G_1 \sqcap G_2$ is strong PFG. $\square$

Theorem 11

Let $G_1=(V_1,E_1,m_{V_1},m_{E_1})$ and $G_2=(V_2,E_2,m_{V_2},m_{E_2})$ are two balanced PFG and $\rho (G_1)=\rho (G_2)$ , then $G_1\sqcap G_2$ is balanced and $\rho (G_1)=\rho (G_2)=\rho (G_1\sqcap G_2)$.

Proof

Let $G_1$ and $G_2$ be two balanced PFG and $\rho (G_1)=\rho (G_2)=(\lambda _1, \lambda _2, \lambda _3)$, where $\lambda _1, \lambda _2, \lambda _3$ are three real numbers belongs to [0,1]. Since, $G_1$ and $G_2$ be two balanced PFG then there exist $(p_i,p_j)\in E_1$ and $(q_i,q_j)\in E_2$ such that,

$$\begin{aligned}&\mu _{E_1}(p_i,p_j)=\lambda _1\min \{\mu _{V_1}(p_i), \mu _{V_1}(p_j)\}\\&\eta _{E_1}(p_i,p_j)=\lambda _2\min \{\eta _{V_1}(p_i), \eta _{V_1}(p_j)\}\\&\nu _{E_1}(p_i,p_j)=\lambda _3\max \{\nu _{V_1}(p_i), \nu _{V_1}(p_j)\}, \text{ for } \text{ all } (p_i,p_j)\in E_1 \text {and}\\&\mu _{E_2}(q_i,q_j)=\lambda _1\min \{\mu _{V_2}(q_i), \mu _{V_2}(q_j)\}\\&\eta _{E_2}(q_i,q_j)=\lambda _2\min \{\eta _{V_2}(q_i), \eta _{V_2}(q_j)\}\\&\nu _{E_2}(q_i,q_j)=\lambda _3\max \{\nu _{V_2}(q_i), \nu _{V_2}(q_j)\}, \text{ for } \text{ all } (q_i,q_j)\in E_2 \\ \end{aligned}$$

Now

$$\begin{aligned} (\mu _{E_1}\sqcap \mu _{E_2})((p_i,q_i)(p_j,q_j))= & {} \min \{\mu _{E_1}(p_i,p_j),\mu _{E_2}(q_i,q_j)\}\\= & {} \min \{ \lambda _1 \min \{\mu _{V_1}(p_i, \mu _{V_1}(p_j)\},\\&\lambda _1\min \{\mu _{V_2}(q_i), \mu _{V_2}(q_j)\}\} \\= & {} \lambda _1\min \{ \min \{\mu _{V_1}(p_i), \mu _{V_2}(q_i)\},\\&\min \{\mu _{V_1}(p_j), \mu _{V_2}(q_j)\}\} \\= & {} \lambda _1\min \{(\mu _{V_1}\sqcap \mu _{V_2})(p_i,q_i), (\mu _{V_1}\sqcap \mu _{V_2})(p_j,q_j)\}\\&\text {for all} (p_i,p_j)\in E_1, (p_j,q_j)\in E_2. \end{aligned}$$

Similarly,

$$\begin{aligned}&(\eta _{E_1}\sqcap \eta _{E_2})((p_i,q_i)(p_j,q_j))=\lambda _1\min \{(\eta _{V_1}\sqcap \eta _{V_2})(p_i,q_i), (\eta _{V_1}\sqcap \eta _{V_2})(p_j,q_j)\} \\&(\nu _{E_1}\sqcap \nu _{E_2})((p_i,q_i)(p_j,q_j))=\lambda _1\min \{(\nu _{V_1}\sqcap \nu _{V_2})(p_i,q_i), (\nu _{V_1}\sqcap \nu _{V_2})(p_j,q_j)\}\\&\text {for all} (p_i,p_j)\in E_1, (p_j,q_j)\in E_2. \end{aligned}$$

Hence, $G_1\sqcap G_2$ is balanced and $\rho (G_1\sqcap G_2)=(\lambda _1, \lambda _2, \lambda _3)$. $\square$

Corollary 3

Let $G_1$, $G_2$ be two PFG such that $\rho (G_1)=\rho (G_2)$ then the density of $G_1\sqcap G_2$ may or may not be equal to density of $G_i$, for $i=1,2$.

Example 5

Let us consider two PFGs $G_1$ and $G_2$ in Fig. 10 where $G_1$ is not balanced and their direct product $G_1\sqcap G_2$ (See Fig. 11). The membership value of all edges of $G_1\sqcap G_2$ are shown in the table below.

Fig. 10

a $G_1$ and b $G_2$

Fig. 11

$G_1\sqcap G_2$

For this example, $\rho (G_1)=\rho (G_2)=(0.6,0.75,0.75)$ and $\rho (G_1\sqcap G_2)=(0.54,0.7,0.75)$. So, for this example, $\rho (G_1)=\rho (G_2)$ but it is not equal to $\rho (G_1\sqcap G_2)$, this holds since $G_2$ is not balanced.

5 Application of balanced picture fuzzy graph

In this section, a novel application of balanced PFG to business alliance in industry is proposed. The principal of the work is to find the business partner those may be allied under certain condition described below.

Here we consider eight companies, Coal India Limited (CIL), Tata Limited(TL), Hindustan limited(HL), Oil and Natural Gas corporation Limited(ONGC), Reliance Industries(RI), Life Insurance corporation (LIC), Infosys Limited(IL), Aditya Birla Group(ABG). Any company may engage in allied business with one or more companies. so we draw a PFG among the eight companies where, each companies are represented as node of the graph and alliance business between two companies are connected by an edge. For example, if Coal India Limited (CIL) alliance with Tata Limited(TL), then there is an edge between CIL and TL. If there is no alliance business between Coal India Limited (CIL) alliance with Tata Limited(TL) then there is no edge between CIL and TL. Now we consider the membership function of nodes and edges as following.

For Nodes:

• 1. The strength and operating style of each companies referred as a positive membership degree of the node.

• 2. The market placement of each companies referred as a neutral membership degree of the node.

• 3. Poor management system of each companies gives the negative membership degree of the node.

For Edges:

• 1. Alliance business between two companies are successfully increasing referred as a positive membership degree of each edges.

• 2. Alliance business between two companies are no growth referred as a neutral membership degree of each edges.

• 3. Alliance business between two companies are to be failure referred as a negative membership degree of each edges.

Also, the membership value of each node and edges are shown in Fig. 12 and the Table 3 respectively. From Fig. 12, we see that the membership value of the nodes CIL and TL are (0.5, 0.3, 0.2) and (0.7, 0.2, 0.1) respectively. Also, from Table 3, we have the membership value of the edge between the nodes CIL and TL is (0.4, 0.14, 0.8). Similarly, we can find the membership values of other nodes and edges from Fig. 12 and the Table 3.

Fig. 12

A PFG corresponding to eight companies

Table 3 Membership values of edges

Here, the business relationship rate (density) of the graph is (0.8, 0.7, 0.4). From the above graph, $S=\{CIL, TL, IS, ONGC, RI\}$ is largest subgraph in which the relationship rate are all equal for every pair of nodes. Hence, the subgraph $S=\{CIL, TL, IS, ONGC, RI\}$ is balanced. Therefore, these five companies namely Coal India Limited, Tata Limited, Infosys Limited, Oil and Natural Gas Corporation Limited, Reliance Industries can be alliance properly. So our example helps to alliance a lot of companies with their strategies described above.

6 Discussion and conclusion

The density of a PFG is defined and showed that each component lie between 0 and 1. If we assign the neutral value to 0, then one can determine the density for IFG. If we assign neutral and negative membership values to 0, then we obtain the density for fuzzy graph. So, the density of PFG generalizes for FG and IFG. If the membership values of all nodes and edges are one then, we get the density for Crisp graph.

In this article, some new terminologies are defined. Some useful properties of PFG are studied. The definition and properties of PFG like, average PFG, balanced PFG, size, order, density of a PFG are given. The direct product of two PFGs is defined and presented some properties. Also, an algorithm is give to test whether a PFG is balanced or not. Beside this an application of balanced PFG to business alliance is presented. This paper will help to the new researchers to extend PFG. In future, we will study picture fuzzy threshold graph, picture fuzzy k-competition graph, picture fuzzy planer graph, etc.