# Chromatic Polynomial of Intuitionistic Fuzzy Graphs Usingα,β-Levels

Authors: V. N. SrinivasaRao Repalle; Lateram Zawuga Hordofa; Mamo Abebe Ashebo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

The article describes a new thought on the chromatic polynomial of an intuitionistic fuzzy graph which is illustrated based onα,β-level graphs. Besides, the alpha-beta fundamental set of an intuitionistic fuzzy graph is also defined with a vivid description. In addition to that, some characterizations of the chromatic polynomial of an intuitionistic fuzzy graph are specified as well verified. Furthermore, some untouched properties of the α,β-level graph are also projected and proved.


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## Body

## 1. Introduction

Graph coloring has a lot of applications in areas such as wireless radio channel assignment [1], timetable scheduling [2], and job scheduling [3]. Graph coloring is described as vertex coloring, edge coloring, and map coloring. The concept of the chromatic polynomial was introduced as the number of distinct ways of performing map coloring [4]. The chromatic polynomials of many families of graphs have been computed so far [5, 6]. In addition, the computations of the chromatic polynomials of certain graphs using the Mobius inversion theorem were reported by Srinivasa Rao et al. [7].The fuzzy set theory was introduced by Zadeh [8]. A fuzzy set communicates an element of a given set with a membership value in [0, 1]. The introduction of fuzzy graph theory was given based on the fuzzy set introduced by Kauffman [9]. The traditional fuzzy set cannot be used to completely describe all evidence in the problems where someone wants to know how much nonmembership values. Such a problem got a solution by Atanassov who introduced an intuitionistic fuzzy set (IFS) [10]. IFS is an extension of Zadeh’s set theory and is described by a membership function, a nonmembership function, and a hesitation function [10]. The concept of an intuitionistic fuzzy graph (IFG) was introduced by Atanassov [11]. Atanassov and Shannon developed certain properties of IFGs [12]. And also, Akram and Davvaz discussed more properties of IFGs [13]. Akram et al. presented strong IFGs and intuitionistic fuzzy line graphs [14]. Akram et al. presented an algorithm for computing the sum distance matrix, eccentricity, radius, and diameter of IFG [15]. Electoral systems [16], human cells clustering [17], and water supply systems [18] are some application areas of IFGs. Also, IFG digraph is applied in medical diagnosis, gas pipelines, and decision-making systems [19].The chromatic polynomials of fuzzy graphs usingα-cuts and their algebraic properties were developed by Abebe and Srinivasa Rao Repalle [20]. The coloring of IFG using α,β-cuts (levels) was introduced by Mohideen and Rifayathali [21]. Some properties of IFG by α,β-levels were presented by Akram [22]. To fill the gap in these articles, the authors aimed to find the chromatic polynomial of IFG using α,β-levels, defineα,β-fundamental set, and discuss more untouched properties ofα,β-level graphs. This paper presents the chromatic polynomial of an IFG using α,β-levels by illustrating it based on α,β-level graphs. In addition, it defines the alpha-beta fundamental set of an IFG. Furthermore, it states and proves some properties of the α,β-level graph and the chromatic polynomial of IFG.This manuscript is organized as follows: Section2 is the preliminaries that are essential for understanding the article. Section 3develops some properties ofα,β-levels of IFGs and proves wherever verification is required. Section 4 defines the notion of the chromatic polynomial of an IFG using α,β-levels and also develops some related concepts on a chromatic polynomial of an IFG. Section 5 is the summary of the article.


## 2. Preliminaries

This section defines related concepts such as chromatic polynomial, IFG, and level graph of IFG. In addition, it provides some useful propositions.Definition 1 (see [20]).
The number of distinct ways of attaining a proper vertex coloring of a simple graphG with at most k colors is called chromatic polynomial of G and is denoted by PG,k.Proposition 1 (see [23]).
The chromatic polynomial of a simple graphG may be found by the formula: PG,k=PG−e,k−PG/e,k, where PG−e,k is the chromatic polynomial of a subgraph of G whose one of its edges e has been removed, and PG/e,k is the chromatic polynomial of a subgraph of G whose one of its edge e has been contracted.Proposition 2 (see [23]).
LetG be a graph containing nonadjacent vertices u and v, and let H be the graph obtained from G the contracting edge u,v. Then, PG,k=PG+u,v,k+PH,k.Proposition 3 (see [23]).
IfG1,G2,…,Gn are pairwise disjoint components of a simple graph G whose union is G, then its chromatic polynomial is computed as(1)PG,k=PG1,k×PG2,k×⋯×PGn,k,Definition 2 (see [24]).
An intuitionistic fuzzy graph is of the formG=V,E, where(i)
V=v1,v2,…,vn such that μ1:V⟶0,1 and γ1:V⟶0,1 denote the degree of membership and the degree of nonmembership of the element vi in V, respectively, and 0≤μ1vi+γ1vi≤1, for every vi∈V, i=1,2,…,n(i)
E⊆V×V where μ2:V×V⟶0,1 and γ2:V×V⟶0,1 are such that μ2vi,vj≤minμ1vi,μ1vj(2)γ2vi,vj≤maxγ1vi,γ1vj.
Satisfy the constraint,0≤μ2vi,vj+γ2vi,vj≤1 for every vi,vj∈E. Where μ2vi,vj and γ2vi,vj are the degree of membership and the degree of nonmembership of the element vi,vj in E respectively.Definition 3 (see [21]).
The chromatic number of IFG:G=V,E is the intuitionistic fuzzy number χG=x,mx,nx|x∈X, where X=1,2,…,V, mx=supα∈0,1|x∈Aα,β, nx=infβ∈0,1|x∈Aα,β, and Aα,β=χ1,0,…,χα,β,α,β∈0,1.Definition 4.
[22]. Let G=V,E be an IFG. Then, for any α,β∈0,1, α,β-level graph of G is defined as the crisp graph Gα,β=Vα,β,Eα,β, where Vα,β=u∈V|μ1u≥α, γ1u≤β and Eα,β=u,v∈V×V|μ2u,V≥α, γ2u,v≤β.


## 3. Theα,β-Fundamental Set and the Properties of α,β-Level Graph of IFGs

This section defines theα,β-fundamental set and illustrates the α,β-level graph of IFGs. In addition, it states and proves some character of the α,β-level graph of IFGs.Definition 5.
LetG=V,E be an IFG. If L=α1,β1,α2,β2,…,αm,βm is such that αi≤αi+1 and βi≥βi+1 for each i=1,2,…,m−1, then F=0,1∪L is said to be a fundamental set of G.Example 1.
Consider an IFG with four intuitionistic fuzzy vertices and five intuitionistic fuzzy edges given in Figure1.
The fundamental set of an IFG in Figure 1 isF= {(0, 1), (0.2, 0.6), (0.3, 0.6), (0.4, 0.5), (0.5, 0.5), (0.8, 0.2)}. To show that the reason why (0.2, 0.5) ∉F, consider the levels (0.2, 0.6), (0.2, 0.5), and (0.3, 0.6). Now, let us determine if a level (0.2, 0.5) satisfies the condition in the definition of fundamental set. It is forward that 0.2 ≤ 0.2, 0.6  0.5, 0.2 ≤ 0.3 but 0.5 ≱ 0.6. This shows that (0.2, 0.5) ∉F.
The level graphs along with their chromatic number are illustrated in Figure2.(3)G0.8,0.2=N1,χG0.8,0.2=1.Figure 1
Intuitionistic fuzzy graph.Figure 2
Illustration on theGα,β of the graph in Figure 1 along with their chromatic numbers.Table 1
The comparison ofVα,β, Eα,β, χGα,β, and Pα,βIG,k of IFG in Figure 1.
α=0,β=1464kk−1k−2k−3α=0.2,β=0.6453kk−1k−22α=0.3,β=0.6432kk−13α=0.4,β=0.5322kk−12α=0.5,β=0.5201k2α=0.8,β=0.2101kFundamental setssEα,βχGα,βPα,βIG,kRemark 1.
(i)
Table1 contains the vertex set, edge set, and chromatic number of Gα,β for each α,β in F(ii)
If there is a vertexv with μ1v=1,γ1v=0, then 1,0-level graph exists(iii)
G0,1=V0,1,E0,1 is K4 as V0,1=v∈V|μ1≥0andγ1≤1 contains all the vertices listed in V and E0,1=u,v∈V×V|μ2≥0andγ2≤1 implies that all the vertices are joined(iv)
The chromatic number of an IFG in Figure1 is given by(4)χG=χGα,β=4,0,13,0.2,0.62,0.4,0.51,0.8,0.2T.


### 3.1.The Properties ofα,β-Level Graph of IFGs

Theorem 1.
LetG be an IFG and Gα1,β1 and Gα2,β2 be level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then,(5)Vα1,β1⊇Vα2,β2,Eα1,β1⊇Eα2,β2.Proof.
LetG be an IFG. Suppose Gα1,β1 and Gα2,β2 are level graphs of G with α1≤α2 and β1≥β2. Now, Gα1,β1 and Gα2,β2 have vertex sets; Vα!,β1=v∈V|μ1v≥α1andγ1v≤β1 and Vα2,β2=v∈V|μ1v≥α2andγ1v≤β2, respectively. To figure out the relation Vα1,β1 has with Vα2,β2, we draw them on the same plane as follows.
Based on Figure3, Vα1,β1 is an area of a polygon EFGB and Vα2,β2 is an area of a polygon ABCD. This shows that Vα2,β2 is contained in Vα1,β1. Thus, Vα1,β1⊇Vα2,β2.
Similarly,Gα1,β1 and Gα2,β2 have the edge sets; Eα1,β1=u,v∈V×V|μ2u,v≥α1,γ2u,v≤β1 and Eα2,β2=u,v∈V×V|μ2u,v≥α2,γ2u,v≤β2, respectively. For comparison, these sets can be interpreted on the same plane as follows.
Based on Figure4, Eα1,β1 is an area of a polygon HLNO and Eα2,β2 is an area of a polygon HIJK. This indicates that Eα2,β2 is contained in Eα1,β1. Thus, Eα1,β1⊇Eα2,β2.
Alternate proof: LetGα1,β1 and Gα2,β2 be level graphs of G with α1≤α2 and β2≤β1. Then,(i)
The vertex sets to these levels areVα1,β1=v∈V|μ1v≥α1andγ1v≤β1 and Vα2,β2=v∈V|μ1v≥α2andγ1v≤β2. Now, let v∈Vα2,β2. Then, μ1v≥α2andγ1v≤β2, and since α2≥α1 and β2≤β1, it implies μ1v≥α1andγ1v≤β1. This indicates v∈Vα1,β1. Thus, Vα1,β1⊇Vα2,β2(ii)
The edge sets ofGα1,β1 and Gα2,β2 are Eα1,β1=u,v∈V×V|μ2u,v≥α1,γ2u,v≤β1 and Eα2,β2=u,v∈V×V|μ2u,v≥α2,γ2u,v≤β2, respectively.
Now, letu,v∈Eα2,β2. Then, μ2u,v≥α2 and γ2u,v≤β2 and α2≥α1 and β2≤β1, and it implies μ2u,v≥α1andγ2u,v≤β1. This shows u,v∈Eα1,β1. Hence, Eα1,β1⊇Eα2,β2.Figure 3
The comparison ofVα1,β1 with Vα2,β2 of an IFG.Figure 4
The comparison ofEα1,β1 and Eα2,β2 of an IFG.Corollary 1.
LetGα1,β1 and Gα2,β2 be level graphs of an IFG, G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then,(i)
Vα2,β2≤Vα1,β1(ii)
Eα2,β2≤Eα1,β1Proof.
Assume an IFG,G. Let Gα1,β1 and Gα2,β2 be two-level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, by applying Theorem 1: Vα1,β1⊇Vα2,β2 and Eα1,β1⊇Eα2,β2. Thus, Vα2,β2≤Vα1,β1 and Eα2,β2≤Eα1,β1 hold.Theorem 2.
LetGα1,β1 and Gα2,β2 be level graphs of an IFG; G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, Gα1,β1 is a super graph of Gα2,β2.Proof.
LetG be an IFG and 0≤α1≤α2≤1 and 1≥β1≥β2≥0 be given. Then, by Theorem 1, Vα1,β1⊇Vα2,β2 and Eα1,β1⊇Eα2,β2. From the fact that if the vertex set and edge set of a certain graph J are a subset of a vertex set and edge set of another graph H, respectively, then H is a super graph of J. Thus, Gα1,β1 is a super graph of Gα2,β2.


## 3.1.The Properties ofα,β-Level Graph of IFGs

Theorem 1.
LetG be an IFG and Gα1,β1 and Gα2,β2 be level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then,(5)Vα1,β1⊇Vα2,β2,Eα1,β1⊇Eα2,β2.Proof.
LetG be an IFG. Suppose Gα1,β1 and Gα2,β2 are level graphs of G with α1≤α2 and β1≥β2. Now, Gα1,β1 and Gα2,β2 have vertex sets; Vα!,β1=v∈V|μ1v≥α1andγ1v≤β1 and Vα2,β2=v∈V|μ1v≥α2andγ1v≤β2, respectively. To figure out the relation Vα1,β1 has with Vα2,β2, we draw them on the same plane as follows.
Based on Figure3, Vα1,β1 is an area of a polygon EFGB and Vα2,β2 is an area of a polygon ABCD. This shows that Vα2,β2 is contained in Vα1,β1. Thus, Vα1,β1⊇Vα2,β2.
Similarly,Gα1,β1 and Gα2,β2 have the edge sets; Eα1,β1=u,v∈V×V|μ2u,v≥α1,γ2u,v≤β1 and Eα2,β2=u,v∈V×V|μ2u,v≥α2,γ2u,v≤β2, respectively. For comparison, these sets can be interpreted on the same plane as follows.
Based on Figure4, Eα1,β1 is an area of a polygon HLNO and Eα2,β2 is an area of a polygon HIJK. This indicates that Eα2,β2 is contained in Eα1,β1. Thus, Eα1,β1⊇Eα2,β2.
Alternate proof: LetGα1,β1 and Gα2,β2 be level graphs of G with α1≤α2 and β2≤β1. Then,(i)
The vertex sets to these levels areVα1,β1=v∈V|μ1v≥α1andγ1v≤β1 and Vα2,β2=v∈V|μ1v≥α2andγ1v≤β2. Now, let v∈Vα2,β2. Then, μ1v≥α2andγ1v≤β2, and since α2≥α1 and β2≤β1, it implies μ1v≥α1andγ1v≤β1. This indicates v∈Vα1,β1. Thus, Vα1,β1⊇Vα2,β2(ii)
The edge sets ofGα1,β1 and Gα2,β2 are Eα1,β1=u,v∈V×V|μ2u,v≥α1,γ2u,v≤β1 and Eα2,β2=u,v∈V×V|μ2u,v≥α2,γ2u,v≤β2, respectively.
Now, letu,v∈Eα2,β2. Then, μ2u,v≥α2 and γ2u,v≤β2 and α2≥α1 and β2≤β1, and it implies μ2u,v≥α1andγ2u,v≤β1. This shows u,v∈Eα1,β1. Hence, Eα1,β1⊇Eα2,β2.Figure 3
The comparison ofVα1,β1 with Vα2,β2 of an IFG.Figure 4
The comparison ofEα1,β1 and Eα2,β2 of an IFG.Corollary 1.
LetGα1,β1 and Gα2,β2 be level graphs of an IFG, G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then,(i)
Vα2,β2≤Vα1,β1(ii)
Eα2,β2≤Eα1,β1Proof.
Assume an IFG,G. Let Gα1,β1 and Gα2,β2 be two-level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, by applying Theorem 1: Vα1,β1⊇Vα2,β2 and Eα1,β1⊇Eα2,β2. Thus, Vα2,β2≤Vα1,β1 and Eα2,β2≤Eα1,β1 hold.Theorem 2.
LetGα1,β1 and Gα2,β2 be level graphs of an IFG; G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, Gα1,β1 is a super graph of Gα2,β2.Proof.
LetG be an IFG and 0≤α1≤α2≤1 and 1≥β1≥β2≥0 be given. Then, by Theorem 1, Vα1,β1⊇Vα2,β2 and Eα1,β1⊇Eα2,β2. From the fact that if the vertex set and edge set of a certain graph J are a subset of a vertex set and edge set of another graph H, respectively, then H is a super graph of J. Thus, Gα1,β1 is a super graph of Gα2,β2.


## 4. The Chromatic Polynomial of IFG Usingα,β-Levels

Here, we define a new idea of the chromatic polynomial of IFG usingα,β-levels, and also, we compute the chromatic polynomial of an IFG using α,β-levels for illustration. Furthermore, we state and prove some properties of the chromatic polynomial of IFG using α,β-levels.Definition 6.
LetG be IFG and F be its fundamental set. Then, the chromatic polynomial of G using α,β-level with at most k given colors is denoted by Pα,βIG,k and is defined as the chromatic polynomial of Gα,β, where α,β. That is, Pα,βIG,k=PGα,β,k for each α,β∈F.Example 2.
Consider the IFG,G in Figure 1. All the level graphs of G have been put in Figure 2, and by Definition 6, the chromatic polynomial of G is the chromatic polynomial of these level graphs. Thus, the chromatic polynomial of G is the chromatic polynomial of the level graphs combined in a single equation and is computed as follows:(i)
Whenα=0 and β=1, we have P0,1IG,k=PG0,1,k, and as G0,1=K4, PG0,1,k is the chromatic polynomial of K4. Hence, PG0,1,k=PK4,k=kk−1k−2k−3.(ii)
Whenα=0.2 and β=0.6, we compute P0.2,0.6IG,k=PG0.2,0.6,k by the addition-contraction method as in Figure 5.Thus,PG0.2,0.6,k=PK4,k+PK3,k=kk−1k−22(iii)
Whenα=0.3 and β=0.6, we have P0.3,0.6IG,k=PG0.3,0.6,k, and it is the chromatic polynomial of P4. Therefore, PG0.3,0.6,k=PP4,k=kk−13.(iv)
Whenα=0.4 and β=0.5, we obtain P0.4,0.5IG,k=PG0.4,0.5,k and PG0.4,0.5,k is the chromatic polynomial of P3. Therefore, PG0.4,0.5,k=PP3,k=kk−12.(v)
Whenα=0.5 and β=0.5, we obtain P0.5,0.5IG,k=PG0.5,0.5,k and PG0.5,0.5,k is the chromatic polynomial of N2. Therefore, PG0.5,0.5,k=PN2,k=k2.(vi)
Whenα=0.8 and β=0.2, we obtain P0.8,0.2IG,k=PG0.8,0.2,k and PG0.8,0.2,k is the chromatic polynomial of N1. Therefore, PG0.8,0.2,k=PN1,k=k.
In general, the chromatic polynomial of IFG in Figure1 is summarized in the following equation:(6)Pα,βIG,k=kk−1k−2k−3,α=0,β=1,kk−1k−22,α=0.2,β=0.6,kk−13,α=0.3,β=0.6,kk−12,α=0.4,β=0.5,k2,α=0.5,β=0.5,k,α=0.8,β=0.2.Figure 5
The computation ofP0.2,0.6IG,k.Remark 2.
(i)
Asα increases the degree of the chromatic polynomial, the chromatic number, the number of edges, and the number of vertices decrease.(ii)
Asβ decreases the degree of the chromatic polynomial, the chromatic number, the number of edges, and the number of vertices also decrease.


### 4.1. Properties of the Chromatic Polynomial of IFGs Usingα,β-Levels

Theorem 3.
LetG be IFG on n intuitionistic fuzzy vertices and let Gα,β be α,β-level graph of G. If α=0 and β=1, then(7)Pα,βIG,k=PKn,k.Proof.
LetG be an IFG. Then, when α=0 and β=1, G0,1 possesses V0,1=x∈V|μ1w≥0andγ1w≤1 and E0,1=w,v∈V×V|μ2x,y≥0andγ2w,v≤1.
For every edge inG, μ2w,v≥0 and γ2w,v≤1. So, every edge in G satisfies E0,1. In addition to edges in G0,1, E0,1 contains edge(s) correspond to edge(s) labeled 0,0 of G. Therefore, every pair of vertices is joined. V0,1=n, as every vertex in G, could satisfy V0,1. This implies that G0,1 and Kn represent the same graph. Hence, P0,1IG,k=PG0,1,k=PKn,k.Theorem 4.
LetG be IFG. If Gα1,β1 and Gα2,β2 are intuitionistic fuzzy levels of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0, then(8)degPα1,β1IG,k≥degPα2,β2IG,k.Proof.
Assume thatGis an IFG. Let Gα1,β1 and Gα2,β2 be the level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, degPα1,β1IG,k=degPGα1,β1,k and degPα2,β2IG,k=degPGα2,β2,k. It is obvious to realize that the degree of the chromatic polynomial of a simple graph equals the number of its vertex set. That is, deg(PGα1,β1,k=Vα1,β1 and deg(PGα2,β2,k=Vα2,β2, and also, since α1≤α2 and β1≥β2, Vα1,β1≥Vα2,β2 (see Corollary 1). Hence, degPα1,β1IG,k≥degPα2,β2IG,k.Theorem 5.
LetH be an IFG and Hc be its underlying graph. If αj≤αi and βi≤βj for each i=1,2,…,m and αi,βi∈I, then Pαj,βjIH,k=PHc,k for some j= 1, 2, ..., m.Proof.
Suppose thatHc is an underlying graph of IFG H. Let 0,1∪α1,β1,α2,β2,…,αm,βm be a fundamental set of H and let αj≤αi and βi≤βj for each i=1,2,…,m and for some j=1,2,…,m. By Theorem 11, Hαj,βj is a super graph of Hαi,βi for each i=1,2,…,m. This shows that Hαj,βj contains all vertices and all edges of Hc as Hc is the underlying crisp graph of H. Hence, Hαj,βj and Hc represent the same graph, and both have the same chromatic polynomial. That is, Pαj,βjIH,k=PHc,k.Theorem 6.
LetH and J be components of an IFG, G, and let α,β be any level such that α>0 and β<1. Then,(9)Pα,βIG,k=Pα,βIH,k×Pα,βIJ,k.Proof.
LetG be IFG and α,β be any of its levels such that α>0 and β<1. Also, let H and J be an intuitionistic fuzzy component of G. Now, since H and J are components of G, Hα,β and Jα,β are also components of Gα,β. So, PGα,β,k=PHα,β,k×PJα,β,k.
Thus,Pα,βIG,k=Pα,βIH,k×Pα,βIJ,k.Corollary 2.
LetF1,F2,…,Fn be intuitionistic fuzzy components of an IFG, F, and α,β-be any level of F. Then,(10)Pα,βIF,k=Pα,βIF1,k×Pα,βIF2,k×…×Pα,βIFn,k.
Proof of this corollary is omitted as it can be forwarded by extending Theorem6.Theorem 7.
IfH and J are intuitionistic fuzzy component graphs of an IFG, G with n intuitionistic fuzzy vertices and m intuitionistic fuzzy vertices, respectively, and Kn+m is a complete graph on n+m vertices, then P0,1IG,k=PKn+m,k.Proof.
Consider an IFG,G=V,E. Now, let H and J be intuitionistic fuzzy component graphs of G containing n and m intuitionistic fuzzy vertices, respectively. Then, 0,1-level graph of G, G0,1 possesses(i)
V0,1=x∈V|μ1w≥0,γ1w≤1 . V0,1=n+m, as every vertex in both H and J, satisfy V0,1(ii)
E0,1=w,v∈V×V|μ2x,y≥0,γ2w,v≤1
Every edge joining any two verticesw and v in both component graphs H and J satisfy μ2w,v≥0 and γ2w,v≤1. And also, in addition to the edges in both H and J, E0,1 contains an edge(s) corresponding to edge(s) labeled 0,0 of G which are actually neither in the crisp graph of H nor in the crisp graph of J.
From (i),0,1-level graph of G has n+m vertices, and from (ii), every pair of vertices of G is joined. In other words, the 0,1-level graph of G is Kn+m. Hence, P0,1IG,k=PKn+m,k.Corollary 3.
LetG be IFG. If H and J are intuitionistic fuzzy component graphs of G with n intuitionistic fuzzy vertices and m intuitionistic fuzzy vertices, respectively, then(11)DegP0,1IG,k=n+m.Proof.
Based on Theorem7, 0,1-level graph forms a complete crisp graph; Kn+m and P0,1IG,k=PKn+m,k. Thus, the corollary holds.


## 4.1. Properties of the Chromatic Polynomial of IFGs Usingα,β-Levels

Theorem 3.
LetG be IFG on n intuitionistic fuzzy vertices and let Gα,β be α,β-level graph of G. If α=0 and β=1, then(7)Pα,βIG,k=PKn,k.Proof.
LetG be an IFG. Then, when α=0 and β=1, G0,1 possesses V0,1=x∈V|μ1w≥0andγ1w≤1 and E0,1=w,v∈V×V|μ2x,y≥0andγ2w,v≤1.
For every edge inG, μ2w,v≥0 and γ2w,v≤1. So, every edge in G satisfies E0,1. In addition to edges in G0,1, E0,1 contains edge(s) correspond to edge(s) labeled 0,0 of G. Therefore, every pair of vertices is joined. V0,1=n, as every vertex in G, could satisfy V0,1. This implies that G0,1 and Kn represent the same graph. Hence, P0,1IG,k=PG0,1,k=PKn,k.Theorem 4.
LetG be IFG. If Gα1,β1 and Gα2,β2 are intuitionistic fuzzy levels of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0, then(8)degPα1,β1IG,k≥degPα2,β2IG,k.Proof.
Assume thatGis an IFG. Let Gα1,β1 and Gα2,β2 be the level graphs of G with 0≤α1≤α2≤1 and 1≥β1≥β2≥0. Then, degPα1,β1IG,k=degPGα1,β1,k and degPα2,β2IG,k=degPGα2,β2,k. It is obvious to realize that the degree of the chromatic polynomial of a simple graph equals the number of its vertex set. That is, deg(PGα1,β1,k=Vα1,β1 and deg(PGα2,β2,k=Vα2,β2, and also, since α1≤α2 and β1≥β2, Vα1,β1≥Vα2,β2 (see Corollary 1). Hence, degPα1,β1IG,k≥degPα2,β2IG,k.Theorem 5.
LetH be an IFG and Hc be its underlying graph. If αj≤αi and βi≤βj for each i=1,2,…,m and αi,βi∈I, then Pαj,βjIH,k=PHc,k for some j= 1, 2, ..., m.Proof.
Suppose thatHc is an underlying graph of IFG H. Let 0,1∪α1,β1,α2,β2,…,αm,βm be a fundamental set of H and let αj≤αi and βi≤βj for each i=1,2,…,m and for some j=1,2,…,m. By Theorem 11, Hαj,βj is a super graph of Hαi,βi for each i=1,2,…,m. This shows that Hαj,βj contains all vertices and all edges of Hc as Hc is the underlying crisp graph of H. Hence, Hαj,βj and Hc represent the same graph, and both have the same chromatic polynomial. That is, Pαj,βjIH,k=PHc,k.Theorem 6.
LetH and J be components of an IFG, G, and let α,β be any level such that α>0 and β<1. Then,(9)Pα,βIG,k=Pα,βIH,k×Pα,βIJ,k.Proof.
LetG be IFG and α,β be any of its levels such that α>0 and β<1. Also, let H and J be an intuitionistic fuzzy component of G. Now, since H and J are components of G, Hα,β and Jα,β are also components of Gα,β. So, PGα,β,k=PHα,β,k×PJα,β,k.
Thus,Pα,βIG,k=Pα,βIH,k×Pα,βIJ,k.Corollary 2.
LetF1,F2,…,Fn be intuitionistic fuzzy components of an IFG, F, and α,β-be any level of F. Then,(10)Pα,βIF,k=Pα,βIF1,k×Pα,βIF2,k×…×Pα,βIFn,k.
Proof of this corollary is omitted as it can be forwarded by extending Theorem6.Theorem 7.
IfH and J are intuitionistic fuzzy component graphs of an IFG, G with n intuitionistic fuzzy vertices and m intuitionistic fuzzy vertices, respectively, and Kn+m is a complete graph on n+m vertices, then P0,1IG,k=PKn+m,k.Proof.
Consider an IFG,G=V,E. Now, let H and J be intuitionistic fuzzy component graphs of G containing n and m intuitionistic fuzzy vertices, respectively. Then, 0,1-level graph of G, G0,1 possesses(i)
V0,1=x∈V|μ1w≥0,γ1w≤1 . V0,1=n+m, as every vertex in both H and J, satisfy V0,1(ii)
E0,1=w,v∈V×V|μ2x,y≥0,γ2w,v≤1
Every edge joining any two verticesw and v in both component graphs H and J satisfy μ2w,v≥0 and γ2w,v≤1. And also, in addition to the edges in both H and J, E0,1 contains an edge(s) corresponding to edge(s) labeled 0,0 of G which are actually neither in the crisp graph of H nor in the crisp graph of J.
From (i),0,1-level graph of G has n+m vertices, and from (ii), every pair of vertices of G is joined. In other words, the 0,1-level graph of G is Kn+m. Hence, P0,1IG,k=PKn+m,k.Corollary 3.
LetG be IFG. If H and J are intuitionistic fuzzy component graphs of G with n intuitionistic fuzzy vertices and m intuitionistic fuzzy vertices, respectively, then(11)DegP0,1IG,k=n+m.Proof.
Based on Theorem7, 0,1-level graph forms a complete crisp graph; Kn+m and P0,1IG,k=PKn+m,k. Thus, the corollary holds.


## 5. Conclusions

To conclude, the coloring of the IFG usingα,β-levels is applied in traffic light control, register allocation, clustering analysis, and decision-making system. The chromatic polynomial of IFG using α,β-levels provides alternative solutions in such application areas. In this work, the notion of the α,β-fundamental set of an IFG has been clearly stated with relevant examples. Apart from that, some characteristics α,β-level graphs have been detailed and proved. Also, the chromatic polynomial of IFG, by using α,β-levels, has been presented. Moreover, the chromatic polynomials of IFG of the different α,β-levels have been derived and compared. Furthermore, certain properties of the chromatic polynomial of an IFG have also been discussed and illustrated in a detailed note.



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*Source: 9320700-2022-06-28.xml*