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Radio number of 2 super subdivision for path related graphs

  • We studied radio labelings of graphs in response to the Channel Assignment Problem (CAP). In a graph G, the radio labeling is a mapping ϖ:V(G){0,1,2,...,}, such as |ϖ(μ)ϖ(μ)|diam(G)+1d(μ,μ). The label of μ for under ϖ is defined by the integer ϖ(μ), and the span under is defined by span(ϖ)=max{|ϖ(μ)ϖ(μ)|:μ,μV(G)}. rn(G)=minϖspan(ϖ) is defined as the radio number of G when the minimum over all radio labeling ϖ of G is taken. G is said to be optimal if its radio labeling is span(ϖ)=rn(G). A graph H is said to be an m super subdivision if G is replaced by the complete bipartite graph Km,m with m=2 in such a way that the end vertices of the edge are merged with any two vertices of the same partite set X or Y of Km,m after removal of the edge of G. Up to this point, many lower and upper bounds of rn(G) have been found for several kinds of graph families. This work presents a comprehensive analysis of the radio number rn(G) for a graph G, with particular emphasis on the m super subdivision of a path Pn with n(n3) vertices, along with a complete bipartite graph Km,m consisting of m v/ertices, where m=2.

    Citation: Baskar Mari, Ravi Sankar Jeyaraj. Radio number of 2 super subdivision for path related graphs[J]. AIMS Mathematics, 2024, 9(4): 8214-8229. doi: 10.3934/math.2024399

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  • We studied radio labelings of graphs in response to the Channel Assignment Problem (CAP). In a graph G, the radio labeling is a mapping ϖ:V(G){0,1,2,...,}, such as |ϖ(μ)ϖ(μ)|diam(G)+1d(μ,μ). The label of μ for under ϖ is defined by the integer ϖ(μ), and the span under is defined by span(ϖ)=max{|ϖ(μ)ϖ(μ)|:μ,μV(G)}. rn(G)=minϖspan(ϖ) is defined as the radio number of G when the minimum over all radio labeling ϖ of G is taken. G is said to be optimal if its radio labeling is span(ϖ)=rn(G). A graph H is said to be an m super subdivision if G is replaced by the complete bipartite graph Km,m with m=2 in such a way that the end vertices of the edge are merged with any two vertices of the same partite set X or Y of Km,m after removal of the edge of G. Up to this point, many lower and upper bounds of rn(G) have been found for several kinds of graph families. This work presents a comprehensive analysis of the radio number rn(G) for a graph G, with particular emphasis on the m super subdivision of a path Pn with n(n3) vertices, along with a complete bipartite graph Km,m consisting of m v/ertices, where m=2.



    In recent years, several fascinating research areas have emerged within the field of study associated with graph theory. Concerning these, graph labeling constitutes a highly intriguing and dynamic field that encompasses numerous applications. There are a number of different scientific fields that utilize graph labeling, such as communication networks, database administration, coding theory, and many others. The concept was initially proposed by [1]. Recently, numerous types of labeling have been extensively studied, such as graceful labeling by [2], (modular) irregular labeling by [3], edge labeling by [4] and so on. All graphs, denoted as G=(V,E), in the presented work are simple undirected, connected, and finite. Here, V represents the vertex set, and E represents the edge set of G. The graph labeling assignment involves assigning integers regarding vertices (v), edges (e), or both, according to specific conditions. Radio labeling is a technique that is used to address the challenge of channel assignment. With graph G, there exist several different radio labeling challenges. One of the foremost significant contributions to this field was conducted by [5], whose primary findings consisted of determining the precise values of rn(G) for paths and cycles. This is one of the most wellknown studies in this area.

    The channel assignment problem, initially proposed by [6] and later revisited by [7], is mainly motivated by the need to comply with regulations on the allocation of channels for FM broadcasting stations. This problem aimed to incorporate the concept of radio labeling of graphs also referred to as multilevel distance labeling. The radio labeling problem is an important topic in discrete mathematics due to its various applications, like in frequency assignment in mobile communication systems, signal processing, circuit design, etc. [8] found the upper boundaries to obtain the radio numbers associated with the cycle. [5] determined an accurate value for the radio number of paths. [9] stated that three sufficient and necessary conditions must be met to attain the lower bound for the radio number of the block graph. Radio numbers of line graphs of trees and block graphs have also been discussed by the same authors. According to [10], the problem of radio channel assignment for the network is modeled by Cayley graphs. A discussion of the radio number of the Cartesian product of two trees can be found in [11]. The study of [12] have introduced supersubdivision of graphs.

    To better illustrate the channel assignment problem scopes of radio labeling and the major contributions of theoretical as well as practical significance of this article, we provide Table 1 for comparison with other research works on the radio number of different graphs.

    Table 1.  Summary of Existing Results.
    Graphs Radio Numbers
    Paths For n4, rn(Pn)={2k2+2if n=2k+1;2k(k1)+1if n=2k. [5]
    Square of any paths For k=n2, rn(P2n)={k2+2,if n0 (mod 4);and n9;k2+1,Otherwise. [13]
    Cube of a path For k=n2, rn(P3n)={n2+126,if n0 (mod 6);n22n+196,if n1 (mod 6);n2+2n+106,if n2 (mod 6);n2+156,if n3 (mod 6);n22n+166,if n4 (mod 6);n2+2n+136,if n5 (mod 6); [14]
    Cross product Pn(P2) For n3, rn(Pn(P2))=n2n+1 [15]
    Corona product of PnWn For n6, rn(PnWn)=2n+5 [16]
    Cartesian product of path and complete For n3 and m3, rn(PnPm)={mn22n+22,if n is even,mn22n+m+22,if n is odd. [17]
    Cartesian product of paths Pn and Petersen Graph P For n3, rn(PnP)={5n2n+1,if n is even,5n22n+m+6,if n is odd. [18]
    Strong Product K3Pn For n2 and k=n2, rn(K3Pn)={2k(3k+2)+1,if n=2k+1;2k(3k1)+1,if n=2k. [19]
    Petersen and star For n3, rn(P5,2K1,n)=10n+27. [20]
    Cycle For n3, rn(Cn)={n22π(n)+1,if n0,2 (mod 4);n12π(n),if n1,3 (mod 4); [5]
    Square of even cycles For n is even, rn(C2n)={2k2+5k12,if n=4k and k is odd; 2k2+3k2,if n=4k and k is even; k2+k5+1,if n=4k+2 and k is odd; k2+45+1,if n=4k+2 and k is even. [21]
    Trees For n3, rn(G)=ni=1li(l1+l2li)+l1l22n2+1 iff ¯l1(l1+l21)2. [22]
    Gear Graph For n4, rn(Gn)=4n+2 [23]
    Middle Graph For n2, rn(M(Pn))={4k21,if n=2k,4k(k+1),if n=2k+1. [24]

     | Show Table
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    This section starts with an overview of the notations that will be applied throughout the paper, listed in Table 2.

    Table 2.  Table of Notations.
    Parameters Description
    SS(G,m) 2-Supersub division of graphs
    vi Original vertex in G
    vj Vertices in SS(G)
    δ The number of vertices in SS(G,m)
    C(SS(G)) Center of the SS(G) graph
    d or diam(G) Diameter
    ε(v) Eccentricity of a vertex v
    (u) Level function
    (G) Total level function.
    rn(G) Radio number.

     | Show Table
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    This study consists of defining essential terms and notations in this section that will be used throughout the entire research paper. One may refer to [25] for all terminologies and notations and [26] for graph labeling. A graph G's shortest path is measured by the distance d(u,v) between two vertices. diam(G)=max{d(u,v):u,vG} is the diameter of a graph G (or simply d for use in equations). Although eccentricity can be defined in many ways, it is usually referred to as the distance between a certain vertex (v) and any other vertex in a graph (G) at which the vertex is at its maximum value. The eccentricity is denoted by ε(v) through this concept. A subgraph is defined here as the vertex whose eccentricity is lowest and which is the center of the graph G, called C(G).

    Definition 3.1. ([7]) In a graph G, the radio labeling is a mapping ϖ:V(G){0,1,2,3,...,}, such as |ϖ(μ)ϖ(μ)|diam(G)+1d(μ,μ). A label of μ under ϖ is defined by the integer ϖ(μ), and the span under is defined by span(ϖ)=max{|ϖ(μ)ϖ(μ)|:μ,μV(G)}. rn(G)=minϖspan(ϖ) is defined as the radio number of G when the minimum over all radio labeling ϖ of G is taken. G is said to be optimal if it's radio labeling is span(ϖ)=rn(G).

    Definition 3.2. ([27]) The 2super subdivision graph SS(G,m) is constructed from the original graph G by changing each of the edges with a complete bipartite graph Km,m where m=2. This substitution involves combining each endpoint of the edge with any two vertices from either partite set X or Y of Km,m when removing the original edge from G.

    Illustration: The following Figures 1 and 2 are the examples for definition (3.2).

    Figure 1.  Graph SS(P9;2).
    Figure 2.  Graph SS(P10;2).

    Lemma 3.1. C(SS(Pn;2))={2K1if n is evenK1otherwise.

    Proof. Let SS(Pn;2) a 2 super subdivision of a path Pn and S be a collection of all vertices in SS(Pn;2) which have the lowest possible eccentricity. The eccentricity for vertex vi (vj) if 1in (1j2n2) can be expressed as follows:

    Case 1: If n is an odd number.

    ε(vi)=(n1)+|2i(n+1)|, if 1in.

    ε(vj)={2nj2,if j=1,3,...,n2,2nj1,if j=2,4,...,n1,j,if j=n,n+2,...,2n3,j1,if j=n+1,n+3,...,2(n1).

    One can see that S contains only vn+12. Thus, by the above function, ε(vn+12) is smaller than the eccentricity of the remaining vertices in SS(Pn;2). The graph indued by S is isomorphic to the complete graph K1 and hence C(SS(Pn;2))=K1.

    Case 2: If n is an even number.

    ε(vi)={2(ni),if 1in2,2(i1),if n+22in.

    ε(vj)={|nj1|+(n1),if j=1,3,...,2n3,|nj|+(n1),if j=2,4,...,2n2.

    One can see that S={vn1,vn}. Since the graph induced by S has the same isomorphic relations as the complete graph 2K1, we can conclude that case (2) is C(SS(Pn;2))=2K1.

    Let us now consider the new denotation of the vertex of SS(Pn;2) with respect to its center. SS(Pn;2)'s vertices are renamed accordingly.

    If n is an odd number, then let n=2k+1

    vi={v1C=iif i=n+12,v1L(n2i+1)if 1in12,v1R(2i(n+1))if n+32in1.

    vj={v1L(nj1)if j=1,3,...,n2,v2L(nj)if j=2,4,...,n1,v1R(jn+1)if j=n,n+2,...,2n3,v2R(jn)if j=n+1,n+3,...,2(n1).

    If n is an even number, then let n=2k

    vi={v1L(n2i+1)if 1in2,v1R(2i(n+1))if n2+1in.

    vj={v1C=jif j=n1,v2C=jif j=n,v1L(nj1)if j=1,3,...,n3,v2L(nj)if j=2,4,...,n2,v1R(jn+1)if j=n+1,n+3,...,2n3,v2R(jn)if j=n+2,n+4,...,2(n1).

    Illustration: In Figures 3 and 4, the 2 super subdivision SS(P9;2) and SS(P10;2).

    Figure 3.  The graph SS(P9;2).
    Figure 4.  The graph SS(P10;2).

    One defines the level function towards V(SS(Pn;2)) as the set of whole numbers W by (μ)=min{d(μ,μ);μV(C(SS(Pn;2)))} for each μV(SS(Pn;2))V(C(SS(Pn;2))). In the graph SS(Pn;2), the maximum level is k=n1. In the graph G, (G) represents the total level, and is defined as: (G)=μV(G)(μ).

    In this section, the 2 super subdivision of the path SS(Pn;2) is determined, and also the radio number of the 2 super subdivision of the path SS(Pn;2) graph is found.

    Observation:

    1. |V(SS(Pn;2))|=δ=3n2.

    2. d(μ,μ)(μ)+(μ).

    3. If μi,μi+1V(SS(Pn;2)), 0iδ2 are on opposite side and d(μi,μi+1)=d(μi+1,μi+2) or d(μi,μi+1)=d(μi+1,μi+2) ±1 or d(μi,μi+1)=d(μi+1,μi+2) ±2.

    Lemma 4.1. Let Pn be a path of length n. The total level function of SS(Pn;2) is (SS(Pn;2))={(3n24n+1)2ifnis an odd.n(3n4)2ifnis an even.

    Proof. Let Pn be the path of length n. If n is an odd number, by Lemma (3.1) C(SS(Pn;2)) is a single vertex.

    μV(SS(Pn;2))(μ)=(ni=1,2,...,d(vC,vi)+2n2j=1,2,...,d(vC,vj))=2(2n2i=1,3,...,i+n1i=2,4,...,i)=2(2n12k=1(2k1)+n12k=12k)=2(4n12k=1k2n12k=11+2n12k=1k)=2(4(n12(n12+1)2)2(n12)+2((n12)(n12+1)2))=(3n24n+1)2.

    Consider the nlength path, that is Pn. If n is an even number, by Lemma (3.1) C(SS(Pn;2)) has 2K1 vertices.

    μV(SS(Pn;2))(μ)=(ni=1,2,...,d(vC,vi)+2n2j=1,2,...,d(vC,vj))μV(SS(Pn;2))(μ)=2(n1i=1,3,...,i+2n2i=2,4,...,i)=2(2n2k=1(2k1)+n22k=12k)=2(4(n2(n2+1)2)2(n2)+2(n22(n22+1)2))=n(3n4)2.

    Lemma 4.2. Let ϖ be an assignment of distinct non-negative integers to V(SS(Pn;2)), and μ1,μ2,μ3,...,μδ be the ordering of V(SS(Pn;2)) such that ϖ(μi)<ϖ(μi+1) defined by ϖ(μ1)=0 and ϖ(μi+1)=ϖ(μi)+d+1d(μi,μi+1). Then, ϖ is a radio labeling if for any 1iδ1, the following holds:

    1. d(μi,μi+1)n+1 if n is odd.

    2. d(μi,μi+1)n+1 and d(μi,μi+1)d(μi+1,μi+2) if n is even.

    Proof. Let ϖ(μ1)=0 and ϖ(μi+1)ϖ(μi)+d+1d(μi,μi+1) for any 1iδ1 and k=n1. For each 1iδ1, let ϖi=ϖ(μi+1)ϖ(μi). We must prove that ϖ is a radio labeling if both of the above relations hold. I.e. we have to show for any ij, |ϖ(μi)ϖ(μj)|d+1d(μi,μj).

    Case 1: When n is odd, we have d=2k, k=n1, we let (1), and we take i>j. then,

    ϖ(μi)ϖ(μj)= ϖj+ϖj+1+ϖj+2+...+ϖi2+ϖi1= (ij)(d+1)d(μj,μj+1)d(μj+1,μj+2)...d(μi1,μi) (ij)(d+1)(ij)k (ij)(d+1k)ϖ(μi)ϖ(μj) d+1d(μi,μj). Since ij0.

    Case 2: When n is even, we have d=2k, n1=k, we let (2), and we take i>j. then,

    ϖ(μi)ϖ(μj)=ϖj+ϖj+1+ϖj+2+...+ϖi2+ϖi1=(ij)(d+1)d(μj,μj+1)d(μj+1,μj+2)...d(μi1,μi).

    If ij is even, then

    ϖ(μi)ϖ(μj) (ij)(d+1)(ij(n2)2)(n)(ij(n2)2)(n1)(ij)(2k+1)(ij(k+12)2)(k+1)(ij(k+12)2)(k)ij2(2k+1)k12d+1k12ϖ(μi)ϖ(μj)d+1d(μi,μj).

    If ij is odd, then

    ϖ(μi)ϖ(μj) (ij)(d+1)(ij(n3)2)(n)(ij(n3)2)(n1)(ij)(2k+1)(ij(k+13)2)(k+1)(ij(k+13)2)(k)ij+12(2k+1)3k12d+13k12ϖ(μi)ϖ(μj)d+1d(μi,μj).

    Since ij0, in both cases ϖ is a radio labeling and hence the result.

    Theorem 4.1 ([15]). The rn(SS(P2;2))=4.

    Proof. The radio number of SS(P2;2) is shown in Figure 5.

    Figure 5.  rn(SS(P2;2))=4.

    Theorem 4.2. If SS(Pn;2) is the 2 super subdivision of Pn(n3), then rn(SS(Pn;2))3n25n+3.

    Proof. To demonstrate the conclusion, we take two cases into consideration.

    Case (1): n is an even number.

    For SS(Pn;2), let,

    ϖ:V(SS(Pn;2)){0,1,2,...,(3n25n+3)} be defined by

    ϖ(μi+1)=ϖ(μi)+d+1((μi)+(μi+1)) for i=1,2,...,δ2 and

    ϖ(μδ)=ϖ(μδ1)+d((μδ1)+(μδ)).

    We first set μ1=vn1, μδ=vn. By observation, δ=3n2 for 2iδ1, and we set μh:=vi, where

    h={3(n2i+1),if 1in2,6(ni)+2,if n2+1in.

    and μh:=vj, where

    h={3(nj)2,if j=1,3,...,n3,3(nj)1,if j=2,4,...,n2,3(2nj1),if j=n+1,n+3,...,2n3,3(2nj)2,if j=n+2,n+4,...,2n2.

    Case (2): n is an odd number.

    For SS(Pn;2), let

    ϖ:V(SS(Pn;2){0,1,2,...,(3n25n+3)} be defined by

    ϖ(μi+1)=ϖ(μi)+d+1((μi)+(μi+1)) for i=1,2,...,δ2 and

    ϖ(μδ)=ϖ(μδ1)+d((μδ1)+(μδ)).

    We first set μ1=vn+12, μδ=vn2. By observation, δ=3n2 for 2iδ1, and we set μh:=vi, where

    h={2(n2i)+3,if 1in12,2(2(ni)+1),if n+32in.

    and μh:=vj, where

    h={2n+j,if j=1,3,...,n4,2n+12j,if j=2,4,...,n1,3nj,if j=n,n+2,...,2n3,4n2j,if j=n+1,n+3,...,2(n1).

    Therefore, the vertices of SS(Pn;2) can be assigned a label equivalent to the lower bound if the requirement of Lemma (4.2) is satisfied. This is true for case (1) and case (2). Hence, ϖ is a radio labeling. Thus, we have

    rn(SS(Pn;2))3n25n+3.

    Theorem 4.3. If SS(Pn;2) is the 2 super subdivision of Pn(n3), then rn(SS(Pn;2))3n25n+3.

    Proof. Let SS(Pn;2) be the 2 super subdivision of the given path with order x and size y. If (Vi,Vj) is the bipartition of the vertex set of K2,2, then X=ni=1|Vi|=n and Y=nj=1|Vj|=2(n1). Thus, SS(Pn;2) has 3n2 vertices and 4(n1) edges. Also, the names of vertices of Vi and Vj are v1,v2,v3,...,vX and v1,v2,v3,...,vY such that X+Y=x.

    For SS(Pn;2), let,

    ϖ:V(SS(Pn;2)){0,1,2,...,(3n25n+3)} be defined by

    ϖ(μi+1)=ϖ(μi)+d+1((μi)+(μi+1)) for i=1,2,...,δ2 and

    ϖ(μδ)=ϖ(μδ1)+d((μδ1)+(μδ)).

    Let μ1=vn+12 and μδ=vn2 if n is odd. Let μ1=vn1 and μδ=vn if n is even.

    Let ϖ be a radio labeling for SS(Pn;2) with a linear order μ1,μ2,μ3,...,μδ of vertices of SS(Pn;2) such that ϖ(μ1)=0<ϖ(μ2)<ϖ(μ3)<ϖ(μ4)<...<ϖ(μδ). Then, ϖ(μi+1)ϖ(μi)d+1d(μi,μi+1), 1iδ1. Incorporating these δ1 on the inequalities, we obtain

    δ1i=1ϖ(μi+1)ϖ(μi)δ1i=1(d+1)δ1i=1d(μi,μi+1)ϖ(μδ)ϖ(μ1)(δ1)(d+1)δ1i=1d(μi,μi+1)rn(SS(Pn;2))=ϖ(μδ)(δ1)(d+1)δ1i=1d(μi,μi+1) (4.1)

    Case 1: n is an even number.

    For SS(Pn;2), i=1,2,...,δ1, by observation, we acquire

    d(μi,μi+1)(μi)+(μi+1)δ1i=1d(μi,μi+1)δ1i=1[(μi)+(μi+1)]=[(μ1)+(μ2)]+[(μ2)+(μ3)]+...+[(μδ1)+(μδ)]=μV(G)(μ)(μδ)+μV(G)(μ)(μ1)=2μV(G)(μ)(μ1)(μδ)2(SS(Pn;2))(μ1)(μδ)

    Since μ1 and μδ are both center vertices, (μ1)=(μδ)=0 and

    δ1i=1d(μi,μi+1)2(SS(Pn;2)) (4.2)

    By substituting (4.2) in (4.1), we get

    rn(SS(Pn;2))(δ1)(d+1)2(SS(Pn;2)).

    For SS(Pn;2), δ=3n2, d=2n2 and (SS(Pn;2))=(n(3n4)2). Substituting these into the above equation, we get

    =(δ1)(d+1)2(n(3n4)2)=((3n2)1)(2n2+1)n(3n4)3n25n+3.

    Case 2: n is an odd number.

    For SS(Pn;2), i=1,2,...,δ1, by observation, we acquire

    d(μi,μi+1)(μi)+(μi+1)δ1i=1d(μi,μi+1)δ1i=1[(μi)+(μi+1)]=[(μ1)+(μ2)]+[(μ2)+(μ3)]+...+[(μδ1)+(μδ)]=μV(G)(μ)(μδ)+μV(G)(μ)(μ1)=2μV(G)(μ)(μ1)(μδ)2(SS(Pn;2))(μ1)(μδ)

    Since (μ1)=0, (μδ)=1 and

    δ1i=1d(μi,μi+1)2(SS(Pn;2))1 (4.3)

    By substituting (4.3) in (4.1), we get

    rn(SS(Pn;2))(δ1)(d+1)(2(SS(Pn;2))1).

    For SS(Pn;2), δ=3n2, d=2n2 and (SS(Pn;2))=(3n24n+12).

    Substituting these into the above equation, we get

    =(δ1)(d+1)(2(3n24n+12)1)=((3n2)1)(2n2+1)(3n24n+1)+13n25n+3.

    Thus, from above two cases we have the desired result.

    Theorem 4.4. If SS(Pn;2) is the 2 super subdivision of Pn(n3), then rn(SS(Pn;2))=3n25n+3.

    Proof. The proof is based on Theorems (4.2) and (4.3), respectively.

    Example 4.1. Figure 6 demonstrates both the arrangement (ordering) of the vertices as well as the optimal radio labeling of SS(P9;2).

    V1CV1R8V2L1V2R7V1L2V1R6V2L3V2R5V1L4V1R4V2L5V2R3V1L6V1R2V2L7V2R1V1L8V1R1V1L7V1R3V1L5V1R5V1L3V1R7V1L1.
    Figure 6.  rn(SS(P9;2))=201.

    Example 4.2. Figure 7 demonstrates both the arrangement (ordering) of the vertices as well as the optimal radio labeling of SS(P10;2).

    V1CV1R9V1L1V2R8V2L2V1R8V1L2V1R7V1L3V2R6V2L4V1R6V1L4V1R5V1L5V2R4V2L6V1R4V1L6V1R3V1L7V2R2V2L8V1R2V1L8V1R1V1L9V2C.
    Figure 7.  rn(SS(P10;2))=253.

    It is challenging to design a wireless network that avoids interference. Our work on this problem is based on 2 Super divisions for paths. The goal of this work is to provide proof of radio labels for 2 super subdivisions of paths and to obtain the exact radio numbers for the 2 super subdivisions of path graphs. In addition, our research focuses on the application of radio labels to 2 super subdivision of path graphs.

    All authors declare no conflicts of interest in this paper.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.



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