# Cordial and Total Cordial Labeling of Corona Product of Paths and Second Order of Lemniscate Graphs

Authors: Ashraf ELrokh; Mohammed M. Ali Al-Shamiri; Shokry Nada; Atef Abd El-hay
Journal: Journal of Mathematics
 (2022)

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## Abstract

A simple graph is called cordial if it admits 0-1 labeling that satisfies certain conditions. The second order of lemniscate graph is a graph of two second order of circles that have one vertex in common. In this paper, we introduce some new results on cordial labeling, total cordial, and present necessary and sufficient conditions of cordial and total cordial for corona product of paths and second order of lemniscate graphs.


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## Body

## 1. Introduction

Labelling methods are used for a wide range of applications in different subjects including coding theory, computer science, and communication networks. Graph labeling is an assignment of positive integers on vertices or edges or both of them which fulfilled certain conditions. Hundreds of research studies have been working with different types of labeling graphs [1–11], and a reference for this purpose is the survey written by Gallian [7]. All graphs considered, in this theme, are finite, simple, and undirected. The original concept of cordial graphs is due to Cahit [2]. He proved the following: each tree is cordial; a complete graph Kn is cordial if and only if n≤3 and a complete bipartite graph Kn,m is cordial for all positive integers n and m [3]. Let G=V,E be a graph, and let f:V⟶0,1 be a labeling of its vertices, and let the induced edge labeling f∗E⟶0,1 be given by f∗uv=fu+fvmod2, where e=uv∈E and u,v∈V. Let v0 and v1 be the numbers of vertices that are labeled by 0 and 1, respectively, and let e0 and e1 be the corresponding numbers of edges. Such a labeling is called cordial if both v0−v1≤1 and e0−e1≤1 hold. A graph is called cordial if it admits a cordial labeling. As an extension of the cordial labeling, we define a total cordial labeling of a graph G with vertex set and edge set as an cordial labeling such that number of vertices and edges labeled with 0 and the number of vertices and edges labeled with 1 differ by at most 1, i.e., v0+e0−e1+v1≤1. A graph with a total cordial labeling is called a total cordial graph. If the vertices of the graph are assigned values subject to certain conditions, it is known as graph labeling. Following three are the common features of any graph labeling problem: (1) a set of numbers from which vertex labels are assigned; (2) a rule that assigns a value to each edge; and (3) a condition that these values must satisfy.A path withn vertices and n−1 edges is denoted by Pn, and a cycle with n vertices and n edges is denoted by Cn [12]. The second power of a lemniscate graph is defined as the union of two second power of cycles where both have a common vertex; it is denoted by Ln,m2≡Cn2♯Cm2 [13]. Obviously, Ln,m2 has n+m−1 vertices and 2n+2m−4 edges. The corona product G1⊙G2 of two graphs Gi (with ni vertices and mi edges), i=1,2, is the graph obtained by taking one copy of G1 and n1 copies of G2 and then joining the ith vertex of G1 with an edge to every vertex in the ith copy of G2. It is easy to show that G1⊙G2 has n11+n2 vertices and m1+n1m2+n1n2 edges [7, 14–18]. In this paper, we study the cordial and total cordial of the corona product Pk⊙Ln,m2 of paths and second power of lemniscate graphs and show that this is cordial and total cordial for all positive integers k,n,m. The rest of the paper is organized as follows. In Section 1, brief summary of definitions that are useful for the present investigations is presented. Terminologies and notations are introduced in Section 2. The main result is presented in Section 3. Finally, the conclusion of this paper is introduced.


## 2. Terminology and Notation

Given a path or a cycle with4r vertices, let L4r denote the labeling 0011… 0011 (repeated r-times) and let L4r' denote the labeling 1100… 1100 (repeated r times). The labeling 1001 1001… 1001 (repeated r times) and 0110… 0110 (repeated r times) is denoted by S4r and S4r'. Let M2r denote the labeling 0101…01, zero-one repeated r−times if r is even and 0101...010 if r is odd. Sometimes, we modify labeling by adding symbols at one end or the other (or both). If G and H are two graphs, where G has n vertices, the labeling of the corona G⊙H is often denoted by A:B1,B2,B3,...,Bn, where A is the labeling of the n vertices of G, and Bi, 1≤i≤n, is the labeling of the vertices of the copy of H that is connected to the ith vertex of G. For a given labeling of the corona G⊙H, we denote vi and ei (i=0,1) to represent the numbers of vertices and edges, respectively, labeled by i. Let us denote xi and ai to be the numbers of vertices and edges labeled by i for the graph G. Also, we let yi and bi be those for H, which are connected to the vertices labeled 0 of G. Likewise, let yi′ and bi′ be those for H, which are connected to the vertices labeled 1 of G. It is easily to verify that v0=x0+x0y0+x1y0′, v1=x1+x0y1+x1y1′, e0=a0+x0b0+x1b0′+x0y1+x1y0′, and e1=a1+x0b1+x1b1′+x0y0+x1y1′. Thus, v0−v1=x0−x1+x0y0−y1+x1y0′−y1′ and e0−e1=a0−a1+x0b0−b1+x1b0′−b1′+x0y0−y1−x1y0′−y1′. In particular, if we have only one labeling for all copies of H, i.e., yi=yi′ and bi=bi′, then v0=x0+ny0, v1=x1+ny1, e0=a0+nb0+x0y1+x1y0, and e1=a1+nb1+x0y0+x1y1. Thus, v0−v1=x0−x1+ny0−y1 and e0−e1=a0−a1+nb0−b1+x1−x0y0−y1, where n is the order of G. Figure 1 illustrates the condition cordial and total cordial labeling of P3⊙L3,7.Figure 1
Cordial and total cordial labeling ofP3⊙L3,7.


## 3. Results and Discussion

In this section, we show that the corona product of paths and second power of lemniscate graphs,Pk⊙Ln,m2, is cordial and also total cordial for all k≥1, n,m≥3.Throughout our proofs, the way of labelingLn,m2 starts always from a vertex that next the common vertex and go further opposite to this common vertex. Before considering the general form of the final result, let us first prove it in the following specific case. Our main theorem is as follows.Theorem 1.
The corona product of paths and second power of lemniscate graphs,Pk⊙Ln,m2, is cordial and also total cordial for all k≥1, n,m≥3.
In order to prove this theorem, we will introduce a number of lemmas as follows.Lemma 1.
Pk⊙L3,m2 is cordial and total cordial for all k≥1 and m≥3.Proof.
Case 1. When m≥3 and k=2r, r≥1, one can choose the labeling M2r;00100,11011,…,r−times for P2r⊙L3,32. Therefore, x0=x1=r,a0=0,a1=2r−1,y0=4,y1=1,b0=2,b1=4,y0′=1,y1′=4,b0′=2, and b1′=4. Hence, v0−v1=0, e0−e1=1 and v0+e0−e1+v1=1. Thus, P2r⊙L3,32, r≥1, is cordial and total cordial.Case 2. When m≥3 and k=2r+1, r≥0, one can choose the labeling M2r+1;00100,11011,00100,11011,…,r−times,11100 for P2r+1⊙L3,32. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=4,y1=1,b0=2,b1=4,y0′=1,y1′=4,b0′=2,b1′=4,y0∗=2,y1∗=3,b0∗=4, and b1∗=2, where yi∗ and bi∗ are the numbers of vertices and edges labeled i in L3,32 that are connected to the last zero in P4r+3. Consequently, it is easy to show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P2r+1⊙L3,32, r≥0, is cordial and total cordial.Case 3. When m≡0mod4 and k≡0mod4, that means, k=4r,r≥1 and m=4t, t>1, then the labeling L4r;0313M4t−4,0313M4t−4,01L4M4t−4',01L4M4t−4',...,r−times for P4r⊙L3,4t2 can be applied. Therefore, x0=x1=2r,a0=2r,a1=2r−1,y0=y1=2t+1,b0=4t+1,b1=4t,y0′=y1′=2t+1,b0′=4t, and b1′=4t+1. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the case P4r⊙L3,42, the labeling L4r;0313,0313,01L4,01L4,...,r−times is sufficient and thus P4r⊙L3,4t2 is cordial and also total cordial.Case 4. When m≡0mod4 and k≡1mod4 that meansk=4r+1,r≥0 and m=4t, t>1, then the labeling L4r0;0313M4t−4,0313M4t−4,01L4M4t−4',01L4M4t−4',…,r−times,03L3M4t−4 for P4r+1⊙L3,4t2 is considered. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=y1=2t+1,b0=4t+1,b1=4t,y0′=y1′=2t+1,b0′=4t, and b1′=4t+1. Hence, v0−v1=1, e0−e1=1, and v0+e0−e1+v1=0. For the case P4r+1⊙L3,42, the labeling L4r0;0313,0313,01L4,01L4,...,r−times,1303 is sufficient and thus P4r+1⊙L3,4t2 is cordial and total cordial.Case 5. When m≡0mod4 and k≡2mod4 that means k=4r+2,r≥0, and m=4t, t>1, then the labeling L4r10;0313M4t−4,0313M4t−4,01L4M4t−4',01L4M4t−4',...,r−times,01L4M4t−4',0313M4t−4 for P4r+2⊙L3,4t2 is applied. Therefore, x0=x1=2r+1,a0=2r+1,a1=2r,y0=y1=2t+1,b0=4t+1,b1=4t,y0′=y1′=2t+1,b0′=4t, and b1′=4t+1. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the case P4r+2⊙L3,42, the labeling L4r10;0313,0313,01L4,01L4,...,r−times,01L4,0313 is sufficient and thus P4r+2⊙L3,4t2 is cordial and total cordial.Case 6. When m≡0mod4 and k≡3mod4 that means k=4r+3,r≥0 and m=4t, t>1, then one can select the labeling L4r001;0313M4t−4,0313M4t−4,01L4M4t−4',01L4M4t−4',…,r−times,0313M4t−4,0313M4t−4,01L4M4t−4' for P4r+3⊙L3,4t2.Therefore,x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=y1=2t+1,b0=4t+1,b1=4t,y0′=y1′=2t+1,b0′=4t, and b1′=4t+1. Hence, one can easily show that v0−v1=1, e0−e1=1 and v0+e0−e1+v1=0. For the case P4r+3⊙L3,42, the labeling L4r001;0313,0313,01L4,01L4...,r−times is sufficient and thus P4r+3⊙L3,4t2 is cordial and total cordial.Case 7. When m≡1mod4 and k≡0mod4 that means k=4r,r≥1 and m=4t+1, t>1, then one can choose the labeling L4r;101L4′0M4t−6'0,101L4′0M4t−6'0,010L41M4t−61,010L41M4t−61,...,r−times for P4r⊙L3,4t+12. Therefore, x0=x1=2r,a0=2r,a1=2r−1,y0=2t+1,y1=2t+2,b0=4t+2,b1=4t+1,y0′=2t+2,y1′=2t+1,b0′=4t+2, and b1′=4t+1. Hence, one can easily show that v0−v1=0, e0−e1=1 and v0+e0−e1+v1=1. For the special case P4r⊙L3,52, the labeling L4r;01L4′0,01L4′0,10L41,10L41,...,r−times is sufficient and thus P4r⊙L3,4t+12 is cordial and total cordial.Case 8. When m≡1mod4 and k≡1mod4 that means k=4r+1,r≥0 and m=4t+1, t>1, then one can select the labeling L4r0;101L4′0M4t−6'0,101L4′0M4t−6'0,010L41M4t−61,010L41M4t−61,...,r−times,101L4′1M4t−6'1 for P4r+1⊙L3,4t+12. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=2t+1,y1=2t+2,b0=4t+2,b1=4t+1,y0′=2t+2,y1′=2t+1,b0′=4t+2, and b1′=4t+1. So, v0−v1=0, e0−e1=0 and v0+e0−e1+v1=0. For the special case P4r+1⊙L3,52, the labeling L4r'0;01L4′0,01L4′0,10L41,10L41,...,r−times,10L41 is sufficient and thus P4r+1⊙L3,4t+12 is cordial and total cordial.Case 9. When m≡1mod4 and k≡2mod4 that means k=4r+2,r≥0 and m=4t+1, t>1, then the labeling L4r10;101L4′0M4t−6'0,101L4′0M4t−6'0,010L41M4t−61,010L41M4t−61,...,r−times,010L41M4t−60,101L4′0M4t−6'1 for P4r+2⊙L3,4t+12 can be applied. Therefore, x0=2r+1,x1=2r+1,a0=2r+1,a1=2r,y0=2t+1,y1=2t+2,b0=4t+2,b1=4t+1,y0′=2t+2,y1′=2t+1,b0′=4t+2, and b1′=4t+1. Hence, v0−v1=0, e0−e1=1 and v0+e0−e1+v1=1. For the special case P4r+2⊙L3,52, the labeling L4r10;01L4′0,01L4′0,10L41,10L41,...,r−times,10L41,01L4′0 is sufficient and thus P4r+2⊙L3,4t+12 is cordial and also total cordial.Case 10. When m≡1mod4 and k≡3mod4 that means k=4r+3,r≥0 and m=4t+1, t>1, then take the labeling L4r021;101L4′0M4t−6'0,101L4′0M4t−6'0,010L41M4t−61,010L41M4t−61,...,r−times,101L4′0M4t−6'1,101L4′0M4t−6'1,,010L41M4t−60 for P4r+3L3,4t+12. Therefore, x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=2t+1,y1=2t+2,b0=4t+2,b1=4t+1,y0′=2t+2,y1′=2t+1,b0′=4t+2, and b1′=4t+1. Hence, v0−v1=0, e0−e1=0, and v0+e0−e1+v1=0. For the special case P4r+3⊙L3,52, the labeling L4r021;01L4′0,01L4′0,10L41,10L41,...,r−times,01L4′0,10L41,10L41 is sufficient and thus P4r+3⊙L3,4t+12 is cordial and total cordial.Case 11. When m≡2mod4 and k even that means m=4t+2, t>1, and k=2r,r≥1, then by taking the labeling M2r;021L4′0M4t−4',120L41M4t−4,...,r−times for P2r⊙L3,4t+22, therefore x0=x1=r,a0=0,a1=2r−1,y0=2t+3,y1=2t+1,b0=4t+2,b1=4t+3,y0′=2t+1,y1′=2t+3,b0′=4t+2, and b1′=4t+3. Hence, v0−v1=0, e0−e1=1 and v0+e0−e1+v1=1. For the case P2r⊙L3,62, the labeling M2r;02L4′01,12L410,...,r−times is sufficient and thus P2r⊙L3,62 is cordial and total cordial.Case 12.When m≡2mod4 and k odd that means m=4t+2, t>1, and k=2r+1,r≥1, then the labeling M2r+1;021L4′0M4t−4',120L41M4t−4,...,r−times,010L41M4t−4 for P2r+1⊙L3,4t+22 is considered. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=2t+3,y1=2t+1,b0=4t+2,b1=4t+3,y0′=2t+1,y1′=2t+3,b0′=4t+2 and b1′=4t+3. So, v0−v1=1, e0−e1=1, and v0+e0−e1+v1=0. For the case P2r+1⊙L3,62, the labeling M2r+1;02L4′01,12L410,...,r−times,01L410 is sufficient and thus P2r+1⊙L3,62 is cordial and total cordial.Case 13. When m≡3mod4 and k≡0mod4 that means m=4t+3, t≥1, and k=4r,r≥1, then the labeling L4r;02M4t+3',02M4t+3',12M4t+3,12M4t+3,...,r−times for P4r⊙L3,4t+32 can be applied. Therefore, x0=x1=2r,a0=2r,a1=2r−1,y0=2t+3,y1=2t+2,b0=4t+3,b1=4t+4,y0′=2t+2,y1′=2t+3,b0′=4t+3, and b1′=4t+4. Hence, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P4r⊙L3,4t+32 is cordial and total cordial.Case 14. When m≡3mod4 and k≡1mod4 that means m=4t+3, t≥1, and k=4r+1,r≥0, then one can select the labeling L4r'0;12M4t+3,12M4t+3,02M4t+3',02M4t+3',...,r−times,12M4t+3 for P4r+1⊙L3,4t+32. Therefore, x0=2r+1,x1=2r,a0=2r+1,a1=2r−1,y0=2t+3,y1=2t+2,b0=4t+3,b1=4t+4,y0′=2t+2,y1′=2t+3,b0′=4t+3,b1′=4t+4,y0∗=2t+2,y1∗=2t+3,b0∗=4t+3, and b1∗=4t+4. So, |v0−v1|=0, |e0−e1|=0, and v0+e0−e1+v1=0. Thus, P4r+1⊙L3,4t+32 is cordial and total cordial.Case 15. When m≡3mod4 and k≡2mod4 that means m=4t+3, t>1, and k=4r+2,r≥0, then one can take the labeling L4r10;0313M4t−4,0313M4t−4,01L4M4t−4',01L4M4t−4',...,r−times,01L4M4t−4',0313M4t−4 for P4r+2⊙L3,4t2. Therefore, x0=x1=2r+1,a0=2r+1,a1=2r,y0=2t+3,y1=2t+2,b0=4t+3,b1=4t+4,y0′=2t+2,y1′=2t+3,b0′=4t+3, and b1′=4t+4. Hence, |v0−v1|=0, |e0−e1|=1, and v0+e0−e1+v1=1. Thus, P4r+2⊙L3,4t+32 is cordial and total cordial.Case 16. When m≡3mod4 and k≡3mod4 that meansm=4t+3, t≥1, and k=4r+1,r≥0, then one can choose the labeling L4r100;02M4t+3',02M4t+3',12M4t+3,12M4t+3,…,r−times,12M4t+3,02M4t+3',12M4t+3 for P4r+3⊙L3,4t+32.
Therefore,x0=2r+2,x1=2r+1,a0=2r+2,a1=2r,y0=2t+3,y1=2t+2,b0=4t+3,b1=4t+4,y0′=2t+2,y1′=2t+3,b0′=4t+3,b1′=4t+4,y0∗=2t+2,y1∗=2t+3,b0∗=4t+3, and b1∗=4t+4. Hence, |v0−v1|=0, |e0−e1|=0, and v0+e0−e1+v1=0. Thus, P4r+3⊙L3,4t+32 is cordial and total cordial.Lemma 2.
Pk⊙Ln,m2 is cordial and total cordial for all k≥1 and m>6.Proof.
Letk=4r+i' (i'=0,1,2,3 and r≥1) or k=2r+j' (j'=0,1 and r≥1), n=4s+i and m=4t+j (i,j=1,2,3 and s,t≥2), then we may use the labeling Ai' or Aj' for Pk as given in Table 1. For a given value of j with 1≤i,j≤3, we may use one of the labeling in the set {Bij, Bij'} for Ln,m, where Bij and Bij' are the labeling of Ln,m2 which are connected to the vertices labeled 0 in Pk, while Bij and Bij' are the labeling of Pm which are connected to the vertices labeled 1 in Pk as given in Table 2. Using Table 3 and the formulas v0−v1=x0−x1+x0.y0−y1+x1.y0′−y1′, e0−e1=a0−a1+x0.b0−b1+x1.b0′−b1′+x0.y0−y1−x1.y0′−y1′, and v0+e0−e1+v1=x0−x1+2x0.y0−y1+a0−a1+x0.b0−b1+x1.b0′−b1′, we can compute the values shown in the last two columns of Table 3. We see that Pk⊙Ln.m2 is isomorphic to Pk⊙Lm,n2. Since all of these values are 1 or 0, the lemma follows.Table 1
Labelling ofPk.
K=4r+i′,i′=0,1,2,3Labelling ofPkx0x1a0a1i′=0A0=L4r2r2r2r2r−1i′=1A1=L4r02r+12r2r2ri′=2A2=L4r102r+12r+12r+12ri′=3A3=L4r0012r+22r+12r+12r+1k=2r+j′,i′=0,1i′=0A4=M2rrr02r−1i′=1A5=M2r+1r+1r02rTable 2
Labelling ofLn,m2.
n=4s+i,m=4t+j,i,j=0,1,2,3Labelling ofLn,m2y0y1y0y1i=0j=0B00=L4′M4s−5L4′L4t−4'2s+2t+12s+2t−14s+4t+24s+4t+2i=0j=0B00'=L4M4s−5'L4M4t−4'2s+2t−12s+2t+14s+4t+24s+4t+2i=0j=1B01=L4M4s−5'M4t−6'1L4′02s+2t2s+2t4s+4t−14s+4t−1i=0j=2B02=L4′M4s−51L4′0M4t−4'2s+2t+12s+2t4s+4t4s+4ti=0j=2B02'=L4M4s−5'0L41M4t−42s+2t2s+2t+14s+4t4s+4ti=0j=3B03=L4M4s−5'M4t+32s+2t+12s+2t+14s+2t+24s+4t+2i=1j=1B11=1L4′0M4s−6'02M4t−60L412s+2t+12s+2t4s+4t4s+4ti=1j=1B11'=0L41M4s−612M4t−6'1L4′02s+2t2s+2t+14s+4t4s+4ti=1j=2B12=0L41M4s−6M4t−6'1L4′02s+2t+12s+2t+14s+4t+14s+4t+1i=1j=3B13=1L4′0M4s−6'M4t+32s+2t+22s+2t+14s+4t+24s+4t+2i=1j=3B13'=0L41M4s−6M4t+3'2s+2t+12s+2t+24s+4t+24s+4t+2i=2j=2B22=M4s−40L41L4′0M4t−4'2s+2t+22s+2t+14s+4t+24s+4t+2i=2j=2B22'=M4s−4'1L4′0L41M4t−42s+2t+12s+2t+24s+4t+24s+4t+2i=2j=3B23=0L41M4s−5M4t+3'2s+2t+22s+2t+24s+4t+34s+4t+3i=3j=3B33=M4s+2M4t+32s+2t+22s+2t+34s+4t+34s+4t+3i=3j=3B33'=M4s+2M4t+32s+2t+32s+2t+24s+4t+34s+4t+3Table 3
Labelling ofPk⊙Ln,m2.
i′/j′ijPkLn,m2v0−v1e0−e1v0+e0−e1+v1000A4B00,B00′011100A5B00,B00′,..,B00,B00′,B00′011001A0B01,B01,B01,B01011101A1B01,B01,B01,B01,...,B01101201A2B01,B01,B01,B01,...,B01,B01011301A3B01,B01,B01B01,...,B01,B01,B01101002A4B02,B02′011102A5B02,B02′,..,B02,B02′,B02′011001A0B03,B03,B03,B03011101A1B03,B03,B03,B03,...,B03101201A2B03,B03,B03,B03,...,B03,B03011301A3B03,B03,B03,B03,...,B03,B03,B03101011A4B11,B11′011111A5B11,B11′,..,B11,B11′,B11′011012A0B12,B12,B12,B12011112A1B12,B12,B12,B12,..,B12101212A2B12,B12,B12,B12,..,B12,B12011312A3B12,B12,B12,B12,..,B12,B12,B12101013A4B13,B13′011113A5B13,B13′,..,B13,B13′,B13′011022A4B22,B22′011122A5B22,B22′,..,B22,B22′,B22′011023A0B23,B23,B23,B23011123A1B23,B23,B23,B23,..,B23101223A2B23,B23,B23,B23,..,B23,B23011323A3B23,B23,B23,B23,..,B23,B23,B23101033A4B33,B33′011133A5B33,B33′,..,B33,B33′,B33′011Lemma 3.
Pk⊙L4,m2 is cordial and total cordial for all k≥1 and m>3.Proof.
Case 1. When m≡0mod4 and k=r, r≥1 that means m=4t, t>1, and k=r, r≥1. Then, take the labeling 1r;100L4M4t−4',...,r−times for Pr⊙L4,4t2. Therefore, x0=0,x1=r,a0=r−1,a1=0,y0′=2t+2,y1′=2t+1,b0′=4t+2, and b1′=4t+2. Hence, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the case Pr⊙L4,42, the labeling 1r;03130,...,r−times is sufficient and thus Pr⊙L4,4t2, r≥1, is cordial and total cordial.Case 2. When m≡1mod4 and k≡0mod4 that means m=4t+1, t>1, and k=4r,r≥1, then the labeling S4r;103L41M4t−61,103L41M4t−61,103L41M4t−61,103L41M4t−61,...,r−times for P4r⊙L4,4t+12 is applied. Therefore, x0=x1=2r,a0=2r−1,a1=2r,y0=2t+2,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+2,b0′=4t+3, and b1′=4t+3. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the case P4r⊙L4,52, the labeling S4r;102L41,102L41,102L41,102L41,...,r−times is sufficient and thus P4r⊙L4,4t+12 is cordial and total cordial.Case 3. When m≡1mod4 and k≡1mod4 that means m=4t+1, t>1, and k=4r+1,r≥0, then one can select the labeling S4r0;103L41M4t−61,103L41M4t−61,103L41M4t−61,103L41M4t−61,...,r−times,103L41M4t−61 for P4r+1⊙L4,4t+12. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=2t+2,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+2,b0′=4t+3, and b1′=4t+3. So, v0−v1=1. e0−e1=0, and v0+e0−e1+v1=1. For the case P4r+1⊙L4,52, the labeling S4r0;102L41,102L41,102L41,102L41,...,r−times,102L41 is sufficient and thus P4r+1⊙L4,4t+12 is cordial and total cordial.Case 4. When m≡1mod4 and k≡2mod4 that means m=4t+1, t>1, and k=4r+2,r≥0, then the labeling S4r01;103L41M4t−61,103L41M4t−61,103L41M4t−61,103L41M4t−61,…,r−times,103L41M4t−61,103L41M4t−61 for P4r+2⊙L4,4t+12 is applied. Therefore, x0=x1=2r+1,a0=2r,a1=2r+1,y0=2t+2,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+2,b0′=4t+3, and b1′=4t+3. Hence, v0−v1=0, e0−e1=1, and v0+e0–e1+v1=1. For the case P4r+2⊙L4,52, the labeling S4r01;102L41,102L41,102L41,102L41,…,r−times,102L41,102L41 is sufficient and thus P4r+2⊙L4,4t+12 is cordial and total cordial.Case 5. When m≡1mod4 and k≡3mod4 that means and m=4t+1, t>1, and k=4r+3,r≥0, then one can take the labeling S4r001;103L41M4t−61,103L41M4t−61,103L41M4t−61,103L41M4t−61,…,r−times,103L41M4t−61,103L41M4t−61,103L41M4t−61 for P4r+3⊙L4,4t+12. Therefore, x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=2t+2,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+2,b0′=4t+3, and b1′=4t+3. So, v0−v1=1, e0−e1=0, and v0+e0–e1+v1=1. For the case P4r+3⊙L4,52, the labeling S4r01;102L41,102L41,102L41,102L41,…,r−times,102L41,102L41,102L41 is sufficient and thus P4r+3⊙L4,4t+12 is cordial and total cordial.Case 6. When m≡2mod4 and k is even that means m=4t+2, t>1, and k=2r, r≥1. Then, one can choose the labeling M2r;103L41M4t−4,013L4′0M4t−4',...,r−times for P2r⊙L4,4t+22. Therefore, x0=x1=r,a0=0,a1=2r−1,y0=2t+3,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+3, and b0′=b1′=4t+4. Hence, one can easily show that v0−v1=0. e0−e1=1 and v0+e0−e1+v1=1. For the special case P2r⊙L4,62, the labeling M2r;102L401,012L4′10,...,r−times is sufficient, and thus P2r⊙L4,4t+22, r≥1, is cordial and total cordial.Case 7. When m≡2mod4 and k is odd that means m=4t+2, t>1, and k=2r+1 where r≥0, then one can choose the labeling M2r+1;103L41M4t−4,013L4′0M4t−4′,...,r−times,013L4′0M4t−4' for P2r+1⊙L4,4t+22. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=2t+3,y1=2t+2,b0=b1=4t+3,y0′=2t+2,y1′=2t+3,b0′=b1′=4t+4,y0∗=2t+2,y1∗=2t+3, and b0∗=b1∗=4t+3, where yi∗ and bi∗ are the numbers of vertices and edges labeled i in L4,4t+22 that are connected to the last zero in P4r+3. Consequently, it is easy to show that v0−v1=0, e0−e1=1 and v0+e0−e1+v1=1. For the special case P2r+1⊙L4,62, the labeling M2r;102L401,012L4′10,...,r−times,012L4′10 is sufficient and thus P2r+1⊙L4,4t+22, r≥0, is cordial and total cordial.Case 8. When m≡3mod4 and k≡0mod4 that means m=4t+3, t>1, and k=4r,r≥1, then one can choose the labeling L4r;103M4t11,103M4t11,103M4t11,103M4t11,...,r−times for P4r⊙L4,4t+32. Therefore, x0=x1=2r,a0=2r−1,a1=2r,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Hence, one can easily show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P4r⊙L4,4t+32 is cordial and total cordial.Case 9. When m≡3mod4 and k≡1mod4 that means m=4t+3, t>1, and k=4r+1,r≥0. Then, the labeling L4r0;103M4t11,103M4t11,103M4t11,103M4t11,...,r−times,103M4t11 for P4r+1⊙L4,4t+32 is considered. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. So, v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. Thus, P4r+1⊙L4,4t+32 is cordial and total cordial.Case 10. When m≡3mod4 and k≡2mod4 that means m=4t+3, t>1, and k=4r+2,r≥0, then one can select the labeling L4r01;103M4t11,103M4t11,103M4t11,103M4t11,...,r−times,103M4t11,103M4t11 for P4r+2⊙L4,4t+32. Therefore, x0=x1=2r+1,a0=2r,a1=2r+1,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Hence, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P4r+2⊙L4,4t+32 is cordial and total cordial.Case 11. When m≡3mod4 and k≡3mod4 that means m=4t+3, t>1, and k=4r+3,r≥0, then one can take the labeling L4r001;103M4t11,103M4t11,103M4t11,103M4t11,...,r−times,103M4t11,103M4t11,103M4t11 for P4r+3⊙L4,4t+12. Therefore, x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Hence, v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. Thus, P4r+3⊙L4,4t+32 is cordial and also total cordial; by this, the lemma was proved.Lemma 4.
Pk⊙L5,m2 is cordial and also total cordial for all k≥1 and m≥3.Proof.
Case 1. When m≡0mod4 that means m=4t, t≥1, since Pk⊙L5,42 isomorphic to Pk⊙L4,52, by Lemma 3, Pk⊙L4,52 is cordial and total cordial and then Pk⊙L5,42 is cordial and total cordial. Also, since Pk⊙L5,4t2, t≥1 isomorphic to Pk⊙L4t,52, by Lemma 3, Pk⊙L4t,52 is cordial and total cordial and then Pk⊙L5,4t2 is cordial and total cordial.Case 2. When m≡1mod4 and k is even that means m=4t +1, t≥1, and k=2r, r≥1, then one can take the labeling M2r;L4′1L4′0M4t−6'0,L40L41M4t−61,...,r−times for P2r⊙L5,4t+12. Therefore, x0=r,x1=r,a0=0,a1=2r−1,y0=2t+3,y1=2t+2,b0=b1=4t+4,y0′=2t+2,y1′=2t+3, and b0′=b1′=4t+4. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the special case P2r⊙L5,52, the labeling M2r;L4L4′0,L4′L41,...,r−times is sufficient and thus P2r⊙L5,4t+12, r≥1, is cordial and total cordial.Case 3.When m≡1mod4 and k is odd that means m=4t +1, t≥1, and k=2r+1, r≥1, then the labeling M2r+1;L4′1L4′0M4t−6'0,L40L41M4t−61,...,r−times,L40L41M4t−61 for P2r+1⊙L5,4t+12 can be applied. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=2t+3,y1=2t+2,b0=b1=4t+4,y0′=2t+2,y1′=2t+3b0′=b1′=4t+4,y0∗=2t+2,y1∗=2t+3, and b0∗=b1∗=4t+4, where yi∗ and bi∗ are the numbers of vertices and edges labeled i in L5,4t+1 that are connected to the last zero in P2r+1. Consequently, it is easy to show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the special case P2r+1⊙L5,52, the labeling M2r+1;L4L4′0,L4′L41,...,r−times,L4′L41 is sufficient and thus P2r+1⊙L5,4t+12, r≥1, is cordial and total cordial.Case 4. When m≡2mod4 and k≡0mod4 that means m=4t +2, t≥1, and k=4r, r≥1, then one can choose the labeling L4r;L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',…,r−time for P4r⊙L5,4t+22. Therefore, x0=x1=2r,a0=2r,a1=2r−1,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Consequently, it is easy to show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the special case P4r⊙L5,62, the labeling L4r;L4′L410,L4′L410,L4′L410,L4′L410,...,r−time is sufficient and thus P4r⊙L5,4t+22 is cordial and total cordial.Case 5. When m≡2mod4 and k≡1mod4 that means m=4t +2, t≥1, and k=4r+1, r≥0, then one can take the labeling L4r0;L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',…,r−time,L40L41M4t−4' for P4r+1⊙L5,4t+22. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. So, v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. For the special case P4r+1⊙L5,62, the labeling L4r0;L4′L410,L4′L410,L4′L410,L4′L410,...,r−time,L4′L410 is sufficient and thus P4r+1⊙L5,4t+22 is cordial and total cordial.Case 6. When m≡2mod4 and k≡2mod4 that means m=4t +2, t≥1, and k=4r+2, r≥0, then one can select the labeling L4r10;L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',...,r−time,L40L41M4t−4',L40L41M4t−4' for P4r+2⊙L5,4t+22. Therefore, x0=x1=2r+1,a0=2r+1,a1=2r,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Hence, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the special case P4r+2⊙L5,62, the labeling L4r10;L4′L410,L4′L410,L4′L410,L4′L410,...,r−time,L4′L410,L4′L410 is sufficient and thus P4r+2⊙L5,4t+22 is cordial and total cordial.Case 7. When m≡2mod4 and k≡3mod4 that means m=4t +2, t≥1, and k=4r+3, r≥0, then the labeling [L4r001;L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',L40L41M4t−4',...,r−time,L40L41M4t−4',L40L41M4t−4',L40L41M4t−4' for P4r+3⊙L5,4t+22 is considered. Therefore, x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=y1=2t+3,b0=b1=4t+5,y0′=y1′=2t+3, and b0′=b1′=4t+5. Consequently, it is easy to show that v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. For the special case P4r+3⊙L5,62, the labeling L4r001;L4′L410,L4′L410,L4′L410,L4′L410,...,r−time,L4′L410,L4′L410,L4′L410 is sufficient and thus P4r+3⊙L5,4t+22 is cordial and also total cordial.Case 8. When m≡3mod4 and k is even that means m=4t +3, t≥1, and k=2r where r≥1, then the labeling M2r;031M4t+3,130M4t+3...,r−times for P2r⊙L5,4t+32 can be applied. Therefore, x0=r,x1=r,a0=0,a1=2r−1,y0=2t+4,y1=2t+3,b0=b1=4t+6,y0′=2t+3,y1′=2t+4, and b0′=b1′=4t+6. Consequently, it is easy to show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P2r⊙L5,4t+32, r≥1, is cordial and total cordial.Case 9. When m≡3mod4 and k is odd that means m=4t +3, t≥1, and k=2r+1 where r≥0, then one can take the labeling M2r+1;031M4t+3,130M4t+3,...,r−times,130M4t+3 for P2r+1⊙L5,4t+32. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=2t+3,y1=2t+2,b0=b1=4t+4,y0′=2t+2,y1′=2t+3b0′=b1′=4t+4,y0∗=2t+3,y1∗=2t+4, and b0∗=b1∗=4t+6, where yi∗ and bi∗ are the numbers of vertices and edges labeled i in L5,4t+3 that are connected to the last zero in P2r+1. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P2r+1⊙L5,4t+32, r≥1, is cordial and total cordial, and by this, the lemma was proved.Lemma 5.
Pk⊙L6,m2 is cordial and total cordial for all m,k.Proof.
Case 1. When m≡0mod4, since Pk⊙L6,42 is isomorphic to Pk⊙L4,62 and Pk⊙L4,62 is cordial and total cordial, then Pk⊙L6,42 cordial and total cordial. Also, since Pk⊙L6,4t2 is isomorphic to Pk⊙L4t,62 and Pk⊙L4t,62 is cordial and total cordial, then Pk⊙L6,4t2 is cordial and total cordial.Case 2. When m≡1mod4, i.e., m=4t+1, t≥1 , since Pk⊙L6,52 is isomorphic to Pk⊙L5,62 and Pk⊙L4,62 is cordial and total cordial, then Pk⊙L5,62 is cordial and total cordial. Also, since Pk⊙L6,4t+12 is isomorphic to Pk⊙L4t+1,62 and Pk⊙L4t+1,62 is cordial and total cordial, then Pk⊙L6,4t+12 is cordial and total cordial.Case 3. When m≡2mod4 and k is even that means m=4t+2, t≥1, and k=2r, r≥1, then one can choose the labeling M2r;L4′01L4′0M4t−4',L410L41M4t−4,...,r−times for P2r⊙L6,4t+22. Therefore, x0=r,x1=r,a0=0,a1=2r−1,y0=2t+4,y1=2t+3,b0=b1=4t+6,y0′=2t+3,y1′=2t+4, and b0′=b1′=4t+6. Consequently, it is easy to show that v0−v1=0 and e0−e1=1. For the special case P2r⊙L6,62, the labeling M2r;L4′0L4′01,L41L410,...,r−times is sufficient and thus P2r⊙L6,4t+22, r≥1, is cordial and also total cordial.Case 4. When m≡2mod4 and k is odd that means m=4t+2, t≥1, and k=2r+1, r≥0, then one can choose the labeling M2r+1;L4′01L4′0M4t−4',L410L41M4t−4,…,r−times,L410L41M4t−4 for P2r+1⊙L6,4t+22. Therefore, x0=r+1,x1=r,a0=0,a1=2r,y0=2t+2t+1,y1=2s+2t,b0=b1=4s+4t,y0=2t+4,y1=2t+3,b0=b1=4t+6,y0′=2t+3,y1′=2t+4,b0′=b1′=4t+6,y0∗=2t+3,y1∗=2t+4, and b0∗=b1∗=4t+6, where yi∗ and bi∗ are the numbers of vertices and edges labeled i in L6,4t+22 that are connected to the last zero in P2r+1. So, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. For the special case P2r+1⊙L6,62, the labeling M2r+1;L4′0L4′01,L41L410,…,r−times,L41L410 is sufficient and thus P2r+1⊙L6,4t+22, r≥1, is cordial and total cordial.Case 5. When m≡3mod4 and k≡0mod4 that means m=4t+3, t≥1, and k=4r, r≥1, then the labeling L4r;L41M4t+3,L41M4t+3,L41M4t+3,L41M4t+3,...,r−time for P4r⊙L6,4t+32 is applied. Therefore x0=x1=2r,a0=2r,a1=2r−1,y0=y1=2t+3,b0=b1=4t+7,y0′=y1′=2t+4 and b0′=b1′=4t+7. Consequently, it is easy to show that v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P4r⊙L6,4t+32 is cordial and total cordial.Case 6. When m≡3mod4 and k≡1mod4 that means m=4t+3, t≥1, and k=4r+1, r≥0, then one can choose the labeling L4r0;L41M4t+3,L41M4t+3,L41M4t+3,L41M4t+3,...,r−time,L41M4t+3 for P4r+1⊙L6,4t+32. Therefore, x0=2r+1,x1=2r,a0=a1=2r,y0=y1=2t+3,b0=b1=4t+7,y0′=y1′=2t+4, and b0′=b1′=4t+7. So, v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. Thus, P4r+1⊙L6,4t+32 is cordial and total cordial.Case 7. When m≡3mod4 and k≡2mod4 that means m=4t+3, t≥1, and k=4r+2, r≥0, then one can take the labeling L4r10;L41M4t+3,L41M4t+3,L41M4t+3,L41M4t+3,...,r−time,L41M4t+3,L41M4t+3 for P4r+2⊙L6,4t+32. Therefore, x0=x1=2r+1,a0=2r+1,a1=2r,y0=y1=2t+3,b0=b1=4t+7,y0′=y1′=2t+4, and b0′=b1′=4t+7. Hence, v0−v1=0, e0−e1=1, and v0+e0−e1+v1=1. Thus, P4r+2⊙L6,4t+32 is cordial and total cordial.Case 8. When m≡3mod4 and k≡3mod4 that means m=4t+3, t≥1, and k=4r+3, r≥0, then one can select the labeling L4r001;L41M4t+3,L41M4t+3,L41M4t+3,L41M4t+3,…,r−time,L41M4t+3,L41M4t+3,L41M4t+3 for P4r+3⊙L6,4t+32. Therefore, x0=2r+2,x1=2r+1,a0=a1=2r+1,y0=y1=2t+3,b0=b1=4t+7,y0′=y1′=2t+4, and b0′=b1′=4t+7. Consequently, it is easy to show that v0−v1=1, e0−e1=0, and v0+e0−e1+v1=1. Thus, P4r+3⊙L6,4t+32 is cordial and total cordial; by this, the lemma was proved, and through the proofs of these lemmas, we have completed the proof of our main theorem.


## 4. Conclusions

In this paper, we test the cordial and total cordial labeling of corona product of paths and second power of lemniscate graphs. We found thatPk⊙Ln,m2 is cordial and also total cordial for all k≥1, n,m≥3. In future work, we can improve this work by using the different graphs with other mathematical operations to prove the cordial and total cordial labeling.



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*Source: 8521810-2022-05-05.xml*