EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 1, Article Number 5470 ISSN 1307-5543 – ejpam.com Published by New York Business Global Cordial Labeling of Corona Product of Paths and Fourth Order of Lemniscate Graphs Atef Abd El-hay1, Khalid A. Alsatami2,Ashraf ELrokh1,∗,Aya Rabie3 1 Mathematics and Computer Science Department, Faculty of Science Menoufia University, Menoufia, Egypt. 2 Department of Mathematics, College of Science, Qassim University, Buraydah, KSA. 3 Department of Planning Technique Center, Institute of National Planning, Cairo, Egypt. Abstract. A graph G = (V,E) is called cordial if it is possible to label the vertex by the function f : V → 0, 1 and label the edges by f∗ : E → 0, 1, where f∗(uv) = (f(u) + f(v))mod2, u, v ∈ V so that |v0−v1| ≤ 1 and |e0−e1| ≤ 1.A lemniscate graph is a plane curve with a characteristic shape, consisting of two loops that meet at a central point as shown below. The curve is also known as the lemniscate of Bernoulli. A fourth order of lemniscate graph is a graph of two fourth order of circles that have two vertex in common. In this paper, we give the conditions that the corona product of paths and fourth order of lemniscate graphs be cordial. 2020 Mathematics Subject Classifications: 05C78,05C15 Key Words and Phrases: Cordial labeling, corona, lemniscate, Social Networking, Network security, Edge Computing 1. Introduction Let G be a graph with p vertices and q edges. All graphs considered here are simple, finite, connected and undirected. The concept of graph labeling was introduced during the sixties’ of the last century by Rosa [16]. Hundreds of researches have been working with different types of labeling graphs [5, 13, 14, 17]. A labeling of a graph G is a process of allocating numbers or labels to the nodes of G or lines of G or both through mathematical functions [1]. Labeling graphs are used for a wide range of applications in different subjects including astronomy, coding theory and communication networks. Cordial labeling is a weaker version of graceful labeling and harmonious labeling introduced by Cahit in [3]. In 1990, Cahit [4], proved the following: each tree is cordial; an Euerlian graph is not cordial if its size is congruent to 2(mod 4) ; 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. Let ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i1.5470 Email address: ashraf.hefnawy68@yahoo.com (A. ELrokh) atef 1992@yahoo.com (A. Abd El-hay), satamy@qu.edu.sa (K. A. Alsatami), aya.ebrahim@inp.edu.eg (A. Rabie) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 2 of 14 G1, G2 respectively be (p1, q1), (p2, q2) graphs. The fourth power of a lemniscate graph is defined as the union of fourth power of cycles where both have a common vertex; it is denoted by L4 n,m ≡ C4 n♯C 4 m as shown in Fig.1. Obviously, L4 n,m has n+m− 1 vertices and 4n+4m− 18 edges. For more details about the cordial labeling and types of labeling, the reader can refer to [2, 6–12, 15]. The corona G1⊙G2 of two graphs G1 (with n1 vertices , Figure 1: The fourth power of a lemniscate graph L4 7,7. m1 edges) and G2 (with n2 vertices , m2 edges) is defined as the graph obtained by taking one copy of G1 and 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 see that the corona G1⊙G2 that has n1 + n1n2 vertices and m1 + n1m2 + n1n2 edges. 2. Terminology and Notation Given a path or a cycle with 4r vertices, We let L4r denote the labeling 0011...0011 (repeated r-times), let L′ 4r denote the labeling 1100...1100 (repeated r times). The labeling 1001 1001...1001 (repeated r times) and 0110...0110 (repeated r times) are denoted by S4r and S′ 4r. 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 i−th 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 y′i and b′i be those for H, which are connected to the vertices labeled 1 of G. It is easily to verify that v0=x0 + x0y0 + x1y ′ 0, v1=x1 + x0y1 + x1y ′ 1, e0=a0 + x0b0 + x1b ′ 0+x0y1+x1y ′ 0 and e1=a1+x0b1+x1b ′ 1+x0y0+x1y ′ 1. Thus v0−v1= (x0−x1)+x0(y0− y1)+x1(y ′ 0−y′1) and e0−e1= (a0−a1)+x0(b0−b1)+x1(b ′ 0−b′1)+x0(y0−y1)−x1(y ′ 0−y′1) In particular, if we have only one labeling for all copies of H, i.e., yi=y′i and bi=b′i, then v0=x0+ny0, v1=x1+ny1, e0=a0+nb0+x0y1+x1y0 and e1=a1+nb1+x0y0+x1y1. Thus v0 − v1= (x0 − x1) + n(y0 − y1) and e0 − e1= (a0 − a1) + n(b0 − b1) + (x1 − x0)(y0 − y1), where n is the order of G. Section one contains a brief literary analysis of the topic of this work, and Section Two deals with the terminology employed throughout. section three examines and study the cordiality of the corona product Pk⊙L4 n,m of paths and fourth power of lemniscate graphs, and show that this is cordial for all positive integers A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 3 of 14 k ≥ 1, n,m ≥ 3. The last section, is the conclusion which summarize the important points of our finding in this paper 3. Main results. In this section, we show that the corona product of paths and fourth power of lemnis- cate graphs, Pk⊙L4 n,m , is cordial for all k ≥ 1, n,m ≥ 3.Throughout our proofs, the way of labelling L4 n,m 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. Theorem 3.1. The corona Pk⊙L4 n,m between paths Pk and fourth power of lemniscate graphs L4 n,m is cordial for al k ≥ 1, n,m ≥ 3. In order to prove this theorem, we will introduce a number of lemmas as follows. Lemma 1. Pk⊙L4 3,m is cordial for all k ≥ 1 and m ≥ 3. Proof. We need to examine the following cases : Case 1. At m=3, we consider the following subcases. subcase 1.1. k is even. Let k= 2r, r ≥ 1. Then, one can choose the labelling[M2r;00100, 11011, ..., (r−times)] for P2r⊙L4 3,3. Therefore x0=x1=r, a0= 0,a1= 2r − 1,y0= 4, y1= 1,b0= 2,b1= 4,y′0= 1,y′1= 4,b′0= 2 and b′1= 4. Hence, it is easy to show that |v0 − v1| = 0 and |e0 − e1| = 1. Thus P2r⊙L4 3,3 ,r ≥ 1 is cordial. subcase 1.2. k is odd. Let k= 2r+1 where r ≥ 0. Then take the labeling [M2r+1;00100, 11011, 00100, 11011, ..., (r− times), 11100] for P2r+1⊙L4 3,3. Therefore x0=r+1,x1=r, a0= 0,a1= 2r, y0= 4,y1= 1,b0= 2,b1= 4, y′0= 1,y′1= 4,b′0= 2,b′1= 4,y∗0= 2,y∗1= 3,b∗0= 4 and b∗1= 2, where y∗i and b∗i are the numbers of vertices and edges labelled i in L4 3,3 that are connected to the last zero in P4r+3. Con- sequently, it is easy to show that |v0 − v1| = 0 and |e0 − e1| = 1. Thus P2r+1⊙L4 3,3, r ≥ 0, is cordial and the lemma follows. Case 2. At m ≡ 0(mod4), we consider the following subcases. subcase 2.1. k ≡ 0(mod4) Let k= 4r, r ≥ 1 and m= 4t, t> 1. Then, the labelling [L4r;03 13M4t−4, 03 13M4t−4, 01L4M ′ 4t−4, 01L4M ′ 4t−4, ..., (r−times)] for P4r⊙L4 3,4t can be applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r − 1,y0=y1= 2t+ 1,b0= 8t− 3,b1= 8t− 3,y′0=y′1= 2t+ 1,b′0= 8t− 3 and b′1= 8t − 3. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r⊙L4 3,4, the labeling [L4r;03 13, 03 13, 01L4, 01L4, ..., (r − times)] is sufficient and thus P4r⊙L4 3,4t is cordial. subcase 2.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t, t> 1. Then, the labelling [L4r0;03 13M4t−4, 03 13M4t−4, 01L4M ′ 4t−4, 01L4M ′ 4t−4, ..., (r − times),03L3M4t−4] for P4r+1⊙L4 3,4t is considered. There- fore x0= 2r+1,x1= 2r, a0=a1= 2r, y0=y1= 2t+1,b0= 8t−3,b1= 8t−3,y′0=y′1= 2t+1,b′0= 8t− 3 and b′1= 8t−3. Hence, |v0−v1| = 1 and |e0−e1| = 0. For the case P4r+1⊙L4 3,4, the label- ing [L4r0;03 13, 03 13, 01L4, 01L4, ..., (r − times), 1303] is sufficient and thus P4r+1⊙L4 3,4t is cordial. subcase 2.3.k ≡ 2(mod4) A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 4 of 14 Let k= 4r+2,r ≥ 0 andm= 4t, t> 1. Then, the labelling [L4r10;03 13M4t−4, 03 13M4t−4, 01L4M ′ 4t−4, 01L4M ′ 4t−4, ..., (r−times), 01L4M ′ 4t−4, 03 13M4t−4] for P4r+2⊙L4 3,4t is applied. Therefore x0=x1= 2r+1,a0= 2r+1,a1= 2r, y0=y1= 2t+1,b0= 8t−3,b1= 8t−3,y′0=y′1= 2t+ 1,b′0= 8t−3 and b′1= 8t−3. So, |v0−v1| = 0 and |e0−e1| = 1. For the case P4r+2⊙L4 3,4, the labeling [L4r10;03 13, 03 13, 01L4, 01L4, ..., (r − times), 01L4, 03 13] is sufficient and thus P4r+2⊙L4 3,4t is cordial. subcase 2.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t, t> 1. Then, one can select the labelling [L4r001;0313M4t−4, 03 13M4t−4, 01L4M ′ 4t−4, 01L4M ′ 4t−4, ..., (r − times), 03 13M4t−4, 03 13M4t−4, 01L4M ′ 4t−4] for P4r+3⊙L4 3,4t. Therefore x0= 2r + 2,x1= 2r + 1,a0=a1= 2r+1,y0=y1= 2t+1,b0= 8t−3,b1= 8t−3,y′0=y′1= 2t+1,b′0= 8t−3 and b′1= 8t−3. Hence, one can easily show that |v0−v1| = 1 and |e0−e1| = 0. For the case P4r+3⊙L4 3,4, the labeling [L4r001;03 13, 03 13, 01L4, 01L4..., (r − times)] is sufficient and thus P4r+3⊙L4 3,4t is cordial. Case 3. At m ≡ 1(mod4), we consider the following subcases. subcase 3.1. k even Let k= 2r, r ≥ 1 andm= 4t+1, t> 1. Then, one can choose the labelling [M2r;10 13L ′ 4t−412, 10 13S4t−402, ..., (r − times)] for P2r⊙L4 3,4t+1. Therefore x0=x1=r, a0= 0,a1= 2r − 1, y0= 2t + 2,y1= 2t + 1,b0= 8t − 1,b1= 8t − 1,y′0= 2t + 1,y′1= 2t + 2,b′0= 8t − 1 and b′1= 8t− 1. Hence, one can easily show that |v0− v1| = 0 and |e0− e1| =1. For the special case P4r⊙L4 3,5and P4r+2⊙L4 3,5, the labeling [L4r; 1403, 1403, 04 13, 04 13, ..., (r − times)] and [L4r01; 1403, 1403, 04 13, 04 13, ..., (r− times), 1403, 04 13, 04 13] is sufficient and thus P2r⊙L4 3,4t+1 is cordial. subcase 3.2. k odd Let k= 2r + 1,r ≥ 1 and m= 4t + 1, t> 1. Then, one can choose the labelling [M2r+1;10 13L ′ 4t−4 12, 10 13S4t−402, ..., (r−times), 10 13S4t−402] for P2r+1⊙L4 3,4t+1. There- fore x0=r+1,x1=r, a0= 0,a1= 2r, y0= 2t+2,y1= 2t+1,b0= 8t−1,b1= 8t−1,y′0=y”0= 2t+ 1,y′1=y”1= 2t + 2,b′0=b”0= 8t − 1 and b′1=b”1= 8t − 1. Hence, one can easily show that |v0−v1| = 0 and |e0−e1| = 1. For the special case P4r+1⊙L4 3,5and P4r+3⊙L4 3,5, the labeling [L4r0; 1403, 1403, 04 13, 04 13, ..., (r−times),04 13] and [L4r010; 1403, 1403, 04 13, 04 13, ..., (r− times), 1403, 04 13, 04 13, 04 13] and thus P4r⊙L4 3,4t+1 is cordial. Case 4. At m ≡ 2(mod4), we consider the following subcases. subcase 4.1. k ≡ 0(mod4) Let k= 4r, r ≥ 1 and m= 4t+2, t> 1. Then, the labelling [L4r;010310 13M4t−6, 010310 13M4t−6, 010310 13M4t−6, 010310 13M4t−6, ..., (r − times)] for P4r⊙L4 3,4t+2 can be ap- plied. Therefore x0=x1= 2r, a0= 2r, a1= 2r− 1,y0=y1= 2t+1,b0=b1= 8t+1,y′0=y′1= 2t+ 1,b′0= 8t + 1 and b′1= 8t + 1. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r⊙L4 3,6, the labeling [L4r;03 120 12, 03 120 12, 03 120 12, 03 120 12, ..., (r − times)] is sufficient and thus P4r⊙L4 3,4t+2 is cordial. subcase 4.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t+2, t> 1. Then, the labelling [L4r0;010310 13M4t−6, 010310 13M4t−6, 010310 13M4t−6, 010310 13M4t−6, ..., (r−times), 010310 13M4t−6] for P4r+1 ⊙L4 3,4t+2 is considered. Therefore x0= 2r+1,x1= 2r, a0=a1= 2r, y0=y1= 2t+1,b0=b1= 8t+1, y′0=y′1= 2t+ A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 5 of 14 1,b′0= 8t+1 and b′1= 8t+1. Hence, |v0−v1| = 1 and |e0−e1| = 0. For the case P4r+1⊙L4 3,6, the labeling [L4r0;03 120 12, 03 120 12, 03 120 12, 03 120 12, ..., (r− times),03 120 12] is suf- ficient and thus P4r+1⊙L4 3,4t+2 is cordial. subcase 4.3.k ≡ 2(mod4) Let k= 4r + 2,r ≥ 0 and m= 4t+ 2, t> 1. Then, the labelling [L4r10;01031013M4t−6, 01031013M4t−6, 01031013M4t−6, 01031013M4t−6, ..., (r−times), 01031013M4t−6, 01031013M4t−6] for P4r+2 ⊙L4 3,4t+2 is applied. Therefore x0=x1= 2r + 1,a0= 2r + 1,a1= 2r, y0=y1= 2t + 1,b0=b1= 8t+1,y′0=y′1= 2t+1,b′0= 8t+1 and b′1= 8t+1. So, |v0−v1| = 0 and |e0−e1| = 1. For the case P4r+2⊙L4 3,6, the labeling [L4r10;0312012, 0312012, 0312012, 0312012, ..., (r − times),0312012, 0312012] is sufficient and thus P4r+2⊙L4 3,4t+2 is cordial. subcase 4.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t+2, t> 1. Then, one can select the labelling [L4r001;010310 13M4t−6, 01031013M4t−6, 010310 13M4t−6, 010310 13M4t−6, ..., (r−times), 010310 13M4t−6, 010310 13M4t−6, 010310 13M4t−6] for P4r+3⊙L4 3,4t+2. Therefore x0= 2r+2,x1= 2r+1,a0=a1= 2r+ 1,y0=y1= 2t+ 1,b0=b1= 8t+ 1,y′0=y′1= 2t+ 1,b′0= 8t+ 1 and b′1= 8t+ 1. Hence, one can easily show that |v0 − v1| = 1 and |e0 − e1| = 0. For the case P4r+3⊙L4 3,6, the labeling [L4r001;03 120 12, 03 120 12, 03 120 12, 03 120 12, ..., (r−times),03 120 12, 03 120 12, 03 120 12] is sufficient and thus P4r+3⊙L4 3,4t+2 is cordial. Case 5. At m ≡ 3(mod4), we consider the following subcases. subcase 5.1.k even Let k= 4r, r ≥ 1 andm= 4t+3, t ≥ 1. Then, the labelling [M2r;010 12L ′ 4t0102, 010 12L ′ 4t0102, 10102S ′ 4t10 12, 10102S ′ 4t10 12, ..., (r − times)] for P2r⊙L2 3,4t+3 can be applied. Therefore x0=x1=r, a0= 0,a1= 2r−1,y0= 2t+3,y1= 2t+2,b0=b1= 8t+3,y′0= 2t+2,y′1= 2t+3,b′0= 8t+ 3 and b′1= 8t+ 4. Hence, |v0 − v1| = 0 and |e0 − e1| = 1. Thus P2r⊙L2 3,4t+3 is cordial. subcase 5.2.k odd Let k= 4r, r ≥ 1 andm= 4t+3, t ≥ 1. Then, the labelling [M2r+1;010 12L ′ 4t0102, 010 12L ′ 4t0102, 10102S ′ 4t10 12, 10102S ′ 4t10 12, ..., (r − times), 10102S ′ 4t10 12] for P2r+1⊙L2 3,4t+3 can be ap- plied. Therefore x0=r+1,x1=r, a0= 0,a1= 2r, y0= 2t+3,y1= 2t+2,b0=b1= 8t+3,y′0=y”0= 2t+ 2,y′1=y1”= 2t+3,b′0=b”0= 8t+3 and b′1=b”1= 8t+4. Hence, |v0− v1| = 1 and |e0− e1| = 0. Thus P2r⊙L2 3,4t+3 is cordial. Lemma 2. Pk⊙L4 n,m is cordial for all k ≥ 1 and m > 6 except at m = n = 7. Proof. Let k = 4r + i′ ( i′ = 0, 1, 2, 3 and r ≥ 0) , n= 4s + i and m = 4t + j ( i, j = 0, 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 0 ≤ i, j ≤ 3, we may use one of the labeling in the set {Bij, B′ ij} for L4 n,m, where Bij and B ′ ij are the labeling of L4 n,m which are connected to the vertices labeled 0 in Pk, while Bij and B′ ij are the labeling of L4 n,m which are connected to the vertices labeled 1 in Pk as given in Table 3.2. Using Table 3.3 and the formulas v0 − v1 = (x0 − x1) + x0.(y0 − y1) + x1.(y ′ 0 − y′1) and e0 − e1 = (a0 − a1) + x0.(b0 − b1) + x1.(b ′ 0 − b′1) + x0.(y0 − y1) − x1.(y ′ 0 − y′1), we can compute the values shown in the last two columns of Table 3.3. We see that Pk⊙L4 n.m is isomorphic to Pk⊙L4 m,n. Since all of these values are 1 or 0, the lemma follows. A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 6 of 14 Table 3.1.Labeling of Pk K = 4r + i′, i′ = 0, 1, 2, 3 labeling of Pk x0 x1 a0 a1 i′ = 0 A0 = L4r A′ 0 = M4r 2r 2r 2r 2r 2r 0 2r − 1 4r − 1 i = 1 A1 = L4r0 A′ 1 = M4r+1 2r + 1 2r + 1 2r 2r 2r 0 2r 4r i′ = 2 A2 = L4r01 A′ 2 = M4r+2 A′′ 2 = L4r10 2r + 1 2r + 1 2r + 1 2r + 1 2r + 1 2r + 1 2r 0 2r + 1 2r + 1 4r + 1 2r i′ = 3 A3 = L4r001 A′ 3 = M4r+3 A′′ 3 = S4r100 2r + 2 2r + 2 2r + 2 2r + 1 2r + 1 2r + 1 2r + 1 0 2r + 1 2r + 1 4r + 2 2r + 1 Table 3.2.Labeling of L4 n,m n = 4s+ i, m = 4t+ j, i, j = 0, 1, 2, 3 labeling of L4 n,m y0 y1 b0 b1 i, j = 0 B00 = S′ 4s12M ′ 4t−603 2s+ 2t 2s+ 2t− 1 8s+ 8t− 9 8s+ 8t− 9 i, j = 0 B′ 00 = L′ 4s02M4t−613 2s+ 2t− 1 2s+ 2t 8s+ 8t− 9 8s+ 8t− 9 i = 0, j = 1 B01 = 13M4s−603L4t−4013 2s+ 2t− 1 2s+ 2t+ 1 8s+ 8t− 6 8s+ 8t− 8 i = 0, j = 1 B01 = 03M ′ 4s−613S4t−4103 2s+ 2t+ 1 2s+ 2t− 1 8s+ 8t− 6 8s+ 8t− 8 i = 0, j = 2 B02 = S′ 4s013M4t−403 2s+ 2t+ 1 2s+ 2t 8s+ 8t− 5 8s+ 8t− 5 i = 0, j = 2 B′ 02 = L′ 4s103M ′ 4t−413 2s+ 2t 2s+ 2t+ 1 8s+ 8t− 5 8s+ 8t− 5 i = 0, j = 3 B03 = 13M4s−603L ′ 4t−413010 2s+ 2t 2s+ 2t+ 2 8s+ 8t− 2 8s+ 8t− 4 i = 0, j = 3 B′ 03 = 03M ′ 4s−613S ′ 4t−403101 2s+ 2t+ 2 2s+ 2t 8s+ 8t− 2 8s+ 8t− 4 i, j = 1 B11 = L4s02L ′ 4t−413 2s+ 2t 2s+ 2t− 1 8s+ 8t− 5 8s+ 8t− 5 i, j = 1 B′ 11 = S4s12S ′ 4t−403 2s+ 2t− 1 2s+ 2t 8s+ 8t− 5 8s+ 8t− 5 i = 1, j = 2 B12 = 13L ′ 4s−4041013M4t−6 2s+ 2t 2s+ 2t+ 2 8s+ 8t− 2 8s+ 8t− 4 i = 1, j = 2 B′ 12 = 03S ′ 4s−4140103M ′ 4t−6 2s+ 2t+ 2 2s+ 2t 8s+ 8t− 2 8s+ 8t− 4 i = 1, j = 3 B13 = 13L ′ 4s−40210L ′ 4t−4 2s+ 2t+ 1 2s+ 2t+ 2 8s+ 8t− 1 8s+ 8t− 1 i = 1, j = 3 B′ 13 = 03S4s−41201S4t−4 2s+ 2t+ 2 2s+ 2t+ 1 8s+ 8t− 1 8s+ 8t− 1 i, j = 2 B22 = 0313L ′ 4s−4021013M4t−6 2s+ 2t+ 1 2s+ 2t+ 2 8s+ 8t− 1 8s+ 8t− 1 i, j = 2 B′ 22 = 1303S ′ 4s−4120103M ′ 4t−6 2s+ 2t+ 2 2s+ 2t+ 1 8s+ 8t− 1 8s+ 8t− 1 i = 2, j = 3 B33 = 031013M4s−60L ′ 4s−41 2s+ 2t+ 2 2s+ 2t+ 2 8s+ 8t+ 1 8s+ 8t+ 1 i = 2, j = 3 B33 = 130103M ′ 4s−61S ′ 4s−40 2s+ 2t+ 2 2s+ 2t+ 2 8s+ 8t+ 1 8s+ 8t+ 1 i, j = 3 B′ 33 = 02M ′ 4s−213S ′ 4t−403101 2s+ 2t+ 3 2s+ 2t+ 2 8s+ 8t+ 3 8s+ 8t+ 3 i, j = 3 B33 = 12M4s−203L ′ 4t−413010 2s+ 2t+ 2 2s+ 2t+ 3 8s+ 8t+ 3 8s+ 8t+ 3 A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 7 of 14 Table 3.3.Labeling of Pk ⊙ L4 n,m i ij Pk L4 n,m |v0 − v1| |e0 − e1| 0 00 A′ 0 B00, B ′ 00, B00, B ′ 00 0 1 1 00 A′ 1 B00, B ′ 00, B00, B ′ 00, ..., B ′ 00 0 1 2 00 A′ 2 B00, B ′ 00, B00, B ′ 00, ..., B00, B ′ 00 0 1 3 00 A′ 3 B00, B ′ 00, B00, B ′ 00, ..., B00, B ′ 00, B ′ 00 0 1 0 01 A0 B01, B01, B ′ 01, B ′ 01 0 1 1 01 A1 B01, B01, B ′ 01, B ′ 01, ..., B01 1 0 2 01 A2 B01, B01, B ′ 01, B ′ 01, ..., B01, B ′ 01 0 1 3 01 A3 B01, B01, B ′ 01, B ′ 01, ..., B01, B01, B ′ 01 1 0 0 02 A′ 0 B02, B ′ 02, B02, B ′ 02 0 1 1 02 A′ 1 B02, B ′ 02, B02, B ′ 02, ..., B ′ 02 0 1 2 02 A′ 2 B02, B ′ 02, B02, B ′ 02, ..., B02, B ′ 02 0 1 3 02 A′ 3 B02, B ′ 02, B02, B ′ 02, ..., B02, B ′ 02, B ′ 02 0 1 0 03 A0 B03, B03, B ′ 03, B ′ 03 0 1 1 03 A1 B03, B03, B ′ 03, B ′ 03, ..., B03 1 0 2 03 A′′ 2 B03, B03, B ′ 03, B ′ 03, ..., B ′ 03, B03 0 1 3 03 A′′ 3 B03, B03, B ′ 03, B ′ 03, ..., B ′ 03, B03, B03 1 0 0 11 A′ 0 B11, B ′ 11, B11, B ′ 11 0 1 1 11 A′ 1 B11, B ′ 11, B11, B ′ 11, ..., B ′ 11 0 1 2 11 A′ 2 B11, B ′ 11, B11, B ′ 11, ..., B11, B ′ 11 0 1 3 11 A′ 3 B11, B ′ 11, B11, B ′ 11, ..., B11, B ′ 11, B ′ 11 0 1 0 12 A0 B12, B12, B ′ 12, B ′ 12 0 1 1 12 A1 B12, B12, B ′ 12, B ′ 12, ..., B12 1 0 2 12 A2 B12, B12, B ′ 12, B ′ 12, ..., B12, B ′ 12 0 1 3 12 A3 B12, B12, B ′ 12, B ′ 12, ..., B12, B12, B ′ 12 1 0 0 13 A′ 0 B13, B ′ 13, B13, B ′ 13 0 1 1 13 A′ 1 B13, B ′ 13, B13, B ′ 13, ..., B ′ 13 0 1 2 13 A′ 2 B13, B ′ 13, B13, B ′ 13, ..., B13, B ′ 13 0 1 3 13 A′ 3 B13, B ′ 13, B13, B ′ 13, ..., B13, B ′ 13, B ′ 13 0 1 A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 8 of 14 i ij Pk L4 n,m |v0 − v1| |e0 − e1| 0 22 A′ 0 B22, B ′ 22, B22, B ′ 22 0 1 1 22 A′ 1 B22, B ′ 22, B22, B ′ 22, ..., B ′ 22 0 1 2 22 A′ 2 B22, B ′ 22, B22, B ′ 22, ..., B22, B ′ 22 0 1 3 22 A′ 3 B22, B ′ 22, B22, B ′ 22, ..., B22, B ′ 22, B ′ 22 0 1 0 23 A0 B23, B23, B ′ 23, B ′ 23 0 1 1 23 A1 B23, B23, B ′ 23, B ′ 23, ..., B23 1 0 2 23 A2 B23, B23, B ′ 23, B ′ 23, ..., B23, B ′ 23 0 1 3 23 A3 B23, B23, B ′ 23, B ′ 23, ..., B23, B23, B ′ 23 1 0 0 33 A′ 0 B33, B ′ 33, B33, B ′ 33 0 1 1 33 A′ 1 B33, B ′ 33, B33, B ′ 33, ..., B ′ 33 0 1 2 33 A′ 2 B33, B ′ 33, B33, B ′ 33, ..., B33, B ′ 33 0 1 3 33 A′ 3 B33, B ′ 33, B33, B ′ 33, ..., B33, B ′ 33, B ′ 33 0 1 Lemma 3. Pk⊙L4 4,m is cordial for all k ≥ 1 and m > 3 . Proof. We need to examine the following cases : Case 1. At m ≡ 0(mod4), we consider the following sub subcases. subcase 1.1.k ≡ 0(mod4) Let k= 4r, r ≥ 1 andm= 4t, t> 1. Then, the labelling [L4r;0 15M4t−603, 0 15M4t−603, 105M ′ 4t−6 13 , 105M ′ 4t−6 13..., (r−times)] for P4r⊙ L4 4,4t is applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r− 1,y0= 2t+1,y1= 2t+2,b0= 8t−1,b1= 8t−2,y′0= 2t+2,y′1= 2t+1,b′0= 8t−1 and b′1= 8t−2. So, |v0−v1| = 0 and |e0−e1| = 1. For the case P4r⊙L4 4,4, the labeling [M4r;010103, 1010 13, 010103, 1010 13, ..., (r − times)] is sufficient and thus P4r⊙L4 4,4t is cordial. subcase 1.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t, t> 1. Then, the labelling [L4r0;015M4t−603, 015M4t−603, 105M ′ 4t−6 13 , 105M ′ 4t−6 13..., (r− times), 0 15M4t−603] for P4r+1⊙ L4 4,4t is applied. There- fore x0= 2r + 1,x1= 2r, a0=a1= 2r, y0= 2t + 1,y1= 2t + 2,b0= 8t − 1,b1= 8t − 2,y′0= 2t + 2,y′1= 2t + 1,b′0= 8t − 1 and b′1= 8t − 2. So, |v0 − v1| = 0 and |e0 − e1| = 0. For the case P4r+1⊙L4 4,4, the labeling [M4r+1;010103, 1010 13, 010103, 1010 13, ..., (r − times), 1010 13] is sufficient and thus P4r+1⊙L4 4,4t is cordial. subcase 1.3.k ≡ 2(mod4) Let k= 4r+2,r ≥ 0 andm= 4t, t> 1. Then, the labelling [L4r10;0 15M4t−603, 0 15M4t−603, 105M ′ 4t−6 13, 105M ′ 4t−6 13..., (r−times), 105M ′ 4t−6 13, 0 15M4t−603] for P4r+2 ⊙ L4 4,4t is ap- plied. Therefore x0=x1= 2r+1,a0= 2r+1,a1= 2r, y0= 2t+1,y1= 2t+2,b0= 8t−1,b1= 8t− 2,y′0= 2t+2,y′1= 2t+1,b′0= 8t−1 and b′1= 8t−2. So, |v0−v1| = 0 and |e0−e1| = 1. For the case P4r+2⊙L4 4,4, the labeling [M4r+2;010103, 1010 13, 010103, 1010 13, ..., (r−times), 1010 13, 010103] is sufficient and thus P4r+2⊙L4 4,4t is cordial. subcase 1.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t, t> 1. Then, the labelling [L4r001;0 15M4t−603, 0 15M4t−603, 105M ′ 4t−6 13, 105M ′ 4t−6 13..., (r−times), 0 15M4t−603, 0 15M4t−603, 105M ′ 4t−6 13] for P4r+3⊙ L4 4,4t is applied. Therefore x0= 2r + 2,x1= 2r + 1,a0=a1= 2r + 1,y0= 2t + 1,y1= 2t + 2,b0= 8t− 1,b1= 8t− 2,y′0= 2t+ 2,y′1= 2t+ 1,b′0= 8t− 1 and b′1= 8t− 2. So, |v0 − v1| = 0 A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 9 of 14 and |e0 − e1| = 0. For the case P4r+3⊙L4 4,4, the labeling [M4r+3;010103, 1010 13, 010103, 1010 13, ..., (r − times), 010103, 1010 13, 1010 13] is sufficient and thus P4r+3⊙L4 4,4t is cor- dial. Case 2. At m ≡ 1(mod4), we consider the following subcases. subcase 2.1.k ≡ 0(mod4) Let k= 4r, r ≥ 1 andm= 4t+1, t> 1. Then, the labelling [M4r;010 12S ′ 4t−403, 10102L ′ 4t 13, 010 12S ′ 4t−403, 10102L ′ 4t 13, ..., (r−times)] for P4r⊙ L4 4,4t+1 is applied. Therefore x0=x1= 2r, a0= 0,a1= 4r − 1,y0= 2t + 3,y1= 2t + 1,b0= 8t, b1= 8t + 1,y′0= 2t + 1,y′1= 2t + 3,b′0= 8t and b′1= 8t + 1. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r⊙L4 4,5, the la- beling [M4r;03 12021,03 12021, 1302 120, 1302 120, ..., (r − times)] is sufficient and thus P4r⊙L4 4,4t+1 is cordial. subcase 2.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t+1, t> 1. Then, the labelling [M4r+1;010 12S ′ 4t−403, 10102L ′ 4t 13, 010 12S ′ 4t−403, 10102L ′ 4t 13, ..., (r − times), 010 13L ′ 4t−403] for P4r+1⊙ L4 4,4t+1 is applied. Therefore x0= 2r+1,x1= 2r, a0= 0,a1= 4r, y0= 2t+3,y1= 2t+1,b0= 8t, b1= 8t+1,y′0= 2t+ 1,y′1= 2t+3,b′0= 8t , b′1= 8t+1,y”0= 2t+2,y”1= 2t+2,b”0= 8t and b”1= 8t+1. So, |v0−v1| = 1 and |e0 − e1| = 1. For the case P4r+1⊙L4 4,5, the labeling [M4r;03 12021,03 12021, 1302 120, 1302 120, ..., (r − times), 010 140] is sufficient and thus P4r+1⊙L4 4,4t+1 is cordial. subcase 2.3.k ≡ 2(mod4) Let k= 4r+2,r ≥ 0 andm= 4t, t> 1. Then, the labelling [M4r+1;010 12S ′ 4t−403, 10102L ′ 4t 13, 010 12S ′ 4t−403, 10102L ′ 4t 13, ..., (r − times), 010 12S ′ 4t−403, 10102L ′ 4t 13] for P4r+2⊙L4 4,4t is applied. Therefore x0=x1= 2r+1,a0= 0,a1= 4r+1,y0= 2t+3,y1= 2t+1,b0= 8t, b1= 8t+ 1,y′0= 2t+1,y′1= 2t+3,b′0= 8t and b′1= 8t+1. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r+2⊙L4 4,5, the labeling [L4r01;03 12021,03 12021, 1302 120, 1302 120, ..., (r− times), 03 12021,03 12021] is sufficient and thus P4r+2⊙L4 4,4t+1 is cordial. subcase 2.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t+1, t> 1. Then, the labelling [M4r+1;010 12S ′ 4t−403, 10102L ′ 4t 13, 010 12S ′ 4t−403, 10102L ′ 4t 13, ..., (r − times), 010 13L ′ 4t−403] for P4r+3⊙ L4 4,4t+1 is applied. Therefore x0= 2r + 2,x1= 2r + 1,a0= 0,a1= 4r + 2,y0= 2t+ 3,y1= 2t+ 1,b0= 8t, b1= 8t+ 1,y′0= 2t+1,y′1= 2t+3,b′0= 8t , b′1= 8t+1,y”0= 2t+2,y”1= 2t+2,b”0= 8t and b”1= 8t+1. So, |v0−v1| = 1 and |e0−e1| = 1. For the case P4r+3⊙L4 4,5, the labeling [M4r;03 12021, 1302 120, 03 12021, 1302 120, ..., (r−times),03 12021, 1302 120, 010 140]is sufficient and thus P4r+3⊙L4 4,4t+1 is cordial. Case 3. At m ≡ 2(mod4), we consider the following sub subcases. subcase 3.1.k ≡ 0(mod4) Let k= 4r, r ≥ 1 andm= 4t+2, t> 1. Then, the labelling [L4r;0 15M4t−60310, 0 15M4t−60310, 105M ′ 4t−6 1301, 105M ′ 4t−6 1301, ..., (r−times)] for P4r⊙L4 4,4t+2 is applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r−1,y0= 2t+2, y1= 2t+3,b0= 8t+3,b1= 8t+2,y′0= 2t+3,y′1= 2t+2,b′0= 8t+3 and b′1= 8t + 2. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r⊙L4 4,6, the labeling [L4r;03 1402, 03 1402, 1304 12, 1304 12, ..., (r− times)] is sufficient and thus P4r⊙L4 4,4t+2 is cordial. subcase 3.2.k ≡ 1(mod4) A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 10 of 14 Let k= 4r+1,r ≥ 0 andm= 4t+2, t> 1. Then, the labelling [L4r0;015M4t−60310, 015 M4t−60310, 105M ′ 4t−6 1301, 105M ′ 4t−6 1301, ..., (r−times), 0 15M4t−60310] for P4r+1⊙L4 4,4t+2 is applied. Therefore x0= 2r+1,x1= 2r, a0=a1= 2r, y0= 2t+2,y1= 2t+3,b0= 8t+3,b1= 8t+2,y′0= 2t+ 3,y′1= 2t + 2,b′0= 8t + 3 and b′1= 8t + 2. So, |v0 − v1| = 0 and |e0 − e1| = 0. For the case P4r+1⊙L4 4,6, the labeling [L4r0;03 1402, 03 1402, 1304 12, 1304 12, ..., (r − times),03 1402] is sufficient and thus P4r+1⊙L4 4,4t+2 is cordial. subcase 3.3.k ≡ 2(mod4) Let k= 4r+2,r ≥ 0 andm= 4t+2, t> 1. Then, the labelling [L4r10;015M4t−60310, 015M4t−60310, 105M ′ 4t−6 1301, 105M ′ 4t−6 1301, ..., (r−times), 105M ′ 4t−6 1301, 0 15M4t−60310] for P4r+2⊙L4 4,4t+2 is applied. Therefore x0=x1= 2r + 1,a0= 2r + 1,a1= 2r, y0= 2t + 2,y1= 2t + 3,b0= 8t + 3,b1= 8t+2,y′0= 2t+3,y′1= 2t+2,b′0= 8t+3 and b′1= 8t+2. So, |v0−v1| = 0 and |e0−e1| = 1. For the case P4r+2⊙L4 4,6, the labeling [L4r01;03 1402, 03 1402, 1304 12, 1304 12, ..., (r − times),03 1402, 1304 12] is sufficient and thus P4r+2⊙L4 4,4t+2 is cordial. subcase 3.4.k ≡ 3(mod4) Let k= 4r + 3,r ≥ 0 and m= 4t+ 2, t> 1. Then, the labelling [L4r021;0 15M4t−60310, 015M4t−60310, 105M ′ 4t−6 1301, 105M ′ 4t−6 1301, ..., (r−times), 0 15M4t−60310, 0 15M4t−60310, 105M ′ 4t−6 1301] for P4r+3⊙ L4 4,4t+2 is applied. Therefore x0= 2r+2,x1= 2r+1,a0=a1= 2r+ 1,y0= 2t+2,y1= 2t+3,b0= 8t+3,b1= 8t+2,y′0= 2t+3,y′1= 2t+2,b′0= 8t+3 and b′1= 8t+2. So, |v0−v1| = 1 and |e0−e1| = 0. For the case P4r+2⊙L4 4,6, the labeling [L4r001; 1304 12, 1304 12, 03 1402, 03 1402, ..., (r−times), 1304 12, 1304 12, 03 1402] is sufficient and thus P4r+3⊙L4 4,4t+2 is cordial. Case 4. At m ≡ 3(mod4), we consider the following sub subcases. subcase 4.1.k is even. Let k= 2r, r ≥ 1 andm= 4t+3, t ≥ 1. Then, the labelling [M2r;103 12L4t−40102, 1010M4t 12, ..., (r−times)] for P2r⊙ L4 4,4t+3 is applied. Therefore x0=x1=r, a0= 0,a1= 2r−1,y0= 2t+ 4,y1= 2t+2,b0= 8t+5,b1= 8t+4,y′0= 2t+2,y′1= 2t+4,b′0= 8t+3 and b′1= 8t+6. Hence, |v0 − v1| = 0 and |e0 − e1| = 1 and thus P2r⊙L4 4,4t+3 is cordial. subcase 4.2.k is odd. Let k= 2r+1,r ≥ 0 and m= 4t+3, t ≥ 1. Then, the labelling [M2r+1;103 12L4t−40102, 1010M4t 12, ..., (r − times), 103M4t 12] for P2r+1⊙L4 4,4t+3 is applied. Therefore x0=r + 1,x1=r, a0= 0,a1= 2r, y0= 2t + 4,y1= 2t + 2,b0= 8t + 5,b1= 8t + 4,y′0= 2t + 2,y′1= 2t + 4,b′0= 8t+3,b′1= 8t+6,y”0= 2t+3,y”1= 2t+3,b”0= 8t+4 and b”1= 8t+5. Hence, |v0−v1| = 1 and |e0 − e1| = 1. Thus P2r+1⊙L4 4,4t+3 is cordial and the lemma follows.■ Lemma 4. Pk⊙L4 5,m is cordial for all k ≥ 1 and m ≥ 3 . Proof . We need to examine the following two cases : Case 1. At m ≡ 0(mod4), i.e m= 4t,t ≥ 1 . We see that Pk⊙L4 5,4 and Pk⊙L4 5,4t,t ≥ 1 are cordial. This is clear since these graphs are isomorphic to Pk⊙L4 4,5 and Pk⊙L4 4t,5 respectively. So, by lemma 3, we conclude that Pk⊙L4 5,4 and Pk⊙L4 5,4t are cordial. Case 2. At m ≡ 1(mod4), we consider the following subcases. subcase 2.1.k ≡ 0(mod4) Let k= 4r, r ≥ 1 andm= 4t+1, t> 1. Then, the labelling [L4r;014M4t−4031, 0 14M4t−4031, A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 11 of 14 104M ′ 4t−4 130, 104M ′ 4t−4 130, ..., (r−times)] for P4r⊙ L4 5,4t+1 is applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r−1,y0= 2t+2,y1= 2t+3,b0= 8t−3,b1= 8t+2,y′0= 2t+3,y′1= 2t+2,b′0= 8t+3 and b′1= 8t + 2. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r⊙L4 5,5, the labeling [M4r;105 13, 0 1503105, 13, 0 1503, ..., (r− times)] is sufficient and thus P4r⊙L4 5,4t+1 is cor- dial. subcase 2.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t+1, t> 1. Then, the labelling [L4r0;0 14M4t−4031, 0 14M4t−4031, 104M ′ 4t−4 130, 104M ′ 4t−4 130, ..., (r − times), 0 14M4t−4031] for P4r+1⊙ L4 5,4t+1 is applied. Therefore x0= 2r+1,x1= 2r, a0=a1= 2r, y0= 2t+2,y1= 2t+3,b0= 8t−3,b1= 8t+2,y′0= 2t+ 3,y′1= 2t + 2,b′0= 8t + 3 and b′1= 8t + 2. So, |v0 − v1| = 0 and |e0 − e1| = 0. For the case P4r+1⊙L4 5,5, the labeling [M4r+1;105 13, 0 1503, 105 13, 0 1503, ..., (r−times), 0 1503] is suf- ficient and thus P4r+1⊙L4 5,4t+1 is cordial. subcase 2.3.k ≡ 2(mod4) Let k= 4r+2,r ≥ 0 andm= 4t+1, t> 1. Then, the labelling [L4r01;0 14M4t−4031, 0 14M4t−4031, 104M ′ 4t−4 130, 104M ′ 4t−4 130, ..., (r−times), 0 14M4t−4031, 104M ′ 4t−4 130] for P4r+2⊙ L4 5,4t+1 is applied. Therefore x0=x1= 2r + 1,a0= 2r, a1= 2r + 1,y0= 2t + 2,y1= 2t + 3,b0= 8t − 3,b1= 8t+ 2,y′0= 2t+ 3,y′1= 2t+ 2,b′0= 8t+ 3 and b′1= 8t+ 2. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P4r+2⊙L4 5,5, the labeling [M4r+2;105 13, 0 1503, 105 13, 0 1503, ..., (r− times), 105 13, 0 1503] is sufficient and thus P4r+2⊙L4 5,4t+1 is cordial. subcase 2.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t+1, t> 1. Then, the labelling [L4r001;0 14M4t−4031, 0 14M4t−4031, 104M ′ 4t−4 130, 104M ′ 4t−4 130, ..., (r−times), 014M4t−4031, 0 14M4t−4031, 104M ′ 4t−4 130] for P4r+3⊙L4 5,4t+1 is applied. Therefore x0= 2r+2,x1= 2r+ 1,a0=a1= 2r+1,y0= 2t+2,y1= 2t+3,b0= 8t−3,b1= 8t+2,y′0= 2t+3,y′1= 2t+2,b′0= 8t+3 and b′1= 8t+ 2. So, |v0 − v1| = 0 and |e0 − e1| = 0. For the case P4r+3⊙L4 5,5, the labeling [M4r+3;105 13, 0 1503, 105 13, 0 1503, ..., (r− times), 0 1503, 105 13, 0 1503] is sufficient and thus P4r+3⊙L4 5,4t+1 is cordial. Case 3. At m ≡ 2(mod4), we consider the following sub subcases. subcase 3.1.k is even. Let k= 2r, r ≥ 1 and m= 4t + 2, t> 1. Then, the labelling [M2r;03 140103M ′ 4t−6, 13040 13M4t−6, ..., (r − times)] for P2r⊙ L4 5,4t+2 is applied. There- fore x0=x1=r, a0= 0,a1= 2r − 1,y0= 2t + 4,y1= 2t + 2,b0= 8t + 4,b1= 8t + 5,y′0= 2t + 2,y′1= 2t + 4,b′0= 8t + 4 and b′1= 8t + 5. So, |v0 − v1| = 0 and |e0 − e1| = 1. For the case P2r⊙L4 5,6, the labeling [M2r;03 1304, 1303 14, ..., (r − times)] is sufficient and thus P2r⊙L4 5,4t+2 is cordial. subcase 3.2.k is odd. Let k= 2r + 1,r ≥ 0 and m= 4t+ 2, t> 1. Then, the labelling [M2r+1;03 140103M ′ 4t−6, 13040 13M4t−6 , ..., (r − times), 103L4t 12] for P2r+1⊙L4 5,4t+2 is applied. Therefore x0=r+1,x1=r, a0= 0,a1= 2r, y0= 2t+4,y1= 2t+2,b0= 8t+4,b1= 8t+ 5,y′0= 2t+2,y′1= 2t+4,b′0= 8t+4,b′1= 8t+5,y”0= 2t+3,y”1= 2t+3,b”0= 8t+5 and b”1= 8t+4. So, |v0 − v1| = 1 and |e0 − e1| = 1. For the case P2r+1⊙L4 5,6, the labeling [M2r+1;03 1304, 1303 14, ..., (r − times), 105 14] is sufficient and thus P2r+1⊙L4 5,4t+2 is cordial. Case 4. At m ≡ 3(mod4), we consider the following sub subcases. subcase 4.1.k ≡ 0(mod4) Let k= 4r, r ≥ 1 and m= 4t + 3, t ≥ 1. Then, the labelling [L4r;0 14M ′ 4t02, 0 14 A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 12 of 14 M ′ 4t02, 104M4t 12, 104M4t 12, ..., (r−times)] for P4r⊙ L4 5,4t+3 is applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r−1,y0= 2t+3,y1= 2t+4,b0= 8t+7,b1= 8t+6,y′0= 2t+4,y′1= 2t+3,b′0= 8t+7 and b′1= 8t+ 6. So, |v0 − v1| = 0 and |e0 − e1| = 1. Thus P4r⊙L4 5,4t+3 is cordial. subcase 4.2.k ≡ 1(mod4) Let k= 4r+1,r ≥ 0 andm= 4t+3, t ≥ 1. Then, the labelling [L4r0;0 14M ′ 4t02, 0 14M ′ 4t02, 104M4t 12, 104M4t 12, ..., (r − times), 0 14M ′ 4t02] for P4r+1⊙ L4 5,4t+3 is applied. Therefore x0= 2r+1,x1= 2r, a0=a1=2r, y0= 2t+3,y1= 2t+4,b0= 8t+7,b1= 8t+6,y′0= 2t+4,y′1= 2t+ 3,b′0= 8t + 7 and b′1= 8t + 6. So, |v0 − v1| = 0 and |e0 − e1| = 0. Thus P4r+1⊙L4 5,4t+3 is cordial. subcase 4.3.k ≡ 2(mod4) Let k= 4r+2,r ≥ 0 andm= 4t+3, t ≥ 1. Then, the labelling [L4r01;0 14M ′ 4t02, 014M ′ 4t02, 104M4t 12, 104M4t 12, ..., (r − times), 0 14M ′ 4t02, 104M4t 12] for P4r+2⊙L4 5,4t+3 is applied. Therefore x0=x1= 2r + 1,a0= 2r, a1= 2r + 1,y0= 2t + 3,y1= 2t + 4,b0= 8t + 7,b1= 8t + 6,y′0= 2t+4,y′1= 2t+3,b′0= 8t+7 and b′1= 8t+6. So, |v0−v1| = 0 and |e0− e1| = 1. Thus P4r+2⊙L4 5,4t+3 is cordial. subcase 4.4.k ≡ 3(mod4) Let k= 4r+3,r ≥ 0 andm= 4t+3, t ≥ 1. Then, the labelling [L4r001;0 14M ′ 4t02, 014M ′ 4t02, 104M4t 12, 104M4t 12, ..., (r − times), 0 14M ′ 4t02, 0 14M ′ 4t02, 104M4t 12] for P4r+3⊙L4 5,4t+3 is applied. Therefore x0=x1= 2r+1,a0=a1= 2r+1,y0= 2t+3,y1= 2t+4,b0= 8t+7,b1= 8t+ 6,y′0= 2t+4,y′1= 2t+3,b′0= 8t+7 and b′1= 8t+6. So, |v0− v1| = 1 and |e0− e1| =0. Thus P4r+3⊙L4 5,4t+3 is cordial and the lemma follows.■ Lemma 5. Pk⊙L4 6,m; is cordial for all m, k. Proof. Let n = 6, then we need to examine the following cases : Case 1. At m ≡ 0(mod4), i.e m= 4t,t ≥ 1 . We see that Pk⊙L4 6,4 and Pk⊙L4 6,4t,t ≥ 1 are cordial. This is clear since these graphs are isomorphic to Pk⊙L4 4,6 and Pk⊙L4 4t,6 respectively. So, by lemma 3, we conclude that Pk⊙L4 6,4 and Pk⊙L4 6,4t are cordial. Case 2. At m ≡ 1(mod4), i.e m= 4t+ 1,t ≥ 1. We see that Pk⊙L4 6,5 and Pk⊙L4 6,4t+1,t ≥ 1 are cordial. This is clear since these graphs are isomorphic to Pk⊙L4 5,6 and Pk⊙L4 4t+1,6 respectively. So, by lemma 4, we conclude that Pk⊙L4 6,5 and Pk⊙L4 6,4t+1 are cordial. Case 3. At m ≡ 2(mod4), we need to examine the following two subcases: subcase 3.1.k even Let k= 2r, r ≥ 1 andm= 4t+3, t ≥ 1. Then, the labelling [M2r; 1205M4t 1201, 1205M4t 1201, 02 15M ′ 4t0210,02 15M ′ 4t0210, ..., (r−times)] for P2r⊙L4 6,4t+2 is applied. Therefore x0=x1=r, a0= 0,a1= 2r− 1,y0= 2t+4,y1= 2t+3,b0=b1= 8t+7,y′0= 2t+3,y′1= 2t+4,b′0= 8t+7 and b′1= 8t+ 7. So, |v0 − v1| = 0 and |e0 − e1| = 1. Thus P2r⊙L4 6,4t+2 is cordial. subcase 3.2.k odd Let k= 2r + 1,r ≥ 0 and m= 4t+ 3, t ≥ 1. Then, the labelling [M2r+1; 1205M4t 1201, 1205M4t 1201, 02 15M ′ 4t0210,02 15M ′ 4t0210, ..., (r−times),02 15M ′ 4t0210] for P2r+1⊙L4 5,4t+3 is applied. Therefore x0=r + 1,x1=r, a0= 0,a1= 2r, y0= 2t + 4,y1=2t + 3,b0=b1= 8t + 7,y′0=y”0= 2t + 4,y′1=y”1= 2t + 3,b′0=b”0= 8t + 7 and b′1=b”1= 8t + 7. So, |v0 − v1| = 0 and A. Abd El-hay et al. / Eur. J. Pure Appl. Math, 18 (1) (2025), 5470 13 of 14 |e0 − e1| = 0. Thus P2r+1⊙L4 6,4t+3 is cordial. Case 4. At m ≡ 3(mod4), we need to examine the following subcases: subcase 4.1. At k ≡ 0(mod4) . Let k= 4r, r ≥ 1. Then, the labeling [L4r;S403M4t0, S403M4t0, S403M4t0, S403M4t0, ..., (r− time)] for P4r⊙L4 6,4t+3 is applied. Therefore x0=x1= 2r, a0= 2r, a1= 2r − 1,y0=y1= 2t + 4,b0=b1= 8t+ 9,y′0=y′1= 2t+ 4 and b′0=b′1= 8t+ 9. Consequently, it is easy to show that |v0 − v1| = 0 and |e0 − e1| = 1. Thus P4r⊙L4 6,4t+3, is cordial. subcase 4.2. At k ≡ 1(mod4) . Let k= 4r+1, r ≥ 0. Then one can choose the labeling [L4r0;S403M4t0, S403M4t0, S403M4t0, S403M4t0, ..., (r−time),S403M4t0] for P4r+1⊙L4 6,4t+3. Therefore x0= 2r+1,x1= 2r, a0=a1= 2r, y0=y1= 2t + 4,b0=b1= 8t + 9,y′0=y′1= 2t + 4 and b′0=b′1= 8t + 9. So, |v0 − v1| = 1 and |e0 − e1| = 0. Thus P4r+1⊙L4 6,4t+3, is cordial. subcase 4.3. At k ≡ 2(mod4) . Let k= 4r+2, r ≥ 0. Then one can take the labeling [L4r10;S403M4t0, S403M4t0, S403M4t0, S403M4t0, ..., (r − time),S403M4t0, S403M4t0] for P4r+2⊙L4 6,4t+3. Therefore x0=x1= 2r + 1,a0= 2r+1,a1= 2r, y0=y1= 2t+4,b0=b1= 8t+9,y′0=y′1= 2t+4 and b′0=b′1= 8t+9. Hence, |v0 − v1| = 0 and |e0 − e1| = 1. Thus P4r+2⊙L4 6,4t+3, is cordial. subcase 4.4. At k ≡ 3(mod4) . Let k= 4r+3, r ≥ 0. Then one can select the labeling [L4r001;S403M4t0, S403M4t0, S403M4t0, S403M4t0, ..., (r−time),S403M4t0, S403M4t0, S403M4t0] for P4r+3⊙L4 6,4t+3. Therefore x0= 2r+ 2,x1= 2r+1,a0=a1= 2r+1,y0=y1= 2t+4,b0=b1= 8t+9,y′0=y′1= 2t+4 and b′0=b′1= 8t+9. Consequently, it is easy to show that |v0 − v1| = 1 and |e0 − e1| = 0. Thus P4r+3⊙L4 6,4t+3, is cordial and the lemma follows.■ 4. Conclusion We proved that the corona Pk⊙L4 n,m between paths Pk and fourth power of lemniscate graphs L4 n,m is cordial for al k ≥ 1, n,m ≥ 3. In the future, we will apply cordial labeling to other types of graphs. 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