EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5569 ISSN 1307-5543 – ejpam.com Published by New York Business Global Upper Bound of Radio Span for Shadow-path Network and Its Mathematical Modeling Nawal M. NourEldeen1,2, Elsayed Badr3,4,5,Hanan M. Shabana6,∗, M. E. Abdel-Aal7 1 Department of Mathematics, College of Science, Taibah University, Madinah, Kingdom of Saudi Arabia 2 Department of Mathematics, Women’s College of Arts, Science and Education, Ain Shams University, Egypt 3 Department of Information Systems, College of Information Technology, Misr University for Science and Technology (MUST), P.O. BOX 77, Giza, Egypt 4 Scientific Computing Department, Faculty of Computers and Artificial Intelligence, Benha University,Benha, Egypt 5 The Egyptian School of Data Science (ESDS), Benha University, Egypt 6 Physics and Engineering Mathematics Department, Faculty of Electronic Engineering, Menoufia University, Menouf, 32952, Egypt 7 Department of Mathematics, Faculty of Science, Benha University, Benha 13518, Egypt Abstract. Motivated by the frequency assignment problem, we investigate radio labeling of graphs. In graph theory and discrete mathematics, radio labeling of graphs has received sig- nificant attention as it is of immense importance for numerous applications to a wide range of areas such as circuit and sensor network, signal processing, design, frequency assignment in mobile communication systems, etc. An assignment of labels satisfying specific constraints to the edges, vertices, or both of graph G is known as graph labeling. Radio labeling of a graph G is a tech- nique of labeling vertices of G by non-negative integers. Hence, radio labeling problem presents an efficient graph modeling for the frequency assignment problem. In radio labeling of a graph G, the maximum label used for labeling vertices is called the span of that radio labeling. The minimum span from all radio labelings of G is known as the radio number of G. That radio number reflects the efficient usage of the available frequencies in frequency assignment for a network modeled by the graph G. This paper contributes the mathematical proof (Theorems 1-4), an integer linear programming model for finding the upper bound of radio number of shadow graphs. It also intro- duces an application of radio labeling of graphs in cryptography. Additionally, a computational study has been conducted wherein it demonstrates that our results (Theorems 1-4) outperform both the mathematical model, and the results published in the literature. 2020 Mathematics Subject Classifications: 05C78, 05C12, 05C15 ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5569 Email addresses: neldeen@taibahu.edu.sa (N. M. NourEldeen), Elsayed.Badr@must.edu.eg (E. Badr), hananshabana22@gmail.com (H. M. Shabana), mohamed.abdelghani@fsc.bu.edu.eg (M. E. Abdel-Aal) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 2 of 25 Key Words and Phrases: Frequency(channel) assignment problem, Radio labeling, Integer linear programing, Radio number, Cryptography. An overview of the notations that will be applied throughout the paper, listed in Table 1. Table 1: Table of Notations. Parameter Description V (G) The set of vertices of the graph G. E (G) The set of edges of the graph G. d (u, v) The distance between two vertices u, v in a graph. diam (G) The diameter of G. Pn Path of n vertices. Cn Cycle with n vertices. rn (G) The radio number of G. FAP Frequency assignment problem D2 (Pn) The shadow path graph ILPM Integer Linear Programming Model 1. Introduction The communication in wireless networks depends on the frequencies (channels) allotted to them. For an effective communication, the frequencies should assigned to transmitters in such a way that avoids the interference. Here, the interference is closely related to the geographical position of the transmitting stations. The interference is increasing whenever the stations are geographical close. To avoid such interference, we have to assign frequen- cies to stations such that difference between the assigned channels must be large enough. The process of assigning a limited number of available frequencies to transmitters in radio networks with avoiding interference is known as the frequency assignment problem (FAP). Wireless networks appear in a variety of services such as T.V. and radio broadcasting, Engineering, Economics, military services, and many other. Due to increasing popularity of wireless services and the limited available frequencies, FAP has a lot of interest from the scientific and the business communities. As the optimizing usage of the available frequencies means higher traffic capacity, more bandwidth and bigger coverage for the existing radio networks. As a result, many researchers and a wide variety of models have investigated FAPs and solution techniques have been proposed. The labeling technique in graph theory has played an important role in solving FAP; thereby the time and cost will be saved. For modeling FAP in graph theory, an interference graph G = (V (G) , E (G))is constructed where V (G) and E (G) are the set of vertices and the set of edges of G respec- tively. This graph represents the interference between transmitters. Each vertex belong to V (G) represents a unique transmitter. In the event of the broadcasting of the two transmitters may interfere, then their corresponding vertices from V (G) are joined by an N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 3 of 25 edge. Positive integers are used as labels for the frequency channels. Therefore, FAP [1] is equivalent to the vertex coloring (labeling) issue of the graph G with certain labeling con- straints. In [2] author examined a comprehensive overview of graph labeling. Griggs and Yeh [3] presented the first graph labeling technique, known as L(2, 1) labeling or distance two labeling to address the frequency assignment problem. The graph labeling L(2, 1) of a simple graph G is a non-negative real-valued function L : V (G) → [0, ∞) where V (G) is the vertex set of the graph G. Whenever u and v are two vertices in G such that the distance between u and v is 1, then | L(u) − L(v) | ≥ 2, and whenever the vertices u and v are of distance 2 from each other, then | L(u) − L(v) | ≥ 1|. Another graph labeling technique called radio Labeling was introduced by Chartrand et al. [4]. The radio labeling problem of a connected graph G = (V (G) , E (G))is denoted as follows. Let u, v ∈ V (G). The distance d (u, v) denote the length of the shortest path between u, v. The diameter diam (G) of G is defined as the maximum distance between any two vertices in G. Thus, diam (G) = max {d (u, v) : u, v ∈ V (G) }. A radio labeling of G is an injective function f from V (G) to N = {0, 1, 2, 3, · · · }, satisfying the following constraints |f (u)− f (v)| ≥ diam (G) + 1− d (u, v). For all u, v ∈ V (G) . The integer f (u) is said to be the color (label) of u under f and, the span of f is denoted by span (f) = max {|f (u) − f (v)| : u, v ∈ V (G)} . The radio number of G, or (rn (G)), is defined as rn (G) = min {span (f)} for all radio labeling f of G. The objective of radio labeling problem of G is finding the value of rn (G). 2. Related Work From viewpoint of complexity, the problem of getting the radio number of a given graph is NP complete [5]. Saha and Panigrahi [6] presented an algorithm which finds the upper bounds of rn ( G) for a given graph G. In [7] Badr and Moussa suggested a developing algorithm for one presented in [6]. Besides that they gave an approach for solving the radio labeling problem using mathematical modeling. Recently radio labeling of graphs and finding the exact or the upper bounds of radio number of graphs has a lot of attention. In [8] an assignment of radio numbers to triangular networks and rhombic honeycomb networks is discussed. The study of the radio labeling of fuzzy graphs is N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 4 of 25 studied in [9]. In addition, the study of the radio labeling for many different types of graphs has been the subject of intellectual research by several authors [10–23]. Many authors have found the radio labeling and radio labeling of path graphs, the middle graph of the path, a strong product of the path graph, the cross product of the path, and the corona product of the path sub division of paths,etc. This study aims to fill the research gaps by determining the radio number for the shadow of path graph. In this paper, we are concerned with finding an upper bound for radio number rn ( G) of networks modeled by the shadow of path graph. We present a theoretical approach for getting an upper bound of radio number of the shadow of path graph. Moreover, modeling the problem of radio labeling of the shadow path graph using integer linear programing is proposed. Besides that, an experimental study is carried out in order to demonstrate our approach’s effectiveness in comparison to previously published results. Overall, our investigation demonstrates that, in terms of the running time and radio numbers’ upper bound, the suggested results perform better than the earlier findings. The rest of the paper is structured as follows. Section 3 introduces our theoretical approach for finding an upper bound of radio number of the shadow of path graph. The modeling of radio labeling of shadow of path graph using integer linear programing is pre- sented in section 4. Section 5, gives the computational study for comparing our theoretical approach, integer programing model and previous algorithms known in the literature. The application of radio labeling technique in cryptography is introduced in Section 6. The conclusion of work is presented in section 7. 3. Theoretical approach for upper bound of radio number of shadow networks Hereafter, we are interested in getting an upper bound of radio number of shadow networks. Definition 1. Let Pn be the path with n vertices. The shadow graph D2(Pn) is constructed by taking two copies of Pn. Join each vertex u in the first path to the neighbors of the corresponding vertex v in the second path as shown in Figure 1. Figure 1: The graph D2(Pn). In D2(Pn), the set of vertices is V (D2(Pn)) = {ui, vi : 1 ≤ i ≤ n} and the set of edges N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 5 of 25 is defined as: E(D2(Pn)) = {uiui+1, vivi+1 vi : 1 ≤ i ≤ n− 1} ∪ {uivi+1, viui+1 vi : 1 ≤ i ≤ n− 1}. For convenience, we divided all vertices of D2(Pn) into two subsets Vu and Vv. Where Vu = {ui : 1 ≤ i ≤ n} and Vv = {vi : 1 ≤ i ≤ n}. Hence, the graph D2(Pn) has a set of 2n vertices; that is |Vu| = |Vv| = n and the diameter of D2(Pn) equals n− 1, n > 2. In the following, we investigate the radio number of D2(Pn). Lemma 1. For a positive integer n = 2, rn(D2(Pn)) ≤ 4. Proof. Let (f(u1), f(u2), f(v1), f(v2)) = (0, 3, 1, 4). Lemma 2. For a positive integer n = 3, rn(D2(Pn)) ≤ 6. Proof. Let (f(u1), f(u2), f(u3), f(v1), f(v2), f(v3)) = (0, 5, 1, 2, 6, 3). Lemma 3. For a positive integer n = 4, rn(D2(Pn)) ≤ 12. Proof. For n = 4, let (f(u1), f(u2), f(u3), f(u4), f(v1), f(v2), f(v3), f(v4)) = (0, 5, 10, 1, 2, 7, 12, 3). Lemma 4. For a positive integer n = 5, rn(D2(Pn)) ≤ 22. Proof. For n = 5 Let (f(u1), f(u2), f(u3), f(u4), f(u5), f(v1), f(v2), f(v3), f(v4), f(v5)) = = (6, 13, 0, 16, 7, 9, 19, 3, 22, 10) In the next theorems, we find upper bound of rn(D2(Pn)) for n ≥ 6. Theorem 1. Let n ≥ 6 be a positive integer such that n ≡ 2(mod4). Then rn(D2(Pn)) ≤ n2 − n 2 . Proof. For k ≥ 1, n ≡ 2(mod4) such that n = 4k+2, let f : V (D2(Pn)) → N define as follows: f(ui) =  (2n+ 3)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k + 1 f(uk+1)− (2n− 1)− (3n− 5)(j − 1) + 2(j − 1)(j − 2), if k + 2 ≤ i ≤ 2k + 1 where 1 ≤ j ≤ k, and i = k + 1 + j f(uk+1+j) + (6k + 1)− 2(j − 1), if 2k + 2 ≤ i ≤ 3k + 1, where 1 ≤ j ≤ k, and , i = 3k + 2− j f(uj) + 2j − 1, if 3k + 2 ≤ i ≤ 4k + 2 = n, where 1 ≤ j ≤ k + 1, and i = 4k + 3− j N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 6 of 25 And for the vertices belongs to Vv, f(vi) =  (n+ 2) + (2n+ 7)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k, f(vk) + (2n+ 1) + (n+ 1)(j − 1) + 2(j − 1)(j − 2), if k + 1 ≤ i ≤ 2k + 1, where 1 ≤ j ≤ k + 1, andi = k + j f(vk+j) + (2k + 1) + 2(j − 1), if 2k + 2 ≤ i ≤ 3k + 2, where 1 ≤ j ≤ k + 1, and i = 3k + 3− j f(vj)− 2j + 1, if 3k + 3 ≤ i ≤ 4k + 2 = n, where 1 ≤ j ≤ k, and i = 4k + 3− j Now we claim to prove that the function f given above is a radio labelling of D2(Pn) Consequently, It is required that f verifies the following condition |f(x)− f(y)| ≥ n− d(x, y) for any two distinct vertices x, y ∈ V (D2(Pn)). case 1 x, y ∈ Vu. If {x, y} = {u1, un} then |f(x)− f(y)| = |f(u1)− f(un)| = |f(u1)− [f(u1) + 2− 1]| = 1 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. case 2 If {x, y} = {u1, uk+1} then |f(x)− f(y)| = |f(u1)− f(uk+1)| = |f(u1)− [(2n+ 3)(k) + 2k(k − 1)]| = | − [(2n+ 3)(k) + 2k(k − 1)]| = (2n+ 3)(k) + 2k(k − 1) ≥ n− k, ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = k case 3 If {x, y} = {u1, u2k+1} then d(x, y) = 2k |f(x)− f(y)| = |f(u1)− f(u2k+1)| = |f(u1)− [f(uk+1)− (2n− 1)− (3n− 5)(k − 1) + 2(k − 1)(k − 2)]| = |f(uk+1)− (2n− 1)− (3n− 5)(k − 1) + 2(k − 1)(k − 2)| = |(2n+ 3)(k) + 2k(k − 1)− (2n− 1)− (3n− 5)(k − 1) + 2(k − 1)(k − 2)| ≥ n− 2k case 4 If {x, y} = {u1, u3k+1} then d(x, y) = 3k |f(x)− f(y)| = |f(u1)− f(u3k+1)| = |f(u1)− [f(uk+2) + (6k + 1)]| = |f(uk+2) + (6k + 1)| N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 7 of 25 = |f(uk+1)− (2n− 1) + (6k + 1)| = |(2n+ 3)(k) + 2k(k − 1)− (2n− 1) + (6k + 1)| ≥ n− 3k case 5 x, y ∈ Vv. If {x, y} = {v1, vn} then |f(x)− f(y)| = |f(v1)− f(vn)| = |f(v1)− [f(v1)− 1]| = 1 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. case 6 If {x, y} = {v1, vk} then |f(x)− f(y)| = |f(v1)− f(vk)| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2)| ≥ n, where n = 4k + 2 ≥ diam(D2(Pn)) + 1− d(x, y) case 7 If {x, y} = {v1, v2k+1} then |f(x)− f(y)| = |f(v1)− f(v2k+1)| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)+ + (2n+ 1) + (n+ 1)(k) + 2(k)(k − 1)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1) + (n+ 1)(k) + 2(k)(k − 1)| ≥ n, where n = 4k + 2, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 8 If {x, y} = {v1, v3k+2} then |f(x)− f(y)| = |f(v1)− f(v3k+2)| = |f(v1)− [f(vk+1) + (2k + 1)]| = |f(v1)− [f(vk) + (2n+ 1) + (2k + 1)]| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)+ + (2n+ 1) + (2k + 1)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1) + (2k + 1)| ≥ n− (3k + 1), where n = 4k + 2, d(x, y) = 3k + 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 9 x ∈ Vu and y ∈ Vv Let x = ui and y = vl , where 1 ≤ i, l ≤ n In this case we have several subcases: N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 8 of 25 Subcase 9.1 If i = 1 , j = k then |f(x)− f(y)| = |f(u1)− f(vk)| = |0− [(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)]| = |(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)| ≥ n− (k − 1), where , d(x, y) = k − 1 ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x)− f(y)| ≥ n. Subcase 9.2 , 1 ≤ i ≤ k + 1 and k + 1 ≤ l ≤ 2k + 1 |f(x)− f(y)| = |f(ui)− f(vl)| = |f(ui)− f(vk+j)|, where 1 ≤ j ≤ k = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [f(vk) + (2n+ 1) + (n+ 1)(j − 1)+ + 2(j − 1)(j − 2)]| = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [(n+ 2) + (2n+ 7)(k − 1)+ + 2(k − 1)(k − 2) + (2n+ 1) + (n+ 1)(j − 1) + 2(j − 1)(j − 2)]|. If i = 1 , l = k + 1 i.e. j = 1 then |f(x)− f(y)| = |f(ui)− f(vl)| = |f(ui)− f(vk+j)|, where 1 ≤ j ≤ k = |0− [f(vk) + (2n+ 1) + (n+ 1)(j − 1) + 2(j − 1)(j − 2)]| = | − [(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1)| = |(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1)| ≥ n, where n = 4k + 2, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Subcase 9.3. Suppose, 1 ≤ i ≤ k + 1 and 2k + 2 ≤ l ≤ 3k + 2 If i = 1 , l = 3k + 2 i.e. j = 1 then |f(x)− f(y)| = |f(u1)− f(v3k+2)| = |f(ui)− f(vn−k)| = |0− f(vk+1) + 2k + 1| = |(n+ 2) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2) + 2k + 1| ≥ n, where n = 4k + 2, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x)− f(y)| ≥ n. For more illustration, Figure 2 shows the radio labeling of D2(P14) according to The- orem 1. N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 9 of 25 Figure 2: Radio labeling of D2(P14) following Theorem 1. Theorem 2. Let n > 6 be a positive integer such that n ≡ 3(mod4). Then rn(D2(Pn)) ≤ n2 − 3. Proof. Let k ≥ 1, n ≡ 3(mod4) such that n = 4k+3, we define f : V (D2(Pn)) → N as follows: f(ui) =  (2n+ 3)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k + 1 f(uk+1)− (2n− 2)− (3n− 6)(j − 1) + 2(j − 1)(j − 2), if k + 2 ≤ i ≤ 2k + 1, where 1 ≤ j ≤ k, and i = k + 1 + j n2 − n− 1, if i = 2k + 2 f(uk+1+j) + (6k + 2)− 2(j − 1), if 2k + 3 ≤ i ≤ 3k + 2 where 1 ≤ j ≤ k, and i = 3k + 3− j) f(uj) + 2j − 1, if 3k + 3 ≤ i ≤ 4k + 3 = n where 1 ≤ j ≤ k + 1, and i = 4k + 4− j And for the vertices belongs to Vv, f(vi) =  (n+ 3) + (2n+ 7)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k f(vk) + (2n− 1) + n(j − 1) + 2(j − 1)(j − 2), if k + 1 ≤ i ≤ 2k + 1,where 1 ≤ j ≤ k + 1, and i = k + j n2 − 3, if i = 2k + 2 f(vk+j) + (2k + 1) + 2(j − 1), if 2k + 3 ≤ i ≤ 3k + 3 where 1 ≤ j ≤ k + 1, and i = 3k + 4− j f(vn+1−j) = f(vj)− 2j, if 3k + 4 ≤ i ≤ 4k + 3 = n where 1 ≤ j ≤ k, and i = 4k + 4− j Now we claim to prove that the function f given above is a radio labelling of D2(Pn) Consequently, It is required that f verifies the following condition with a span equal to the desired number. So, we need only to check it for n ≥ 6 , n = 4k + 3 where k ≥ 1. It remains to verify that |f(x)− f(y)| ≥ diam(D2(Pn)) + 1− d(x, y) for any two distinct vertices x, y of D2(Pn). We distinguish several cases: case 1 x, y ∈ Vu. If {x, y} = {u1, un} then |f(x)− f(y)| = |f(u1)− f(un)| = |f(u1)− [f(u1) + 2− 1]| = 1 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 10 of 25 case 2 If {x, y} = {u1, uk+1} then |f(x)− f(y)| = |f(u1)− f(uk+1)| = |f(u1)− [(2n+ 3)(k) + 2k(k − 1)]| = | − [(2n+ 3)(k) + 2k(k − 1)]| = (2n+ 3)(k) + 2k(k − 1) ≥ n ≥ diam(D2(Pn)) + 1− d(x, y) case 3 If {x, y} = {u1, u2k+1} then |f(x)− f(y)| = |f(u1)− f(u2k+1)| = |f(u1)− [f(uk+1)− (2n− 2)− (3n− 6)(k − 1) + 2(k − 1)(k − 2)]| = |f(uk+1)− (2n− 2)− (3n− 6)(k − 1) + 2(k − 1)(k − 2)| = |(2n+ 3)(k) + 2k(k − 1)− (2n− 2)− (3n− 6)(k − 1) + 2(k − 1)(k − 2)| ≥ n− 2k ≥ diam(D2(Pn)) + 1− d(x, y) case 4 If {x, y} = {u1, u2k+2} then |f(x)− f(y)| = |f(u1)− f(u2k+2)| = |f(u1)− [n2 − n− 1]| = |n2 − n− 1| ≥ n ≥ diam(D2(Pn)) + 1− d(x, y) case 5 If {x, y} = {u1, u3k+2} then |f(x)− f(y)| = |f(u1)− f(u3k+2)| = |f(u1)− [f(uk+2) + (6k + 2)]| = |f(uk+2) + (6k + 2)| = |(n2 − n− 1) + (6k + 2)| ≥ n ≥ diam(D2(Pn)) + 1− d(x, y) case 6 x, y ∈ Vv. If {x, y} = {v1, vn} then |f(x)− f(y)| = |f(v1)− f(vn)| = |f(v1)− [f(v1)− 2]| = 2 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 11 of 25 case 7 If {x, y} = {v1, vk} then |f(x)− f(y)| = |f(v1)− f(vk)| = |(n+ 3)− [(n+ 3) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2)| ≥ n, where n = 4k + 3, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 8 If {x, y} = {v1, v2k+1} then |f(x)− f(y)| = |f(v1)− f(v2k+1)| = |(n+ 3)− [(n+ 3) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)+ + (2n− 1) + (n)(k) + 2(k)(k − 1)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n− 1) + (n)(k) + 2(k)(k − 1)| ≥ n, where n = 4k + 3, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 9 If {x, y} = {v1, v2k+2} then |f(x)− f(y)| = |f(v1)− f(v2k+2)| = |f(v1)− [n2 − 3]| = |n2 − 3| ≥ n ≥ diam(D2(Pn)) + 1− d(x, y) case 10 If {x, y} = {v1, v3k+3} then |f(x)− f(y)| = |f(v1)− f(v3k+3)| = |f(v1)− [f(vk+1) + (2k + 1)]| = |f(v1)− [f(vk) + (2n− 1) + (2k + 1)]| = |(n+ 2)− [(n+ 3) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1)+ + (2k + 1)]| = |(2n+ 7)(k − 1) + 2(k − 1)(k − 2) + (2n+ 1) + (2k + 1) + 1| ≥ n ≥ diam(D2(Pn)) + 1− d(x, y) case 11 x ∈ Vu and y ∈ Vv Let x = ui and y = vl , where 1 ≤ i, l ≤ n In this case we have several subcases: Subcase 11.1 If i = l , 1 ≤ i, l ≤ k then |f(x)− f(y)| = |f(ui)− f(vi)|, 1 ≤ i ≤ k N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 12 of 25 = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [(n+ 3) + (2n+ 7)(i− 1)+ + 2(i− 1)(i− 2)]| = | − [(n+ 3) + (4)(i− 1)]| = |(n+ 3) + (4)(i− 1)|, 1 ≤ i ≤ k ≥ n ≥ diam(D2(Pn)) + 1− d(x, y). Subcase 11.2 If i = l = 2k + 2 |f(x)− f(y)| = |f(ui)− f(vi)|, i = 2k + 2 = |n2 − n− 1− [n2 − 3]| = | − [(n− 2)]| = |(n− 2)|,where d(x, y) = 2 ≥ diam(D2(Pn)) + 1− d(x, y). Subcase 11.3 If i = l , 3k + 4 ≤ i, l ≤ 4k + 3 = n then |f(x)− f(y)| = |f(ui)− f(vi)|, 3k + 4 ≤ i, l ≤ 4k + 3 = n = |f(uj) + 2j − 1− [f(vj)− 2j]|, 1 ≤ j ≤ k = |f(uj)− f(vj) + 4j − 1|, 1 ≤ j ≤ k ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x)− f(y)| ≥ n. Subcase 11.4 Assume i ̸= l. Let, 1 ≤ i ≤ k + 1 and k + 1 ≤ l ≤ 2k + 1 |f(x)− f(y)| = |f(ui)− f(vl)| = |f(ui)− f(vk+j)|, where 1 ≤ j ≤ k = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [f(vk) + (2n− 1)+ + (n)(j − 1) + 2(j − 1)(j − 2)]|, = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [(n+ 3) + (2n+ 7)(k − 1)+ + 2(k − 1)(k − 2) + (2n− 1) + (n)(j − 1) + 2(j − 1)(j − 2)]| If i = 1 , l = 2k + 1 i.e. j = k + 1 then |f(x)− f(y)| = |f(ui)− f(vl)| = |f(u1)− f(v2k+1)| = |0− [f(vk) + (2n− 1) + (n)(j − 1) + 2(j − 1)(j − 2)]| = | − (n+ 3) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)+ = |(n+ 3) + (2n+ 7)(k − 1) + 2(k − 1)(k − 2)+ + (2n− 1) + (n)(k) + 2(k)(k − 1)| N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 13 of 25 ≥ n, where n = 4k + 3, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x)− f(y)| ≥ n. If i = 4k + 3 and l = 2k + 2 then |f(x)− f(y)| = |f(ui)− f(vl)| = |f(u4k+3)− f(v2k+2)| = |f(u1) + 1− f(v2k+2)| = |0 + 1− [n2 − 3]| = | − [n2 − 3− 1]| = |n2 − 4| ≥ n,where n = 4k + 3, k ≥ 1, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x) − f(y)| ≥ n. The radio condition is |f(x) − f(y)| ≥ n satisfied by every pair of vertices Figure 3. , illustrates the proof of this case, presents the labeling D2(P15) According Theorem 2. Figure 3: Radio labeling of D2(P15) following Theorem 2. Theorem 3. Let n > 6 be a positive integer such that n ≡ 0(mod4). Then rn(D2(Pn)) ≤ n2 − n 2 . Proof. For any positive integer n ≡ 0(mod4), such that n = 4k, k ≥ 2, let f : V (D2(Pn)) → N define as follows: f(ui) =  (2n+ 3)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k f(uk) + 2n− 3 if i = k + 1 f(uk)− 2n+ 2− (3n− 7)(j − 1) + 2(j − 1)(j − 2), if k + 2 ≤ i ≤ 2k, where 1 ≤ j ≤ k − 1, and i = k + 1 + j f(uk+1+j) + (6k − 3)− 2(j − 1), if 2k + 1 ≤ i ≤ 3k − 1 where 1 ≤ j ≤ k − 1, and i = 3k − J f(uj) + 2j − 1, then 3k ≤ i ≤ 4k = n where 1 ≤ j ≤ k + 1 and i = 4k + 1− j N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 14 of 25 And for the vertices belongs to Vv, f(vi) =  (n+ 2) + (2n+ 7)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k − 1 f(vk−1) + 2n− 1 if i = k f(vk) + (2n+ 1) + (n+ 3)(j − 1) + 2(j − 1)(j − 2), if k ≤ i ≤ 2k where 1 ≤ j ≤ k, and i = k + j f(vk−1+j) + (2k − 1) + 2(j − 1), if 2k + 1 ≤ i ≤ 3k + 1 where 1 ≤ j ≤ k + 1and i = 3k + 2− j f(vj)− (2j − 1), if 3k + 2 ≤ i ≤ 4k = n where 1 ≤ j ≤ k − 1, and i = 4k + 1− j Hereafter, we show that the function f satisfies the radio labeling condition for any two distinct vertices x, y of D2(Pn) for n > 6, n = 4k where k ≥ 2. case 1 x, y ∈ Vu. If {x, y} = {u1, un} then |f(x)− f(y)| = |f(u1)− f(un)| = |f(u1)− [f(u1) + 2− 1]| = 1 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. case 2 If {x, y} = {u1, uk} then |f(x)− f(y)| = |f(u1)− f(uk)| = |f(u1)− [(2n+ 3)(k − 1) + 2(k − 1)(k − 2)]| = | − [(2n+ 3)(k − 1) + 2(k − 1)(k − 2)]| = (2n+ 3)(k − 1) + 2(k − 1)(k − 2) ≥ n, where n = 4k ≥ diam(D2(Pn)) + 1− d(x, y) case 3 If {x, y} = {u1, uk+1} then |f(x)− f(y)| = |f(u1)− f(uk+1)| = |f(u1)− [f(uk) + 2n− 3]| = |[f(uk) + 2n− 3]| = |(2n+ 3)(k − 1) + 2(k − 1)(k − 2) + 2n− 3| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 4 If {x, y} = {u1, u2k} then |f(x)− f(y)| = |f(u1)− f(u2k)| = |f(u1)− [f(uk)− 2n+ 2− (3n− 7)(k − 2) + 2(k − 2)(k − 3)]| = |f(uk)− 2n+ 2− (3n− 7)(k − 2) + 2(k − 2)(k − 3)| = |(2n+ 3)(k − 1) + 2(k − 1)(k − 2)− 2n+ 2− (3n− 7)(k − 2)+ N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 15 of 25 + 2(k − 2)(k − 3)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 5 If {x, y} = {u1, u3k+1} then |f(x)− f(y)| = |f(u1)− f(u3k−1)| = |f(u1)− [f(uk+2) + (6k − 3)]| = |f(uk+2) + (6k − 3)| = |f(uk)− 2n+ 2 + (6k − 3)| = |(2n+ 3)(k − 1) + 2(k − 1)(k − 2)− 2n+ 2 + (6k − 3)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 6 x, y ∈ Vv. If {x, y} = {v1, vn} then |f(x)− f(y)| = |f(v1)− f(vn)| = |f(v1)− [f(v1)− 1]| = 1 ≥ diam(D2(Pn)) + 1− d(x, y), where d(x, y) = n− 1. case 7 If {x, y} = {v1, vk−1} then |f(x)− f(y)| = |f(v1)− f(vk−1)| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3)]| = |(2n+ 7)(k − 2) + 2(k − 2)(k − 3)| ≥ n, where n = 4k ≥ diam(D2(Pn)) + 1− d(x, y). case 8 If {x, y} = {v1, v2k−1} then |f(x)− f(y)| = |f(v1)− f(v2k−1)| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3)+ + (2n+ 1) + (n+ 1)(k − 1) + 2(k − 1)(k − 2)]| = |(2n+ 7)(k − 1) + 2(k − 2)(k − 3) + (2n+ 1) + (n+ 1)(k − 1)+ + 2(k − 1)(k − 2)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 9If {x, y} = {v1, v3k−1} then |f(x)− f(y)| = |f(v1)− f(v3k−1)| N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 16 of 25 = |f(v1)− [f(vk) + (2k − 1)]| = |f(v1)− [f(vk−1) + (2n+ 1) + (2k − 1)]| = |(n+ 2)− [(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3) + (2n+ 1)+ + (2k − 1)]| = |(2n+ 7)(k − 2) + 2(k − 2)(k − 3) + (2n+ 1) + (2k − 1)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y) case 10 x ∈ Vu and y ∈ Vv Let x = ui and y = vl , where 1 ≤ i, l ≤ n In this case we have several subcases: Subcase 10.1 If i = 1 , j = k − 1 then |f(x)− f(y)| = |f(u1)− f(vk−1)| = |0− [(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3)]| = |(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Otherwise, |f(x)− f(y)| ≥ n. Subcase 10.2. Suppose, 1 ≤ i ≤ k and k ≤ l ≤ 2k − 1 |f(x)− f(y)| = |f(ui)− f(vl)| = |f(ui)− f(vk−1+j)|, where 1 ≤ j ≤ k − 1 = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [f(vk−1) + (2n+ 1) + (n+ 1)(j − 1)+ + 2(j − 1)(j − 2)]| = |(2n+ 3)(i− 1) + 2(i− 1)(i− 2)− [(n+ 2) + (2n+ 7)(k − 2)+ + 2(k − 2)(k − 3) + (2n+ 1) + (n+ 1)(j − 1) + 2(j − 1)(j − 2)]|. Subcase 10.3. If i = 1 , l = k i.e. j = 1 then |f(x)− f(y)| = |f(ui)− f(vl)| = |f(ui)− f(vk−1+j)|, where 1 ≤ j ≤ k − 1 = |0− [f(vk−1) + (2n+ 1) + (n+ 1)(j − 1) + 2(j − 1)(j − 2)]| = | − (n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3) + (2n+ 1)| = |(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3) + (2n+ 1)| ≥ n, where n = 4k, d(x, y) ≥ 1 ≥ diam(D2(Pn)) + 1− d(x, y). Subcase 10.4. Suppose, 1 ≤ i ≤ k and 2k ≤ l ≤ 3k − 1 |f(x)− f(y)| = |f(ui)− f(vl)| N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 17 of 25 = |f(ui)− f(vn−(k+j−2))|, where 1 ≤ j ≤ k. Subcase 10.5. If i = 1 , l = 3k − 1 i.e. j = 1 then |f(x)− f(y)| = |f(u1)− f(v3k−1)| = |f(ui)− f(vn−k−1)| = |0− f(vk) + 2k − 1| = |(n+ 2) + (2n+ 7)(k − 2) + 2(k − 2)(k − 3) + 2k − 1| ≥ n, where n = 4k, d(x, y) ≥ 1. Otherwise, |f(x)−f(y)| ≥ n. Similarly, proved if k+2 ≤ i ≤ 2k+1 and k+1 ≤ l ≤ 2k+1, for this , the radio condition is |f(x)− f(y)| ≥ n satisfied by every pair of vertices. Figure 4. , illustrates the proof of this case, presents the labeling D2(P16) According Theorem 3. Figure 4: Radio labeling of D2(P16) following Theorem 3. Theorem 4. Let n > 6 be a positive integer such that n ≡ 1(mod4). Then rn(D2(Pn)) ≤ n2 − 2. Proof. For any positive integer n ≡ 1(mod4), such that n = 4k + 1, k ≥ 2, let f : V (D2(Pn)) → N define as follows: f(ui) =  (2n+ 3)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k f(uk) + 2n− 3 if i = k + 1 f(uk)− (2n− 3)− (3n− 8)(j − 1) + 2(j − 1)(j − 2), if k + 2 ≤ i ≤ 2k where 1 ≤ j ≤ k − 1, and i = k + 1 + j n2 − 2, if i = 2k + 1 f(uk+1+j) + (6k − 2)− 2(j − 1), if 2k + 2 ≤ i ≤ 3k, where 1 ≤ j ≤ k − 1, and i = 3k + 1− j f(uj) + 2j − 1, if 3k + 1 ≤ i ≤ 4k + 1 = n where 1 ≤ j ≤ k + 1 and i = 4k + 2− j and N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 18 of 25 f(vi) =  (n+ 3) + (2n+ 7)(i− 1) + 2(i− 1)(i− 2), if 1 ≤ i ≤ k − 1 f(uk−1) + 2n− 2 if i = k f(vk) + (2n) + (n+ 2)(j − 1) + 2(j − 1)(j − 2), if k + 1 ≤ i ≤ 2k, where 1 ≤ j ≤ k, and i = k + j n2 − 2, if i = 2k + 1 f(vk−1+j) + (2k + 1) + 2(j − 1), if 2k + 2 ≤ i ≤ 3k + 2 where 1 ≤ j ≤ k + 1, and i = 3k + 3− j f(vj)− 2j, if 3k + 3 ≤ i ≤ 4k + 1 = n where 1 ≤ j ≤ k − 1, and i = 4k + 2− j The proof is left to the reader. We omit the proof, but Figure 5, presents the labeling D2(P13) illustrates Theorem 4 namely, n ≡ 1 (mod 4), if n = 4k + 1, k = 3. Figure 5: Radio labeling D2(P13) following Theorem 4. 4. Integer Linear Programming Model (ILPM ) For resolving optimization issues, mathematical programming is a crucial instrument. The references [18–20] provide specific information regarding the process of conceptualizing real-life problems as mathematical models. For a connected graph G of order n. Let V (G) = {x1, x2, · · · , xn } . We construct the distance matrix D = [dij] of G where dij = d (xi, xj) for 1 ≤ i, j ≤ n. For 1 ≤ i ≤ n, assume that yi is the radio label of the vertex xi. Now, we can use integer programming to present the mathematical model for the radio labeling problem. The function F is denoted as F = y1 + y2 + · · ·+ yn The claim is minimizing F subject to the ( n 2 ) constraints |yi − yj | ≥ n− d (xi, xj) for 1 ≤ i ≤ n− 1; 2 ≤ j ≤ n and i < j where y1, y2, · · · , yn are integer numbers. Therefore, rn (G) = max1≤i≤n {yi} . 5. Experimental Study Hereafter, a set of experiments have been performed in order to compare the perfor- mance of algorithms presented in [6] with the obtained upper bounds by Theorems 1-4. Furthermore, a comparison is carried out between the findings of those theorems and the mathematical model that was presented in Section 4. In our experiments we have used a N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 19 of 25 PC with Core i7 processor with 2.8 GHz CPU and 8 GB of RAM. Our implementations are done with MATLAB R2016a and MS Windows 7 Professional system. According to Table 2, the results presented in Lemma 4 demonstrate best performance compared to those in [6] for n = 5. For n = 2, 3, and 4, similar outcomes are observed. Conversely, the outcomes in Lemma 1, 2, 3, and 4 outperform those in [7] across all values of n. In terms of time complexity table 2 shows that the computations of Lemmas 1, 2, 3, and 4 give superior performance compared to those in [6, 7]. Table 2: Comparison among our approach, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 2, 3, 4, 5. n 2n Lemma 1,2,3,4 Saha [6] ILPM [7] rn CPU Time rn CPU Time rn CPU Time 2 4 4 O(1) 4 0.002396 5 0.003654 3 6 6 O(1) 6 0.004862 9 0.004854 4 8 12 O(1) 12 0.005359 19 0.019494 5 10 22 O(1) 23 0.005539 33 0.034559 Table 3: Comparison among Theorem 1, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 2(mod) 4. n 2n Theorem 1 Saha [6] ILPM [7] rn CPU Time rn CPU Time rn CPU Time 6 12 33 O(1) 33 0.042993 51 0.036009 10 20 95 O(1) 95 0.043329 163 0.100475 14 28 189 O(1) 189 0.043968 343 0.150742 18 36 315 O(1) 315 0.048544 579 0.151717 22 44 473 O(1) 473 0.054947 883 0.162229 26 52 663 O(1) 663 0.082347 1251 0.182440 30 60 885 O(1) 885 0.083501 1683 0.190866 34 68 1139 O(1) 1139 0.096042 2179 0.218329 38 76 1425 O(1) 1425 0.096626 2739 0.222084 42 84 1743 O(1) 1743 0.117833 3363 0.264229 46 92 2093 O(1) 2093 0.118883 4051 0.278491 50 100 2475 O(1) 2475 0.121298 4803 0.301669 54 108 2889 O(1) 2889 0.128889 5619 0.420457 58 116 3335 O(1) 3335 0.131909 6499 0.475030 62 124 3813 O(1) 3813 0.148007 7443 0.477373 64 128 4064 O(1) 4064 0.178655 7939 0.527846 68 136 4590 O(1) 4590 0.198579 8979 0.635992 72 144 5148 O(1) 5148 0.20673 10083 0.720457 76 152 5738 O(1) 5738 0.213292 11251 0.875030 80 160 6360 O(1) 6360 0.231493 12483 0.976116 Based on the upper bound of the radio number of shadow graph with n = 2 (mod) 4, it can be observed from Table 3 and Figure 6 that the outcomes presented in Theorem 1 coincide with those in [6]. Conversely, the results from Theorem 1 surpass the findings in [7] for all values of n. Table 3 indicates that the results from Theorem 1 have a time N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 20 of 25 complexity of O(1), while those in [6] have a complexity of O(n3). Furthermore, the results from Theorem 1 require less time compared to the results in [7]. Figure 6: Comparison among Theorem 1, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 2(mod) 4. Table 4: Comparison among Theorem 2, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 3(mod)4. n 2n Theorem 2 Saha [6] ILPM [7] rn CPU Time rn CPU Time rn CPU Time 7 14 46 O(1) 46 0.043355 73 0.089586 11 22 118 O(1) 118 0.048763 201 0.090475 15 30 222 O(1) 222 0.050551 393 0.130588 19 38 358 O(1) 358 0.050987 649 0.161792 23 46 526 O(1) 526 0.051813 969 0.220337 27 54 726 O(1) 726 0.081522 1353 0.305778 31 62 958 O(1) 958 0.086948 1801 0.207713 35 70 1222 O(1) 1222 0.092306 2313 0.266207 39 78 1518 O(1) 1518 0.095106 2889 0.288857 43 86 1846 O(1) 1846 0.102625 3529 0.368033 47 94 2206 O(1) 2206 0.116411 4233 0.486650 51 102 2598 O(1) 2598 0.121707 5001 0.530145 55 110 3022 O(1) 3022 0.124297 5833 0.544023 59 118 3478 O(1) 3478 0.140879 6729 0.551847 63 126 3966 O(1) 3966 0.15269 7689 0.635992 67 134 4486 O(1) 4486 0.164255 8713 0.720457 71 142 5038 O(1) 5038 0.190907 9801 0.829255 75 150 5622 O(1) 5622 0.206634 10953 0.972498 79 158 6238 O(1) 6238 0.224507 12169 0.998945 83 166 6886 O(1) 6886 0.273817 13449 1.034510 87 174 7566 O(1) 7566 0.351691 14793 1.293168 N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 21 of 25 Table 4 and Figure 7 demonstrate that the results presented in Theorem 2 align with the findings in [6], considering the upper bound of the radio number of shadow graph with n = 3(mod) 4. Additionally, it is worth noting that the results obtained in Theorem 2 outperform the results in [7] for all values of n. In terms of running time, Table 4 indicates that the results derived from Theorem 2 have a constant time complexity of O(1), whereas the results in [6] have a time complexity of O(n3). Furthermore, the results from Theorem 2 require less time compared to the results in [7]. Figure 7: Comparison among Theorem 2, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 3(mod)4 Figure 8: Comparison among Theorem 3, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 0(mod)4. Table 5 and Figure 8 demonstrate that the results presented in Theorem 3 are identical to the results in [6], considering the upper bound of the radio number of shadow graph with n = 0 (mod) 4. Additionally, it is worth noting that the results obtained in The- orem 3 outperform the results in [7] for all values of n. In terms of running time, Table 5 indicates that the results derived from Theorem 3 have a constant time complexity of O(1), whereas the results in [6] have a time complexity of O(n3). Furthermore, the results from Theorem 3 require less time compared to the results in [7]. N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 22 of 25 Table 5: Comparison among Theorem 3, Saha and ILPM [7] for the radio number of Shadow graph with n = 0(mod)4. n 2n Theorem 3 Saha [6] ILPM [7] rn CPU Time m CPU Time rn CPU Time 8 16 60 O(1) 60 0.036009 99 0.033053 12 24 138 O(1) 138 0.046854 243 0.089839 16 32 248 O(1) 248 0.052537 451 0.097359 20 40 390 O(1) 390 0.053037 723 0.102315 24 48 564 O(1) 564 0.054454 1059 0.124432 28 56 770 O(1) 770 0.082983 1459 0.171284 32 64 1008 O(1) 1008 0.086663 1923 0.185284 36 72 1278 O(1) 1278 0.099965 2451 0.218490 40 80 1580 O(1) 1580 0.102013 3043 0.235134 44 88 1914 O(1) 1914 0.123509 3699 0.254231 48 96 2280 O(1) 2280 0.123897 4419 0.287511 52 104 2678 O(1) 2678 0.134616 5203 0.451393 56 112 3108 O(1) 3108 0.141561 6051 0.466374 60 120 3570 O(1) 3570 0.149171 6963 0.477208 64 128 4064 O(1) 4064 0.178655 7939 0.527846 68 136 4590 O(1) 4590 0.198579 8979 0.635992 72 144 5148 O(1) 5148 0.20673 10083 0.976116 76 152 5738 O(1) 5738 0.213292 11251 0.983264 80 160 6360 O(1) 6360 0.231493 12483 0.998945 84 168 7014 O(1) 7014 0.252509 13779 1.034510 Figure 9: Comparison among Theorem 4, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 1(mod)4. Table 6 and Figure 9 demonstrate that the results presented in Theorem 4 align with the results in [6], considering the upper bound of the radio number of shadow graph with n = 1 (mod) 4. Additionally, it is worth noting that the results in Theorem 4 outperform the results in [7] for all values of n. In terms of running time, Table 6 indicates that the results in Theorem 4 have a constant time complexity of O(1), whereas the results in [6] N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 23 of 25 have a time complexity of O(n3). Furthermore, the results in Theorem 4 require less time compared to the results in [7]. Table 6: Comparison among Theorem 4, Saha [6] and ILPM [7] for the radio number of Shadow graph with n = 1(mod)4. n 2n Theorem 4 Saha [6] ILPM [7] rn CPU Time rn CPU Time rn CPU Time 9 18 79 O(1) 79 0.033053 129 0.091875 13 26 167 O(1) 167 0.046598 289 0.124002 17 34 287 O(1) 287 0.04728 513 0.134112 21 42 439 O(1) 439 0.049498 801 0.158377 25 50 623 O(1) 623 0.052472 1153 0.205898 29 58 839 O(1) 839 0.08865 1569 0.218329 33 66 1087 O(1) 1087 0.089514 2049 0.222084 37 74 1367 O(1) 1367 0.092204 2593 0.264229 41 82 1679 O(1) 1679 0.100206 3201 0.278491 45 90 2023 O(1) 2023 0.102479 3873 0.305885 49 98 2399 O(1) 2399 0.114788 4609 0.308508 53 106 2807 O(1) 2807 0.133461 5409 0.451393 57 114 3247 O(1) 3247 0.148361 6273 0.466374 61 122 3719 O(1) 3719 0.166031 7201 0.477208 65 130 4223 O(1) 4223 0.16886 8193 0.527846 69 138 4759 O(1) 4759 0.21307 9249 0.644822 73 146 5327 O(1) 5327 0.241707 10369 0.720457 77 154 5927 O(1) 5927 0.260753 11553 0.829255 81 162 6559 O(1) 6559 0.299115 12801 0.937023 85 170 7223 O(1) 7223 0.307929 14113 1.069794 6. Application of radio labeling of graph in cryptography Recently, information technology has been widely used in various fields of life. Hence, information security becomes a vital issue. Cryptography is a significant method for information security. In cryptographic algorithms, keys used for encryption and decryption are lengthy number sequences which are created using random number generators. Such generators follow a uniform distribution that can be is easily determined by hackers, as any number has an equal chance of being generated. Using radio labeling of graphs, Keys can be generated in more faster and effective manner. The radio labeled graph is used as the cipher graph. At the receiver end the encrypted message is received in the form of edge or vertex sequence of this cipher graph. Consequently, it is difficult for an opponent to guess and hack, as the radio labeling problem is an NP-hard problem. The radio number of a graph may also be used to generate invertible matrices that serve for keys generation. In this manner, the radio labeling considered an efficient tool for cryptography as the problem of finding the radio number of a graph is an NP-complete problem. We refer the reader to [24–26] for applications of radio labeling in cryptography N. M. NourEldeen et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5569 24 of 25 7. Conclusions Effective allocating of frequency resources in wireless networks is a challenging prob- lem. The claim is avoiding the interference among stations and minimizing the usage of the available frequencies. Here we consider the radio graph labeling as a graph-theoretical approach to modeling the problem of frequency assignment. In such modeling an inter- ference graph G is built. The vertices of the interference graph represent the stations. Two vertices are adjacent if their corresponding stations may interfere. Radio labeling of the interference graph is done to avoid the interference and moreover emphasizing on minimization the using of frequency resources. Then the minimum value of the maximum assigned frequency (radio label) among all radio labeling of G , called the radio number of the graph is obtained, this radio number represents the highest frequency should be used to avoid interference. In this paper, the radio labeling and the upper bound of radio number of networks as shadow of path graph are investigated. Wherein, a theoretical approach for finding such number is presented and mathematical model for finding upper bound of radio number of shadow of path graph is proposed. In order to validate the effectiveness of our approach, an experimental study is conducted, comparing our results to previous findings. The results of the study demonstrate that our proposed approach surpasses the previous results in terms of both the running time and the upper bound of the radio number. 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