Ratio Mathematica Volume 45, 2023 Radio Mean Labeling of Digraphs Palani K* Sabarina Subi S S† Abstract Let 𝐷 be a strong digraph and let 𝑑(𝑒, 𝑣) denote the distance between any two vertices in 𝐷. A radio mean labeling is a one-to-one mapping 𝑓 from 𝑉(𝐷) to 𝑁 satisfying the condition 𝑑(𝑒, 𝑣) +⌈ 𝑓(𝑒)+𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + π‘‘π‘–π‘Žπ‘š(𝐷) for every 𝑒, 𝑣 ∈ 𝑉(𝐷). The span of a labeling 𝑓 is the maximum integer that 𝑓 maps to a vertex of𝐷. The radio mean number of 𝐷, π‘Ÿπ‘šπ‘› (𝐷) is the lowest span taken over all radio mean labelings of the graph 𝐷. In this paper, we analyze radio mean labeling for some newly defined digraphs. Keywords: Radio Mean, Radio Mean Number, Radio Mean Labeling, Digraphs. AMS Subject Classification: 05C78‑. *Associate Professor, PG & Research Department of Mathematics (A.P.C. Mahalaxmi College for Women, Thoothukudi-628 002, Tamilnadu, India); palani@apcmcollege.ac.in. †Research scholar (Reg No.20112012092001), A.P.C. Mahalaxmi College for Women, Thoothukudi-628 002, Tamilnadu, India. sabarin203@gmail.com. (Affiliated to Manonmaniam Sundaranar University, Abishekapatti, Tirunelveli - 627 012, Tamil Nadu, India). ‑Received on July 29, 2022. Accepted on October 15, 2022. Published on January 30, 2023. doi: 10.23755/rm.v45i0.1023. ISSN: 1592-7415. eISSN: 2282-8214. Β©The Authors. This paper is published under the CC-BY licence agreement. 248 mailto:palani@apcmcollege.ac.in Palani K and Sabarina Subi S S 1. Introduction The graph labeling problem is one of the recent developing area in graph theory. Alex Rosa first introduced this problem in 1967[10]. Radio labeling is motivated by the channel assignment problem introduced by W. K. Hale in 1980[4].In 2001, Gary Chartrand defined the concept of radio labeling of 𝐺[2].Liu and Zhu first determined the radio number in 2005[5].Ponraj et al.[8] introduced the notion of radio mean labeling of graphs and investigated radio mean number of some graphs [9]. In this paper, we introduce a new definition for radio mean labeling of digraphs and also we study radio mean number of some newly defined digraphs. Radio Labeling is used for X-ray, crystallography, coding theory, network security, network addressing, channel assignment process, social network analysis such as connectivity, scalability, routing, computing, cell biology etc., The following results are used in the subsequent section. Definition 1.1.Let 𝐷 be a strong digraph and let 𝑑(𝑒, 𝑣) denote the distance between any two vertices in𝐷. A radio mean labeling is a one-to-one mapping 𝑓 from 𝑉(𝐷) to 𝑁 satisfying the condition 𝑑(𝑒, 𝑣) +⌈ 𝑓(𝑒)+𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + π‘‘π‘–π‘Žπ‘š(𝐷) for every 𝑒, 𝑣 ∈ 𝑉(𝐷). The span of a labeling 𝑓 is the maximum integer that 𝑓 maps to a vertex of𝐷. The radio mean number of𝐷, π‘Ÿπ‘šπ‘› (𝐷) is the lowest span taken over all radio mean labelings of the graph𝐷. Definition 1.2. Consider a globe. Let 𝑒, 𝑣 be the vertices of degree𝑛. Orient the edges of all but one 𝑒 βˆ’ 𝑣 path of length two in same direction. Orient the left out 𝑒 βˆ’ 𝑣 path in opposite direction. It is strongly connected and is called a Diglobe. It is denoted as 𝐺𝑙(𝑛).βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— Figure 1.1. Diglobe Definition 1.3. If the edges of all 𝑒 βˆ’ 𝑣 paths are in a single common direction it is not a strong digraph. We name it as sole diglobe (𝑆𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—). Remark 1.4. If there are atleast two paths are oriented in opposite directions, the diglobe becomes strongly connected. 249 Radio Mean Labeling of Digraphs Remark 1.5. Orient the edges of the globe in such a way that the two edges in each 𝑒 βˆ’ 𝑣 path of length 2 get opposite directions. The resulting digraph is called alternate diglobe and is denoted as 𝐴𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—. 𝐴𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— is not a strong digraph. Definition 1.6. Consider a book with 𝑛 pages sharing a common edge. The common edge is called the spine or base of the book. Orient all the edges except the spine in the one single direction and the spine in opposite direction. The resulting digraph is strongly connected and is called as directed book. The directed book is denoted as 𝐡(π‘š, 𝑛).βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— Remark 1.7. The ditriangular book with n-pages is defined as n copies of cycle C3 sharing a common edge in a directed book. The ditriangular book is denoted as 𝐡(3, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— . Figure 1.2. Ditriangular Book Remark 1.8. The diquadrilateral book with n-pages is defined as 𝑛 copies of cycle 𝐢4 sharing a common edge in a directed book. The diquadrilateral book is denoted as 𝐡(4, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—. Figure 1.3. Diquadrilateral Book 250 Palani K and Sabarina Subi S S 2. Main Results Theorem 2.1.The radio mean number of diglobe(𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— ) is less than or equal to 𝑛 + 3 for3 ≀ 𝑛 ≀ 4. Proof. Let 𝐷 be a diglobe. Let 𝑉(𝐷) = {𝑣1, 𝑣2, 𝑣3, … … . , 𝑣𝑛 , 𝑒, 𝑣} and 𝐴(𝐷) = {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/1 ≀ 𝑖 ≀ ⌊ 𝑛 2 βŒ‹} βˆͺ {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/ ⌊ 𝑛 2 βŒ‹ + 2 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣 ⌊ 𝑛 2 βŒ‹+1 𝑒⃗⃗ βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— /1 ≀ 𝑖 ≀ ⌊ 𝑛 2 βŒ‹} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— / ⌊ 𝑛 2 βŒ‹ + 2 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑣 ⌊ 𝑛 2 βŒ‹+1 }βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— The diameter of diglobe is 4. Define 𝑓: 𝑉(𝐷) β†’ 𝑁 by 𝑓(𝑣𝑖 ) = 𝑖 + 1 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑒) = 𝑛 + 2 𝑓(𝑣) = 𝑛 + 3 Claim. 𝑓 is a valid radio mean labeling. Since the diameter is 4, to prove 𝑓 is a radio mean labeling, it is enough to prove that, 𝑑(π‘₯, 𝑦) + ⌈ 𝑓(π‘₯)+𝑓(𝑦) 2 βŒ‰ β‰₯ 5 … … … … (1)for every pair of vertices (π‘₯, 𝑦) where π‘₯ β‰  𝑦. Equivalently, it is enough to prove (1) for pair of vertices with minimum 𝑓 values and minimum 𝑑(π‘₯, 𝑦)values. Hence, the proof involves the following cases Case a. Consider the pairs(𝑒, 𝑣𝑖 ). Here, 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) + ⌈ 𝑓(𝑒) + 𝑓(𝑣𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 2 + 𝑖 + 1 2 βŒ‰ β‰₯ 5 Case b. Consider the pairs(𝑣𝑖 , 𝑣). Here, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑖 + 1 + 𝑛 + 3 2 βŒ‰ β‰₯ 5 Case c. Consider the pairs (𝑒, 𝑣) π‘Žπ‘›π‘‘ 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) = 2 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) + ⌈ 𝑓(𝑒) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 2 + ⌈ 𝑛 + 2 + 𝑛 + 3 2 βŒ‰ > 5 Case d. Consider the pairs (𝑣𝑖 , 𝑣𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) = 2, then, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣𝑖+1) 2 βŒ‰ β‰₯ 2 + ⌈ 𝑖 + 1 + 𝑖 + 2 2 βŒ‰ > 5 Hence, by all the above cases, the radio mean condition is satisfied by 𝑓. Further, 𝑓 attains its maximum corresponding to 𝑣 and therefore π‘Ÿπ‘šπ‘›(𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— ) ≀ 𝑛 + 3 for 3 ≀ 𝑛 ≀ 4. Theorem 2.2. The radio mean number of diglobe(𝐺𝑙(𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— ) is 𝑛 + 2 for n > 4. Proof. Let 𝐷 be a diglobe. Let 𝑉(𝐷) = {𝑣1, 𝑣2, 𝑣3, … … . , 𝑣𝑛 , 𝑒, 𝑣} and 𝐴(𝐷) = {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/1 ≀ 𝑖 ≀ ⌊ 𝑛 2 βŒ‹} βˆͺ {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/ ⌊ 𝑛 2 βŒ‹ + 2 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣 ⌊ 𝑛 2 βŒ‹+1 𝑒⃗⃗ βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— /1 ≀ 𝑖 251 Radio Mean Labeling of Digraphs ≀ ⌊ 𝑛 2 βŒ‹} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— / ⌊ 𝑛 2 βŒ‹ + 2 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑣 ⌊ 𝑛 2 βŒ‹+1 }βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— The diameter of diglobe is 4. Define 𝑓: 𝑉(𝐷) β†’ 𝑁 by 𝑓(𝑣𝑖 ) = 𝑖 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑒) = 𝑛 + 1 𝑓(𝑣) = 𝑛 + 2 Claim.𝑓 is a valid radio mean labeling. Since the diameter is 4, to prove 𝑓 is a radio mean labeling, it is enough to prove that, 𝑑(π‘₯, 𝑦) + ⌈ 𝑓(π‘₯)+𝑓(𝑦) 2 βŒ‰ β‰₯ 5 … … … … (1) for every pair of vertices (π‘₯, 𝑦) where π‘₯ β‰  𝑦. Equivalently, it is enough to prove (1) for pair of vertices with minimum 𝑓 values and minimum 𝑑(π‘₯, 𝑦)values. Hence, the proof involves the following cases Case a. Consider the pairs (𝑒, 𝑣𝑖 ). Here, 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) + ⌈ 𝑓(𝑒) + 𝑓(𝑣𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 1 + 𝑖 2 βŒ‰ β‰₯ 5 Case b. Consider the pairs (𝑣𝑖 , 𝑣). Here, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑖 + 𝑛 + 2 2 βŒ‰ β‰₯ 5 Case c. Consider the pairs (𝑒, 𝑣) π‘Žπ‘›π‘‘ 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) = 2 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) + ⌈ 𝑓(𝑒) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 2 + ⌈ 𝑛 + 1 + 𝑛 + 2 2 βŒ‰ > 5 Case d. Consider the pairs (𝑣𝑖 , 𝑣𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) = 2, then, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣𝑖+1) 2 βŒ‰ β‰₯ 2 + ⌈ 𝑖 + 𝑖 + 1 2 βŒ‰ > 5 Hence, by all the above cases, the radio mean condition is satisfied by 𝑓. Further, 𝑓 attains its maximum corresponding to 𝑣 and is 𝑛 + 2 for 𝑛 > 4. Since 𝐷 contains only 𝑛 + 2 vertices, 𝑛 + 2 is the minimum of the maximum integer that could be assigned to the vertices of 𝐷. Hence π‘Ÿπ‘šπ‘›(𝐷) = 𝑛 + 2 for𝑛 > 4. Theorem 2.3. The radio mean number of ditriangular book(𝐡(3, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—) is less than or equal to 𝑛 + 3 for𝑛 = 2. Proof. Let 𝐷 be a ditriangular book. Let 𝑉(𝐷) = {𝑣1, 𝑣2, 𝑣3, … … . , 𝑣𝑛 , 𝑒, 𝑣} and 𝐴(𝐷) = {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— /1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑒}βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— The diameter of ditriangular book is 3. Define 𝑓: 𝑉(𝐷) β†’ 𝑁 by 𝑓(𝑣𝑖 ) = 𝑖 + 1 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑒) = 𝑛 + 2 𝑓(𝑣) = 𝑛 + 3 Claim.𝑓 is a valid radio mean labeling. 252 Palani K and Sabarina Subi S S Since the diameter is 3, to prove 𝑓 is a radio mean labeling, it is enough to prove that, 𝑑(π‘₯, 𝑦) + ⌈ 𝑓(π‘₯)+𝑓(𝑦) 2 βŒ‰ β‰₯ 4 … … … … (1)for every pair of vertices (π‘₯, 𝑦) where π‘₯ β‰  𝑦. Equivalently, it is enough to prove (1) for pair of vertices with minimum 𝑓 values and minimum 𝑑(π‘₯, 𝑦)values. Hence, the proof involves the following cases Case a. Consider the pairs (𝑒, 𝑣𝑖 ). Here, 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) + ⌈ 𝑓(𝑒) + 𝑓(𝑣𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 2 + 𝑖 + 1 2 βŒ‰ β‰₯ 4 Case b. Consider the pairs (𝑣𝑖 , 𝑣). Here, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑖 + 1 + 𝑛 + 3 2 βŒ‰ > 4 Case c. Consider the pairs (𝑒, 𝑣) π‘Žπ‘›π‘‘ 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) = 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) + ⌈ 𝑓(𝑒) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 2 + 𝑛 + 3 2 βŒ‰ > 4 Case d. Consider the pairs (𝑣𝑖 , 𝑣𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) = 3, then, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣𝑖+1) 2 βŒ‰ β‰₯ 3 + ⌈ 𝑖 + 1 + 𝑖 + 2 2 βŒ‰ > 4 Hence, by all the above cases, the radio mean condition is satisfied by 𝑓. Further, 𝑓 attains its maximum corresponding to 𝑣 and therefore π‘Ÿπ‘šπ‘›(𝐡(3, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—) ≀ 𝑛 + 3 for 𝑛 = 2. Theorem 2.4.The radio mean number of ditriangular book(𝐡(3, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—) is 𝑛 + 2 for 𝑛 > 2. Proof. Let 𝐷 be a ditriangular book. Let 𝑉(𝐷) = {𝑣1, 𝑣2, 𝑣3, … … . , 𝑣𝑛 , 𝑒, 𝑣} and 𝐴(𝐷) = {𝑒𝑣𝑖⃗⃗⃗⃗⃗⃗⃗/1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— /1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑒}βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— The diameter of ditriangular book is 3. Define 𝑓: 𝑉(𝐷) β†’ 𝑁 by 𝑓(𝑣𝑖 ) = 𝑖 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑒) = 𝑛 + 1 𝑓(𝑣) = 𝑛 + 2 Claim. 𝑓 is a valid radio mean labeling. Since the diameter is 3, to prove 𝑓 is a radio mean labeling, it is enough to prove that, 𝑑(π‘₯, 𝑦) + ⌈ 𝑓(π‘₯)+𝑓(𝑦) 2 βŒ‰ β‰₯ 4 … … … … (1)for every pair of vertices (π‘₯, 𝑦) where π‘₯ β‰  𝑦. Equivalently, it is enough to prove (1) for pair of vertices with minimum 𝑓 values and minimum 𝑑(π‘₯, 𝑦)values. Hence, the proof involves the following cases Case a. Consider the pairs (𝑒, 𝑣𝑖 ). Here, 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣𝑖 ) + ⌈ 𝑓(𝑒) + 𝑓(𝑣𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 1 + 𝑖 2 βŒ‰ β‰₯ 4 Case b. Consider the pairs (𝑣𝑖 , 𝑣). Here, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑖 + 𝑛 + 2 2 βŒ‰ β‰₯ 4 253 Radio Mean Labeling of Digraphs Case c. Consider the pairs (𝑒, 𝑣) π‘Žπ‘›π‘‘ 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) = 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑣) + ⌈ 𝑓(𝑒) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑛 + 1 + 𝑛 + 2 2 βŒ‰ > 4 Case d. Consider the pairs (𝑣𝑖 , 𝑣𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) = 3, then, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣𝑖+1) 2 βŒ‰ β‰₯ 3 + ⌈ 𝑖 + 𝑖 + 1 2 βŒ‰ > 4 Hence, by all the above cases, the radio mean condition is satisfied by𝑓. Further, 𝑓 attains its maximum corresponding to 𝑣 and is 𝑛 + 2 for 𝑛 > 2. Since 𝐷 contains only 𝑛 + 2 vertices, 𝑛 + 2 is the minimum of the maximum integer that could be assigned to the vertices of𝐷. Hence π‘Ÿπ‘šπ‘›(𝐷) = 𝑛 + 2 for𝑛 > 2. Theorem 2.5. The radio mean number of diquadrilateral book(𝐡(4, 𝑛)βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ—) is 2𝑛 + 2 for 𝑛 > 3. Proof. Let 𝐷 be a diquadrilateral book. Let 𝑉(𝐷) = {𝑣1, 𝑣2, 𝑣3, … … . , 𝑣𝑛 , 𝑒1, 𝑒2, 𝑒3, … … . , 𝑒𝑛 , 𝑒, 𝑣} and 𝐴(𝐷) = {𝑒𝑒𝑖⃗⃗⃗⃗⃗⃗ βƒ—/1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑒𝑖𝑣𝑖⃗⃗ βƒ—βƒ— βƒ—βƒ—βƒ—βƒ— /1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑖𝑣⃗⃗ βƒ—βƒ— βƒ—βƒ— /1 ≀ 𝑖 ≀ 𝑛} βˆͺ {𝑣𝑒}βƒ—βƒ— βƒ—βƒ— βƒ—βƒ— βƒ— The diameter of diquadrilateral book is 5. Define 𝑓: 𝑉(𝐷) β†’ 𝑁 by 𝑓(𝑒𝑖 ) = 𝑖 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑣𝑖 ) = 2𝑛 βˆ’ 𝑖 + 1 , 1 ≀ 𝑖 ≀ 𝑛 𝑓(𝑒) = 2𝑛 + 1 𝑓(𝑣) = 2𝑛 + 2 Claim. 𝑓 is a valid radio mean labeling. Since the diameter is5, to prove 𝑓 is a radio mean labeling, it is enough to prove that, 𝑑(π‘₯, 𝑦) + ⌈ 𝑓(π‘₯)+𝑓(𝑦) 2 βŒ‰ β‰₯ 6 … … … … (1)for every pair of vertices (π‘₯, 𝑦) where π‘₯ β‰  𝑦. Equivalently, it is enough to prove (1) for pair of vertices with minimum 𝑓 values and minimum 𝑑(π‘₯, 𝑦)values. Hence, the proof involves the following cases Case a. Consider the pairs (𝑒, 𝑒𝑖 ). Here, 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑒𝑖 ) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑒, 𝑒𝑖 ) + ⌈ 𝑓(𝑒) + 𝑓(𝑒𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 2𝑛 + 1 + 𝑖 2 βŒ‰ β‰₯ 6 Case b. Consider the pairs (𝑣𝑖 , 𝑣). Here, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) β‰₯ 1 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣) 2 βŒ‰ β‰₯ 1 + ⌈ 2𝑛 βˆ’ 𝑖 + 1 + 2𝑛 + 2 2 βŒ‰ > 6 Case c. Consider the pairs (𝑒𝑖 , 𝑣𝑖 ) π‘Žπ‘›π‘‘ 𝑑⃗⃗⃗ βƒ—(𝑒𝑖 , 𝑣𝑖 ) = 1 𝑑⃗⃗⃗ βƒ—(𝑒𝑖 , 𝑣𝑖 ) + ⌈ 𝑓(𝑒𝑖 ) + 𝑓(𝑣𝑖 ) 2 βŒ‰ β‰₯ 1 + ⌈ 𝑖 + 2𝑛 βˆ’ 𝑖 + 1 2 βŒ‰ > 6 Case d. Consider the pairs (𝑒𝑖 , 𝑒𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑒𝑖 , 𝑒𝑖+1) = 4, then, 𝑑⃗⃗⃗ βƒ—(𝑒𝑖 , 𝑒𝑖+1) + ⌈ 𝑓(𝑒𝑖 ) + 𝑓(𝑒𝑖+1) 2 βŒ‰ β‰₯ 4 + ⌈ 𝑖 + 𝑖 + 1 2 βŒ‰ β‰₯ 6 254 Palani K and Sabarina Subi S S Case e. Consider the pairs (𝑣𝑖 , 𝑣𝑖+1) and 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) = 4, then, 𝑑⃗⃗⃗ βƒ—(𝑣𝑖 , 𝑣𝑖+1) + ⌈ 𝑓(𝑣𝑖 ) + 𝑓(𝑣𝑖+1) 2 βŒ‰ β‰₯ 4 + ⌈ 2𝑛 βˆ’ 𝑖 + 1 + 2𝑛 βˆ’ 𝑖 2 βŒ‰ > 6 Hence, by all the above cases, the radio mean condition is satisfied by 𝑓. Further, 𝑓 attains its maximum corresponding to 𝑣 and is 2𝑛 + 2 for 𝑛 > 3. Since 𝐷 contains only 2𝑛 + 2 vertices, 2𝑛 + 2 is the minimum of the maximum integer that could be assigned to the vertices of 𝐷. Hence, π‘Ÿπ‘šπ‘›(𝐷) = 2𝑛 + 2 for 𝑛 > 3. 3. 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