EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 3, 2022, 1331-1343 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Planarity of a Directed Pathos Total Digraph of Some Special Arborescence Graphs Jill Maegan B. Pamplona1, Imelda S. Aniversario2 1 Mathematics Department, Northwestern Mindanao State College of Science and Technology, Tangub City, Misamis Occidental 2 Department of Mathematics and Statistics, College of Science and Mathematics, Mindanao State University - Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. An arborescence graph is a directed graph in which, for a vertex u called the root, and any other vertex v, there is exactly one directed path from u to v. The directed pathos of an arborescence Ar is defined as a collection of minimum number of arc disjoint open directed paths whose union is Ar. In [6], for an arborescence Ar, a directed pathos total digraph Q = DPT (Ar) has vertex set V (Q) = V (Ar) ∪ A(Ar) ∪ P (Ar), where V (Ar) is the vertex set, A(Ar) is the arc set, and P (Ar) is a directed pathos set of Ar. The arc set A(Q) consists of the following arcs: ab such that a, b ∈ A(Ar) and the head of a coincides with the tail of b; uv such that u, v ∈ V (Ar) and u is adjacent to v; au(ua) such that a ∈ A(Ar) and u ∈ V (Ar) and the head (tail) of a is u; Pa such that a ∈ A(Ar) and P ∈ P (Ar) and the arc a lies on the directed path P ; PiPj such that Pi, Pj ∈ P (Ar) and it is possible to reach the head of Pj from the tail of Pi through a common vertex, and it is also possible to reach the head of Pi from the tail of Pj . In this paper, the concept of planarity of the directed pathos total digraph (that is, as an acyclic directed graph which can be drawn with non crossing arcs oriented in one direction) is being dis- cussed and applied to a directed pathos total digraph of an arborescence Ar (DPT (Ar)). Further, the internal vertices of these directed pathos total digraph of Ar are cconsidered. Finally, the planarity of an arborescence resulting from the vertex-gluing of two directed paths is presented and corresponding internal vertex number is obtained. 2020 Mathematics Subject Classifications: 05C20 Key Words and Phrases: Directed path, arborescence graph, directed pathos total digraph of an arborescence, internal vertex number of a directed pathos total digraph of an arborescence DOI: https://doi.org/10.29020/nybg.ejpam.v15i3.4492 Email addresses: jillmaegan.pamplona@g.msuiit.edu.ph (J.M.E. Pamplona), imelda.aniverasrio@g.msuiit.edu.ph (I.S. Aniverasrio) https://www.ejpam.com 1331 © 2022 EJPAM All rights reserved. Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1332 1. Introduction There are many graph valued functions (or graph operators) for which one can con- struct a new graph from a given graph, such as the line graphs, the total graphs, and their generalizations. The line graph of a graph G, written L(G), is the graph whose vertices are the edges of G, with two vertices of L(G) adjacent whenever the corresponding edges of G have a common vertex [8]. Harary and Norman [5] extended the concept of line graph of a graph and introduced the concept of line digraph of a directed graph. The line digraph L(D) of a digraph D has the arcs of D as vertices. There is an arc from D-arc pq towards D-arc uv if and only if q = u. Behzad [1] introduced the concept of total graph of a graph. The total graph of a graph G, written T (G), is the graph whose vertices can be put in one-to-one correspondence with the vertices and edges of G in such a way that two vertices of T (G) are adjacent if and only if the corresponding elements of G are adjacent, where the vertices and edges of G are called its members. Gary Chatrand and James Stewart [2] extended the concept of total graph of a graph to the directed case thereby introducing the total digraph. The total digraph of a directed graph D, written T (D), is the digraph whose vertices are in one-to-one correspondence with the vertices and arcs of D and such that the vertex u is adjacent to the vertex v in T (D) if and only if the element corresponding to u is adjacent to the element corresponding to v in D. The concept of pathos of a graph G was introduced by Harary [4] as a collection of minimum number of edge disjoint open paths whose union is G. The path number of a graph G is the number of paths in any pathos. The path number of a tree T equals k, where 2k is the number of odd degree vertices of T . Stanton and Cowan [7] calculated the path number of certain classes of graphs like trees and complete graphs. Gudagudi [3] extended the concept of pathos of graphs to trees thereby introducing the concept called pathos line graph of a tree. A pathos line graph of a tree T , written PL(T ), is a graph whose vertices are the edges and paths of a pathos of T , with two vertices of PL(T ) adjacent whenever the corresponding edges of T are adjacent or the edge lies on the corresponding path of the pathos. Since the pattern of pathos for a tree is not unique, the corresponding pathos line graph is also not unique. The present study is on the directed pathos of total arborescence graphs denoted by DPT (Ar), where an arborescence graph Ar is a directed graph for which from an initial vertex u there is only one directed path going to another vertex v. 2. Preliminary Concepts and Results In this section, some concepts relating to directed pathos total digraph of an arbores- cence are defined and DPT (Ar) is discussed. Definition 2.1. A vertex u ∈ V (D) of a directed graph D is a root vertex if u is only an initial vertex, that is, d−(u) = 0. Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1333 Definition 2.2. [6] An arborescence, denoted by Ar, is a directed graph in which, from a root vertex u and for any other vertex v, there is exactly one directed path from u to v. Example 2.3. The graph T in Figure 1 is an example of an arborescence graph where vertex a is its root, that is, d−(a) = 0 and there is exactly one directed path from a to other vertices b, c, d, e, f, g, h, i, j, k, l, thus, T = Ar. Figure 1: An arborescence graph T Definition 2.4. [4] The pathos of a graph G is a collection of minimum number of edge disjoint open paths whose union is G. The path number of a graph G is the number of paths in a pathos. Definition 2.5. The directed pathos of an arborescence Ar is defined as a collection of minimum number of arc disjoint open directed paths whose union is Ar. Definition 2.6. [6] For an arborescence Ar, a directed pathos total digraph Q = DPT (Ar) has vertex set V (Q) = V (Ar) ∪ A(Ar) ∪ P (Ar), where V (Ar) is the vertex set, A(Ar) is the arc set, and P (Ar) is a directed pathos set of Ar. The arc set A(Q) consists of the following arcs: (i) ab such that a, b ∈ A(Ar) and the head of a coincides with the tail of b; (ii) uv such that u, v ∈ V (Ar) and u is adjacent to v or an arc from u to v exists; (iii) au(ua) such that a ∈ A(Ar) and u ∈ V (Ar) and the head (tail) of a is u; (iv) Pa such that a ∈ A(Ar) and P ∈ P (Ar) and the arc a lies on the directed path P ; and (v) PiPj such that Pi, Pj ∈ P (Ar) and it is possible to reach the head of Pj from the tail of Pi through a common vertex, but it is possible to reach the head of Pi from the tail of Pj. Definition 2.7. A vertex v ∈ V (G) is said to be an inner vertex of a planar digraph G if vertex v does not belong to the boundary of the exterior region in any embeddings of G in the plane. The inner vertex number i(G) is the maximum number of inner vertices of a planar digraph G. Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1334 Example 2.8. Consider the graphs in Figure 2 where the directed pathos total digraph of Ar is shown. Also, i(Q) = 2. Figure 2: Directed pathos total digraph of Ar In the context of directed graphs, a digraph G is outerplanar if i(G) = 0 and it is mini- mally non outerplanar if i(G) = 1. We now enumerate some of the characterizations of the planarity of the DPT (Ar) [6]. Theorem 2.9. Every DPT (Ar) is either strictly unilateral or strictly weak. Theorem 2.10. A directed pathos total digraph DPT (Ar) of an arborescence Ar is planar if and only if the underlying graph of Ar is a star graph K1,n on n ≤ 3 vertices. Theorem 2.11. A directed pathos total digraph DPT (Ar) of an arborescence Ar is out- erplanar if and only if Ar is either P⃗2 or P⃗3. Theorem 2.12. A directed pathos total digraph DPT (Ar) of an arborescence Ar is max- imal outerplanar if and only if Ar is P⃗3. Theorem 2.13. A directed pathos total digraph DPT (Ar) of an arborescence Ar is min- imally non outerplanar if and only if Ar is P⃗4. Theorem 2.14. A directed pathos total digraph DPT (Ar) of an arborescence Ar has crossing number one if and only if the underlying graph of Ar is K1,4. 3. Main Results This section presents some properties and characterizations of a directed pathos total digraph of an arborescence graph. The first result is intended for the arborescence which is a directed path P⃗n. Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1335 Theorem 3.1. If Ar = P⃗n, then the directed pathos total digraph DPT (Ar) of Ar is pla- nar. Proof: Suppose that Ar = P⃗n. Let V (P⃗n) = {v1, v2, v3, . . . , vn} and let A(P⃗n) = {e1, e2, e3, . . . , en−1} such that v1 and e1 = (v1, v2) are the root and root arc of P⃗n, re- spectively and ei = (vi, vi+1) for 2 ≤ i ≤ n − 1. Then v1, v2, . . . , vn, e1, e2, . . . , en−1 are the vertices of T (Ar). Also, (vi, vi+1), (vi, ei), (ei, vi+1), (ei, ei+1), for 1 ≤ i ≤ n − 1 are the arcs of T (Ar). Let P (Ar) = {P1} be a directed pathos of Ar such that P1 lies on the arcs e1 = (v1, v2), e2 = (v2, v3), . . . , en−1 = (vn−1, vn). The directed pathos vertex P is a neighbor of the vertices e1, e2, . . . , en−1. This shows that the crossing number of DPT (P⃗n) is zero, that is, cr(DPT (P⃗n)) = 0 (see Figure 3). Hence, DPT (P⃗n) is planar. Figure 3: Directed pathos total digraph of P⃗n Theorem 3.2. For an arborescence graph P⃗n with n ≥ 2 vertices and n − 1 arcs,∣∣A(DPT (Ar)) ∣∣ = 5(n− 1)− 1 = 5n− 6. Proof: We do this by induction. For n = 2, note that DPT (P⃗2) consists of vertices v1, v2, e1, and P and arcs (v1, v2), (v1, e1), (e1, v2), and (P, e1). (See Figure 4). Thus,∣∣A(DPT (P⃗3)) ∣∣ = 4 = 5(2− 1)− 1 = 5(n− 1)− 1. Figure 4: Digraph P⃗2 and its directed pathos total digraph Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1336 For n = 3, DPT (P⃗3) consists of vertices v1, v2, v3, e1, e2, and P and arcs (v1, v2), (v2, v3), (v1, e1), (v2, e2), (e1, v2), (e2, v3), (e1, e2), (P, e1), and (P, e2). (See Figure 5). Thus, ∣∣A(DPT (P⃗3)) ∣∣ = 9 = 5(3− 1)− 1 = 5(n− 1)− 1. Figure 5: Digraph P⃗3 and its directed pathos total digraph Assume that for P⃗n−1 with n− 1 vertices, ∣∣A(DPT (P⃗n−1)) ∣∣ = 5 ( (n− 1)− 1 ) − 1. That is,∣∣A(DPT (P⃗n−1)) ∣∣ = 5((n− 1)− 1)− 1 = 5(n− 2)− 1 = 5n− 10− 1 = 5(n− 1)− 6 = 5n− 11. That is, A ( DPT (P⃗n−1) ) consists of the arcs (v1, v2), (v2, v3), . . . , (vn−2, vn−1), (v1, e1), (v2, e2), . . . , (vn−2, en−1), (e1, v2), (e2, v3), . . . , (en−2, vn−1), (e1, e2), (e2, e3), . . . , (en−3, en−2), (P, e1), (P, e2), . . . , (P, en−2). (See Figure 6). Figure 6: Digraph P⃗n−1 and its directed pathos total digraph Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1337 Adding one vertex to P⃗n−1 results into a directed path P⃗n, with the additional arc (vn−1, vn). (See Figure 7). Figure 7: Digraph P⃗n and its directed pathos total digraph Hence, DPT (P⃗n) contains the arcs in DPT (P⃗n−1) and the arcs (vn−1, vn), (en−1, vn), (vn−1, en−1), (en−2, en−1), and (P, en−1). Therefore, |A(DPT (P⃗n))| = 5n− 11 + 5 = 5n− 6 = 5n− 5− 1 = 5(n− 1)− 1. We present a closely similar result from [6] in the next theorem using the usual notation of a path and taking into consideration a directed path as an arborescence. Theorem 3.3. The i(DPT (P⃗n)) = n− 3 if and only if n ≥ 4. Proof: Suppose that i(DPT (P⃗n)) = n − 3, where n < 4. Suppose n = 3 and Ar = P⃗3. Thus we have v1, v2, v3 as the vertices of Ar and e1 = (v1, v2) and e2 = v2, v3 as the arcs of Ar. Then the vertices of T (Ar) are {v1, v2, v3, e1, e2} and the arcs are (v1, v2), (v2, v3), (v1, e1), (e1, v2), (v2, e2), (e2, v3), (e1, e2). Let P (Ar) = {P2} where P2 = [v1v2, v2v3]. Therefore, DPT (Ar) is an outerplanar. A contradiction since DPT (Ar) should contain an internal vertex. (See Figure 5). Conversely, suppose that Ar = P⃗n for n ≥ 4. We will show i(DPT (P⃗n)) = n − 3 by induction. Let Ar = P⃗4 and let V (P⃗4) = {v1, v2, v3, v4}. Thus, V (DPT (P⃗4)) = {v1, v2, v3, v4, e1, e2, e3, P} where e1 = (v1, v2), e2 = (v2, v3), and e3 = (v3, v4) as the arcs of P⃗4, and P is the pathos of P⃗4. Hence, A(DPT (P⃗4)) = {(v1, v2), (v2, v3), (v3, v4), (v1, e1), (e1, v2), (v2, e2), (e2, v3), (v3, e3), (e3, v4), (e1, e2), (e2, e3), (P, e1), (P, e2), (P, e3)}. (See Figure 8). Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1338 Figure 8: DPT (P⃗4) of P⃗4 So e2 is the only internal vertex of DPT (P⃗4). That is, i(DPT (P⃗4)) = 1 = 4− 3 = n− 3. Assume that for n > 4, i(DPT (P⃗n−1)) = n − 1 − 3 = n − 4. That is, V (DPT (P⃗n−1)) = {v1, v2, . . . , vn−1, e1, e2, . . . , en−2, P} where e1 = (v1, v2), e2 = (v2, v3), . . . , en−2 = (vn−2, vn−1) and P is the pathos of P⃗n−1. Also, A(DPT (P⃗n−1)) = {(v1, v2), (v2, v3), . . . , (vn−2, vn−1), (v1, e1), (e1, v2), (v2, e2), (e2, v3), . . . , (vn−2, en−2), (en−2, vn−1), (e1, e2), (e2, e3), . . . , (e1, e2), (e2, e3), . . . , (en−3, en−2), (P, e1), (P, e2), . . . , (P, en−2)}. (See Figure 9). Figure 9: DPT (P⃗n−1) of P⃗n−1 It follows that e2, e3, . . . , en−3 are the internal vertices of DPT (P⃗n−1). That is, i(DPT (P⃗n−1)) = n − 3 − 2 + 1 = n − 4. Now, adding one vertex to the right side of vn−1 of P⃗n−1 to obtain P⃗n, we will have V (P⃗n) = {v1, v2, . . . , vn} and A(P⃗n) = {e1, e2, . . . , en−1} where e1 = (v1, v2), e2 = (v2, v3), . . . , en−2 = (vn−2, vn−1), en−1 = (vn−1, vn). Thus, an addition of the vertices vn and en−1 to the V (P⃗n) results in addi- tional arcs (vn−1, vn), (vn−1, en−1), (en−1, vn), (en−2, en−1), and (P, en−1). (See Figure 10). Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1339 Figure 10: DPT (P⃗n) of P⃗n Therefore, the vertices e2, e3, . . . , en−2 are the internal vertices of DPT (P⃗n). That is, i(DPT (P⃗n)) = n− 2− 2 + 1 = n− 3. In view of Theorem 3.3, a directed pathos total digraph DPT (P⃗n) is outerplanar for n = 3 and minimally non outerplanar for n = 4. Theorem 3.4. For an arborescence P⃗n, DPT (P⃗n) is strictly weak. Proof: Suppose that Ar = P⃗n. Let V (P⃗n) = {v1, v2, v3, . . . , vn} be the vertex set and let A(P⃗n) = {e1, e2, e3, . . . , en−1} be the arc set of P⃗n such that v1 and e1 = (v1, v2) are the root and root arc of P⃗n, respectively, and ei = (vi, vi+1) for 2 ≤ i ≤ n − 1. Then v1, v2, . . . , vn, e1, e2, . . . , en−1, P are the vertices of T (Ar) where P (Ar) = P is the directed pathos of Ar such that P lies on the arcs x1 = (v1, v2), x2 = (v2, v3), . . . , xn−1 = (vn−1, vn). Also, (vi, vi+1), (vi, ei), (ei, vi+1), (ei, ei+1), for 1 ≤ i ≤ n − 1 are the arcs of T (Ar) and since P lies on the arcs x1, x2, . . . , xn−1, the directed pathos vertex P is a neighbor of the vertices x1, x2, . . . , xn−1. Note that from v1, there is a semi-directed path to vertices v2, v3, . . . , vn and from v1, also there is a semi-directed path to vertices e1, e2, . . . , en−1. However, there is no semi-directed path from v1 to P . From Theorem 2.9, a directed pathos total digraph of an arborescence is either strictly unilateral or strictly weak. Thus, for any arborescence directed planar graph, its DPT (Ar) is strictly weak. Corollary 3.5. For any arborescence graph Ar containing K1,4, its DPT (Ar) is non- planar. Proof: This follows from Theorem 2.14. Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1340 Theorem 3.6. For an arborescence graph Ar = S⃗1,2(n), the directed pathos total digraph DPT (Ar) is nonplanar if and only if n ≥ 3. Proof: Suppose Ar = S⃗1,2(n), where n ≤ 2. Then Ar is just a path. By Theorem 3.1, DPT (Ar) is planar. Conversely, suppose that Ar = S⃗1,2(n), where n ≥ 3. For n = 3, let V (Ar) = {v1, v2, v3, v4, v5, v6, v7} be the vertex set and A(Ar) = {e1 = (v1v2), e2 = (v2v3), e3 = (v3v4), e4 = (v4v5), e5 = (v3v6), e6 = (v6v7)} be the arc set of Ar such that v1 and e1 = (v1, v2) are the root and root arc of Ar, respectively. Then we have the fol- lowing vertices for T (Ar), that is V (T (Ar)) = {v1, v2, . . . , v7, e1, e2, e3, . . . , e6} and arcs (vi, ei) for 1 ≤ i ≤ 4, 6, (v3, e5), (ei, vi+1) for 1 ≤ i ≤ 6, (vi, vi+1) for 1 ≤ i ≤ 4, 6, (v3, v6), (e1, e2), (e2, e3), (e3, e4), (e5, e6), (e2, e5). Let P (Ar) = {P1, P2} be a directed pathos set of Ar such that P1 lies on the arcs (v1, v2), (v2, v3), (v3, v4), (v4, v5); P2 lies on (v3, v6), (v6, v7). Thus the directed pathos vertex P1 is a neighbor of the vertices v1v2, v2v3, v3v4, v4v5; P2 is a neighbor of v3v6, v6v7. This shows that cr(DPT (Ar)) = 1, a nonplanar. For n ≥ 4, it is nonplanar since K1,4 is its subdigraph. This shows that cr(DPT (Ar)) ̸= 0 by Theorem 2.14. Hence, DPT (Ar) is nonplanar. Theorem 3.7. For an arborescence P⃗n, DPT (P⃗n) contains a K1,n−1 graph. Proof: From Theorem 3.1, note that P lies on the arcs e1e2, e2e3, . . . , en−2en−1. Thus, P is a neighbor of e1e2, e2e3, . . . , en−2en−1. Therefore, K1,n−1 is a subdigraph of DPT (P⃗n). Theorem 3.8. For an arborescence graph Ar which is an n-pan, the directed pathos total digraph DPT (Ar) has cr(n-pan) = 1 if and only if n ≥ 3. Proof: Suppose that Ar is an n-pan with n < 2 and cr(Ar) = 1. Let V (Ar) = {v1, v2, v3} be the vertex set and A(Ar) = {e1, e2} be the arc set of Ar such that v1 and e1 = (v1, v2) are the root and root arc of Ar, respectively. Thus, Ar ∼= P⃗3. By Theorem 3.1, all path graphs are planar, thus cr(Ar) = 0, a contradiction. Conversely, suppose that Ar is an n-pan graph with n ≥ 3 vertices. We consider the following cases. Case 1: Suppose that Ar is an n-pan graph with n = 3. Then V ( T (Ar) ) = {v1, v2, v3, v4, e1, e2, e3, e4} is the vertex set of T (Ar) and its arcs are (vi, vi+1), (ei, ei+1), (ei, vi+1) for 1 ≤ i ≤ 3, (vi, ei) for 1 ≤ i ≤ 4, (v4, v2), and (e4, e2). Let P (Ar) = {P1} be a directed pathos set of Ar such that P1 lies on the arcs (v1, v2), (v2, v3), (v3, v4), (v4, v2). Then the directed pathos vertex P1 is a neighbor of the vertices v1v2, v2v3, v3v4, v4v2. This shows that the crossing number of DPT (Ar) is one, that is, cr(DPT (Ar)) = 1 where (e4, e2) crosses (P1, e1). Case 2: Suppose that the underlying graph of Ar is an n-pan graph with n = 4. Then V ( T (Ar) ) = {v1, v2, . . . , vn, e1, e2, . . . , en} is the vertex set of T (Ar) and (vi, vi+1), (ei, ei+1), (ei, vi+1), (vi, ei), (v4, v2), and (e4, e2) are the arcs. Let P (Ar) = {P1} be a di- rected pathos set of Ar such that P1 lies on the arcs (v1, v2), (v2, v3), (v3, v4), . . . , (vi, vi+1), Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1341 (v4, v2) for 1 ≤ i ≤ n− 1. Then the directed pathos vertex P1 is a neighbor of the vertices v1v2, v2v3, v3v4, . . . , vivi+1, v4v2 for 1 ≤ i ≤ n− 1. This shows that the crossing number of DPT (Ar) is one, that is, cr(DPT (Ar)) = 1. For all n, (e4, e2) crosses (P1, e1). Theorem 3.9. If Ar = P⃗m(x1) • P⃗n(y1) then the directed pathos total digraph DPT (Ar) of Ar is planar where x1 and y1 are the initial vertices of P⃗m(x1) and P⃗n(y1) , respectively. Proof: Suppose that Ar = P⃗m • P⃗n. Let V (P⃗m) = {x1, x2, x3, . . . , xm} and let A(P⃗m) = {a1, a2, a3, . . . , am−1} such that x1 and a1 = (x1, x2) are the root and root arc of P⃗m, respectively and ai = (xi, xi+1) for 2 ≤ i ≤ m − 1. Then x1, x2, . . . , xm, a1, a2, . . . , am−1 are the vertices of T (P⃗m). Also, (xi, xi+1), (xi, ai), (ai, xi+1), (ai, ai+1), are the arcs of T (P⃗m). Let V (P⃗n) = {y1, y2, y3, . . . , yn} and let A(P⃗n) = {b1, b2, b3, . . . , bn−1} such that y1 and b1 = (y1, y2) are the root and root arc of P⃗n, respectively and bj = (yj , yj+1) for 2 ≤ j ≤ n − 1. Then y1, y2, . . . , yn, b1, b2, . . . , bn−1 are the vertices of T (P⃗n). Also, (yj , yj+1), (yj , bj), (bj , yj+1), (bj , bj+1), are the arcs of T (P⃗n). Let P (Ar) = {P1, P2} such that P1 lies on the arcs a1 = (x1, x2), a2 = (x2, x3), . . . , am−1 = (xm−1, xm) and P2 lies on the arcs b1 = (y1, y2), b2 = (y2, y3), . . . , bn−1 = (yn−1, yn). The directed pathos vertex P2 is a neighbor of the vertices b1, b2, . . . , bn−1. Note that x1 and y1 are the initial vertices of graphs P⃗m and P⃗n, respectively, so x1 = y1 in P⃗m • P⃗n. This shows that the cr(DPT (P⃗n)) = 0 (see Figure 11). Hence, DPT (P⃗n) is planar. Figure 11: Directed pathos total digraph of P⃗m(x1) • P⃗n(y1) Theorem 3.10. For an Ar = P⃗m(x1) • P⃗n(y1) , i(DPT (Ar)) = (m + n) − 6 if and only if m,n ≥ 4. Proof: Suppose that i(DPT (Ar)) = (m+ n)− 6, where m,n < 4. Let m,n = 3. Thus, we have v1, v2, v3, v4, v5, v6 as the vertices of Ar and e1 = (v1, v2), e2 = (v2, v3), e3 = (v1, v4) e4 = (v4, v5) as the arcs of Ar. Then the vertices of T (Ar) are {v1, v2, v3, v4, v5, e1, e2, e3, e4} Jill Maegan B. Pamplona, Imelda S. Aniversario / Eur. J. Pure Appl. Math, 15 (3) (2022), 1331-1343 1342 and the arcs are (v1, v2), (v2, v3), (v1, v4), (v4, v5), (v1, e1), (e1, v2), (v2, e2), (e2, v3), (v1, e3), (e3, v4), (v4, e4), (e4, v5). Let P (Ar) = {P1, P2} where P1 = [v1v2, v2v3] and P2 = [v1v4, v4v5]. Therefore, DPT (Ar) is an outerplanar. A contradiction since DPT (Ar) should contain an internal vertex. Conversely, suppose thatAr = P⃗m(x1) •P⃗n(y1) form,n ≥ 4. We will show i(DPT (P⃗m(x1) • P⃗n(y1) )) = (m + n) − 6 by induction. Let Ar = P⃗4(x1) • P⃗4(y1) and let V (P⃗4(x1) • P⃗4(y1) ) = {v1, v2, v3, v4, v5, v6, v7}. Thus, V (DPT (P⃗4(x1) •P⃗4(y1) )) = {v1, v2, v3, v4, v5, v6, v7, e1, e2, e3, e4, e5, e6, P1, P2} where e1 = (v1, v2), e2 = (v2, v3), e3 = (v3, v4), e4 = (v1, v5), e5 = (v5, v6), and e6 = (v6, v7) as the arcs of P⃗4(x1) • P⃗4(y1) , and {P1, P2} is the pathos set of P⃗4(x1) • P⃗4(y1) . Hence, A(DPT (P⃗4(x1) • P⃗4(y1) ))={(v1, v2), (v2, v3), (v3, v4), (v1, v5), (v5, v6), (v6, v7), (v1, e1), (e1, v2), (v2, e2), (e2, v3), (v3, e3), (e3, v4), (v1, e4), (e4, v5), (v5, e6), (e6, v7), (e1, e2), (e2, e3), (e4, e5), (e5, e6), (P1, e1), (P1, e2), (P1, e3), (P2, e4), (P2, e5), (P2, e6)}. So e2 and e5 are the only internal vertices of DPT (P⃗4(x1) • P⃗4(y1) ). That is, i(DPT (Ar)) = 2 = 4 + 4− 6 = m+ n− 6. Assume that form,n > 4, i(DPT (Ar)) = (m−1)+(n−1)−6. That is, V (DPT (Ar)) = {x1, x2 ,x3, . . . , xm−1, y1, y2, y3, . . . , yn−1, a1, a2, a3, . . . , am−2, b1, b2, b3, . . . , bn−2, P1, P2}. Also, A(DPT (Ar)) = {(x1, x2), (x2, x3), . . . , (xm−2, xm−1), (y1, y2), (y2, y3), . . . , (yn−2, yn−1), (x1, a1), (a1, x2), . . . , (xm−2, am−2), (am−2, xm−1), (y1, b1), (b1, y2), . . . , (yn−2, bn−2), (bn−2, yn−1), (a1, a2), . . . , (am−3, am−2), (b1, b2), . . . , (bn−3, bn−2), (P1, a1), . . . , (P1, am−2), (P2, b1), . . . , (P2, bn−2)}. It follows that a2, a3, . . . , am−3 and b2, b3, . . . , bn−3 are the internal vertices of DPT (Ar). That is i(DPT (P⃗m−1(x1) • P⃗n−1(y1) )) = m−3−1+n−3−1 = m+n−4. Now, adding one vertex to the right sides of xm−1 and yn−1 of P⃗m−1 and Pn−1 to obtain P⃗m•P⃗n, we will have V (DPT (P⃗m(x1) •P⃗n(y1) )) = {x1, x2, x3, . . . , xm−1, xm, y1, y2, y3, . . . , yn−1, yn, a1, a2, a3, . . . , am−2, am−1, b1, b2, b3, . . . , bn−2, bn−1, P1, P2}. Also, A(DPT (Ar)) = {(x1, x2), (x2, x3), . . . , (xm−2, xm−1), (xm−1, xm), (y1, y2), (y2, y3), . . . , (yn−2, yn−1), (yn−1, yn), (x1, a1), (a1, x2), . . . , (xm−2, am−2), (am−2, xm−1), (y1, b1), (b1, y2), . . . , (yn−2, bn−2), (bn−2, yn−1), (a1, a2), . . . , (am−3, am−2), (am−2, am−1), (b1, b2), . . . , (bn−3, bn−2), (bn−2, bn−1), (P1, a1), . . . , (P1, am−2), (P1, am−1), (P2, b1), . . . , (P2, bn−2), (P2, bn−1)}. Therefore, the vertices a2, a3, . . . , am−2 and b2, b3, . . . , bn−2 are the internal vertices of DPT (Ar). That is, i(DPT (P⃗m(x1) • P⃗n(y1) )) = m − 3 + n − 3 = m+ n− 6. Acknowledgements This research is funded by the Department of Science and Technology - Accelerated Sci- ence and Technology Human Resource Development Program (DOST-ASTHRDP), Philip- pines. REFERENCES 1343 References [1] M Behzad. Graphs and their chromatic numbers. Doctoral thesis, Michigan State University, 1967. [2] G Chartrand and MJ Stewart. Total digraph. Canadian Math. Bull., 9:171–176, 1966. 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