Sharp Upper and Lower Bounds of VDB Topological Indices of Digraphs

: A vertex-degree-based (VDB, for short) topological index ϕ induced by the numbers (cid:110) ϕ ij (cid:111) was recently deﬁned for a digraph D , as ϕ ( D ) = 1 2 ∑ uv ϕ d + u d − v , where d + u denotes the out-degree of the vertex u , d − v denotes the in-degree of the vertex v , and the sum runs over the set of arcs uv of D . This deﬁnition generalizes the concept of a VDB topological index of a graph. In a general setting, we ﬁnd sharp lower and upper bounds of a symmetric VDB topological index over D n , the set of all digraphs with n non-isolated vertices. Applications to well-known topological indices are deduced. We also determine extremal values of symmetric VDB topological indices over OT ( n ) and O ( G ) , the set of oriented trees with n vertices, and the set of all orientations of a ﬁxed graph G , respectively.


Introduction
A digraph D is a finite nonempty set V called vertices, together with a set A of ordered pairs of distinct vertices of D, called arcs. If a = (u, v) is an arc of D, then we write uv and say that the two vertices are adjacent. Given a vertex u of G, the out-degree of u is denoted by d + u and defined as the number of arcs of the form uv, where v ∈ V. The in-degree of u is denoted by d − u and defined as the number of arcs of the form wu, where w ∈ V. A vertex u in D is called a sink vertex (resp. source vertex) if d + u = 0 (resp. d − u = 0). We denote by q = q(D) the number of vertices of D which are sink vertices or source vertices. If d + u = d − u = 0, then u is an isolated vertex. The set of digraphs with n non-isolated vertices is denoted by D n .
One special class of digraphs is the oriented graphs. A pair of arcs of a digraph D of the form uv and vu are called symmetric arcs. If D has no symmetric arcs, then D is an oriented graph. We note that D can be obtained from a graph G by substituting each edge uv by an arc uv or vu, but not both. In this case, we say that D is an orientation of G. For example, in Figure 1 we show the directed path − → P n and the directed cycle − → C n , orientations of the path P n and cycle C n , respectively. A sink-source orientation of a graph G is an orientation in which every vertex is a sink vertex or a source vertex. Clearly, when we reverse the orientations of all arcs in a sink-source orientation, we obtain a sink-source orientation again. For instance, the digraphs − → K 1,n−1 and − → K n−1,1 in Figure 1 are sink-source orientations of the star S n . Note that − → K n−1,1 is obtained by reversing all arcs of − → K 1,n−1 . Let D 1 = (V 1 , A 1 ) and D 2 = (V 2 , A 2 ) be digraphs with no common vertices. The direct sum of digraphs D 1 and D 2 , denoted by D 1 ⊕ D 2 , is the digraph with vertex and arc sets V 1 ∪ V 2 and A 1 ∪ A 2 , respectively. In general, ⊕ k i=1 D i denote the direct sum of the digraphs Orientations of P n , C n , and S n .
The following notation and concepts were introduced in [1]. Let D ∈ D n . Let us denote by n + i (resp. n − i ) the number of vertices in D with out-degree (resp. in-degree) i, for all 0 ≤ i ≤ n − 1. For every 1 ≤ i, j ≤ n − 1, define the set The cardinality of A ij is denoted by a ij . Clearly, where a is the number of arcs D has. A VDB topological index is a function ϕ induced by real numbers ϕ ij , where 1 ≤ i, j ≤ n − 1, defined as [1] Equivalently, When ϕ ij = ϕ ji for all 1 ≤ i, j ≤ n − 1, we say that ϕ is a symmetric VDB topological index. In this case, the expression given in (2) can be simplified. In fact, let for all 1 ≤ i, j ≤ n − 1, and for all i = 1, . . . , n − 1. Then where In particular, when D = G is a graph, it was shown in [1] that Formula (6) reduces to where m ij is the number of edges in G which join vertices of degree i and j. So we recover the degree-based-topological indices of graphs, a concept which has been, and currently is, extensively investigated in the mathematical and chemical literature [2][3][4]. For recent results, we refer to [5][6][7][8][9][10][11][12]. This paper is organized as follows. In Section 2, in a general setting (Theorem 1), we find sharp lower and upper bounds of a symmetric VDB topological index over the set D n . As a byproduct, we obtain over D n , sharp upper and lower bounds of well-known VDB topological indices, which include the First Zagreb index M 1 (ϕ ij = i + j) [13], the Second Zagreb index M 2 (ϕ ij = ij) [13], the Randić index χ (ϕ ij = 1/ ij) [14], the Harmonic index H (ϕ ij = 2/(i + j)) [15], the Geometric-Arithmetic GA (ϕ ij = 2 ij/(i + j)) [16], the Sum-Connectivity SC (ϕ ij = 1/ i + j) [17], the Atom-Bond-Connectivity ABC (ϕ ij = (i + j − 2)/ij) [18], and the Augmented Zagreb AZ (ϕ ij = (ij/(i + j − 2)) 3 ) [19].
In Section 3, based on Theorem 2, we give sharp upper and lower bounds of symmetric VDB topological indices over the set OT (n), the set of oriented trees with n vertices. In particular, we deduce sharp upper and lower bounds for the well-known indices mentioned above over OT (n). Finally, in Section 4, we consider the problem of finding the extremal values of a symmetric VDB topological index among all orientations in O(G), the set of all orientations of a fixed graph G. In order to do this, we define strictly nondecreasing (resp. nonincreasing) symmetric VDB topological indices and show that for these indices, the value of any orientation at G is not greater (resp. smaller) than half the value at G. Moreover, equality occurs, and only if the orientation is a sink-source orientation of G. In particular, when G is a bipartite graph, we show that the sink-source orientations of G attain extremal values.

Bounds of VDB Topological Indices of Digraphs
From now on, when we say that ϕ is a symmetric VDB topological index, we mean that ϕ is induced by the numbers ϕ ij , where (i, j) ∈ K, and it is defined as in the equivalent definitions (2), (3), or (6). In the first part of this section, we generalize several results of [20] to digraphs.
Let ϕ be a symmetric VDB topological index. Consider the function f ij = ijϕ ij i+j defined over the set K. For each (r, s) ∈ K, consider the subset of K Recall that q is the number of vertices which are sink or source vertices of a digraph D.

Lemma 1.
Let ϕ be a symmetric VDB topological index and D ∈ D n . Let (r, s) ∈ K. Then Proof. The numbers p ij defined in (4) in (5) satisfy the relation (see (10) in [1]) Note that q = n + 0 + n − 0 . By (7), which implies On the other hand, Now, substituting (8) in (9), we deduce Let ϕ be a symmetric VDB topological index with associated function f ij = ijϕ ij i+j . Define the sets Theorem 1. Let ϕ be a symmetric VDB topological index and D ∈ D n . Then Moreover, equality on the left occurs, and only if p xy = 0 for all (x, y) ∈ K c min ( f ). Equality on the right occurs, and only if p xy = 0 for all (x, y) ∈ K c max ( f ).
Proof. Assume that f rs = max On the other hand, since The proof of the left inequality (and the equality condition) is similar.
So by Theorem 1, in order to find extremal values of a VDB topological index ϕ over i+j . Fortunately, these were computed for the main VDB topological indices in [21] (see Table 1). Table 1. K min ( f ) and K max ( f ) for some VDB topological indices.

VDB Index
Notation An important class of digraphs which occur frequently as extremal values of VDB topological indices are the arc-balanced digraphs, which we define as follows.
for every arc uv of D, and q = 0.
A regular digraph is a digraph D such that d + u = d − u = r, for all vertices u in D, where r is a positive integer. Clearly, every regular digraph is arc-balanced. Figure 2 are arc-balanced but not regular digraphs.  Table 1. The following result is clear.

Example 1. The digraphs in
for some nonnegative integers k 1 and k 2 . 2.
p ij = 0 for all (i, j) ∈ K such that i < j and q = 0 ⇔ D is an arc-balanced digraph; 3.

Lemma 3.
Assume that n is odd. Let D ∈ D n . If q = n, then p 11 ≤ n−3 2 .
Proof. Every vertex of D is a sink vertex or a source vertex. Consequently, where p 11 (E) = 0. In particular, n = n(E) + 2p 11 .

Corollary 1.
Let D ∈ D n . Then (a) Equality on the left occurs ⇔ n is even and D = n 2 − → P 2 or n is odd, and D = Equality on the right occurs ⇔ D = K n .

n 4
if n even (a) Equality on the left occurs ⇔ n is even and D = n 2 − → P 2 or n is odd and D = n−3 2

− →
Equality on the right occurs ⇔ D = K n .
(a) Equality on the left occurs Equality on the right occurs ⇔ D is an arc-balanced digraph.

5.
(n − 1) Equality on the right occurs ⇔ D = K n . Equality on the right occurs ⇔ D = K n .

(a) Equality on the left occurs
− → C n j , for some nonnegative integers Equality on the right occurs ⇔ D = K n .

(a) Equality on the left occurs
Equality on the right occurs ⇔ D = K n .
Proof. Recall that f ij = ijϕ ij i+j is the associated function of the symmetric VDB topological index ϕ. The expressions for f ij are shown in Table 2.
Since 0 ≤ q ≤ n, we easily deduce the result from Theorem 1 and Lemma 2. We only have to separately consider M 1 and M 2 when n is odd. By Theorem 1, Since n is odd, 2M 1 (D) > n, and so 2M 1 (D) ≥ n + 1. Equivalently, For the equality condition, it is clear that M 1 Conversely, suppose that M 1 (D) = n+1 2 . Then by (11), which implies q ≥ n − 1. So there are only two possibilities: q = n − 1 and q = n. If q = n, then by Lemma 3, p 11 ≤ n−3 2 . On the other hand, by Lemma 1 applied to (r, s) = (1, 2), Thus, which implies p 11 ≥ n−1 2 , a contradiction. Hence, q = n − 1. Consequently, It follows from Theorem 1 that p ij = 0 for all (i, j) = (1, 1). Finally, by Lemma 2, The case of M 2 when n is odd is similar.
In the case of the ABC index, note that ϕ ij = 0 if, and only if (i, j) = (1, 1). Then it is clear that Remark 1. Using a linear programming modeling technique, the authors in [22] find some of the extremal values given in Corollary 1.
Now we give bounds of VDB topological indices in terms of the number of arcs. Let ϕ be a symmetric VDB topological index. Let us define The complements in K are denoted by L c max and L c min , respectively.

Theorem 2.
Let ϕ be a symmetric VDB topological index. If D is a digraph with a arcs, then Equality on the left occurs if, and only if p ij = 0 for all (i, j) ∈ L c min . Equality on the right occurs if, and only if p ij = 0 for all (i, j) ∈ L c max .
Proof. From (2) and (1), If ϕ(D) = 1 2 a max K ϕ ij , then by (12) for all (i, j) ∈ K. Hence, if (i, j) ∈ L c max , then ϕ ij − max K ϕ ij = 0 and so p ij = 0. Conversely, if p ij = 0 for all (i, j) ∈ L c max , then The proof of the left inequality (and equality) is similar.

Bounds of VDB Topological Indices of Tree Orientations
The set of oriented trees with n vertices is denoted by OT (n). It is our interest in this section to determine the extremal values of a VDB topological index over OT (n). Clearly, a = n − 1 for every T ∈ OT (n). Hence, by Theorem 2 we deduce the following.

Corollary 2. Let T ∈ OT (n). Then
Equality on the left occurs if, and only if p ij = 0 for all (i, j) ∈ L c min . Equality on the right occurs if, and only if p ij = 0 for all (i, j) ∈ L c max .
Now we can obtain a first list of sharp upper and lower bounds for some VDB topological indices over OT (n).  3.
(n−1)  Proof. The inequalities on the left (and equality conditions) are immediate consequence of Corollary 1. The inequalities on the right of 1-4 are consequence of Corollary 2 having in mind Table 3. Table 3. L max and max K ϕ ij for χ, H, GA, and SC.
Proof. The inequalities on the left of 1-3 (and equality conditions) are a consequence of Corollary 2, having in mind Table 4.

VDB Index
And the fact that T ∈ OT (n) is such that p ij = 0 for all (i, j) = (1, 1) if, and only if On the other hand, the right inequality in 1 holds again by Corollary 2, bearing in mind Table 5.  The only extremal values we have not determined yet are the maximal values of M 1 , M 2 , and AZ over OT (n). The problem in these indices is that L max = (n − 1, n − 1), and there is no oriented tree such that p ij = 0 for all (i, j) = (n − 1, n − 1). In the next section we will show that the maximum value of M 1 and M 2 over OT (n) is attained in − → K 1,n−1 or − → K n−1,1 (see Theorem 6). We propose the following problem.

Problem 1.
Find the maximum value of AZ over OT (n).

Bounds of VDB Topological Indices over Orientations of a Fixed Graph
Let ϕ be a symmetric VDB topological index and G a graph. Let O(G) be the set of orientations of the graph G. Our main concern now is to determine the extremal values of a symmetric VDB topological index over O(G). In order to do this, let us define a partial order over K as follows: if (i, j), (k, l) ∈ K, then (i, j) (k, l) ⇔ i ≤ k and j ≤ l.

Example 2.
Consider the generalized Randić index χ α induced by the numbers (ij) α , where α ∈ R, α = 0. Clearly, χ α is strictly nondecreasing when α > 0, and strictly nonincreasing when α < 0. In particular, the Randić index χ is strictly nonincreasing and the second Zagreb index M 2 is strictly nondecreasing. Additionally, the harmonic index and the sum-connectivity index are strictly nonincreasing, and the first Zagreb M 1 is strictly nondecreasing.
Theorem 5. Let ϕ be a strictly nondecreasing (resp. nonincreasing) symmetric VDB topological index and G a graph. Let D be any orientation of G. Then Equality holds if, and only if D is a sink-source orientation of G.
Proof. We will assume that ϕ is strictly nondecreasing, and the other case is similar. Note that for every vertex u of G. Hence, for any arc uv of D, (d + u , d − v ) (d u , d v ). It follows by the nondecreasing property of ϕ and (3), If D is a sink-source orientation of G, then d + u = 0 or d − u = 0, for all vertices u of V. If vw is an arc of D then d + v = 0 and d − w = 0. Hence, d − v = 0 and d + w = 0, which implies by (13) that d v = d + v and d w = d − w . Hence, Conversely, assume that ϕ(D) = 1 2 ϕ(G). Then by (14), for every uv ∈ A Now since ϕ is strictly nondecreasing, (d + u , d − v ) = (d u , d v ) for every uv ∈ A. Finally, by (13), d − u = 0 and d + v = 0. This clearly implies that D is a sink-source orientation of G.

Corollary 3.
Let ϕ be a strictly nondecreasing (resp. nonincreasing) symmetric VDB topological index and G a bipartite graph. Then the maximal (resp. minimal) value of ϕ over O(G) is attained in a sink-source orientation of G.
Proof. We assume that ϕ is strictly nondecreasing, and the other case is similar. Since G is a bipartite graph, G has a sink-source orientation which we call E [23]. Let D be any orientation of G. Then by Theorem 5,