ISSN 2148-838Xhttps://doi.org/10.13069/jacodesmath.935980 J. Algebra Comb. Discrete Appl. 8(2) • 107–118 Received: 12 July 2020 Accepted: 5 January 2021 Journal of Algebra Combinatorics Discrete Structures and Applications General degree distance of graphs Research Article Tomáš Vetrík Abstract: We generalize several topological indices and introduce the general degree distance of a connected graph G. For a, b ∈ R, the general degree distance DDa,b(G) = ∑ v∈V (G)[degG(v)] aSbG(v), where V (G) is the vertex set of G, degG(v) is the degree of a vertex v, SbG(v) = ∑ w∈V (G)\{v}[dG(v, w)] b and dG(v, w) is the distance between v and w in G. We present some sharp bounds on the general degree distance for multipartite graphs and trees of given order, graphs of given order and chromatic number, graphs of given order and vertex connectivity, and graphs of given order and number of pendant vertices. 2010 MSC: 05C35, 05C12, 92E10 Keywords: Degree distance, Chromatic number, Vertex connectivity 1. Introduction We denote the vertex set and edge set of a graph G by V (G) and E(G), respectively The number of vertices of G is called the order. For v ∈ V (G), the degree of v, degG(v), is the number of vertices adjacent to v. The distance between two vertices v and w in G, denoted by dG(v,w), is the number of edges in a shortest path between v and w. We denote the complete graph and the star of order n by Kn and Sn, respectively. Topological indices are molecular descriptors which have been studied due to their extensive appli- cations. These graph invariants play an important role in engineering, materials science, pharmaceutical sciences and especially in chemistry, since they can be correlated with many chemical and physical prop- erties of molecules. Graph theory can be used to characterize these chemical structures. One of the most well-known distance-based topological indices is the degree distance. The degree distance of a connected graph G, DD(G) = ∑ {v,w}⊆V (G) (degG(v) + degG(w))dG(v,w), Tomáš Vetrík; Department of Mathematics and Applied Mathematics, University of the Free State, Bloemfontein, South Africa (email: vetrikt@ufs.ac.za). 107 https://orcid.org/0000-0002-0387-7276 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 was introduced independently by Dobrynin and Kochetova [5] and Gutman [6]. Bounds on the degree distance for graphs with given vertex connectivity were obtained in [2], bounds for graphs with given edge connectivity in [1], bounds for graphs of minimum degree in [13], bounds for cacti in [20] and bounds for bicyclic graphs in [3]. The degree distance for unicyclic graphs with prescribed matching number was investigated in [10] and graph products were studied in [19] and [16]. Relations between the degree distance and the eccentric distance sum were investigated in [9] and relations between the degree distance and the Gutman index in [4]. The reciprocal degree distance RDD(G) = ∑ {v,w}⊆V (G) degG(v) + degG(w) dG(v,w) of a connected graph G has been widely studied too. Bounds on the reciprocal degree distance for graphs with cut edges or cut vertices were given in [12], bounds for bipartite graphs and outerplanar graphs in [11]. The reciprocal degree distance of graph products was studied in [15] and the Steiner reciprocal degree distance in [17]. The generalized degree distance was first presented in [8] and studied for example in [7] and [14]. For a,b ∈ R, we introduce the general degree distance of a connected graph G as DDa,b(G) = ∑ v∈V (G) ( [degG(v)] a ∑ w∈V (G)\{v} [dG(v,w)] b ) = ∑ {v,w}⊆V (G) ([degG(v)] a + [degG(w)] a)[dG(v,w)] b. Let SbG(v) = ∑ w∈V (G)\{v}[dG(v,w)] b. Then we can write DDa,b(G) = ∑ v∈V (G) [degG(v)] aSbG(v). If a = 1, then DD1,b(G) is the generalized degree distance. If a = 1 and b = 1, we get the classical degree distance. If a = 1 and b = −1, we get the reciprocal degree distance. If a = 0 and b = 1, then DD0,1(G) = 2W(G), where W(G) is the Wiener index. If a = 0 and b = −1, then DD0,−1(G) = 2H(G), where H(G) is the Harary index. We present several bounds on the general degree distance of graphs. 2. Preliminary results Lemmas 2.1 and 2.2 are used in the proofs of some main results. Note that (a,b) 6= (0,0) means that not both a and b are 0. Lemma 2.1. Let G be any connected graph such that u1 and u2 are non-adjacent vertices in G. Then for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0), DDa,b(G + u1u2) < DDa,b(G). Proof. Let G′ be the graph G + u1u2. Then for any two vertices v,w ∈ V (G), we get dG′(v,w) ≤ dG(v,w) and [dG′(v,w)]b ≤ [dG(v,w)]b, where b ≥ 0. Therefore, SbG′(v) ≤ S b G(v) for each v ∈ V (G), where b ≥ 0. For v ∈ V (G) \ {u1,u2}, we have degG′(v) = degG(v), thus [degG′(v)]a = [degG(v)]a and [degG′(v)] aSbG′(v) ≤ [degG(v)] aSbG(v) for a ≤ 0 and b ≥ 0. Now we consider the vertices u1 and u2. Since 1 = dG′(u1,u2) < dG(u1,u2), we obtain [dG′(u1,u2)] b < [dG(u1,u2)] b for b > 0. Thus SbG′(ui) < S b G(ui) for i = 1,2. Note that if b = 0, then SbG′(ui) = S b G(ui). 108 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 For the degrees, we have degG′(ui) = degG(ui) + 1. For a < 0, we obtain [degG′(ui)]a < [degG(ui)]a, and for a = 0, we have [degG′(ui)]a = [degG(ui)]a = 1. It follows that [degG′(ui)]a SbG′(ui) < [degG(ui)] aSbG(ui) for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0). Thus DDa,b(G ′) = 2∑ i=1 [degG′(ui)] aSbG′(ui) + ∑ v∈V (G′)\{u1,u2} [degG′(v)] aSbG′(v) < 2∑ i=1 [degG(ui)] aSbG(ui) + ∑ v∈V (G)\{u1,u2} [degG(v)] aSbG(v) = DDa,b(G ′). Lemma 2.2. Let G be any connected graph such that u1 and u2 are non-adjacent vertices in G. Then for a ≥ 0 and b ≤ 0, where (a,b) 6= (0,0), DDa,b(G + u1u2) > DDa,b(G). Proof. Let G′ be the graph G + u1u2. Then for any two vertices v,w ∈ V (G), we get dG′(v,w) ≤ dG(v,w) and [dG′(v,w)]b ≥ [dG(v,w)]b, where b ≤ 0. Therefore, SbG′(v) ≥ S b G(v) for each v ∈ V (G), where b ≤ 0. For v ∈ V (G) \ {u1,u2}, we have degG′(v) = degG(v), thus [degG′(v)]a = [degG(v)]a and [degG′(v)] aSbG′(v) ≥ [degG(v)] aSbG(v) for a ≥ 0 and b ≤ 0. Now we consider the vertices u1 and u2. Since 1 = dG′(u1,u2) < dG(u1,u2), we obtain 1 = [dG′(u1,u2)] b > [dG(u1,u2)] b > 0 for b < 0. Thus SbG′(ui) > S b G(ui) for i = 1,2. Note that if b = 0, then SbG′(ui) = S b G(ui). For the degrees, we have degG′(ui) = degG(ui) + 1. For a > 0, we obtain [degG′(ui)]a > [degG(ui)]a, and for a = 0, we have [degG′(ui)]a = [degG(ui)]a = 1. It follows that [degG′(ui)]a SbG′(ui) > [degG(ui)] aSbG(ui) for a ≥ 0 and b ≤ 0, where (a,b) 6= (0,0). Thus DDa,b(G ′) = 2∑ i=1 [degG′(ui)] aSbG′(ui) + ∑ v∈V (G′)\{u1,u2} [degG′(v)] aSbG′(v) > 2∑ i=1 [degG(ui)] aSbG(ui) + ∑ v∈V (G)\{u1,u2} [degG(v)] aSbG(v) = DDa,b(G ′). The proof is complete. Lemma 2.3 was presented in [18]. We use it in the next section to compare the DDa,b indices of some graphs. Lemma 2.3. Let 1 ≤ x < y and c > 0. For a > 1 and a < 0, we have (x + c)a −xa < (y + c)a −ya. 3. Main results From Lemma 2.1, we know that among graphs of order n, the complete graph Kn is the graph having the smallest general degree distance for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0). Similarly, by Lemma 2.2, among graphs of order n, Kn is the graph having the largest general degree distance for a ≥ 0 and b ≤ 0, where (a,b) 6= (0,0). 109 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 For an integer k ≥ 2, a k-partite graph is a graph such that we can divide its vertices into k disjoint sets, where any vertices in the same set are non-adjacent. A 2-partite graph is called a bipartite graph. The complete k-partite graph with partite sets of orders n1,n2, . . . ,nk is denoted by Kn1,n2,...,nk. Any two vertices which are not in the same partite set of Kn1,n2,...,nk are adjacent. If ni and nj differ by at most 1 for every i,j, where 1 ≤ i < j ≤ k, then the graph Kn1,n2,...,nk of order n is called the Turán graph and it is denoted by T(n,k). We show that T(n,k) has the smallest DDa,b index among k-partite graphs with n vertices. Theorem 3.1. Let G be any k-partite graph of order n, where 2 ≤ k ≤ n. Then for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0), we have DDa,b(G) ≥ DDa,b(T(n,k)) with equality only if G is the Turán graph T(n,k). Proof. Let G′ be a graph having the smallest DDa,b index among k-partite graphs of order n. By Lemma 2.1, any two vertices in different partite sets must be adjacent. Thus G′ is Kn1,n2,...,nk for some positive integers n1,n2, . . . ,nk. We prove by contradiction that ni and nj differ by at most 1 for every i,j, where 1 ≤ i < j ≤ k. Assume that ni and nj differ by at least 2 for some i,j, where 1 ≤ i < j ≤ k. Without loss of generality, assume that n1 ≥ n2 + 2. We compare the DDa,b indices of G′ = Kn1,n2,...,nk and G ′′ = Kn1−1,n2+1,...,nk. For every vertex v from the first partite set and v′ from the second partite set of Kn1,n2,...,nk, we have degG′(v) = n−n1, SbG′(v) = 1 b(n−n1) + 2b(n1 −1) and degG′(v ′) = n−n2, SbG′(v ′) = 1b(n−n2) + 2b(n2 −1). For every vertex w from the first partite set and w′ from the second partite set of Kn1−1,n2+1,...,nk, we have degG′′(w) = n− (n1 −1), SbG′′(w) = 1 b(n− (n1 −1)) + 2b(n1 −2) and degG′′(w ′) = n− (n2 + 1), SbG′′(w ′) = 1b(n− (n2 + 1)) + 2bn2. For any other vertex z, we have degG′(z) = degG′′(z) and SbG′(z) = S b G′′(z). For b > 0, SbG′(v)−S b G′′(w) = 2 b −1 > 0, SbG′′(w)−S b G′′(w ′) = 2b(n1 −n2 −2)−n1 + n2 + 2 = (2b −1)(n1 −n2 −2) ≥ 0, SbG′′(w ′)−SbG′(v ′) = 2b −1 > 0, therefore 0 < SbG′(v ′) < SbG′′(w ′) ≤ SbG′′(w) < S b G′(v). For b = 0, we obtain SbG′(v ′) = SbG′′(w ′) = SbG′′(w) = S b G′(v) = n−1. Since n1 −1 ≥ n2 + 1, we have 0 < (n−n2)a < (n−n2 −1)a ≤ (n−n1 + 1)a < (n−n1)a 110 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 for a < 0. Obviously, for a = 0, we get (n−n2)a = (n−n2 −1)a = (n−n1 + 1)a = (n−n1)a = 1. Thus, for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0), we have DDa,b(G ′)−DDa,b(G′′) = n1(n−n1)aSbG′(v) + n2(n−n2) aSbG′(v ′) − (n1 −1)(n−n1 + 1)aSbG′′(w)− (n2 + 1)(n−n2 −1) aSbG′′(w ′) = (n1 −1)(n−n1)aSbG′(v) + (n2 + 1)(n−n2) aSbG′(v ′) + (n−n1)aSbG′(v)− (n−n2) aSbG′(v ′) − (n1 −1)(n−n1 + 1)aSbG′′(w)− (n2 + 1)(n−n2 −1) aSbG′′(w ′) > (n1 −1)(n−n1)aSbG′(v) + (n2 + 1)(n−n2) aSbG′(v ′) − (n1 −1)(n−n1 + 1)aSbG′′(w)− (n2 + 1)(n−n2 −1) aSbG′′(w ′) = (n1 −1)(n−n1)aSbG′′(w) + (n2 + 1)(n−n2) aSbG′′(w ′) + (n1 −1)(n−n1)a[SbG′(v)−S b G′′(w)] − (n2 + 1)(n−n2)a[SbG′′(w ′)−SbG′(v ′)] − (n1 −1)(n−n1 + 1)aSbG′′(w)− (n2 + 1)(n−n2 −1) aSbG′′(w ′) ≥ (n1 −1)(n−n1)aSbG′′(w) + (n2 + 1)(n−n2) aSbG′′(w ′) − (n1 −1)(n−n1 + 1)aSbG′′(w)− (n2 + 1)(n−n2 −1) aSbG′′(w ′) = (n1 −1)SbG′′(w)[(n−n1) a − (n−n1 + 1)a] + (n2 + 1)S b G′′(w ′)[(n−n2)a − (n−n2 −1)a] ≥ (n2 + 1)SbG′′(w ′)[(n−n1)a − (n−n1 + 1)a] + (n2 + 1)S b G′′(w ′)[(n−n2)a − (n−n2 −1)a] ≥ 0, since for a = 0, we obtain (n−n1)a−(n−n1 + 1)a = 0 and (n−n2)a−(n−n2 −1)a = 0, and for a < 0, by Lemma 2.3, (n−n2)a − (n−n2 −1)a > (n−n1 + 1)a − (n−n1)a, thus (n−n2)a − (n−n2 −1)a + (n−n1)a − (n−n1 + 1)a > 0. Hence DDa,b(G′)−DDa,b(G′′) > 0 and DDa,b(G′) > DDa,b(G′′). So G′ is not a graph with the smallest DDa,b index and we have a contradiction. Therefore, ni and nj differ by at most 1 which means that G′ is the Turán graph T(n,k). We can use k = 2 in Theorem 3.1 to get the following corollary for bipartite graphs. Corollary 3.2. Let G be any bipartite graph of order n, where n ≥ 2. Then for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0), we have DDa,b(G) ≥ DDa,b(Kdn 2 e,bn 2 c) with equality only if G is Kdn 2 e,bn 2 c. The chromatic number of a graph G is the minimum number of colors needed to color the vertices of G so that no two adjacent vertices have the same color. Let us present a bound on the general degree distance for graphs with given chromatic number. Theorem 3.3. Let G be any connected graph of order n and chromatic number χ, where 2 ≤ χ ≤ n. Then for a ≤ 0 and b ≥ 0, where (a,b) 6= (0,0), we have DDa,b(G) ≥ DDa,b(T(n,χ)) with equality only if G is the Turán graph T(n,χ). 111 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 Proof. Let G′ be any graph having the smallest DDa,b index in terms of order n and chromatic number χ. There is no edge between the vertices in the same color class, therefore G′ must be a χ-partite graph. Then, by Theorem 3.1, G′ is T(n,χ). The union H = H1 ∪ H2 and join F = H1 + H2 of the graphs H1 and H2 have the vertex sets V (H) = V (F) = V (H1) ∪ V (H2). The edge set E(H) = E(H1) ∪ E(H2). The set E(F) contains the edges in E(H) and the edges connecting each vertex in V (H1) and each vertex in V (H2). We show that (Kn−κ−1∪K1)+Kκ has the extremal DDa,b(G) index for graphs of given vertex connectivity, where a ≥ 1 and b ≤ 0. The vertex connectivity of G is the minimum number of vertices whose removal disconnects G. Theorem 3.4. Let G be any connected graph of order n and vertex connectivity κ, where 1 ≤ κ ≤ n−2. Then for a ≥ 1 and b ≤ 0, DDa,b(G) ≤ DDa,b((Kn−κ−1 ∪K1) + Kκ) with equality only if G is (Kn−κ−1 ∪K1) + Kκ. Proof. Let G′ be any graph with the largest DDa,b index with respect to order n and vertex connectivity κ. So there is a set A ⊂ V (G′) of cardinality κ, such that G′−A is disconnected (where G′−A is obtained from G′ be the removal of the vertices in A and the removal of each edge of G′ incident with a vertex in A). We can divide the vertices in V (G′) \A into the sets A1 and A2, where no vertex in A1 is adjacent to a vertex in A2. By Lemma 2.2, there is an edge connecting each pair of vertices in A1, each pair of vertices in A2 and the degree of each vertex in A is n−1 in G′. Let |A1| = n1 and |A2| = n2. Without loss of generality, assume that n1 ≥ n2 ≥ 1. We get n−κ = n1 + n2 and G′ is (Kn1 ∪Kn2) + Kκ. We prove that n2 = 1. Assume to the contrary that n2 ≥ 2 (where n1 ≥ n2). We compare the DDa,b indices of G′ = (Kn1 ∪Kn2) + Kκ and G′′ = (Kn1+1 ∪Kn2−1) + Kκ. For each z ∈ A, we get degG′(z) = degG′′(z) = n − 1 and SbG′(z) = S b G′′(z) = n − 1. For each v ∈ V (Kn1), degG′(v) = κ + n1 −1 and SbG′(v) = 1 b(κ + n1 −1) + 2bn2. For each v′ ∈ V (Kn2), we obtain degG′(v ′) = κ + n2 −1 and SbG′(v ′) = 1b(κ + n2 −1) + 2bn1. For each w ∈ V (Kn1+1), we have degG′′(w) = κ + n1 and SbG′′(w) = 1 b(κ + n1) + 2 b(n2 −1). For each w′ ∈ V (Kn2−1), we get degG′′(w ′) = κ + n2 −2, and SbG′′(w ′) = 1b(κ + n2 −2) + 2b(n1 + 1). For b ≤ 0, SbG′′(w ′)−SbG′(v ′) = 2b −1 ≤ 0, SbG′(v ′)−SbG′(v) = 2 b(n1 −n2)−n1 + n2 = (2b −1)(n1 −n2) ≤ 0, SbG′(v)−S b G′′(w) = 2 b −1 ≤ 0, therefore 0 < SbG′′(w ′) ≤ SbG′(v ′) ≤ SbG′(v) ≤ S b G′′(w). 112 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 Note that 0 < (κ + n2 −2)a < (κ + n2 −1)a ≤ (κ + n1 −1)a < (κ + n1)a. for a ≥ 1. Thus for a ≥ 1 and b ≤ 0, we have DDa,b(G ′)−DDa,b(G′′) = n1(κ + n1 −1)aSbG′(v) + n2(κ + n2 −1) aSbG′(v ′) − (n1 + 1)(κ + n1)aSbG′′(w)− (n2 −1)(κ + n2 −2) aSbG′′(w ′) = n1(κ + n1 −1)aSbG′(v) + n2(κ + n2 −1) aSbG′(v ′) − n1(κ + n1)aSbG′′(w)−n2(κ + n2 −2) aSbG′′(w ′) − (κ + n1)aSbG′′(w) + (κ + n2 −2) aSbG′′(w ′) < n1(κ + n1 −1)aSbG′(v) + n2(κ + n2 −1) aSbG′(v ′) − n1(κ + n1)aSbG′′(w)−n2(κ + n2 −2) aSbG′′(w ′) = n1(κ + n1 −1)aSbG′(v) + n2(κ + n2 −1) aSbG′(v ′) − n1(κ + n1)aSbG′(v)−n2(κ + n2 −2) aSbG′(v ′) − n1(κ + n1)a[SbG′′(w)−S b G′(v)] + n2(κ + n2 −2) a[SbG′(v ′)−SbG′′(w ′)] ≤ n1(κ + n1 −1)aSbG′(v) + n2(κ + n2 −1) aSbG′(v ′) − n1(κ + n1)aSbG′(v)−n2(κ + n2 −2) aSbG′(v ′) = n1S b G′(v)[(κ + n1 −1) a − (κ + n1)a] + n2S b G′(v ′)[(κ + n2 −1)a − (κ + n2 −2)a] ≤ n1SbG′(v)[(κ + n1 −1) a − (κ + n1)a] + n1S b G′(v)[(κ + n2 −1) a − (κ + n2 −2)a] ≤ 0, since for a = 1, we obtain (κ + n1 − 1)a − (κ + n1)a + (κ + n2 − 1)a − (κ + n2 − 2)a = 0 and for a > 1, by Lemma 2.3, (κ + n2 −1)a − (κ + n2 −2)a < (κ + n1)a − (κ + n1 −1)a, thus (κ + n1 −1)a − (κ + n1)a + (κ + n2 −1)a − (κ + n2 −2)a < 0. Hence DDa,b(G′) −DDa,b(G′′) < 0 and DDa,b(G′) < DDa,b(G′′). So G′ is not a graph with the largest DDa,b index and we have a contradiction. We obtain n2 = 1 and consequently, n1 = n−κ−1. Thus G′ is (Kn−κ−1 ∪K1) + Kκ. A pendant vertex is a vertex of degree one. Let Kn−k ? Sk+1 be obtained by joining k pendant vertices to one vertex of Kn−k. We study the DDa,b index for graphs with pendant vertices and show that Kn−k ? Sk+1 is the extremal graph. Theorem 3.5. Let G be any connected graph having order n and k pendant vertices, where 1 ≤ k ≤ n−3. Then for a ≥ 1 and b ≤ 0, where (a,b) 6= (1,0), DDa,b(G) ≤ DDa,b(Kn−k ? Sk+1) with equality only if G is Kn−k ? Sk+1. Proof. Let G′ be a graph with the largest DDa,b index with respect to order n and k pendant vertices. Let A be the set of pendant vertices of G′. By Lemma 2.2, there is an edge connecting each pair of vertices in V (G′) \ A. So G′ contains Kn−k as a subgraph. We prove that one vertex of that Kn−k is adjacent to all the k pendant vertices in G′. 113 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 Suppose to the contrary that Kn−k contains two vertices v and w, such that each of them is adjacent to a pendant vertex in G′. Let us denote the pendant neighbors of v by vi where i = 1,2, . . . ,n1, and the pendant neighbors of w by wj where j = 1,2, . . . ,n2. Clearly, n1,n2 are positive integers and n1 +n2 ≤ k. Without loss of generality, assume that n1 ≥ n2. We compare the DDa,b indices of the graphs G′ and G′′ having the same vertex sets, where E(G′′) = {vw1,vw2, . . . ,vwn1}∪E(G′)\{ww1,ww2, . . . ,wwn1}. We obtain degG′(z) = degG′′(z) and SbG′(z) = S b G′′(z) if z is not v,w,vi,wj, where i = 1,2, . . . ,n1 and j = 1,2, . . . ,n2. Let s = n − k − 1. Then s ≥ 2. We have degG′(v) = s + n1, degG′(w) = s + n2, degG′′(v) = s + n1 + n2 and degG′′(w) = s. We obtain SbG′(v) = (s + n1) + 2 b(k −n1), SbG′(w) = (s + n2) + 2 b(k −n2), SbG′′(v) = (s + n1 + n2) + 2 b(k −n1 −n2), SbG′′(w) = s + 2 bk, SbG′(vi) = 1 + 2 b(s + n1 −1) + 3b(k −n1), SbG′(wj) = 1 + 2 b(s + n2 −1) + 3b(k −n2), SbG′′(vi) = S b G′′(wj) = 1 + 2 b(s + n1 + n2 −1) + 3b(k −n1 −n2). For b ≤ 0, SbG′(vi)−S b G′′(vi) = n2(3 b −2b) ≤ 0, SbG′(wj)−S b G′′(wj) = n1(3 b −2b) ≤ 0. For b < 0, SbG′′(w)−S b G′(w) = n2(2 b −1) < 0, SbG′(w)−S b G′(v) = (n2 −n1) + 2 b(n1 −n2) = (2b −1)(n1 −n2) ≤ 0, SbG′(v)−S b G′′(v) = n2(2 b −1) < 0, therefore 0 < SbG′′(w) < S b G′(w) ≤ S b G′(v) < S b G′′(v). If b = 0, obviously 0 < SbG′′(w) = S b G′(w) = S b G′(v) = S b G′′(v). Note that 0 < sa < (s + n2) a ≤ (s + n1)a < (s + n1 + n2)a for a ≥ 1. Thus, for a ≥ 1 and b ≤ 0, we obtain DDa,b(G ′)−DDa,b(G′′) = [degG′(v)]aSbG′(v)− [degG′′(v)] aSbG′′(v) + [degG′(w)] aSbG′(w)− [degG′′(w)] aSbG′′(w) + n1∑ i=1 ([degG′(vi)] aSbG′(vi)− [degG′′(vi)] aSbG′′(vi)) + n2∑ j=1 ([degG′(wj)] aSbG′(wj)− [degG′′(wj)] aSbG′′(wj)) = (s + n1) aSbG′(v)− (s + n1 + n2) aSbG′′(v) + (s + n2) aSbG′(w) − saSbG′′(w) + n1[S b G′(vi)−S b G′′(vi)] + n2[S b G′(wj)−S b G′′(wj)] = (s + n1) aSbG′(v)− (s + n1 + n2) aSbG′(v) + (s + n2) aSbG′(w) − saSbG′(w) + (s + n1 + n2) a[SbG′(v)−S b G′′(v)] + sa[SbG′(w)−S b G′′(w)] + 2n1n2(3 b −2b) ≤ (s + n1)aSbG′(v)− (s + n1 + n2) aSbG′(v) 114 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 + (s + n2) aSbG′(w)−s aSbG′(w) = SbG′(v)[(s + n1) a − (s + n1 + n2)a] + SbG′(w)[(s + n2) a −sa] ≤ SbG′(v)[(s + n1) a − (s + n1 + n2)a] + SbG′(v)[(s + n2) a −sa] ≤ 0, since for a = 1, we have (s + n1)a − (s + n1 + n2)a + (s + n2)a −sa = 0 and for a > 1, by Lemma 2.3, (s + n2) a −sa < (s + n1 + n2)a − (s + n1)a, thus (s + n1) a − (s + n1 + n2)a + (s + n2)a −sa < 0. So DDa,b(G′)−DDa,b(G′′) < 0 for a > 1 and b ≤ 0. For a ≥ 1 and b < 0, we have SbG′(v)−S b G′′(v) = S b G′′(w)−S b G′(w) = n2(2 b −1) < 0, thus (s + n1 + n2) a[SbG′(v)−S b G′′(v)] + s a[SbG′(w)−S b G′′(w)] < 0 and we again get DDa,b(G′)−DDa,b(G′′) < 0. Therefore, DDa,b(G′) < DDa,b(G′′) for a ≥ 1 and b ≤ 0, where (a,b) 6= (1,0). Thus G′ is not a graph with the largest DDa,b index and we have a contradiction. Hence G′ is Kn−k ? Sk+1. The problem studied in the previous theorem is trivial for a = 1 and b = 0. All graphs G with n vertices and m edges have the same DD1,0(G) index. We obtain DD1,0(G) = ∑ v∈V (G) ( [degG(v)] 1 ∑ w∈V (G)\{v} [dG(v,w)] 0 ) = (n−1) ∑ v∈V (G) degG(v) = 2m(n−1). So, by Lemma 2.2, each graph containing Kn−k and k pendant vertices is a graph with the largest DD1,0 index with respect to order n and k pendant vertices. Finally, we consider connected graphs without cycles called trees. In the following proof, we show that the diameter of the extremal tree T of given order is at most 2, which means that T is a star. Note that the distance between any two furthest vertices v and w is called the diameter of T and a shortest path between v and w is a diametral path. Theorem 3.6. Let T be any tree of order n ≥ 4. Then for a ≥ 1 and b ≤ 0, where (a,b) 6= (1,0), we have DDa,b(T) ≤ DDa,b(Sn) with equality only if T is Sn. Proof. Let T ′ be a tree of order n with the largest DDa,b index. We prove that T ′ is Sn. If n ≤ 3, clearly T is Sn, so we study trees for n ≥ 4. Assume to the contrary that T ′ is not Sn. Thus the diameter of T ′ is d ≥ 3. We denote a diametral path of T ′ by u0u1u2 . . .ud, where degT′(u1) = n1 and degT′(u2) = n2. Clearly, n1,n2 ≥ 2. So u1 is adjacent to n1 −1 pendant vertices, say v1,v2, . . . ,vn1−1 (one of them is u0). We compare the DDa,b indices of the trees T ′ and T ′′ having the same vertex sets, while the edge set E(T ′′) = {u2v1, u2v2, . . . ,u2vn1−1}∪E(T ′)\{u1v1, u1v2, . . . ,u1vn1−1}. We have degT′′(u1) = 1 and degT′′(u2) = n1 + n2 −1. For i = 1,2, . . . ,n1 −1, dT′(u1,vi) = dT′′(u2,vi) = 1 and dT′′(u1,vi) = dT′(u2,vi) = 2, 115 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 thus for b ≤ 0, SbT′(u1)−S b T′′(u1) = (n1 −1)(1 b −2b) ≥ 0, SbT′(u2)−S b T′′(u2) = (n1 −1)(2 b −1b) ≤ 0. Since dT′(u1,z) = dT′′(u2,z) + 1 for each z ∈ V (T ′)\{u1,u2,v1,v2, . . . ,vn1−1}, we get [dT′(u1,z)] b < [dT′′(u2,z)] b and SbT′(u1) < S b T′′(u2) for b < 0. Obviously, [dT′(u1,z)] b = [dT′′(u2,z)] b and SbT′(u1) = S b T′′(u2) for b = 0. For any z other than u1,u2 and vi, where i = 1,2, . . . ,n1 − 1, we have degT′(z) = degT′′(z) and dT′(z,x) ≥ dT′′(z,x), where x ∈ V (T ′). For b ≤ 0, we obtain [dT′(z,x)]b ≤ [dT′′(z,x)]b and SbT′(z) ≤ S b T′′(z), therefore [degT′(z)] aSbT′(z) ≤ [degT′′(z)] aSbT′′(z). Similarly, [degT′(vi)]aSbT′(vi) ≤ [degT′′(vi)] aSbT′′(vi). Then, for a ≥ 1 and b ≤ 0, DDa,b(T ′)−DDa,b(T ′′) ≤ [degT′(u1)]aSbT′(u1)− [degT′′(u1)] aSbT′′(u1) + [degT′(u2)] aSbT′(u2)− [degT′′(u2)] aSbT′′(u2) = na1S b T′(u1)−1 aSbT′′(u1) + n a 2S b T′(u2)− (n1 + n2 −1) aSbT′′(u2) = na1S b T′(u1)−S b T′(u1) + n a 2S b T′′(u2)− (n1 + n2 −1) aSbT′′(u2) + [SbT′(u1)−S b T′′(u1)] + n a 2[S b T′(u2)−S b T′′(u2)] ≤ na1S b T′(u1)−S b T′(u1) + n a 2S b T′′(u2)− (n1 + n2 −1) aSbT′′(u2) = SbT′(u1)[n a 1 −1] + S b T′′(u2)[n a 2 − (n1 + n2 −1) a] ≤ SbT′′(u2)[n a 1 −1] + S b T′′(u2)[n a 2 − (n1 + n2 −1) a] ≤ 0, since for a = 1, we have na1 −1 + na2 − (n1 + n2 −1)a = 0 and for a > 1, by Lemma 2.3, na1 −1 a < (n1 + n2 −1)a −na2, thus na1 −1 + n a 2 − (n1 + n2 −1) a < 0. So DDa,b(T ′)−DDa,b(T ′′) < 0 for a > 1 and b ≤ 0. For a ≥ 1 and b < 0, we have SbT′(u1) < S b T′′(u2), thus SbT′(u1)[n a 1 −1] < S b T′′(u2)[n a 1 −1] and we again get DDa,b(T ′)−DDa,b(T ′′) < 0. Therefore, DDa,b(T ′) < DDa,b(T ′′) for a ≥ 1 and b ≤ 0, where (a,b) 6= (1,0), which is a contradic- tion. Hence G′ is Sn. Let us note that if a = 1 and b = 0, then all trees T of order n have the same DD1,0 index, since DD1,0(T) = ∑ v∈V (T) ( [degT (v)] 1 ∑ w∈V (T)\{v} [dT (v,w)] 0 ) = (n−1) ∑ v∈V (T) degT (v) = 2(n−1)2. 116 T. Vetrík / J. Algebra Comb. Discrete Appl. 8(2) (2021) 107–118 4. Conclusion We presented some bounds on the general degree distance for multipartite graphs and trees of given order, graphs of given order and chromatic number, graphs of given order and vertex connectivity, and graphs of given order and number of pendant vertices. There is a huge space for further research, since one can study lower and upper bounds on the DDa,b index for general graphs or special classes of graphs for various invariants of graphs. Let us state the following problems. Problem 4.1. 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