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Mathematics 2019, 7(4), 324; https://doi.org/10.3390/math7040324

Article
The Bounds of Vertex Padmakar–Ivan Index on k-Trees
1
Department of Mathematics and Physics, Texas A&M International University, Laredo, TX 78041, USA
2
Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
3
School of Mathematics and Physics, Anhui Jianzhu University, Hefei 230601, China
4
Department of Mathematics, The University of Mississippi, University, MS 38677, USA
*
Author to whom correspondence should be addressed.
Received: 26 January 2019 / Accepted: 11 March 2019 / Published: 1 April 2019

Abstract

:
The Padmakar–Ivan ( P I ) index is a distance-based topological index and a molecular structure descriptor, which is the sum of the number of vertices over all edges u v of a graph such that these vertices are not equidistant from u and v. In this paper, we explore the results of P I -indices from trees to recursively clustered trees, the k-trees. Exact sharp upper bounds of PI indices on k-trees are obtained by the recursive relationships, and the corresponding extremal graphs are given. In addition, we determine the P I -values on some classes of k-trees and compare them, and our results extend and enrich some known conclusions.
Keywords:
extremal values; PI index; k-trees; distance

1. Introduction

Let G be a simple connected non-oriented graph with vertex set V ( G ) and edge set E ( G ) . The distance d ( x , y ) between the vertices x , y V ( G ) is the minimum length of the paths between x and y in G. The oldest and most thoroughly examined molecular descriptor is Wiener index or path number [1], which was first considered in trees by Wiener in 1947 as follows: W ( G ) = { x , y } V ( G ) d ( x , y ) . Compared to Wiener index, Szeged index was proposed by Gutman [2] in 1994 that, given x y E ( G ) , let n x y ( x ) be the number of vertices w V ( G ) such that d ( x , w ) < d ( y , w ) , S z ( G ) = x y E ( G ) n x y ( x ) n x y ( y ) . Based on the considerable success of Wiener index and Sz index, Khadikar proposed a new distance-based index [3] to be used in the field of nano-technology, that is edge Padmakar–Ivan (PIe) index, P I e ( G ) = x y E ( G ) [ n e ( x ) + n e ( y ) ] , where n e ( x ) denotes the number of edges which are closer to the vertex x than to the vertex y, and n e ( y ) denotes the number of edges which are closer to the vertex y than to the vertex x, respectively.
It is easy to see that the above concept does not count edges equidistant from both ends of the edge e = x y . Based on this idea, Khalifeh et al. [4] introduced a new PI index of vertex version that P I ( G ) = P I v ( G ) = x y E ( G ) [ n x y ( x ) + n x y ( y ) ] . Note that, in order to obtain a good recursive formulas, we do not consider the vertices x , y for n x y ( x ) and n x y ( y ) . Thus, n x y ( x ) + n x y ( y ) n 2 .
Nowadays, Padmakar–Ivan indices are widely used in Quantitative Structure–Activity Relationship (QSAR) and Quantitative Structure–Property Relationship (QSPR) [5,6], and there are many interesting results [5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] between graph theory and chemistry. For instances, Klavžar [27] provided PI-partitions and arbitrary Cartesian product. Pattabiraman and Paulraja [28] presented the formulas for vertex PI indices of the strong product of a graph and the complete multipartite graph. Ilić and Milosavljević [29] established basic properties of weighted vertex PI index and some lower and upper bounds on special graphs. Wang and Wei [30] studied vertex PI index on an extention of trees (cacti). In [31], Das and Gutman obtained a lower bound on the vertex PI index of a connected graph in terms of numbers of vertices, edges, pendent vertices, and clique number. Hoji et al. [32] provided exact formulas for the vertex PI indices of Kronecker product of a connected graph G and a complete graph. Since the tree is a basic class of graphs in mathematics and chemistry, and these results indicate that either the stars or the paths attain the maximal or minimal bounds for particular chemical indices, then a natural question is how about the situations for vertex Padmakar–Ivan index?
Because P I index is a distance-based index and not very easy to calculate, we first consider the bipartite graph G with n vertices. Note that the tree is a subclass of bipartite graphs which have no odd cycles. By the definition of P I ( G ) and the assumption that we do not consider the vertices x , y for n x y ( x ) and n x y ( y ) , one can obtain that every edge of G has the P I -value as n 2 . Thus, the following observation is obtained.
Obervation 1.
For a bipartite graph G with n vertices and m edges, P I ( G ) = ( n 2 ) m . In particular, if G is a tree, then P I ( G ) = ( n 1 ) ( n 2 ) .
Next, we will consider the graphs with odd cycles. In particular, the general tree, k-tree, contains a lot of odd cycles. Then, we are going to consider the PI indices of k-trees and figure out whether or not a k-star or a k-path attains the maximal or minimal bound for P I -indices of k-trees. Our main results are as follows: Theorems 1 and 2 give the exact P I -values of k-stars, k-paths and k-spirals (see Definitions 1–5 below).
Theorem 1.
For any k-star S n k and k-path P n k with n = k p + s vertices, where p 0 is an integer and s [ 2 , k + 1 ] , we have
( i ) P I ( S n k ) = k ( n k ) ( n k 1 ) , ( i i ) P I ( P n k ) = k ( k + 1 ) ( p 1 ) ( 3 k p + 6 s 2 k 4 ) 6 + ( s 1 ) s ( 3 k s + 2 ) 3 .
Theorem 2.
For any k-spiral T n , c k with n k vertices, where c [ 1 , k 1 ] , we have
P I ( T n , c k ) = ( n k ) ( n k 1 ) ( 4 k n + 2 ) 3 i f n [ k , 2 k c ] , 3 c ( n 2 k + c 1 ) ( n 2 k + c ) + ( k c ) ( 2 c 2 + 3 n c 4 k c + 3 k n 4 k 2 6 k + 3 n 2 ) 3 i f n 2 k c + 1 .
Theorem 3 proves that k-stars achieve the maximal values of P I -values for k-trees, and Theorem 4 shows that k-paths do not arrive the minimal values and certain P I -values of k-spirals are less than that of k-paths.
Theorem 3.
For any k-tree T n k with n k 1 , we have P I ( T n k ) P I ( S n k ) .
Theorem 4.
For any k-spiral T n , c k with n k 1 , we have
( i ) P I ( P n k ) P I ( T n , c k ) i f c [ 1 , k + 1 2 ) , ( i i ) P I ( P n k ) P I ( T n , c k ) i f c [ k + 1 2 , k 1 ] .

2. Preliminary

In this section, we first give some notations and lemmas that are crucial in the following sections. As usual, G = ( V , E ) is a connected finite simple undirected graph with vertex set V = V ( G ) and edge set E = E ( G ) . Let | G | or | V | be the cardinality of V. For any S V ( G ) and F E ( G ) , we use G [ S ] to denote the subgraph of G induced by S, G S to denote the subgraph induced by V ( G ) S and G F to denote the subgraph of G obtained by deleting F. w ( G S ) is the number of components of G S and S is a cut set if w ( G S ) 2 . For any u , v V ( G ) , P u v is a path connecting u and v, d ( u , v ) is the distance between u and v, N ( v ) = N G ( v ) = { w V ( G ) , v w E ( G ) } is the neighborhood of v and N [ v ] = N ( v ) { v } . For any integers a , b with a b , the interval [ a , b ] is the set of all integers between a and b including a , b . In addition, let [ a , b ) = [ a , b ] { b } and ( a , b ] = [ a , b ] { a } . In particular, [ a , b ] = ϕ for a > b . f ( x ) is a derivative of any differentiable function f ( x ) , where x is the variable. x is the largest integer that is less than or equal to x; x is the smallest integer that is greater than or equal to x. It is clear that d is from 0 to the diameter of graphs. Other undefined notations are referred to [33].
It is commonly known that a chordal graph G with at least three vertices is a triangulated graph and contains a simplicial vertex, whose neighborhood induces a clique. During recent decades, there are many interesting studies related to chordal graphs. In 1969, Beineke and Pippert [7] gave the definition of k-trees, which is a significant subclass of chordal graphs. Now, we just give some definitions about k-trees below.
Definition 1.
For positive integers n , k with n k , the k-tree, denoted by T n k , is defined recursively as follows: The smallest k-tree is the k-clique K k . If G is a k-tree with n k vertices and a new vertex v of degree k is added and joined to the vertices of a k-clique in G, then the obtained graph is a k-tree with n + 1 vertices.
Definition 2.
For positive integers n , k with n k , the k-path, denoted by P n k , is defined as follows: starting with a k-clique G [ { v 1 , v 2 v k } ] . For i [ k + 1 , n ] , the vertex v i is adjacent to vertices { v i 1 , v i 2 v i k } only.
Definition 3.
For positive integers n , k with n k , the k-star, denoted by S n k , is defined as follows: Starting with a k-clique G [ { v 1 , v 2 v k } ] and an independent set S with | S | = n k . For i [ k + 1 , n ] , the vertex v i is adjacent to vertices { v 1 , v 2 v k } only.
Definition 4.
For positive integers n , k , c with n k and c [ 1 , k 1 ] , let v 1 , v 2 , , v n c be the simplicial ordering of P n c k c . The k-spiral, denoted by T n , c k , is defined as P n c k c + K c , which is, V ( T n , c k ) = { v 1 , v 2 , , v n } and E ( T n , c k ) = E ( P n c k c ) E ( K c ) { v 1 v l , v 2 v l , , v n c v l } , for l [ n c + 1 , n ] .
Definition 5.
Let v V ( T n k ) be a vertex of degree k whose neighbors form a k-clique of T n k , then v is called a k-simplicial vertex. Let S 1 ( T n k ) be the set of all k-simplicial vertices of T n k , for n k + 2 , and set S 1 ( K k ) = ϕ , S 1 ( K k + 1 ) = { v } , where v is any vertex of K k + 1 . Let G 0 = G , G i = G i 1 v i , where v i is a k-simplicial vertex of G i 1 , then { v 1 , v 2 , v n } is called a simplicial elimination ordering of the n-vertex graph G.
In order to consider the P I -value of any k-tree G, let G = G { u } be a k-tree obtained by adding a new vertex u to G. For any v 1 , v 2 V ( G ) , let d ( v 1 , v 2 ) be the distance between v 1 and v 2 in G, d ( v 1 , v 2 ) be the distance between v 1 and v 2 in G . Now, we define a function that measures the difference of P I -values of any edge relating a vertex from G to G as follows: f : { w V ( G ) , x y E ( G ) } to { 1 , 0 } as follows:
f ( w , x y ) = 0 , if w = u and d ( x , w ) = d ( y , w ) , 0 , if w V ( G ) and d ( x , w ) d ( x , w ) = d ( y , w ) d ( y , w ) , 1 , if o t h e r w i s e .
Using the construction of k-trees, we can derive the following lemmas. Note that P I ( x y ) = n x y ( x ) + n x y ( y ) and P I ( x y ) n 2 .
Lemma 1.
Let x y be any edge of a k-tree G with at least n k + 1 vertices, then P I ( x y ) n k 1 .
Proof. 
Since every vertex of any k-tree G with at least k + 1 vertices must be in some ( k + 1 ) -cliques, which is, | N ( x ) N ( y ) | k 1 for any x y E ( G ) , we have P I ( x y ) n ( k 1 ) 2 = n k 1 .  □
Lemma 2.
Let x y be any edge of a k-tree G with n vertices and G = G { u } be a k-tree obtained by adding u to G. If w V ( G ) , then f ( w , x y ) = 0 .
Proof. 
By adding u to G, since G is a k-tree, we can get that the distance of any pair of vertices of G will increase at most 1, then f ( w , x y ) 1 . If w V ( G ) , then there exists a shortest path P x w or P y w such that u V ( P x w ) or V ( P y w ) , that is, f ( w , x y ) = 0 . □
Lemma 3.
For any k-path G with n vertices, where n k + 2 , let S 1 ( G ) = { v 1 , v n } and { v 1 , v 2 , , v n } be the simplicial elimination ordering of G, then d ( v i , v j ) = j i k , for i < j and i , j [ 1 , n ] . Furthermore, if n = k p + s with p 1 , s [ 2 , k + 1 ] , then
d ( v , v k p + s ) = p + 1 i f v { v 1 , v 2 , , v s 1 } , p i i f v { v k i + s , v k i + s + 1 , , v k ( i + 1 ) + s 1 } , i [ 0 , p 1 ] .
Proof. 
If j i k , then v i , v j must be in the same ( k + 1 ) -clique of G, and we have d ( v i , v j ) = 1 ; if j i k + 1 , then P v i v j = v i v i + k v i + 2 k v i + ( j i k 1 ) k v i + j i k k v j is one of the shortest paths between v i and v j . Thus, d ( v i , v j ) = j i k and Lemma 3 is proved. □
Lemma 4.
For any k-spiral T n , c k with n vertices and v i , v j V ( T n , c k ) for i < j ,
d ( v i , v j ) = 1 , i f j i k c , i , j [ 1 , n c ] , 1 , i f i o r j [ n c + 1 , n ] , 2 , i f j i k c + 1 , i , j [ 1 , n c ] .
Proof. 
If j i k c with i , j [ 1 , n c ] , by Definition 4, we can get that v i , v j must be in the same ( k + 1 ) -clique of G and d ( v i , v j ) = 1 ; If i or j [ n c + 1 , n ] , without loss of generality, say v i such that i [ n c + 1 , n ] , then N [ v i ] = V ( T n , c k ) , that is, d ( v i , v j ) = 1 ; If j i k c + 1 with i , j [ 1 , n c ] , then v i N ( v j ) and P v i v j = v i v n v j is one of the shortest paths between v i and v j , that is, d ( v i , v j ) = 2 . Thus, Lemma 4 is proved. □

3. Main Proofs

In this section, we give the proofs of main results by inductions. For a k-tree T n k , if n = k or k + 1 , then T n k is a k or (k + 1)-clique, that is, P I ( T n k ) = 0 . Thus, all of the theorems are true and we will only consider the case when n k + 2 below.
Proof of Theorem 1.
For ( i ) , let V ( S n k ) = { u 1 , u 2 , , u n } , G [ { u 1 , , u k } ] be a k-clique and N ( u l 0 ) = { u 1 , u 2 , , u k } for l 0 k + 1 . Just by Definition 3, we can get that for i , j [ 1 , k ] , N [ u i ] = N [ u j ] = V ( S n k ) , then P I ( u i u j ) = n u i u j ( u i ) + n u i u j ( u j ) = 0 ; for i [ 1 , k ] and l 0 [ k + 1 , n ] , | N [ u i ] N [ u l 0 ] | = n k 1 , then P I ( u i u l ) = n k 1 . Thus, we can get P I ( S n k ) = i , j [ 1 , k ] P I ( u i u j ) + i [ 1 , k ] , l 0 [ k + 1 , n ] P I ( u i u l 0 ) = k ( n k ) ( n k 1 ) .
For ( i i ) , we will proceed it by induction on | P n k | = n k + 2 . If n = k + 2 , let { v 1 , v 2 , , v k + 2 } be the simplicial elimination ordering of P k + 2 k . By Lemma 3, we can get that P I ( v 1 v i ) = 1 , P I ( v i v i ) = 0 and P I ( v i v k + 2 ) = 1 for i , i [ 2 , k + 1 ] . Thus, P I ( P k + 2 k ) = i = 2 k + 1 P I ( v 1 v i ) + i = 2 k + 1 P I ( v i v k + 2 ) = 2 k . Assume that Theorem 1 is true for a k-path with at most k p + s 1 vertices, where p 1 , 2 s k + 1 . Let P n k be a k-path such that | P n k | = k p + s , V ( P n k ) = { v 1 , v 2 , , v k p + s } and { v 1 , v 2 , , v k p + s } be the simplicial elimination ordering of P n k . Set P n 1 k = P n k { v k p + s } , then { v 1 , v 2 , , v k p + s 1 } is the simplicial elimination ordering of P n 1 k and for any edge v i v j E ( P n k ) , d ( v i , v j ) or d ( v i , v j ) is the distance of v i and v j in P n 1 k or P n k , respectively.
Let α = [ k ( k + 1 ) ( p 1 ) ( 3 k p + 6 s 2 k 4 ) 6 + ( s 1 ) s ( 3 k s + 2 ) 3 ] [ k ( k + 1 ) ( p 1 ) ( 3 k p + 6 s 2 k 10 ) 6 + ( s 2 ) ( s 1 ) ( 3 k s + 3 ) 3 ] = p k 2 + p k k 2 3 k + 2 k s s 2 + 3 s 2 . If we can show that by adding v k p + s to P n 1 k , P I ( P n k ) = P I ( P n 1 k ) + α , then Theorem 1 is true.
Set w = v k p + s , A 1 = { v 1 v s , v 1 v s + 1 , , v 1 v k + 1 } , A 2 = { v 2 v s , , v 2 v k + 2 } , , A s 1 = { v s 1 v s , , v s 1 v k + s 1 } and B 1 = { v 1 v 2 , v 1 v 3 , , v 1 v s 1 } , B 2 = { v 2 v 3 , , v 2 v s 1 } , , B s 2 = { v s 2 v s 1 } , B s 1 = ϕ . By Definition 2 and Lemma 3, we have d ( v 1 , v k p + s ) = p + 1 , d ( v s , v k p + s ) = p and d ( v 1 , v k p + s ) = p + 1 , d ( v 2 , v k p + s ) = p + 1 , that is, d ( v 1 , v k p + s ) d ( v s , v k p + s ) and d ( v 1 , v k p + s ) = d ( v 2 , v k p + s ) . Thus, f ( w , v 1 v s ) = 1 and f ( w , v 1 v 2 ) = 0 . Similarly, for any edge v h 1 v h 2 i = 1 s 1 A i with h 1 < h 2 , we have d ( v h 1 , v k p + s ) d ( v h 2 , v k p + s ) , that is, f ( w , v h 1 v h 2 ) = 1 ; For v h 1 v h 2 i = 1 s 1 B i with h 1 < h 2 , we have d ( v h 1 , v k p + s ) = d ( v h 2 , v k p + s ) , that is, f ( w , v h 1 v h 2 ) = 0 . Thus, we can get that
f ( v k p + s , x y ) = 1 , if x y i = 1 s 1 A i , 0 , if x y i = 1 s 1 B i .
For t [ 0 , p 2 ] , set A k t + s = { v k t + s v k ( t + 1 ) + s } , A k t + s + 1 = { v k t + s + 1 v k ( t + 1 ) + s , v k t + s + 1 v k ( t + 1 ) + s + 1 } , , A k ( t + 1 ) + s 1 = { v k ( t + 1 ) + s 1 v k ( t + 1 ) + s , v k ( t + 1 ) + s 1 v k ( t + 1 ) + s + 1 , , v k ( t + 1 ) + s 1 v k ( t + 2 ) + s 1 } , and B k t + s = { v k t + s v k t + s + 1 , , v k t + s v k ( t + 1 ) + s 1 } , B k t + s + 1 = { v k t + s + 1 v k t + s + 2 , , v k t + s + 1 v k ( t + 1 ) + s 1 } , , B k ( t + 1 ) + s 2 = { v k ( t + 1 ) + s 2 v k ( t + 1 ) + s 1 } , B k ( t + 1 ) + s 1 = ϕ . For t = 0 and by Lemma 3, we have d ( v s , v k p + s ) = p , d ( v k + s , v k p + s ) = p 1 and d ( v s , v k p + s ) = p , d ( v s + 1 , v k p + s ) = p , that is, d ( v s , v k p + s ) d ( v k + s , v k p + s ) and d ( v s , v k p + s ) = d ( v s + 1 , v k p + s ) . Thus, f ( w , v s v k + s ) = 1 and f ( w , v s v s + 1 ) = 0 . similarly, for any edge v h 1 v h 2 i = k t + s k ( t + 1 ) + s 1 A i with h 1 < h 2 , we have d ( v h 1 , v k p + s ) d ( v h 2 , v k p + s ) , that is, f ( w , v h 1 v h 2 ) = 1 ; for v h 1 v h 2 i = k t + s k ( t + 1 ) + s 1 B i with h 1 < h 2 , we have d ( v h 1 , v k p + s ) = d ( v h 2 , v k p + s ) , that is, f ( w , v h 1 v h 2 ) = 0 . Thus, we can get that
f ( v k p + s , x y ) = 1 , if x y i = k t + s k ( t + 1 ) + s 1 A i , 0 , if x y i = k t + s k ( t + 1 ) + s 1 B i .
Next, we consider the edges in the ( k + 1 ) -clique P n k [ N [ v k p + s ] ] . For any edge v h 1 v h 2 with h 1 , h 2 [ k ( p 1 ) + s , k p + s 1 ] , we have d ( v h 1 , v k p + s ) = d ( v h 2 , v k p + s ) = 1 , that is, f ( w , v h 1 v h 2 ) = 0 . For any edge v h v k p + s with h [ k ( p 1 ) + s , k p ] , by Lemma 3, we can obtain that d ( v 1 , v h ) = p , d ( v 1 , v k p + s ) = p + 1 , d ( v h k , v h ) = 1 , d ( v h k , v k p + s ) = 2 and when h k ( p 1 ) + s , d ( v k ( p 1 ) + s , v h ) = 1 , d ( v k ( p 1 ) + s , v k p + s ) = 1 , that is, d ( v 1 , v h ) d ( v 1 , v k p + s ) , d ( v h k , v h ) d ( v h k , v k p + s ) and d ( v k ( p 1 ) + s , v h ) = d ( v k ( p 1 ) + s , v k p + s ) . Similarly, we get that for j [ 1 , p 1 ] , j [ 1 , p ] and l h ,
d ( v l , v h ) d ( v l , v k p + s ) if l [ 1 , s 1 ] [ h j k , k ( p j ) + s 1 ] , d ( v l , v h ) = d ( v l , v k p + s ) if l [ k ( p j ) + s , h j k + k 1 ] [ h + 1 , k p + s 1 ] .
Thus, if v h = v k ( p 1 ) + s , then d ( v l , v k ( p 1 ) + s ) d ( v l , v k p + s ) with l [ 1 , s 1 ] { j = 1 p 1 [ k ( p 1 ) + s j k , ( p j ) k + s 1 ] } = [ 1 , ( p 1 ) k + s 1 ] and d ( v l , v k ( p 1 ) + s ) = d ( v l , v k p + s ) with l [ ( p 1 ) k + s + 1 , k p + s ] , that is, P I ( v k ( p 1 ) + s v k p + s ) = ( p 1 ) k + s 1 ; similarly, we can obtain that P I ( v k ( p 1 ) + s + 1 v k p + s ) = ( p 1 ) ( k 1 ) + s 1 ; P I ( v k ( p 1 ) + s + 2 v k p + s ) = ( p 1 ) ( k 2 ) + s 1 ; ; P I ( v k p v k p + s ) = ( p 1 ) s + s 1 .
For any edge v h v k p + s with h [ k p + 1 , k p + s 1 ] , by Lemma 3, we can obtain that d ( v h k , v h ) = 1 , d ( v h k , v k p + s ) = 2 and d ( v k ( p 1 ) + s , v h ) = 1 , d ( v k ( p 1 ) + s , v k p + s ) = 1 , that is, d ( v h k , v h ) d ( v h k , v k p + s ) and d ( v k ( p 1 ) + s , v h ) = d ( v k ( p 1 ) + s , v k p + s ) . Similarly, we get that for j [ 1 , p ] and l h ,
d ( v l , v h ) d ( v l , v k p + s ) if l [ h j k , k ( p j ) + s 1 ] , d ( v l , v h ) = d ( v l , v k p + s ) if l [ k ( p j ) + s , h j k + k 1 ] [ h + 1 , k p + s 1 ] .
Thus, if v h = v k p + 1 , then d ( v l , v k p + 1 ) d ( v l , v k p + s ) for l j = 1 p [ k p + 1 j k , k ( p j ) + s 1 ] and d ( v l , v k p + 1 ) = d ( v l , v k p + s ) with l { j = 1 p [ k ( p j ) + s , k ( p + 1 j ) ] } [ h + 1 , k p + s 1 ] , that is, P I ( v k p + 1 v k p + s ) = ( s 1 ) p ; similarly, we have P I ( v k p + 1 v k p + s ) = ( s 2 ) p ; ; P I ( v k p + s 2 v k p + s ) = 2 p ; P I ( v k p + s 1 v k p + s ) = p .
Set w V ( P n 1 k ) , if x y E ( P n k ) with x or y v k p + s , by Lemma 2, we have f ( w , x y ) = 0 . Thus,
P I ( P n k ) P I ( P n 1 k ) = x y i = 1 k ( p 1 ) + s 1 ( A i B i ) f ( w , x y ) + P I ( v k ( p 1 ) + s v k p + s ) + P I ( v k ( p 1 ) + s + 1 v k p + s ) + + P I ( v k p + s 1 v k p + s ) = [ ( k + 2 s ) + ( k + 3 s ) + + k ] + ( 1 + 2 + + k ) ( p 1 ) + [ k ( p 1 ) + s 1 ] + [ ( k 1 ) ( p 1 ) + s 1 ] + [ ( k 2 ) ( p 1 ) + s 1 ] + + [ s ( p 1 ) + s 1 ] + [ ( s 1 ) p + ( s 2 ) p + + 2 p + p ] = p k 2 + p k k 2 3 k + 2 k s s 2 + 3 s 2 = α .
Thus, P I ( P n k ) = k ( k + 1 ) ( p 1 ) ( 3 k p + 6 s 2 k 4 ) 6 + ( s 1 ) s ( 3 k s + 2 ) 3 , for | P n k | = k p + s and Theorem 1 is proved. □
Proof of Theorem 2.
We will proceed with it by induction on n k + 2 . If n = k + 2 , by Definition 4, we have T n , c k is also a k-path, that is, P I ( T n , c k ) = 2 k . If n k + 3 , assume that Theorem 2 is true for the k-spiral with at most n 1 vertices, we will consider T n , c k with n vertices. Let T n , c k be a k-spiral with V ( T n , c k ) = V ( T n 1 , c k ) { v } and E ( T n , c k ) = E ( T n 1 , c k ) { v v n 1 , v v n 2 , , v v n k } such that v 1 , v 2 , , v n c 1 is the simplicial ordering of P n c 1 k c , where T n 1 , c k = P n c 1 k c + K c with V ( T n 1 , c k ) = { v 1 , v 2 , , v n 1 } and E ( T n 1 , c k ) = E ( P n c 1 k c ) E ( K c ) { v 1 v l , v 2 v l , , v n c 1 v l } for l [ n c , n 1 ] . For any edge v i v j E ( T n , c k ) , d ( v i , v j ) or d ( v i , v j ) is the distance of v i and v j in T n 1 , c k or T n , c k , respectively.
For k + 2 n 2 k c , let γ = ( n k ) ( n k 1 ) ( 4 k n + 2 ) 3 ( n k 1 ) ( n k 2 ) ( 4 k n + 3 ) 3 = ( n k 1 ) ( 3 k n + 2 ) . If we can show that by adding v to T n 1 , c k , P I ( T n , c k ) = P I ( T n 1 , c k ) + γ , then Theorem 2 is true.
Set w = v and let l [ n c , n 1 ] , by Lemma 4, we have d ( v l , v ) = 1 and d ( v i , v ) = 2 for i [ 1 , n k 1 ] , that is, f ( w , v l v i ) = 1 ; d ( v l , v ) = d ( v i , v ) = 1 for i [ n k , n 1 ] , that is, f ( w , v l v i ) = 0 . Set C 1 = { v 1 v 2 , v 1 v 3 , , v 1 v n k 1 } , C 2 = { v 2 v 3 , v 2 v 4 , , v 2 v n k 1 } , , C n k 2 = { v n k 2 v n k 1 } , C n k 1 = ϕ , D 1 = { v 1 v n k , v 1 v n k + 1 , , v 1 v k c + 1 } , D 2 = { v 2 v n k , v 2 v n k + 1 , , v 2 v k c + 2 } , , D n k 1 = { v n k 1 v n k , v n k 1 v n k + 1 , , v n k 1 v n c 1 } . By Lemma 4, we have d ( v 1 , v ) = d ( v 2 , v ) = 2 and d ( v n k , v ) = 1 , that is, f ( w , v 1 v 2 ) = 0 and f ( w , v 1 v n k ) = 1 . Similarly, for v h 1 v h 2 i = 1 n k 1 C i with h 1 < h 2 , we have d ( v h 1 , v ) = d ( v h 2 , v ) = 2 , that is, f ( w , v h 1 v h 2 ) = 0 ; for v h 1 v h 2 i = 1 n k 1 D i with h 1 < h 2 , we have d ( v h 1 , v ) = 2 and d ( v h 2 , v ) = 1 , that is, f ( w , v h 1 v h 2 ) = 1 . Set C n k = { v n k v n k + 1 , v n k v n k + 2 , , v n k v n c 1 } , C n k + 1 = { v n k + 1 v n k + 2 , v n k + 1 v n k + 3 , , v n k + 1 v n c 1 } , , C n c 2 = { v n c 2 v n c 1 } . By Lemma 4, we have d ( v n k , v ) = d ( v n k 1 , v ) = 1 , that is, f ( w , v n k v n k 1 ) = 0 . Similarly, for v h 1 v h 2 i = n k n c 2 C i with h 1 < h 2 , we have d ( v h 1 , v ) = d ( v h 2 , v ) = 1 , that is, f ( w , v h 1 v h 2 ) = 0 .
Set E 1 = { v v i , i [ n k , n c 1 ] } , by Lemma 4, we have d ( v i , v ) = 2 , d ( v i , v n k ) = 1 for i [ 1 , n k 1 ] and d ( v j , v ) = d ( v j , v n k ) = 1 for i [ n k + 1 , n ] . Thus, P I ( v n k v ) = n k 1 . Similarly, P I ( v n k + 1 v ) = P I ( v n k + 2 v ) = = P I ( v k c + 1 v ) = n k 1 . In addition, by Lemma 4, we have d ( v i , v ) = 2 , d ( v i , v k c + 2 ) = 1 for i [ 2 , n k 1 ] , d ( v 1 , v ) = d ( v 1 , v k c + 2 ) = 2 and d ( v j , v ) = d ( v j , v k c + 2 ) = 1 for j [ n k , n ] . Thus, P I ( v k c + 2 v ) = n k 2 . Similarly, we have P I ( v k c + 3 v ) = n k 3 , P I ( v k c + 4 v ) = n k 4 , , P I ( v n c 1 v ) = 1 . Set E 2 = { v v l , l [ n c , n 1 ] } , since N [ v l ] N [ v ] = n k 1 , we have P I ( v v l ) = n k 1 . Set E 3 = { v i v l , i [ 1 , n c 1 ] , l [ n c , n 1 ] } , by Lemma 4, we have d ( v i , v ) = 2 for i [ 1 , n k 1 ] , d ( v i , v ) = 1 for i [ n k , n c 1 ] , d ( v l , v ) = 1 for l [ n c , n 1 ] . Thus, f ( w , v i v l ) = 1 for i [ 1 , n k 1 ] and f ( w , v i v l ) = 0 for i [ n k , n c 1 ] .
Set w V ( T n k ) { v } , if x y E ( T n , c k ) with x or y v , by Lemma 2, we have f ( w , x y ) = 0 . Thus,
P I ( T n k ) P I ( T n 1 k ) = x y i = 1 n c 2 C i f ( w , x y ) + x y i = 1 n k 1 D i f ( w , x y ) + x y E 1 E 2 P I ( x y ) + x y E 3 f ( w , x y ) = 0 + [ ( 2 k n c + 2 ) + ( 2 k n c + 3 ) + + ( k c ) ] + [ 1 + 2 + + ( n k 2 ) + ( n k 1 ) ( 2 k n c + 2 ) ] + c ( n k 1 ) + c ( n k 1 ) = ( n k 1 ) ( 3 k n + 2 ) = γ ,
and Theorem 2 is proved.
For n 2 k c + 1 , let σ = 3 c ( n 2 k + c 1 ) ( n 2 k + c ) + ( k c ) ( 2 c 2 + 3 n c 4 k c + 3 k n 4 k 2 6 k + 3 n 2 ) 3 3 c ( n 2 k + c 2 ) ( n 1 2 k + c ) + ( k c ) ( 2 c 2 + 3 ( n 1 ) c 4 k c + 3 k ( n 1 ) 4 k 2 6 k + 3 n 2 ) 3 = k 2 4 k c + c 2 + 2 n c 3 c + k . If we can show that by adding v to T n 1 , c k , P I ( T n , c k ) = P I ( T n 1 , c k ) + σ , then Theorem 2 is proved.
Set w = v , by Lemma 4, we have d ( v l , v ) = 1 for l [ n c , n 1 ] , d ( v i , v ) = 2 for i [ 1 , n k 1 ] and d ( v j , v ) = 1 for j [ n k , n c 1 ] . Thus, f ( w , v l v i ) = 1 and f ( w , v l v j ) = 0 . Set C 1 = { v 1 v 2 , v 1 v 3 , , v 1 v k c + 1 } , C 2 = { v 2 v 3 , v 2 v 4 , , v 2 v k c + 2 } , , C n 2 k + c 1 = { v n 2 k + c 1 v n 2 k + c , v n 2 k + c 1 v n 2 k + c + 1 , , v n 2 k + c 1 v n k 1 } , C n 2 k + c = { v n 2 k + c v n 2 k + s + 1 , v n 2 k + c v n 2 k + s + 2 , , v n 2 k + c v n k 1 } , C n 2 k + c + 1 = { v n 2 k + c + 1 v n 2 k + c + 2 , v n 2 k + c + 1 v n 2 k + c + 3 , , v n 2 k + c + 1 v n k 1 } , . , C n k 1 = ϕ , D n 2 k + c = { v n 2 k + c v n k } , D n 2 k + c + 1 = { v n 2 k + c + 1 v n k , v n 2 k + c + 1 v n k + 1 } , , D n k 1 = { v n k 1 v n k , v n k 1 v n k + 1 , , v n k 1 v n c 1 } .
By Lemma 4, we can get that d ( v 1 , v ) = d ( v 2 , v ) = 2 and d ( v n k , v ) = 1 , that is, f ( w , v 1 v 2 ) = 0 and f ( w , v 1 v n k ) = 1 . Similarly, for v h 1 v h 2 i = 1 n k 1 C i with h 1 < h 2 , we have d ( v h 1 , v ) = d ( v h 2 , v ) = 2 , that is, f ( w , v h 1 v h 2 ) = 0 ; for v h 1 v h 2 i = n 2 k + c n k 1 D i with h 1 < h 2 , we have d ( v h 1 , v ) = 2 and d ( v h 2 , v ) = 1 , that is, f ( w , v h 1 v h 2 ) = 1 . Set C n k = { v n k v n k + 1 , v n k v n k + 2 , , v n k v n c 1 } , C n k + 1 = { v n k + 1 v n k + 2 , v n k + 1 v n k + 3 , , v n k + 1 v n c 1 } , , C n c + 2 = { v n c 2 v n c 1 } . By Lemma 4, we can get that d ( v n k , v ) = d ( v n k + 1 , v ) = 1 , that is, f ( w , v n k v n k + 1 ) = 0 . Similarly, for v h 1 v h 2 i = n k n c 2 C i with h 1 < h 2 , we have d ( v h 1 , v ) = d ( v h 2 , v ) = 1 , that is, f ( w , v h 1 v h 2 ) = 0 .
Set E 1 = { v v i , i [ n k , n c 1 ] } , by Lemma 4, we have d ( v , v n k 1 ) = 2 , d ( v n c 1 , v n k 1 ) = 1 , d ( v , v i ) = d ( v n c 1 , v i ) = 1 for i [ n k , n c 2 ] [ n c , n 1 ] and d ( v , v j ) = d ( v n c 1 , v j ) = 2 for j [ 1 , n k 2 ] . Thus, P I ( v v n c 1 ) = 1 . Similarly, we have P I ( v v n c 2 ) = 2 , P I ( v v n c 3 ) = 3 , , P I ( v v n k ) = k c . Set E 2 = { v v l , l [ n c , n 1 ] } , since N [ v l ] N [ v ] = n k 1 , we have P I ( v v l ) = n k 1 . Set E 3 = { v i v l , i [ 1 , n c 1 ] , l [ n c , n 1 ] } , by Lemma 4, we have d ( v , v i ) = 2 , d ( v , v l ) = 1 for i [ 1 , n k 1 ] and d ( v , v i ) = d ( v , v l ) = 1 for i [ n k , n c 1 ] . Thus, f ( w , v i v l ) = 1 for i [ 1 , n k 1 ] and f ( w , v i v l ) = 0 for i [ n k , n c 1 ] .
Set w V ( T n k ) { v } , if x y E ( T n , c k ) with x or y v , by Lemma 2, we have f ( w , x y ) = 0 . Thus,
P I ( T n k ) P I ( T n 1 k ) = x y i = 1 n c 2 C i f ( w , x y ) + x y i = n 2 k + c n k 1 D i f ( w , x y ) + x y E 1 E 2 P I ( x y ) + x y E 3 f ( w , x y ) = 0 + [ 1 + 2 + 3 + + ( k c ) ] + [ 1 + 2 + 3 + + ( k c ) ] + c ( n k 1 ) + c ( n k 1 ) = k 2 4 k c + c 2 + 2 n c 3 c + k = σ ,
and Theorem 2 is proved. □
Proof of Theorem 3.
For n k + 2 , we will proceed it by introduction on | T n k | = n . If n = k + 2 , T n k is also a k-path, that is, P I ( T n k ) = 2 k . If n k + 3 , assume that Theorem 3 is true for the k-tree with at most n 1 vertices, let v S 1 ( T n k ) and T n 1 k = T n k v , by the induction hypothesis, we have P I ( T n 1 k ) P I ( S n 1 k ) = k ( n k 1 ) ( n k 2 ) . By adding back v, let N ( v ) = { x 1 , x 2 , , x k } and w = v . Since T n k [ v , x 1 , x 2 , , x k ] is a ( k + 1 ) -clique, we have f ( w , x i x j ) = 0 for i , j [ 1 , k ] . By Lemmas 1 and 2, we can obtain that P I ( v x i ) n k 1 with i [ 1 , k ] and f ( w , x y ) 1 for any edge x y E ( T n k ) E ( T n k [ v , x 1 , x 2 , , x k ] ) . Next, set w V ( T n k ) { v } , by Lemma 2, if x y E ( T n k ) with x or y v , we have f ( w , x y ) = 0 . Since | E ( T n k ) E ( T n k [ v , x 1 , x 2 , , x k ] ) | = k ( n k 1 ) , we have
P I ( T n k ) = P I ( T n 1 k ) + x y E ( T n k { v x i , i [ 1 , k ] } ) f ( w , x y ) + i = 1 k P I ( v x i ) P I ( S n k k ) + k ( n k 1 ) + k ( n k 1 ) = k ( n k 1 ) ( n k 2 ) + k ( n k 1 ) + k ( n k 1 ) = k ( n k ) ( n k 1 ) = P I ( S n k ) .
Thus, this finishes the proof of Theorem 3. □
Proof of Theorem 4.
For k = 1 and by Fact 1, we can get that every tree with the same vertices has the same P I -value; then, Theorem 4 is obvious; for k 2 , if k + 2 n 2 k c , let n = k p + s with p = 1 and s = n k , by Theorem 1, we have P I ( P n k ) P I ( T n , c k ) = ( s 1 ) s ( 3 k s + 2 ) 3 ( n k ) ( n k 1 ) ( 4 k n + 2 ) 3 = ( n k 1 ) ( n k ) [ 3 k ( n k ) + 2 ] 3 ( n k ) ( n k 1 ) ( 4 k n + 2 ) 3 = 0 , and Theorem 4 is true. If n 2 k c + 1 , p = n s k and by Theorems 1 and 2, define the new functions as follows: for z 2 k c + 1 , 1 c k 1 and 2 s k + 1 ,
g ( z ) = ( k + 1 ) ( z s k ) ( 3 z + 3 s 2 k 4 ) 6 + s ( s 1 ) ( 3 k s + 2 ) 3 , h ( z , c ) = c ( z 2 k + c 1 ) ( z 2 k + c ) + ( k c ) ( 2 c 2 + 3 z c 4 k c + 3 k z 4 k 2 6 k + 3 z 2 ) 3 , l ( z , c ) = g ( z ) h ( z , c ) = ( k 2 + 1 2 c ) z 2 + ( c 2 + 2 c + 4 k c 11 k 2 6 5 k 2 2 3 ) z + k s 2 2 k 2 s 6 k s 2 + 5 k 3 3 + 5 k 2 3 + s 2 2 + 4 k 3 s 3 3 6 k 2 c + 3 k c 2 c 3 3 4 k c + c 2 2 c 3 , l z ( z ) = l z ( z , c ) = ( k + 1 2 c ) z c 2 + 2 c + 4 k c 11 k 2 6 5 k 2 2 3 .
Then, it is enough to determine whether or not l ( z , c ) 0 is true. By some calculations, we can obtain the following claim:
Claim 1.
z 1 = 2 k c + 1 , z 2 = 2 k c + 2 are the two roots of l ( z , c ) = 0 with c k + 1 2 .
Proof. 
For any c [ 1 , k 1 ] , let z 1 = 2 k c + 1 , z 2 = 2 k c + 2 , and we have l ( 2 k c + 1 , c ) = 0 , l ( 2 k c + 2 , c ) = 0 . If c k + 1 2 , then Claim is true. □
For fixed c [ 1 , k + 1 2 ) , that is, k 2 + 1 2 c > 0 , then the function of l ( z , c ) about z is open up. Since z is an integer and by Fact 2, we have l ( z , c ) 0 for z 2 k c + 1 and Theorem 4 is true; if c = k + 1 2 and k 1 , we have l z ( z ) = 1 k 2 12 0 , that is, l ( z , k + 1 2 ) is decreasing about z. By the proof of Fact 2, we have l ( 2 k c + 1 , c ) = 0 . For z 2 k c + 1 , we can get that l ( z , k + 1 2 ) l ( 2 k c + 1 , k + 1 2 ) = 0 and Theorem 4 is true; for fixed c ( k + 1 2 , k 1 ] , that is, k 2 + 1 2 c < 0 , then the function of l ( z , c ) about z is open down. Since z is an integer and by Claim, we can obtain that l ( z , c ) 0 for z 2 k c + 1 and this finishes the proof of Theorem 4. □

4. Conclusions

We can see that the k-stars attain the maximal values of P I -values for k-trees. One of the guesses is that the k-paths attain the minimal values. Actually, it is not the case and some P I -values of k-spirals is even smaller than that of k-paths. Meanwhile, not all P I -values of k-spirals are less than the values of all other k-trees. This fact indicates an interesting problem—which type of k-trees will achieve the minimum P I -value?

Author Contributions

S.W. contributes for supervision, methodology, validation, project administration and formal analysing. S.W., Z.S., J.-B.L., B.W. contribute for resources, investigation some computations and wrote the initial draft of the paper which were investigated and approved by S.W. and B.W. wrote the final draft.

Funding

This research was funded by Natural Science Foundation of Guangdong Province under grant 2018A0303130115.

Conflicts of Interest

The authors declare no conflict of interest.

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