On MinimumWiener Polarity Index of Unicyclic Graphs with Prescribed Maximum Degree

The Wiener polarity index of a connected graph G is defined as the number of its pairs of vertices that are at distance three. By introducing some graph transformations, in different way with that of Huang et al., 2013, we determine the minimum Wiener polarity index of unicyclic graphs with any givenmaximum degree and girth, and characterize extremal graphs.These observations lead to the determination of the minimum Wiener polarity index of unicyclic graphs and the characterization of the extremal graphs.


Introduction
All the graphs considered in this paper are connected and simple.The distance between two vertices  and V of a graph , denoted by   (, V) or (, V) for short, is the length of a shortest path connecting these two vertices.The Wiener polarity index of a graph  is defined as the number of its unordered pairs of vertices that has distance three [1].Namely, the Wiener polarity index of a molecular graph  is   () =     {{, V} :   (, V) = 3, , V ∈  ()}     .
In 1998, by using the Wiener polarity index, Lukovits and Linert in [2] demonstrated quantitative structure-property relationships in a series of acyclic and cycle-containing hydrocarbons.Besides, a physicochemical interpretation of   () was found by Rouvray and King [3].M. Liu and B. Liu described the relations between Wiener polarity index, Zagreb index, Wiener index, and hyper-Wiener index in [4]; they also determined there the first two smallest Wiener polarity indices among all unicyclic graphs.Recently, Du et al. obtained the smallest and largest Wiener polarity indices together with the corresponding graphs among all trees with any given number of vertices [5].In [6], Deng and Xiao, determined the maximum Wiener polarity index of trees with  pendants and characterized the extremal graphs.In [7], the authors determined the Wiener polarity of fullerenes and hexagonal systems.In [8], Hou et al. determine the maximum wiener polarity index of unicyclic graphs and characterizes the corresponding extremal graphs.This work determines in differrent way with that of [9] the minimum Wiener polarity index of unicyclic graphs with any given maximum degree and girth; these observations lead to the determination of minimum wiener polarity index and the characterization of the corresponding extremal graphs.
Before proceeding, let us introduce some more symbols and terminology.Denote by   (), or () for short, the neighborhood of vertex  in graph , the set of vertices adjacent to .Let   () = () = |  ()| represent the degree of vertex  in graph .If () = 1, vertex  is called a pendent vertex.The girth () of a unicyclic graph  is the length of its unique cycle.As usual, let   ,   , and   represent a cycle, a star, and a path on  vertices, respectively, and  ,Δ the set of unicyclic graphs with  vertices and maximum degree Δ.For any unicyclic graph  and any vertex V in its cycle , the union of vertex V and the component of  − () that has a vertex  adjacent to V induces a tree in , which is called a hanging tree on vertex V and is denoted by The union of all these hanging trees [V] is denoted by [V].For other symbols and terminology not specified herein, we follow that of [10].

Property of Graph Transformations
Let  1 ,  2 ≥ 0 be two integers.After joining  1 and  2 vertices, each by an edge, to the two endpoints of a path   ,, respectively, we obtain a new graph and denote it by    1 , 2 [11].For example, the path on two vertices may be regarded as  2 0,0 , and the star   with  ≥ 4 may be regarded as  1 1,−2 .Denote by  ,Δ the set of trees with order  ( vertices) and maximum degree Δ.
a cycle with  ≥ 3 and let  1 ,  2 , . . .,   be nonnegative integers.We attach at first   isolated vertices to vertex V  for every  ∈ {2, 3, 4, . . ., }.And then past a 1-degree vertex of a star that has  1 + 2 vertices to vertex V 1 to obtain a graph denoted by   1 , 2 ,...,  ,,Δ,1 , where  denotes the number of vertices of the obtained graph; or attach  1 isolated vertices directly to vertex V 1 to obtain graph   1 , 2 ,...,  ,,Δ,0 ; or past the 1-degree vertex of Finally, let ,,Δ, 1 − 1 −1 denote the graph obtained by attaching at first  2 isolated vertices to vertex V 1 and then pasting the 1-degree vertex of For clarity, we depict some of these graphs in Figure 1.
In what follows, for clarity, we denote by the unique cycle of any graph  ∈  ,Δ with girth .Lemma 3. Let  ∈  ,Δ and [V  ] be one of its hanging tree.
If a unicyclic graph  has cycle , then its wiener polarity index can be expressed as (refer to [4] for example) Notice that if  is a tree, then its wiener polarity index can be expressed as   () = ∑ V∈() (() − 1)((V) − 1).According to the differences between  and [V  ], it is not difficult to deduce (for simplicity the common terms are deleted, this method is also employed in the proof of other leammas) that Since Δ  ≤   − 2, it follows that   (   ) = 2. Combining this observation with   (V  ) =    (V  ),  [V  ] (  ) ≥ 2, Lemmas 1 and 2 and the fact that among trees with maximun degree Δ  =   − 2 and order   graph  2 1,Δ  −1 has maximum wiener polarity index, we deduce that Again by Lemmas 1 and 2, the inequalities in above formula become equalities if and only if for some positive integer  ≤ Δ  − 1 and  [V  ] (  ) = 2. So,  = 1 and the lemma follows.
Lemma 4. Let  ∈  ,Δ and [V  ] be one of its hanging tree.
According to the differences between  and [V  ] and formula (2), it is not difficult to deduce that Since and at least one of these two inequalities strictly holds.By formula (6), we also have   (  ) −   () < 0. Hence, the lemma follows.

If its hanging trees are 𝑇
on any vertex V  with   (V  ) = 1 and the neighbor of V  in  having degree two.
Proof.Assume without loss of generality that Δ 1 = Δ and  1 is a hanging tree on V 1 .After tranforming every hanging tree By induction on the number of hanging trees on vertex V 1 , it follows that if we denote by  4 the graph obtained from  2 by transforming then   ( 4 ) <   ( 2 ) when  2 has at least two hanging trees on vertex V 1 , where In the same way, one can show with ease that if one hanging tree  = V  ⋅ ⋅ ⋅  of  4 on vertex V  has at least three vertices, then the graph  4 − V 1  +  has less wiener polarity index than that of  4 , where  is the neighbor of V 1 in   .And so, we leave its proof to the readers.Now, we consider the case when every hanging tree of  4 on vertices V  ,  ̸ = 1, is a path of order two.Let   4 (V  ) =   +2,  ≥ 2 and assume without loss of generality that   ≥ 2, where  ≥ 2. Denote by  5 the graph obtained from  4 by transforming [V  ] to a path of order   .Then In conclusion, if  6 is the graph obtained from  4 by transforming ) is not less than the total contribution of them to   ( 7 ), furthermore, the contribution of other vertex to   ( 6 ) is not less than the contribution of its corresponding vertex to   ( 7 ), it follows that   ( 7 ) ≤   ( 6 ).And so, the lemma follows.Proof.Assume without loss of generality that Δ 1 = Δ and  1 is a hanging tree on V 1 .After tranforming every hanging tree   ̸ =  1 of  to a path    , we obtain a new graph  2 ∈  ,Δ .By Lemma 4 we have   ( 2 ) ≤   (). Let When  = 1, since   3 ( 1 ) = 2 and   2 ( 11 ) =   3 ( 11 ), we have By induction on the number of hanging trees on vertex V 1 , it follows that if we denote by  4 the graph obtained from  2 by transforming [V 1 ] to   =    −Δ 1,Δ−1 then   ( 4 ) <   ( 2 ) when  2 has at least two hanging trees on vertex V 1 , where In the same way, one can show with ease that if one hanging tree  = V  ⋅ ⋅ ⋅  of  4 on vertex V  has at least three vertices, then the graph  4 − V 1  +  has less wiener polarity index than that of  4 , where  is the neighbor of V 1 in   .And so, we leave its proof to the readers.Now, we consider the case when every hanging tree of  4 on vertices V  ,  ̸ = 1, is a path of order two.Let  = V   be a hanging tree of The second equality holds since   4 () =   5 () and   5 () = 2.And so, the lemma follows.
Proof.Assume without loss of generality that  1 is a hanging tree on V 1 .Consider at first the case when  has at least two hanging trees on vertex and that when  = 1, since The above inequality becomes equality if and only if By induction on the number of hanging trees on vertex V 1 , it follows that if we denote by  2 the graph obtained from  by transforming Consider secondly the case when  1 is the unique hanging tree of  on vertex V 1 .By similar reasoning as above, one can show with ease that if one hanging tree of  on vertex V  has at least three vertices, then the graph obtained from  by transforming [V  ] to a path of order |[V  ]| has less wiener polarity index than that of .And so, we assume in what follows that every hanging tree of  on vertices V  ,  ̸ = 1, is a path of order two. Let And so, the lemma follows.
Lemma 8. Let  ∈  ,Δ .If every hanging tree is a path then   (  ) <   (), where   is obtained from  by transforming [V  ] to Proof.Assume without loss of generality that  = 1.Let (V 1 ) = {V 2 , V  ,  11 , . . .,  1 }, and let And so, the lemma follows by induction on the number of hanging trees on vertex V 1 that has length of at least three.

Minimum Index and Extremal Graphs with Any Given Girth
This section charaterizes the extremal unicyclic graphs with any given girth and minimum wiener polarity index.To this end, we first consider the case when the maximum degree vertices are contained in the cycle and then compare the results with those obtained in the previous section.Since there is only one unicyclic graph that has maximum degree Δ, girth , and  = Δ +  − 2 vertices, we need to only consider the case when a unicyclic graph has at least  ≥ Δ+−1 vertices.

Lemma 9.
Let  ∈  ,Δ be a unnicylic graph with girth  = 3 and  ≥ Δ + 3 vertices.If (V 1 ) = Δ and every hanging tree of  is a path then   () ≥  − 3, with equality if and only if Proof.The lemma is clearly true when  = Δ + 3 since there are only five nonisomorphic unicyclic graphs in this subcase.So, assume in what follows that  ≥ Δ + 4.
Let us consider at first the case when  = 1.In this case, every hanging tree on V 1 is a path on two vertices.To prove the lemma, we may assume in what follows that  is an extremal unicyclic graph with minimum wiener polarity index.If the total number of hanging trees on vertices V 2 and V 3 is at least two, assume without loss of generality that  1 = V 2  21 ⋅ ⋅ ⋅  2 and  2 = V 3  31 ⋅ ⋅ ⋅  3 are two of these hanging trees.Let  1 =  − V 3  31 +  31  2 .When  or  ≥ 2, say  ≥ 2, by formula (2) we have Since   ( 31 ) =   1 ( 31 ) =   1 ( 2 ) = 2 and   (V 2 ),   (V 3 ) ≥ 3, it follows that   () −   ( 1 ) ≥ 2 in this subcase, which is a contradiction since  is an extremal graph.By induction on the total number of hanging trees on vertices V 2 and V 3 , we deduce that  =  1,Δ−2,0 ,3,Δ,−Δ−2 in this subcase.
Proof.Assume that  is an extremal graph with minimum wiener polarity index and properties postulated in the lemma.By Lemma 4 we may assume at first that every hanging tree of  is a path.And so, by Lemma 8, every [V  ] of  is either a path or we move all the hanging trees on vertex V 2 to V 1 to obtain graph  1 .Since   (V 3 ) = Δ ≥   (V 4 ), by the reasoning as is employed in the first three paragraphs of the proof of Lemma 6, one can deduce from formula (2) that   ( 1 ) ≤   ().And so, we may assume in what follows that  1 ≥ 2.
If   (V 1 ) +   (V 3 ) <  then with the equality holding if and only if the union of hanging trees of  on V 1 is a path.Now, combining this observation with Lemma 4 we deduce that the first statement of Lemma 11 is true since   (V 1 ) +   (V 3 ) <  implies Δ ≤ ( − 1)/2.
If   (V 1 ) +   (V 3 ) = , which implies Δ ≥ /2, then And so, the second statement of the lemma is also true.