给定悬挂点个数的分子树的ISDD指数的极值

Extremal ISDD Index of Molecular Trees with Fixed Number of Pendent Vertices

设G=(V (G),E (G))为n阶连通图,其顶点集为V (G),边集为E (G),用deg (x)表示顶点x的度,则图G的反对称分割指数为ISDD(G)=∑_(xy∈E(G))(deg(x)·deg(y)/deg(x)^(2)+deg(y)^(2))。本文主要采用不等式和分类讨论法对具有固定悬挂点的分子树的ISDD指数进行了研究,分别讨论了悬挂点个数为偶数和悬挂点个数大于等于3时分子树的ISDD指数的极值,分子树是指顶点度不超过4的树。首先,确定了当悬挂点个数为偶数时,分子树中反对称分割指数为最小值,此时,ISDD(MT)=1/2n-31/85p-1/10;其次,确定了当悬挂点个数大于等于3时,分子树中反对称分割指数为最大值,此时,ISDD(MT)=1/2n-9/65p-1/2,并刻画了达到ISDD指数极值的分子树。

Let G=(V(G),E(G))be a connected graph of order n,V(G)be vertex set of G,E(G)be edge set of G,deg(x)be degree of the vertex x.The inverse symmetric division deg index of G is ISDD(G)=∑_(xy∈E(G))(deg(x)·deg(y)/deg(x)^(2)+deg(y)^(2)) .Inequality and classification discussion are used to study the+deg()y ISDD(G)of molecular tree with fixed number of pendent vertices,respectively,the extreme value of the ISDD index of the molecular tree with the number of pendent vertices is even number and the number of pendent vertices is more than or equal to 3 are discussed,the tree whose vertex degree is less than 4 is called molecular tree.Firstly,the minimum value of the inverse symmetric division deg index of G is determined when the number of pendent vertices is even,that is ISDD(MT)=1/2n-31/85p-1/10.Sec‐ondly,when the number of pendent vertices is greater than or equal to 3,the maximum value of the inverse symmetric division deg index of G in the molecular tree is determined,that is ISDD(MT)=1/2n-9/65p-1/2,and the molecular tree of ISDD index reaching the extreme value is described.