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Computing the Determinant of the Distance Matrix of a Bicyclic Graph

机译:计算双环图距离矩阵的行列式

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LetGbe a connected graph with vertex setV= {v1, ...,vn}. The distanced(vi,vj) between two verticesviandvjis the number of edges of a shortest path linking them. The distance matrix ofGis then×nmatrix such that its (i,j)-entry is equal tod(vi,vj). A formula to compute the determinant of this matrix in terms of the number of vertices was found when the graph either is a tree or is a unicyclic graph. For a byciclic graph, the determinant is known in the case where the cycles have no common edges. In this paper, we present some advances for the remaining cases; i.e., when the cycles share at least one edge. We also present a conjecture for the unsolved cases.
机译:设顶点为setV = {v1,...,vn}的连通图。两个顶点之间的distance(vi,vj)是连接它们的最短路径的边数。然后,Gis的距离矩阵×n矩阵,使其(i,j)项等于d(vi,vj)。当图是树或单环图时,找到了一个根据顶点数计算该矩阵行列式的公式。对于周期图,在循环没有公共边的情况下,行列式是已知的。在本文中,我们介绍了其余案例的一些进展。也就是说,当循环共享至少一个边沿时。我们还提出了一个未解决案件的猜想。

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