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On Characterizations of Rigid Graphs in the Plane Using Spanning Trees On Characterizations of Rigid Graphs in the Plane

机译:使用生成树表征平面中的刚性图表征平面中的刚性图

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We study characterizations of generic rigid graphs and generic circuits in the plane using only few decompositions into spanning trees. Generic rigid graphs in the plane can be characterized by spanning tree decompositions [5,6]. A graph G with n vertices and 2n − 3 edges is generic rigid in the plane if and only if doubling any edge results in a graph which is the union of two spanning trees. This requires 2n − 3 decompositions into spanning trees. We show that n − 2 decompositions suffice: only edges of G − T can be doubled where T is a spanning tree of G. A recent result on tensegrity frameworks by Recski [7] implies a characterization of generic circuits in the plane. A graph G with n vertices and 2n − 2 edges is a generic circuit in the plane if and only if replacing any edge of G by any (possibly new) edge results in a graph which is the union of two spanning trees. This requires (2n-2)(((n) || 2) - 1)+1(2n-2)({nchoose 2} - 1)+1 decompositions into spanning trees. We show that 2n − 2 decompositions suffice. Let e1,e2,¼,e2n-2e_1,e_2,ldots,e_{2n-2} be any circular order of edges of G (i.e. e0 = e2n-2e_0 = e_{2n-2}). The graph G is a generic circuit in the plane if and only if G + ei - ei-1G + e_i - e_{i-1} is the union of two spanning trees for any i = 1,2, ¼, 2n-2i = 1,2, ldots, 2n-2. Furthermore, we show that only n decompositions into spanning trees suffice.
机译:我们仅使用很少的分解成生成树来研究平面中的一般刚性图和一般电路的特征。平面中的一般刚性图可以通过生成树分解来表征[5,6]。当且仅当将任何一个边加倍导致图是两个生成树的并集时,具有n个顶点和2n-3个边的图G在平面中是一般刚性的。这需要2n-3分解为生成树。我们证明n-2分解就足够了:只有G-T的边缘可以加倍,其中T是G的生成树。Recski [7]在张力框架上的最新结果暗示了平面中通用电路的特征。当且仅当用任意(可能是新的)边替换G的任意边会导致图是两个生成树的并集时,具有n个顶点和2n-2个边的图G是平面中的通用电路。这需要将(2n-2)((((n)|| 2)-1)+1(2n-2)({nchoose 2}-1)+1分解为生成树。我们证明2n − 2分解就足够了。令e 1 ,e 2 ,¼,e 2n-2 e_1,e_2,ldots,e_ {2n-2}是任何循环顺序G的边数(即e 0 = e 2n-2 e_0 = e_ {2n-2})。当且仅当G + e i -e i-1 G + e_i-e_ {i-1}是并集时,图G是平面中的通用电路任意i = 1,2,¼,2n-2i = 1,2,ldots,2n-2的两个生成树的数量。此外,我们证明只有n个分解为生成树就足够了。

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