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

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- Received:
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DOI: 10.1007/s00373-008-0836-2

- Cite this article as:
- Bereg, S. Graphs and Combinatorics (2009) 25: 139. doi:10.1007/s00373-008-0836-2

## Abstract

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 2*n* − 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 2*n* − 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 2*n* − 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 \((2n-2)({n\choose 2} - 1)+1\) decompositions into spanning trees. We show that 2*n* − 2 decompositions suffice. Let \(e_1,e_2,\ldots,e_{2n-2}\) be any circular order of edges of *G* (i.e. \(e_0 = e_{2n-2}\)). The graph *G* is a generic circuit in the plane if and only if \(G + e_i - e_{i-1}\) is the union of two spanning trees for any \(i = 1,2, \ldots, 2n-2\). Furthermore, we show that only *n* decompositions into spanning trees suffice.