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Why is this graph not generically globally rigid? - MathOverflow

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Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up. Let us assume that we are given a connected, undirected graph. Under the assumption that no three points are collinear, such a graph is uniquely realizable in the plane iff we can certify that it is generically globally rigid. A graph is generically globally rigid iff it is (i) generically redundantly rigid, and (ii) 3-connected (Laman, Hendrikson, Jackson and Jordan). The graph below is 3-connected and generically rigid in the plane. However, it is not redundantly rigid (the removal of edge [4,3] permits nodes [1,2,5,6] to shear). Therefore, by definition it cannot be generically globally rigid. However, I cannot see any local or global degrees of freedom in this graph. A lack of ge

Why is this graph not generically globally rigid? - MathOverflow Why is this graph not generically globally rigid? Ask Question Asked 14 years, 6 months ago Modified 9 years ago Viewed 781 times 2 $\begingroup$ Let us assume that we are given a connected, undirected graph. Under the assumption that no three points are collinear, such a graph is uniquely realizable in the plane iff we can certify that it is generically globally rigid. A graph is generically globally rigid iff it is (i) generically redundantly rigid, and (ii) 3-connected (Laman, Hendrikson, Jackson and Jordan). The graph below i

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