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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 1, 2011 at 20:45 history edited j.c. CC BY-SA 3.0
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May 1, 2011 at 19:27 comment added user14324 @jc, one final really naive question... how can we be sure that (5) insures a generic framework? What if I said I was going to randomly and densely place vertices in cells of a finite matrix?
May 1, 2011 at 19:22 vote accept user14324
May 1, 2011 at 19:22 comment added user14324 @jc, oh right, I see what you mean. I meant that the coordinates should be chosen with uniform random probability, under the distance constraint, across a two-dimensional interval.
May 1, 2011 at 19:01 comment added user14324 @jc, by (5) I only mean to imply that I know how the set of two-dimensional coordinates were initially decided, but I don't have access to the exact values. Is that fair?
May 1, 2011 at 16:50 comment added j.c. In (5), I suppose I should interpret you choosing the x- and y- coordinates in the way you've described -- it's written in a way that could be interpreted as putting all of the vertices in a 1-D interval. (5) now contradicts the first sentence of your post where you say you have known edge lengths but unknown vertex coordinates. But if you stick to just (4) and (5) (interpreted in the way I've described), then yes, the result of Jackson and Jordán guarantees that these are (almost surely) globally rigid embeddings of graphs.
May 1, 2011 at 8:03 comment added user14324 @jc, I updated (4) and added a further specification for $G$, (5), that should hopefully imply algebraic independence of vertex coordinates. Can we now say that $G$ is globally rigid?
May 1, 2011 at 7:49 comment added user14324 Am I correct in assuming that the corollary in Jackson and Jordán's paper let's me change (4) to specify the vertices to be at least 6-connected, and so long as my graph is generic, I can be confident about global rigidity?
Apr 30, 2011 at 17:14 history answered j.c. CC BY-SA 3.0