Timeline for Extensions of the Koebe–Andreev–Thurston theorem to sphere packing?
Current License: CC BY-SA 3.0
8 events
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Jan 13, 2012 at 20:11 | comment | added | Will Jagy | Igor, this seems to be the induction step in your book's answer. I am still uncertain about graphs on $n$ vertices that are not complete, in $\mathbb R^{n-1}$ or possibly $R^{n-2}.$ Steve Carnahan feels that there is enough room in $\mathbb R^{n-1}$ to slightly perturb the positions of the balls (so as to erase edges from a complete graph and realize the graph we are actually given), perhaps keeping all radii the same. Note that we can do all graphs with 5 vertices in $\mathbb R^3,$ if not complete then planar, if complete then a regular tetrahedron with a smaller ball at center. | |
Jan 13, 2012 at 12:29 | vote | accept | Joseph O'Rourke | ||
Jan 13, 2012 at 12:25 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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Jan 13, 2012 at 9:21 | comment | added | Igor Rivin | Well, by Ian's argument, you can send two spheres to tangent hyperplanes, at which point you are asking if you can place a $K_{n-2}$ in $\mathbb{R}^{n-3},$ which you can, by the magic of regular simplices. | |
Jan 13, 2012 at 7:00 | comment | added | Will Jagy | I see, part (c) , page 191, answer page 395, exactly what Ian came up with. Do you think we can always place a graph with $n$ vertices in $\mathbb R^{n-2}?$ | |
Jan 13, 2012 at 6:28 | comment | added | Igor Pak | Yeah, this is Exc. 20.12 (and a solution at the end) in my book math.ucla.edu/~pak/geompol8.pdf | |
Jan 13, 2012 at 6:12 | comment | added | Will Jagy | Also, $K_6$ cannot be a subgraph, from Ian's answer. | |
Jan 13, 2012 at 6:01 | history | answered | Igor Pak | CC BY-SA 3.0 |