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Oct 18, 2021 at 4:16 history edited David Roberts CC BY-SA 4.0
fixed arxiv front-end links
Aug 30, 2018 at 9:37 comment added M. Winter Unfortunatly all of your links broke down. Can you include updated links, or better, the full references for these papers? They seem to be very interesting.
Jul 25, 2010 at 16:27 comment added Gil Kalai I see, i suppose you still insist on a realization with line intervals for edges and plane triangles for 2-faces. It is true that Tarski's algorithm applies but this algoithm is highly inpracticle.
Jul 25, 2010 at 2:57 comment added Vinayak Pathak I am talking about the generalization of the Czaszar polyhedron. It's a polyhedron whose skeleton is $K_7$. For n = 6, 7, 8, 9, 10, 11, there cannot be a polyhedron whose skeleton is $K_n$, which can be shown using the Euler characteristic. So the next candidate is $K_12$, and that's still open. I had assumed that 3-polytopes are the same as polyhedra. But it seems they are the same as "convex" polyhedra?
Jul 25, 2010 at 1:18 comment added Gil Kalai Graphs of 3-polytopes are planar so even $K_5$ cannot be realized as the graph of a 3-polytope. you must mean something else.
Jul 24, 2010 at 17:53 comment added Vinayak Pathak I meant $K_{12}$.
Jul 24, 2010 at 17:52 comment added Vinayak Pathak I have heard that it's still not known if the complete graph on 12 vertices can be realized as the 1-skeleton of a 3-polytope. But according to what you have said, it should have been decidable. So can't we just use Tarski's algorithm once on $K_12$ and check? Or is the input size really really big for this case?
Jul 1, 2010 at 15:23 history edited Gil Kalai CC BY-SA 2.5
added 10 characters in body
May 17, 2010 at 16:56 history edited Gil Kalai CC BY-SA 2.5
added 219 characters in body
May 3, 2010 at 19:15 history edited Gil Kalai CC BY-SA 2.5
added 391 characters in body
May 1, 2010 at 18:37 history edited Gil Kalai CC BY-SA 2.5
typo
Apr 30, 2010 at 12:38 history answered Gil Kalai CC BY-SA 2.5