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Posted 08 February 2008 - 05:29 PM
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Posted 10 February 2008 - 05:44 AM
Quote:
Original post by Vorpy
Each vertex must have exactly 3 edges. It is not possible for a vertex to have less than 3, and if there are more than 3 then we can split the vertex into two vertices with an edge between them, creating a shape with more vertices.
Posted 10 February 2008 - 09:17 AM
Posted 10 February 2008 - 10:46 AM
Posted 11 February 2008 - 06:13 AM
Quote:
Original post by Vorpy
I believe that whenever more than 3 edges meet at a single vertex, it is always possible to adjust the bounding planes of the convex polyhedron to split the vertex into two vertices connected by an edge, each of which will have at least 2 of the edges that were previously going to the original vertex. No degenerate edges or triangles are needed. The number of vertices and edges increases by 1, and this can be repeated until every vertex is connected to 3 edges.
Posted 11 February 2008 - 06:16 AM
Quote:
Original post by alvaro
Let G be the average number of edges that meet at a vertex (over all vertices). Clearly, G>=3. Now we know that
G*V = E/2
Posted 11 February 2008 - 01:14 PM
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Posted 11 February 2008 - 01:51 PM
Quote:
Original post by Vorpy
I think we might be trying to solve two or three different problems.
Quote:
Original post by VorpyI tried to describe my solution using graph theory because it is easier to visualize than polyhedra. The vertices of the polyhedron correspond to nodes in the graph, the edges correspond to edges, and the faces correspond to regions (one of the faces corresponds to the region that is "outside" the graph).
There are some mathematical theories that every convex polyhedra has a corresponding planar, 3 connected graph, and vice versa. So it's possible to just work with the 3 connected planar graphs instead of polyhedra to find the solution. The problem is then: What is the maximum number of vertices a planar 3 connected graph with N regions can contain?
As I described above, working with the graphs, you can always split vertices with more than four edges to get a new planar 3 connected graph with the same number of regions and more vertices.
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Posted 11 February 2008 - 05:36 PM
Quote:
Original post by Vorpy
I think we might be trying to solve two or three different problems.
Alvero's formula can be fixed by multiplying instead of dividing by two. It should be G*V = E*2.
Posted 14 February 2008 - 05:41 AM
Quote:
Original post by samgzman
For my problem, the polyhedron's edges are defined by the intersections of a finite set of aribitrarily oriented pairs of parallel planes. The polyhedron is defined by the intersection of all spaces between these pairs. The faces of the polyhedron are therefore coincident with these planes. I assume that any valid configuration establishes a closed manifold.
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