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I have two faces ($S_1$ and $S_2$), each bounded by a series of Vertex $V$. These two faces may or may not intersect. In addition to that there are two vertical faces $S_3$ and $S_4$, that connect between $S_1$ and $S_2$ vertically, as shown in the below diagram. The question is how to find the polyhedrons that are formed by the bounding of the four faces?

An example of this is shown below: alt text

In the above example, there should be two polyhedrons returned, one above $S_2$, another below $S_1$.

Both $S_1$ and $S_2$ are defined by 3D coordinates. Let's say the coordinates for $S_1$ is ($P_1$,$P_2$, $P_3$, $P_4$) and $S_2$ is $P_5$,$P_6$, $P_7$, $P_8$, and the intersected points are $P_9$ and $P_{10}$. I want a program that returns me two polyhedrons:




One way to do this is to find the intersecting line of the two faces, and then construct polyhedron from there. But is there a more elegant method?

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What kind of polygons should the faces be? – J. M. Dec 10 '10 at 8:30
@J.M., what do you mean by what kind of polygons? – Graviton Dec 10 '10 at 8:31
What about the polyhedron $P_1, P_2, P_9, P_{10}, P_8, P_7$? Should the program return that one as well? What if the intersection of $S_1$ and $S_2$ doesn't meet the edges of $S_2$? What if $S_1$ and $S_2$ are disjoint? (By the way - it would really help if you labelled the vertices of your figure.) – Sam Nead Dec 10 '10 at 11:50
This may be a numerically unstable problem. Imagine $S/1$ and $S/2$ such that (for one assignment of vertex matching) all three lines go through a point. If one vertex moves slightly, the configuration will change. If your vertices are rational, you can do this exactly, but if they are reals, no. – Ross Millikan Dec 10 '10 at 14:01
I see six planes in the figure, not four. – TonyK Jun 9 '11 at 17:18

There is a large literature about how to take a drawing in the plane which tries to give a 3-dimensional view of a polyhedron and extract a polyhedron from this drawing, in contrast to applying Steinitz's Theorem which gives necessary and sufficient conditions that a graph be the vertex-edge graph of a convex 3-dimensional polyhedron:

Here is a sample of the other kind of work:

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I think you misunderstand my point; I have the coordinates of the planes ( or faces) in 3D format, so it's not a case of extracting polyhedron from the drawing. All I need is an efficient algorithm to come up with the polyhedron bounded by two faces. – Graviton Dec 11 '10 at 15:24

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