Given a set of planes that define a convex polyhedron. I would need to quickly find an arbitrary point inside the polyhedron. Not necessarily, but ideally the centroid. I want to do this without any plane intersections if possible.

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# Point inside convex polyhedron defined by planes

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#2
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Posted 21 February 2013 - 03:16 AM

You need to prove that your ~~planes~~ halfspaces define a non-empty convex polyhedron first; I don't think there are practical ways which don't involve intersecting planes and convex polyhedra and building explicit faces, edges and vertices (from which computing the centroid is easy).

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#3
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Posted 21 February 2013 - 06:02 AM

Your plane equations are: Ni*x-Di, with Ni normalized row vectors, Di are scalars, and x is a column vector.

Now minimize the sum of squared distances with respect to x: S=sum[(Ni*x-Di)^2]

Obtain: x=((sum[Ni^T*Ni])^+) * sum[Di*Ni^T]

I'm almost sure that the pseudo-inverse is not required (and you can take the regular inverse), because this reminds me a lot of SVD. However, this is just a hunch.

EDIT: Yup, the matrix should be invertible for any closed polyhedrons. No need for pseudo-inverse.

**Edited by max343, 21 February 2013 - 06:52 AM.**

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#4
Prime Members - Reputation: **684**

Posted 21 February 2013 - 08:10 AM

You just need to check if the point is behind (ie half-spaces) every plane defining your volume. If so, the point is inside your volume

EDIT Sorry if it's not the point of your question

**Edited by Tournicoti, 21 February 2013 - 11:21 AM.**

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#5
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Posted 21 February 2013 - 10:25 AM

You just need to check if the point is behind (ie half-spaces) every plane defining your volume. If so, the point is inside your volume

Sure, but the question was how to find such a point that satisfies this criteria.

Thanks max343. I was thinking that this is a minimization problem, but I am bit rusty here. I will look into this!

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#6
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Posted 21 February 2013 - 07:53 PM

1. Find x0 as I previously described.

2. Solve the weighted minimization problem with the weights: Wi = |Ni*x0-Di|

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#7
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Posted 21 February 2013 - 08:39 PM

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#9
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Posted 22 February 2013 - 03:53 AM

I don't think that Simplex algorithm is the best choice here, since it searches for maximum on the boundary, while OP wanted some interior point.This sounds similar to finding a feasible solution to a linear-programming problem, so I am sure there is literature about it. This seems to be called "Phase I" in the simplex algorithm. I read the description in Wikipedia, but I am not sure I understand it.

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#10
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Posted 22 February 2013 - 06:07 AM

Please, read my post carefully. The simplex algorithm has two phases: Phase I finds a feasible configuration, and phase II improves the configuration to find the optimum value of the objective function. Phase I of the simplex algorithm is exactly the same problem the OP has.

I don't think that Simplex algorithm is the best choice here, since it searches for maximum on the boundary, while OP wanted some interior point.This sounds similar to finding a feasible solution to a linear-programming problem, so I am sure there is literature about it. This seems to be called "Phase I" in the simplex algorithm. I read the description in Wikipedia, but I am not sure I understand it.

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#11
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Posted 22 February 2013 - 01:28 PM

OP wanted something close to the centroid, this is distinctly an interior point.

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#12
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Posted 22 February 2013 - 01:39 PM

If you have a point at the boundary, finding one in the interior is pretty trivial. The wording of the first post says that he would be satisfied with that.

Maybe I missed it, but I don't think a better solution has been suggested so far... What you suggested returns the same answer if you flip one of the inequalities, so it can't possibly return something in the interior in both cases.

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#13
Members - Reputation: **340**

Posted 22 February 2013 - 03:15 PM

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#14
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Posted 22 February 2013 - 03:32 PM

I don't have inequalities in my solution. What I suggested disregards direction because it already assumes that the feasible region of the half spaces defines a convex polyhedron. If we'd flip one of the inequalities, the feasible region would be empty.

No, that's not how it works. Imagine you have a bunch of half-spaces that already define a convex polyhedron. Now take a plane that cuts that convex polyhedron through the middle, and either add one half-space or the other to the list. In both situations the feasible region is not empty, and your proposed solution will return the same answer.

**Edited by Álvaro, 22 February 2013 - 03:33 PM.**