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I'm currently working on 3D mesh segmentation.

And I want the segmented parts obtained to be homeomorphic to a disk (I mean easily unfoldable, so that we can map it as a 2D part without any points overlaping).

I have read a lot of papers on the subject, most of them talk about this homeomorphism to 2D disk, but none of them show how to prove it, or how to do it.

SO I really don't know what can I use on this surface to check if it is conformal to what I want.

Is there any formula using the points, the erea or the normals, or any criteria that shows that a given surface is topologycaly equivalent to disk?

Thank you all.

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Doesn't it suffice to check that your mesh has only one boundary curve? –  lhf Apr 28 '11 at 19:16
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1 boundary component + Euler characteristic equal to $1$ –  user8268 Apr 28 '11 at 19:23
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How about connected and Euler-Characteristic 1 –  Alexander Thumm Apr 28 '11 at 19:26
    
thanks all of you guys ;) –  Gatsu Apr 28 '11 at 20:27

1 Answer 1

The comments are right... this is easy. You first need to find the (directed) boundary edges of the mesh (which can be done by iterating over all faces of the mesh and canceling out those edges that appear twice in opposite directions, the remaining ones will be on the boundary). Secondly, you search for closed loops formed by boundary edges (build a lookup table that maps every boundary vertex to its successor, based on the list of boundary edges you already have, and follow this chain until you meet a vertex that you've encountered before). If there is only one loop, the patch is disk-like.

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Thanks a lot guys, for your answers. I tough I got it at one moment, but I still need to fix a last point. For example, if we consider a sphere built with meshes, and a (somehow) small hole on the surface of the sphere. We can say that there is only one boundary curve on this structure, and it will be the contour of the hole. Can we still say that this mesh is homeomorphic to disk? Because I think to map it, I'll need to cut it anyway. Thank you again ;) –  Gatsu Apr 28 '11 at 20:16
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Doesn't a torus with a hole in it also have only one boundary component? I think you need to check Euler characteristic as well. @Gatsu: Yes, a sphere with a hole in it is homeomorphic to a disk. –  MartianInvader Apr 28 '11 at 20:20
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Ok, but since the example you gave might occur, I'd better to check the Euler Characteristic as well, right? And this last has to be equal to 1 to be topologically equal to a disk. Thank you a lot, I think it will be enough for my problem. –  Gatsu Apr 28 '11 at 20:26

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