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June 26

Simplex

Given a simlex of n+1 vertices, slice it with hyperplanes along its axes of symmetry - that is, for each pair of vertices, construct a hyperplane that connects the point halfway between those two with all other vertices. How many chunks will this slice the simplex into, and will they all have the same volume? Black Carrot 04:12, 26 June 2007 (UTC)[reply]

Well, the number of edges of a simplex is a triangular number. So you'll have

E={{n+1} \choose {2}}

edges, and that should give you

2^{E}

chunks. I don't know about the equal volume idea though. - Rainwarrior 06:07, 26 June 2007 (UTC)[reply]

Er, that was an overestimate. I think all of the planes will be intersecting at one point. So each new division would really only be passing through two of the existing spaces (adjacent to the edge, and adjacent to the other (n-1) vertices. My guess would be 2E instead. - Rainwarrior 06:44, 26 June 2007 (UTC)[reply]

Actually, thinking about it in terms of Barycentric coordinates, I think you're right that the volumes should be equal. Planes connecting (n-1) points to the midpoint of each edge should be at the barycentre, which means the simplex created by (n-1) points and this centre point should have the same volume as every other simplex from every other group of (n-1) points and this centre, and finally each of these simplexes is divided exactly in two by the remaining plane, which should make your hypothesis true. - Rainwarrior 06:32, 26 June 2007 (UTC)[reply]

Yeah, I was pretty sure about the symmetry. The way I figured it, any set of (n-1) planes is mapped to itself by reflection across the remaining plane, it having been chosen as an axis of symmetry. Since the planes are the only thing defining the boundaries of each section other than the surface of the simplex, which is also symmetric across each axis, chunks map to chunks. So, given any starting chunk, it's symmetric with each chunk it's adjascent to by reflection, and etc. The harder part is the actual number of chunks. In 3 and fewer dimensions, I can just draw it, which gives 1, 2, 6, 12, but I'm not sure how to generalize that. Black Carrot 07:48, 26 June 2007 (UTC)[reply]

Okay, sorry, I miscounted. The last one should be 24, which suggests factorials or something similar. In general, it seems that the chunks of each simplex can be made by chunking each of its one-less-dimensional faces, and extruding those faces to the center, which would prove that factorial idea nicely. Does anyone support that notion? Black Carrot 07:57, 26 June 2007 (UTC)[reply]

Is it really 24? Mentally it comes out to 12 for me, but maybe I'm wrong. - Rainwarrior 05:11, 27 June 2007 (UTC)[reply]

For each pair of vertices (A,B), the bisectrix hyperplane divides hyperspace into the points closer to A and closer to B. Therefore there is a bijection between the permutations of the vertices and the chunks into which the hyperplanes partition space. The interior points of each chunk share the closeness ranking of the vertices. --Lambiam ^Talk 09:36, 26 June 2007 (UTC)[reply]

I'm afraid I don't follow you. Black Carrot 02:31, 27 June 2007 (UTC)[reply]

I think I get it now. Given each permutation of n+1 vertices (A,B,C...), a set of n divisions (A,B) (B,C) (C,D)... can be defined. Each bunch of divisions can be put into correspondence with a chunk by picking the one on the first side of each - The one on the A side of (A,B), the B side of (B,C), etc. Is that about right? Black Carrot 03:01, 27 June 2007 (UTC)[reply]

If there are n vertices, there are n(n–1)/2 pairs, each giving a division; for example, for n = 3, the vertices {A,B,C} give rise to the divisions (A,B), (A,C) and (B,C). Except on the bisectrix for (B,C), each point is closer to B or closer to C. In the ASCII art below, the chunk labelled ABC corresponds to the points for which A is closest, then B, and finally C. And for chunk CAB, C is closest, followed by A, with B coming in last. The orderings in which B precedes C (ABC, BAC and BCA) are on one side of the (B,C) bisectrix; the others are on the other side.

                       A
                       .
                      /:\
                     / : \
                    /  :  \
                   /   :   \
                  /    :    \
                 /     :     \
                /  ABC : ACB  \
               /..     :     ..\
              /   ...  :  ...   \
             /       ..:..       \
            /  BAC  ...:...  CAB  \
           /     ...   :   ...     \
          /   ...      :      ...   \
         / ...         :         ... \
        /..     BCA    :    CBA     ..\
       /_______________:_______________\
      B                                 C

In general, there are n! orderings, each of which gives a different chunk. --Lambiam ^Talk 13:29, 27 June 2007 (UTC)[reply]

Thanks. That's helped a lot. Black Carrot 08:14, 29 June 2007 (UTC)[reply]

What do Tesseracts have to do with Time Travel?

I was just curious, but what do Tesseracts have to do with time travel?68.105.139.161 05:29, 26 June 2007 (UTC)[reply]

I'm not sure they do. Tesseracts are four-dimensional analogs of cubes. Since most people are only familiar with physical space, which appears to be 3-dimensional, a convenient way to visualize the fourth dimension is to think of it as a time dimension. Other than that, I don't think they are in any way related to time or time travel. -- Meni Rosenfeld (talk) 07:30, 26 June 2007 (UTC)[reply]

(after edit confict) In our article on Tesseracts, the secrtion Tesseracts in art and literature mentions several references to the notion or term in science fiction. Presumably, the connection is that tesseracts are 4D objects, and in some fictional-scientific way help to achieve time travel – after all, is time not the mysterious fourth dimension? --Lambiam ^Talk 07:34, 26 June 2007 (UTC)[reply]

For some reason the words "Time Cube" came to mind. Confusing Manifestation 13:14, 26 June 2007 (UTC)[reply]

If you consider space-time as a four dimensional vector space, then the shape a 3-D cube forms moving forward through time is a Tesseract (the fourth-dimensional length depends on the length of time that elapses). Dugwiki 15:30, 26 June 2007 (UTC)[reply]

Nothing, except in the book A Wrinkle in Time. --71.146.129.86 17:20, 26 June 2007 (UTC)[reply]

Question about Xfig

I do not know whether this is the right place to ask, or whether it should be the computer desk.

With Xfig, I used ARC drawing by specifying three points. I drew four arcs to form a simple closed curve. Then I used Edit, fill color with blue, fill style with filled, clicked apply and clicked done. I managed to paint the area of EACH ARC with blue color. I did RTFM but probably overlooked something.

Question: How do I paint the interior of this simple closed curve with blue color? Thank you in advance. Twma 05:56, 26 June 2007 (UTC)[reply]

Rather than ARC you'll need to define your curve as a "closed spline", this will define a closed region. While the four arcs appear to define a closed region to the program they are seperate objects, so fillings not possible. I've only got a windows port so I don't know if theres a way to join the arcs together so they create a closed spline. --Salix alba (talk) 12:06, 26 June 2007 (UTC)[reply]

Thank you. In four color theorem, the arcs are shared by neighboring regions although I over simplify the problem in the above setting. Twma 02:05, 27 June 2007 (UTC)[reply]

I still need help. Thank you in advance. Twma 02:36, 28 June 2007 (UTC)[reply]

Digitizing volumes

Is there any open code available for digitizing arbitrary 3d shapes (using a uniform grid)? Or else, any algorithms that I could implement? The volume (input) is known in terms of several randomly located points on its surface. The output will be a set of points located on a uniform grid. Thanks! deeptrivia (talk) 16:40, 26 June 2007 (UTC)[reply]

Possibly the convex hull of the points would give a polgonization of the surface, you could then intersect to polygons with the grid lines to give a regular set. The Qhull algorithm, is very efficient. While solving the problem this would be less than ideal as it would square off the surface. An alternative would be to consider algorithms for isosurface and Level set and meshing may help solve the more genaral problem. There is a very an extensive litrature on all these topics and source forge seems to have a good few project devoted to meshing. --Salix alba (talk) 23:17, 26 June 2007 (UTC)[reply]

Your problem statement confuses me. Are you trying to begin with sparsely sample surface points, infer a surface (maybe smooth, maybe not), and generate a "uniform" grid on that inferred surface (for some meaning of uniform)? Depending on the initial sampling and what you know about the surface, this could be difficult. I'd suggest visiting the comp.graphics.algorithms newsgroup, browsing, then posting a more detailed request there. --KSmrq^T 01:30, 27 June 2007 (UTC)[reply]