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Partitioning high-dimensional space

  1. Dec 10, 2006 #1
    Hello to all,
    I know (as almost anyone) how to divide an square in let's say 4 equal parts, a cube in 8 or more, i.e. sub-squares and sub-cubes respectively. But a high dimensional space??? :confused: . ..I mean where dimensions are >>3, for example 100 or 1000?. In fact I don't know if the space is "hyper-rectangle" or "hypersphere". Probably this second part (to know or determine the shape of the hyperspace) is another question...

    I apologise if this is not the correct forum, then I'll apreciate if you can suggest another forum to post this/these questions :redface:

    thanks
     
  2. jcsd
  3. Dec 10, 2006 #2

    HallsofIvy

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    The only way I can make sense of your question is to assume you are asking about partitioning an n-dimensional "parallelapiped".

    If you divide each edge into n parts, then you divide the entire figure into:
    dim 2: n2 parts
    dim 3: n3 parts
    dim 4: n4 parts
    .
    .
    .
    dim 1000: n1000 parts.

    Get the idea?

    You have to be dividing into parallelopipeds (your "hyper-rectangles"). Spheres and "hyper-spheres" won't partition space.
     
  4. Dec 11, 2006 #3
    First than all thanks for helping to clarify this subject.

    Let's see, if I have a dim=1000 space where each dimension is 100 units length, is it possible to divide it in 4 (or any arbitrary number of) independent and equally sized parts? and if so, how can I reference each of these "parts".

    I mean, is it correct to say

    part 1: (0,25) in all 1000 dimmensions
    part 2: (25,50) in all 1000 dimmensions
    part 3: (50,75) in all 1000 dimmensions
    part 4: (75,100) in all 1000 dimmensions

    just as it would happens in 2 and 3 d cubes.
    :confused:
     
    Last edited by a moderator: Dec 11, 2006
  5. Dec 11, 2006 #4
    May be it will help.

    Let we have n-dimensional "parallelepiped".

    part 1: [tex]0< x_1< 1, 0< x_2< 1,... 0< x_{n-1}< 1, 0< x_n< 1[/tex],

    part 2: [tex]0< x_1< 1, 0< x_2< 1,... 0< x_{n-1}< 1, 1< x_n< 2[/tex],

    part 3: [tex]0< x_1< 1, 0< x_2< 1,... 1< x_{n-1}< 2, 0< x_n< 1[/tex],

    part 4: [tex]0< x_1< 1, 0< x_2< 1,... 1< x_{n-1}< 2, 1< x_n< 2[/tex],

    part 5: [tex]0< x_1< 1, 0< x_2< 1,... 1< x_{n-2}< 2, 0< x_{n-1}< 1, 0< x_n< 1[/tex],
    ...

    part [tex]n^{999}[/tex]: [tex]0< x_1< 0, 999< x_2< 1000,... 999< x_{n-1}< 1000, 999< x_n< 1000[/tex],
    ...

    part [tex]n^{1000}[/tex] -1: [tex]999< x_1< 1000, 999< x_2< 1000,... 999< x_{n-1}< 1000, 998< x_n< 999[/tex]

    part [tex]n^{1000}[/tex]: [tex]999< x_1< 1000, 999< x_2< 1000,... 999< x_{n-1}< 1000, 999< x_n< 1000[/tex]


    You can replace "<" by "<=".
     
    Last edited: Dec 11, 2006
  6. Dec 11, 2006 #5

    HallsofIvy

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    It doesn't make sense to say "each dimension is 100 units length". I assume you meant to say "there is an edge of the parallelopiped in the direction of each axis having length 100 units.
    Now, when you say " divide it in 4 (or any arbitrary number of) independent and equally sized parts?" do you mean divide the parallelopiped in to 4 equal parts? That's how I would interpret it but your "part 1, part 2, part 3, part 4" seems to be dividing each edge into 4 parts: that divides the parallelopiped into 41000 equal parts.

    In particular I am puzzled by your "just as it would happens in 2 and 3 d cubes.". It is true that, since 22= 4, you can divide a square into 4 equal parts by dividing each side into 2 equal parts. Since n3= 4 does not have any integer solution, there is no way to divide a 3-d cube into 4 identical parts by dividing each edge into equal parts. You might well be able to do it by dividing different edges into different numbers of (non-equal) parts. The same is true for 10000 dimensional space.
     
  7. Dec 12, 2006 #6
    Thanks again,
    Effectively a "hyper-figure" can be divided in equally sized parts by using 2^n parts, where n is the dimensionality.
    So a 2d square can be divided into 2^4 equally sized parts, a 3d cube in 2^3 that is 8 little equally sized cubes. But is is partially true, as a 2d square can be divided into just 2 equally sized rectangles (or subspaces or sub areas), and a 3d squared space or cube can be divided into 2 big subcubes and also in 4 3d rectangles o (prisms?).
    Example:
    For a cube with one vertice in the origin, the exteme coordinates for each edge are in (100, 100, 100) for x, y,z axis, it can be divided into 4 equally sized parts. To help me explain I made an illustration. I know how to handle each one by its coordinates, ie. I know the region in the "delimited space" they occupy.

    So I wonder if for higher dimensions it is still possible to make something as illustrated and hence to be able to handle each part individually and how :smile:

    I apologise if the vocabulary is not technically acceptable, I should tell I am not expert in this subject. But as I want to learn and dissipate ignorance I ask :rolleyes: , and appreciate your time to answer.
     

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  8. Dec 12, 2006 #7
    I admit that in the example each edge is not equally divided, otherwise it would be imposible to divide a 3d into 4 parts as HallsofIvy said.
    But any way it is possible to divide the 3d space into arbitrary number of regions.
     
  9. Dec 12, 2006 #8
    For example you want to divide n-dimensional "parallelepiped" in s parts.

    part 1: [tex]0<x<\frac{1000}{s},0<y<1000,0<z<1000,...,0<t<1000[/tex]
    part 2: [tex]\frac{1000}{s}<x<\frac{2000}{s},0<y<1000,0<z<1000,...,0<t<1000[/tex]
    ...
    part s: [tex]1000-\frac{1000}{s}<x<1000,0<y<1000,0<z<1000,...,0<t<1000[/tex]

    If you want then you can divide x,y dimensions or y,z dimensions (your figure).
     
    Last edited: Dec 12, 2006
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