Partitioning high-dimensional space

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In summary, a high dimensional space can be partitioned into "parallelopiped" shapes where each dimension is 100 units length.
  • #1
jblade
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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
 
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  • #2
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.
 
  • #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:
 
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  • #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 "<=".
 
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  • #5
jblade said:
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:
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 into 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.
 
  • #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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  • #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.
 
  • #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).
 
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1. What is partitioning high-dimensional space?

Partitioning high-dimensional space is a method used in data analysis to divide a large dataset into smaller, more manageable subsets. This is done in order to simplify the analysis process and make it more efficient.

2. Why is partitioning high-dimensional space important?

Partitioning high-dimensional space is important because it allows for easier and more accurate analysis of large datasets. By dividing the data into smaller subsets, it becomes easier to identify patterns and relationships within the data.

3. What are some common techniques used for partitioning high-dimensional space?

Some common techniques for partitioning high-dimensional space include k-means clustering, hierarchical clustering, and principal component analysis. These techniques use different methods to group data points based on their similarities and differences.

4. How do I know which partitioning technique to use?

The choice of partitioning technique depends on the type of data and the specific goals of the analysis. It is important to understand the strengths and limitations of each technique and choose the one that best fits the data and analysis objectives.

5. Are there any challenges or limitations to partitioning high-dimensional space?

Yes, there are some challenges and limitations to partitioning high-dimensional space. One challenge is determining the optimal number of partitions or clusters, as this can greatly impact the results of the analysis. Additionally, some techniques may not work well with certain types of data or may produce biased results if the data is not properly preprocessed.

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