Hiki Dimensional Split Guess
1. 1-dimensional space divided by 0-dimensional space (point segmentation straight)
When the number of 0-dimensional spaces that do not repeat is 0, 1, 2, 3, 4, 1 dimension can be divided into 1, 2, 3, 4, 5
2. 2-dimensional space divided by 1-dimensional space (straight separation surface)
When the number of 1-dimensional space between 1-dimensional space is 0, 1, 2, 3, 4, 2-dimensional space can be divided into 1, 2, 4, 7, 11
3. 3-dimensional space is divided by 2-dimensional space (three-dimensional side segmentation)
When the number of 2-dimensional spaces that do not repeat is 0, 1, 2, 3, 4, 3-dimensional space can be divided into 1, 2, 4, 8, 15
------------------- Passing, find a law --------------------
a. If there is no accident, it should be changed according to 1, 2, 4, ..., 2 ^ (n 1), n is the number of space for segmentation.
accidents
B1. 1 dimension is divided by 2 0-dimensional space, up to 2 ^ 2, but 2 ^ 2-1
B2. 2D is divided by 3 1-dimensional space, up to 2 ^ 3, but 2 ^ 3-1
B3. 3D is divided by 4 2-dimensional space, up to 2 ^ 4, but 2 ^ 4-1
So derive
Bn. N-dimensional is divided by N 1 N-1-dimensional space, up to 2 ^ (n 1), but 2 ^ (n 1) -1
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Push
4. 4-dimensional space divided by 3-dimensional space
When the number of 3-dimensional spaces that do not repeat is 0, 1, 2, 3, 4, 5, 4-dimensional space can be divided into 1, 2, 4, 8, 16, 31
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理 happy ;-)
