April 28th, 2002

sleepy

dimensions

0-dimensional item: point.
Is a 0-dimensional vertex.

1-dimensional item: line.
Has two 0-dimensional vertices.
Is a 1-dimensional edge.

2-dimensional item: square.
Has four 0-dimensional vertices.
Has four 1-dimensional edges.
Is a 2-dimensional face.

3-dimensional item: cube.
Has eight 0-dimensional vertices.
Has twelve 1-dimensional edges.
Has six 2-dimensional faces.
Is a 3-dimensional cube.

Theory:
f( id, pd ) = f( id - 1, pd ) * 2 + f( id - 1, pd - 1 )
"id" is "item dimension"
"pd" is "part dimension"

thus f( 2, 1 ) is "how many 1-dimensional whatevers does a 2-dimensional solid have"

f( n, m ) = 0 for n < 0 or m < 0
f( n, n ) = 1 for all n >= 0
f( 0, n ) = 0 for all n > 0

rationale: Taking an item from dimension n to dimension n + 1 in this method involves duplicating the item in the n+1th dimension and extruding it through that dimension. Two duplicates of the original is what yields the f( id - 1, pd ) * 2. The extrusion - taking every part of dimension m and turning it into a part of dimension m + 1 - is what yields the f( id - 1, pd - 1 ).

Predictions:

4-dimensional item: hypercube.
Has 16 0-dimensional vertices.
Has 32 1-dimensional edges.
Has 24 2-dimensional faces.
Has 8 3-dimensional solid facets.
Is a 4-dimensional hypercube.

5-dimensional item: hypercube 5.
Has 32 0-dimensional vertices.
Has 80 1-dimensional edges.
Has 80 2-dimensional faces.
Has 40 3-dimensional solid facets.
Has 10 4-dimensional hypercube facets.
Is a 5-dimensional hypercube.

6-dimensional item: hypercube 6.
Has 64 0-dimensional vertices.
Has 192 1-dimensional edges.
Has 240 2-dimensional faces.
Has 160 3-dimensional solid facets.
Has 60 4-dimensional hypercube facets.
Has 12 5-dimensional hypercube facets.
Is a 6-dimensional hypercube.

. . . now I want to draw one :P
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