# 185 geometry/hypercube.p

## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 185 geometry/hypercube.p

How many vertices, edges, faces, etc. does a hypercube have?

geometry/hypercube.s

Take any vertex of the hypercube, and ask how many k-V's it

participates in. To make a k-V it needs to combine with k adjacent and

orthogonal vertices, and there are (nCk) distinct ways of doing this

[that is, choose k directions out of n possible ones]. Then multiply

by 2^n, the total number of vertices. But this involves multiple

counting, since each k-V is shared by 2^k vertices. So divide by 2^k,

and this yields the answer: (nCk)*2^{n-k}.

For example, 12d hypercube:

0-v: 4,096
1-v: 24,576
2-v: 67,584
3-v: 112,640
4-v: 126,720
5-v: 101,376
6-v: 59,136
7-v: 25,344
8-v: 7,920
9-v: 1,760
10-v: 264
11-v: 24
12-v: 1

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