How Many Edges Are in an 11D Hypercube?

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The discussion centers on determining the number of edges in an 11-dimensional hypercube. The initial estimate suggests there are 11,264 edges, derived from a recurrence relation that defines the number of edges (E(N)) in terms of the edges of the previous dimension (E(N-1)) and the number of vertices (2^N). The recurrence relations for points, edges, and faces are presented, indicating a systematic approach to calculating the properties of hypercubes as dimensions increase. One participant expresses interest in the derivation of these relations, while another shares an alternative method based on conceptualizing how dimensions are added, emphasizing the doubling of existing elements and the addition of new elements through stretching. The correctness of the edge count is acknowledged, reinforcing the validity of the calculations presented.
maze
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How many edges (1 dimensional facets) are there in a 11 dimensional hypercube?

Here is a diagram for 1, 2, and 3 dimensions:
http://img151.imageshack.us/img151/3056/hypercubefacets2ye6.png
 
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maze said:
How many edges (1 dimensional facets) are there in a 11 dimensional hypercube?

I would imagine there ought to be 11264 edges. In thinking about it, for a given dimension N there's:

P(N) = # of points = 2^N
E(N) = # of edges = 2*E(N-1) + 2^(N-1)
F(N) = # of faces = 2*F(N-1) + E(N-1)

Which would mean the next step (the one I can't mentally imagine) would be:

S(N) = # of solids = 2*S(N-1) + F(N-1)

And so forth.

DaveE
 
The number is correct (nice), so I assume your stuff is right. I'm not sure I understand where those recurrence relations come from though. Perhaps you could explain more, as I'm interested to know. I solved it in a completely different way, with formula below in the spoiler.

let n be the dimension of the cube and s the dimension of the facet, then the number of facets is
(n choose n-s) 2^(n-s)
 
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maze said:
The number is correct (nice), so I assume your stuff is right. I'm not sure I understand where those recurrence relations come from though. Perhaps you could explain more, as I'm interested to know.

I solved it just by thinking about how each element is generated. It could probably use some simplification, since it's obviously reducible to some degree (as your formula demonstrates).

Anyway, the idea I had was basically that when you add a dimension you start by doubling whatever it was you already had. So if you're going from 2 dimensions to 3 dimensions, you're taking the square you already had and making another one, which will be connected to the first. So start by doubling.

But that's not all, obviously. You're also adding new elements by stretching existing elements. Each point stretches into a line, each line into a face, and each face into a solid, etc. So look at how many elements you had previously, and that's how many elements of 1D higher that you'll be adding to the next iteration.

DaveE
 

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