How Many Edges Are in an 11D Hypercube?

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Discussion Overview

The discussion revolves around determining the number of edges in an 11-dimensional hypercube, exploring the mathematical relationships and recurrence relations that govern the properties of hypercubes in various dimensions.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the number of edges in an 11-dimensional hypercube and provides a diagram for lower dimensions.
  • Another participant proposes that there are 11264 edges and presents recurrence relations for points, edges, and faces in hypercubes, suggesting a pattern based on dimensionality.
  • Some participants express interest in the recurrence relations and request clarification on their derivation.
  • One participant describes their own reasoning process for calculating the number of edges, emphasizing the doubling of existing elements when adding dimensions and the stretching of elements from lower dimensions.

Areas of Agreement / Disagreement

While there is agreement on the proposed number of edges, participants express differing understandings of the recurrence relations and the reasoning behind them, indicating that the discussion remains unresolved regarding the derivation of these relations.

Contextual Notes

The discussion includes various assumptions about the properties of hypercubes and their dimensional relationships, but these assumptions are not explicitly defined or agreed upon by all participants.

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)
 
Last edited:
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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