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Edges in 11D Hypercube

  1. Dec 7, 2008 #1
    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 [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 8, 2008 #2
    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.

  4. Dec 8, 2008 #3
    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: Dec 8, 2008
  5. Dec 8, 2008 #4
    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.

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