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How many ways can you color the edges of a hexagon in two colors?

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    How many ways can you color the edges of a hexagon in two colors? It is assumed two colorings are identical if there is a way to flip or rotate the hexagon.

    2. Relevant equations

    Orbit Stabilizer Lemma and Burnside's Lemma


    3. The attempt at a solution

    This, implements the Orbit Stabilizer Lemma and Burnside's Lemma (think necklace permutations) however, is there anything special or different to computing this because you are now dealing with the faces/edges rather than the vertices (or beads of a necklace)?

    Thanks.
     
  2. jcsd
  3. Feb 29, 2012 #2
    To answer my own question: it does not matter.

    Di6 symmetry with order 12.
    6 rotation symmetries
    6 reflection symmetries

    By Burnside's Lemma:

    ƒ([itex]n[/itex]) = [itex]\frac{1}{12}[/itex]⋅(2⋅[itex]n[/itex] + 2⋅[itex]n^{2}[/itex] + 4⋅[itex]n^{3}[/itex] + 3⋅[itex]n^{4}[/itex] + [itex]n^{6}[/itex])

    Where
    [itex]n[/itex] := # of colors
    ƒ([itex]n[/itex]) := # of unique colorings

    [itex]n[/itex] = 2
    ƒ([itex]2[/itex]) = 13


    [itex]n[/itex] = 3
    ƒ([itex]3[/itex]) = 92

    [itex]n[/itex] = 4
    ƒ([itex]4[/itex]) = 430

    [itex]n[/itex] = 5
    ƒ([itex]5[/itex]) = 1505


    [itex]\cdots[/itex]
     
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