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Cube and square doubles

  1. Aug 25, 2009 #1
    I think I've managed to construct a symmetric set of functions, that lets a [tex] x, x^2, x^3\;[/tex] double an algebraic number.

    I have a series that goes {6,9},{27,54} already. If I set x = 3, then I replace these terms with [tex]\; \{ 2x, x^2 \} ; \{ x^3, 2x^3 \} [/tex]
    I have the series and a symmetric set of functions; can I put these in a matrix, and what happens if I do, I'm going to need matrix functions?

    This is just something that fell out of a posted list of the first two sets of combinations for the pocket Rubik's cube - is it the "doubler" I need though? I mean, there it is...

    ed: whoops, bracketed as it should be.
     
    Last edited: Aug 25, 2009
  2. jcsd
  3. Aug 26, 2009 #2
    Ok, I had a thunk about this, and the problem per se, is about composition and decomposition.

    If I start with a cube, I can see 3 faces of it; This is the first d/dx = 3 square products from a cube (product). Next d/dx on 3 squares (now disjoint) gets 6x, then x. That is, 3 squares decompose into 6 edges or nodes, then into a single "X" crossing remains in G.

    Take 1 x and square it; double the square, and so on to re-integrate. The 2-sliced cube is obviously algebraic, there are [tex] (8-1)!3^7 [/tex] digits in the number, each is a squares product in a subgroup of squares.

    But the next duple is {120,321} or {321,120} since the order of the "results" is irrelevant (each is independent functionally), and so is the operator order, except for the asymmetry: [tex] 3x^2 \;> 2x^3 [/tex], which implies a subtraction.

    This makes sense because the outer function "looks ahead"; you can see in the chart for G that the two generators are out-of-phase at n > 0. The f o g must borrow states from q; q is the inner "algorithmic series" which is the most descent; f is least. Assume the first 15+m states for f and q are distinct, up to some value for n where borrowing means the f-list has connections to q's algebra and copies states. The f and q lists are also copies so they double |G|.

    The physical cube isn't doubled its orientations are (for each crossing in X).
     
  4. Aug 26, 2009 #3
    This is the list in question, which I believe is a density chart of the rationals in |G| = N-1.

    There are 15 intervals (n) which is the same # as the values for f and q added together. The doubling of Q the quotient is seen in the two functions (full and quarter turns are "double" and "split" moves), that are out-of-phase from n = 1. These are two "path operators" because you can switch from f to q during a graph-walk. The connections between the individual quotients, and how this is ordered by n is what I'm interested in. The f gets ahead of the q function so it must borrow states from Q, or the inner quotient is generated in smaller intervals; at what point is the first congruence in f and q?

    This is pursuant to reading up some papers, including Fowler's, some of Gardner's and Baez' online editorial about the symmetry groups and Rubik's especially. I realize this is only one aspect of a much larger domain.

    Code (Text):
    The number f of positions that require n full twists and number q of positions that require n quarter turn twists are:
    n   f            q
    0   1            1
    1   9            6
    2   54           27
    3   321          120
    4   1847         534
    5   9992         2256
    6   50136        8969
    7   227536       33058
    8   870072       114149
    9   1887748      360508
    10  623800       930588
    11  2644         1350852
    12                  782536
    13                  90280
    14                  276
    .
     
     
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