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Thread: calculating quantity of shapes

  1. #1 calculating quantity of shapes 
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    Oct 2017
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    consider n points on a cicrcle evenly spaced, n vertices connecting those points (each point has exactly 2 vertices connected to it)
    the total number (m) of different shapes which can be constructed this way would be:
    m=(n-1)!/2
    then concider the subclass devision of these shapes per symmetry level:
    m1:full symmetry, when the shape is rotated n steps only 1 shape emerge
    m2:when the shape is rotated n steps, 2 shapes emerge
    ...
    mn:when the shape is rotated n steps,n shapes emerge

    if n is a prime number then m1=(n-1)/2 and mn=((n-1)!-n+1))/2

    so for n=5:
    m=12
    m1=2
    m5=2
    12=1x2+5x2

    for n=6 (experimently found):
    m=60
    m1=1
    m2=1
    m3=5
    m6=7
    60=1x1+2x1+3x5+6x7

    for n=7:
    m=360
    m1=3
    m7=51
    360=1x3+7x51

    for n=8:
    m=2520 (i am a bit reluctant to go in experimental mode here)

    so any idea on how to calculate:
    -the number of subclasses (has to do with the divisors of n)
    -the number of shapes within a subclass

    help much appreciated!


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