Originally posted by ua41
That's exactly what I mean-
so you'd have some redundant combinations because they are just "flipped" images of each other so 2^64 is too many.
Just think of one the corner squares as black, the rest as white. This accounts for 4 different sets, flip it 90 degrees and you'll end up with a different corner as black. This one coloring scheme accounts for 4 combinations
To solve the problem when rotation of the board is allowed but flipping (i.e. mirror images) are not:
Count the number of fixed boards with 90 degree rotational symmetry. One corner of the board with 16 squares determines the rest: 2^16.
Count the number of fixed boards with 180 degree rotational symmetry that don't also have 90 degree rotational symmetry. One half of the board with 32 squares determines the rest, but subtract out the ones counted for 90 degree rotation: 2^32 - 2^16. So that we don't count each of these twice when we remove rotations in the final tally (once with no rotation and again with a 90 degree rotation), divide this number by 2.
Count the number of fixed boards with 360 degree rotational symmetry that don't also have 90 degree or 180 degree rotational symmetry: 2^64 - (2^32 - 2^16) - (2^16). So that we don't count each of these four times when we remove rotations in the final tally (once with no rotation and again with 90, 180, and 270), divide this number by 4.
Add up the three terms and simplify:
2^16 + (2^32 - 2^16)/2 + (2^64 - 2^32)/4
2^16 + 2^31 - 2^15 + 2^62 - 2^30
2^62 + (2^30)(2-1) + (2^15)(2-1)
2^62 + 2^30 + 2^15 = 4,611,686,017,353,613,312
This is essentially the same as ignoring the relatively few positions with rotational symmetry and doing (2^64)/4 = 2^62 = 4,611,686,018,427,387,904 which is identical up to the 10th digit.
This method seems sound to me, but I might have miscounted somewhere.