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If on a chess board each square could be black or white - how many combinations of boards could there be?

Bebop5

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2 to the 64th power ? (whatever that comes to) reminds me of the grains of rice story

u
Sharp Edge

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Do we not have to compensate for some repeats? I mean, you can flip the board 90 degrees and it can end up being 4 different sets

d

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No, because if you do EVERY single combination facing 1 way, the others oon the other sides are already done.

2^64 is correct.

greenpawn34

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18,446,744,073,709,551,615 grains of rice weighing 461,168,602,000 tons. (thank you Wiki).

Now this is Swiss Gambit territory.

At the start of a game using the beginner formula.

Queen = 9
Rook = 5
Bishop = 3
Knight = 3
Pawn = 1

After winning a bet the loser askes you to contruct a game saying he wiil
give you £1.00 in value for every piece left on the board, both Black and White
at the time of checkmate.

You agree. What is the most amount of money you can score?

Basically how many possible pawn promotions (thus making £1 = £9)
can you squeeze out of a game of chess?

BK

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Originally posted by greenpawn34
18,446,744,073,709,551,615 grains of rice weighing 461,168,602,000 tons. (thank you Wiki).

Now this is Swiss Gambit territory.

At the start of a game using the beginner formula.

Queen = 9
Rook = 5
Bishop = 3
Knight = 3
Pawn = 1

After winning a bet the loser askes you to contruct a game saying he wiil
give you £1.00 in value for every ...[text shortened]... how many possible pawn promotions (thus making £1 = £9)
can you squeeze out of a game of chess?
so the guy will give you a dollar for every point?

greenpawn34

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A Dollar, a Pound, A Euro, what every currency you want.
He will give one for every point remaining on the board.

u
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Originally posted by dogfish44
No, because if you do EVERY single combination facing 1 way, the others oon the other sides are already done.

2^64 is correct.
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

S
Caninus Interruptus

2014.05.01

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Originally posted by greenpawn34
18,446,744,073,709,551,615 grains of rice weighing 461,168,602,000 tons. (thank you Wiki).

Now this is Swiss Gambit territory.

At the start of a game using the beginner formula.

Queen = 9
Rook = 5
Bishop = 3
Knight = 3
Pawn = 1

After winning a bet the loser askes you to contruct a game saying he wiil
give you £1.00 in value for every ...[text shortened]... how many possible pawn promotions (thus making £1 = £9)
can you squeeze out of a game of chess?
You can promote every pawn on the board, but you must capture all the minor pieces. 9 x 9 Queens + 2 x 2 Rooks = 91 points. This multiplied by 2 armies = 182 points = £182

greenpawn34

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Cheers SG. £182.00 that will do.

There must be a proof game. Would not mind seeing it.

Campaigner

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No the board must stay 'up' no flips or turns.

AP

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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.

S
Caninus Interruptus

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Originally posted by greenpawn34
Cheers SG. £182.00 that will do.

There must be a proof game. Would not mind seeing it.

Bebop5

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Originally posted by Arctic Penguin
[quote]Originally posted by ua41
[b]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 corne ...[text shortened]... he 10th digit.

This method seems sound to me, but I might have miscounted somewhere.
My God, why must everything be over-analyzed and complicated?? Do you flip the board or rotate it when you play chess? NO. Then just solve the problem as if it was a chessboard glued to the table in front of you. Jeeeez.

AP

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Originally posted by Bebop5
My God, why must everything be over-analyzed and complicated?? Do you flip the board or rotate it when you play chess? NO. Then just solve the problem as if it was a chessboard glued to the table in front of you. Jeeeez.
Most chess boards I've seen don't specify which end is for black and which end is for white, so yes it probably does get rotated between games. The puzzle is a bit too simple if the board is just fixed to the table, don't you think?

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