BrainKing - Diskusní kluby (Chess variants (10x8))

Chess variants (10x8)

Sam has closed his piano and gone to bed ... now we can talk about the real stuff of life ... love, liberty and games such as
Janus, Capablanca Random, Embassy Chess & the odd mention of other 10x8 variants is welcome too

For posting:
- invitations to games (you can also use the New Game menu or for particular games: Janus; Capablanca Random; or Embassy)
- information about upcoming tournaments
- disussion of games (please limit this to completed games or discussion on how a game has arrived at a certain position
... speculation on who has an advantage or the benefits of potential moves is not permitted while that particular game is in progress)
- links to interesting related sites (non-promotional)

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Mód: Každý může psát

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16. října 2004, 19:46:20

Chessmaster1000

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Subjekt: CardinalFlight......

Změněno uživatelem Chessmaster1000 (16. října 2004, 19:48:25)

<>What works well with the program is sticking the king and rooks where they usually belong >and randomize the other pieces. This gives 740 different setups.

How did you come up with that number? My calculations give 720.

The number of initial positions where the Rooks and the King is on their "normal" places AND the Bishops are on different colors is:

(7!/4) - (5!/2)·(6+3) = 1260 - 540 = 720

>If you randomize all the pieces except that bishops are on opposite colors and king between >rooks there are 84,000 different ways.

I agree on that. The number F we are looking is:

F = (Σ[{i=1,8} (i+1)·i/2])·(7!/4) - (5!/2)·9·((Σ[{k=1,8} (k+1)·k/2])-6·3-4·5-2·7) - 6·a-4·b-2·c

where:
a = 2·(5!/2)·(6+3) + (5!/2)·(10+1)
b = 3·(5!/2)·(6+3) + 2·(5!/2)·(10+1)
c = 4·(5!/2)*(6+3) + 3·(5!/2)·(10+1)

And it really results in 84000.

How did you calculated that number? You used the following method or something different? I hope you didn't count all positions by hand:-)