BrainKing - 掲示板 (Chess variants (8x8))

管理人: WhisperzQ

, Mort

, Bwild

Chess variants (8x8)

including Amazon, Anti, Atomic, Berolina, Corner, Crazy Screen, Cylinder, Dark, Extinction, Fischer Random, Fortress, Horde, Knight Relay, Legan, Loop, Maharajah, Screen, Three Checks

For posting:
- invitations to games (you can also use the New Game menu)
- information about upcoming tournaments
- discussion 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)
- links to interesting related sites (non-promotional)

Community Announcements:
- Nasmichael is helping to co-ordinate the Fischer Random Chess Email Chess (FRCEC) Club and can set up quad or trio games if you send him a PM here.

掲示板表
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モード: 誰でも投稿可能

14. １１月 2007, 19:47:49

Herlock Sholmes

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件名: Random Cheversi

when I started to think about improving Cheversi I cannot stop, lol ... one thing is for sure with random placement of the pieces ... someone will have an advantage from the very beginning ... does this advantage really mean much ? Look at Ludo for example ... the other thing is, that with random placement White (starting color) has usual advantage as a first player (leader) like in Five in a row, or Reversi ... but Black has the last move which balances this advantage by White ...
To play well random placement Cheversi is a real art and I think that computer power should be emplyed in order to solve this game ... by the way, can you calculate how many different placemet of 16 pieces can be on a board with 64 squares ... ?
This is how many starting positions we may have.
In our lifetimes we will never encounter the same position. And this makes Random Cheversi an exciting game.
and that's it for now.
Andy.

14. １１月 2007, 20:51:01

AbigailII

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件名: Re: Random Cheversi

AbigailII (14. １１月 2007, 20:52:53)に変更されました。

dicepro: can you calculate how many different placemet of 16 pieces can be on a board with 64 squares

That's fairly trivial. Not counting rotations and reflections of the board, the number is (64

50*49) / (2^6) which equals 159708538424128885551360000. (You might want to turn of the stupid smileys).