15x1 tourny

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pe
Posts: 39
Joined: Fri Sep 27, 2002 5:08 pm

15x1 tourny

Post by pe » Mon Jan 03, 2005 5:11 pm

Greg,

i'm playing in a 15 (players) x 1 (game) tourny at the moment. i just noticed that one player has 2x white and 12 times black. this is randomly set ?

is it hard to make it 7/7 for everyone ?

cheers,
peter

slowblunder
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Joined: Thu Aug 04, 2005 4:23 pm

Post by slowblunder » Sat Sep 03, 2005 4:35 am

It is not hard to implement that each player has 7 games with White and 7 with Black - neither in theory nor in reality. The following algorithm is used in OTB blitz tournaments and works fine for any number of players (odd or even).

Odd number (example 15):
All 15 players are sitting at a long table with 7 chessboards, 7 players on the "North" side, 7 on the "South" side and the 15th at one end ("East" side is okay). Although it doesn´t really matter, the seven boards´ orientations (for players 1-14) usually are alternatively, that means: if you play White, your two neighbours both play Black.

After the first round all players move one place clockwise (including the starting "East" player). So another player is sitting on "East" enjoying a little break, and the others play a game with the opposite colour than the preceeding round. And so on ...

The tournament ends as soon as every player has occupied each seat once (15 rounds), the "East" seat does count, too. It is obvious that all participants will have played seven White and 7 Black games plus a round without a game ("East" position). It is not obvious that each participant meets each other participant exactly once, but it works!
If the boards´ orientation is alternatively you have the nice effect that all play White-Black-White-Black-...

Even number:
Having an even number of players you don´t use the "East" place, but one player doesn´t rotate during the whole tournament, HE MUST SIT ON THE SAME CHAIR ALL THE TOURNAMENT. This player usually is sitting at one end of the long table. So his right neighbour will become his next opponent if this non-moving player is sitting on the "South-West" corner. Again, each player will meet each other player exactly once!
After each round the non-moving player has to turn his board so that he also plays White-Black-White-Black...
Of course, in the end 50% will have one more game with Black, the other 50% have one more game with White. And it may happen that you play two Black-games or two White-games in sequence - but both is unavoidable!


In Germany this system is called "bicycle-chain-algorithm" which may be a better description than all explanations above. "Round robin" in English?

gmiller
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Post by gmiller » Sun Sep 04, 2005 9:09 am

It's not so much a problem of the algorithm as it is changing the way the whole system of creating a match works. As of now the colors are determind for each individual game as they're created, not when the match is created. It's not impossible, but certainly not a little change either. I've always had this pretty low on the priorities list because the only way to really be fair is to play an even number of games against each opponent. Because even when you do get an even number of games as black and white, it's still feasible you'll get white against the easy players, and black against the better players.

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