Author: Dann Corbit
Date: 10:57:48 11/30/99
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On November 30, 1999 at 13:34:21, José de Jesús García Ruvalcaba wrote: [snip] >Hi Dann, > what you say does not make any sense. You can not have a four-player swiss >tournament of several rounds. The simple reason is that in a swiss tournament >you can not face the same opponent twice. With four players, that makes for >three rounds at most, and in that case you have actually played a non-standard >round robin. I was thinking of just running the formheap operation and playing the top. R1: 4 players R2: 4 players reverse roles of white/black R3: 2 players R4: 2 players reverse roles of white/black I was not aware that the players were not allowed to face each other again by definition of the format. I was simply thinking of the data structure itself. [snip] > But two or three rounds more than log(number of players), rounded up to the >next integer, is not. > Even if the strongest entry is upsetted in the first round, if it is really >stronger it has a chance to win the tournament. It also has a good chance to lose it (or not come in first at least). Why two or three rounds more? Why not 12? What is the exact mathematical crossover where it becomes fair? Do we really know without an accurate seeding? To accurately establish an ELO actually requires hundreds of games. To imagine that we can fairly determine the [true] winner in: k*ceil(log(n)/log(2)) [for some small k] is not good math. This is more than ever the case with computer chess programs. Every contest there are new programs or programs that are largely rewritten or have multi-threading added or whatever. If we just want to *name* a winner, we could run a single game with each program and count points. >>On the other hand, the programmers are reluctant to play a lot of games. So >>perhaps having a 3rd or 4th place program win does not really matter. It's >>still a neat title to hang on the wall and it gives the underdogs a fighting >>chance to win.
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