Author: José de Jesús García Ruvalcaba
Date: 12:12:58 11/30/99
Go up one level in this thread
On November 30, 1999 at 13:57:48, Dann Corbit wrote: >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 > Usually, the number of players in a swiss tournament does not decrease, unless somebody drops out. Playing a match for every round of the tournament is posible. Actually, that is the way team tournaments are played (only that all the games are played simultaneously). >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. > Maybe you should read a description of the swiss system. One can be found at FIDE pages (I do not have the URL handy) , but it is extensive and detailed. >[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? I do not know, but it is an interesting problem. Unfortunately I am too busy to try to give a mathematically justified answer any time soon. > Do we really know without an accurate seeding? > It is posible and valid to play a swiss tournament with a random seeding, given enough rounds. I have played in some tournaments exactly that way, it is used when all or almost all players are unrated (and I prefer a drawing of lots rather than an alphabetical ranking). >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. Here it is the big difference. Producing a reliable rating list and conducting a tournament are very different tasks, for which very different solutions are required. In computer chess, the main difficulty comes from tournaments and matches in which the programs are operated by independent testers (and not by team members), and they should be very careful to play meaningful games. >>>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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