Author: John Merlino
Date: 12:05:10 05/22/03
Go up one level in this thread
On May 22, 2003 at 00:53:37, Stephen A. Boak wrote: >On May 21, 2003 at 08:50:15, Uri Blass wrote: >> >>That tells us that testers were simply lazy at that time and did not try all >>combination of changing one parameter. >> >>Ed found that reducing the chess knowledge of Rebel from 100 to 25 helps but no >>tester was able to find it. >> >>Uri > >Let's think. > >How many combinations of single parameters & values are there? >How many single parameters can be adjusted on CM9000? >How many possible numerical settings on each parameter can be used? >Multiply the figures. Call this result 'A'. I can answer that! I'll even edit the ranges of ones that people would reasonably stay within: Style values Attacker/Defender = 200 values Strength of Play = 1 value (nobody would change this below the maximum) Material/Positional = 200 values Randomness = 1 value (nobody would normally change this below the value of 0) Selective Search = 16 values (normally, about half of these are used by testers) Contempt for Draw = 21 values (I'll assume that testers would not use values outside the range of -1.0 to 1.0) Transposition Table = Using the CM GUI, depending on how much RAM the computer has, a typical tester will have about 6 values to choose from) Positional values King Safety, Mobility, Pawn Weakness, Control of Center, Passed Pawns = 200 values for BOTH self and opponent, meaning there are 10 settings with 200 values each in this section. And I have known users to test both the minimum and maximum values, so the entire range of all of these is "valid". Material values All pieces, both self and opponent, literally have 150 values, although I don't think any user would deviate more than 20% or so from the default. So, let's approximately say: Pawn = 5 values each Knight = 12 values each Bishop = 12 values each Rook = 20 values each Queen = 40 values each So, 12 settings with 200 values each = 200^12 = ((2 x 10^2) ^ 12) = (2^12 x 10^24) = appr. (4 x 10^27) .... uh, do I need to go on? Ok, fine. The remainder of the values are (16 x 21 x 6 x 5 x 5 x 12 x 12 x 12 x 12 x 20 x 20 x 40 x 40) = (6 x 10^14) so (4 x 10^27) x (6 x 10^14) is a really frickin' big number.... jm
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.