Author: Dann Corbit
Date: 19:41:09 11/20/03
Go up one level in this thread
On November 20, 2003 at 22:27:14, Dann Corbit wrote: >On November 20, 2003 at 22:21:46, Dann Corbit wrote: > >>On November 20, 2003 at 19:50:12, Uri Blass wrote: >> >>>On November 20, 2003 at 18:57:24, J. C. Boco wrote: >>> >>>>Doubling the processor speed or the time for thinking adds about 80 points to >>>>the SSDF-computer-ratings. But PC's are already running so fast that doubling >>>>their speed might mean searching at 15ply instead of 14ply. >>> >>>How do you get it? >>> >>>Based on the ssdf list even being 3 times faster does not give 80 elo. >>>A1200 is often more than 3 times faster than K6-450 >>> >>>Uri >> >>This query: >>SELECT a.engine, (a.Elo - b.Elo)/(1200.0/450.0) AS EloDiff >>FROM Ssdf AS a, Ssdf AS b >>WHERE a.Engine = b.Engine >> >>And >> a.Hardware like "*1200 MHz*" >> >>And >> b.Hardware like "*450 Mhz*" >> >>Order by (a.Elo - b.Elo)/(1200.0/450.0) Desc >> >>Returns this result set: >> >>engine EloDiff >>Gandalf 4.32h 50.625 >>Crafty 18.12/CB 50.25 >>Gandalf 5.0 42 >>Fritz 7.0 42 >>Deep Fritz 7.0 41.25 >>Rebel Century 4.0 40.125 >>Hiarcs 8.0 36.75 >>Chess Tiger 14.0 CB 32.25 >>Gambit Tiger 2.0 27.375 >>Deep Fritz 25.125 >>Junior 7.0 22.875 >>Chess Tiger 15.0 20.625 >>Shredder 5.32 20.25 >> >>And this query: >> >>SELECT sum((a.Elo-b.Elo)/(1200/450))/count((a.Elo-b.Elo)/(1200/450)) AS >>EloDiffAverage >>FROM Ssdf AS a, Ssdf AS b >>WHERE a.Engine=b.Engine And a.Hardware Like "*1200 MHz*" And b.Hardware Like >>"*450 Mhz*"; >> >>Gives this result: >>34.7307692308 >> >>~35 ELO per doubling seems like what we might expect. Give or take a large >>error bar. > >This query: >SELECT sum((a.Elo-b.Elo)/(450/200))/count((a.Elo-b.Elo)/(450/200)) AS >EloDiffAverage >FROM Ssdf AS a, Ssdf AS b >WHERE a.Engine=b.Engine And a.Hardware Like "*450 MHz*" And b.Hardware Like >"*200 MMX*"; > >Gives this result: >32.0888888889 > >This query: > >SELECT sum((a.Elo-b.Elo)/(200/90))/count((a.Elo-b.Elo)/(200/90)) AS >EloDiffAverage >FROM Ssdf AS a, Ssdf AS b >WHERE a.Engine=b.Engine And a.Hardware Like "*200 MMX*" And b.Hardware Like "*90 >Mhz*"; > >Gives this result: >36.18 > >Looks like it holds pretty steady. Now this query: SELECT sum((a.Elo-b.Elo)/(90/66))/count((a.Elo-b.Elo)/(90/66)) AS EloDiffAverage FROM Ssdf AS a, Ssdf AS b WHERE a.Engine=b.Engine And a.Hardware Like "*90 MHz*" And b.Hardware Like "*66 Mhz*"; Gives this result: 51.3333333333 (perhaps transition from 486 to Pentium was more dramatic)
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