Author: Uri Blass
Date: 19:40:57 11/20/03
Go up one level in this thread
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. 35 is too little. (a-b)/(1200/450) is not what program earns from doubling even if we assume that the speed difference is really 1200/450 and it is more than it. By your logic if the difference was hardware that is twice faster we get (a-b)/2 that is half of the improvement. Uri
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