Author: Dann Corbit
Date: 19:44:28 11/20/03
Go up one level in this thread
On November 20, 2003 at 22:41:09, Dann Corbit wrote: >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) How about this one: SELECT sum((a.Elo-b.Elo)/(66/33))/count((a.Elo-b.Elo)/(66/33)) AS EloDiffAverage FROM Ssdf AS a, Ssdf AS b WHERE a.Engine=b.Engine And a.Hardware Like "*66 MHz*" And b.Hardware Like "*33 Mhz*"; Gives this result: 20.5 So it may be that rather than speed dropping off as CPU power exponentially rises, the architecture of the chip changing is even more important than the raw MHz speed. I suspect that a pure MIPS ratio would give a much more steady number. But I have not tried that.
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