Author: Robert Hyatt
Date: 12:01:55 09/04/02
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On September 04, 2002 at 14:49:53, Vincent Diepeveen wrote: >On September 04, 2002 at 14:24:47, Robert Hyatt wrote: > >With YOUR method it is very easily possible to get always >> 2.0. > >The easy case is a program that gets a speedup of about 1.9 >and profits just a bit more than 5% from a filled hashtable. > >That's by definition > 2.0 then. No, that is by hand-waving > 2.0. Because the serial search would _also_ profit by the filled hash table and run faster. Try again... > >I need to add that you also have to search in the time of the >opponent. > >Easy scenario. Suppose you take 24 positions. 1 position you >get say 1.9 speedup (cleaned hashtable). The other 23 positions >you mispredict the move. However the move is a transposition to >the search tree the program plays. > >Then the moves get played and you start a new search with that hashtable. > >So you get like > 10 ply out of hashtable directly. So? So does the one-processor test and it all recalibrates nicely... > >That saves n minutes calculation where you calculate n minutes 50% of >your search is for free then. I don't need to mention speedup is >about 2x 1.9 = 3.8 then at 2 processors when compared to someone using >a cleaned hashtable each time getting a 1.9 speedup. Nobody does that. My speedups are calculated consistently, unlike the mish-mash crap you do above. If the one processor test uses exactly the same approach as the 4 processor test, then the four processor test doesn't get any "gain" from hashing that the one processor test doesn't also get... Stop waving your hands. You look absolutely ridiculous. > >The average speedup is then : (23 * 3.8) + 1.9 = 3.72 >this at 2 processors. If I compare the speed of a two processor search to the speed at which a stump-grinder can grind stumps, I get a speedup of > 1,000,000. Not that the comparison makes much sense. But it makes as much sense as your rambling above...
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