Author: martin fierz
Date: 23:57:41 09/25/04
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On September 26, 2004 at 00:39:43, Sergei S. Markoff wrote: >How much root move ordering based on nodes count for each move improves playing >strength? Does anyone have results on it? > >I was in thought about this method of root moves ordering and was disguisted >when discovered that Dr.Hyatt invented it before me :)) wow sergei, you have even more questions than i do :-) i can't answer this question, because ed schröder told me i was doing crappy testing when i tried to measure this. he told me to count the number of nodes needed in test sets to check move ordering. root MO based on node count works about as well as ordering the new best move to the top of the list and shifting the rest to the right for me - node count gives less than 1% more nodes in my test set (significant? probably not). now, i have invented (and patented LOL) a new method for root move ordering. it is a combination of the two above. first, i shift new best moves to the top of the list. then i order the list based on node count but i only swap two moves at i,i+1 if the condition nodecount[i+1] > X*nodecount[i] holds. note with X=1 you have the normal node-based move ordering. i use X=3 (tested 2,3,4). why is this better? previous best moves (which are likely to be good) remain closer to the top of the list, and will only be removed from the top of the list if another move generates a significantly larger subtree => random fluctuations of subtree sizes will not remove the previous best moves from the top of the list. i got a 2% node reduction in my test set with this. this translates to a gain of 1 or 2 elo points!! wow!!!! i need your results to compare :-) as for all your other questions, i wish i was far enough with my program to test all your interesting ideas, but i'm not :-( cheers martin
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