Author: Tony Werten
Date: 00:13:07 04/29/04
Hi all, a while ago we had some discussions about diminishing returns in search for chess. My opinion was that you can't prove that with programs searching d vs d+1 ply depth because the advantage of the d+1 program gets smaller. ie at d=1 it has a 100% depth advantage, at d=2 it's 50% etc. Some people claimed that you can't compare it that way because bla bla exponential something bla :) Well, I found an easier way to explain it. A few assumption: The easiest way to win is when you see a trick, your opponent doesn't see. The depth that needs to be searched to see a trick is equally divided. ie there are as many tricks hidden 1 ply away as there are tricks at 2 ply ( it doesn't really matter but it's easier to visualize ) w is player d+1 b is player d d=1: b sees tricks 1 ply away, w sees ply 1 and 2 => w sees 2.0x as many tricks d=2: b:1,2 w:1,2,3 => w: 1.5x d=3: b:1,2,3 w:1,2,3,4 => w: 1.3x ... d=10: b: 1..10 w: 1..11 => w:1.1 x Conclusion: There may or may not be diminishing returns in chess, but d vs d+1 are not going to prove it, because those matches by itself are a clear example of diminishing returns regardless what game is played. disclamer: I know chess isn't only about tricks, but it is an advantage to see more of them then your opponent. Clearly the win percentage is depending on other (random) stuff as well. BUT When you see less more, the advantage becomes less. Tony
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