Author: Bruce Moreland
Date: 12:32:55 12/18/02
Go up one level in this thread
On December 18, 2002 at 12:13:56, Sune Fischer wrote: >On December 18, 2002 at 11:38:41, Gian-Carlo Pascutto wrote: > >>On December 18, 2002 at 07:47:15, Sune Fischer wrote: >> >>>On December 17, 2002 at 19:42:10, Bruce Moreland wrote: >>> >>>>We have his new version, and it gets to the same depth more slowly, and finds >>>>more answers, than R=3. This proves nothing. I could make a program where the >>>>eval function incorporates a 2-ply search. It would take longer to search 9 >>>>plies, but it would get a lot more right. This is the same result that Omid >>>>got. Did he just prove that my hypothetical program is better? Of course not. >>>> >>>>If you accept his method as proof, he did prove that VR=3 is better than R=2, I >>>>point out. But he should have tackled R=3, too, if he is going to present that >>>>data. >>> >>>If you want to compare _search_ algorithms, you shouldn't go and change the >>>evaluation or completely redefine the word "node" from one program to the next. >>> >>>The whole assumption here is that they are identical, except for changes in the >>>search parameters. >> >>I don't see how this affects Bruce's point. >> >>His point is that searching slower to a certain depth but getting more >>solutions is no proof that the algorithm is better. >> >>Are you arguing this is wrong? > >Yes I am. >Bruce is saying that a node is a node. >This is not the case if you pile a bunch of evaluation on some nodes and not on >others, obviously those heavy nodes are going to a better job than the >"lightweight" nodes. > >In _this_ case you can't compare nodes. > >A node is meant to be unit of search, change the unit and you change the >everything. I proposed a silly example, and you've gotten stuck on that. I'll propose a more sensible one and see where that gets us. Take a program, call it A. A uses R=2, and finds 50 solutions through ply 9. Take another program, call it B. B does not use null move at all, and it finds 55 solutions through ply 9. Is B better than A? You won't argue that B is better than A, because you'll tell me that of course B takes much longer to produce its results. But what you are telling me really is that you have some notions, and you are willing to let these notions bridge a few gaps: You believe that if they were given equal time, that R=2 would find more than the version with no null move. But my example doesn't say that. For all you know, if you give them equal time, the R=2 version will find 52 solutions. The assumption that it will find (well) in excess of 55 solutions is a leap of faith. This is what is happening in Omid's paper. R=3 finds 65 solutions in one unit of time. R=2 finds 66 solutions in 2.2 units of time. Does this support the conventional logic that R=2 is better than R=3? That would be a very poor leap of faith. VR=3 finds 71 solutions 1.6 units of time. VR=3 is shown to be better than R=2, but is it shown to be better than R=3? This is a better leap of faith, but it is still a leap of faith. bruce
This page took 0.04 seconds to execute
Last modified: Thu, 07 Jul 11 08:48:38 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.