Author: Dann Corbit
Date: 13:35:29 03/29/04
Go up one level in this thread
On March 29, 2004 at 16:22:21, martin fierz wrote: >On March 29, 2004 at 15:50:10, Dann Corbit wrote: > >>On March 29, 2004 at 10:17:18, martin fierz wrote: >> >>>aloha! >>> >>>i was discussing this somewhere in a thread, but thought i'd like to make this >>>question more visible in the hope of getting a good answer: >>> >>>everybody knows that with plain alpha-beta, a fixed number of moves N per node, >>>and perfect move ordering a search to depth D needs >>> >>>nodes(depth) = sqrt(N)^(D/2) nodes. >>> >>>with absolutely imperfect move ordering it needs >>> >>>nodes(depth) = N^(D) nodes. >>> >>>a typical chess program gets something like 90% move ordering in the sense that >>>if a cutoff move exists, it will search it as first move in 90% of all cases. >>>here's my question: >>> >>>can anybody give an estimate for nodes(depth) as function of this move ordering >>>parameter? obviously, this would also depend on when you find the best move in >>>those cases where you don't find it first. any kind of model is acceptable, e.g. >>>you always find it on 2nd, always on sqrt(N)th, always last, at a random number, >>>whatever. i'm just interested in the general behavior of nodes(depth) as a >>>function of the cutoff-%age. >>> >>>i'd be extremely surprised if nobody ever estimated this, so: has any of you >>>ever seen or calculated such numbers, and if yes, what do they look like? >>> >>>and is there any theory how this would apply to a modern chess program with >>>nullmove and extensions instead of the plain A/B framework above? >>> >>>basically this question of course means: do you really gain anything tangible >>>when improving your MO from say 90% to 92%? >> >>I have not done the math, but I am guessing no matter what king of move ordering >>you have (purely randome or the pv move every time) you will get something like >>this: >> >>nodes = some_constant * sqrt(mini_max_nodes) >> >>If you have random move ordering, then the constant will be very large. >>If you have perfect move ordering, then the constant will be very small. >> >>You will never get worst case unless you try very hard to achieve it. >>It might be possible to degenerate to mini-max (or very close to it) but you >>will have to choose the worst possible move at every single turn except the >>leaves. I doubt if anyone can do it. >>;-) > >i disagree with your formula. it is definitely not some_constant. it is a >constant between 1...sqrt(N) taken to the power of D/2. else there would be no >point in improving move ordering, or at least, not as much as there is :-) What if the constant is one trillion? If you choose the node at random, you will still find the right one by random chance on the first try 1/n times (where n is the number of moves) and on the second try 1/(n-1) times, etc.. On average, we won't have to try more than half of the nodes to find it (the best one). This will cause a huge reduction in the number of nodes. As you can see, you would have to put forth a stupendous effort to cause minimax behavior.
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