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Subject: Re: move ordering and node count

Author: Dann Corbit

Date: 13:58:58 03/29/04

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On March 29, 2004 at 16:35:29, Dann Corbit wrote:

>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.

Consider also that the 2nd best nodes will prevent lots of searches (all except
the best node), as well as the 3rd best (all except the top 2 nodes), etc.  So
alpha-beta even with very bad move ordering will still cause a huge number of
cutoffs.



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