Computer Chess Club Archives


Search

Terms

Messages

Subject: Re: move ordering and node count

Author: Uri Blass

Date: 12:50:41 03/29/04

Go up one level in this thread


On March 29, 2004 at 15:49:40, Uri Blass wrote:

>On March 29, 2004 at 15:43:38, Uri Blass wrote:
>
>>On March 29, 2004 at 15:22:50, martin fierz wrote:
>>
>>>On March 29, 2004 at 14:30:17, Dieter Buerssner 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.
>>>>
>>>>This has not so much to do with your question, but I doubt the last part of your
>>>>sentence. I believe, it will be impossible to become as bad as the minimax tree,
>>>>even when by purporse ordering the moves "perfectly wrong". You will still have
>>>>plenty of situations, where many different moves give a cutoff. In a previous
>>>>experiment, I got 50% beta cutoffs for the first move, when randomizing the move
>>>>order. Note that this is far away, from a minimax tree (where you would need
>>>>100% beta cutoffs in the last tried moves - there are in general many more move,
>>>>you try before).
>>>>
>>>>Regards,
>>>>Dieter
>>>
>>>hi dieter,
>>>
>>>hmm, everybody writes this that making A/B MO as bad as possible you return to
>>>minimax. somewhere below gerd just made the same point as you did here. and if
>>>two experienced programmers like you say so, i am of course afraid to contradict
>>>you :-)
>>>but i have to contradict you all the same. in a perfectly misordered tree you
>>>will *never* fail high. which also means that the case that you and gerd were
>>>thinking of never happens.
>>>
>>>cheers
>>>  martin
>>
>>I think that it is impossible never to fail high.
>>What do you do when all legal moves cause a fail high?
>>
>>Uri
>
>Let take for example the following position
>
>[D]7k/7p/8/8/7q/8/7P/6QK b - - 0 1
>
>When you search Qxh2+ what do you search for white in order not to fail high.
>The answer is that there is no move for white that does not fail high so your
>tree is clearly bigger than the tree when you use minimax.
>
>Uri

I mean clearly smaller.



This page took 0 seconds to execute

Last modified: Thu, 15 Apr 21 08:11:13 -0700

Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.