Computer Chess Club Archives


Search

Terms

Messages

Subject: Re: move ordering and node count

Author: Dieter Buerssner

Date: 11:17:01 03/29/04

Go up one level in this thread


On March 29, 2004 at 13:55:22, Gerd Isenberg wrote:

>Node = 2 * moves^(depth/2) * (1/MoveOrdering)^(depth/2)
>Help me! What's the minimum of MoveOrdering to get a minimax tree?

Good question. Answer for the easier formula without factor of 2 above:
MoveOrdering=1/moves. Does the minimax tree have a factor of 2, too? If not, and
if I calculated correctly, MO = log_depth(4)*1/moves.

I think, it shows the the formula cannot be generally valid, although it might
give a good estimate for typical cases. It is clear, that the formula can
predict more nodes than the minimax tree for very bad move ordering.

Regards,
Dieter





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.