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Subject: Re: Beating MTD(n,f) isn't MTD-best a candidate. (NT)

Author: Gian-Carlo Pascutto

Date: 13:17:36 06/05/01

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On June 05, 2001 at 15:55:46, Dan Andersson wrote:

>MTD-best doesn't resolve the value of the best move if it can prove that all
>other moves are worse. It is described in-depth in Aske Plaats:
> Research Re: search & Re-search
> http://www.cs.vu.nl/~aske/Papers/abstr-ks.html

The site is unreachable for me right now.

How can it prove that all other moves are worse? We already know
they are most likely equal.

--
GCP



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