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

Author: Dan Andersson

Date: 12:55:46 06/05/01

Go up one level in this thread

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

• Re: Beating MTD(n,f) isn't MTD-best a candidate. (NT) Gian-Carlo Pascutto 13:17:36 06/05/01
  • Re: Beating MTD(n,f) isn't MTD-best a candidate. (NT) Dan Andersson 13:21:22 06/05/01

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