Author: Robert Hyatt
Date: 12:35:47 09/07/04
Go up one level in this thread
On September 07, 2004 at 14:38:10, Michael Henderson wrote: >On September 07, 2004 at 10:17:08, Robert Hyatt wrote: > >>On September 05, 2004 at 23:16:30, Michael Henderson wrote: >> >>>On September 05, 2004 at 17:32:57, Alessandro Scotti wrote: >>> >>>>On September 05, 2004 at 03:24:56, Michael Henderson wrote: >>>> >>>>>How was 6...Bxg1 possibly stored as the best move at that node? Once you search >>>>>all the moves, to whatever depth, the best move Qh4# is stored along with score >>>>>returned. Is this something characteristic of MTD(f)? >>>> >>>>It was stored there because it was explored deeper. The search returned Qh4# as >>>>the best move, but then it wasn't stored in the hash table because the >>>>replacement policy (deeper is better) preferred the entry that was already >>>>there. >>> >>>In MTD(f) all searches use null window? If that is true how is it possible to >>>get an exact score/move stored in hash table? >> >> >>On the last "2 cycles" the same PV will fail low and high. Since you need to >>store both bounds, you can always keep a "best move" even when you fail low (the >>best move comes from the next-to last or last search that failed high here...) >> >>There are other ways to do the same thing, all taking advantage of the same >>basic property of the mtd(f) search... > >Does this mean that the temporary PV's you extract are actually constructed from >beta moves? That seems to work out logically... That's all there are. But if you store both bounds in one entry, when you store an alpha bound you can still keep the beta bound's best move, so that you can get most of the PV after the search ends, except when a key entry gets overwritten of course...
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