Author: Gerd Isenberg
Date: 11:58:53 03/29/04
Go up one level in this thread
<snip> >>Sorry Martin for my mathematical ignorance, > >gerd! you and mathematical ignorance?? gerd, the master of the bitwise operator >universe want to claim mathematical ignorance?! > >>but what if you sort white moves >>(even ply) perfect and black moves (odd ply) worst case? >> >>Or all-nodes perfect but cut-nodes worst (there a sill pv-nodes)? > >i never said i knew anything about this, instead i was looking for people who do >know about it to answer my questions! >let me try to answer, but i'm not so sure about this... > >i would think that the second part of your question is easy to answer: there you >get the same worst case as usual. ordering all-nodes is useless (which is why >some engines don't order moves any further after the first few, assuming they >are all-nodes). ordering the cut-nodes worst will obviously give you no cutoffs >either. Inside some cut nodes each move fails high. > and the "still pv-nodes" thing doesn't matter either, because pv-nodes >are also "all-nodes" in the sense that you have to search every move there (i >assume with all-nodes you talked about those which fail low for all moves, else >pv-nodes are simply a part of the all-node-set). No i want to exclude pv-nodes from all-nodes: Each pv's sibling is a cut-node. All-node are childs of cut-nodes. Every child of an all-node is a cut-node. But of course i forgot the smily ;-) > >as for your first question, i think that as you will always be looking at the >best move for white, therefore you will, as white, always be either in a pv >node, or in a cut-node (if black plays bad moves, as he can under your >assumption). you cannot go into a fail-low-node (except if you use aspiration >search and fail low there, but let's forget about that for now...). which means >that black will always be in either pv nodes or fail-low-nodes, and therefore it >doesn't matter - in this case, you have the same as perfect move ordering for >both sides. > >perhaps i get this wrong, you correct me. on second thought, it seems to me that >you want to discredit my formula because that of course is useless for such >cases. of course there i am assuming some sensible move ordering where both >sides are ordering with equal efficiency. > No sorry - i had to study your formula for a while and i got Steve's formula earlier - pattern matching ;-) Cheers, Gerd <snip>
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