Author: Robert Pope
Date: 06:44:39 08/31/00
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On August 31, 2000 at 04:27:38, Andreas Stabel wrote: >Here is what I have calculated: >Table from opening position: > | | Unique nodes | Unique nodes II Unique nodes | Factor | >Ply| Total # nodes | ep = pawn two | ep = oppositeII ep = Only if | prev. | > | | | pawn can hit II ep is legal | row | >---+---------------+---------------+--------------II---------------+--------| > 0 | 1 | 1 | 1 II 1 | | > 1 | 21 | 21 | 21 II 21 | 21.00 | > 2 | 421 | 421 | 421 II 421 | 20.05 | > 3 | 9323 | 8023 | 5783 II 5783 | 13.74 | > 4 | 206604 | 109262 | 77796 II 77796 | 13.45 | > 5 | 5072213 | 1351950 | 898812 II 898812 | 11.55 | > 6 | 124132537 | 15334851 | 10281864 II 10281862 | 11.44 | > 7 | 3320034397 | 160373323 | 106193912 II 106193643 | 10.33 | > 8 | 88319013353 | | II | | > 9 | 2527849247520 | | II | | > >I started calculating ply 8 and it ran for 2-3 weeks before I had to stop it. >Sadly this program is not restartable, so all was lost - it needed approx. a >week more to finish. But it seemed that the expected approx. 10^9 unique nodes >on ply 8 was what it was approaching when I terminated it. The problem with >this calculation is that I need a fast machine with at least 11 giga byte >free space to run for nearly a month to finish, so if anybody want to try >tell me ! > >Best regards >Andreas Stabel Thanks! Just to make sure I am reading this correctly: These are total # of nodes traversed, not total # of ending positions? So to get total ply n positions, I would do Nodes(n)-Nodes(n-1). E.g. Positions reachable in two plies = 421-21=400.
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