Author: Heiner Marxen
Date: 14:05:13 01/02/01
Go up one level in this thread
On January 02, 2001 at 04:45:47, Andreas Stabel wrote: >On December 29, 2000 at 10:26:06, Steven J. Edwards wrote: > >>Briefly, I have calculated the number of distinct positions after N ply for N >>equals zero upto nine. Can someone extend this series, or has anyone already >>done so? >> >>1,20,400,5362,72078,822518,9417683 >> >>See: >> >>http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=057745 >> >>-- Steven >> >>chessnotation@earthlink.net > >I have earlier posted the following table which, unfortunately, does not >match your numbers after ply 3. > >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 | | > > >Best regards >Andreas Stabel Andreas, obviously you are summung up the numbers of Steven, i.e. 21 = 1+20 421 = 1+20+400 5783 = 1+20+400+5362 I did not check further. (column opposite pawn can hit) Heiner
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.