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Subject: Nolot #5 Solved yet again

Author: Joshua Lee

Date: 08:54:16 06/11/01


I wanted to try one of the hardest Nolot Positions also LCT  with more memory
now that i have an aditional 256MB on top of the original 256MB for a total of
512MB in which the max i can use is 384 but i don't think this is a good setting
as powers of two are 128, 256 and 512  also i think even powers of two are
better than odd powers of two 2^6 better than 2^7  which would explain why most
people got better results with 64MB for hash tables under fritzmark than 128 but
above 64MB the fritzmark is similar which i do not understand.

My system is an 800Mhz Athlon with 512MB PC-133 and next time i upgrad it will
have to be much faster than dual 1.7Ghz Xeons

position - LCTCMB12

[D] r2qrb1k/1p1b2p1/p2ppn1p/8/3NP3/1BN5/PPP3QP/1K3RR1 w - - 0 1

Analysis by Fritz 6:

1.Rxf6--
  µ  (-1.25)   Depth: 1/3   00:00:00
1.Rxf6-- Qxf6
  -+  (-1.81)   Depth: 1/3   00:00:00
1.Nxe6!
  -+  (-1.78)   Depth: 1/5   00:00:00
1.Qg6!
  ³  (-0.47)   Depth: 1/5   00:00:00
1.Qg6!
  ²  (0.28)   Depth: 1/5   00:00:00
1.Qg6 Rc8
  =  (0.03)   Depth: 2/6   00:00:00
1.Qg6 Rc8 2.h4
  =  (0.03)   Depth: 3/11   00:00:00
1.Qg6 Rc8 2.h4 Qc7
  =  (-0.09)   Depth: 4/15   00:00:00  1kN
1.Nce2!
  =  (-0.06)   Depth: 4/15   00:00:00  4kN
1.Nce2 Qe7 2.Nf4 Qf7 3.Ng6+ Kh7
  =  (0.06)   Depth: 5/17   00:00:00  9kN
1.Nce2 e5 2.Nf5 Bc6 3.Nc3 Qc7
  =  (0.16)   Depth: 6/23   00:00:00  31kN
1.Nce2 e5 2.Nf5 Bxf5 3.Rxf5 Rc8 4.Bf7 Re7
  =  (0.16)   Depth: 7/25   00:00:00  104kN
1.Nce2 e5 2.Nf5 Bxf5 3.Rxf5 Rc8 4.Bd5 Qc7 5.Nc3
  =  (0.25)   Depth: 8/25   00:00:00  230kN
1.Nf3!
  ²  (0.28)   Depth: 8/25   00:00:01  528kN
1.Nf3! Bc6 2.Nh4 Kg8 3.Ng6 Qc7 4.Nxf8 Kxf8
  ²  (0.31)   Depth: 8/26   00:00:01  663kN
1.Nf3 Qa5 2.Nh4 Qg5 3.Qxg5 hxg5 4.Rxg5 Bc6 5.Ng6+ Kh7
  ²  (0.28)   Depth: 9/27   00:00:03  1616kN
1.Nce2!
  ²  (0.31)   Depth: 9/27   00:00:03  1834kN
1.Nce2! e5 2.Nf5 Bxf5 3.Rxf5 Rc8 4.Bf7 Qc7
  ²  (0.38)   Depth: 9/27   00:00:04  2019kN
1.Nce2 e5 2.Nf5 Bxf5 3.Rxf5 Rc8 4.Nc3 Qe7 5.Nd5 Nxd5
  ²  (0.34)   Depth: 10/27   00:00:05  2530kN
1.Nf3!
  ²  (0.38)   Depth: 10/28   00:00:07  3379kN
1.Nf3!
  ²  (0.69)   Depth: 11/31   00:00:23  12152kN
1.Nf3 Ng8 2.Ne2 Bc6 3.Nfd4 Qg5 4.Qxg5 hxg5 5.Nxc6 bxc6
  ±  (0.91)   Depth: 12/34   00:01:05  35740kN
1.Nf3 Qa5 2.Nh4 Qg5 3.Ng6+ Kh7 4.Qf3 Qa5 5.Qg3 Nh5
  ±  (0.91)   Depth: 13/37   00:02:13  75670kN
1.Nf3 Qa5 2.e5 dxe5 3.Nh4 Bc6 4.Qg3 Nh5 5.Ng6+ Kg8
  ±  (1.00)   Depth: 14/37   00:05:18  183551kN
1.Nf3 Bc6 2.Nh4 g5 3.Ng6+ Kh7 4.Nxf8+ Rxf8 5.Bxe6 Qe7
  ±  (1.00)   Depth: 15/38   00:13:26  473329kN
1.Nf3 Bc6 2.Nh4 g5 3.Ng6+ Kh7 4.Nxf8+ Rxf8 5.Bxe6 Qe7
  ±  (1.09)   Depth: 16/39   00:31:07  1108241kN
1.Nf3 Bc6 2.Nh4 g5 3.Ng6+ Kh7 4.Nxf8+ Rxf8 5.Bxe6 Bd7
  ±  (1.00)   Depth: 17/42   00:56:38  2041451kN
1.Nce2!
  ±  (1.03)   Depth: 17/42   01:33:07  3406053kN
1.Nce2! e5 2.Nf5 Bxf5 3.Rxf5 Qe7 4.Nc3 Rac8 5.Rgf1 d5
  ±  (1.16)   Depth: 17/45   01:51:36  4096070kN
1.Nce2 e5 2.Nf5 Bxf5 3.Rxf5 d5 4.Nc3 Qd6 5.Nxd5 Nxd5
  ±  (1.09)   Depth: 18/45   03:08:18  6944239kN
1.Nf3!
  ±  (1.13)   Depth: 18/48   04:22:58  9675215kN
1.Nf3 Bc6 2.Nh4 g5 3.Ng6+ Kh7 4.Nxf8+ Rxf8 5.Bxe6 Bd7
  ±  (1.06)   Depth: 19/47   09:28:31  21078954kN
1.Nce2!
  ±  (1.09)   Depth: 19/47   11:00:50  24651154kN
1.Nce2! g5 2.h4 g4 3.Nf4 Bg7 4.Ng6+ Kg8 5.e5 dxe5
  ±  (1.22)   Depth: 19/47   15:22:25  34640168kN
1.e5!
  ±  (1.25)   Depth: 19/47   18:57:21  42797794kN

(Lee, Pensacola,Fl 11.06.2001)


With less memory it took between 26-28 hours but screwey enough this may be
faster with just 64MB for hash tables don't ask me why.




I think if anyone ever plans to renew the nolot positions they should add the
position from below as no program has solved this yet now would fritz without
30-90 hours i think! it is from a recent post

[D] r1qr2k1/pb2bpp1/1p2pn1p/2p4P/3P1B2/2PB1N2/PP2QPP1/1K1R3R w - - 0 1



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