Author: Robert Hyatt
Date: 08:15:22 05/01/04
Go up one level in this thread
On May 01, 2004 at 06:00:12, Uri Blass wrote: >On May 01, 2004 at 00:12:43, Robert Hyatt wrote: > >>On May 01, 2004 at 00:03:41, Andrew Wagner wrote: >> >>>On April 30, 2004 at 23:47:06, Robert Hyatt wrote: >>> >>>>On April 30, 2004 at 23:26:55, Andrew Wagner wrote: >>>> >>>>>[snip] >>>>>> >>>>>>Black is already winning in position 2: >>>>>> >>>>> >>>>>Right, I just meant the sac was losing, not the position. In fact, the first >>>>>position is also winning without the sac, quite easily. My point with the second >>>>>position was that where in the first position the two connected passed pawns >>>>>made up for the difference of the rook, in the second position they weren't. I'm >>>>>thinking it would be tricky to create a non-expensive eval term for such a >>>>>complicated static evaluation. I mean, a human can look at it and quickly see >>>>>"the king's not close enough", but that's not so easy for a computer to see >>>>>without calculating. >>>> >>> >>>Hmm? You can't be doing too much with it if you're not solving this until 13 >>>ply. >> >> >>Look at the position _carefully_. This is a _one_ tempo position. One white >>pawn prevents the white king from being able to stop the two connected black >>pawns. I am conservative there. Move the king one file toward the h file and >>you get this: >> 10 0.14 -2.54 1. ... c3 2. bxc3 Rxc3 3. Ra2 Rc2 4. >> Ra1 Re2 5. Rd1 Rxe3+ 6. Kf2 Rh3 7. >> Ke1 >> 10-> 0.20 -2.54 1. ... c3 2. bxc3 Rxc3 3. Ra2 Rc2 4. >> Ra1 Re2 5. Rd1 Rxe3+ 6. Kf2 Rh3 7. >> Ke1 (s=2) >> 11 0.26 -2.58 1. ... c3 2. bxc3 Rxc3 3. Ra2 Rc2 4. >> Ra1 Re2 5. Kf3 Rxh2 6. Rd1 Rh3+ 7. >> Kf2 Rh2+ 8. Kf3 >> 11-> 0.51 -2.58 1. ... c3 2. bxc3 Rxc3 3. Ra2 Rc2 4. >> Ra1 Re2 5. Kf3 Rxh2 6. Rd1 Rh3+ 7. >> Kf2 Rh2+ 8. Kf3 (s=3) >> 12 0.68 -2.37 1. ... c3 2. bxc3 Rxc3 3. Kh4 Rxa3 >> 4. Kh5 Ke6 5. Kg5 Rc3 6. Kh6 Kd5 7. >> Kxh7 a3 (s=2) >> 12 0.91 +1 1. ... Rxb2!! >> 12 1.19 -3.61 1. ... Rxb2 2. Rxb2 c3 3. Rb6+ Ke7 >> 4. Rb7+ Kd6 5. Rb6+ Kc5 6. Rb8 d2 7. >> >> >>Finds it one ply quicker. But the problem here is the threat of repetition by >>white. Takes a bit of a search to see that the black king finally escapes all >>the rook checks. >> >>Move the white king to the h-file and you get this: >> >> 11 0.30 +1 1. ... Rxb2!! >> 11 0.39 -3.52 1. ... Rxb2 2. Rxb2 c3 3. Rb6+ Ke7 >> 4. Rb7+ Kd6 5. Rb6+ Kc5 6. Rb7 d2 7. >> Rxh7 Kd6 8. Rf7 d1=Q 9. Rxf5 (s=12) >> 11-> 0.47 -3.52 1. ... Rxb2 2. Rxb2 c3 3. Rb6+ Ke7 >> 4. Rb7+ Kd6 5. Rb6+ Kc5 6. Rb7 d2 7. >> Rxh7 Kd6 8. Rf7 d1=Q 9. Rxf5 (s=2) >> 12 0.66 -3.79 1. ... Rxb2 2. Rxb2 c3 3. Rb6+ Ke7 >> 4. Rb7+ Kd6 5. Rb6+ Kc5 6. Rb8 d2 7. >> Rc8+ Kb6 8. Rd8 c2 9. Rxd2 c1=Q 10. >> Rd3 >> 12-> 0.91 -3.79 1. ... Rxb2 2. Rxb2 c3 3. Rb6+ Ke7 >> 4. Rb7+ Kd6 5. Rb6+ Kc5 6. Rb8 d2 7. >> Rc8+ Kb6 8. Rd8 c2 9. Rxd2 c1=Q 10. >> Rd3 >> >>So you can get the idea. The white pawn at e3 is the key here. Remove it and >>move it to g3: >> >> 12-> 1.95 -1.77 1. ... Ke6 2. Ke3 Kd5 3. h3 h5 4. Rf2 >> d2+ 5. Kxd2 Rxb2+ 6. Ke1 c3 7. Re2 >> c2 8. Re5+ Kd4 9. Kd2 (s=2) >> 13 2.54 -1.83 1. ... Ke6 2. Ke3 Kd5 3. h3 h5 4. h4 >> Rb6 5. Kf2 Re6 6. Kg2 Ke4 7. Kf2 Re7 >> 13-> 3.27 -1.83 1. ... Ke6 2. Ke3 Kd5 3. h3 h5 4. h4 >> Rb6 5. Kf2 Re6 6. Kg2 Ke4 7. Kf2 Re7 >> (s=2) >> 14 7.31 -1 1. ... Ke6 >> 14 11.82 -1.42 1. ... Ke6 2. h3 c3 3. bxc3 Rxc3 4. >> Ke3 Rxa3 5. Rxd3 Rb3 6. Kd4 h5 7. Rd2 >> a3 8. Re2+ Kd6 9. Rg2 >> 14-> 14.96 -1.42 1. ... Ke6 2. h3 c3 3. bxc3 Rxc3 4. >> Ke3 Rxa3 5. Rxd3 Rb3 6. Kd4 h5 7. Rd2 >> a3 8. Re2+ Kd6 9. Rg2 >> >> >>The sac doesn't work. That is why this is _really_ a tactical problem, not an >>evaluation problem... > >I do not see it as a proof that it is a tactical problem because evaluation >function can consider also the pawn at e3 > >Uri The entire game can be solved by positional evaluation alone. But it is not easy.
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