Author: Robert Hyatt
Date: 21:12:43 04/30/04
Go up one level in this thread
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...
>>It is not that hard either. :) I already do it.
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