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Subject: Re: Endgame position

Author: Bruce Moreland

Date: 09:52:42 05/12/98

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On May 12, 1998 at 12:23:37, Kai Skibbe wrote:

>On May 12, 1998 at 07:05:53, Howard Exner wrote:
>
>>On May 12, 1998 at 01:17:08, Kai Skibbe wrote:
>>
>>>The following position is from a game between
>>>Rebel 9.0 28 MB HT and Fritz 5.03 64MB HT on
>>>a PC with a AMD K6-200.
>>>
>>>WKh3, Rh6, Pg3,g2
>>>BKc5, Rb6, Ph7
>>>White to move
>>>
>>>Rebel 9.0 as white was unable to find the winning
>>>exchange Rxb6 in 5 minutes. He played Rxh7 and after
>>>some moves the game became a draw.
>>>I have tested the same position with Shredder 2.0 64MB
>>>and Genius 5.0 32 MB. Both programs didn´t see the winning
>>>exchange in 10 minutes and played the same move as
>>>Rebel (Rxh7).
>>>I´m not sure if the game could be a win for white after
>>>Rxh7, but after Rxb6 the win should be clear.
>>>
>>>Does any chessprogram play the move Rxb6 and see the win ?
>>>The evaluation should be at least 4 pawn, if the program
>>>has understood the position.
>>>
>>>Why is this position so difficult for programs ?
>>
>>Both moves, Rxb6 and Rxh7 will win. In the Rxh7 line the
>>black king is too far away and will be cut off by white's
>>Rook (Rd7).
>>
>>Could you post the remainder of the game ? I'd like to see
>>how Fritz managed to draw this. I've seen Rebel routinely
>>win these kinds of positions when the king is cut off.
>
>Here is the rest of the game after the position
>Rebel 9.0 28MB - Fritz 5.03 64MB
>59. Rxh7	Kd5
>60. g4		Ke5
>61. g5		Kf5
>62. Kh4		Rb1
>63. Rf7+	Kg6
>64. Rf6+	Kg7
>65. Rd6		Rh1+
>66. Kg3		Rg1
>67. Rb6		Kf7
>68. Kf3		Rc1
>69. Kg4		Rd1
>70. Rc6		Kg7
>71. Ra6		Rc1
>72. g3		Rf1
>73. Ra7+	Kg6
>74. Ra4		Rd1
>75. Ra6+	Rg7
>
>At this point I stopped the game. IMO there is no win for white.
>
>I have tested the position after 59. Rxb6  Kxb6. From this position
>all programs play the correct moves ( 60. Kg4 or g4 ).
>
>BTW CM 5500 and CM5555 find the move Rxb6 instantly, but the evaluation
>is only
>about +1.8 pawns. So they both have not seen a promoted pawn.

If you remove white's g2 pawn, and force the move 1. Rxh7, the position
is still won.  Since the g2 pawn is probably not going to hurt white, I
conclude that the position with two pawns up is also won.  There is a
chance that I'm wrong, but probably not.

The rest of this post assumes that the white g2 pawn has been
disappeared.

If I force 1. Rxh7 Kd5 2. g4 Ke5 3. g5, 3. ... Kf5 draws.

Here is the win after 1. Rxh7 Kd5 2. g4 Ke5.

    a   b   c   d   e   f   g   h
  +---+---+---+---+---+---+---+---+
8 |   |///|   |///|   |///|   |///| 8
  +---+---+---+---+---+---+---+---+
7 |///|   |///|   |///|   |///| R | 7
  +---+---+---+---+---+---+---+---+
6 |   |/r/|   |///|   |///|   |///| 6
  +---+---+---+---+---+---+---+---+
5 |///|   |///|   |/k/|   |///|   | 5
  +---+---+---+---+---+---+---+---+
4 |   |///|   |///|   |///| P |///| 4
  +---+---+---+---+---+---+---+---+
3 |///|   |///|   |///|   |///| K | 3
  +---+---+---+---+---+---+---+---+
2 |   |///|   |///|   |///|   |///| 2
  +---+---+---+---+---+---+---+---+
1 |///|   |///|   |///|   |///|   | 1
  +---+---+---+---+---+---+---+---+
    a   b   c   d   e   f   g   h

8/7R/1r/4k/6P/7K w

1. Kh4 Rb3 (Rb2, Rb1) 2. g5 (Kg5) Kf5 3. Rh6! Rb2 (Rb1, Rb8, Ke5) 4.
Rf6+! Ke5
5. Kh5 Rh2+ 6. Kg6 Rg2 (Rd2, Rc2, Rb2, Ra2) 7. Rf1 (Rb6, Ra6, Kh6) Rg3
(Rg4,
Rd2, Rc2, Rb2, Ra2, Ke6) 8. Re1+ (Kh6) Kd4 (Kd6, Kd5) 9. Kf6 (Kh6) Rf3+
10. Kg7
Rg3 (Kd5) 11. g6 Kd5 12. Kf7 (Kh7) Rf3+ 13. Kg8 Rg3 (Rh3, Kd6) 14. g7
Kd6 15.
Rh1 Ke7 (Ke6) 16. Rh7 (Kh7, Kh8) Rg2 (Rg1, Rg4, Rg5, Rg6, Kd6, Kf6, Ke6)
17.
Kh8 Ke6 18. g8=Q+ Rxg8+ 19. Kxg8! Kd5 20. Rh5+ (Rh4, Rh1, Rh8, Kf7, Kg7,
Kf8)
Kc4 (Ke4, Kd4) 21. Rh4+ (Kf7) Kb3 (Kd3, Kd5, Kc3) 22. Kf7 (Kg7) Kc2
(Kb2, Kc3)
23. Rh3 (Ke6) Kb2 (Kd2) 24. Ke6 (Kg6, Kf6) Kc2 25. Kd5 (Kf5, Ke5) Kd2
26. Ke4
Kc2 27. Rd3 (Kd4) Kc1 (Kb2) 28. Ke3 Kc2 29. Ke2 Kb1 (Kc1) 30. Kd1 (Kd2)
Kb2 31.
Kd2 Ka1 (Kb1) 32. Rb3 (Kc1, Kc2) Ka2 33. Kc2 Ka1 34. Ra3#

It is interesting to note that after 1. Rxh7 Kd5, the attempt to cut off
black's king with 2. Re7 fails to 2. ... Rb8! =

bruce




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