English Chess Forum

This was one of three award winning endgame studies given by CHESS magazine in its May 1963 (no 434) issue.

Don't know how to do diagrams ( ) so here is the position:

WHITE (to play and win) Kc2, Rb8, Bc1, Nd1, Pa2 and g3;

BLACK Ka4, Ba6 and h4, Pb4 c3 and e2.

Could find no (promised) solutions in the subsequent issues, but the other two were solvable by Fritz - here, though it gives nothing better than 1Nb2+ =

Any ideas?

1R6/8/b7/8/kp5b/2p3P1/P1K1p3/2BN4 w - - 0 1

I set the position up in ChessBase, copied the position and pasted it into the tag.<br> <br> I'm also getting a draw after Nb2+ or Nxc3+.

What was the stipulation (i.e. win or draw)? It's usual to have one for studies.

As stated, it is White to play and win. Which begs the question of why the chess programs only show equality.........

An unsound study? Or did Kasparyan see something the computers still can't??

This Kasparyan study, without the pawn on g3, won 1st Prize in New Statesman, 1962-1963.

It is intended to be a draw, and, in 1963, it was. However, advances in endgames theory have shown it to have no solution. The composer's solution is:

1. Nb2+ Ka5 (1... cxb2 2. Bd2 b1=Q+ (2... Bd3+ 3. Kxb2 Bf6+ 4. Kc1 Bc3 5. Re8)
(2... Bf6 3. Rxb4+ Ka3 4. Rb3+ Kxa2 5. Rb6 Bc4 6. Rb4 Ka1 7. Be1 Be5 8. Ra4+
Ba2 9. Rb4 b1=Q+ 10. Rxb1+ Bxb1+ 11. Kd2) 3. Kxb1 e1=Q+ 4. Bxe1 Bxe1 5. Ra8 Kb5
6. a3) 2. Re8 e1=Q (2... cxb2 3. Bxb2 e1=Q 4. Rxe1 Bxe1 5. a3) 3. Rxe1 Bxe1 4.
a3 (4. Nd3 Bxd3+ 5. Kxd3 Ka4 6. Bg5 (6. Kc2 Bd2) 6... Ka3 7. Be7 Kb2 $1
8. Bxb4 c2 9. Ba3+ Kb1 10. Kc4 Kxa2) (4. Nd1 $2 Bc4 5. Kb1 Bd3+ 6. Ka1 Bg3 7.
Ne3 Be5) 4... bxa3 5. Nd1 a2 (5... Bd3+ 6. Kb3) 6. Nxc3 a1=Q (6... Bxc3 7.
Bb2 Bb4 8. Bh8 Bc4 (8... Ba3 9. Kb3) 9. Kb2) 7. Bb2 Bd3+ (7... Bxc3 8. Bxc3+) 8. Kb3 Bc4+ 9. Kc2 Bd3+ 10. Kb3 1/2-1/2

The issue is that 7...Qxb2+ leads to N v BB, which is now known to be a win for the bishops.

John Roycroft, in his collection of studies by Kasparyan, writes of this:

The study can still be considered 'correct' if one accepts the reasonable proposition that a composer cannot be expected to know better than the endgame theory current at the time of first publication. In 1963 no one would have contested a composer's reliance on this ending being a general draw.

Where did the pawn at g3 come from? Perhaps someone's attempt at a correction?

Thanks for that - it is quite clearly (and obviously erroneously) stated as a "White to win" study in CHESS. No wonder they never published the solutions

Brian Stephenson wrote: The issue is that 7...Qxb2+ leads to N v BB, which is now known to be a win for the bishops.

58 moves according to the Shredder site though, so a draw by the 50 move rule over the board with correct play.

Roger de Coverly wrote:

Brian Stephenson wrote: The issue is that 7...Qxb2+ leads to N v BB, which is now known to be a win for the bishops.

58 moves according to the Shredder site though, so a draw by the 50 move rule over the board with correct play.

I think you need to tell it that it's Black to move after 8.Kxb2 - then you'll get mate in 35. With White to move it's 58 moves to win. The tablebases on my computer agree with the Shredder website in both cases.

Ian Kingston wrote:

Roger de Coverly wrote:

Brian Stephenson wrote: The issue is that 7...Qxb2+ leads to N v BB, which is now known to be a win for the bishops.

58 moves according to the Shredder site though, so a draw by the 50 move rule over the board with correct play.

I think you need to tell it that it's Black to move after 8.Kxb2 - then you'll get mate in 35. With White to move it's 58 moves to win. The tablebases on my computer agree with the Shredder website in both cases.

I took a stroll with Houdini in this, and it found 7...Bd3+! 8.Kb3 Qxb2+ 9.Kxb2 when it is only 31 moves to mate, and against tablebase defence it still won the night at move 29 - which is really what matters, because then there are another 50 moves to mate, and the win is trivial.

Brian Stephenson wrote:This Kasparyan study, without the pawn on g3, won 1st Prize in New Statesman, 1962-1963.

It is intended to be a draw, and, in 1963, it was. However, advances in endgames theory have shown it to have no solution. The composer's solution is:

1. Nb2+ Ka5 (1... cxb2 2. Bd2 b1=Q+ (2... Bd3+ 3. Kxb2 Bf6+ 4. Kc1 Bc3 5. Re8)
(2... Bf6 3. Rxb4+ Ka3 4. Rb3+ Kxa2 5. Rb6 Bc4 6. Rb4 Ka1 7. Be1 Be5 8. Ra4+
Ba2 9. Rb4 b1=Q+ 10. Rxb1+ Bxb1+ 11. Kd2) 3. Kxb1 e1=Q+ 4. Bxe1 Bxe1 5. Ra8 Kb5
6. a3) 2. Re8 e1=Q (2... cxb2 3. Bxb2 e1=Q 4. Rxe1 Bxe1 5. a3) 3. Rxe1 Bxe1 4.
a3 (4. Nd3 Bxd3+ 5. Kxd3 Ka4 6. Bg5 (6. Kc2 Bd2) 6... Ka3 7. Be7 Kb2 $1
8. Bxb4 c2 9. Ba3+ Kb1 10. Kc4 Kxa2) (4. Nd1 $2 Bc4 5. Kb1 Bd3+ 6. Ka1 Bg3 7.
Ne3 Be5) 4... bxa3 5. Nd1 a2 (5... Bd3+ 6. Kb3) 6. Nxc3 a1=Q (6... Bxc3 7.
Bb2 Bb4 8. Bh8 Bc4 (8... Ba3 9. Kb3) 9. Kb2) 7. Bb2 Bd3+ (7... Bxc3 8. Bxc3+) 8. Kb3 Bc4+ 9. Kc2 Bd3+ 10. Kb3 1/2-1/2

The issue is that 7...Qxb2+ leads to N v BB, which is now known to be a win for the bishops.

John Roycroft, in his collection of studies by Kasparyan, writes of this:

The study can still be considered 'correct' if one accepts the reasonable proposition that a composer cannot be expected to know better than the endgame theory current at the time of first publication. In 1963 no one would have contested a composer's reliance on this ending being a general draw.

Where did the pawn at g3 come from? Perhaps someone's attempt at a correction?

[FEN "1R6/8/b7/8/kp5b/2p5/P1K1p3/2BN4 w - - 0 1"] [SetUp "1"] 1. Nb2+ Ka5 (1... cxb2 2. Bd2 b1=Q+ (2... Bd3+ 3. Kxb2 Bf6+ 4. Kc1 Bc3 5. Re8) (2... Bf6 3. Rxb4+ Ka3 4. Rb3+ Kxa2 5. Rb6 Bc4 6. Rb4 Ka1 7. Be1 Be5 8. Ra4+ Ba2 9. Rb4 b1=Q+ 10. Rxb1+ Bxb1+ 11. Kd2) 3. Kxb1 e1=Q+ 4. Bxe1 Bxe1 5. Ra8 Kb5 6. a3) 2. Re8 e1=Q (2... cxb2 3. Bxb2 e1=Q 4. Rxe1 Bxe1 5. a3) 3. Rxe1 Bxe1 4. a3 (4. Nd3 Bxd3+ 5. Kxd3 Ka4 6. Bg5 (6. Kc2 Bd2) 6... Ka3 7. Be7 Kb2 $1 8. Bxb4 c2 9. Ba3+ Kb1 10. Kc4 Kxa2) (4. Nd1 $2 Bc4 5. Kb1 Bd3+ 6. Ka1 Bg3 7. Ne3 Be5) 4... bxa3 5. Nd1 a2 (5... Bd3+ 6. Kb3) 6. Nxc3 a1=Q (6... Bxc3 7. Bb2 Bb4 8. Bh8 Bc4 (8... Ba3 9. Kb3) 9. Kb2) 7. Bb2 Bd3+ (7... Bxc3 8. Bxc3+) 8. Kb3 Bc4+ 9. Kc2 Bd3+ 10. Kb3 { 1/2-1/2 }