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A very cute puzzle, though not all that difficult.
It is the stalemate that is the biggest hindrance to white. If white plays Re3, then black plays e1Q, and white cannot take at e1 without stalemating black. So, how does white untangle this knot? By threatening mate:
1. Bd7! e1Q (Kd7 2.Re3+-)
2. Bb5!
And black will have to give up the queen at a5 just to prevent Re8 being mate on the next move.
Bf5-d7-b5 seems to set up a mating net, the new queen could only check on a5 and not guard e8
Bd7 seals the deal
Bd7. If Kxd7 Rxe3 and it is easy mate after capturing another pawn. But if e1 and queening. Bb5 and Queen cannot give any sort of threat to the king…Re8# – MAHENDRAKUMAR (Hyderabad, India)
Bd7. If Kd7 then Rxe3 and the rest is easy. But if e1 and Queening, the Bb5. Queen cannot give any further threats to the King and where ever he keeps his Queen it is Re8# – MAHENDRAKUMAR (Hyderabad, India)
Bd7 ensures that the King moves away from stalemate position or face mate by Re8. Once the King takes the Bishop then the rook can have the pawns
Initial FEN, full references and solution:
Fen:8/3k4/1K6/8/4BR2/4p3/4p3/8 w – – 0 1
Reti, Kölnische Volkszeitung, 1928
correction by Rinck, Bohemia, 1935
win
The first hard choice is right here: 1.Bc6+ or 1.Bf5+?
1.Bc6+ is a try but not the solution: 1…Kd6! 2.Rd4+ Ke5 3.Re4+ Kd6! and white cannot take the pawn e3: 4.Rxe3 e1Q 5.Rxe1 stalemate.
1…Kd8
2.Rd4+ Ke7
3.Re4+ Kd8
Here we come to the posted puzzle:
Black’s defense is based on a similar stalemate trick that may occur earlier:
4.Rxe3? e1Q
5.Rxe1 stalemate
Instead, 4.Bd7! wins by letting Black promote his pawn!
4…e1Q
5.Bb5! and the threat Re8# costs Black his new-born queen.
Themes: echo-manoeuver between try and solution, stalemate avoidance.
Note: The corrector of this study, Henri Rinck is himself a great study composer. His last book is titled “1414 fins de partie”, 1414 own endgames in one book!!!
Cortex,
It is a pity we did not get the full puzzle.
The posted problem seems to be a subset of original problem as told to us by Cortex.it hardly is a correction.
1Bf5+ Kd6
2Rd4+ Ke7
3Re4+ Kd8 and we get the posted position.
An interesting study, and an interesting history!
Disclaimer: I have no merit to mention the following, all the researches were done by John Beasley and Timothy Whitworth, work I summarize clumsily here!
First, the study was thus:
8/3k4/1K6/4R3/4B3/4p3/4p3/8 w – – 0 1
Intended solution:
1.Bf5+ Kd8
2.Bd3 e1Q
3.Bb5
but 2.Bd7! is a beautiful cook.
Who was the first to discover that cook? We don’t know, but later, another study from Prokop published a mirror position after the introduction play
Prokop, Shakmatny Listok, 10.7.1929, #377, 6th place
4b3/7K/2kPR3/8/4n3/B1pp4/3P4/8 w – – 0 1
win
1.d7+ Kxd7
2.Rxe4 Bg6+ (getting rid of his extra piece to close a stalemate net)
3.Kxg6 cxd2
4.Rd4+ Ke8!
and now, in this somewhat mirrored position, only the “dual”
5.Be7 wins.
Then the definitive and elegant correction of Rinck, changing slightly the Rook position in Reti’s original setup removed the main line and promoted the cook as the only solution
8/3k4/1K6/8/4BR2/4p3/4p3/8 w – – 0 1
Taken and summarized from:
John Beasley, “The Chess Endgame Studies of Richard Réti”, 2012 on his website here
The source of John Beasley is some Shakmatny Listok 1929 excerpts related to this precise study sent to him by Timothy Whitworth
Dear Cortex,
My comments were based on misunderstanding.I thought the Fen given by you in 1st post as original Reti and one given by Susan as correction by Rinck.From your 2nd post I realised that problem by Reti was indeed different and had a cook. So you are right.I also appreciate your deep study of studies.