The solution must be unique.
If more solutions are specified, their content must be homostrategic (must follow the same plan).
(Problem 112) Dmitry Pronkin, First Prize, “Die Schwalbe”, 1985 Shortest Proof Game, 12.5 moves, (2 solutions). 12.5 SPG (15+14), 2 solutions | |
[2k4r/1bpp1ppp/1p1r1p2/2b5/4s3/q7/P1PPPPPP/RSBQKBSR] |
Dmitry Pronkin has created a theme, which is named after him.
Theme Pronkin : A promoted piece goes to the initial square of a similar piece, which is already captured. |
The two solutions of problem-112, which are very interesting and it is good for you to study them on your chessboard, are the following :
1.b4 Sf6 2.Bb2 Se4 3.Bf6 exf6 4.b5 Qe7 5.b6 Qa3
6.bxa7 Bc5 7.axb8=B Ra6 8.Ba7 Rd6 9.Bb6 Kd8 10.Ba5 b6
11.Bc3 Bb7 12.Bb2 Kc8 13.Bc1
1.Sc3 Sf6 2.Sd5 Se4 3.Sf6+ exf6 4.b4 Qe7 5.b5 Qa3
6.b6 Bc5 7.bxa7 b6 8.axb8=S Bb7 9.Sa6 0-0-0 10.Sb4 Rde8
11.Sd5 Re6 12.Sc3 Rd6 13.Sb1
Partial Retrograde Analysis
In some problems the Partial Retrograde Analysis (PRA) is applicable. With this method we cannot specify the complete history of the moves with certainty, but each of the alternative histories demands different solution.
(Problem 113) W. Langstaff, Chess Amateur, 1922 White plays and mates in 2 moves. (Partial retroanalysis). #2 PRA (5+3) | |
[4k2r/8/5B1P/3R1KpP/8/8/8/8] |
The problem-113, by W. Langstaff, published in 1922 on the magazine “Chess Amateur”, is a relatively simple example. The stipulation is “White plays and mates in 2 moves”. It is not possible for the solver to decide which was the last black move, but there are two choices, and the partial retroanalysis gives two different solutions :
(a) Black may have moved their King or their Rook, losing the right for a castling move, so the solution is :
Key : 1.Ke6! ~ 2.Rd8#
(b) Black may have moved the Pawn g7-g5 (not g6-g5, because a Pawn in g6 would give check), so the white has the right to capture it en-passant and the solution is :
Key : 1.hxg6 e.p. [2.Rd8#] 0-0 2.h7#
'These two solutions' or 'not these two solutions'?
(Problem 156) Pal Benko, British Chess Magazine, 1971 Helpmate in 3 moves. (Retroanalysis). h#3 retro (7+8) 2.1.1.1.1.1 | |
[8/8/8/8/s3k3/p3pb1r/p1q1P2P/RS2K1SR] |
This problem is accompanied with a story. I do not know if the story is true.
The composer Benko has created it aiming to fool a top Grandmaster, Bobby Fischer. Fischer, (who has died recently in the age of 64), seeing the white Rooks and the white King in their respective initial positions, gave instantly two solutions, which are incorrect :
1.Bxe2 Sxe2 2.Qc4 Sbc3+ 3.Kd3 0-0-0#
1.Qxe2 Sxe2 2.Bg4 Sbc3+ / Sec3+ / Sg3+ 3.Kf3 0-0#
In helpmates black plays first. But which was the last move of white? Observing the diagram we see that the last move has been made by the white King, thus white has lost the right for castling.
The correct solutions are:
Key : 1.Bxe2! Sxh3 2.Kf3 Rg1 3.Qe4 Rg3#
Key : 1.Qxe2+! Sxe2 2.Kd3 Sxa3 3.Be4 Rd1#
(This post in Greek language).
1 comment:
Prokin's composition is amazing!
About Benko's problem. On his auto-biography (My Life, Games and Compositions) Benko says: "I created this problem with the express purpose of fooling Bobby Fischer, just like Loyd did to Steinitz. Bobby fell right into my trap solutions: 1. Q:e2+ Ne2 2. Bg4 Nc3+ 3. Kf3 0-0# and 1. B:e2 N:e2 2. Qc4 Nbc3+ 3. Kd3 0-0-0#. Both solutions are illegal because White's last move had to be with his King, thus voiding any thought of castling!".
Benko doesn't say anything about how fast Bobby "solved" the problem though.
Cheers,
Roberto Stelling
Post a Comment