1 problem(s) found in 1813 milliseconds (displaying 1 problem(s)). [PROBID IN 'P0002321'] [download as LaTeX]

1 - P0002321

Le Sphinx 1866

(16+16)

Welches ist die kürzeste Partie, die mit Patt endet?

**Samuel Loyd**Le Sphinx 1866

(16+16)

Welches ist die kürzeste Partie, die mit Patt endet?

1. e3 a5 2. Dh5 Ta6 3. Dxa5 h5 4. Dxc7 Tah6 5. h4 f6 6. Dxd7+ Kf7 7. Dxb7 Dd3 8. Dxb8 Dh7 9. Dxc8 Kg6 10. De6=

**Keywords:**Construction task, Non-Unique Proof Game

**Genre:**Retro

**FEN:**rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR

**Reprints:**American Chess Journal , p. 18, 15/06/1876

335 Chess Strategy (Loyd) , p. 174, 1878

Ruy Lopez 3 03/1897

61-IV Sam Loyd and his Chess Problems , p. 58, 1913

61/IV Sam Loyd und seine Schachaufgaben 1926

132 Ultimate Themes 1938

6-5 De Waag 03/10/1941

R424 feenschach 21 04/1974

162 Shortest Proof Games 11/1991

Best Problems 45 2008

**Input:**Gerd Wilts, 1995-06-03

**Last update:**A.Buchanan, 2020-05-28 more...

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The problems of this query have been registered by the following contributors:

Gerd Wilts (1)
in 'Ruy Lopez' abgedruckt in der Rubrik "Partidas de fantasia"

AB: Not quite unique alas. 4 Dxc7 & 5 h4 can be transposed. (2001-10-17)Kostas Prentos: Besides the minor transposition, there's another totally different solution:1.c4 d5 2.Db3 Lf5 3.Dxb7 h5 4.Dxa7 Lh7 5.Dxb8 Ta6 6.h4 Th6 7.Dxc7 f6 8.Dxd8+ Kf7 9.Dxd5+ Kg6 10.De6= with many duals, too (2002-07-03)

AB: Yes, I was referring to uniqueness of proof game to reach the diagram position. KP responds to the task which appears here as the stipulation. (2002-07-07)Yoav Ben-Zvi: What is the most simple stipulation that makes the solution unique?My candidate: "Shortest game ending in Stalemate, given that White has not moved a Pawn after his fourth move"

I would add the length of the composer's solution as a courtesy to solvers: "Game in less than 10.0 moves ending in Stalemate, given that White has not moved a Pawn after his fourth move" (2012-08-27)

A.Buchanan: Hi YBZ: the proof game that Kostas gave 5bnr/4p1pb/4Qpkr/7p/7P/2P5/PP1PPPP1/RNB1KBNR has 12 solutions. 1.c3 gives another 12 solutions. And there's 2 solutions for Loyd. Out of these 26 indeed only one has the property that Wh gets all pawn moves out of the way before 4th move. But it's just occurred to me that there's a variant of Loyd which begins 1.c3 and wQ develops via a4 & a5, ending with 5bnr/4p1pq/4Qpkr/7p/7P/2P5/PP1PPPP1/RNB1KBNR this has 15 solutions, and in 6 of these, Wh plays both pawn moves as moves 1 or 2. Similarly if 1.c4. This only differs from Prentos' solution in the occupant of h7. So the suggested stipulation alas wouldn't work. I am pretty sure i have seen a longer version which is sound as a PG. (2020-05-28)A.Buchanan: See P1004131 for a longer sound PG. (2020-05-28)Yoav Ben-Zvi: The cook noted above means that a more elaborate secondary condition is needed. My suggestion:"PG that is shortest to end in Stalemate (less than 10.0 moves) and White plays 2 consecutive captures that are each longer than 2 squares".

An alternative is to accept the transposition:

"PG that is shortest to end in stalemate (less than 10.0 moves) and wPc2 does not move, 2 variants". (2020-06-07)

A.Buchanan: Is it known that there are no h=n.5 (or h=n half-duplex) from the game array for n9? (2022-07-13)A.Buchanan: I read Loyd’s stipulation more as a hard task for composers than a fair ask of solvers. Loyd was a genius so of course he came up with a great answer. Adding a latter-day secondary condition to give an arbitrary uniqueness is no help to solvers, and has an aesthetic price. I imagine that Loyd’s withering reaction - as he generally favoured conceptual elegance over rigorous dual elimination. Non-unique PGs are a bit old-fashioned, but they still have their place and James Malcom proves a master at their efficient construction. I wish that YBZ was still with us to possibly contest my point, RIP. (2022-07-13)more ...

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