Originally posted by JirakonThe best way to present a modified composition is:
I believe this is from Chess Mysteries of Sherlock Holmes. So, unless your name is Raymond Smullyan, you should credit the author when you post his work (especially in a thread asking for "homemade" problems).
Nope, his puzzle was a bit off. The way it was presented in that book, this puzzle was unsolvable. It had a white pawn in place of the bisho ...[text shortened]... was able to determine that if that pawn were a bishop instead, the problem would be solvable.
"R. Smullyan 1979"
"version by Jirakon"
But even this correction is already known:
http://janko.at/Retros/Misc/SmullyanErrata.htm
Thoughts of a logician, p.103
Originally posted by artplayerThe puzzle's concept is not quite unique. Here is a far older example than Smullyan's:
I see. I appreciate your notification that you weren't plagiarizing, I believe your puzzle's concept is unique, even if the position is derived from another puzzle. If you don't mind me asking, is this a philosophical question or is there is a concrete solution? From what I can tell there is no definitve mating lines with the "special moves" variable, in which case I'm interested in figuring out a good answer to your question.
William A. Langstaff
Chess Amateur 1922
White mates in 2
Either Black has just played ...g7-g5, or he has just moved the King or Rook. If the former, 1.hxg6ep [threat 2.Rd8] 1...0-0 2.h7#, and if the latter, 1.Ke6 and 2.Rd8#.
Originally posted by Jirakon4, 3, 3.
If you want something completely original, I started making these about 8 months ago on a plane trip to Haiti. I've yet to even see this concept, so this is as original as it gets:
[fen]4k3/8/8/8/8/8/PPPPPPPP/RNBQKBNR[/fen]
What's the smallest number of [b]consecutive moves that White needs to checkmate Black? (White can't check Black until the final move) What if the Black king were on a4? What if it were on h4?[/b]
White can castle. This is another relatively easy catch for someone who's read Smullyan's book (which I highly recommend, by the way).
First, consider the black bishop on h3. It must be promoted, since the one on c8 never moved. So it was originally the e- or f- pawn, and it must have promoted on b1, or else it couldn't have gotten back out. That requires six captures: six by the f-pawn, or five by the e-pawn and one to put the f-pawn on e6.
Which pieces were captured? White is missing seven pieces, but one of them, the f1-bishop, never moved. So all the missing white pieces are accounted for by those two black pawns. Now, the bishop on a8 is clearly promoted. The h-pawn couldn't get there, and the f-pawn would have had to make five captures on black squares, but Black is only missing five pieces and the c8-bishop never moved. So the bishop on a8 was the d-pawn originally, and it made three captures. But that only leaves one black piece available for pawn captures, and we need that one to get a pawn on g4. So there was never a white pawn in range of the black pawn on its way to promotion. How was it captured, then? It must have been promoted first -- so it was the f-pawn, and it checked the black king on its way up, forcing it to move. Therefore Black can't castle, and since we're told that *somebody* can, it must be White.
Originally posted by JirakonI agree with CZeke's solution, with one minor caveat. The e7 pawn must have promoted, because only it has a chance to capture White's dark-square Bishop.
Another homemade one:
Black 11
[fen]Bn2k2r/1p1p3p/p1p1p1p1/8/6P1/P6b/1PP1P1P1/R3K3[/fen]
White 9
One side can castle. Which side?