In his article “The Horse Concoction", Tim Krabbe mentions retrograde analysis problem that involve placing 10 rooks, knights, and bishops to the board for a mate in 1. The 10 rooks problem shown (it also also on site). However, the other two by Henrik Juel and Guus Rol, are only mentioned. Who made which is unclear, although suspect that it goes respectively.

Here is the relevant portion of the article.

"In the last few years this old idea, where all the pawns of one or even both sides promote to one kind of piece, had been taken up again by the Dane Henrik Juel and the Dutchman Guus Rol - first with Bishops, then with Knights. And when Goldsteen saw Rol's latest version of an earlier composition by Juel, something unusual happened to him: he was jealous. He was just pissed off that he hadn't thought of Juel's and Rol's ten-knight problem (add ten black Knights to a given position so that White can mate in 1)."

Can these problems be found online? I would love to see them out of curiosity and to learn more about retrograde analysis.


Thank to Bojan Bascis, who responded on Matplus.net, I now know what the knight problem, but the bishop problem remains unseen, however.

This is Henrik's Juel predecessor problem.

[Title "Henrik Juel, Probleemblad 1/1999, Add a king and 9 Black knights for a mate in one"]
[FEN "n7/7P/P7/8/8/1PKPP1P1/2P2P2/1N1N4 w - - 0 1"]

And here is the improved problem that Guus Rol created.

[Title "Henrik Juel & Guus Rol, Probleemblad 05/1999, 4th Honorable Mention, Add a king and 10 Black knights for a mate in one"]
[FEN "8/7P/8/P7/6K1/1P1PP1P1/2P2P2/2NN4 w - - 0 1"]

Lastly, although it is not the 10 bishops problem, this problem is a nice one that captures the essence of it wherever it may be.

[Title "Henrik Juel. Probleemblad, 5/1997, Add a mating bishop"]
[FEN "b3B3/1k1B3B/7b/1B3b1B/B6b/3b1K1b/7B/1BbBbbB1 w - - 0 1"]
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  • 5
    No, the answer is a black bishop on d5. You should use logical reasoning to work out the answer, not gut feel based on the "difficulty" of a problem. As you have worked out, the answer should be a black light bishop. So we have 8 light + 2 dark white bishops, and 6 light + 4 dark black bishops. White therefore promoted 7 light and 1 dark bishop; this needs 5 captures (e.g. bxa, dxc, fxe, and a cross-capture gxh+hxg.) Black promoted 5 light and 3 dark bishops; together with white's captures, this needs 5 more captures (e.g. axb, cxd, exf, dxc and fxe; g- and h-pawns went straight.) (continued) – Remellion Jun 12 '19 at 8:48
  • 4
    (continued) Therefore all 10 missing units were captured by pawns trying to avoid each other, which means that the captures cannot have occurred on the 1st or 8th ranks. So ...gxh1B# as the last move is impossible. Incidentally, an 11th capture is out of the question too (so not a black bishop on e4). The solution is therefore to add a black bishop on d5. – Remellion Jun 12 '19 at 8:50
  • 3
    Also with regards to the problems, I suspect these are still not the problems referenced by Krabbé. The first one you were given (+9 knights) seems to be the "earlier composition by Juel" that was later improved by Guus Rol into an add-10-knights problem. The second problem you were given is not the same theme of problem (only adding 1 bishop) but amusing nevertheless. – Remellion Jun 12 '19 at 8:54

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