3

In his article on “The Horse Concoction", Tim Krabbe mentions retrograde analysis questions that center around checkmate and involve 10 rooks, knights, and bishops. The 10 rooks problem is one that he shows. It can also be found here on CSE.

However, he does not show the problems with 10 bishops and 10 knights by Dane Henrik Juel and the Dutchman Guus Rol, although who made which is unclear. I suspect that is goes respectively though. My question is this: can these by found somewhere online? I would like to take a look at them, out of curiosity and so I can learn more about retrograde analysis.

This is the exact quote from the article that contains the aforementioned information. As far as I can tell, the problems revolve around checkmate, to reiterate:

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).

0

Thank to Bojan Bascis, who answered my same question on Matplus, we now know what these formerly mysterious knight and bishop problems are!

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

[Title "Henrik Juel. Probleemblad, 5/1997, Add a mating bishop"]
[FEN "b3B3/1k1B3B/7b/1B3b1B/B6b/3b1K1b/7B/1BbBbbB1 w - - 0 1"]
  • 3
    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 at 8:48
  • 3
    (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 at 8:50
  • 2
    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 at 8:54
  • Ah ok. But you are right that it is amusing. – Rewan Demontay Jun 12 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.