Outrageous Chess – Solutions

by Frederic Friedel
9/21/2020 – Last week we gave you a series of chess problems which cannot be solved by computers, taken from a book called "Outrageous Chess" by Burt Hochberg. Were you able to find the out-of-the-box answers? And do you like this kind of problem? In any case, today we reveal the correct solutions and explain the ideas behind the problems.

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Solutions to the selection of outrageous problems

 

You had to add a pawn and mate in two. The only way to do this is to place a white pawn on h2 for the two-move mate: 1.h4 gxh3 e.p. 2.Bxg6#.

 
 

The wonderful solutions to these twins: 1.d8=B Kd6 2.c8=R Ke6 3. Rc6# – underpromotions to a bishop and a rook. With the white king on e3 the solution to the second position is 1.d8=Q Ke5 2.c8=N# – promotions to queen and knight.

Note that in the above diagrams you can conduct underpromotion by moving the pawn to the eighth rank. With the left mouse key still depressed, move the cursor to the piece which you wish to take.

 
 

With White to play the solution is 1.Qxh4 0-0 2.Qh7#. But if Black must start there is no waiting move. So a different strategy is required: 1.Rh7 0-0 2.Re7 Qc8# (or in more traditional notation: 1...Rh7 2.0-0 Re7 3.Qc8#). You can enter the moves on the diagrams to see it all unfold.

 

There seem to be two possible solutions: 1.0-0 N~ 2.Qa1# or 1.Kf2 N~ 2.Qa1#? Bu only one of these is correct, for the following reason: we need to ask ourselves: how did the black king come to a2? What route did it take? It could not have come from a3 or b3, because these squares have always been attacked by the pawns on b2 and c2. So the black king must have crossed d2 or d1. But that means that the white king must have left the square e1, and consequently castling is not legal. So only 1.Kf2 N~ 2.Qa1# is legal.

One more? Here's one in similar vein: sometimes problemists deliberately break the rules to make a point – or just for fun.

 

Solution: The above position is illegal, because there is no way the black king could have reached a2. We must therefore assume that Black cheated and simply placed his king on that square. So we replace it on the square from where it came. But which square is that? Turns out that every legal square for the black king allows mate in one by White. Isn't that crazy? 

Hope you enjoyed these unusual problems. Please just tell us whether you were able to solve the problems (which ones?) and whether you enjoyed the challenge. I will continue in this vein if you did.

Links

Outrageous Chess

 




Editor-in-Chief emeritus of the ChessBase News page. Studied Philosophy and Linguistics at the University of Hamburg and Oxford, graduating with a thesis on speech act theory and moral language. He started a university career but switched to science journalism, producing documentaries for German TV. In 1986 he co-founded ChessBase.
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Frits Fritschy Frits Fritschy 9/22/2020 07:01
There are many compositions where a bishop promotion is necessary to win, as a queen promotion would lead to stalemate. Tim Krabbé's website (I sadly can't give the link as I did, my message was automatically deleted) even gives a few game examples where bishop promotion was necessary to win. For instance Kholmov-Ehlvest (Volgodonsk 1983): white Kb6 Rh1 pawn c7 / black Kc8 Rh5 pawns e6 and h2. After 72 Rh1-a1, black found the only winning move, which should be clear to anyone now (and also why it is the only winning move).
JNorri JNorri 9/22/2020 10:22
Regarding the Fabel, Dawson had already published in The Chess Amateur 1920 the far more interesting position with white king on f2, pawn already on h4, add Black pawn on f3. Stipulation: retract one move, then mate in two. Solution: back h4-h2, forward h2-h4. Same position, but no longer stalemate.
Michael Jones Michael Jones 9/22/2020 02:06
genem, I don't know of any examples with a bishop, but in a blitz game I once promoted to a rook. I suspect there were probably alternative winning lines, but I didn't have time to calculate them so, once I had noticed that promoting to a queen would give stalemate, the rook promotion was the most obvious move.
genem genem 9/21/2020 08:48
The underpromotion to bishop theme is interesting, because it always seems to require a caveat, such as here the "mate in three" caveat, not simply to forcibly win in any legal way. I am unable to find any underpromotion to bishop that is Necessary to win, which would be cool.
PWintershoven PWintershoven 9/21/2020 02:30
I really enjoyed the puzzles. I could not solve the first one, the rest I managed. Keep it coming.
pwcca pwcca 9/21/2020 01:01
Nice! . For the NGG van Dijk, Ostlendigen 1959, an other argument would be to assume the problem must have one and only one solution (because chess problems must not have doublons) and only castling can be an illegal move so the other solution must be right. It's similar to the riddle of the three daughters.
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