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.


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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