Christmas puzzles: solutions

by Frederic Friedel
1/2/2021 – From Christmas Day, December 25th 2020, until New Year 2021, we published four installments of chess puzzles for you to solve. They were mostly computer resistant, which meant you couldn't simply start and engine to work them out. Today we bring you the solutions of the first three puzzle pages, with the fourth to follow soon. How many did you solve?

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Checkless Chess

Let me remind you: in this chess variant checks are illegal, unless it's checkmate. This means that you cannot make a checking move, even if it would stop the opponent's mate.

Checkless Chess – Helpmate in two

In the first position Black (who starts) needs to play 1.d1=R. White takes a queen: 1...a8=Q. And now comes the unusual mate: 2.Rf1 Qg2#. The black king cannot capture the queen or move to the e-file because that would expose the white king to a discovered check, which is illegal.

In the second position Black plays 1.d1=B and White plays 1...a1=N. There follows 2.Bb3 Nc6#. Black is mated since he cannot move his king off the a2-g8 diagonal. The bishop check would be illegal. Amazing, all four promotions with just four pieces.

Here are the solutions to the two Checkless Chess puzzles from our December 25 puzzle page:

 

Add-a-king

The first two problems are by T.R. Dawson and were composed in 1927. 1946. Your task was to add a white queen in each diagram, on a legal square, and then stalemate Black in one move.

The solutions are +wQe1 and 1.f4 stalemate; +wQa1 and 1.a4 stalemate.

The second set is by Bengt Giöbel. Once again you have to add a white queen, but this time mate in one move.

The solutions are +wQc1 and 1.d3# and +wQa1 and 1.b3#.

In addition, you were asked to figure out why, in diagrams that are perfect mirror images, the solutions are quite different. The answer is that with all the white pawns on their original squares the white queen could not have crossed over the right side of the board. So it must always be placed to the left of the white king.

This is a problem by Rafael M. Kofman, published in Vecherny, Leningrad 1968, where it won the 3rd prize. You are required to replace the white king on the board and then mate in two moves. 

Solution: add the white king on e1, and then 1.0-0-0!  (threatening 2.Rd8#) 2.0-0 2.Rg1#.

Take that back!

The first position is a classic – it was composed by Bruno Sommer in 1910. White takes back his last move and plays something different, mating instead in one move. Solution: his last move was exf6 e.p. (Black has played ...f7-f5 before that). So with a white pawn on e5 (and a black pawn on f5) White delivers mated by playing 1.e8=Q#.

The second position is by Werner Keym, who provided me with a number of puzzles for the Christmas pages. It was published in 2010 and is the most economical form of this kind of retractor. White should take back his previous move and instead mate in two moves. Black is stalemated. Solution: White retracts the move 1.g2-g4 and plays 1.g2-g4. This makes the en passant capture legal, and Black is forced to play 1...hxg3 e.p. and White can play 2.hxg3 mate.

And finally a more recent retractor, a problem by J. Coakley, Chess Cafe 2014: White takes back his previous move and instead mates in one. This has the same theme as the two above, but is quite intricate.

Solution: White cannot castle in the problem position, because both f1 and g1 are attacked. But after taking back 1.bxc6 e.p. the pawns on b5 and c5 block the bishop diagonals. Now White can castle and mate: 1.O-O#.

In retractor problems, an en passant capture is allowed as backward or forward moves, unless they can be proven illegal. 

I'll force you to mate!

The subject of our December 29 puzzle page was the selfmate. In this form of chess problem White must force Black to mate him, while the opponent will do everything he can to avoid that happening. We also included a (festive) self-stalemate that required White to stalemate himself against the will of Black. Here are the annotated solutions to replay:

 

That should suffice for the second day of a hopefully much better new year. I will publish the solution to our final Christmas Puzzle instalment on Monday. The train problem included there has been straining many minds around the world. I have 2700+ GMs begging for the solution, and a math professor, a quantum physicist, the head of a project group at Intel, and others still struggling. They have asked for some clarification:

My question to the young Anish Giri in 2010 was: why did the overhead contact wire that supplies power to the train we were about to take not follow a straight line? Why did it have a zig-zag pattern, as shown in this picture?

Now, after ten years and a small hint he has got it. One reader, albitex, posted the correct solution in our feedback section. I took it down and promised to switch it back on when I had published the solution. In the meantime Albitex tells me "Frederic, I am an electronic technician and I worked for a railway signalling company." Stand to reason!

And now to celebrate the New Year, nobody did a better job than Seattle's Space Needle observation tower. Do not ask me how they and do something like this. Boffins please explain!

 


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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enfant enfant 1/9/2021 11:57
For the Dawson Checkless Chess mate problem,
after 1. f6 Qf8 (or Qh8) White has 2. Qf5 mate!
baquet baquet 1/5/2021 01:10
Wonderful Christmas Gift those problems ! So fun ! Many thanks Frederic!
baquet baquet 1/5/2021 01:07
hELLO ALL, I think in the Coakley retractor there is another solution : retract Re2 to e eight instead play Rh1 to f1 mate !
charlesthegreat charlesthegreat 1/4/2021 09:52
There's another solution in the first take back puzzle. White just captured a black rook on f6, so instead, this move is taken back and e8=Q mates.
MLoftus MLoftus 1/3/2021 02:02
On the overhead wires is not the zig zag pattern to ensure even ewear on the ppantograph?
Joshua Green Joshua Green 1/3/2021 01:32
@Frederic, the solution looks good now, so maybe it was just a cache issue. Thanks!
Frederic Frederic 1/3/2021 08:14
@ARiddles: The number of teeth on a gear is normally prime? Wow, I'm going to spend the day thinking! Hope you are not being sarcastic or anything. BTW have you heard about prime number cicadas? I saw the ones that come out every 17 years! Hard to believe: https://frederic-38110.medium.com/prime-number-cicadas-really-96cdeca5c15d.

@Joshua Green: I changed it to what you correctly suggested, but there are some disagreements between our replayer and the browser cache. The replayer should show you the moves of christmas-solutions02d.pgn.

@Anonychess: Yes, one of my highest-ranking boffin friends told me that as well. So it is just computer graphics?! Well, there goes reality!
ARiddles ARiddles 1/3/2021 05:32
I'm a structural engineer so luckily I know the answer. However my musings "hint" would be "why are the number of teeth on a gear normally prime (or at least co prime) numbers" 😉
Joshua Green Joshua Green 1/3/2021 02:46
My correction to the Flood s#3 solution was misinterpreted above. Here's what the solution should look like:

1. Ra4! (1. Ra5? c5! 2. Rca4 g4+! (2. ... c4? 3. Ra8 g4#)) c6 2. Rca5 c5 3. c4 g4#
rgorn rgorn 1/3/2021 02:14
"Abrasion" might have been the hint Giri got.
Anonychess Anonychess 1/3/2021 02:09
Frederic, it was not a laser show. It was just animation for tv.
Joshua Green Joshua Green 1/2/2021 09:31
Since you asked, I'll answer that I solved all of the ones that are answered here. The most fun was the s#6 by Schinkman, as it required some amount of logical thinking to determine the final position.

I do have two corrections to what you've written above.
- In the s#3 by Flood, the refutation of 1. Ra5? is 1. ... c5! 2. Rca4 g4+! (rather than 2. ... c4? 3. Ra8 g4#).
- The problem by Rubin is a s#5, not a h#5.
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