Christmas Puzzles 2013 – solutions (2)

2/4/2014 – The winners of our 2013 Christmas puzzle contest will be announced after the Zürich Chess Challenge (where their prizes will be signed). The puzzles were selected by problem expert GM John Nunn, who gave us the solutions to the first four, and now provides his wonderfully lucid explanations for the remaining problems. In addition we give you the solutions to our brain teasers.

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ChessBase Christmas Puzzles 2013 – solutions (2)

Our 2013 Christmas puzzles were supplied by multiple world champion solver John Nunn. They contained a varied collection ranging from a mate in three to a seriesselfmate in nine. Apart from one tricky puzzle, the selection was fairly easy. You can review the problems by clicking on the above link or any of the images below, which will take you to specific problems. The solutions to the first four were published here.


January 1

Puzzle 5 – December 29, 2013

Yochanan Afek, The Problemist, 1997

White to play and mate in five moves

At first sight White can mate in just four moves by 1 Be3 (or any other square) 1...g1Q 2 b8Q+ Qg2 when any waiting move by White forces mate next move. Why doesn’t this work, and what is the correct solution? Again, please don’t use your engine, it’s not a difficult problem.

[Event "The Problemist"] [Site "?"] [Date "1997.12.19"] [Round "?"] [White "Afek, Yochanan"] [Black "Mate in five"] [Result "*"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "B6K/1P6/8/8/8/8/6pp/2N3Bk w - - 0 1"] [PlyCount "9"] [EventDate "1997.??.??"] 1. Ba7 $1 {is indeed the correct move, because it allows the bishop to be cut odd by the rook on b6. Play continues} (1. Be3 {or any other square on the same diagonal} g1=Q 2. b8=Q+ Qg2 {when any waiting move forces mate next move. Before you can find the solution, you have to discover why this doesn't work.}) (1. Be3 $2 {is met by} g1=R $1 {and after} 2. b8=Q+ (2. b8=R+ Rg2 3. Rb7 {but Black has two ways to refute this:} Ra2 {or} (3... Rg8+ 4. Kxg8 Kg2 {so how about moving the bishop somewhere else at move 1? If})) 2... Rg2 {there's no convenient way to lift the stalemate. The next idea is to try}) ({White to play and mate in five. At first sight White can mate in just four moves by 1 Be3 (or any other square) 1...g1Q 2 b8Q+ Qg2 when any waiting move by White forces mate next move. Why doesn't this work, and what is the solution? Again, please don't use your engine, it's not a difficult problem.} 1. Ba7 g1=R 2. b8=R+ Rg2 3. Rb7 {, which certainly cuts out the ...Ra2 defence, but unfortunately does nothing about the ...Rg8+ defence. This is where some imagination is required. The idea of playing onto the long diagonal to relieve the stalemate is a natural one (and indeed many problems are based on this type of idea), but here it's the adjacent diagonal that needs to be blocked. Once this idea is found, the rest is easy.}) 1... g1=R 2. b8=R+ Rg2 3. Rb6 Kg1 4. Rd6+ $1 Kf1 (4... Rf2 {and}) (4... Kh1 {are also met by} 5. Rd1#) 5. Rd1# { revealing that the knight is not only there to prevent mates on playing rook or queen to the first rank, but has an active role too.} *

Puzzle 6 – December 30, 2013

Michael Barth, 1st Prize, Arnsdorf TT 2013

Series-selfmate in nine

This means that White plays nine consecutive moves (without Black moving at all) in order to reach a position in which Black is forced to give mate in one. White is not allowed to give check except possibly on the last move of his sequence. While seriesselfmates might be unfamiliar to many readers, this is a relatively easy example and you should certainly try to solve it.

There aren’t many ways a seriesselfmate can end. The most common is for White’s last move to be a check which gives Black only one possible reply, which delivers mate. It’s also possible for White’s last move to put Black in zugzwang and force him to mate, but that is obviously unlikely here in view of the numerous mobile Black units. So what might the checking move be that forces Black to give mate? The most obvious candidate is Rxe3+ forcing ... Qxe3#, the only problem being that it’s not mate since the white king can move to f1.

If only the white knight could be transferred to f1 then this problem would be solved, so at first it seems easy: play the rook to e1, then transfer the knight by Ne3-f1 and finish up with R(x)e3+. However, even with six moves to play with, there’s no way for the rook to reach e1 without disrupting the mate, for example by taking the pawn on g3. Remember that White is not allowed to give check except possibly on the last move, so he cannot just play Rxe3+ and then Re1. Then how about playing the queen to e1 instead? In this case, the queen would have to return to h1 or else the final ...Qxe3 will not be mate, but the same sort of problem seems to arise in that it’s not easy for the queen to reach e1 without taking the g3-pawn. However, in this case there is a solution; play Qh2, Kh1 and Qg1. Now the queen can move to e1, but it’s not necessary since the knight is already unpinned. Then reversing the king-queen swap does the trick. Thus the solution runs:

[Event "1st Prize, Arnsdorf TT"] [Site "?"] [Date "2013.??.??"] [Round "?"] [White "Barth, Michael"] [Black "Seriesselfmate in 9"] [Result "*"] [SetUp "1"] [FEN "8/8/8/5p2/2pnRP2/3kp1p1/2pq2P1/2rN2KQ w - - 0 1"] [PlyCount "18"] [SourceDate "2014.01.27"] 1. Qh2 -- 2. Kh1 -- 3. Qg1 -- 4. Nxe3 -- 5. Nf1 -- 6. Qh2 -- 7. Kg1 -- 8. Qh1 -- 9. Re3+ Qxe3# *

Puzzle 7 – December 31, 2013

Michel Caillaud, Prize, Nunspeet TT 2004

This is the position after White’s 12th move in a game.
All the moves were legal (even if not very good). What was the game?

A useful tip for solving puzzles of this type is to use the fact that the solution is totally unique. Thus the game cannot start 1 e4 e6 2 h4 because it could equally well have started 1 h4 e6 2 e4.

In my view, the trickiest puzzle of the set. Even if you guess that the bishop on b1 is a promoted Black queenside pawn, it’s not at all easy to find the solution, mainly because there are several ways a pawn can reach b1. A number of white pieces have to disappear, and the task is easier if you guess that most of these will be taken by the black pawn on the way to the eighth rank. Indeed, it seems likely that the idea the composer was trying to show was a staircase by the black b-pawn, making captures on a6, b5, a4, etc., on its way to b1. Even now it’s not simple. The h1-rook presents a special problem as there’s no really efficient way for Black to take the rook; for example ...Bd6xh2 and then ...Bh2-d6 costs two moves.

The key idea is to arrange for the h1-rook to be captured by the b-pawn. We’ll assume that Black takes the pawn on h2 with his bishop from d6. The f1-bishop and queen can in one move reach a6 and a4 respectively to provide more fodder for the b-pawn. Thus White can play e4, Ba6, Rh5-b5, c4, Qa4, b3 and Bb2-e5-h2. By now Black’s pawn will be on b1 (having taken the a2-pawn) and White can finish up with Ra7-b7, arranging all his pieces in the required 12 moves and provide captures for the b-pawn all the way to b1. Black’s moves will be the six with the b-pawn, ...e6, ...Bd6xh2-d6 and ...Bb7 (the last to allow the white rook to take the bishop on b7). These 11 moves are just the right number and it only remains to decide on the move-order. It’s rather surprising that there’s only one which works, the main point being that Black must take on h2 as soon as possible to free the rook on h1. The exact sequence is:

[Event "T.T. Nunspeet"] [Site "?"] [Date "2004.12.20"] [Round "?"] [White "Caillaud, Michel"] [Black "Prize"] [Result "*"] [ECO "C06"] [Annotator "Nunn,John"] [PlyCount "23"] [EventDate "2004.??.??"] [SourceDate "2013.12.20"] 1. e4 e6 2. Ba6 Bd6 3. c4 Bxh2 4. Qa4 Bd6 5. Rh5 bxa6 6. Rb5 axb5 7. b3 bxa4 8. Bb2 axb3 9. Be5 bxa2 10. Bh2 axb1=B 11. Rxa7 Bb7 12. Rxb7 *

Dr John Nunn (born April 25, 1955) is one of the world’s best-known chess players and authors. He showed early promise by winning the British Under-14 Championship at the age of twelve. In 1970 he entered Oxford University at the unusually early age of 15 and in 1978 he achieved a double success by gaining both the GM title and a doctorate (with a thesis in algebraic topology).

In 1981 John abandoned academic life for a career as a professional chess player. In 1984 he gained three individual gold medals at the Thessaloniki Olympiad, two for his 10/11 performance on board two for England and one for winning the problem-solving event held on a free day.

John's best period for over-the-board play was 1988-91. In 1989 he was ranked in the world top ten, and in the same year he finished sixth in the GMA World Cup series, which included virtually all the world’s top players. He also won the famous tournament at Wijk aan Zee outright in 1990 and 1991, to add to a previous tie for first place in 1982.

John became a successful chess author in the late 1980s and 1990s, and has three times won the prestigious British Chess Federation Book of the Year prize. In 1997 he (together with Murray Chandler and Graham Burgess) founded Gambit Publications, which now has more than 200 chess books in print. When he effectively retired from over-the-board play in 2003, he revisited an early interest in chess problems and in 2004 won the World Chess Problem Solving Championship, at the same time adding a GM solving title to his earlier over-the-board title. In 2007 he again won the World Chess Problem Solving Championship, and in 2010 had his best year to date, winning both the European and World Chess Problem Solving Championships.

In 1995 he married the German chess player Petra Fink. They have one son, Michael, currently aged fifteen, who also plays chess.

Brain teasers

To the Christmas 2013 chess puzzles we add three general brain teasers:

Brain puzzle 1: Jack loves Mary; Mary loves Peter. Jack is married, Peter is single. In this group does a married person love an unmarried person?

Solution: If Mary is single, then a married person (Jack) loves a single person (Mary); if Mary is married, then, too, a married person (Mary) loves an unmarried person (Peter). So the answer is yes – we can be sure that a married person loves an unmarried person, even though we do not know which one.

Brain puzzle 2: It's the nineteenth century. A man owns a grandfather clock, which he meticulously winds up each day. One day he forgets, and the next morning the clock has stopped. He knows that in a not too distant village there is a church clock that has the accurate time. How does he reset his clock with the correct time? No, he cannot see or hear the village clock from his house, and neither does he have a pocket watch to consult.

Solution: The man winds up his clock, sets it running at, say, 12:00 o'clock. He then walks to the village, looks at the church clock, walks back home and looks at his clock. He then knows the time it took him to make the trip. He sets his clock to the time he saw on the church clock plus half the time his clock is showing. Example: the church clock showed 2:30 p.m. When he got back home his clock showed 12:40, i.e. the round trip took him 40 minutes. Then the right time is 2:30 + 20 min = 2:50 p.m.

Brain puzzle 3: Why must you be extremely suspicious if someone is using a deck of playing cards with backs that look like this:

The back of the card above is an autostereogram: if you focus your eyes behind the card you can clearly see the that it is the king of hearts. If you do not know how to view these images here are some instructions. The card deck is available under the name Magic 3D Playing Cards and is produced by a company called Cheatwell.

Once you have mastered the trick of autostereo focussing you will be able to see the above chessboard in vivid 3D. And if you get really good at it you can view this beautiful 3D stereogram of penguins, or even try these animated autostereograms of a galloping horse and a Young Rival song, Hooked? Then you should visit this gallery by 3Dimka, who has, believe it or not, even produced a 3D Tetris game.

The winners of our Christmas Puzzle contest will receive ChessBase software with autographs of top players – at least one with a personal dedication and the signatures of all the participants of the 2014 Zurich Chess Challenge. But that will logically happen after the players have gathered in Switzerland. We will announce the winners and provide some interesting reader feedback in the second week of February.

Frederic Friedel

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