Christmas Puzzles 2013 – solutions (1)

by ChessBase
1/22/2014 – The prize contest is now closed, we have received your entries and will announce the winners after the Zürich Chess Challenge (where the prizes will be signed) at the beginning of February. The seven chess puzzles were selected by GM John Nunn, who now gives us the solutions to the first four, with extensive explanations that make them interesting even to readers who are not problem fans.

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ChessBase Christmas Puzzles 2013 – solutions (1)

Our 2013 Christmas puzzles were supplied by multiple world champion solver John Nunn. They contained the a varied collection ranging from a mate in three to a seriesselfmate in nine. Apart from one tricky puzzle, the selection, with one exception, 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.


January 1

Puzzle 1 – December 25, 2013

William Shinkman, Offiziers-Schachzeitung 1905

White to play and mate in three

Black’s king is stalemated so White’s first move must release the king. Please don’t use your engine to solve this. It will do so in an instant, but you will deprive yourself of the fun of finding the solution yourself. Come on, it’s only four pieces!

[Event "Offiziers-Schachzeitung"] [Site "?"] [Date "1905.12.19"] [Round "?"] [White "Shinkman, William"] [Black "White to play and mate in 3"] [Result "*"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "2k5/8/8/BQ6/4K3/8/8/8 w - - 0 1"] [PlyCount "5"] [EventDate "1905.??.??"] 1. Qb2 $1 ({It's tempting to try something like} 1. Bb4 Kd8 ({but Black can improve by} 1... Kc7 {and there is no mate in 3. Other tempting possibilities such as 1 Qd5 also don't lead to mate, and while it's possible to solve the problem by trial and error, it's better to use some logic. Once you have seen that bishop moves and 1 Qd5 don't work, it becomes clear that White's first move must be with the queen and that it must release either b7 or d7 for the king. On b7 the prospects for a mate in two are not at all promising as it the king moves to a7 or b8 there's no obvious mate. It gradually becomes clear that it's tough to mate the king when it is on a dark square, because the bishop is less effective then. Thus allowing 1...Kd7 is much more promising, because c7 and d8 are already controlled and if White can cover d6 and e7 with his second move, a mate looks feasible. This leads to the idea of meeting 1... Kd7 with 2 Qe5, only allowing the king access to the light squares c6 and c8; sure enough, 2...Kc8 can be met by 3 Qc7# and 2...Kc6 by the attractive 3 Qd5#. It only remains to find a suitable square for the queen at move one which keeps control of b7 and allows Qe5 at move 2. Thus the solution is}) 2. Qb7 Ke8 3. Qe7#) 1... Kd7 2. Qe5 Kc6 (2... Kc8 3. Qc7#) 3. Qd5# *

Puzzle 2 – December 26, 2013

Anatoly Styopochkin, Special Prize, 59th T.T. SuperProblem, 2012

White to play and selfmate in five.
a) Diagram position – b) Remove the pawn on f6

In a selfmate, White is trying to force Black to deliver mate, while Black is doing his best to avoid this. The above problem has two parts; in the second, just remove the pawn on f6 and again selfmate in five.

If it were Black to move he would be forced to give mate at once by ...fxe4#, but White has no waiting move. The idea here is to find a way to return to something like the initial position, but having wasted a tempo. There is a connection between the solutions to the two parts of the problem.

[Event "59th T.T. SuperProblem"] [Site "?"] [Date "2012.12.19"] [Round "?"] [White "Styopochkin, Anatoly"] [Black "Selfmate in five"] [Result "*"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "8/8/5p2/3k1p2/4RB2/2QKP3/2PPP3/8 w - - 0 1"] [PlyCount "10"] [EventDate "2012.??.??"] {White has to keep control because if the black king is granted too much freedom, it will be impossible to drive it back to d5. This was perhaps the toughest problem this year, as it's hard to pinpoint a chain of logic leading to the solution. The basic in the first part is to eliminate the f4-pawn and force Black to deliver the ...fxe4# mate using the rear f-pawn. The solution runs} 1. Bh2 f4 2. Qc4+ Kd6 3. Qc8 $1 {The king must not be allowed to move to d7; note that if White had played 1 Bg3? then Black would be able to take the bishop here.} f5 (3... Kd5 4. Bxf4 f5 5. Qc3 fxe4# {is the same}) 4. Bxf4+ Kd5 5. Qc3 fxe4# *

[Event "Special Prize, 59th T.T. SuperProblem, "] [Site "?"] [Date "2012.12.19"] [Round "?"] [White "Styopochkin, Anatoly"] [Black "Selfmate in five"] [Result "*"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "8/8/8/3k1p2/4RB2/2QKP3/2PPP3/8 w - - 0 1"] [PlyCount "10"] [EventDate "2013.??.??"] {In the second part the rear f-pawn has disappeared, so White cannot afford to let Black play ...f4. On the plus side, the white queen has more freedom and this provides the key to the solution:} 1. Rb4 $1 Ke6 2. Qg7 {as in the first soltuino, the king must be kept confined} Kd5 3. Qf6 Kc5 4. Qc3+ {the choice of b4 with the rook was to prevent Black's king fleeing to b5 here} Kd5 5. Re4 fxe4# *

Puzzle 3 – December 27, 2013

M.Matouš, 1st Prize, Ceskoslovensky Sach 2005

White to play and win

The material balance is roughly equal, so White must pursue an attack. It’s important to know that two bishops always win against a knight except if there is an immediate draw for the knight. In endgame studies there is no 50-move rule, so you need not be concerned that a 2B vs N win might require more than 50 moves.

This study is by Mario Matous, the Czech study composer who died earlier this year. ChessBase already published an appreciation of him, but did not give this study, which is one of his simpler compositions.

[Event "1.p Ceskoslovensky Sach#98"] [Site "?"] [Date "2005.??.??"] [Round "?"] [White "Matous=M"] [Black "White to play and win"] [Result "1-0"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "7n/4N3/4q1N1/8/B7/5k1P/3B4/6K1 w - - 0 1"] [PlyCount "13"] [EventDate "2005.??.??"] 1. Bc6+ {The only check that makes sense since after} (1. Bd1+ Kg3 {White has no follow-up.}) 1... Kg3 {Forced, because} (1... Ke2 {loses the queen at once after.} 2. Nf4+) 2. Nf5+ $1 {This spectacular knight sacrifice is the only way to continue the attack.} (2. Bf4+ $2 Kxh3 3. Bg2+ Kg4 4. Kh2 {fails for several reasons, the simplest being} Nxg6 5. Bh3+ Kxf4 6. Bxe6 Nxe7) 2... Qxf5 (2... Kxh3 {fails to} 3. Nf4+) 3. Be1+ {Now that Black's queen has been deflected from e6, this check becomes available.} Kxh3 {White's last pawn disappears and the check on g2 leads to nothing, so it seems that White has shot his bolt, but now comes the first really surprising point of the study.} 4. Ne5 $1 {This quiet move traps the Black king on h3 and threatens Bg2#. It's surprising that this move is so effective, especially since Black has a check with his queen.} Qg5+ {Black is forced to check, as otherwise he cannot prevent both Bg2 and Bd7.} (4... Qg4+ 5. Nxg4 Kxg4 6. Kf2 {mates in a further 47 moves.}) 5. Kh1 {Black has no more useful checks, but now his queen is able to cover g2 and d7.} Qg7 {After the move played, White's continuation is not at all obvious since the bishop cannot move to b7, while if it goes to d5 or e4 Black can simply take the knight.} (5... Qg3 6. Bd7+ Kh4 7. Nf3+ {wins the queen for nothing, while after}) (5... Qg4 6. Nxg4 Kxg4 7. Be8 {White mates in another 68 moves (hence my comments about the 50-move rule) .}) 6. Bf2 $1 { This waiting move is the key, since it places Black in a deadly zugzwang. Black's queen cannot move so as to retain control of both d7 and g2, while a knight move interrupts one of the queen's lines of defence.} Ng6 ({Or} 6... Nf7 7. Bd7+ {followed by mate.}) (6... Qg4 7. Nxg4 Kxg4 8. Kg1 {mates in 69 more moves.}) 7. Bg2# 1-0

Puzzle 4 – December 28, 2013

Viktoras Paliulionis, 1st Prize, Jubilee T. Ilievski-60, Orbit 2012

Helpmate in 6.5 – with set play

In a helpmate the two sides cooperate to help White mate Black. The stipulation means that there are two solutions to this problem. In the first solution, Black starts and White mates Black on his sixth move (so both White and Black make six moves). In the second solution White moves first and mates Black on his seventh move (so White plays seven moves and Black six).

You might wonder why the second solution cannot just be the first solution plus a White waiting move somewhere, but that’s part of the fun of the problem. Remember that both sides are playing to help White deliver mate, and good luck!

[Event "Jubilé T. Ilievski-60, Orbit"] [Site "?"] [Date "2012.12.19"] [Round "?"] [White "Paliulionis, Viktoras"] [Black "Helpmate in 6.5 – with set play"] [Result "*"] [Annotator "Nunn,John"] [SetUp "1"] [FEN "4N3/7k/8/8/8/5q2/5r2/1K6 w - - 0 1"] [PlyCount "13"] [EventDate "2012.??.??"] {In a helpmate, it's usually a good start to try to work out a plausible mating position. It's impossible to mate in the h8-corner because the white king is too far away, so mate must take place on the side of the board. There are only two possible types of mating position. In one, the white king faces the black king, for example with the white king on f7, Black could have queen on h8 and rook on h6, with mate by the knight on g5. The other formation has the white king on the edge of the board, for example with the white king on h5, black queen on h8 and rook on g7 White could play Nf6#. This particular position takes seven white moves, so it can only be a candidate for the second part of the problem. Taking the solution in which Black moves first, White has only six moves, so there is no time for the white king to reach the h-file and move the knight to mate, so it must be the first formation. It appears natural to play the king to f5 (taking four moves), while Black plays ...Kh6-h5, queen to h4 and rook to h7 (taking six moves), allowing Ng7#. This even leaves a spare move for White, so it seems easy to arrange. However, when you look for the precise sequence of moves difficulties start to appear. Black has no spare moves, so he cannot start with ...Rf1+, the only move to allow the white king to move to the second rank immediately. This explains the missing white move. However, the white king must move to the second rank by move 2 or it will not reach f5 in time, so Black's rook must quit the second rank by move two. It seems that the first move must be with the queen, and must allow the queen to move directly to h4 later. This leaves e4, g4, g3, h3 and h1 as possibilities. However, all these run into the same problem; after, for example, 1...Qe4+ 2 Kc1 Rf6 3 Kd2 the white king's passage to f5 is cut off by the black queen. At this point it's worth noting an anomaly about the above logic. The solution must be unique, but apparently Black can play either ...Kh6-h5 or ...Kg6-h5. Why should he have to choose one rather than another? This provides the clue: the rook has another route to h6, via h2, and this would force the king to adopt the route via g6, removing the ambiguity. Now we can find the solution:} 1. -- ({Regarding the second part, the first question is why we cannot use the first part and add a white waiting move. Try, for example} 1. Kc1 Rh2 2. Kb1 Rh6 3. Kc2 Qh1 4. Kd3 Kg6 {and the reason appears; the white king cannot move to e4. This reveals that Kc1 in the first solution wasn't a waiting move; it was the essential to nudge the king towards the kingside as to provide a route to f5 that does not cross the e4-square.}) ({We have already noted a possible mate by playing the king to h5 and Nf6#, but this leaves White with no spare moves, so the first move must be} 1. Kc1 {Now Black only needs four moves to get his queen to h8 and rook to g7, so he has two spare moves. However, he must get his queen and rook out of the way to allow the king to reach h5, and the rook is a particular problem as it cannot just move ...Rg2-g7 or ...Rf7-g7, as this will cut the white king off along a file. Indeed, trying to play something such as 1...Rf1+ 2 Kd2 Q moves 3 Ke3 Rd2 (say) 4 Kf4 Rd7 5 Kg5 fails because the queen now needs to move to h8, but from f3 the only possible squares allowing immediate access to h8 are c3, f6 and f8, and all of these obstruct the white king. It follows that White's next move cannot be Kd2 and so must be Kd1, with his remaining moves being Ke2, Kf3, Kg4, Kg5 and Nf6#. The problem here is the f3-square, which will remain guarded by the black queen. The only solution is to move the black rook off the f-file, while at the same time cutting off the black queen's guard of f3. Black can only afford to waste one move with his rook, so it's path is ...Rf7, ...R somewhere cutting off the black queen, and then ...Rg7. This gives us the clue we need and the pieces can be assembled to give} Qa8 2. Kd1 Rf7 3. Ke2 Rb7 4. Kf3 Qa1 5. Kg4 Qh8 6. Kh5 Rg7 7. Nf6# {A really beautiful problem in which the black queen visits all four corners of the board during the course of the two parts.} ) 1... Rh2 2. Kc1 {White has only one tempo to spare so he cannot play Ka1 or make a knight move, which would cost two tempi.} Rh6 {Freeing the king.} 3. Kd2 Qh1 {The key idea; Black could not play this at move one because of mate, but now it is viable and indeed the only route to h4 that does not cut off the white king.} 4. Ke3 Kg6 {Black must transfer his king now or the white king will not be able to move to f5 later} 5. Kf4 Kh5 6. Kf5 Qh4 7. Ng7# *

– Part two will follow soon –

Prize winners

Two winners will be chosen at random from all eligible submissions – we like to reward participation, not getting everything meticulously right. A third prize winner will be selected on the basis of the nicest message (in our opinion) that we receive. Closing date for all entries is January 15, 2014. Note that the arrival date will not influence the selection.

And the prizes? They will consist of the most recent ChessBase software, with autographs of top players – at least one a World Champion or ex. And one of the prizes will have a personal dedication and the signatures of all the participants of the 2014 Zurich Chess Challenge – if nothing comes in the way of us getting them. Definitely worth a shot at winning, don't you think?

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