December 25, 2008 – The problem with stalemate
Friedrich Zlak, The Problemist, 2008
White to play and mate in eight
Any move by the knight lifts the stalemate, but it is essential to choose the correct square otherwise the stalemate problem just reappears at a later stage. The first move is 1 Nf1!. Now 1...gxf1Q+ leads to an early mate after 2 Qxf1 g3 3 Qh3 gxf2+ 4 Kh2 f1N+ 5 Kg1, so Black has to play 1...g3. There follows the subtle 2 Bd4! which is necessary for the final mating position. After 2...gxf1Q+ (2...gxf1R+ is the same) 3 Qxf1 g2 White once again faces the task of lifting the stalemate, but thanks to his earlier moves he is now in a position to give up his queen to give Black’s king some freedom: 4 Qa6! bxa6 5 f7 Kb7 6 f8Q Kc6. Now Black seems to be escaping but thanks to White’s accurate second move the king can’t flee far and White finishes by 7 Qa8+ Kxd6 8 c8N#. Curiously, White sacrifices a knight and a queen to lift the stalemate, but these pieces are reborn via two pawn promotions later on.
December 26, 2008 – Help mate with the knight
F. Abdurahmanovic and B. Ellinghoven
2nd Prize, idee & form 2000-1
Helpmate in five
The basic mating pattern involves Black playing ...Kf8, ...e1R, ...Re8 and ...Bg8. In the meantime White plays Kf6 and mates with his knight on d7 or g6. This involves 4 moves by White and 4 by Black, leaving one spare move for each side. The problem, of course, is that the e6-pawn must be removed to allow ...Re8 and ...Bg8. The most obvious idea is to take it with the knight while it is on its way to d7 or g6. However, this fails for reasons of move-order; for example 1...Kf8 2 Nxe6+ forces Black to move his king again, while 1...Kg8 2 Nxe6 e1R+ 3 Kf6 leaves Black without a constructive third move. Therefore it is the white king which removes the e6-pawn, the tempo-loss Kxe6-f6 accounting for the missing White move. But how is it possible to remove the bishop’s guard of e6 to allow Kxe6, without wasting more than one move? This is the key idea of the problem and the only solution is for Black to play ...Ba2 to allow Nb3. Then everything is determined: 1...Ba2 2 Nb3 Kf8 3 Kxe6 e1R+ 4 Kf6 Re8 5 Nc5 Bg8 6 Nd7#.
December 27, 2008 – An extraordinary win in a precarious situation
Y. Bazlov, Joint 1st Prize
Corus-70 Jubilee Tourney 2008
White to play and win
The solution runs 1 Bh4+! (1 Qxc8 Qa4+ 2 Kd5 Qd1+ gives Black an easy perpetual) 1...Qxh4 2 Rf7+! Kxf7 3 Qf5+ Qf6 (after 3...Nf6 4 h8N+ Kf8 5 Qxc8+ Ne8 6 Ng6+ Black loses his queen) 4 Qxf6+ exf6 (4...Nxf6 5 h8Q Bf4 6 Qg7+ Ke6 7 h7 Nxh7 8 Qxh7 Nd6 9 Qg6+ Ke5 10 Kd7 picks up the e-pawn, after which White wins the resulting Q v B+N ending) 5 h8Q (now White’s h-pawn would normally be decisive, so Black’s only hope is to block in the white queen) 5...Bf8 6 h7 (6 Qh7+ Bg7 7 hxg7 Nxg7 is a drawn ending of Q v 2N+P) 6...Ne7+ 7 Kd7 Ng7 (mission successful ... or is it?) 8 Qg8+! Nxg8 9 h8N#. An incredible finish with the black king mated mid-board by White’s last unit.
December 28, 2008 – Forcing your opponent to mate you
W. Shinkman (source unknown)
Selfmate in six
All Black’s moves are forced, so the problem is easier than it might have been. White’s first two moves are knight sacrifices, designed to give Black’s king a free tempo: 1 Nxb5 axb5 2 Na6 bxa6. Note that White can’t play these moves in the other order, as 1 Nxa6? bxa6 2 Nxb5 may be met by 2...Kb7!. Now White a free tempo and he plays 3 Kd4, heading towards his eventual destination. After 3...Kb7 White arranges to block Black’s bishop in by 4 Qd5+ Kc8 5 b7+ Kc7 leading to the finale 6 Kc5 Ba7#.
December 29, 2008 – Avoid the diabolical trap!
Nikolai Ryabinin, 1st Prize
Zadachi i etyudy 2006
White to play and win
The first move is no surprise: 1 Qh5+ Kg8. Now it’s tempting to continue 2 Kg6?, but this falls into a diabolical trap: 2...Ne5+! 3 fxe5 Rxg4+ 4 Bxg4 Be4+! 5 dxe4 Qxg4+ 6 Qxg4 stalemate. White’s remarkable winning idea is to return to the same position, but without his rook, so that Black will not be stalemated at the end of this line. He can do this by 2 Qh7+ Kf7 3 Qg6+ Kg8 4 Rh8+! Kxh8 5 Qh5+ Kg8 and only now 6 Kg6. However, that is not the end of the story as Black can still play for stalemate by 6...Ne5+! 7 fxe5 Rxg4+ 8 Bxg4 Be4+ 9 dxe4 Qh3!, when White cannot take the queen with either piece. The only winning move is the astonishing 10 Qh8+!!, after which 10...Kxh8 11 Bxh3 Kg8 12 Bxe6+ Kh8 13 Kf7 wins easily enough, while 10...Qxh8 is met by 11 Bxe6#.
December 30, 2008 – Serieshelpstalemate
Zoltan Laborczi, 1st Prize
Bakcsi-75 Jubilee Tourney 2008
Serieshelpstalemate in eleven
Normally, I select the Christmas puzzles from entertaining positions I have seen during the previous year, but this is an exception. When I was at the World Chess Problem Solving Championship in Jurmala during September, the composer of this problem approached me and said that he had composed a problem suitable for the ChessBase Christmas puzzles. After I had solved it, I agreed with him and this is it.
The solution contains a large dash of humour, as Black’s pieces unpin and are pinned themselves in a frenzied whirlpool of activity: 1 Re5 2 Bg5 (Black cannot play Nc4 at once at this would be check) 3 Nc4 4 Be4 5 Rd5 6 Nce5 7 Bg6 8 Rf5 9 Be7 10 Nf6 11 Be8 and after dxe8Q Black is stalemated thanks to no less than five pins. A neat point of this problem is that you can reach the same position by starting with 1 Ne5 (the idea is 2 Bg5 3 Nbc4 4 Be4 5 Ng4 6 Rd5 7 Nce5 8 Bg6 9 Rf5 10 Be7 11 Nf6 12 Be8 and again dxe8Q) but this takes one move too long.
December 31, 2008 – How to force him to do it?
Saturnin Limbach, 1st Prize, Szachista 1938
Selfmate in five
If we assume that Black’s five moves are ...e3 ...e2, ...e4, ...e3 and ...exf2 mate (in some order) then there must be a white piece blocking d2 or the final position will not be mate. In this case why did Black not play ...exd2+ rather than ...exf2#? The reason can only be that White had just played Qf2+, forcing the capture on f2. The piece on d2 must then be the bishop. We can imagine a situation (with Black to play) in which White’s bishop is on d2 and his queen is on g1 or h2, with Black’s pawns on e2 and e4. Then ...e3 is the only move, and White forces mate by Qf2+. One idea to reach this position is by 1 Ba3 e3 2 Bc1 e2 3 Qg1 e4 4 Bd2 e3 5 Qf2+, but Black can play 2...e4!, when Qg1 is not possible as Black can play his king to f4. It turns out that it is not easy for White to reach f2 with his queen; if he plays Qg1, then Black can usually reply ...Kf4, while after Qh4 Black can move his king to g2. The key idea is to use the bishop to control f4, so as to free the queen to move to g1. White must take into account Black’s plan of ...e3 and then ...e4, which prevents the bishop controlling f4 from c1. The only solution is to control f4 from h6, giving the solution 1 Bf8! e3 2 Bh6 e4 (2...e2 3 Qg1 e4 is the same) 3 Qg1 e2 4 Bd2 e3 5 Qf2+ exf2#.
January 1, 2009 – What was the game?
Michel Caillaud, 1st Prize, Messigny 1997
Position after White's ninth move. What was the game?
There are two surprising things about the solution to this puzzle. Firstly, although the position is symmetrical, the play leading up to it is anything but. Secondly, the ‘obvious’ sequence of moves leads to Black lacking a tempo move at the end of the sequence. Therefore Black has to make his tempo move earlier (see Black’s 6th move in the solution). Anyhow, by now you probably want to see the solution, so here it is: 1 a4 c6 2 a5 Qb6 3 axb6 axb6 4 Ra3 Rxa3 5 h4 Rb3 6 cxb3 h6 7 Qc2 h5 8 Qxc6 Nxc6 9 Nc3.
I hope you all enjoyed this year’s puzzles and are looking forward to the 2009 set!
John Nunn
The Brazilian Championship 2008
