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Solutions to our
2008 Christmas Puzzles |
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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
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