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Christmas Puzzles 2005
January 1st, 2006 |
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Solutions
Before we start we would like to thank the many readers
who sent us feedback for their valuable input. We might
mention that unfortunately a number of messages may have
gone missing, since the feedback form on the left of the
puzzle pages were not working properly. A special thanks
goes to Patrick Schönbach, the musician,
programmer and chess fan, who informed us that the Midi
file for the December 26th puzzle was defective. Apparently
the drum part was being replaced by a piano sound, which
produced subtle dissonances. Patrick fixed the file and
we replaced it. You can compare the old
version to the new
one, if you are interested. Incidentally the music does
not play automatically in a Firefox browser.
Christine and Patrick
You may remember Patrick Schönbach from a very interesting
2004
article. We are especially pleased to learn that Patrick
has found a partner for life, Christine, who comes from
Leipzig, where the two plan to live after they get married
later this year. Congratulations!
First we come to the famous Behting study, which we understand
was first published in the magazine Bohemia in
1906.
K. K. Behting, Baltische Schachblätter
1908
White to play and draw
When the solution of this problem is presented to you,
we said, it would be in the form of just three moves, and
then a short explanation, which would suffice to convince
any rational chess player that the position is indeed a
draw. We also asked you to make liberal use of chess engines
to try and work out the solution. And finally we predicted
that even with today's super-powerful machines there would
not be one which finds the correct first move and displays
a 0.00 evaluation (indicating a draw). Apparently this is
correct, since we did not receive any claims to the contrary.
Well, here's the solution: 1.Kc6!! Truly
staggering that only this move secures the draw. The reason
will become clear shortly. 1...g1Q. This
is the only reasonable attempt for Black to win. In the
attached analysis we show why 1...h3 or 1...Kg5 only draw.
2.Nxh4! Qh1+. Black has only this one check,
which is the reason that 1.Kc6!! was required in the starting
position. 3.Nhf3 ½-½.
Take a look at the position. The black king has been completely
immobilised by the two knights, which protect each other
and the pawn on d2. White will simply leave these three
pieces in place and move around with his king. Black can
only give checks with the queen, and White only needs to
avoid the king getting stalemated in a corner. The position
is an iron draw.
Click
to replay the above solution in a
separate window. Note that in the
JavaScript replay page you can follow the analysis by clicking
on the moves on the right.
It is interesting to see why the other tries in the starting
position fail: 1.Nxh4, 1.Nf3, 1.Ng7+, 1.Ng3+ or a different
king move. We have included a lot of analysis for these
lines in our replay section. You can download the position
and continue analysing it with Fritz. It would also be of
interest to know, over the years, if any chess engine finds
the draw (with a draw evaluation), and how long it takes
to work it out.
Jacob Mintz, The Problemist 1982 – I
Helpmate in three (Black to play helps White
to deliver mate in three moves)
For this we already gave the solution: 1.a1Q+ Ne1
2.Qb2 Nc2 3.Qb5 Ra3#. Here are the solutions for
the rest:
Jacob Mintz, The Problemist 1982 – II
 
Helpmate in three. Solution: 1.a1R+ Rc1
2. Rb1 Nd4 3.Rb4 Ra1#
Jacob Mintz, The Problemist 1982
– III
 
Helpmate in three. Solution: 1.a1B Rb3
2.Bc3 Ne6 3.Ba5 Nc5#
Jacob Mintz, The Problemist 1982 – IV
 
Helpmate in three. Solution: 1.a1N Rc1
2.Kb3 Rb1+ 3.Ka2 Rb2#
The first prize was awarded to this quartette for the elegant
allumwandlung, the promotion to all three pieces
to achieve the solutions. Click
to replay.
Viktor Zheglov, Suomen Tehtaevaeniekat 1998-99
Helpmate in 12 (Black to play helps White to deliver
mate in twelve moves)
We gave you a
lot of help for this puzzle, and indeed showed you the
final mating position. You had to find the moves to achieve
it. The remaining problems were that White cannot play his
king to f2 until Black plays ...Kh1, and Black can’t
play ...Kh1 until the white bishop leaves the long diagonal,
and the white bishop can’t leave the long diagonal
until White plays Kf2 (because the bishop is pinned) –
a circular piece of logic which showed that this plan is
a dead end.
The key was to use the black king to block the fifth rank
so to allow the white king to c4; then the white bishop
is used on e2 to block the e-file to allow the white king
to reach f2 via e1. The fact that the black king must block
the fifth rank means that the black king's route is uniquely
determined (Kb3-c4-d5 and then diagonally to h1).
So the solution is: 1.Kc4 Ba6+ 2.Kd5 Kb7 3.b3 Kb6
4.b2 Kb5 5.Ke4+ Kc4 6.Kf3 Kd3 7.Kg2 Kd2 8.Kh1 Be2 9.b1N+
Ke1 10.Nd2 Kf2 11.Nf3 Bf1 12.Nh2 Bg2#. It is remarkable
that the order of all the moves is uniquely determined.
A truly beautiful problem. Click
to replay.
Noam D. Elkies, Havard 2002
Selfmate in two
The solution is 1.Qf7! and now:
1...a1Q#
1...a1R#
1...a1B 2.Qg6 Bb2#
1...a1N 2.Qg6 Nb3# Click
to replay.
The subtlety was that 1.Qg6? does not work, since 1...a1N!
2.Qf7 Nb3+ is not mate (3.Qxb3).
Edgar Holladay, British Chess Magazine
1965
Selfmate in 10
Here, too, we gave you some
help with John Nunn's relatively difficult submission.
It is fairly clear that if several White pieces disappeared
from the board then there would be a mate with Black’s
rook on a6. One might go on to guess that the problem will
end with Black’s rook on a7 and White’s rook
on a6, with Black forced to play ...Rxa6#.
The difficulty with the problem is that White’s first
move disrupts this cosy scenario by moving the knight away
from b5. It must go to a7 in order to allow a later Bb7+
to deflect Black’s rook to b7, from where it can easily
be forced to a7 ready for the denouement. However, at the
moment Bb7 is mate, hence the knight move to block the a-file.
The forcing finale is combined with a curious ratchet manoeuvre
along the eighth rank in which the black rook is forced
to take the white rook.
So the solution is: 1.Na7 Rf8 2.Rg8 Re8 3.Rf8 Rd8
4.Re8 Rc8 5.Rd8 Rb8 6.Rc8 Rxc8 7.Qb8+ Rxb8 8.Bb7+ Rxb7 9.Nb5+
Ra7 10.Ra6 Rxa6#. Click
to replay.
The problem by Hasselkus was, as we subsequently learned,
cooked, i.e. there was a second unintended solution. The
next
day we published another famous selfmate with no less
than eight rook underpromotions. But this too was immediately
cooked by a reader. Our Puzzle
Contest has two new selfmates in a similar vein, both
of which this time will hopefully prove sound.
We proceed to the first very long direct mate problem,
where it was your task to complete the solution, which we
had partially given. We wanted to know exactly how many
moves are required for the mate.
Karl Fabel, Die Schwalbe 1938
White mates in how many moves?
The correct answer is 64 moves: 1.Ke1 Bg4 2.Kd2
Bh3 3.Kd1 Bg4 4.Ke1 Bh3 5.Kf1 d3 6.Ke1 Bg4 7.Kd2 Bh3 8.Kxd3
Bg4 9.Kd2 Bh3 10.Kd1! Bg4 11.Ke1 Bh3 12.Kf1 d4 13.Ke1 Bg4
14.Kd2 Bh3 15.Kd1 Bg4 16.Ke1 Bh3 17.Kf1 d5 18.Ke1 Bg4 19.Kd2
Bh3 20.Kd1 Bg4 21.Ke1 Bh3 22.Kf1 d6 23.Ke1 Bg4 24.Kd2 Bh3
25.Kd1 Bg4 26.Ke1 Bh3 27.Kf1 d3 28.Ke1 Bg4 29.Kd2 Bh3 30.Kxd3
Bg4 31.Kd2 Bh3 32.Kd1 Bg4 33.Ke1 Bh3 34.Kf1 d4 35.Ke1 Bg4
36.Kd2 Bh3 37.Kd1 Bg4 38.Ke1 Bh3 39.Kf1 d5 40.Ke1 Bg4 41.Kd2
Bh3 42.Kd1 Bg4 43.Ke1 Bh3 44.Kf1 d3 45.Ke1 Bg4 46.Kd2 Bh3
47.Kxd3 Bg4 48.Kd2 Bh3 49.Kd1 Bg4 50.Ke1 Bh3 51.Kf1 d4 52.Ke1
Bg4 53.Kd2 Bh3 54.Kd1 Bg4 55.Ke1 Bh3 56.Kf1 d3 57.Ke1 Bg4
58.Kd2 Bh3 59.Kxd3 Bg4 60.Kd2 Bh3 61.Kd1 Bg4 62.Ke1 Bh3
63.Kf1 Bg4 64.Bxg2#. Click
to replay.
Karl Fabel, Die Welt 1947
How many moves to mate?
Here, too, we gave you all the help you needed to find
the solution. If it were Black to move he would have to
play 1...g3 and allow a forced mate. So White has to lose
a move, which he could do by triangulating with his king
(Kg1-h1-h2-g1). But the plan does not work since Black moves
his king to b1, c1, etc. escaping from the trap on the a-file.
White has to first prevent the black king from moving to
b1, which is achieved with the help of the other knight.
1.Na6 Ka2 2.Nc7 Ka1 3.Ne8 Ka2 4.Ng7 Ka1 5.Nh5
Ka2 6.Ng3 Ka1 7.Nf1 Ka2 8.Nd2 Ka1 9.Nb1 Ka2 10.Na3 Ka1.
We now have the same position as in the diagram above, except
that the white knight has been transferred from b4 to a3,
and the black king can no longer escape via b1. White can
now triangulate: 11.Kd2 Ka2 12.Ke1 Ka1 13.Kf1 Ka2
14.Kg1 Ka1 15.Kh1 Ka2 16.Kh2 Ka1 17.Kg1 Ka2 18.Kf1 Ka1 19.Ke1
Ka2 20.Kd2 Ka1 21.Kc1. Again we have the same position
(with the knight on a3), but White has lost a move. Time
to return the knight to its original position. 21...Ka2
22.Nb1 Ka1 23.Nd2 Ka2 24.Nf1 Ka1 25.Ng3 Ka2 26.Nh5 Ka1 27.Nf6
Ka2 28.Ne8 Ka1 29.Nc7 Ka2 30.Na6 Ka1 31.Nb4. Now
we have exactly the same position as in the diagram above,
except that it is Black to move. And that is deadly: 31...g3
32.fxg3 f2 33.Nc2+ bxc2 34.Nxf2 Ka2 35.Kxc2 Ka1 36.g4 Ka2
37.g5 Ka1 38.g6 Ka2 39.g7 Ka1 40.g8Q Ka2 41.Qa8#.
Click
to replay.
Note that the December 31st puzzle is part of our Puzzle
Contest, to which you may now return.
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