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Christmas Puzzle Contest
January 2003
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Solutions
On this page you will find the correct solutions to our
Christmas puzzles, in traditional diagram-and-notation form.
At the bottom of the page there is a link to a Javascript
replay board, where you can also download the positions
and solutions as a PGN file. You will also find a link to
a page with a selection of the letters we received from
our readers.
Prize puzzle 1
T. R. Dawson
White to play
In the above White has wagered that he can lose the game.
But every legal move by both sides leads to a checkmate
of the black king. That was the problem. Many readers came
up with a desperate solution: turn the chessboard around
by 90 degrees. Ah, but apart from the fact that it is not
normal in chess to suddenly switch sides, what about the
notation around the board? White is obviously playing upwards,
towards the top of the diagram.
Another interesting try was the following: "Mr. White
is playing with the black pieces, and Mr. Black with the
white ones. After 1.g7+ Qg7+ 2.hg7#, Mr White loses
the game!" (Uberto Delprato of Italy and many others).
Quite clever, actually, but not the intended (and simpler)
solution.
Some tried bending the rules, e.g. "White deliberately
touches the bishop, is forced to move the piece, and as
a consequence Black's check on the white king stands and
he wins!" (Jaywant Keshav Pai, Mumbai). Read the rules
again: White would have to make a legal move.
The solution is much simpler and really quite straightforward.
In fact a large number of readers got it right. Just read
the critical lines carefully: "White insists: 'I'll
bet you $100 that I can lose this game!' So the two seal
the bet and White actually manages to lose." The solution:
White of course loses the bet, not the game! (Ducking
and running for cover).
Prize puzzle 2
J. Moravec,
La Strategie, 1913
White to play and win
In this beautiful Moravec study White must play the astonishing
1.Kh7! leaving the pawn standing. Now Black has two
obvious alternatives: 1...h4 2.Kg6 h3 3.Kg5 h2 4.Kg4
h1Q 5.Kg3. White wins, since Black cannot prevent the
immediate checkmate without giving up his queen. And now
we see why White had to leave the black pawn standing on
g7. If it is not there Black would have 5...Qh8, protecting
the mating square a1!
We must also consider the defence 4...h1N. White does not
win immediately and must be careful not to let the knight
our of the corner. There are many lines, a possible principle
variation is 5.Kf3 g6 6.Ra5 Kh2 7.Rg5 Kh3 8.Rxg6 Kh4 9.Kg2
Kh5 10.Rg8 Kh6 11.Kxh1 and wins.
What about the other main defence: 1...g5 2.Kg6 g4 3.Kg5!
Incredibly this time White must carefully circle around
the other pawn and leave it standing. 3...g3 4.Kh4 g2
5.Kh3 Kh1 6.Rxg2 and mate in three. This is only possible
because Black has a move with the h-pawn. If White had taken
it on move three Black would be stalemated!
Prize puzzle 3
Samuel Loyd,
Boston Globe, 16.08.1876
White to play and mate in three moves
This vintage Loyd puzzle never ceases to amaze. With so
little material on the board, in such an simple, clear position,
the great American problem composer has managed to show
us how White can mate with knights alone – in four
different variations: 1.e8N+! Kxh8 2.d8N! Ne6 3.Nf7#
(three knights against one); 1...Kh6 2.d8N Ne6 3.Ndf7#
(four knights against one); 1...Kf8 2.d8N Ne6 3.Ng6#
(four against one); or 2...Ke7 3.Ng6# (four against
one). Naturally any other moves or promotions by White fail
to mate in three.
Prize puzzle 4
Frederic Friedel,
Computerschach & Spiele, February 2002
Helpmate in five moves
The first question was: how does Black help White to mate
him (Black) in five moves? The solution is 1.Rc5 h5 2.Re6
h6 3.Kc6 h7 4.Kd6 h8Q 5.Rc6 6.Qd4 mate.
 This
helpmate is valid and valuable because:
1. it produces a pleasing and unsual mating position; and
2. because there is only one unique set of moves that fulfils
the condition and leads to mate in the required number of
moves. There is not a single diviation that is possible,
no change in move order, no transposition. That is what
is required of a proper helpmate.
Now to the second question: originally the author had placed
the black king on b4 (instead of b7), which is in some ways
more harmonious to the solution of the puzzle. Why did he
have to discard this version and instead use the one shown
in the diagram?
 This
helpmate has the solution 1.Rc5 h5 2.Re6 h6 3.Rcc6 h7
4.Kc5 h8Q 5.Kd6 Qd4, just as in the previous version,
with the same final mating position.
But there is also a second solution: 1.Rb5 h5 2.Rhb6
h6 3.Ka3 h7 4.Rb3 h8Q 5.R6b4 Qa1, with the final position
given on the left, which is also an interesting mating position.
Helpmates may have more than one solution as long as these
contitions are fulfilled. So far so good.
 Examining
the position more carefully we find another solution: 1.Ra5
h5 2.Rha6 h6 3.Ra2 h7 4.R6a4 h8Q 5.Ka3 6.Qc3.
That is a third solution with a different mate. Still very
good, since it is a different motif (the famous "epaulette
mate").
All of this has so far only served to make the helpmate
composition more valuable than the original one given above
(with the black king on b7). A helpmate with three different,
interesting mates. However on further examination we discover
that there are a number of deviations possible in some of
the above solutions:
The first mate can also be reached with the moves 1.Re5
h5 2.Rc6 h6 3.Kc5 h7 4.Kd6 h8Q 5.Re6 6.Qd4; and even
in this line we can deviate with 3.Ree6 h7 4.Kc5 h8Q 5.Kd6
Qd4. And on the third mate we can go 1.Ra5 h5 2.Rha6
h6 3.Ra2 h7 4.R6a4 h8Q 5.Ka3 Qc3 with the additional
deviation 4.Ka3 h8Q 5.Ra4 6.Qc3#. Only the second mate has
no deviations or alternative moves.
All this of course ruins a potentially beautiful three
solution helpmate. It is concievable that we could add some
pieces to solve the problems, but then the economy of the
position would be lost. So the best way to go is to sacrifice
the two additional mates and go for the purity of the published
solution.
Prize puzzle 5
K. Hannemann,
Dagens Nyheder, 1933
White to play and mate in 1, 2, 3 and 4
The stipulation of mating in exactly one, two, three and
four moves makes it thankfully very difficult to use computer
assistance with this problem. Except of course in part one,
which Fritz will solve in a small fraction of a millisecond.
For the others you have to use human brain.
1.e8Q#
1.e8R+ Kd7 2.Re7#
1.e8B d5 2.Kc6 dxe4 3.Bd7#
1.e8N Kd7 2.Ng7 d5 3.e5 d4 4.e6#
(or 1...d5 2.Kc6 dxe4 3.Ng7#)
Now isn't that just unbelievable: the author has achieved
an "allumwandlung", all four promotions,
in this problem. In each part White must promote to a different
piece to achieve the goal. There is no alternative, White
cannot play any other moves to solve the problem.
Prize puzzle 6
The background music on the 1st of January puzzle page
(and the current page, incidentally) is an imaginative improvisation
on the famous The Blue Danube by Johann Strauss;
or if I crib from the many correct answers we received I
could also say: it is the 5th Waltz, Opus 314 "An der
schönen blauen Donau", composed in 1867 by Johann
Strauss II or "The younger" (25.10.1825 - 3.6.1899,
Vienna, Austria). "Boy I am I glad I watched '2001:
A Space Odyssey' recently!" wrote Vincent Leslie Carpenter
IV of Lee's Summit, MO, as did a number of other readers
in similar vein.
Finally Arie Haenel of Jerusalem identified the piece most
completely: "I immediately recognized that the tune
was based on it, but it took me quite a long time to figure
out what was the name and the author of this beautiful variation.
The piece is the Arabesques on themes by The Beautiful
Blue Danube by Adolf Schulz-Evler (1852-1952)."
Thanks Arie, now we know everything.
Winners
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1. The general prize.
The copy of our top program Fritz,
signed by a world chess champion, was won by Massimiliano
Benotti of Rome, Italy.
2. The professional prize.
This signed copy of the book "Kasparov
against The World", autographed by Garry
Kasparov, goes to Cornel Pacurar of Mississauga, Ontario,
Canada. Cornel's submission, like a dozen others,
was quite impeccable. His name was selected by lot
from all perfect submissions.
Frederic Friedel
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