| |
 |
ChessBase Christmas Puzzles
December 25, 2006 |
|
| Click
"Stop" on your browser to stop the music
and "Refresh" to start it. |
 John
Nunn's 2006 Christmas puzzles
The 2006 Christmas puzzles, which were provided by mathematician,
chess grandmaster, problemist, author and publisher Dr John
Nunn, apparently contributed to a large number of readers
keeping their minds alert during the festive season. We
feel confident in saying this because of the many hundreds
of letters we received during the Christmas week. All the
puzzles were solved by at least some readers, the final
problem understandably having the smallest number of successful
solutions.
At the bottom of this page we are
providing a fairly random selection of letters – actually
almost always excerpts thereof. But first the impatiently
awaited solutions. Note that there is a replay button at
the end of this section where you can, where appropriate,
replay the solution on a JavaScript board.
Solutions
You were required to find a position with just four pieces
on the board, including the kings, in which White mates
with the move Kc6-d8 mate.
Eric Angelini, Europe Echecs, 1988
This is the only starting position with the given material
that allows 1.Nc6-d8 mate.
John Beasley, 6th HM, The Problemist,
1997
Losing chess – White to play and win
The only winning line is 1.Ne2! and now:
Nenad Petrovic, 1st-2nd Prize, T.T.
Problem 1953-54
This is the starting position that we sought. The game
went 1...f5 2.exf6+ Kxf6 3.d5+ e5 4.dxe6+,
reaching the position of the puzzle:
Milan Velimirovic, 2nd Prize, Mat
Plus, 1997
Selfmate in four moves
Let’s worry about White’s first move later
and suppose that it is Black to play. The key idea is to
block the h8-a1 diagonal with the white rook, while at the
same time hiding the rook behind Black’s rook so that
White cannot interpose on the first rank. The way this works
may be seen in the line 1...Rxa6 2.Rc7+ Rg6 3.Rg7 (only
this square works; the white rook cannot return to the first
rank, and also cuts off the bishop’s attack on a1)
3...Kc1 4 Qc2+ Kxc2#. Black’s three other moves leads
to similar variations: 1...Ra4 2.Rc5+ Re4 3.Re5 Kc1 4.Qc2+
Kxc2#, 1...Rxb5 2.Rc6+ Rf5 3.Rf6 Kc1 4.Qc2+ Kxc2# and 1...Ra3
2.Rc4+ Rd3 3.Rd4 Kc1 4.Qc2+ Kxc2#.
Thus there is a line prepared for all Black’s moves,
and White only needs to make a waiting move to solve the
problem. However, it turns out to be not so easy to find
one. Moving one of the bishops interferes with one of the
lines given above, for example 1 Bg7? Rxa6! and White can
no longer play his rook round to g7, or 1 Be5? Rxb5! and
the e5-square is blocked. The only waiting move is 1 Qf2!
and then everything works as in the lines given above:
1.Qf2 [1.Bg7? Rxa6; 1.Be5? Rxb5] 1...Ra3
[1...Ra4 2.Rc5+ Re4 3.Re5 Kc1 4.Qc2+; 1...Rxb5 2.Rc6+ Rf5
3.Rf6 Kc1 4.Qc2+; 1...Rxa6 2.Rc7+ Rg6 3.Rg7 Kc1 4.Qc2+]
2.Rc4+ Rd3 3.Rd4 Kc1 4.Qc2+ Kxc2 mate.
With the following material:
you were asked to construct a position in which it was
Black to move and he was lost, even though he wasn't in
check. The solution:
Source unknown
Apart from a reflection (with the black king on b8) this
is the only position that satisfies the condition of the
problem.
M. Nesic, 5th Commendation, Schach-Echo
1970
Helpmate in two moves
At first sight it should be easy to mate along the b1-h7
diagonal, for example if White plays Re4, Black plays ...Rg7
and White mates by Rh4# (double check!). All we need is
a waiting move for Black to start with and this works perfectly.
The problem is that there is no waiting move. All Black’s
moves either disrupt the mate or check the white king. The
solution is very surprising as it involves a totally different
mating pattern in which White switches diagonals with his
bishop. Black plays ...Qa8, White replies Bb3, Black plays
...Rh6 and White mates by Rc7#. The move ...Qa8 is necessary
as this is the only way to clear the queen out of the way
so that it does not interfere with the mate.
So the solution is: 1.Qa8 Bb3 2.Rh6 Rc7 mate.
A very simple question: construct a legal game that
ends with (the black move) 5...Rh1 mate. Note that
the last move is not a capture.
Indeed, the question is simple, but the answer quite hard.
Unfortunately we have misplaced the solution. Somehow we
seem to recall that a special move – en passant or
castling – was involved. And that the sequence of
moves was entirely unique, i.e. each move had to be played
in exactly the same order.
So please continue to search for the solution – there
are just five moves to find. If you become really desperate
(but not before February 1st 2007) you can write to us,
using the feedback button in the left navigation. In really
urgent cases we will try to dig out the answer and send
it to you.
Puzzle
8 – January 1st, 2007
Andrei Kornilov, commended, Thèmes-64
1985-86
The problem was, knowing only that the position is legal,
to work out which pieces were white and which black. One
can even deduce which player was to move.
This is quite a challenging puzzle, if only because you
need two chess sets to put up the position. The solution
is somewhat daunting, but fascinating to go through carefully,
step by step. The dizziness you will feel at the end is
not at all unpleasant. So here goes:
-
First of all, the king on g6 can’t be black (when
the king on g4 would be white), because then the pawns
on f5 and h5 would have to be the same colour (or else
both kings would be in check) and then one or other
king would be in an impossible double check.
-
Thus the king on g6 is white and the king on g4 black.
Then the pawns on f7 and h7 must be white. Look at the
pawns on the g- and h-files. There must have been at
least one pawn capture here (gxh). Now look at the pawns
on the c- to f-files. Although we don’t know which
are white and which black, we know that originally there
were only two pawns on each file. Thus these pawns must
have made at least 14 captures. Together with gxh this
makes 15 captures, exactly balancing the number of missing
pieces. Thus there can be ‘extra’ pawn captures
and in particular all the pawn captures must have been
from left to right.
-
Looking at the diagram, Black’s a-pawn cannot
still be on the board. Although it could reach b6 or
f2 by capturing, for example, it cannot reach any of
the currently occupied squares. Thus Black’s a-pawn
must have promoted, with the promoted piece later being
captured by a white pawn. It follows that of the 15
pawns on the board, 8 are white and 7 black. White’s
pawns have made a total of 8 captures and Black’s
pawns have made 7 captures.
-
We know that the pawns on the g- and h-files can only
have made one capture between them. Thus the pawns on
h3 and h5 can’t both be white (since the tripled
h-file pawns would need more than one capture) and they
can’t both be black (because then either White
and Black both played gxh, or Black played ...gxh twice).
Thus there are two white pawns on the h-file and one
must have originally come from the g-file. It follows
that the g7-pawn is black otherwise there would have
been doubled g-pawns, necessitating an extra pawn capture.
Now we can see that the h5-pawn is black (if the h3-pawn
were black, white would have to play gxh twice to get
the pawns to h5 and h7 behind the h3-pawn) and so the
h3-pawn is white. At least now we know that Black is
in check in the diagram, and therefore it is his turn
to move.
-
The f3-pawn must be black or else the black king would
be in an illegal double check. The pawn on e7 must be
white, because if it were black the f8-bishop could
never have escaped from its original square to be captured
by a white pawn.
-
Now we need to look at the white pawn captures more
closely. There are eight of these, including gxh7, so
the white pawns on the c- to f- must have made seven
captures, all from left to right. The more the white
pawns are weighted to the left of this bunch, the fewer
captures are needed for them to reach their present
positions. The pawn on c7 cannot be black, because then
the fewest white pawn captures would be required when
White has two d-pawns, three e-pawns and the f7-pawn
(which we already know to be white); but this needs
eight pawn captures, which is too many, so c7 is white.
Now suppose that one d-pawn is white and one black (both
can’t be black as this again requires too many
white pawn captures). If white then has three e-pawns
and one f-pawn, the e-pawns require all seven captures
between them, leaving none for the f7-pawn. Thus this
pawn must have advanced to f7 and all three black f-pawns
captured to move behind it. However, this takes too
many black pawn captures. Thus both d-pawns are white.
-
We are left only with the cluster of pawns on e4, e5,
e6, f5 and f6, of which just one is white. The white
pawn can’t be on the e-file, as then the black
pawns would require eight captures, so the white pawn
is at f5 or f6. It can’t be at f6, as this requires
an extra capture to get the f6-pawn behind Black’s
f5-pawn. Thus f5 is white and we have the solution:
the white pawns are the ones at c7, d7, e7, f7, h7,
d6, f5 and h3 and the white king is at g6.
So this is the correct (legal) position:
In case you are doubtful that the position is legal at
all, here’s how in might arise in a ‘game’:
1.g4 Nc6 2.f4 Nd4 3.Kf2 Nb3 4.axb3 a5 5.Ra4 h5 6.Rc4 a4
7.Kf3 f6 8.g5 Rh7 9.g6 a3 10.gxh7 Kf7 11.Nc3 a2 12.Rc6 a1Q
13.Ke3 bxc6 14.Bg2 Kg6 15.Bd5 Kf5 16.Nf3 Kg4 17.Ke4 Kh3
18.Kf5 Qa4 19.Kg6 cxd5 20.Ne4 dxe4 21.Rg1 exf3 22.Rg5 Qc4
23.bxc4 Ra6 24.Rd5 Kg4 25.Rd6 cxd6 26.e4 Nh6 27.e5 Rc6 28.e6
Nf7 29.exf7 d5 30.Qe2 Qc7 31.Qe4 dxe4 32.d4 e6 33.d5 Be7
34.d6 Bb7 35.dxe7 d6 36.b4 Rc5 37.Bb2 Kh3 38.Be5 dxe5 39.bxc5
Bd5 40.cxd5 Kg4 41.c6 Qd7 42.cxd7 Kh3 43.d6 Kg4 44.f5 Kh3
45.c3 Kg4 46.c4 Kh3 47.c5 Kg4 48.c6 Kh3 49.c7 Kg4 50.h3+.
Doubtless there are many shorter methods.
Click
to replay all problems on our JavaScript board
Feedback from our readers
Bjorn Verstraate & Bjorn Verstraate, Spijkenisse,
Netherlands
Me and a friend just saw your puzzle page and we decided
to give it a go. Amazed by position 2, and incredibly happy
when we finally solved it.
Ling Feng Ye, Montreal, Canada
Thanks a lot for the puzzles. I found the 3rd one particularly
interesting. It took me quite a while to find the idea,
which involves two en passant captures. Keep up the good
work!
Ben Rodriguez, New Jersey, USA
I do not like this kind of puzzles at all. In my opinion
they do not contribute in anything to the develop of young
players, are boring to amateurs like me and who know what
professional players can say about them. You should publish
examples of puzzles where the setting is clear and the pieces
are in place. I do not think that in a tournament my oponent
is going to let me imagine what will happens in a position
where half of the pieces or some of them are not on the
board. It is stupy as well the selfmate puzzles. I would
like to know if Dr. Nunn practice this kind of stupy selfmate
in tournaments, because that will be a great opportunity
for me to get my first GM norm. There are many beutiful
thing in and about chess and one of them is reality and
logic. Get real and logic, stop the nonsense.
Parag Samant, Mumbai, India
Thanks for the great fun that your series provides us every
year!
Paul Leyderman, Chicago, IL
Thank you so much for the holiday puzzles!
Hui Shao, Beijing, China
Due to the seequake near Taiwai, the Internet connection
from mainland China to Europe was lost yesterday. It nearly
kept me awake until midnight cause I really wanted to see
puzzle 3. Luckily, now I can see it and puzzle 4! Puzzle
3 makes me think about becoming a detective. Thank you very
much.
Gerson Berlinger, Bad Friedrichshall, Germany
I just solved today's chess problem (like the three before).
A very nice composition! This year for the first time I
used a chess board for solving, though I didn't really need
it. A few moments after having set up the position, everything
was clear to me. Strange, eh?
JP de Ruiter, Nijmegen, Netherlands
I'm a researcher in the field of Cognitive Science from
the Netherlands. John Nunn is really an evil genius when
it comes to chess problems. I solved number three relatively
quickly with my two main rules for solving retrograde analysis:
Rule 1: If it looks impossible, it is an en-passant capture.
Rule 2: If it still looks impossible, it is another en-passant
capture.
Piotr Waniurski, Lublin, Poland
Fantastic Puzzle #5. I spent a few hours and found two symmetrical
solutions (I suppose they are unique).
Thomas Rehmeier, Jefferson City, Missouri, USA
Thank you for putting these puzzles online it has brought
much fun. I really enjoyed Puzzle 7, after a little while
I came up with a solution. [Correct solution given].
Marcel, Hug, Herrliberg, Zürich Switzerland
Number 7 is another nice one. [Correct solution given]
Arie, Haenel, Jerusalem, Israel
Magic! Here is the soluion. [Correct solution given]
Roberto Stelling, Rio de Janeiro, Brazil
Puzzle #7 is really cool. [Correct solution given] I've
done that without a board to avoid any optical illusions.
Cheers and wish you all a Puzzling New Year. A special praise
from this Brazilian fan goes to the the double GM (over
the board game and problem solving) John Nunn.
Andy Jenkins, Stockport, UK
I fondly remember the "different" knights that
were in fact the same shade of grey. The coloured "Rubic's
cube" is even better. I used the Photoshop "ink
dropper" tool to prove to my daughter that the "brown"
and "yellow" squares were the same. I'm less impressed
by "dimples or pimples". This technique has been
used in graphical operating systems for years to give the
appearance of buttons that go in or come out. I think we'd
be in trouble if our brains didn't work that way.
Nathan Claus, Indianapolis, IN
Why would the optical illusions cause us to 'lose all faith
in our visual senses'? Our mind and senses are constructed
to perceive reality, not fantasy flash animations that don't
obey physical law. In actuality, I believe it's a triumph
for human beings to be able to analyse something beyond
their physical senses and come to the conclusion that their
senses are lying.
Gerrie Bloothoofd, Holland
I love your incredible section on puzzles this year. John
Nunn's puzzle number 7 is especially inspiring. It starts
to make you more creative. You start to see amazing mates,
but always with one move too many, or with a capture, or
with the checkmate on g1. However, Nunn's 5...Rh1# is unfindable.
Incredible. I'm looking forward to the answer of this.
Yigal Nochomovitz, Berkeley, CA
The solution to the 5. ... Rh1# is nice. [Correct solution
given]
Dr. Manikumar, Chennai, India
Puzzle #7 [Correct solution given. Puzzle #8 is an excellent
retro analysis. Simply superb. Really made me happy that
finally I could find it. [Slightly flawed solution given].
Thanks a lot for the entertaining puzzles. Nothing makes
me more happy than such problems...
David Messham, Congleton, UK
The "help" for the December 30th helpmate was
sneaky, reminding us of the Sam Loyd problem with its double
check solution. This just drew my attention to the fact
that in the Dec 30th problem as set, if it were White to
play there would be a similar solution (1.Re4 Rg7 2.Rh4#).
And this is no help whatsoever in finding the real solution.
No wonder it's taken me three days to solve.
Albert Frank, Brussels, Belgium
Puzzle 7: [Correct solution given]. Beautiful!
Roberto Stelling, Rio de Janeiro, Brazil
Loved puzzle 8 Figuring out the kings colors and pawns f7
and h7 colors was trivial, almost as easy as figuring out
which side had 8 and which had 7 pawns. Then the real task
begun. Using capture counts it was not hard to proof that
c7 had to be white and that d6 and d7 couldn't be both black.
After some more pawn tossing I realized that both d6 and
d7 had to be white, otherwise there was no way of white
pawns reaching f7 and h7. (on an almost possible scheme,
capture count wise, black would end up with pawns on g7
and e7, leaving white with one less capture).
Then the only possible way has having d7 and d6 white with
two white captures to go: exf7 and gxh7. [Full solution
and a shorter proof game follows]. Cheers and thanks for
a great solving week! Too bad the other 51 weeks of the
year are not as tasty as this one!
Jacques Labelle, Math professor, Montreal, Canada
Problem #7 [Correct solution given]. A fantastic problem!
My nephew Francois Labelle (Computer science at Berkeley)
found and showed me that problem several (3 or 4) years
ago; even so it took me 30 minutes to remember the solution.
His computer took initialy 10 days to discover it. Francois
is an expert on that kind of problems: chess games uniquely
determined by there last move.
Marc Boulé, Montréal, Canada
Once again thank you for the very interesting Christmas
chess puzzles for what is becoming a great holiday season
tradition. Mr. Nunn has outdone himself, and has me scratching
my head not only when trying to solve the problems, but
also for wondering how he comes up with them to begin with.
I believe I have solved all eight, and I will be eagerly
awaiting the solutions to see how well (or bad) I did. I
very much enjoyed puzzle number 7, which brought back good
memories of a certain 1999 Christmas puzzle... Being half
programmer and half chess enthusiast, I resorted to computer
assistance for solving puzzles 4 to 8. For me, trying to
figure-out algorithms to make such puzzles computable in
short time by my machine is as much fun as good old fashion
brain solving. I tip my hat to the real puzzlers who solve
these things without any outside assistance.
Ibrahim Shehab, Kuwait
Waiting for the solutions!
Joshua Green, Princeton, NJ, USA
Thanks for all the puzzles (especially the last one!). I
would call this a "color the pieces problem" rather
than a "monochromatic puzzle" since the latter
is something completely different (but is also sometimes
used in retrograde problems). This problem is quite interesting,
and it's nice that a few subtleties (trapping the Black
dark-squared Bishop on f8 in one setup, getting pawns around
each other in others) are necessary to unravel the position
– simply counting captures isn't good enough! [Correct
solution and proof game given]
Morgon Mills, Toronto, Canada
I only attempted one puzzle this year, and found the solution
after about 45 minutes. My first attempt to construct a
game ending in 5...Rh1# was 1.e3 b6 2.Bd3 Ba6 3.Bg6 hxg6
4.Ne2 Rxh2 5.Kf1 Rxh1#, but then I realized that the final
move could not be a capture. Finally I hit upon the right
idea and after some tinkering with move orders quickly discovered
the solution. [Correct solution given]
Zalmen Kornin, Curitiba, Brazil
A smoothly assorted smorgasbord of chessical hors-d'oeuvre
(if I'm expressing myself properly). We demand no prize
for hearing good music, or reading good poetry, so chess
problems are also enjoyable without payment. A message for
Frederic Friedel: it was really spectacular to see again
the Dover cover for the Loyd book. Mine disapeared in the
seventies, while the book itself is now hard-covered in
Black, and still amongst my favorites.
Joost de Heer, Geleen, The Netherlands
In the description of 'losing chess' in the puzzle of Dec
26, John Nunn wrote: "To clarify the rules of losing
chess: if more than one capture is possible, the side to
play can choose which one to make. A player who promotes
a pawn may choose which piece to promote to, just as in
normal chess. You cannot promote to a king." This isn't
entirely correct. In losing chess, the king is just an ordinary
piece, and you can promote to king, and possibly have multiple
kings on the board.
Christopher Wong, Douglaston, New York
Puzzle 7, December 31: [Correct solution given]. After lots
of tries I accidentally found the solution. Wow. And for
the record, I consider myself gosh awful at solving chess
problems in general.
Bob Banerian, Milwaukee, WI, USA
Nice puzzles this year - nice and tough. Has John Nunn ever
tried to tackle the knight excelsior problem?
Karthik Samuthiram, San Diego, CA, USA
Puzzle #8: [Correct solution and logic given]. I wrote a
program and fed in certain constraints. I did not get an
unique solution. I had add more constraints before I got
an unique solution.
Kiyoshi Takahashi, Orsay, France
Hello, hello!! Thank you for the amazing puzzles. I like
#7 very much. Happy new year and see you next year!
David Messham, Congleton Cheshire
Solutions please! More than a week has passed. In particular,
I really want the solution to Dec 29, which I can neither
solve nor forget. Go on - email it to me if you aren't going
to post it on the web site!
Stork Chen, Nijmegen, Netherlands
This year we really had some great Christmas puzzles. I
could only solve half of them. Great. But are you going
to enlighten us with the solutions? If I remember correctly,
you said they would be published last week. I'm dying to
see them!
A
selection of earlier letters was given here.
|
