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Solutions to our
2012 Christmas Puzzles – Part 2
By John Nunn |
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December 30, 2012 – Serieshelpmate
A. Ivunin, 2nd Prize, Redkie shanry
plus 1996
Serieshelpmate in 34 moves
The first step in such problems is to try to identify the
mating position. Black can easily take the b4- and d5-pawns
and then promote all four pawns, so he has five pieces available
for blocking squares. However, none of them can be a dark-squared
bishop, so a mate with the king on h1 and bishops on g1
and h2, for example, is impossible. This basically rules
out all mates in which the white king does not participate,
since there is no mating position possible without a dark-squared
bishop. Note that the black king cannot end up on f5, because
in addition to the five blocking pieces on g4, f4, e5, e6
and f6, there would have to be one on g6 for White to take
with his final move. Therefore there is only one possibility:
the black king must be on f7, with mating move Bg6 and blocking
pieces on f6,e6, e7, f8 and g8. The f6-piece must be a knight,
the g8-piece must be a bishop and the f8-piece must be a
rook. There is more flexibility with the pieces on e6 and
e7.
Thus far it has been easy, but now some choices have to
be made. It takes 18 moves to promote all the black pawns
and four moves for the king to reach f7, leaving 12 moves
to take the pawns on b4 and d5 and move the black pieces
into position. To simplify matters, let’s assume that the
knight on e6 stays unmoved throughout the solution. Black’s
main problem is how to take the white pawns without wasting
too much time. It’s tempting to promote to a bishop on h1
(a queen is impossible as this would give check) and use
that to take on d5, but there are two problems with this.
Firstly, the bishop cannot move to g8 without wasting two
moves playing the knight away from e6 and back again (and
why should the knight move to one square rather than another?)
and secondly there is then no natural way to take the b4-pawn
without further loss of time.
Therefore it makes sense to promote to a knight on h1.
Then by moving to f2 and d3, the two white pawns can be
taken one after the other. The knight can then move from
d5 to its final square on f6. This takes five moves, so
now we are left with seven moves to get the remaining three
pieces into position. The main difficulty is getting the
bishop to g8 and since this takes four moves from d1 but
only three from b1 (via g6 and f7) we can assume that the
b-pawn will promote to a bishop. Now we are down to four
moves to get the two pieces created at d1 into position,
which sounds feasible. Let’s see how the move-order works
out: 1...Kg4 (releasing the h-pawn, which is the
first black pawn to promote; moving the king to g3 would
cost a move) 2...h4 3...h3 4...h2 5...h1N 6...Nf2 7...Nd3
8...Nxb4 9...Nxd5 (Black cannot play ...Nf6 until his
king has crossed that square to reach f7, so for the moment
the d-pawns are blocked; the next step is for Black to get
his bishop to g6 to allow the black king to cross the f5-square)
10...b4 11...b3 12...b2 13...b1B 14...Bg6 15...Kf5 16...Kf6
(the bishop has done its shielding duty on g6 and must
now transfer to g8 to allow Black’s king to occupy f7) 17...Bf7
18...Bg8 19...Kf7 20...Nf6 (at long last freeing the
d-pawns) 21...d5 22...d4 23...d3 24...d2 25...d1Q (promoting
to a queen and then moving to e7 is the only way to get
to a suitable square in two moves) 26...Qd6 27...Qe7
28...d5 29...d4 30...d3 31...d2 32...d1R 33...Rd8 34...Rf8
and finally White gets to play Bg6#. Each of
the four black pawns promoted a different piece.
Click to
replay
December 31, 2011 – Proof game
B. Gräfrath, Die Schwalbe
2009 – Dedicated to Cedric Lytton
This is the position after Black’s 12th move.
What was the game?
This is the toughest puzzle of the 2011 set and requires
considerable imagination to solve. In the diagram Black
has definitely made the six moves ...d5-d4, ...e6, ...Bd6,
...Ne7 and ...0-0 (not necessarily in this order) and some
moves with his queen. There are six white pieces missing.
Now it’s theoretically possible that a move such as ...Bd6
could be a capture, but this looks unlikely as White would
have to spend three moves getting a knight to d6 to be taken.
Instead, it appears much more likely that all the missing
white pieces were taken by Black’s queen. This means that
every black queen move must have been a capture.
At first sight, White has several spare moves as it seems
that the only moves he has definitely played are e3, f3,
0-0-0 and Rhf1, leaving him with eight spare moves which
are presumably going to be used to line his pieces up to
be taken by the black queen. However, if we note that the
c1-bishop cannot have moved and must have been captured
on its original square, then its pretty clear that the black
queen must have been extremely agile to get in and out of
c1. A possible scenario is that the black queen took the
pawn on a2, then the knight on b1 and then the bishop, before
coming out again. However, with every queen move having
to be a capture, what could it have taken on b1 and a2 on
the way back?
On the other hand, if the queen came in via d1 and c1,
then it must have checked the white king, making it impossible
to castle. Something doesn’t add up here and this is where
imagination comes in. Perhaps, after all, White did not
castle. That makes the black queen’s task much easier, as
it could take on d1, c1, b1 and a2, all captures, thus allowing
the a1-rook to move. The rook couldn’t move directly to
d1, as the king has to be allowed to reach c1, but it could
go to e1 and then back to d1. So we could imagine White
playing e3, f3, Kf2-g3 to allow Black’s queen to move along
White’s first rank, then Rae1, Kf2-e2-d1-c1 and Rd1. That’s
ten moves; adding in one for Rhf1 leaves one spare.
This presents a new problem. If all Black’s queen moves
are captures, then the queen’s first move, which could be
...Qd6, ...Qd5, ...Qg5 or ...Qh4 (for example) must have
been a capture. But no white piece can reach one of these
squares in one move, so we have run out of white moves.
Evidently the king manoeuvre takes too long. You can try
only moving the king to f2, but this also doesn’t work because
then the black queen cannot take on f1 or g1 without forcing
the king to move again, and moving these pieces out takes
too much time. Second leap of the imagination: suppose White
castles kingside. Now the white moves are 0-0, Rae1,
Kf2-e2-d1-c1 and Rd1. This saves four moves over the previous
route, since the king takes three fewer moves, and one more
is saved because the rook is moved from h1 to f1 without
loss of time.
Now things start to look promising, since if we add in
e3 and f3, that’s only nine moves, leaving three to get
White’s pieces into position for the black queen captures.
The most natural way to use these moves is by Bc4-d5 and
Nf3, freeing White for kingside castling and allowing ...Qxd5
and ...Qxf3, when the queen is perfectly positioned to take
on d1 to start its meal of White’s queenside forces. After
this it’s really only a question of move-order: 1 e3
(White must start with this as his bishop must be on
d5 by move three) 1...d5 2 Bc4 d4 3 Bd5 Qxd5 4 Nf3 Qxf3
5 0-0 (just in time to prevent White’s king being checked
when the enemy queen arrives on d1) 5...Qxd1 6 f3 (White
must wait for his a1-rook to be freed, and he uses the tempi
to edge his king nearer c1) 6...Qxc1 7 Kf2 Qxb1 8 Ke2
Qxa2 (Black’s queen is in position, and while White
is arranging his forces, Black plays the remaining moves
with his kingside pieces) 9 Rae1 e6 10 Kd1 Bd6 11 Kc1
Ne7 12 Rd1 0-0 reaching the required position.
It’s remarkable that despite White’s king and rook signalling
a possible queenside castling, the only way to reach the
position is to castle on the opposite side of the board.
Click to
replay
Click
to download all seven problems in PGN
January 1, 2012 – Benko's letter problems
In each of the following positions it is
White to play and mate in three.
1. Rd8 Kxf7 (1... Kxh6 2. f8=Q+ Kg6 3.
Qxf6#)
2. Rd7+ Ke6 (2... Kg6 3. Nf8#) 3. Nf8#
1-0
1. Bd4 Bh4 (1... Bh3 2. Bxf2 Bg2
3. Re1#)
2. Re1+ Bxe1 3. Nc1# 1-0 |
1. Rg7 Ke6 (1... Kf5 2. Rg5+ Ke6
3. Re7#)
2. Rde7+ Kf5 3. Rg5# 1-0 |
| |
|
1. Bf5 Bg6 2. Nd7+ Kf7 3. Be6# 1-0
|
1. Nab6 Kb5 2. Nc8+ Kxa6
(2... Kc6 3. Ne7#) 3. Rb6# 1-0 |
| |
|
1. Rd8 Kf5 2. Rg8 Ke5 (2... e5 3.
Qf3#)
(2... Nc3 3. Rg5#) 3. Rg5# 1-0 |
1. Nb1 Kc4 (1... e4 2.e3+ Kc4 3.Na3#)
(1...c5 2. e3+ Kc4 3. Na3#) 2. Na3+ Kd4 3.
e3# 1-0 |
Click to
replay or download
the PGN
Solution to the logical puzzle
[A reminder: you meet three people, a liar, a truth
teller and an idiot. The latter answers yes or no completely
at random, whatever you ask him. You have two yes-no questions
to find the path to the village.]
There are various solutions to the puzzle. The key point
is to ensure that the second question is not directed to
the idiot and the universal
question ("What would you say if I asked you which
was the way to the village?") works. So you ask one
of the men, while pointing to one of the others: "Of
those two, is he more likely to answer a question correctly?"
If the answer is yes, ask the universal question to the
person you didn't point to; if the answer is no, ask the
question to the person you did point to.
Lars Rasmussen, an old friend from Denmark and one of the
early pioneers of endgame databases, sent us the following
analysis:
Let's call the three men A, B and C. Direct the first
question at A: "If I ask you whether B is unpredictable,
will you say yes?" If the answer is yes, direct the
second question at C, otherwise ask B. The second question
is: "If I ask you whether this road leads to the
village, will you say yes?"
I need to emphasize that one of the three men needs to
be a random generator – the beauty of the problem
depends on it! Otherwise there would be no need to ask
two questions. A random generator is a much more troublesome
person because you can extract exactly zero bits of information
from him, like flipping a coin or looking at a horoscope.
But that is exactly the beauty of the problem. You have
no way to avoid asking the first question to a person
who will give you zero bits of information, but after
you have heard the answer you know something you did not
know before. That sounds paradoxical, but in reality you
will only hit the random generator 1/3 of the time, so
on average you will gain 2/3 bits of information, just
enough to identify a reliable person (truth teller or
liar) for the second question.
Incidentally I remember visiting you [Frederic Friedel]
21 years ago. I still have a copy of a magazine with a
picture of a two-board simul in which John crushed Matthias
and me. I played the Marshall, to which you commented:
"Don't you know that John is a leading expert in
this opening?"
Lars Rasmussen and the first five-piece ending for PCs
Looking back at the article (in German) we see that Lars
had visited us in 1990 to show us some pioneer work he had
done in endgame databases. He had got hold of some state-of-the-art
hardware: an 80386 computer with 4 MB of onboard RAM and
a super-fast – now hang on to your hats – 638
MB hard drive! Using that he had calculated all positions
that can occur in the five-piece ending rook and bishop
vs rook. After eliminating all symmetries he ended up with
115,500,000 positions, which occupied 130 MB on the hard
drive.
Naturally nobody else could use such gigantic files, so
Lars started compressing the data, and actually got it down
to a size – 937 KBytes – that would fit on a
standard floppy disk. How did he do it? By strategically
storing a very small number of key positions, and using
clever algorithms to reconstruct the missing positions from
them on the fly.
In 1990 Lars Rasmussen (left) showed Matthias
Wüllenweber of ChessBase and
John Nunn his endgame program. Later John played a simul
against both.
Lars showed John Nunn the program, and he immediately got
to work checking the standard endgame literature with Lars'
program. After some hours he reported that there were "many
errors." "In the program," Lars asked anxiously?
"No, in classical endgame theory," John replied.
The result of the encounter was that ChessBase published
the Rasmussen program, and with it the very first five-piece
ending that was accessible to chess fans on their home computers
(a previous five-piece endgame, two bishops vs knight, had
been calculated by German programmer Hans Zellner, but had
to be stored on 59 floppy disks). The program ran on MS-DOS
computers ("PC, XT, AT, 386" the specs said) and
could be delivered on a 5.25" (1.2 MB) or 3.5"
(1.44 MB) floppy. It was able to run directly from the diskette,
thus saving valuable hard disk space.
The above screen shot in the German magazine shows a postion
in which Lars' program announces a win in 59 moves, starting
with 1.Bf5 – every other legal move only draws. We
tested the program against GM visitors ranging on the Elo
scale between 2500 and 2800, and none was able to win against
the computer in difficult positions. John Nunn was so frustrated
by his failed attempts that he decided to calm his nerves
by walloping the two programmers shown above in simultaneous
five-minute games. After that the world was restored to
its natural order for him.
Frederic Friedel
Prize winners
The four prize winners are
- Bojan Basic Serbia
- Karthik Samuthiram USA
- Ryo Shiomi Japan
- Omer Friedland Israel
who will kindly send us their postal addresses. These readers,
like a number of others, got every puzzle more or less correct
and were chosen from all entries that were similarly accurate
using a random number generator provided by Random.org.
You can try it out yourself using the widget on the right. |
