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Solutions to our
2007 Christmas Puzzles |
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December 25, 2007 – The Missing Kings
Add the two missing kings in such a
way that White,
who is on the move, can deliver immediate mate, i.e. mate
in one move.
A number of readers (and some good friends in the chess
world) gleefully sent us a solution with the white king
on f6 and the black king on h8. Doesn't work, as Bryan Lamb,
a chessmaster from Canada, shows in his analysis:
First of all we know that with only two bishops, there
is only one option available for the black king to be
mated, that it has to be placed in one of the four corners.
Let us take each in turn:
- Black king on a1: the only mating move from a white
king is Kc3-c2, which does not work because the black
queen guards c2.
- Black king on a8: The only mating move possible here
is Kc6-b6, which does not work because the black queen
guards b6.
- Black king on h8: the only move possible here is Kf6-g6.
This is more promising because it is a legal move but
it is not mate in 1 because the black queen blocks on
f6.
- Black king on h1 (the only solution possible). Since
the white king’s correct final position happens
to be f2 it is clear that the White king must have started
on f3. [If it was on g3, e3, e2, e1 or f1 the black
king would have been in illegal check]. Thus, on the
move prior to the starting position White’s king
is on f3. Black plays Qf2+ for reasons we do not understand
and white then plays Kxf2 #. This problem may be difficult
to see psychologically because of the placing of the
white king on a square right near the black queen.
So this is the required position, and
White plays 1.Kxf2#
December 26, 2007 – Serial moves to selfsacrifice
E. Bartel and A.H. Kniest, 1965
Series helpmate in four – a: diagram; b: bKg1–>h1
(Black moves four times in a row and then White plays one
move to mate him)
Noam D. Elkies and others drew our attention to the fact
that this problem has a "twin", with the black
king at h1 instead of g1. So Black moves four times in a
row and then White plays one move to mate him. In one case
the black king is on g1, as in the diagram above, in the
twin problem the black king is on h1. The solutions are
quite different.
Bryan Lamb wrote:
We can logically deduce from the pieces present on the
board that in both problems the final position must have
the black king on h1, and the rook delivering mate on
either the rank or the file.
In the diagram position it is the file. This means the
pawn on g2 stays put and the h2 pawn promotes and acts
as a blocking piece on g1. The promoted black piece can
only be a rook (it is the only piece which is not capable
of blocking a check on the h file from g1). So we get
the solution 1.h1R 2.Kh2 3.Rg1 4.Kh1 then
Rh4#.
In the twin position (with the black king on h1) the
mate is on the rank. This time the h2 pawn is a block,
and the g2 pawn promotes and returns to g2 to act as a
blocking piece. This piece can only be a knight, as any
other piece will block on g1 or capture on f1. So the
solution is 1.g1N 2.Nf3 3.Nh4 4.Ng2 then
Rf1#.
The comment that this solution depends on the specific
positions of the king and rook means that there is no
other way to bring the knight from g1 to g2 without it
either checking the king on c2 (via e1) or moving through
the rook’s square (such as e2/h3/f4).
J.M. Rice, 1970
Series helpmate in eight
Black moves eight times in a row and then White plays one
move to mate him.
Bryan Lamb's train of thought:
We need the white rook to checkmate along the first rank.
Since it is already on the first rank, it will have to
capture a black piece to mate.
Black’s king should remain on a1 since otherwise
we need three black pieces to block its escape to the
second rank. That is impossible. If the black king goes
to b1 this takes one move. It takes a move to get a rook
to b2, and at least two more moves to get either a bishop
or a second rook to a2. that is 4 moves so far. Since
the only way this can happen also necessitates getting
the knight to c2 (which takes four more moves, with none
left) and allowing the captured piece to be the rook on
d1, we have used up all eight moves to get the pieces
in position, except that there is no time to move the
bishop on g1 away, or to block the diagonal to white’s
king to allow the f2 rook to move.
Therefore, black’s king must stay on a1. A rook
must block it on a2, and a knight must block it on b2.
The white rook must capture a black piece on c1. So the
8 necessary moves are
1. move a rook to a2
2 to 4. moving the knight from h3 to b2
5. block the a7-g1 diagonal to allow the f2 rook to move
6. move the bishop away from g1
7. place a black piece on c1, and last but not least...
8. place a piece on the a-file to shield the white king
from check from the black rook which must move to a2.
This outlines all the necessary moves. Now all we need
is the correct order. Clearly, the only piece which can
block the a file in one move is the bishop, so 1.Ba6
Now we can’t move the rook to a2 yet without first
blocking the a7-g1 diagonal (to prevent an illegal check
by the Bg1), therefore 2. Rd4
Now it is possible to move the f2 rook, which has to
precede the knight maneuver and the bishop leaving g1.
So 3. Ra2
The bishop on g1 cannot move until it is unpinned by
the knight getting to d1 so 4.Nf2, 5.Nd1
Now the bishop can place itself on c1 to free the knight:
6.Be3 7.Bc1
And last but not least 8.Nb2, followed
by Rxc1# by White.
These moves cannot be interchanged because of self interferences
or because of checks to the white or black king. This
is why the white king is placed on a7. Its only role there
is to eliminate duals.
December 27, 2007 – Difficult or easy, that
is the question
Jacob Mintz, Prize, Ideal-Mate
Review 2000
Series helpmate in 21
Black moves twenty-one times in a row and then White
plays one move to mate him.
A fair number of solvers, amongst them a very, very
strong grandmaster, suspected that we had made an error
and that it was a series helpmate in 22. Bryan Lamb explains
why they were wrong:
The explanation given as Nunn’s as to why the Nc7
mate solution does not work is not sufficient. it doesn't
work, that is true, but there is a solution in 22 moves
which is quite similar to the real solution, and it is
instructive to see both.
Clearly if the final move is to be Nc7 mate, a black
rook must be on b6 because a knight will check the king
and a queen or bishop can capture on c7. Black must also
have a blocking piece on a7, which can only be a bishop
for similar reasoning. Black’s third pawn must also
promote and block on a5. This piece could be anything
but is most likely a queen which would get to a5 more
quickly.
We can now consider whether this is a possible solution
that can be arrived in 21 moves. I tried to do it, but
could only get such a position in 22 moves. For example
(trial attempt for the Nc7 solution): 1.a4 2.a3 3.a2 4.a1R
5.Rg1 6.Rxg4 7.Rg6 8.Rxb6 9.g5 10.g4 11.g3 12.g2 13.g1Q
14.Qxe3 15.Qxc3 16.Qa5 17.c3 18.c2 19.c1B 20.Bf4 21.Bb8
22.Bxa7, followed by Nc7 mate. This takes 22 moves and
thus is not the solution.
Attempting to accomplish this position is a shorter number
of moves won’t work as the minimum number of moves
to get each promoted piece to the correct square without
checking white’s king is already in effect. Therefore
we look for an alternate solution. The only other way
white can check Black in one move is axb8N. This is mate
as long as Black blocks a5 and b5 with two pieces and
gives up a third promoted pawn on b8.
Most of the ideas in the correct solution are similar
to the first attempt, such as getting a queen to a5 and
the motion of the promoted rook (it ends up on b5 instead
of b6 but in other respects it does the same things along
the way). There is one key difference which makes this
solution work. In the correct solution, since black gives
up a promoted piece on b8, it needs one less move than
it would to get to a7 and act as a block. In all other
ways the number of required moves for the other parts
of the problem are the same.
Each move of the solution is unique. The ones which are
not obvious are given with explanation.
1.a4 2.a3 3.a2 4.a1R. It can’t
be a queen, which would capture on g4 with check. A knight
takes too long to move around to capture or to block.
It can’t be a bishop because in the possible continuation
5. Bxc3 6. Be5, Black’s c-pawn has no correct piece
to promote to since a queen or rook will check the white
king and a second bishop won’t be able to capture
the g4-pawn, and a knight is too slow. 5.Rg1 6.Rxg4
7.Rg5 8.Rb5 (the rook must block on b5 because
a5 must be left for black’s other piece which is
coming from c3) 9.g5 (this pawn couldn’t
move until after Black’s previous move) 10.g4
11.g3 12.g2 13.g1Q (since the subsequent three
moves require horizontal and diagonal moves,
only a queen will work here) 14.Qxe3
(the pawn has to be taken in order to allow the future
promoted bishop to get to f4. this protects the uniqueness
of the solution by giving the black queen only one possible
route to a5 through the two captures) 15.Qxc3
16.Qa5 17.c3 18.c2 19.c1B (queen doesn’t
work because it is check) 20.Bf4 21.Bb8,
followed by a7xb8N mate.
December 28, 2007 – Revisiting the retractor
Dr. J. Sunyer, Chess Amateur 1923
White and Black retract their last moves,
then Black plays a move that allows White to mate in one.
Bryan Lamb wrote:
I have seen this problem before and know the solution,
but I give it here including reasons why this is the only
solution possible.
If White is to mate, white needs one additional piece.
Therefore Black must retract a capture.
If Black‘s retraction is the capture of a white
piece on e8, we should examine possibilities. Since the
goal is to mate Black, there can be no use in this line
for White to retract a capture, since a black piece on
h5 is irrelevant if black retracts an e8 capture.
If White retracts Kg6-h5 and Black retracts the capture
on e8, it must be a queen or rook (as bishop or knight
are insufficient to mate). However, if Black retracts
from the left side (such as Kd8xQe8) and plays a different
move, there is no mate possible. Similarly if black retracts
Ke7xQe8 there is no mate as Black’s alternate move
takes him away from the side of the board.
If Black retracts from the right side, such as Kf8xQe8
or Kf8xRe8 we find that Black has no other possible alternate
move to play! The only way to allow Black an alternate
move is to have White retract Kg5-Kh5 at the start, but
again Black’s alternate move in this scenario would
have to be away from the edge, and thus no mate is possible
on the next move.
The result of this deduction is that a black capture
retraction on e8 is not the solution. This means that
Black must retract the capture of a white piece on h5,
with the piece of his that the white king will also retract
the capture of on h5.
This means that the white piece which delivers mate starts
on h5 and must mate in one. This, combined with the fact
that the black king is on e8, means that the black king
is too far away to get into a position to be mated by
making one move, except if it makes a “double
jump”, which is only possible in castling.
This means that the solution must be that White retracts
Kg6xRh5 and Black retracts Rh8xQh5. As has already been
explained the necessary forward move for Black is castling,
and the only mate White has is Qh7#, which explains why
White must retract his capture from g6, in order to guard
h7. Clearly a white rook is insufficient for the mate
and clearly Black’s piece must be a rook to allow
castling, thus showing that each move is unique.
In summary, the solution is that White retracts
Kg6xRh5 and Black retracts Rh8xQh5 to get the
following starting position:
Now Black plays 1.0-0, allowing White
to play Qh7#.
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