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Solutions to our
2007 Christmas Puzzles |
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December 29, 2007 – The Philanthropist
Dr. Karl Fabel (Source
unknown)
White to play and not win!
Once again we hand over, for a discussion of the solution,
to Bryan Lamb, the chess master from Canada who has taken
upon himself the tast to illuminate our Christmas problems
to the full. Here his analysis:
White has six legal moves from the initial position:
Nxf6, Ne7, Nxb6, Nc7, e4 and c4. Of these, only c4+ is
not mate. Therefore this must be White’s first move.
After 1.c4+ Rxc4, White has two possible
moves which are not mate, namely Nc7+ and e4+.
Consider 2.Nc7+ Rxc7. Then, White is forced into the
line 3.Ne7+ Rxe7 4.e4+ Rxe4 5.fxe4#, which is not a non-win
and therefore is not the solution.
Therefore the second move must be 2.e4+ Rxe4.
it is fairly plain to find the next two moves, both of
which are forced as the only non-mating moves namely 3.Ne7+
Rxe7 4.Nc7+ Rxc7 stalemate.
One way to summarize this problem is that Black’s
rook cannot end up on e4, since White can capture it here.
It begins with a capture on c4, and so it must either
end up on e4 or c7 by going “around the rectangle”
c4/c7/e7/e4. As e4 is not an allowable final square it
must be c7 and therefore it captures counterclockwise
“around the rectangle”.
Hans Kluver, Karl
Fabel, 1947
Losing Chess: White to play and "win"
In Losing Chess capturing is compulsory – whenever
a player can capture, he has to do so (if he has different
choices to capture, he may choose any one of them). There
is no check or checkmate, and the king plays no special
role in the game. It can be captured like any other piece.
Bryan Lamb explains the solution:
White must play to a square that black cannot move adjacent
to on his first move. This happens to be automatic, so
the main theme is for White to be able to sacrifice the
king to the queen regardless of where the queen moves.
So: 1.Ke4.
Now any black king move, including Ke6, is met by Kf5.
So the black queen must move. If it moves along the c-file,
White can easily sacrifice the king by either moving next
to the queen or to a square the queen attacks. If Black
plays the queen to b7, the king stays on the e4-h1 diagonal.
If Black plays it to a6, White plays Kd3.
If the black queen moves to any square on the back rank,
White simply places or leaves the king on a diagonal or
file the queen attacks along. This is possible regardless
of where the queen goes, as can be easily seen by trying
all the moves. One possible defence is 1...Qd8.
To this White replies 2.Kd4! and wins.
In the diagram position, when we consider the necessity
of being able to get to both d3 and f5 (to defeat Qa6
and Ke6 respectively) we now see why only Ke4 will work
as a solution, since this is the only square that is adjacent
to both d3 and f5 by a king move.
December 30, 2007 – Kling conditionals
H. (J.) Kling, 1849
White to mate in three – with his pawn!
This time let's ask someone else, Eduard Shamilov of Dupont,
USA:
First of all, one has to cut the king off from g8, by
either Qa2 or Qc4 to prevent it from running from the
pawn. Following Ke8, it then transpires that to mate with
the pawn on e7, which would take up the remaining two
moves, White would need to have control of e8, which he
does not, so the mate must be delivered on e6, for which
White must take e8, d8, c8 and c6 under control while
driving the king to d7, and the only viable square for
the queen for this job is a8 (c8 would mate), which is
why only 1.Qa2 works. The game continues
1...Ke8 2.Qa8+ Kd7 3.e6#.
H. (J.) Kling,
1849
White to play, mates in six, only moving his king
Back to Bryan Lamb, who describes the solution thus:
Since white can only move the king, the mate must be
a discovered check and mate. In order to discover mate
in this position, the black pawn has to be out of the
way because it will either be able to block the mate by
being pinned or by moving into a pin. Thus white must
force the black pawn to advance: 1. Kf7+ d4 2.
Kf8! dxc3. The white queen is a redundant piece
because since the discovered mate is along the diagonal
the bishop on a1 is all White needs to do this.
Now, to avoid a stalemate white must release the black
king. 3.Ke8!! Kg7 4.Ke7. White is only
allowed to move the king, so must play partly as if it
is a king and pawn ending: lose a move and gain the opposition
to force Black’s king back! 4…Kh8
5. Kf6 c2 6. Kf7 mate.
December 31, 2007 – The wandering king
Otto Wurzburg, 1919
Only the white king moves. How many legal
moves does it take to get him to the square f4?
Eduard Shamilov tells us how to do it:
To get to the f4 square, white must first eliminate the
g-pawn, for which he must first capture the black knight,
for which he must first knock out the h1 rook, but not
before he gets to the cleric on a8, which in turn is guarded
by the rook at h8, whose protector is on a1, shielded
from attack by the knight at c1.
White can only reach the knight through, tracing the
route retrospectively, d1, e1, f2, g3, g4, f5, g6, f7,
e7, d7, c7, b6, so the first mission is to take out the
a-pawn. Once the steed at c1 is taken care of, White then
backtracks all the way to a2 to remove the bishop, then
back again for the h8 rook, followed by a detour to the
left to gain access to the light squares, a glide along
the long diagonal, an upwards march to clear away the
last defender, and then final leg to f4 through g5.
Here's a full route: 1.Ka4, a5, a6, xa7, b6, c7, d7,
e7, f7, g6, f5, g4, g3, f2, e1, d1, xc1, d1, e1, f2, g3,
g4, f5, g6, f7, e7, d7, c7, b6, a5, a4, a3, a2, xa1, a2,
a3, a4, a5, b6, c7, d7, e7, f7, g7, xh8, g8, f7, e7, d7,
c7, b8, xa8, b7, c6, d5, e5, f5, g4, f3, g2, xh1, g2,
g3, g4, h5, h6, xh7, g6, xg5, f4! That took a grand total
of 70 moves!
January 1st, 2008 – Constructions past and
present
Sam Loyd, 1866
(i) Shortest discovered checkmate
(ii) Shortest perpetual check
(iii) Shortest reflected mate
There are three problems to solve:
Problem 1: Find the shortest sequence
of moves that lead to a checkmate by discovered check. Naturally
both sides are collaborating to try to achieve this goal.
There were many submission in five and more moves. Bryan
Lamb reasoned:
The discovered mate must be by a pawn (if by a piece,
the piece must get in place for the discovery which takes
at least two or more moves, and since also a pawn must
move anyway to open a line to the king it is unnecessary
to use a piece for the discovery.
The shortest number of moves to get a king in line for
a discovered check by a pawn’s initial move is four
(one to move a pawn and three more to get the king in
position). Therefore, no solution is possible in less
than four moves.
I present a solution which is four moves long and it
will be known to be the quickest: 1.f3 h5 2.Kf2
e5 3.Kg3 h4+ 4.Kg4 d5 (or d6) mate. The solution
is not unique as the e5 and h5 moves can be interchanged.
Problem 2: Find the shortest sequence
of moves after which one side can force a perpetual. In
this case both sides cooperate until the position is reached;
after that one side has no choice but to accept perpetual.
Note that the side delivering perpetual check prefers it
to checkmate, which he may also deliver. Master Lamb:
Reaching a position where a perpetual is unavoidable
means there can be no effective way of blocking a check.
The king in its initial position has too many blocking
methods, so the king must come out. It has to be at the
third rank, in front of the pawns, at least, which takes
three moves including one pawn move.
Thus the minimum number of moves where a perpetual is
evident is three, and I present a solution with three
moves. Thus it must be the shortest. 1.f4 e5 2.Kf2
Qf6 3.Kg3 Qxf4+. At this stage a perpetual check
can be declared and it is demonstrable by the repetition
Qh6-f4-h6+, which White cannot avoid. The “mate”
black might choose would come as a result of 4.Kh3 d5+
5.g4 Qxg4#, but Black prefers the perpetual, we are told.
Note that White has alternate moves with the king such
as 4.Kh3 Qh6+ 5.Kg4 Qf4+ 6.Kh5 Qh6+ but white is stuck
on the squares h5, g4, h3 and g3 so eventually there will
be a three-fold repetition.
Problem 3: Black has agreed to play reflected
moves as long as this is possible, e.g. 1.e4 e5 2.Nf3 Nf6
3.Nxe5 Nxe4 etc. (so that's where the Petroff comes
from!). What is the shortest mate White can deliver if Black
faithfully reflects his moves? Note that there are three
solutions. Bryan Lamb:
First it should be pointed out that the ubiquitous Scholar’s
mate does not work because when white’s queen leaves
d1, black’s queen leaves d8 thus giving black’s
king a flight square.
How can this mate be accomplished in three ways? Well,
to take advantage of the black queen vacating d8 (which
is necessary since clearly it is the white queen which
can deliver the quickest mate), we observe that the c8
bishop is left undefended. Sequences ending in Qxc8# should
be sought.
I present three such solutions (which is the number of
solutions asked for). There are three ways to get to c8:
via the file and via the c8-h3 diagonal. As it turns out
there are two different ways of doing this on the c8–h3
diagonal, namely via f5 and via h3. So the solutions are:
1.c4 c5 2.Qa4 Qa5 3.Qc6 Qc3 4.Qxc8#
1.d4 d5 2.Qd3 Qd6 3.Qf5 Qf4 4.Qxc8#
1.d4 d5 2.Qd3 Qd6 3.Qh3 Qh6 4.Qxc8#
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