Shortest Proof Games
– Part 1
Peter's Problem World with FIDE Master of Chess
Composition
Peter Wong

Note that Peter's articles, follow a chess problem
convention in using ‘S’ to represent the knight
(from the German word,
Springer). ‘N’ is reserved for a fairy piece called the nightrider.
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The
shortest proof game (SPG) is a type of problem in which the task is to
reconstruct a legal game.
Starting from the opening array, the solver
has to find the shortest possible game that leads to the diagram position.
White and Black thus effectively cooperate to achieve this, and that the
moves would not be sensible in a competitive game is considered irrelevant.
SPGs are also characterised by their exact
play – the move order in each solution is unique (otherwise the problem is
unsound). This requirement for a single move order may serve as a
pointer to the correct sequence when you solve such a problem, and the
precise play also contributes to a SPG’s artistic quality.
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The
stipulation of a SPG gives the length of the solution in pairs of
single moves. Hence in Problem 97, “SPG in 9” indicates
the position is reached after Black’s 9th move.
Solving a SPG typically begins with a
count of the number of “visible” moves made by the pieces no longer on
their starting squares. Here, White’s queen, bishop, and two
pawns have used up at least four moves, leaving five spare moves for
other purposes.
Black takes five moves to promote the
f-pawn, and the remaining time is taken up by the new queen to capture
White’s missing pawns, before its return to f7. Because of the
obstructing white pawn on the f-file, the black pawn must promote on
e1, after capturing White’s knight on e2.
The promotion also obliges the white
king to move out of the way, and the only viable method is a
surprising castling-and-return manoeuvre. |
97. Markus
Ott
Thémes-64
1981
1st Hon. Mention
       
       
       
       
       
       
       
       
SPG in 9 |
1.e4
f5 2.Bd3 f4 3.Se2 f3 4.0-0 fxe2 5.f4 e1(Q) 6.Qe2 Qxd2 7.Kf2 Qa5 8.Rh1 Qxa2
9.Ke1 Qf7.

1.e3 h5 2.Qxh5 a6
3.Qd1 Rxh2 4.Bxa6 Rh8 5.Rxh8 Sf6 6.Rxf8+ Kxf8 7.Bf1 Ke8. The white queen
and bishop execute switchbacks to their original squares, and so do
the black king and rook.

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In Problem 99, most of White’s
moves and their order are fairly obvious from the diagram: Pe3, Ba6,
Pc4, and then Qa4. That leaves White two spare moves to sacrifice
the missing pawn and bishop to an opposing knight.
Implementing this plan proved to be
difficult, however, because Black’s knight from g8, after making the two
captures, has insufficient time to get itself captured, e.g. 1.d4 Sc6
2.Bh6 Sxh6 3.e3 Sf5 4.Ba6 Sfxd4 5.c4 Sb8 6.Qa4 Sd-?
The surprising answer is to sacrifice the
knight from b8 instead, followed by its replacement by the “sibling”
piece: 1.d4 Sa6 2.Bh6 Sxh6 3.e3 Sf5 4.Bxa6 Sxd4 5.c4 Sc6 6.Qa4 Sb8.
The
thematic
twinning shifts the knight back to g8, creating another SPG
position. |
99. Michel
Caillaud
Problemesis
2001
       
       
       
       
       
       
       
       
SPG in 6
(b) Nb8 to g8 |
Amusingly enough, the second part sees an encore of the Sibling theme,
but done in reverse: 1.d4 Sh6 2.Bxh6 Sc6 3.e3 Sxd4 4.Ba6 Sf5 5.c4 Sxh6
6.Qa4 Sg8.

Problem
100 calls for a SPG in 6.5, indicating the diagram is attained after
White’s 7th move.
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100.
Richard Müller
Rochade
1985
       
       
       
       
       
       
       
       
SPG in 6.5 |
Both sides are missing only two pieces,
but straightforward attempts to remove them would not work, e.g. 1.e4 d6
2.Ba6 Bf5 3.Bxb7 Bxe4 4.Bxa8 Bd5 5.Bb7 Bxa2 6.Ba6 Be6 7.Bf1 Bc8 – a
single move too long.
The absent a2-pawn, b7-pawn and a8-rook,
from adjacent files, are a telltale sign that White may have promoted,
so let’s begin with 1.a4 d6 2.a5 Bg4 3.a6 Bxe2 4.axb7, and now if
Black were to try to capture the (soon) promoted piece, the play would
still take too long, e.g. 4…Bf3 5.bxa8(Q) Bxa8 6.Sh3 Bb7 7.Sg1 Bc8.
Instead, Black has to capture something
else, a piece that is replaceable by means of White’s promotion:
4…Bxd1 5.bxa8(Q) Bg4 6.Qf3 Bc8 7.Qd1.
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This is
called the Pronkin theme: the substitution of a captured piece
on its initial square by its promoted counterpart.


Problem 102 is for you to
solve. There are two distinct, but related, sequences of moves that arrive
at the same position.
102. Mark Kirtley
Problem Paradise
1999
       
       
       
       
       
       
       
       
SPG in 7
2 solutions
Solution
to Problem 102 (To display, hold down your mouse button and select the
text below)
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>1.b3
Nf6 2.Bb2 Nh5 3.Bf6 gxf6 4.c3 Bh6! 5.Qc2 Bg7
6.Qg6 hxg6 7.d3 Rh7, and 1.c3 Nf6 2.Qc2 Nh5
3.Qg6 hxg6 4.d3 Rh6! 5.Bg5 Rh7 6.Bf6 gxf6 7.b3
Bg7. In the first solution, the black bishop
makes a tempo move – a precise way to
use up the extra time available, while in the
second solution, it’s the black rook that
carries out the tempo loss.>
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