Proof Games – Part 1
No.23 | by Peter Wong
The proof game (PG) 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. PGs 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 PG’s artistic quality.
133. Markus Ott
Thémes-64 1981, 1st Hon. Mention
PG in 9
The stipulation of a PG gives the length of the solution in pairs of single moves. Hence in Problem 133, “PG in 9” indicates the position is reached after Black’s 9th move. Solving a PG 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. 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.
134. Mark Kirtley & Gianni Donati
PG in 7
Problem 134 shows a position where most of the pieces are on their initial squares, meaning that few clues remain as to what has occurred. In such a situation we focus mainly on the missing pieces, i.e. try to work out what captured them and where the captures took place. Here Black’s dark-squared bishop is clearly taken on its original square (it was trapped by the pawns on e7 and g7), and the likely capturer is White’s rook from h1. Therefore an objective may be to promptly clear the h-file of its two pawns, to enable the rook to sweep down to the 8th rank. Also, Black is missing four pieces, so out of only six white moves to be determined (not counting the visible Pe3), we know that four of them are captures, a fact that considerably narrows down the choices of play. 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.
135. Michel Caillaud
PG in 6, (b) Sb8 to g8
In Problem 135, 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 PG position. 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.
136. Richard Müller
PG in 6½
Problem 136 calls for a PG in 6½, indicating the diagram is attained after White’s 7th move. 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. This is called the Pronkin theme: the substitution of a captured piece on its initial square by its promoted counterpart.
137. Joost de Heer
PG in 6
The diagram of Problem 137 exemplifies a pleasing homebase position, in which every piece stands on its game array square. The main point of this PG, however, lies in another visual effect, a kind of deceptive symmetry. In spite of White and Black’s symmetrical arrangement in the final position, the play leading to it is non-symmetrical. 1.Sf3 e5 2.Sxe5 Qe7 3.Sg6 Qxe2+ 4.Qxe2+ Se7 5.Qxe7+ Bxe7 6.Sf8 Bxf8
138. Mark Kirtley
Problem Paradise 1999
PG in 7, 2 solutions
Problem 138 is for you to solve. There are two distinct, but related, sequences of moves that arrive at the same position.
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.