Proof Games – Part 2
No.24 | by Peter Wong
The genre of proof games (PG) was introduced in the previous article. Here we will look further into this problem type by considering some lengthier examples with more elaborate ideas. Many of the popular PG themes are paradoxical in nature, involving surprising manoeuvres or deceptive diagram positions. Therefore it helps to “expect the unexpected” when dealing with this kind of composition. And naturally it is also useful to become familiar with these oft-seen themes, so as to know what sorts of effects are achievable in this type.
139. Mark Kirtley & Michel Caillaud
StrateGems 1999, Commendation
PG in 7, 2 solutions
In Problem 139, Black has made at least two king moves and three pawn moves, out of the seven available. So only two moves remain for use by the missing rook, bishop, and pawn, to assist in their own captures. White’s choice of capturer cannot be any first-rank piece, because it would have insufficient time to return to its original square after the captures. Therefore the main role falls on the missing g-pawn. To facilitate its own capture, this pawn also has to promote first, and such an occurrence – the sacrifice of a promoted piece – is termed the Frolkin theme. 1.g4 h5 2.g5 Rh6 3.gxh6 e5 4.hxg7 Ke7 5.gxf8=Q+ Kf6! 6.Qg7+ Ke6 7.Qg6+ fxg6. Instead of an immediate 5…Ke6, the black king executes a tempo move with 5…Kf6!, motivated by the need to use up the extra time available. In the second solution, it’s White who plays a tempo move, and the Frolkin promotion also changes. 1.g3! e5 2.g4 Be7 3.g5 Bf6 4.gxf6 h5 5.fxg7 Ke7 6.gxh8=S Ke6 7.Sg6 fxg6.
140. Pascal Wassong
PG in 12
All of Black’s moves in Problem 140 are visible from the diagram, and their order is strictly determined already, not being affected by White’s play. Since no captures are involved, it seems that White can play 1.f3, and then simply mark time by shuffling a piece back and forth, e.g. Sh3-g1, for the remainder of the game. However, such a game would take a single move too long, because the shuffling piece will finish on the wrong square after Black’s final move. Instead, the solution requires White to make a tempo-losing trek, which is akin to the triangulation manoeuvre familiar in endgames. 1.f3 e5 2.Kf2 Bc5+ 3.Kg3 d6 4.Qe1 Kd7 5.Qf2 Kc6 6.Qd4 Kb5 7.Qe3 Ka4 8.Qf2 b5 9.Qe1 Bb6 10.Qd1 c5 11.Kf2 Sc6 12.Ke1 Rb8. After shifting the king to make room, White plays the thematic queen seven times only to return it to its original square, for no purpose other than to use up an odd number of moves.
141. Dmitry Pronkin
Die Schwalbe 1985, 1st Prize
PG in 12½, 2 solutions
Black’s doubled pawns in position 141 indicate that a capture has occurred on f6, but White’s solely missing b-pawn could not have made so many captures to reach that square directly. Therefore this pawn has promoted, and the new piece either sacrificed itself on f6, or replaced another piece that did. The former option would be too slow to be viable, because …exf6 has to be played early to allow Black’s pieces to develop. So White must have sacrificed an original piece, and promoted to an equivalent one, before the latter moved to its counterpart’s initial square as an impostor. This is the Pronkin theme, impressively rendered with a knight in one solution, and a bishop in the other. 1.Sc3 Sf6 2.Sd5 Se4 3.Sf6+ exf6 4.b4 Qe7 5.b5 Qa3 6.b6 Bc5 7.bxa7 b6 8.axb8=S Bb7 9.Sa6 0-0-0 10.Sb4 Rde8 11.Sd5 Re6 12.Sc3 Rd6 13.Sb1, and 1.b4 Sf6 2.Bb2 Se4 3.Bf6 exf6 4.b5 Qe7 5.b6 Qa3 6.bxa7 Bc5 7.axb8=B Ra6 8.Ba7 Rd6 9.Bb6 Kd8 10.Ba5 b6 11.Bc3 Bb7 12.Bb2 Kc8 13.Bc1. Black also varies the play nicely across the two phases, especially in the change from castling to non-castling.
142. Andrey Kornilov & Andrey Frolkin
Die Schwalbe 1988, 3rd Prize
PG in 14
Problem 142 effects a splendid exchange of places between two like pieces. If we count the moves required by Black’s pieces to reach their current positions, we find that all of the stipulated fourteen moves are accounted for. That implies the missing d7-pawn and a8-rook had no time to move and were captured on their initial squares. Also, Black’s development depends on an early removal of the d7-pawn (to free the queen), a fact that determines White’s starting moves, 1.Sf3 f5 2.Se5 f4 3.Sxd7 f3 4.Sb6 Qd5. Black must wait for a white knight to return to g1 via h3, before closing the door on that option with …Bh3. And now if White were to capture the rook with 5.Sxa8, the knight would take too long to get back: 5…Qh5 6.Sb6 g5 7.Sd5 Bg7 8.Sf4 Bc3 9.Sh3 Bxh3?? To resolve this issue of White needing to (1) allow Black to play …Bh3 in time and (2) capture the rook with the b6-knight, White brings the other knight into play. 5.Sc3 Qh5 6.Scd5 g5 7.Sf4 Bg7 8.Sh3 Bc3 9.Sg1 Bh3 10.Sxa8 e6 11.Sb6 Se7 12.Sc4 Rf8 13.Sa3 Rf4 14.Sb1 Rc4. Thus the two knights have swapped their identities in the diagram position.
143. Thomas Brand
Die Schwalbe 1995, 2nd Hon. Mention
PG in 13½
White apparently has castled queen-side to arrive at position 143, but appearances can be deceiving… Black needs to promote the a-pawn, shift the new queen to h5, and also capture five white pieces along the way, including the trapped c1-bishop. These objectives largely dictate White’s play as well for the first half of the game, 1.Sf3 a5 2.Sd4 a4 3.Sb3 axb3 4.a3 Ra4 5.Ra2 bxa2 6.e3 axb1=Q 7.Bd3 Qxc1 8.0-0 Qxd1 9.Re1 Qh5. So White has in fact castled on the king-side, to enable the black queen to get to h5 as soon as possible. White now sets out to reach the queen-side castled position artificially, and the king’s trip is assisted in turn by Black, with a well-timed interference on g4, 10.Kf1 Rg4 11.Ke2 d6 12.Kd1 Rg3+ 13.Kc1 Bg4 14.Rd1.
144. Martin Hoffmann
Die Schwalbe 1990, 6th Hon. Mention
PG in 8
Problem 144 is for you to solve. This short game involves a type of clearance manoeuvre that is quite hard to visualise. Hint: White’s missing rook is captured by Black’s queen on f1.
1.e3 d6 2.Ba6 Bh3 3.Qg4 Qd7 4.Sf3 Qb5 5.Rf1 Qxf1+ 6.Bxf1 Sa6 7.d3 Rc8 8.Qxc8+ Bxc8. Both 2.Ba6 and 2…Bh3 are clearance moves, designed to allow another piece (the opposing queen in each case) to travel along the same line that was traversed by the clearing piece.