Proof Game Shorties IV

by Mark Kirtley, Probleemblad 6, 2002

From: "Joost de Heer" <>
Date: Tue, 25 Dec 2001 15:27:39 +0100
Subject: [Retros] Proofgame Shorties IV

Here are the solutions with their comments (with one additional comment from me [Joost]):

1. Caillaud

a) 1. d4 Na6 2. Bh6 Nh6 3. e3 Nf5 4. Ba6 Nd4 5. c4 Nc6 6. Qa4 Nb8

b) 1. d4 Nh6 2. Bh6 Nc6 3. e3 Nd4 4. Ba6 Nf5 5. c4 Nh6 6. Qa4 Ng8

Yes, the stipulated twinning in the second part is a broad and thematic hint to the solution! A knight is diagrammed in it's partner's home square, and the challenge answered here was for this theme to be doubled. A further challenge: can an SPG be composed in which a Knight enters and leaves that square, without that knight capturing any unit?

2. de Heer

1. Nf3 e5 2. Ne5 Qe7 3. Ng6 Qe2 4. Qe2 Ne7 5. Qe7 Be7 6. Nf8 Bf8

Non-symmetric play to an all-units-home position with Black/White symmetry across the horizontal mid-line of the board. A jewel! Can a composer show symmetry across the mid-point of the board, as in

1. d3 e6 2. Bh6 Ba3 3. Qd2 Qe7 4. Kd1,

but with non-symmetric play?

3. Donati

1. a4 b5 2. ab5 Ba6 3. Ra6 Qc8! 4. Rh6 gh6 5. e3! Bg7 6. e4 Bd4 7. e5 Qd8

Answers the challenge for a tempo switchback by one unit, and a tempo loss of any sort by another.

4. Wong

1. e3 Na6 2. Ke2 Nc5 3. Kf3 Ne4 4. Kf4! Ng3 5. Kg4 Nh1 6. g3 Rb8! 7. Bg2 Ra8

Check avoidance is used to prevent the black Knight at g8 from doing the switchback.

5. Donati

1. d3 e6 2. Be3 Ne7 3. Ba7 Nec6 4. Bb8 Ra2 5. Nf3 Rb2 6. Nfd2 Rb1 7. Nb1 Nb8

This answers in a single line of play the same challenge as No.1 does. Gianni asks if somehow a problem with two such imposter pieces, but Rooks instead of Knights, or a Rook and a Knight, might be composed.

6. Kirtley & Donati

1. e3 h5 2. Qh5 a6 3. Qd1! Rh2 4. Ba6 Rh8! 5. Rh8 Nf6 6. Rf8 Kf8 7. Bf1! Ke8!

Four returns to home. There are as may as ten home returns in an SPG of 31.0 moves, by Unto Heinonen (Probleemblad 1999-5).

Comment of JdH: Around the same time as the composition date of this problem, I made the following composition, with a more attractive end position, but a halfmove too long:


(14+13) SPG 7.5


7. Donati

1. a4 c6 2. a5 Qb6 3. ab6 Na6 4. Ra6 ba6 5. b7 Rb8 6. bc8=N Rb6 7. Nb6 ab6

A pair of cross-captures and a Ceriani-Frolkin promotion.

8. de Heer

1. d4 e5 2. de5 Ba3 3. Qd7 Qd7 4. ba3 Qd3 5. ed3 Be6 6. Ke2 Bb3 7. ab3

Two pairs of cross captures on one side.

9. de Heer & Caillaud

1. e4 d5 2. Bc4 de4 3. Be6 Qd2 4. Nd2 fe6 5. Ndf3 ef3 6. Qd6 ed6

Cyclic cross-captures!

10. Caillaud

a) 1. e4 f5 2. e5 Nf6 3. ef6 ef6 4. Bd3 Ke7 5. Be4 fe4 6. f3 f5

b) 1. e4 f6! 2. Bc4 f5 3. Bg8 e6 4. Bf7 Ke7 5. Bg6 fe4 6. Bf5 ef5 7. f3

Cross-captures with reciprocal change: one black pawn captures the white bishop, and the other captures the white Pawn, but this arrangement is switched in the second phase.

11. Surkov

a) 1. a4 b5 2. ab5 Na6 3. Ra6 Bb7! 4. Ra1 Ba6 5. ba6

b) 1. a3 b5 2. Ra2! b4 3. ab4 Ba6 4. Ra6 Nc6 5. Ra1 Na5 6. ba5

The slight twinning change creates a package of paradox: a tempo reversal between the two sides, reciprocal change between capturing and captured units, and a longer solution for the phase in which the white pawn is less advanced! Note that there is no change for the square on which the black Bishop is captured, but there is such a change for the black Knight. Is it possible to compose a two-phase problem that preserves such reciprocal change, but in which the square for neither captured unit changes?

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