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The following question deserves to be asked!

"I see a lot of proof game problems, where I am given a diagram and have to find the unique way to reach that position in the specified number of moves. They look fun, but difficult. Can I have some tips or strategy for solving them, please?"

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Although solving proofgames is not that difficult, the large number of moves involved may seem impressive at first. Let me share you a few tips by showing how I solve a proofgame.

[Title "Étienne Dupuis, Problemesis 2001, PG in 13.5"]
[FEN "rn2k1nr/ppp3pp/4p3/8/4q3/1N1bb2P/PPPP1P1P/RNBQKB1R w KQq - 0 1"]

In this diagram, it is not obvious at first sight how white has spent 14 moves, as we count 3 for the Nb3, and 1 for Ph3. For black, we count a minimum of 2 moves for the queen, 2 for each bishop, and one for Pe6, for a total of 7 moves.

We need to find a hint, and here the hint is given by the white pawn h3. It must have captured a black piece, but which one? Since black is only missing pawns in the diagram and that no black pawn can reach h3, we must deduce that black has made a promotion.

Let us then enumerate various possibilities:

  1. Black promoted a pawn on e1, and the promoted piece was captured on h3. This requires seven extra moves by black, yielding a total of 14 moves, one too many, hence impossible.
  2. White pawn h3 captured a black piece and black promoted a pawn on e1 or g1 to replace this captured piece. This is the correct solution. No other promotion squares are possible since black can only capture one white piece.

Which black piece is an impostor?

  1. The rooks and knights are too far away; it would take too many moves to reach h3 and bring back the promoted piece.
  2. The dark square bishop cannot be captured on light colored square h3.
  3. The light square bishop could, but cannot be promoted on dark squares e1 or g1.
  4. The queen is the only candidate. It requires two moves to bring the queen to h3, and six moves to promote pawn d7 or f7 to a queen on e1 and bring it to e4.

At this stage, we have accounted for all thirteen black moves. We are not exactly sure if the queen played Qd8-h4-h3 or Qd8-d7-h3 or even Qd8-d3-h3 but we'll figure it out later. We also do not know which black pawn promoted, but we know the one of the two did not move at all and was captured at home.

Turning to white's side, we must find moves to

  1. Capture the d7 or f7 pawn.
  2. Get the king out of the way for the black promotion to occur.

Getting the king out of the way not as easy as it looks like. Keeping in mind that black will promote a queen on e1, and then play Qe1-e4, the king must step at least two squares away and he cannot come back through e2. Hence he must come back from f1, indicating the Bf1 must also move out of the way. This requires at least 6 moves and the king would go at least to g1 or g2.

Who captured the black pawn?

  1. The white pawn e2 can capture the d7 pawn, and then black would have played Qd8xPd7 before Qd7-h3 to get rid of the white pawn. This requires four white moves. However, it's not possible, as the f7 pawn would have no one to capture to reach e1.
  2. A white knight can capture both. It would require two extra moves for Nb3 to capture on d7 and four extra moves to capture on f7.
  3. White bishop and queen cannot capture as it would give check and black has no moves to spare.
  4. White rook h1 could but would required eight moves. Impossible.

The missing black pawn was thus captured by the white knight.

It is now time to take out a chessboard and move the pieces along to try finding a solution keeping in minds all the deductions we have made.

Try moves, in any order. You will quickly stumble on an additional difficulty. To bring back the white king, we must shield the check from the Bd3. This bishop must be present on that square before black plays Qe1-e4, as the bishop played Bc8-f5-d3.

Who shields?

  1. This shielding could be performed by the white bishop, in which case we need at least one extra move, if the bishop plays Bf1-a6-e2-f1 for example. But wait, Bd3 is in the way again. Hence if it's the bishop, it's from the other side and we need two extra moves.
  2. Shielding could be performed by the white queen, also two extra moves.
  3. Shielding can also be provided by the Nd3 for free, assuming it played Ng1-e2-d4-b3. However, this looks impossible, as who would capture the black pawn on d7 or f7?

With this additional observation, we get to eight moves for moving the white king out of the way (four for the king, two to move the Bf1 somewhere, and two for shielding on e2, by the queen or the bishop). This implies that the missing black pawn was captured by the white knight and that white Pe2 has not moved at all!

If it has not moved, white started by playing with the knight and black has no choice but to wait for the Pd7 to be captured; he cannot push Pe6 before playing Bc8-f5.

The solution is quickly found to be:

[Title "Étienne Dupuis, Problemesis 2001, PG in 13.5"]
[FEN ""]

1. Nf3 f5 2. Ne5 f4 3. Nxd7 f3 4. Nc5 Qd7 5. Nb3 Qh3 6. gxh3 Bf5 7. Bg2 Bd3 8. Kf1 fxe2+ 9. Kg1 e1=Q 10. Bf1 Qe4 11. Be2 e6 12. Kf1 Bc5 13. Ke1 Be3 14. Bf1

The reasoning made to find the solution gives us also a high confidence that the solution is unique, it cannot be something else.

Hopefully this extensive explanation will help you enjoy solving proof games!

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  • That's very insightful and is the right grade of difficulty for analysis. Maybe you can point out some of the aesthetic features of the problem which are discovered during the solution? Switchbacks, Ceriani-Frolkin, Phoenix etc? – Laska Apr 29 at 7:19
  • 1
    The main feature of this problem is the double switchback of the white bishop f1, but for the solver I assume the fun is understanding which piece the g pawn must capture, as at first sight it is not obvious that we can promote a queen on e1. – Étienne Apr 29 at 21:33

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