Full solution attempt:
Last move obviously was exZf1=Q+
, with yet unspecified piece Z
. If that was original Bf1
, a2
is promoted. Impossible, as a2
can only promote on b8
given that it can only promote by a6xQb7-b8
(the black f pawn promoted on f1, as already established, leaving only the black queen as victim for the white a pawn) - wrong color. Thus Ba2
is original and a2
promoted, again by a6xQb7-b8
. We thus already know the promoted figure Z is either Q
or N
(R
can't get out again). Here is a N
attempt, I have a good hunch that Q
is faster (Q-b7-a6-f1
) and even the moves are unique.
[FEN "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"]
1. a4 b6 2. a5 Bb7 3. a6 Bc6 4. e4 Qc8 5. Bc4 Qb7 6. axb7 Na6 7. b8=N Bb7 8. Nc6 f5 9. d3 f4 10. Be3 fxe3 11. f3 e2 12. Nd4 Nb8 13. Nf5 Bc8 14. Ng3 Bb7 15. Ba2 Bc8 16. Nf1 exf1=Q+
A simple counting of White moves proves that a N
solution can't be improved. Q
could attain 14 moves, here Black also needs 14 moves minimally - so this could work out if timed carefully. But I don't see a trick to actually save the one black move: with a square z5, I could play Bz5 and avoid wasting the white/black tempo, so I am at 15.
[FEN "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"]
1. a4 b6 2. a5 Ba6 3. e4 Bb5 4. Bc4 Qc8 5. a6 Qb7 6. axb7 Nc6 7. Bb3 Ba6 8.
b8=Q+ Bc8 9. Qb7 f6 10. Qa6 f5 11. Qf1 f4 12. d3 Nb8 13. Be3 fxe3 14. Ba2 e2
15. f3 exf1=Q+
Optimality proof attempt: White obviously does need 14 moves (one tempo was wasted on Bb3). But then let's retro-move further from the final diagram: White can only take back "neutrally" f3
. But another move pair is needed before Be3
, which forces d3
before that, and the white queen can't get to f1 anymore! White has no such move, obviously, so a tempo like Bb3
must be inserted.
5 minutes for the logic :-)