I found that the book "Amusements In Chess," by Henry Ernest Dudney, has been put onto the web in the form of a website. In the section "Various Chess Puzzles, I came across this particular puzzle that intrigues me.

The author gives this legal position and asks for a way to reach it. The White king is on c7, and the Black king is on g1.

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On the linked solution page near below it, it is said that: "The order of the moves is immaterial, and this order may be greatly varied. But, although many attempts have been made, nobody has succeeded in reducing the number of my moves."

The author provides his solution in 43 moves, which I have transcribed from descriptive notation, replayable here.

  1. f4 c6 2. Kf2 Qa5 3. Ke3 Kd8 4. f5 Kc7 5. Qe1 Kb6 6. Qg3 Na6 7. Qb8 h5 8. Nf3 Rh6 9. Ne5 Rg6 10. Qxc8 Rg3+ 11. hxg3 Kb5 12. Rh4 f6 13. Rd4 fxe5 14. b4 exd4+ 15. Kf4 h4 16. Qe8 h3 17. Nc3+ dxc3 18. Ba3 h2 19. Rb1 h1=Q 20. Rb2 cxb2 21. Kg5 Qg1 22. Qh5 Ka4 23. b5 Rc8 24. b6 Rc7 25. bxc7 b1=B 26. c8=R Qc7 27. Bd6 Nb4 28. Kg6 Ka3 29. Ra8 Kb2 30. a4 Qgb6 31. a5 Kc1 32. axb6 Kd1 33. bxc7 Ke1 34. Kf7 Nh6+ 35. Ke8 Ba2 36. f6 Bg8 37. f7 Kxf1 38. c8=B Nd5 39. Bb8 Nc7+ 40. Kd8 Ne8 41. fxe8=R Nf7+ 42. Kc7 Nd8 43. Qf7+ Kg1

However, is this truly the best possible? I do not know. So, the question here should be obvious. Can lower than 43 moves, or 86 ply, be achieved, or can it be proved as optimal?


We can do

a semimove better:
[FEN ""]
1.f4 c6 2.Kf2 h5 3.Ke3 Rh6 4.Qe1 Rg6 5.f5 Na6 6.Qg3 f6 7.Qb8 Rg3+ 8.hxg3 Kf7 9.Rh4 Qa5 10.Rd4 h4 11.Nf3 h3 12.Ne5+ fxe5 13.f6 exd4+ 14.Kf4 Ke6 15. b4 Kd5 16.Qxc8 h2 17.Nc3+ dxc3 18.Rb1 h1=Q 19.Rb2 cxb2 20.Qe8 b1=B 21.Ba3 Qg1 22.b5 Kc4 23.Bd6 Rc8 24.Kg5 Rc7 25.Qh5 Qc5+ 26.Kg6 Kb4 27.Kf7 Ka3 28. Ke8 Kb2 29.a4 Ba2 30.b6 Nh6 31.bxc7 Bg8 32.c8=R Qc7 33.a5 Q5b6 34.axb6 Kc1 35.Ra8 Kd1 36.bxc7 Ke1 37.c8=B Kxf1 38.Kd8 Kg1 39.Bb8 Nc7 40.f7 Ne8 41. fxe8=R Nf7+ 42.Kc7 Nd8 43.Qf7 *


Sketch of proof of optimality:

Observe that because the black Bf8 cannot have moved, the two white rooks must be promoted pieces as must be the white Bc8. As white is only lacking three pawns it must be them, in particular, the a- and b- pawns must have captured at least 3 black pieces to get to the c-file and the f-pawn must have captured on e8. Together with the hxg3 capture and the capture of the black Bc8 (which cannot have left its initial position) this accounts for all 6 missing black pieces.
Next, the black Bg8 must also be a promoted pawn and the only rank 1 light square which has a path to g8 is b1. Therefore the black f-pawn must have captured 4 white pieces to get to the b-file. Together with the capture of the white Bf1 this accounts for all 5 missing white pieces.

Now let's count the minimum number of white moves.
3 promotions = 15 moves + 1 R move to a8
lining up pieces for the black f-pawn to capture = 7 moves
hxg3 = 1 move
get K to c7 (only possible via f7) = 9 moves
get B to b8 = 3 moves
only the Q remains. The shortest way from d1 to f7 are 3 moves. If we require the Q to capture c8 before and to then get to f7 without blocking the white K's path and the promotion on e8 then 7 moves are needed which gives a total of 43. Thus, if we can show that there is no way of getting rid of black Bc8 that costs fewer than 4 white moves, we are done.

Who could have captured the black Bc8? It cannot have been the white Ra8 or Bc8 because their promotions required c8 to be unoccupied. It couldn't have been the white K either, because the black Ra8 couldn't have moved beyond b8 while the Bc8 was still there. Therefore the black Nb8 would have to still be there just to allow the K to legally capture on c8. But with the white K on c8 or c7 the Nb8's only move (to a6) would be check forcing the K to capture on b7 or d7, so that's not possible. The Re8 is also not an option because it must be created after the white K has entered d8/c7. Any other piece if it could get there at all would take too long.

Note that we can obtain 42 1/2 moves also by a minor modification of the solution in OP:
[FEN ""]
1.f4 c6 2.Kf2 Qa5 3.Ke3 Kd8 4.f5 Kc7 5.Qe1 Kb6 6.Qg3 Na6 7.Qb8 h5 8.Nf3 Rh6 9.Ne5 Rg6 10.Qxc8 Rg3+ 11.hxg3 Kb5 12.Rh4 f6 13.Rd4 fxe5 14.b4 exd4+ 15.Kf4 h4 16.Qe8 h3 17.Nc3+ dxc3 18.Ba3 h2 19.Rb1 h1=Q 20.Rb2 cxb2 21.Kg5 Qg1 22.Qh5 Ka4 23.b5 Rc8 24.b6 Rc7 25.bxc7 b1=B 26.c8=R Qc7 27.Bd6 Nb4 ?? {Wastes two tempi instead of one} ( 27...Qc5 ! 28.Kg6 Ka3 29.Ra8 Kb2 30.a4 Q5b6 31.a5 Kc1 32.axb6 Kd1 33. bxc7 Ke1 34.Kf7 Nh6+ 35.Ke8 Ba2 36.f6 Bg8 37.f7 Kxf1 38.c8=B Kg1 39. Bb8 Nc7+ 40.Kd8 Ne8 41.fxe8=R Nf7+ 42.Kc7 Nd8 43.Qf7 ) 28.Kg6 Ka3 29.Ra8 Kb2 30.a4 Qgb6 31.a5 Kc1 32.axb6 Kd1 33.bxc7 Ke1 34.Kf7 Nh6+ 35.Ke8 Ba2 36.f6 Bg8 37.f7 Kxf1 38.c8=B Nd5 39.Bb8 Nc7+ 40.Kd8 Ne8 41.fxe8=R Nf7+ 42.Kc7 Nd8 43.Qf7+ Kg1 *


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