I have an odd idea for this challenge. It involves a very specific proof game construction. So pay attention!

Given that:

White and Black cannot move more than one pawn each throughout the entire game from the starting position. No captures are allowed whatsoever.


A position whose proof game takes the longest shortest path to reach under the requirements.

Here is an example position.

enter image description here

  1. a4 h5 2. Ra3 Rh6 3. Rg3 Rb6 4. Rg6 Rb3 5. Rh6 Ra3 6. Rh8 Ra1

During the proof game, only a pawn per side moves and no captures occur. 6 moves is provably the optimal move amount to reach the final position.

Please leave a comment for any needed clarifications. Have a load of thought solving this!

Possible Hint:

RNBQKbNr/p1pppppp/8/1p6/6P1/8/PPPPPP1P/RnBkqbnr is one of my ideas. Finding the shortest proof game for it or a similar position may yield an a superb result.

  • 1
    $\begingroup$ If I find some time (probably never) I might ask the same question but for positions with a unique shortest proof game - seems funnier to investigate, and maybe more chalenging. $\endgroup$
    – Evargalo
    Commented Jun 8, 2021 at 8:52

3 Answers 3


Final position:

enter image description here

Lower bound for shortest proof game :

1. that each black N must have entered its respective corner before the nearby P can move.
2. that the Ks cannot enter their final destinations when the Bs are already there. But something must shield the Ks from the Rs. Therefore the Rs must be one square further away and the white Ns be in between Ks and Rs. Once the Ks are in the Bs can follow and then the white Ns can move on and the Rs take their final positions.

with that in mind we find that
overall white moves are limiting
on top of that there is a "tempo" constraint in the beginning becaus black has to make 9 N plies before the two Ps can move
we'll need at the very least the following white plies:

11 plies to move the K from e1 via b2 and h6 to g8
1 ply to move the P
4 plies for Rs: Rh1-g1 and Ra1-b1-e1-d1
17 plies for Ns: We need to move a N to d1 but Nb1-c3-d1 does not work because the of the following necessary sequence of evens: N has to leave b1 to allow Ra1-b1 which is required to allow a black N to a1 which must happen before Pb2-b3 which is required to allow the white K out which must move through c3 or waste two plies. But if the N moves from c3 to d1 to make room for the K it will prevent the black K to enter via b2. Therefore we know that getting a white N to d1,e8,h1 and a8 will take at least four plies each, plus one because of parity.
5 plies to get the Q from d1 to h6 (Qd1-c1-b2-g7-h6 is not possible because the Q has to leave b2 before the white K can move out but may not enter g7 before the black K has moved out)
4 plies to get the B out (to let in the black K) and back
Total: 42 white plies or 41 1/2 moves

Proof game with minimal number of moves:

1.Nc3 Na6 2.Nf3 Nc5 3.Rb1 Nb3 4.Na4 Na1 5.b3 Nh6 6.Ba3 Ng4 7.Bd6 Ne5 8.Qc1 Rg8 9.Qa3 Ng6 10.Kd1 Nh8 11.Kc1 g6 12.Kb2 Bh6 13.Kc3 Kf8 14.Kd4 Be3+ 15.Ke4 Kg7 16.Qc5 Kf6 17.Qh5 Bc5 18.Kf4 Ke6 19.Kg5 Qf8 20.Rg1 Qg7 21.Re1 Qd4 22.Kh6 Rd8 23.Ng5+ Kd5 24.Ne4+ Kc6 25.Nf6 Kb5 26.Kg7 Kb4 27.Qh6 Qc4 28.Ne8 Ka3 29.Nc3 Kb2 30.Nd1+ Kb1 31.Kg8 Bd4 32.Ba3 Bg7 33.Bc1 Bf8 34.Nf6 Qa4 35.Ne4 Qa3 36.Ng3 Re8 37.Nh1 Rb8 38.Ne3 Qa6 39.Nd5 Qd6 40.Nb6 Qb4 41.Rd1 Qa3 42.Na8 *

  • $\begingroup$ I think you should add the total number of plies for your lower bound. The reader can make the sum for themselves, sure, for people are lazy or don't trust their maths... $\endgroup$
    – Evargalo
    Commented Jun 8, 2021 at 8:49
  • $\begingroup$ @Evargalo Neither do I trust mine ;-) It's just meant as a stop gap. I'm still hoping to find and prove a minimal proof game. $\endgroup$ Commented Jun 8, 2021 at 8:57
  • $\begingroup$ This is creative! $\endgroup$
    – justhalf
    Commented Jun 9, 2021 at 10:13

First note that

no matter which pawn we move, at least one bishop will be trapped on either side, so we can move at most six non-pawn pieces per side to the other side – the trapped bishop also traps the corresponding rook, albeit in a two-cell jail.

With that in mind, here is a position that can be proved to take

30 moves (60 plies): QNB1Rbrn/ppp1pppp/8/K2p4/4P2k/8/PPPP1PPP/NRBr1bnq

Given the pawn structure, the minimum numbers of plies needed to accomplish the necessary piece moves are:
Free rooks: 7 (white) + 7 (black)
Knights and trapped rooks: 10 + 10
(Free) bishops: 3 + 3
Kings: 5 + 5
Pawns: 1 + 1
Queens: 3 + 4
Total: 59

However, if the proof game had 59 plies, black would have one more ply than white, which is impossible. Hence the proof game has 60 plies, and this is achievable:
1. e4 d5 2. Nc3 Nf6 3. Rb1 Rg8 4. Na4 Nh5 5. Nc5 Nf4 6. Nb3 Ng6 7. Na1 Nh8 8. Ke2 Kd7 9. Kd3 Ke6 10. Kc3 Kf6 11. Kb4 Kg5 12. Ka5 Kh4 13. Ne2 Nd7 14. Nd4 Ne5 15. Qe2 Qd7 16. Qb5 Qg4 17. Qe8 Qd1 18. Bb5 Bg4 19. Re1 Rd8 20. Re3 Rd6 21. Rf3 Rc6 22. Rf6 Rc3 23. Rd6 Re3 24. Qa8 Qh1 25. Rd8 Re1 26. Re8 Rd1 27. Nc6 Nf3 28. Nb8 Ng1 29. Bd7 Be2 30. Bc8 Bf1


1. b4 g5 2. Bb2 Bg7 3. Bd4 Be5 4. Na3 Kf8 5. Rb1 Kg7 6. Rb3 Kf6 7. Rh3 Ke6 8. Qa1 Nf6 9. Rh6 Bg3 10. Kd1 Kd6 11. Kc1 Kc6 12. Nc4 Kb5 13. Bb6 Rg8 14. Nf3 Nh5 15. Qh8 Rg6 16. Kb2 Ka4 17. Kc3 Qg8 18. Kd3 Qg7 19. Ke4 Qa1 20. Kf5 Rd6 21. Kg4 Bh4 22. Rg1 Ng3 23. Ba5 Nh1 24. Kh5 Rd3 25. Rg6 Rb3 26. Nb6+ Ka3 27. Kh6 Nc6 28. Nd4 Kb2 29. Kg7 Kc1 30. Kf8 Kd1 31. Ke8 Ke1 32. Nf5 Nd4 33. Nh6 Nb5 34. Kd8 Qd1 35. Qe8 Rb1 36. Rg8 Rc1 37. Rf8 Rb8 38. Na8 Na3 39. Ng8 Nb1

enter image description here

each side: rooks:7+1,knights:4+5,bishop:4,queen:3,king: 13/14,pawn: 1
77 moves minimum / +1 for queens interfering with each other


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