# Who are these rooks?

Here is my newest retrograde chess puzzle, perhaps a bit on the trickier side once again:

Can you determine the origin of each of the four rooks in this position?

• (joke answer) a1, h1, a8, h8 Commented Jul 2 at 13:13
• @kscherrer Interesting theory! Did you have a go to see if it's correct? :) Commented Jul 2 at 14:31
• absolutely not x) All I needed to know is that up to 1 white rook could have been promoted until I gave up. The whole pawn situation is too messed up for me (looking at you, pawn on f7). Maybe with a lot of captures? but theres so few pieces missing.. My brain wants to abandon ship immediately. Commented Jul 2 at 14:40
• Yes, very nice work, it should be published... Warmly upvoted! Commented Jul 4 at 1:05
• A good task. I liked it. For publication, it is better to simplify the question: "The History of Rooks". Commented Jul 14 at 11:58

Here are some thoughts, that restrict the road to a possible solution.

• Some pieces never moved, acting as a wall for the story, they are: ♗f1, ♙c2,d2,e2,g2; ♟c7,d7,e7.
• Black misses exactly one piece, it is the ♝f8.
• White misses exactly two pieces, the ♗c1 first of all, and a pawn.
• The pawn is either ♙f2 or ♙h2 (originally).
• The ♙f7 thus comes either from f2, or from h2. Since there is at most one white capture, particularly also with a pawn, the ♙f7 originates from f2.
• In order to get from f2 to f7, this original ♙f2 needs an even number of captures. So it did not capture anything, and its road was (with other moves in between) ♙f2-(♙f3-)♙f4-♙f5-♙f6-♙f7.
• Previously to ♙f7 the place f7 must have been vacated from ♟f7. Then it must leave the f-file to evacuate the walk of its f-counterpart. This can be possible only by a capture ♟f7xg6.
• The final positions of the pawns originally ♟f7, ♟g7, ♟h7 are also on the same lines. At most two captures are possible. One of them is ♟f7xg6. Finally there is also a ♟ on f6. So the other must-be move is ♟g6xf6. This shows that ♟h7 finally lands as ♟h2 with no capture.
• What are the captured white pieces? There are exactly two missing pieces for white. The missing ♗c1 must thus have been captured, this is possible only with the move ♟g7x♗f6.
• What is the other capture? Well, it is a piece on a light square g6. The ♗f1 is excluded, being blocked all the time by the own pawns on e2,g2. So it is one of the white knights, white rooks, white queen, or a promoted ♙h2 to something that can reach a light square, well to anything.
• On the board there are all pieces that may have been used to offer themselves on g6 for the move ♟f7x♗g6. So that kind of piece must also appear through a promotion at some point. So there is a promotion of ♙h2.
• How can we promote ♙h2? It must avoid the walk of ♟h7-(♟h6-)♟h5-♟h4-♟h3-♟h2. It can do so using at most one capture, avoiding the wall of the ♟f7-g6, ♟g7-f6, ♟h?. The promotion square must be g8, to avoid the file of ♟h7-h2, else two captures are needed.
• ♚e8 must reach a6. The a8-corner and all the heavy pieces around are in there before the ♚a6 reaches a6. The ♚-path is either through c8 or outside the wall ♟c7,d7,e7. Well, the last move of ♚a6 was ♚b5-a6, so anyways, the ♚e8 must reach somehow b5. I will take the path through c8, b7, c6, the outside path ♚e8-f7-e6 cannot be realized.
• There is a problem with the knights in the corners. They must reach their positions before a pawn blocks the entry squares.
• Things must thus be done in order: White offers a piece on g6, ♟f7xg6, ♚e8 escapes through f7 outside the last rank, ♙f2 goes to f7 - but before f7 gets occupied we need the moves ♜h8-g8, ♞f7-h8, then ♗c1 finds its way to f6, but before we realize ♗c1-b2-f6 (or ♗c1-a3-...-f6), the ♖a1 must escape from the a1 corner, and there should be a ♘a1 placed in there, now we have the move ♗f6, black evacuates g7 via ♟g7xf6, and now ♙h2 moves to h6, must take the only missing piece on g7 after ♝f8-g7, ♙h6xg7, then the square g8 must be free and we play one of the moves ♙g7-g8, promotion into the piece offered long time ago on g6.
• In the above chain of moves, some moves are critical. They induce rigidity in the position, blocking wide areas or certain key pieces. One such move is the move ♙a4. After this move there is no chance for a ♚a6 to land there or to leave this place. So ♚a6 is reached before a4. When ♚a6 reaches that square, the three white pieces on a8, b8, b7 must be there. In particular, ♖b7 is not originating from ♖a1. (It is either ♖h1 or ♙h2 at the start of the game.) But only after ♙a4 we may move ♙b3, bring the ♗c1 to f6 and thus make room for a promotion. For short, ♖b7 is initially the ♖h1.
• Which is the piece offered on g6? Not the ♕d1, which must find her place on ♕b8 before ♚a6, and ♙a4. Also not ♘b1, ♘g1, since they must be on ♘a1, ♘a8 before b6 offers a chance to the white pieces in the a8-corner, followed by ♚a6, followed by ♙a4, by the ♖a1 escape, then by ♙b3. Only ♖a1 has a chance.
• Finally the position has two ♖s, so the second rook, ♖g1, comes from the promotion g7-g8♖.
• We also have the order of the black rooks reaching their final destinations.
• ♜h8 cannot leave the h8-corner and the last rank before ♟h7 is moved. So it happens after the promotion, after ♙a4, ♙b3, thus after a black rook invades the a1 corner. This means that the ♜a8 finally reaches ♜b2, so that ♜h8 hides some time on e8 (f8), waits for the promotion, and gets out to the final ♜h1, of course after its cousin reaches ♖g1.

So far we have cleared which rook in the final table is which piece from the initial one. It remains to get a realization of this scenario. Below there is a pgn formatted solution.

The above link illustrates the solution on the lichess server.

[Event "PUZZLE STACK EXCHANGE 127248: Chapter 1"]
[Site "https://lichess.org/study/uLnJP6Qc/vr5env4q"]
[Annotator "https://lichess.org/@/dan_fulea"]
[UTCDate "2024.07.04"]
[UTCTime "00:39:08"]

1. h4 a5 2. Rh3 Ra6 3. Rb3 Re6 4. Rb6 Nc6 5. Ra6 Ne5 6. Nc3 Nf6 7. Nd5 Rg8 8. Nb6 Ng6 9. Na8 { A first key position is reached, the rook from h1 and one knight are inside the cage around a8. } 9... Nh8 10. Nf3 { This knight points to a1, Ra1 will make a tiny space for it. Note that a4 is not available till the Ke8 reaches a6. And to reach a6, there will be some shuffle in that a8 corner. } 10... b6 { Rien ne va plus in the cage around a8. } 11. a3 Bb7 12. Nd4 Bd5 13. Nb3 Qb8 14. Ra2 Kd8 15. Na1 Kc8 16. f4 Kb7 17. Kf2 Kc6 18. Kg3 Kb5 19. Qe1 Bb7 { Not optimal, but it is late here, the idea is to exchange the places of Qe1 and Qb8 in the long run. } 20. Qf2 Bf3 21. Qd4 Qb7 22. Ra7 Qe4 23. Kf2 Qf5 24. Qe4 Qh3 25. Qb7 Qh2 26. Qb8 Bb7 27. h5 Bc8 28. Rb7 Ka6 29. a4 { After a4 now the a8 cage is completely closed. The next step is to bring Ra2 to g6, so that fg6 is possible, thus making room for the march of the white f-pawn. } 29... Rd6 { We are wasting some black moves... } 30. Ra3 Nd5 31. Rg3 Re6 32. Rg6 fxg6 33. f5 Re3 34. f6 Ra3 35. f7 Ra2 36. b3 { Now the black rook is also inside its final cage delimited by the pawns a4, b3, c2, d2, e2, and Bf1. } 36... Nf6 37. Bb2 Nd5 38. Bf6 { A must be move for white, it must come after the white pawn sits on f7, and is - combined with f7xRg6 the only chance to get the final structure of black pawns on f6, g6. } 38... gxf6 39. h6 Bg7 40. hxg7 Rd8 41. g8=R h5 42. Rg7 Rg8 43. Rh7 Nf4 44. Ke3 Rb2 45. Kd4 Rb1 46. Kc3 Rd1 47. Kb2 Qh4 48. Ka2 Qe1 49. Ka3 h4 50. Ka2 Rb1 51. Ka3 Rg7 { Now the two touching rooks will have a march together to the final places g1 and h1. } 52. Rh5 Rh7 53. Rg5 Rh5 54. Rg3 Rg5 55. Rh3 Rg3 56. Rh1 Rh3 57. Rg1 Rh1 58. Ka2 h3 59. Ka3 h2 { It remains to get the cage around the a1 corner in its final stage. } 60. Ka2 Nd3 61. Ka3 Nb2 62. Ka2 Qc1 63. Ka3 Nd1+ 64. Ka2 Rb2+ 65. Ka3 Qb1 *
• Thanks for the solution and the kind words, very well done! You uncovered every little thing about the position :) Commented Jul 4 at 7:26

A brief retro solution.

Balance

Black (2): g7xBf6, f7xg6

White (1): h6xBg7

Before you do anything, you need to give tempo moves to white. 1. ... Qc1-b1 2. Ka2-a3 Rb1-b2+ 3. Ka3-a2 Nb2-d1+

We have set free wK, bQ, bN. bR on a2.

For further actions, Bishop c1 must be returned home. You can get it on f6. To do this, you need to return the piece turned from a pawn. It can only be the rook g1.

bh2-h4; bRh1-f8; wRg1-g8; wg7-g8=R; bRf8-g8; bh4-h7; wh6xBg7; bBg7-f8; bg7xBf6; wBf6-c1; wb2-b3.

bRa2 and wNa1 is free. Now we need to bring the white rook to a1. And you can only take it on the g6.

wf7-f3; bf7xRg6; wRg6-a1; wa2-a4.

bK and wQ free; wRb7-a6; bBc7-d5; bKe8; bQd8; bBc8; b7-b6.

Everyone goes home.

In total, the rooks maneuvered like this: wRa1 died on g6; wRh1 on b7; wPh2-Rg8-g1; bRa8-b2; bRh8-h1.