Retrograde Chess Problems
by Karl Scherer 1997 / published here 2001


How far can you "unplay" the following chess positions?
In other words, find out how many of the previous chess moves are uniquely determined in each of the positions below?
(Problem 1 has also been published at:  Ed Pegg's Retro Pages)

Of course the chess position itself must be "legal", i.e. it must be possible to reach it from the starting position.
Solutions see below

 
 
Retro problem 1
Retro problem 2

Retro problem 3

The beauty of problem 3 lies in the fact that it uses only the standard set of chess pieces,
but that the analysis of the problem will show that three of the actors on the board are actually promoted pieces!



Solutions:

Solution to Retro problem 1
The game can be unplayed for 11 plies. The 6 white and 5 black forced last moves are:
 

White Black
42 Be6-f7 Pf6-f5
43 Rd6-f6 Pf5-f4
44 Rg5-f5 Pg6-g5
45 Qh5-g6 Pg5-g4
46 Bh6-g5 Pg4xf3
47 Bf7-e6
which brings us to the problem position, in which Black has only Pe7xRf6 as a reply.
According to Harry Nelson there is a game with 41 moves which lead to the chess position our table starts with.
Note that move 41 is not forced:  41 Nf6-g8 Pf7-f6 or  41 Bf5-e6+ Kf7-f8

Uniqueness:
We can prove, for example, that White's last move was not Pf7xg8N as follows:
Since Black's original g-pawn captured something white to get to the f-file, Black's original f-pawn cannot have left the f-file and then returned as this would require two more captures and there are only two missing white men.
Thus, a white pawn on f7 could not be White's original f-pawn since Black's queen's rook never left the three corner squares a7, a8 and/or b8, and so the f-pawn could not have taken two black men to get to f7. Neither could a white pawn on f7 have been the original g-pawn, since it would have had to take the black knight to get to the f-file, which means that the knight was not on g8 (and the rook was not there either). Together this implies that one of the white knights must have been a pawn which promoted on a8 after taking the black rook on a7, and is either the original a-pawn or b-pawn.

 


Solution to Retro problem 2
The game can be unplayed for 19 plies! The 10 white and 9 black forced last moves are:
 

White Black
50 Ng4-f6+ Kg8-f8
51 Nf6-g8 Pf7-f6
52 Be6-f7 Pf6-f5
53 Rd6-f6 Pf5-f4
54 Rf5-g5 Pg6-g5
55 Qh5-g6 Pg5-g4
56 Bh6-g5 Ph7-h6
57 Qg6-h7 Ph6-h5
58 Ng8-h6 Pf4-f3
59 Nh8-g6#
which brings us to the problem position in which Black has been checkmated.
Note that move 49 is not forced: 49 ...Kf8-g8 or 49 ...Rf8-e8

Despite the fact that there are more forced retro moves than in problem 1, most of them are easy to prove:

Note 1: None of the white pieces are missing. Hence Black cannot have made a capture in this game.
Hence all of Black's pawns are on their original file, and the missing black a-pawn was captured on the a-file.

Note 2: Two of White's pawns have been promoted to knights during the game.
None of the other pieces on the board can be a promoted piece.
The missing black knight has been captured by a white pawn  before this white pawn promoted to a knight.

Proof of Note 2:
Apart from the missing black a-pawn (which never left the a-file according to note 1), only one black piece is missing, namely a knight. Since the black pawns never left their files and seven are still on the board, in order to promote two white pawns,
one of these white pawns (namely White's f-pawn) must have captured the missing black knight, the other (namely White's a-pawn) must have walked straight forward on its file to promotion.

We now examine every move step my step:

59 Nh8-g6#
The alternate checking move Nf4-g6 would leave Black without a previous move;
remember that Pg6x?h5 is not possible since no white pieces are amiss according to note1.
Also the black king does not have a previous move Kg8-f8 because of the double check on g8.

58 ....Pf4-f3
With no captures by Black, there is no alternative.

58 Ng8-h6
White's move has to give Black a previous move, which cannot be Kg8-f8 or Kf7-f8 because of double
checks on f7 and g8. This leaves Kg7-f8 to consider, triggered by 58 Ne6-g7.
But the move 58 Ne6-g7 would have started off with Black in check (by the knight on e6 !), which is not possible.

57 ....Pf5-f4
With no captures by Black and the king totally blocked, there is no alternative. Same for the the other moves:

57 Qg6-h7
56 Bh6-g5 Ph7-h6
55 Qh5-g6 Pg5-g4
54 Rf5-g5 Pg6-g5
53 Rd6-f6 Pf5-f4
52 ...    Pf6-f5

52 Be6-f7
Note that neither 52 Nf5-g7 nor Ne6-g7 are alternative moves, since both start with Black in check (by Bh6). 

51 ...Pf7-f6

51 Nf6-g8
Here Nf5-g7 is not an alternative, because the move
starts with Black in check (by Bh6). 

50 Ng4-f6+ Kg8-f8
Both moves are obviously forced.

Now we demonstrate a 49 move game which leads to this position:

 1 Nf3 Nf6  11 e4  a6   21 h4  Bd4  31 Kc3  Kg8  41 Rf1 Kg8
 2 Nh4 Nh5  12 Bxa Kf8  22 Bc4 Bc5  32 Pb5  Kf8  42 Rf5 Kf8
 3 Ng6 Rg8  13 a4  Kg7  23 Be6 Bd4  33 Qh5  Kg8  43 Rg5 Kg8
 4 Nh8 g6   14 a5  Re8  24 Nd5 Nb6  34 Ng4  Kf8  44 Kb4 Kf8
 5 f4  Bg7  15 Bf1 Na6  25 Ra6 Na8  35 Nde3 Kg8  45 c6  Kg8
 6 f5  Be5  16 b4  Nc5  26 Rd6 Ba7  36 Nf5  Kf8  46 Nc3 Kf8
 7 f6  Ng7  17 a6  Rb8  27 c4  Kg8  37 Ng7  Kg8  47 Ne4 Kg8
 8 fxN Rf8  18 a7  Kf8  28 c5  Kf8  38 Bh6  Kf8  48 b6  Kf8
 9 g8N Bd5  19 a8N Na4  29 d4  Kg8  39 d5   Kg8  49 Kb5 Kg8
10 Nh6 Rg8  20 Nb6 Bc5  30 Kd2 Kf8  40 e5   Kf8
 


Solution to Retro problem 3
The game can be unplayed for 4 plies.
The 2 white and 2 black forced last moves are:
 

White Black
45 ... Pb7-b6
46 Bc8-b7+ Ka8xBb7
47 Pc7-c8Q #

Yes, these are only 2 moves each, but it is quite difficult to prove that these are the only ones possible:

The move 47. Pc7-c8Q # is obviously the only way how the queen could get to c8 without Black's Kb7
being in check at the start of White's move.

46. ...Ka8xBb7
Only the king can have moved, coming from a8, b8 or c8 and hence out of check.
But on b8 it would have been in double check by Pc7 and Rd8, which is impossible to create by White in one move.
On c8 Black's king would have been in check by Rd8, but the rook had no place to come from.
Hence Black's king came from a8.
To create the check by Rd8, a piece must have moved away from b8 or c8, discovering the rook on d8.
No piece on the board could have done it, therefore the king move must have captured it.
Hence it must have been a chess piece capable of moving from c8 to b7 in the first place, hence a white bishop or queen.
Well, the move Qc8-b7 would have started off with Black in check, which is not possible.
This leaves Ka8xBb7 as the only possibility.

46. Bc8-b7+
The only alternative possibility is that the white bishop captured a piece during this move:
46. Bc8xb7?+  Ka8xBb7. This case is not easly refuted, and we have to analyze carefully.
Let us first take a few basic notes on the situation:

Note 1: We already know that 46. ... Ka8xBb7 is true.
This means that white has a (second) white bishop at b7, which must be a promoted piece.
The white rook d8 must be also be a promoted piece because of the black pawn structure.
Hence no white pawns were ever captured in the game!

Note 2: Four black pieces are missing, namely the white-squared bishop, two knights and a rook.

Note 3: We will now prove that no black pawn has ever left its file (and hence no black pawn has ever captured a white piece).

All we have to show is that both captures b7x?c6 and c7x?b6 are not possible in the same game.
Only two major white pieces are now missing, namly the (original, unpromoted) queen and the white
king's rook which never left g1/h1/h2 (because Bf1 never left its place) until it was captured.
Hence for any captures above the second rank the white king's rook never was available.
Therefore only one white piece was available for the two captures b7x?c6 and c7x?b6. q.e.d

Note 4: We now prove that the promoted white rook on d8 came to pass via Pf6x?e7 and Pf7-e8R,
and that the promoted white bishop on b7 came to pass via Pb6 (or Pd6) x ?c7 and Pc7-c8B.
For any other captures only Black's white-squared bishop was available.

We already know that the two pawns missing must have promoted into Rd8 and Bb7.
According to note 3 the black pawns never changed file, hence each promoting pawn must have
captured at least one black piece before. But the white Pc7 must also have captured a black piece to get to c7,
which leaves three of the four black pieces as possibble captures for the two promoting pawns,
but each can only have captured one black piece, because we assumed 46. Bc8xb7?+  which snaps up
the last black piece available for capture.
In the case of the f-pawn this leaves only the possiblity that Pf6x?e7 and Pf7-e8R happened at some stage.
Note that Pf6xBe7 is not possible because e7 is a black square.
Similarly, Pc7 must have arrived there by capturing the missing black rook or a knight, but not the bishop.

The white Bb7 cannot have been promoted at a8, because this would spend a surplus capture on
Pb7x?a8, but getting to b7 in the first place would have cost a second surplus capture for any missing pawn.
Hence Bb7 was promoted at c8, and for the same reason as above the promoting move did not start at b7.
For the same reason the Pa4 must be the original a-pawn.
This shows that Pb6 (or Pd6) x ?c7 and Pc7-c8B must have happened, and the piece captured on c7
was not the (white-squared) black bishop.

Note 4: We now prove that that 46. Bc8xb7?+ is not possible.

We know from note 4 that only the white-squared black bishop is left available for capture.
So why is 46. Bc8xb7?+ not possible? Well, the problem is that Black would not have a previous move:
The king's corner is totally blocked, and White cannot unblock it within one move.
The only unblocking move can be executed by the bishop e7. But tis implies that the white king was put in check
 by Bf8 before, which is impossible. Black could not have put White in check by Pe7-e6 either
because the pawn on f7 would have been checking the white king already before it started moving! q.e.d.

45....Pb7-b6 is easy to see because of a lack of previous moves for White in all other cases.

Now we demonstrate a game which leads to this position:

 1 Na3 Na6  11 Bg5 Kb8  21 f6   Nb8  31 Ne8 Kb8  41 Kd4 Kb8
 2 Nc4 c6   12 Ne3 Ka8  22 cxR  Nd5  32 a5  Nc7  42 Kc5 Ka8
 3 Nb6 e6   13 a4  Qf8  23 Pc8B Ne7  33 Rb3 Ka8  43 Kd6 Kb8
 4 NxB Bd6  14 Qd4 Bd6  24 PxN  Ne6  34 Rb6 Kb8  44 Nd5 Ka8
 5 Nb6 Qe7  15 Qf6 NxQ  25 Pe8R Kb8  35 Ra6 Bd6
 6 h3  Rc8  16 Kd2 Qf8  26 Rd8  Ka8  36 b4  Be7
 7 Rh2 Rc7  17 Ra3 Qg8  27 Nf3  Kb8  37 b5  Bf8
 8 d4  BxR  18 f4  Bf8  28 Nh4  Ka8  38 b6  Ka8
 9 d5  Kd8  19 d6  Nd5  29 Nf5  Kb8  39 Be7 Kb8
10 Nc4 Kc8  20 f5  Ne7  30 Nd6  Ka8  40 Kd3 Ka8

This proves retro problem 3.


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