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
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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!
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 |
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# |
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.