Retractors are problems where it is asked to retract
some "last" moves. A typical stipulation being "White retracts
his last move and then mate in one".
This is more than just asking for the last
move(s) because there can be a forward play and/or some retro-strategy.
One way to look at retractors is to consider they are fairy problems
where the moves happen to be retractions.
These problems have a retro-flavor because only legal last moves can
be retracted, but they also have the usual, forward, combinatorial flavor
because you have to pick the right retraction, the one that will allow
e.g. to mate in one.
The retro-analytical content of retractors is not necessarily crucial
(or deep). But it may be, as in the following example:
1st Prize Die Schwalbe 1934
15+10. White retracts one move, then mates
Solution: White retracts -1. O-O-O !!,
reaching a position where Black last move must have been g7-g5, so that
forward mate with 1. h5xg6 e.p. is legal. See
full solution for details.
When longer retractions are asked for, the problems also have a cooperative
helpmate flavor because you must pick the right retractions
for both sides.
Dr. Julio Sunyer
The Chess Amateur, 1923
1+1. White and Black retract 1 move, then
helpmate in one
Solution: retract -1. Kg6xRh5 Rh8xQh5 and play forward
1. O-O Qh7#.
These problems include the very weird (my subjective opinion)
By contrast, there exists defensive retractors ...
In these problems, you retract both white and black moves. Black is
defending by trying to retract the moves that will
forbid (or delay) the achievement of White's goal. The canonical example
of a defensive retractor is the Proca-retractor stipulation:
A Proca-retractor in 5 (say) asks for White to retract 5 moves and
then mate in 1 with a forward move. After every White retraction (but
the last) Black retracts what he wants, trying to defend against the
forward mate. White retractions must end in a forward mate against any
retractive defense by Black. As in direct problems, White is allowed
to mate forward in 1 at any time (after a retraction) and this may happen
if Black picks a poor defense. Also, Black may defend by mating White
through a forward move if such an opportunity exists after one of his
5+10. Retract 7 moves, then mate in one (Proca retractor)
White retracts -1. Kd2xBe1!, forcing -1...
e2-e1=B+. In the rest of the solution, White goes on retracting
moves, putting his King in check and forcing Black to retract the only
possible last move: -2. Kc3-d2 e4xPd3 e.p.+ -3. d2-d4
(forced, after the retraction of an en-passant capture) e5-e4+
-4. Kd3xPc3 b4xPc3 e.p.+ -5. c2-c4 b5-b4+ -6. Kc4xRd3 c6xPb5+
(forced, see below) -7. Kc5-c4 and forward
1. b6 mate.
We let the reader check that all moves and intermediary positions
are legal. A key element in the solution is that the uncapture of a
wPb5 by Black is forced (see full
solution) because of the inventory and balance (the wPb2 could not
leave its column). If White had retracted -4. Kc4-d3?? without uncapturing
a R, then Black could have legally retracted e.g. -6 ... c6xBb5+! with
no forward mate.
604 - T. Le Gleuher
Europe Echecs 433, 04/1995
12+11. Retract 52 moves; then mate in one (Pacific
-1. Kg5-g4 c5-c4+ -2. Kh4-g5 Bh2-g3+ -3. Kg3-h4 Bg1-h2+ -4. Rf1-f2
Bd2-e1+ -5. Kf2-g3 Bh2-g1+ -6. Kg1-f2 Bg3-h2+ -7. Rf2-f1 Be1-d2+ -8.
Kh2-g1 Bh4-g3+ -9. Kg3-h2 Bg5-h4+ -10. Kh4-g3 Bh6-g5+ -11. Kg5-h4 Bg7-h6+
-12. Kg6-g5 Bg8-f7+ -13. Kh7-g6 Bf7-g8+ -14. Kh8-h7 Bh6-g7+ -15. Kg7-h8
Bg5-h6+ -16. Kh6-g7 Bh4-g5+ -17. Kg5-h6 Bg3-h4+ -18. Kh4-g5 Bh2-g3+
-19. Kg3-h4 Bg1-h2+ -20. Rf1-f2 Bd2-e1+ -21. Kf2-g3 Bh2-g1+ -22. Kg1-f2
Bg3-h2+ -23. Rf2-f1 Be1-d2+ -24. Kh2-g1 Bh4-g3+ -25. Kg3-h2 Bg5-h4+
-26. Kh4-g3 Bh6-g5+ -27. Kg5-h4 Bg7-h6+ -28. Kg6-g5 Bg8-f7+ -29. Kh7-g6
Bf7-g8+ -30. Kh8-h7 Bh6-g7+ -31. Kg7-h8 Bg5-h6+ -32. Kh6-g7 Bh4-g5+
Now a third cycle would lead to three-fold repetition and draw. So
that we must first move the bBf7:
-33. Kg6-h6 Bg8-f7+ -34. Kg5-h6 Bg3-h4+ -35. Kh4-g5 Bh2-g3+ -36. Kg3-h4
Bg1-h2+ -37. Rf1-f2 Bd2-e1+ -38. Kf2-g3 Bh2-g1+ -39. Kg1-f2 Bg3-h2+
-40. Rf2-f1 Be1-d2+ -41. Kh2-g1 Bh4-g3+ -42. Kg3-h2 Bg5-h4+ -43. Kh4-g3
Bh6-g5+ -44. Kg5-h4 Bg7-h6+ -45. Kh6-g5 Bh8-g7+ -46. Kh7-h6 Bf7-g8+
-47. Kg6-h7 Bg8-f7+ -48. Kf7-g6 Bh7-g8+ -49. Ke8-f7 Bc8-d7+ -50. Kd7-e8
Bb7-c8+ -51. Nd5-c3
and now, because of the fifty-moves rule, Black must retract a pawn
move !! Only possibility is -51 ... a7-a6. White goes on with -52. Nb6-d5
and plays forward 1. c7 mate.
Here after any retraction by Wh. or Bl., the other side may decide
that the retracted move has been a capture, and decide which piece will
be resurrected. Of course, legality must be preserved at all times.
Here is an example:
Tidskrift för Schack, 1924
5+1. Retract 2 moves; then mates in one (Høeg-retractor)
The solution starts with White retracting -1. e6-e7 !!.
This cannot be a capture, so Black does not get a chance to choose which
bl. man was captured.
Now Black must retract a King move. The King could not have come from
d4 or e5 (illegal double check). But whichever other square Black chooses,
allows White to choose a wh. man that Black just captured, and to retract
another wh. move such that he instead can mate Black with a forward
move. It turns out that Black's choice of which bl. man white captured
(if any) on White's second retraction makes no difference:
(Usually retracted moves are written like Kd5e4(R) to emphasize
that Black chooses the square d5, after which White chooses the captured
man, here a wR. But we shall just write Kd5xRe4.)
-1. e6-e7 Kd5xRe4 -2. Be7-f6, and forward 1. Qa8#
-1. - Kf3xRe4 -2. Bh4-f6, 1. Qh1#
-1. - Kf5xRe4 -2. h5-h6, 1. Qe5#
-1. - Ke3xBe4 -2. Be5-f6, 1. Qe1#
-1. - Kd3xRe4 -2. Qa2-a1 (or other moves), 1. Qe2#
-1. - Kf4xBe4 -2. Qa3-a1 (or other moves), 1. Qf3#
It is interesting, and indeed the point of the problem, that White
should never choose that Black captured the strongest wh. man: a Queen.
With another white Queen it is easy to mate, but Black chooses to resurrect
a man that prevents the mate, e.g. -1. e6-e7 Kd5xQe4 (wQ??) -2. Qc2xQ/Re4
(bQR!!) and there is no mate.
There is hardly any retrograde analysis in this problem, but it demonstrates
well the richness of variations that the
Proca type rarely shows.
(Many thanks to Henrik Juel for suggesting this example and spelling
out the solution.)