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This week's problem is something of an experiment for me, since I have not previously
put up any fairy problems, and I know some solvers are allergic to them. But if you
can't have a fairy at Christmas, when can you? So this is a helpmate in 2, which
has four solutions, but with the Circé condition: with the exception of kings,
a captured unit, of either side, is instantly reborn on its original game-array square,
if that square is free (if the square is not free, the captured piece just disappears
from the board in the usual way). Knights and rooks are reborn according to the colour
square on which they are captured, eg. if the bRb8 in the diagram is captured on that
square, it would be reborn on h8 (a dark square), but if it were to move to a white
square and then be captured there, it would be reborn on a8. Pawns are reborn on the
starting square of the file on which they are captured, so the wP on a7, if captured
there, would be reborn on a2, etc.
A slightly long explanation, but I hope your solving journey will be less lengthy!
As might have been expected, we have an Allumwandlung, with all four
possible promotions for the white pawn, as follows:
a) 1.Bg3 axb8Q (bR–h8) 2.Bxb8 (wQ–d1), Qxe1 mate. Note that Black cannot
play 3.Kxe1, because the wQ would instantly re-appear on d1, so Black would still
be in check – the Circé effect!
b) 1.Re2 a8R 2.Qxa8 (wR–h1) Rxg1 mate. Once again, 3.Kxg1 is illegal,
because the wR reappears on a1, giving check.
c) 1.Qf3 a8B 2.Qe2 Bg2 mate, Again the bishop is immune because 3.Kxg2
results in the wB being reborn on f1.
d) 1.e2 a8S 2.Rxa8 (wS–b1) Sd2 mate.
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