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Chess Problem 1990c5c701


feenschach, 1990


LEO a3
PAOs b2; b6
VAO d6
MAOs c2, d4

1.LEa1!  ()

1...MAb4 2.LEa5#  (MAOc2 obstructs pb4)
1...b4   2.LEc1#  (pb4 obstruct MAOc2)
1...MAc6 2.LExa7# (MAOd4 obstructs PAOb6)
1...PAc6 2.LEg1#  (PAOb6 obstructs MAOd4)

Four hurdles are incarcerated leading to four anti-battery mates by the white LEO. Within these four variations are two pairs of mutual incarcerations, as indicated in the solution.

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