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or directly on the move in the solution

1 - 2#

Trillon, JM.

white Bb8b7 Kd5 Qb4 Pf5g6h7f2f3g3b5 Sf1d3 Rf4h6 black Pa7b6c5g7e5g2 Kh8


2 - h#4,5 avec camel & zebra

Camel : hopper (3,1) (quick reminder: the Knight is a hopper (2,1))
Zebra : hopper (3,2)

Pachl, F.

white Kf3 CAf1 black Bh8a8 CAa6f2 Qb7 Pc3d5b5c2 Sf5b1 Rh1h2 Zf6 Kb3
f6: Zebra
f1, a6, f2: Camel


3 - h#2 3 solution anti-circe

anti-circe : after a capture, the capturing piece returns to its original square if it is free, otherwise the capture is illegal

Tritten, P.

white Qb1 Kg7 Rd1 Sh1 black Be4h2 Kf5 Qc2 Pg5g4e6d2 Sd7 Rb7
h#2 anti-circe(4+10)
3 solutions



4 - h=2 2 sol with princess

Princess : piece with both the Knight and the Bishop's march

Bedoni, R & Kerhuel M

white Pd7e7f6 Kf3 black PRf8 Kd6
2 solutions


5 - 2# masand

masand : all units attacked by the checking piece change colour. Except of course the King.

Gockel, H.

white Bd5 Ka4 Qc1 Pd7b6h4h3 Se5f4 Rd6e1 black Pc6a6b3c3f6h7h5 Kf5 Rc7 Sd4 Bg8
#2 masand(11+11)


6 - 2# isardam

Isardam : two opposing units of the same nature cannot observe each other: both Qxf5 and ...cxd5 & ...Rxb7 are illegal.

Gockel, H.

white Qc2 Pb7c5c4 Kb8 Rc7d5 Sc3e7 black Bg8 Ke6 Qg6 Pc6e3f5f6 Sf7 Rb6g7
#2 Isardam(9+10)


7 - hs#3 eiffel

Eiffel the units paralyse according to the P-N-B-R-Q-P cycle.
The author of this problem is also the inventor of the Eiffel condition.
helped selfmate : the 2 sides collaborate until the penultimate move after which white plays and black is forced to mate

Petkov, P.

white Pd7g3 Be5 Rg7 Kh1 black Pe7d4e3e2c2h3 Kh6
hs#3 Eiffel(5+7)
b) +PNh5


8 - 2# optional replacement

optional replacement: a side that captures a unit can replace it wherever it wants (except for self-check) as in Super-Circe,
except for two differences:
Bishops must respawn on squares of the same colour, pawns cannot respawn on the 1st or 8th row.

Beasley, J.

white Qe4 Pd7e6a2h7f5d4 Sb1 Rg2 Ka5 black Pc6g5 Sf8 Be2 Kc4
optional replacement


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