Single Comment

For two Camels, two Knights or a Wildebeest we can prove that the longest
forced mate is one move, because the geometry allows only a few mating
patterns. This two-move example with two cannons is not trivial. Is it
possible to prove that two moves is the maximum. Known is that the mating
patterns is always with the Black King on the side (X1), White's King on
X3 or on b3 against when X = a. And White's cannons are at Y1 and Z1 with
Y between X and Z and Y not adjacent to Z. The last move is a vertical move
by a cannon (C YA-Y1 or C ZA-Z1). Blacks last move is a horizontal King
move, which can only be forced if the end file is involved (Second rank
squares can only be covered by the White King), so this must be Ka1-b1,
which means that X = b. One retromove by a cannon later, c1 must be
covered. This is impossible. With two (Cannon + passive Bishop)-pieces
(passive Bishop is a Bishop that does not capture), it should work (from:
White CmB on c4 and c5, White King King b3, Black King b1, Black to move)
1. ...,Kb1-a1 2. CmB c4-f1, Ka1-b1 3. CmB c5-g1#. Probably (not certainly,
it should be possible to force this with two CmB's and a King against a
lone King.

It would be interesting to prove this (and of course the
King+Cannon+Knight against King)