Single Comment

There is no forced mate with Wildebeest + King against lone King:

There are two different checkmates with a Wildebeest and King against a
lone King (not counting mirror images or rotated images):
a. White Wildebeest d2, White King b3 and Black King a1
b. White Wildebeest d2, White King a3 and Black King a1
The Wildebeest moved last. Before that move, the positions were
a. White Wildebeest X, White King b3 and Black King a1
b. White Wildebeest X, White King a3 and Black King a1
X is a position from where a Wildebeest can move to d2
Blacks last move is Kb1-a1. Before that move, the positions were
a. White Wildebeest X, White King b3 and Black King b1
b. White Wildebeest X, White King a3 and Black King b1
Blacks Kb1-a1 must be forced, so all other squares must be covered. Square
c1 can't be covered by the White King, so c1 must be covered by the
Wildebeest, so X is a position that both covers d2 and c1. There is only
one solution: X = b3. So possibility a is impossible, because both the
Wildebeest and the King must be on b3. This leaves possibility b, but
there, the Black King can move to c2