Comments/Ratings for a Single Item

Can anyone find a forced mate in this position: White King c3, White Wildebeest d4, Black King a1? What if we remove the Wildebeest and add two White Camels to d4 and e4? Running Zillions at 3 minutes per move for 20 moves in these positions did not result in a mate.

Glenn E. Overby II writes, on the Recognized Chess Variants page, 'The Wildebeest is markedly superior to a Rook, although not quite the Queen's equal.' I am inclined to value the Wildebeest (Gnu) halfway between the Rook and the Queen. Perhaps less than that, if there is no forced mate against the lone King.

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

By the Way: The possibility with both the Wildebeest and the King on b3
would be stalemate anyway.

For the Camels: the Checkmate positions are:

White King (a3 or b3), White Camel (b4 or d2) White Camel (a4, c4 or e2)
Black King a1
White's (b4 or d2)-Camel moved last, so before that move, the Camel was
on (c1, e3, e5, c7, a7, a3, c5, g3 or g1) (No a1 because of the Black
King).
Black moved Kb1-a1:
Position before that:
White King (a3 or b3), White Camel (c1, e3, e5, c7, a7, a3, c5, g3 or g1)
White Camel (a4, c4 or e2) Black King b1.
Blacks Kb1-a1 must be forced, so c1 must be covered by a Camel (The King
doesn't cover it). So there must be a Camel on b4, d4 or f2, but there
isn't. So no mate with King + two Camels against lone King.

Endgames are my favorite part of chess - but not when they end in a draw due to insufficient mating material! I am unhappy with the limitations of the Wildebeest. Ralph Betza's Half-Duck is a fascinating leaper that does have the ability to mate in the endgame. Under the name 'Lion', it can be found in Greg Strong's Opulent Chess, along with Daniel C. Macdonald's Wizard (Camel+Ferz).

(loss-condition (White Black) (checkmated King))

So Zillions was not even trying to stalemate the Black King. When this ZRF has been corrected, we should be able test the endgame King and Wildebeest against King.

Can we force a stalemate victory in the ending King and Silver General versus King? I once tried running a Zillion - Zillions test game for over 250 moves on an 8x8 board, at one minute per move, with no success. Perhaps the problem here is the fact that the Silver General cannot lose a tempo (triangulate). Neither can the Knight, which alternates colors each move - stalemate wins probably happen in rare situations where the lone King is already trapped in the corner. Even the ending King and Bishop versus King seems to be a draw - I do not see any way to force the lone King to the edge of the board - I have a K+B versus K test game (325 moves) on my hard drive.

The linked game page already has several Excellent ratings. I will also give this variant an Excellent rating, assuming that, in general, King and Wildebeest versus King leads to a stalemate victory. The Wildebeest, while a useful piece, should be no stronger than a Rook on 10x10 and larger boards. It is much weaker than a Gryphon, which is halfway between a Rook and a Queen - see Ralph Betza's analysis of bent riders.

I suspect that the longest possible forced stalemate [K and N versus K] is four moves, as in Joost Brugh's [2006-04-03] comment.

Strangely enough, the Bison (Camel + Zebra) does not share the Wildebeest's checkmating problem (covered in Joost Brugh's earlier comment). See Comments/Ratings for a forced mate in six on the BISON page.