Jörg Knappen wrote on Thu, Nov 21, 2013 08:54 AM UTC:
I think a got a proof for the hex geometry.
We orient the hexes such that there is a horizontal line of rook movement,
and denote that direction by 1. The other directions of rook movement are
denoted by \omega and (\omega-1) [the use of the letter \omega is
inspired by Eisenstein numbers]. The centre of a hex is given by a+b\omega
with a,b integer numbers.
First step is a drawing: When we go horizontally firs and vertically as a
hex bishop second, we can reach only one half of the hexes (a+2b\omega).
We repeat this for the other rook directions and mark the hexes
accordingly. They fall in two classes: (i) hexes which can be reached in
one way only (ii) hexes that can be reached in all three way.
The second class forms a grid described by 2a+2b\omega (both coordinates
must be even.
Finally we map these to rook and bishop moves. The path to a three-way
reachable hex (2a+2b\omega) using horizontal and vertical moves
(elementary vertical bishop step: (2\omega -1)) consists of b bishop steps
and b+2a rook steps. Therefore the number of rook and bishop steps are both
odd or both even, giving an even SOLL.
The other direction: Take r rook steps and s bishop steps and demand that
r+s is even. Then we go to r+s*(2\omega -1) = (r-s) +2s\omega. This is a
three-way reachable square again, because (r+s) even implies (r-s) even.