Each directed edge in the graph represents a term in a big Hamiltonian (or, more precisely, i times the Hamiltonian) that looks like
[ 0 1 ] [ -1 0 ]
(with the direction of the arrow asymmetrically picking out which vertex gets the +1 and which gets the -1) and also there's undirected edges that represent
[ 0 i ] [ i 0 ]
[ 0 -i ] [ -i 0 ]
as indicated. The vertical columns of arrows always run up, and the horizontal rows of arrows alternate left and right as you go up, and the i/-i self-edges alternate checkerboardily. This guarantees that the sign of (for instance) an up-right path is always the negative of a right-up path: so those two possibilities cancel out. Also, a [self-loop then take a step] cancels out a [take a step then self-loop]. The only two-step paths that don't cancel in this way are
(a) two steps in the same direction
(b) a step forward and then back in the same direction
(c) two self-loops
When I try to compute the double-time derivative of the wave-function ψtt, (a) and (b) together yield a ∇2 term, and (c) yields -ψ. So I wind up with a Klein-Gordon □2ψ = -ψ.
The picture on the left is an implementation of this, and indeed it looks more properly circular than my last attempt!