There are 18 quadrilateral circuits, 4 per vertex and 2 per edge; we can take them to be faces of a (topological) polyhedron, with Euler characteristic zero. It follows that the graph is embeddable either on a torus or on a Klein bottle. See http://en.wikipedia.org/wiki/Euler_characteristic Given how the graph was constructed, I would have expected the latter. However, there exist 6 distinct closed zonal "collars" each comprising 6 faces, and all untwisted. Together these cover the polyhedron doubly, in (locally) perpendicular directions; so it's hard to imagine how it could fail to be orientable, and hence toroidal! Whichever, is there now some way to reverse the previous identification, and lift the embedding back to 36 vertices? Fred Lunnon On 4/15/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 4/15/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I have posted a diagram, showing as much of the symmetry of this graph as planar representation allows, at https://www.dropbox.com/s/aeea6bqlrt23tck/semi_6x6.png Note the 4-fold diagonal coincidences --- whereby hangs another tale entirely!
Ugh --- grotesquely aliased lines --- see instead https://www.dropbox.com/s/8b0stotr8343mq2/semi_6x6.pdf
The external edges obviously form a tour (closed Hamiltonian path); the remaining diagonals also form a tour, isomorphic to the first.
Fred Lunnon [15/04/14]
On 4/14/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Since you dispatched 4x4 so readily, perhaps you can manage something about the graph genus 6x6 knight moves on a torus instead?
In the 4x4 case the vertices had valency 4 instead of 8 . The 6x6 case degenerates as well, less predictably: vertices opposite across the torus have all neighbours in common; identification yields 18 vertices and 36 edges and valency 4 immersed on a Klein bottle. The symmetry group has order 144 .
Fred Lunnon