On 4/15/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Oh, sorry, I didn't see that you needed it to be an embedding. It can't be embedded on the torus, since it has 36 vertices, 144 edges and an Euler characteristic of at most:
36 - 144 + 96 = -12
So, if you are to embed it on a surface without crossings, you'll need to have at least seven holes.
Whoa! ... where does your 96 come from? WFL __________ There exist two distinct planar immersions of the smaller graph with maximal (D_6) symmetry: to accompany the diagram of the first at https://www.dropbox.com/s/8b0stotr8343mq2/semi_6x6.pdf I have posted the second at https://www.dropbox.com/s/gyispnh717ayz86/semi0_6x6.pdf In the latter case, the complement of the exterior tour splits into 3+1 smaller circuits, diagonals skipping 3 unequal intervals (rather than 2). Less elegant perhaps; but 4-circuit faces and 6-circuit zones are more readily discernable. Fred Lunnon
----- Original Message ----- From: Fred Lunnon Sent: 04/15/14 11:31 PM To: math-fun Subject: Re: [math-fun] Aknother knight's surprise?
But why is the result an embedding? WFL
On 4/15/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
[...]
Whichever, is there now some way to reverse the previous identification, and lift the embedding back to 36 vertices?
Yes! You implied the existence of such a procedure yourself, when you said "vertices opposite across the torus have all neighbours in common".
Specifically, for every vertex v in the 18-vertex graph, create vertices v_1 and v_2 in the new graph. Then connect u_i with v_j in the 36-vertex graph iff u is connected to v in the 18-vertex graph.
The 36-vertex graph should have 2^18 times more symmetries than the 18-vertex graph, given by the wreath product C_2 wr G, where G is the automorphism group of the 18-vertex graph (as a subgroup of S_18) -- the one you claim (probably correctly) has order 144.
Sincerely,
Adam P. Goucher
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