Re: [math-fun] Aknother knight's surprise?
[...]
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
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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
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participants (2)
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Adam P. Goucher -
Fred Lunnon