Re: [math-fun] Aknother knight's surprise?
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 are 144 edges. The maximum number of faces is achieved when every face is a triangle, in which case there are (2/3).144 = 96 triangles. Sincerely, Adam P. Goucher
On 4/16/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Whoa! ... where does your 96 come from? WFL
There are 144 edges. The maximum number of faces is achieved when every face is a triangle, in which case there are (2/3).144 = 96 triangles.
Sincerely,
Adam P. Goucher
Ah, but there are no triangles: both graphs are bipartite, with girth 4 . So the number of faces equals 72 maximum; now via your reasoning, the genus of the larger graph equals at least (144 + 2 - 36 -72)/2 = 17 . Also the genus is bounded above by that of the complete bipartite graph, which via Ringel's (1955) theorem equals Ceiling( (18-2)^2/4 ) = 64 . Not such a nice surprise ... Fred Lunnon
Er --- (144 + 2 - 36 - 72)/2 = 19 . In fact, this almost certainly is the exact genus; and extends trivially to genus = n^2/2 + 1 for the n x n toric board, at any rate when n even. Fred Lunnon On 4/16/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 4/16/14, Adam P. Goucher <apgoucher@gmx.com> wrote:
Whoa! ... where does your 96 come from? WFL
There are 144 edges. The maximum number of faces is achieved when every face is a triangle, in which case there are (2/3).144 = 96 triangles.
Sincerely,
Adam P. Goucher
Ah, but there are no triangles: both graphs are bipartite, with girth 4 . So the number of faces equals 72 maximum; now via your reasoning, the genus of the larger graph equals at least (144 + 2 - 36 -72)/2 = 17 .
Also the genus is bounded above by that of the complete bipartite graph, which via Ringel's (1955) theorem equals Ceiling( (18-2)^2/4 ) = 64 .
Not such a nice surprise ...
Fred Lunnon
participants (2)
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Adam P. Goucher -
Fred Lunnon