Re: Continued training with octahedron.
- From: "mina_world" <mina_world@xxxxxxxxxxx>
- Date: Fri, 4 Jan 2008 14:42:58 +0900
"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
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"mina_world" <mina_world@xxxxxxxxxxx> wrote in message
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Hello sir~
I showed icosahedral graph by octahedron.
http://mathworld.wolfram.com/IcosahedralGraph.html
There is Hamiltonian cycle count 2560 in the below table of link.
Maybe, this is equal to edge's cycle in octahedron.
(Namely, One ant is on a vertex of octahedron.
This ant moves on edge of octahedron.
This ant visits each vertex exactly once and also returns to the starting
vertex.
This count is 2560.)
Anyway, my question is...How can you calculate it(2560) ?
Sorry. modify : octahdron ==> icosahedron.
For reference,
Tetrahedral graph ==> Hamiltonian cycle count 6
Cubical graph ==> Hamiltonian cycle count 12
Octahedral graph ==> Hamiltonian cycle count 32
Dodecahedral graph ==> Hamiltonian cycle count 60
Isosahedral graph ==> Hamiltonian cycle count 2560
Maybe, It looks like direct counting.
http://board-2.blueweb.co.kr/user/math565/data/math/hem001.jpg
and It seems that these diagrams are equivalence in solid figure.
so,
One ant is on a vertex of icosahedron.
This ant moves on edge of icosahedron.
This ant visits each vertex exactly once and also returns to the starting
vertex.
I think...
If I igonore the rotation permutation of solid,
this ant's pass(cycle) is unique.
so, maybe...
answer is...(with rotation group)
tetrahedron ==> 12
cube ==> 24
octahedron ==> 24
dodecahedron ==> 60
icosahedron ==> 60
Of course, my idea is intuitive.
so, I need exact explanation.
.
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