Re: Continued training with octahedron.

mina_world wrote:
"mina_world" <mina_world@xxxxxxxxxxx> wrote in message news:flisvi$cn$1@xxxxxxxxxxxxxxxxxxx
"mina_world" <mina_world@xxxxxxxxxxx> wrote in message news:flireg$p$1@xxxxxxxxxxxxxxxxxxx
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

Here I think this means there are 6 Hamiltonian
cycles, where cycles have a direction, which
can be shown by arrows on the edges that are visited.

For example, Wikipedia has a figure of a directed cycle
here:
< http://en.wikipedia.org/wiki/Path_%28graph_theory%29 >

Also, I think in graph theory for the graphs
of the icosahedron and so on, "rotating in
space" is not allowed and rotating in the
plane is not allowed when counting
Hamiltonian cycles. It's a convention
of graph theory.



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)

I don't think rotations count. The graph
sits in the plane in a fixed position. An
ant walks on the edges and returns, having visited
every vertex exactly once. How many ways, if clockwise
and anticlockwise are considered different?

That's my impression ...

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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Relevant Pages

  • Re: Continued training with octahedron.
    ... I showed icosahedral graph by octahedron. ... There is Hamiltonian cycle count 2560 in the below table of link. ... this is equal to edge's cycle in octahedron. ... One ant is on a vertex of icosahedron. ...
    (sci.math)
  • Re: Continued training with octahedron.
    ... I showed icosahedral graph by octahedron. ... There is Hamiltonian cycle count 2560 in the below table of link. ... this is equal to edge's cycle in octahedron. ... One ant is on a vertex of icosahedron. ...
    (sci.math)
  • Continued training with octahedron.
    ... I showed icosahedral graph by octahedron. ... There is Hamiltonian cycle count 2560 in the below table of link. ... this is equal to edge's cycle in octahedron. ... This ant moves on edge of octahedron. ...
    (sci.math)
  • Re: Continued training with octahedron.
    ... I showed icosahedral graph by octahedron. ... There is Hamiltonian cycle count 2560 in the below table of link. ... this is equal to edge's cycle in octahedron. ... This ant moves on edge of octahedron. ...
    (sci.math)
  • Re: Labelling polyhedron faces
    ... Graph Theory. ... You're looking for a Hamiltonian cycle in the icosohedron ... To get the other condition (n is opposite n+6), ...
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