- View SourceHello All,

Is there a formula for giving the number of possible configurations for

connecting sequentially all twelve faces of a dodecahedron

so that each face is crossed no more than once? This probably can be

expressed more simply by connecting the vertexes of the Icosahedron

instead. I think the result will be an odd number.

I believe the number for sequentially connecting the twenty vertexes of

the dodecahedron so that each vertex is crossed no more than once is

thirteen basic configurations with thirteen more as endomorphs. I

obtained these figures with models, string, and time, but I won't stake

my life on the correctness and am hoping for someone could provide a

formula to confirm or deny those figures and avoid the tediousness of

now figuring out its dual.

Thanks!

Russ Kinter - View Source
What you need to do is go to:

On a related note, you might also want to try:

(Those were hemi-hedra, by the way.)

--- In Polytopia@yahoogroups.com, "pyth7" <pyth7@...> wrote:

> Hello All,

> Is there a formula for giving the number of possible configurations for

> connecting sequentially all twelve faces of a dodecahedron

> so that each face is crossed no more than once? This probably can be

> expressed more simply by connecting the vertexes of the Icosahedron

> instead. I think the result will be an odd number.> I believe the number for sequentially connecting the twenty vertexes of

> the dodecahedron so that each vertex is crossed no more than once is

> thirteen basic configurations with thirteen more as endo-morphs. I

> obtained these figures with models, string, and time, but I won't stake

> my life on the correctness and am hoping for someone could provide a

> formula to confirm or deny those figures and avoid the tediousness of

> now figuring out its dual.> Thanks!

> Russ Kinter