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Re: Hamiltonian cycles on directed graphs
Posted:
Jun 5, 2010 7:32 AM
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OOps I don't know what happened there on my screen it looks like a->b not a(r)b (NB it does not imply b->a). I'm sorry about that I think it must have been the incompatability of fonts.
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From: mathgroup-adm@smc.vnet.net [mathgroup-adm@smc.vnet.net] On Behalf Of Murray Eisenberg [murray@math.umass.edu]
Sent: Friday, June 04, 2010 1:02 PM
Subject: Re: Hamiltonian cycles on directed graphs
Please explain your notation: does something like a(r)b mean you have
both a->b and b->a ?
On 6/3/2010 5:40 AM, King, Peter R wrote:
> My understanding is that, currently, the Hamiltonian cycles function does not work properly on directed graphs. Is there a way round this? For example I have the directed graph
>
> z == {a(r)b,a(r)d,b(r)c, b->i,c (r)a,c (r)k,d(r)e,d(r)g,e(r)f,e(r)c,f(r)d,f(r)l,g(r)h,g(r)a,h(r)i,h(r)f,i(r)g,i(r)j,j(r)l,j(r)b,k(r)j,k(r)e,l(r)k,l(r)h};
>
> for which I would like the Hamiltonian cycles (can I specify which vertex I start from - although this is unimportant in this example).
>
> One small point, I can plot this in 2D and get arrows to show the directionality but in 3D I am struggling.
>
> Thanks
>
--
Murray Eisenberg murrayeisenberg@gmail.com
80 Fearing Street phone 413 549-1020 (H)
Amherst, MA 01002-1912
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