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Re: Hamiltonian cycles on directed graphs

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  • Subject: [mg110149] Re: Hamiltonian cycles on directed graphs
  • From: Murray Eisenberg <murray at>
  • Date: Fri, 4 Jun 2010 08:02:06 -0400 (EDT)

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 at
80 Fearing Street                    phone 413 549-1020 (H)
Amherst, MA 01002-1912

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