Re: Hamiltonian cycles on directed graphs

*To*: mathgroup at smc.vnet.net*Subject*: [mg110149] Re: Hamiltonian cycles on directed graphs*From*: Murray Eisenberg <murray at math.umass.edu>*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 gmail.com 80 Fearing Street phone 413 549-1020 (H) Amherst, MA 01002-1912