
Re: Hamiltonian cycles on directed graphs
Posted:
Jun 5, 2010 7:32 AM


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. ________________________________________ From: mathgroupadm@smc.vnet.net [mathgroupadm@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 5491020 (H) Amherst, MA 010021912

