Comments on: The random graph, 1

By: Peter Cameron

Peter Cameron — Fri, 17 Aug 2012 09:29:11 +0000

Let me answer the second question first. If z is always outside U and V, then there are infinitely many z that work (since having found any. finite number we can add them to U and find another one), so we can always avoid any finite number of previously chosen vertices.

As to the first question, it is easiest just to assume that z is outside both U and V. Certainly it will automatically be outside U because we are considering graphs with no loops. In fact you can prove that even if you don’t assume it’s outside V, there will be a suitable z outside V that works.

By: Rupert

Rupert — Thu, 16 Aug 2012 14:30:05 +0000

In the definition of “correctly joined”, does z have to lie outside of U and V? Or is it somehow automatic that this can always be done?

I’m wondering why the image you find for a_{n+1} in the proof that a graph satisfying (*) is universal is guaranteed to not be among the preceding a_i’s.

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By: Peter Cameron

Peter Cameron — Tue, 14 Sep 2010 09:21:29 +0000

Not sure whether this helps with the intuition, but the more symmetry an object has, the fewer different ways you can draw it, and so the fewer copies there are lying around for the random search to find.

By: Jason Orendorff

Jason Orendorff — Mon, 13 Sep 2010 23:03:50 +0000

Never mind — inversely proportional is right. For some reason I can’t put my finger on, I find this strangely counterintuitive.

By: Jason Orendorff

Jason Orendorff — Mon, 13 Sep 2010 22:58:04 +0000

“the probability of a graph is inversely proportional to its symmetry (the order of its automorphism group)”

proportional, not inversely proportional, right?

(I think this is great! Thanks for the post; it’ll take me weeks of spare moments to unpack it…)