User talk:GaneshVanahalli

I don't think that your proof for Cayley's formula is valid, hence I removed this from the Wikipedia page.

When you say 'STEP: choose one of the n vertices and create an edge', this is completely ambiguous and imprecise.

We're dealing with a labelled graph here. We can't freely create an edge, because it could create a cycle (and you can't simply say that the final answer does not count non-trees, so such a cycle-forming edge is not created. That is a circular argument... you can't assume the theorem itself to prove the theorem).

Even beyond that, why would such a sequence, with such freedom of edge-choice, form a unique labelled tree? You have not counted the number of ways to choose edges? So how is the unique tree being formed by doing this randomly? And how is choosing vertex v and adding an edge to w not the same as choosing vertex w and adding an edge to v? This would introduce double-counting?

If the choice of first vertex introduces a normalisation factor, as you claim, then each subsequent vertex choice also introduces a normalisation factor, by symmetry/induction. The whole process is iterative. Pitman's rigorous proof gets around this by considering rooted trees (in a formal way) and then directing edges away from the roots. But that proof already exists.

In the end, the proof is more of a 'proof by wishful thinking'. It was also poor in formality, language etc., and it had typos. Please consider more carefully before unintentionally compromising an important mathematical source for my graph theory students.

Here's an example of a desperate proof attempt that I could just as easily write: "At each step, choose one of the k vertices already added to the tree, and connect it by an edge to one of the n-k vertices not yet added, for k = 1,2,...,n-1. This gives a total of ((n-1)!)^2 constructions." Although this is indeed the number of such constructions, there is no obvious 'normalisation factor' to recover the number of unique labelled trees thus created. I cannot get from ((n-1)!)^2 to n^(n-2) in any obvious way. So an argument for uniqueness is absolutely vital. I cannot just say that the normalisation factor is n^(n-2)/((n-1)!)^2 to help me get the answer I want? So my proof attempt is not getting me anywhere. Similarly, you cannot say that the initial choice of vertex is the only choice that affects uniqueness (unless you can prove that this is the case).

Dr N Korpelainen ( http://www.derby.ac.uk/staff/nicholas-korpelainen/ )

Nkorppi (talk) 11:23, 13 March 2017 (UTC)