Langbahn Team – Weltmeisterschaft

Talk:Tutte polynomial

Isn't the bit about the Jones polynomial all wrong?

The current text about the hyperbola xy=1 is completely at odds with Propos 1 of [ref], which in a later example clearly states that V_{D(G)}( s ) = ( s^{-1/2} + s^{1/2} )^{|E|-|V|+1} s^{1-|E|-|V|} T_G( s^2, (s^2+1)/(s^2-s) ).

[ref] Francois Jaeger, Tutte Polynomials and Link polynomials, Proc. Am. Math. Soc. Vol 103, No 2, June 1988 —Preceding unsigned comment added by 82.33.119.244 (talk) 18:40, 20 February 2010 (UTC)[reply]

Typesetting

The typesetting in this article needs fixing. 86.145.57.232 (talk) 11:33, 24 January 2014 (UTC)[reply]

Looks ok to me. What, specifically, do you see wrong? —David Eppstein (talk) 17:07, 24 January 2014 (UTC)[reply]

Wrong formula

I think this formula is wrong: Isn't this should be  ? 133.11.30.75 (talk) 16:27, 24 August 2014 (UTC)[reply]

Well-defined?

Why even mention well-definedness in the definition? There is no choice of representatives being made nor anyting that would potentially make the definition ambigous. Instead, I suggets it should be mentioned that the has only finitely many summands (hence indeed produces a polynomial) because the graph is finite. Or maybe this was what was meant with well-definedness in this context?--Hagman (talk) 12:02, 31 August 2014 (UTC)[reply]

There are several ways of defining the Tutte polynomial, for example one might start from the contraction-deletion definition. For some of these definitions there are choices, for example, the order of edges to operate on, which make it non-obvious that the answer is well-defined, in the sense of being independent of the various choices. Starting from the Whitney rank polynomial definition makes it clear that it is well-defined, because there are no choices to make, hence the comment in the text. Of course it is now not obvious that it is a contraction-deletion invariant. Deltahedron (talk) 12:36, 31 August 2014 (UTC)[reply]

Specialization to polynomials

In several places it is stated that the Tutte polynomial "specializes to" certain other polynomials. But according to the formulas which then follow, it doesn't seem to exactly specialize to these polynomials, but rather to other polynomials from which the relevant polynomials can be obtained by some modifications. This seems a bit misleading to me, but perhaps I'm just misunderstanding the terminology? However, the accompanying illustrations just claim to show these polynomials in the Tutte plane, which makes me further suspect that the statements are indeed contradictory. Perhaps Thore Husfeldt, who seems to have written these sections way back in 2008, could clarify? —St.Nerol (talk, contribs) 16:13, 11 December 2024 (UTC)[reply]

Quelle: https://en.wikipedia.org/wiki/Talk:Tutte_polynomial