Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was wondering what were the state-of-the-art methods to simulate low temperature configurations of Potts-like models that exhibit a discontinuous phase transition. For models with a continuous phase transition, annealing methods (parallel tempering, simulated tempering and their friends) work pretty well: nevertheless, for discontinuous phase transition, these methods are typically not satisfying because a huge number of intermediate temperatures are needed.

Is there any other approach that has proven efficient?

share|improve this question
add comment

1 Answer

Hi Alekk,

There is a paper from several years ago by Bhatnagar and Randall on torpid mixing of simulated tempering on the Potts model. In it, they mention that they also have a proof of rapid mixing for a related tempering approach on the same model (I don't remember if this is actually in the paper; they haven't yet published everything that they know on the subject).

I've talked to a few people who were experts on this sort of thing in the intervening time, and they have mostly claimed that there has been very little progress on proving efficiency for anything that isn't a mean-field model. Of course, the people who I've been talking to tend towards proving 'good' (i.e. small-order polynomial in size of the lattice) for discrete models, and in L^{1} distance; its possible that other areas of the literature have analyses in other metrics or for other models.

Also, I don't know the etiquette here, but I've been thinking about a few simple related models for a little while now. I'd certainly be interested in talking & sharing results/simulations/etc if you are in fact working in the area (my guess from your blog is that you're a graduate student as well?). My (slightly embarrassing and very old) email is cauchie.p@gmail.com.

Cheers,

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.