Potts model and Monte Carlo Slow Down

A simple model of interacting particles

The mean field Potts model is extremely simple: there are {N} interacting particles {x_1, \ldots, x_N} and each one of them can be in {q} different states {1,2, \ldots, q}. Define the Hamiltonian

\displaystyle  H_N(x) = -\frac{1}{N} \sum_{i,j} \delta(x_i, x_j)

where {x=(x_1, \ldots, x_N)} and {\delta} is the Kronecker symbol. The normalization {\frac{1}{N}} ensures that the energy is an extensive quantity so that the mean energy per particle {h_N(x) = \frac{1}{N} H_N(x)} does no degenerate to {0} or {+\infty} for large values of {N}. The sign minus is here to favorize configurations that have a lot of particles in the same state. The Boltzman distribution at inverse temperature {\beta} on {\{1, \ldots, q\}^N} is given by

\displaystyle P_{N,\beta} = \frac{1}{Z_N(\beta)} e^{-\beta H_N(x)}

where {Z_N(\beta)} is a normalization constant. Notice that if we choose a configuration uniformly at random in {\{1, \ldots, q\}^N}, with overwhelming probability the ratio of particles in state {k} will be close to {\frac{1}{q}}. Also it is obvious that if we define

\displaystyle  L^{(N)}_k(x) = \frac{1}{N} \, \Big( \textrm{Number of particles in state }k \Big)

then {L=(L^{(N)}_1, \ldots, L^{(N)}_q)} will be close to {(\frac{1}{q}, \ldots, \frac{1}{q})} for a configuration taken uniformly at random. Stirling formula even says that the probability that {L} is close to {\nu = (\nu_1, \ldots, \nu_q)} is close to {e^{-N \, R(\nu)}} where

\displaystyle R(\nu) = \nu_1 \ln(q\nu_1) + \ldots + \nu_q \ln(q\nu_q).

Indeed {(\frac{1}{q}, \ldots, \frac{1}{q}) = \textrm{argmin} \, R(\nu)}. The situation is quite different under the Boltzman distribution since it favorizes the configurations that have a lot of particles in the same state: this is because the Hamiltonian {H_N(x)} is minimized for configurations that have all the particles in the same state. In short there is a competition between the entropy (there are a lot of configurations close to the ratio {(\frac{1}{q}, \ldots, \frac{1}{q})}) and the energy that favorizes the configurations where all the particles are in the same state.

With a little more work, one can show that there is a critical inverse temperature {\beta_c} such that:

  • for {\beta < \beta_c} the entropy wins the battle: the most probable configurations are close to the ratio {(\frac{1}{q}, \ldots, \frac{1}{q})}
  • for {\beta > \beta_c} the energy effect shows up: there are {q} most probable configurations that are the permutations of {(a_{\beta},b_{\beta},b_{\beta}, \ldots, b_{\beta})} where {a_{\beta}} and {b_{\beta}} are computable quantities.

The point is that above {\beta_c} the system has more than one stable equilibrium point. Maybe more important, if we compute the energy of these most probable states

\displaystyle  h(\beta) = \lim \frac{1}{N} H_N(\textrm{most probable state})

then this function has a discontinuity at {\beta=\beta_c}. I will try to show in the weeks to come how this behaviour can dramatically slow down usual Monte-Carlo approach to the study of these kind of models.

Hugo Touchette has a very nice review of statistical physics that I like a lot and a good survey of the Potts model. Also T. Tao has a very nice exposition of related models. The blog of Georg von Hippel is dedicated to similar models on lattices, which are far more complex that this mean field approximation presented here.

MCMC Simulations

These is extremely easy to simulate this mean field Potts model since we only need to keep track of the ratio {L=(L_1, \ldots, L_q)} to have an accurate picture of the system. For example, a typical Markov Chain Monte Carlo approach would run as follows:

  • choose a particle {x_i} uniformly at random in {\{1,2, \ldots, N\}}
  • try to switch its value uniformly in {\{1,2, \ldots, q\} \setminus \{x_i\}}
  • compute the Metropolis ratio
  • update accordingly.

If we do that {10^5} times for {q=3} states at inverse temperature {\beta=1.5} and for {100} particles (which is fine since we only need to keep track of the {3}-dimensional ratio vector) and plot the result in barycentric coordinates we get a picture that looks like:
Potts Model MCMC

Here I started with a configuration where all the particles were in the same states i.e ratio vector equal to {(1,0,0)}. We can see that even with {10^5} steps, the algorithm struggles to go from one most probable position {(a,b,b)} to the other two {(b,a,b)} and {(b,b,a)} – in this simulation, one of the most probable state has even not been visited! Indeed, this approach was extremely naive, and this is quite interesting to try to come up with better algorithms. Btw, Christian Robert’s blog has tons of interesting stuffs related to MCMC and how to boost up the naive approach presented here.

Advertisements

Challenge – Announcement

Have you already heard about the 17×17 Challenge ? It all started 7 months ago, and still no solution has been found. I have the feeling though that no really serious attempt at solving the problem has been done, even if very encouraging results have been found. It could be very interesting to keep track of the progress made so that results will be published here. Send what you have reached so far!

In the next few weeks, I will try to write about the Monte Carlo approaches described in the excellent book “Monte Carlo Strategies in Scientific Computing” by Jun Liu: I do not think that one can easily solve the problem with a direct implementation of these methods, but I want to know more about it. These are all more or less Markov Chain Monte Carlo Methods: I have tried the “parallel tempering” method, which gives results that are not that bad, and was planning to check the “flat histogram” method. “Simulated annealing” methods have succesfully been used to solve the Sudoku problem though: see Christian Robert blog for example. I am also planning to continue my reading of the fascinating book “Information, Physics, and Computation” by Marc Mezard: methods described in this book (belief propagation, k-SAT problem, etc…) are more adapted to the 17×17 challenge I think, and I am quite optimistic about it.

 

17x17_4_anonymous

Best Solution so far, by Rohan Puttagunta: only 4 monochromatic rectangles!

Read also here,here,here,here and there – and have a try with this on-line simulator !