Re: Simulated Annealing
From: Randy Poe (poespam-trap_at_yahoo.com)
Date: 09/03/04
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Date: 3 Sep 2004 08:19:30 -0700
dab722@mail.usask.ca (Bert) wrote in message news:<367911ba.0409022135.7510638@posting.google.com>...
> Hello,
>
> In the metropolis approach to simulated annealing the condition for
> accepting a positive differential cost between two configurations is
> if
>
> P[0,1) < exp(-diffCost/k_b*T)
>
> I can accept this. It is not clear to me what the mathematical
> statement means.
The thing on the right is the probability of a change.
>From the context below I take it that the thing on
the left is a number drawn from the uniform distribution
U[0,1).
> I read it as tossing the dice and if the Boltzmann
> has a higher probability then pure chance (uniform distribution) then
> accept the change.
In Monte Carlo simulation, if you want to simulate the
occurrence of an event that occurs with probability p,
then you draw a uniform random number X and the event
occurs if X < p.
For example, suppose p = 0.9. 90% of the time your
uniform random number will be < 0.9. Thus, if you
make the event happen whenever x < 0.9, then the event
will occur the desired proportion of time. This technique
is just a special case (two possible outcomes) of how
to draw from a discrete (n-outcome) probability
distribution. It doesn't matter what particular distribution
is used to generate p on the right hand side.
>
> Am I missing anything?
>
One minor wrinkle is that the expression on the right
is not quite a valid distribution since it can exceed
1. But the inequality still has the desired effect: if
the rhs >1, then P[0,1) is guaranteed to be less than
the rhs, and the transition occurs with certainty.
- Randy
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