Stochastic Relaxation, Simulating Annealing, Global Minimizers.
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Transcript of Stochastic Relaxation, Simulating Annealing, Global Minimizers.
Stochastic Relaxation,
Simulating Annealing,Global Minimizers
Different types of relaxation
Variable by variable relaxation – strict minimization
Changing a small subset of variables simultaneously – Window strict minimization relaxation
Stochastic relaxation – may increase the energy – should be followed by strict minimization
Complex landscape of E(X)
How to escape local minima? First go uphill, then may hit a lower basin In order to go uphill should allow increase in E(x) Add stochasticity: allow E(x) to increase with
probability which is governed by an external temperature-like parameter T
The Metropolis Algorithm (Kirpartick et al. 1983)
Assume xold is the current state, define xnew to be a neighboring state and delE=E(xnew)-E(xold) then
If delE<0 replace xold by xnew
else choose xnew with probability P(xnew)=
and xold with probability P(xold)=1- P(xnew)
TdelEe /
The probability to accept an increasing energy move
The Metropolis Algorithm As T 0 and when delE>0 : P(xnew) 0 At T=0: strict minimization High T randomizes the configuration away
from the minimum Low T cannot escape local minimaStarting from a high T, the slower T is
decreased the lower E(x) is achieved The slow reduction in T allows the material to
obtain a more arranged configuration: increase the size of its crystals and reduce their defects
Fast cooling – amorphous solid
Slow cooling - crystalline solid
SA for the 2D Ising
E=-ijsisj , i and j are nearest neighbors
++
++
Eold=-2
SA for the 2D Ising
E=-ijsisj , i and j are nearest neighbors
++
++
+
++
Eold=-2 Enew=2
SA for the 2D Ising
E=-ijsisj , i and j are nearest neighbors
++
++
+
++
Eold=-2 Enew=2
delE=Enew- Eold=4>0
P(Enew)=exp(-4/T)
SA for the 2D Ising
E=-ijsisj , i and j are nearest neighbors
++
++
+
++
Eold=-2 Enew=2
delE=Enew- Eold=4>0
P(Enew)=exp(-4/T) =0.3
=> T=-4/ln0.3 ~ 3.3
Reduce T by a factor , 0<<1: Tn+1=Tn
Exc#7: SA for the 2D Ising (see Exc#1)Consider the following cases:
1. For h1= h2=0 set a stripe of width 3,6 or 12 with opposite sign
2. For h1=-0.1, h2=0.4 set -1 at h1 and +1 at h2
3. Repeat 2. with 2 squares of 8x8 plus spins with h2=0.4 located apart from each other
Calculate T0 to allow 10% flips of a spin surrounded by 4 neighbors of the same sign
Use faster / slower cooling scheduling
a. What was the starting T0 , E in each case
b. How was T0 decreased, how many sweeps were employed c. What was the final configuration, was the global minimum
achievable? If not try different T0
d. Is it harder to flip a wider stripe?e. Is it harder to flip 2 squares than just one?