Numeric.GSL.Minimization

Minimization of a multidimensional function Minimization of a multidimensional function using some of the algorithms described in:

http://www.gnu.org/software/gsl/manual/html_node/Multidimensional-Minimization.html

Documentation

minimizeConjugateGradient

Arguments

:: Double

initial step size

-> Double

minimization parameter

-> Double

desired precision of the solution (gradient test)

-> Int

maximum number of iterations allowed

-> ([Double] -> Double)

function to minimize

-> ([Double] -> [Double])

gradient

-> [Double]

starting point

-> ([Double], Matrix Double)

solution vector, and the optimization trajectory followed by the algorithm

The Fletcher-Reeves conjugate gradient algorithm gsl_multimin_fminimizer_conjugate_fr. This is the example in the GSL manual:

minimize = minimizeConjugateGradient 1E-2 1E-4 1E-3 30
f [x,y] = 10*(x-1)^2 + 20*(y-2)^2 + 30
  --
df [x,y] = [20*(x-1), 40*(y-2)]
   --
main = do
    let (s,p) = minimize f df [5,7]
    print s
    print p
  --
> main
[1.0,2.0]
 0. 687.848 4.996 6.991
 1. 683.555 4.989 6.972
 2. 675.013 4.974 6.935
 3. 658.108 4.944 6.861
 4. 625.013 4.885 6.712
 5. 561.684 4.766 6.415
 6. 446.467 4.528 5.821
 7. 261.794 4.053 4.632
 8.  75.498 3.102 2.255
 9.  67.037 2.852 1.630
10.  45.316 2.191 1.762
11.  30.186 0.869 2.026
12.     30.    1.    2.

The path to the solution can be graphically shown by means of:

Graphics.Plot.mplot $ drop 2 (toColumns p)

minimizeNMSimplex

Arguments

:: ([Double] -> Double)

function to minimize

-> [Double]

starting point

-> [Double]

sizes of the initial search box

-> Double

desired precision of the solution

-> Int

maximum number of iterations allowed

-> ([Double], Matrix Double)

solution vector, and the optimization trajectory followed by the algorithm

The method of Nelder and Mead, implemented by gsl_multimin_fminimizer_nmsimplex. The gradient of the function is not required. This is the example in the GSL manual:

minimize f xi = minimizeNMSimplex f xi (replicate (length xi) 1) 1e-2 100
  -- 
f [x,y] = 10*(x-1)^2 + 20*(y-2)^2 + 30
  --
main = do
    let (s,p) = minimize f [5,7]
    print s
    print p
  --
> main
[0.9920430849306285,1.9969168063253164]
0. 512.500    1.082 6.500    5.
 1. 290.625    1.372 5.250    4.
 2. 290.625    1.372 5.250    4.
 3. 252.500    1.372 5.500    1.
 4. 101.406    1.823 2.625 3.500
 5. 101.406    1.823 2.625 3.500
 6.     60.    1.823    0.    3.
 7.  42.275    1.303 2.094 1.875
 8.  42.275    1.303 2.094 1.875
 9.  35.684    1.026 0.258 1.906
10.  35.664    0.804 0.588 2.445
11.  30.680    0.467 1.258 2.025
12.  30.680    0.356 1.258 2.025
13.  30.539    0.285 1.093 1.849
14.  30.137    0.168 0.883 2.004
15.  30.137    0.123 0.883 2.004
16.  30.090    0.100 0.958 2.060
17.  30.005 6.051e-2 1.022 2.004
18.  30.005 4.249e-2 1.022 2.004
19.  30.005 4.249e-2 1.022 2.004
20.  30.005 2.742e-2 1.022 2.004
21.  30.005 2.119e-2 1.022 2.004
22.  30.001 1.530e-2 0.992 1.997
23.  30.001 1.259e-2 0.992 1.997
24.  30.001 7.663e-3 0.992 1.997

The path to the solution can be graphically shown by means of:

Graphics.Plot.mplot $ drop 3 (toColumns p)