Gradient descent: Difference between revisions
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[https://books.google.co.uk/books?id=dFHvBQAAQBAJ&pg=PA543&lpg=PA543&dq=c%23+steepest+descent+method+to+find+minima+of+two+variable+function&source=bl&ots=TCyD-ts9ui&sig=ACfU3U306Og2fOhTjRv2Ms-BW00IhomoBg&hl=en&sa=X&ved=2ahUKEwitzrmL3aXjAhWwVRUIHSEYCU8Q6AEwCXoECAgQAQ#v=onepage&q=c%23%20steepest%20descent%20method%20to%20find%20minima%20of%20two%20variable%20function&f=false This book excerpt] shows sample C# code for solving this task. |
[https://books.google.co.uk/books?id=dFHvBQAAQBAJ&pg=PA543&lpg=PA543&dq=c%23+steepest+descent+method+to+find+minima+of+two+variable+function&source=bl&ots=TCyD-ts9ui&sig=ACfU3U306Og2fOhTjRv2Ms-BW00IhomoBg&hl=en&sa=X&ved=2ahUKEwitzrmL3aXjAhWwVRUIHSEYCU8Q6AEwCXoECAgQAQ#v=onepage&q=c%23%20steepest%20descent%20method%20to%20find%20minima%20of%20two%20variable%20function&f=false This book excerpt] shows sample C# code for solving this task. |
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<br><br> |
<br><br> |
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=={{header|Fortran}}== |
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'''Compiler:''' [[gfortran 8.3.0]]<br> |
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The way a FORTRAN programmer would do this would be to automatically differentiate the function using the diff command in Maxima: |
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(%i3) (x-1)*(x-1)*exp(-y^2)+y*(y+2)*exp(-2*x^2); |
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2 2 |
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2 - y - 2 x |
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(%o3) (x - 1) %e + %e y (y + 2) |
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(%i4) diff(%o3,x); |
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2 2 |
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- y - 2 x |
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(%o4) 2 (x - 1) %e - 4 x %e y (y + 2) |
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(%i5) diff(%o3,y); |
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2 2 2 |
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2 - y - 2 x - 2 x |
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(%o5) (- 2 (x - 1) y %e ) + %e (y + 2) + %e y |
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and then have it automatically turned into statements with the fortran command: |
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(%i6) fortran(%o4); |
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2*(x-1)*exp(-y**2)-4*x*exp(-2*x**2)*y*(y+2) |
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(%o6) done |
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(%i7) fortran(%o5); |
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(-2*(x-1)**2*y*exp(-y**2))+exp(-2*x**2)*(y+2)+exp(-2*x**2)*y |
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(%o7) done |
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The optimization subroutine GD sets the reverse communication variable IFLAG. This allows the evaluation of the function and the gradient to be done separately. |
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<lang Fortran> SUBROUTINE EVALFG (N, X, F, G) |
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IMPLICIT NONE |
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INTEGER N |
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REAL X(N), F, G(N) |
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F = (X(1) - 1.)**2 * EXP(-X(2)**2) + |
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$ X(2) * (X(2) + 2.) * EXP(-2. * X(1)**2) |
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G(1) = 2. * (X(1) - 1.) * EXP(-X(2)**2) - 4. * X(1) * |
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$ EXP(-2. * X(1)**2) * X(2) * (X(2) + 2.) |
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G(2) = (-2. * (X(1) - 1.)**2 * X(2) * EXP(-X(2)**2)) + |
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$ EXP(-2. * X(1)**2) * (X(2) + 2.) + |
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$ EXP(-2. * X(1)**2) * X(2) |
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RETURN |
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END |
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*----------------------------------------------------------------------- |
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* gd - Gradient descent |
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* G must be set correctly at the initial point X. |
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* |
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*___Name______Type________In/Out___Description_________________________ |
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* N Integer In Number of Variables. |
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* X(N) Real Both Variables |
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* G(N) Real Both Gradient |
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* TOL Real In Relative convergence tolerance |
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* IFLAG Integer Both Reverse Communication Flag |
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* 0 on input: begin |
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* 0 on output: done |
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* 1 compute G |
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*----------------------------------------------------------------------- |
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SUBROUTINE GD (N, X, G, TOL, IFLAG) |
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IMPLICIT NONE |
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INTEGER N, IFLAG |
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REAL X(N), G(N), TOL |
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REAL ETA |
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PARAMETER (ETA = 0.3) ! Learning rate |
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INTEGER I |
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REAL GNORM ! norm of gradient |
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IF (IFLAG .EQ. 1) GO TO 20 |
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10 DO I = 1, N ! take step |
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X(I) = X(I) - ETA * G(I) |
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END DO |
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IFLAG = 1 |
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RETURN ! let main program evaluate G |
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20 GNORM = 0. ! convergence test |
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DO I = 1, N |
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GNORM = GNORM + G(I)**2 |
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END DO |
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GNORM = SQRT(GNORM) |
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IF (GNORM < TOL) THEN |
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IFLAG = 0 |
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RETURN ! success |
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END IF |
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GO TO 10 |
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END ! of gd |
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PROGRAM GDDEMO |
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IMPLICIT NONE |
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INTEGER N |
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PARAMETER (N = 2) |
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INTEGER ITER, J, IFLAG |
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REAL X(N), F, G(N), TOL |
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X(1) = -0.1 ! initial values |
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X(2) = -1.0 |
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TOL = 1.E-6 |
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IFLAG = 0 |
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DO J = 1, 1 000 000 |
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CALL GD (N, X, G, TOL, IFLAG) |
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IF (IFLAG .EQ. 1) THEN |
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CALL EVALFG (N, X, F, G) |
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ELSE |
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ITER = J |
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GO TO 50 |
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END IF |
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END DO |
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STOP 'too many iterations!' |
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50 PRINT '(A, I7, A, F10.6, A, F10.6, A, F10.6)', |
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$ 'After ', ITER, ' steps, found minimum at x=', |
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$ X(1), ' y=', X(2), ' of f=', F |
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STOP 'program complete' |
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END |
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</lang> |
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{{out}} |
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<pre>After 14 steps, found minimum at x= 0.107627 y= -1.223260 of f= -0.750063 |
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STOP program complete |
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</pre> |
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=={{header|Go}}== |
=={{header|Go}}== |
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Revision as of 11:56, 13 May 2020
Gradient descent (also known as steepest descent) is a first-order iterative optimization algorithm for finding the minimum of a function which is described in this Wikipedia article.
- Task
Use this algorithm to search for minimum values of the bi-variate function:
f(x, y) = (x - 1)(x - 1)e^(-y^2) + y(y+2)e^(-2x^2)
around x = 0.1 and y = -1.
This book excerpt shows sample C# code for solving this task.
Fortran
Compiler: gfortran 8.3.0
The way a FORTRAN programmer would do this would be to automatically differentiate the function using the diff command in Maxima:
(%i3) (x-1)*(x-1)*exp(-y^2)+y*(y+2)*exp(-2*x^2);
2 2
2 - y - 2 x
(%o3) (x - 1) %e + %e y (y + 2)
(%i4) diff(%o3,x);
2 2
- y - 2 x
(%o4) 2 (x - 1) %e - 4 x %e y (y + 2)
(%i5) diff(%o3,y);
2 2 2
2 - y - 2 x - 2 x
(%o5) (- 2 (x - 1) y %e ) + %e (y + 2) + %e y
and then have it automatically turned into statements with the fortran command:
(%i6) fortran(%o4);
2*(x-1)*exp(-y**2)-4*x*exp(-2*x**2)*y*(y+2)
(%o6) done
(%i7) fortran(%o5);
(-2*(x-1)**2*y*exp(-y**2))+exp(-2*x**2)*(y+2)+exp(-2*x**2)*y
(%o7) done
The optimization subroutine GD sets the reverse communication variable IFLAG. This allows the evaluation of the function and the gradient to be done separately. <lang Fortran> SUBROUTINE EVALFG (N, X, F, G)
IMPLICIT NONE
INTEGER N
REAL X(N), F, G(N)
F = (X(1) - 1.)**2 * EXP(-X(2)**2) +
$ X(2) * (X(2) + 2.) * EXP(-2. * X(1)**2)
G(1) = 2. * (X(1) - 1.) * EXP(-X(2)**2) - 4. * X(1) *
$ EXP(-2. * X(1)**2) * X(2) * (X(2) + 2.)
G(2) = (-2. * (X(1) - 1.)**2 * X(2) * EXP(-X(2)**2)) +
$ EXP(-2. * X(1)**2) * (X(2) + 2.) +
$ EXP(-2. * X(1)**2) * X(2)
RETURN
END
- -----------------------------------------------------------------------
- gd - Gradient descent
- G must be set correctly at the initial point X.
- ___Name______Type________In/Out___Description_________________________
- N Integer In Number of Variables.
- X(N) Real Both Variables
- G(N) Real Both Gradient
- TOL Real In Relative convergence tolerance
- IFLAG Integer Both Reverse Communication Flag
- 0 on input: begin
- 0 on output: done
- 1 compute G
- -----------------------------------------------------------------------
SUBROUTINE GD (N, X, G, TOL, IFLAG)
IMPLICIT NONE
INTEGER N, IFLAG
REAL X(N), G(N), TOL
REAL ETA
PARAMETER (ETA = 0.3) ! Learning rate
INTEGER I
REAL GNORM ! norm of gradient
IF (IFLAG .EQ. 1) GO TO 20
10 DO I = 1, N ! take step
X(I) = X(I) - ETA * G(I)
END DO
IFLAG = 1
RETURN ! let main program evaluate G
20 GNORM = 0. ! convergence test
DO I = 1, N
GNORM = GNORM + G(I)**2
END DO
GNORM = SQRT(GNORM)
IF (GNORM < TOL) THEN
IFLAG = 0
RETURN ! success
END IF
GO TO 10
END ! of gd
PROGRAM GDDEMO
IMPLICIT NONE
INTEGER N
PARAMETER (N = 2)
INTEGER ITER, J, IFLAG
REAL X(N), F, G(N), TOL
X(1) = -0.1 ! initial values
X(2) = -1.0
TOL = 1.E-6
IFLAG = 0
DO J = 1, 1 000 000
CALL GD (N, X, G, TOL, IFLAG)
IF (IFLAG .EQ. 1) THEN
CALL EVALFG (N, X, F, G)
ELSE
ITER = J
GO TO 50
END IF
END DO
STOP 'too many iterations!'
50 PRINT '(A, I7, A, F10.6, A, F10.6, A, F10.6)',
$ 'After ', ITER, ' steps, found minimum at x=',
$ X(1), ' y=', X(2), ' of f=', F
STOP 'program complete'
END
</lang>
- Output:
After 14 steps, found minimum at x= 0.107627 y= -1.223260 of f= -0.750063 STOP program complete
Go
This is a translation of the C# code in the book excerpt linked to above and hence also of the first Typescript example below.
For some unknown reason the results differ from the other solutions after the first 4 decimal places but are near enough for an approximate method such as this. <lang go>package main
import (
"fmt" "math"
)
func steepestDescent(x []float64, alpha, tolerance float64) {
n := len(x) h := tolerance g0 := g(x) // Initial estimate of result.
// Calculate initial gradient. fi := gradG(x, h)
// Calculate initial norm.
delG := 0.0
for i := 0; i < n; i++ {
delG += fi[i] * fi[i]
}
delG = math.Sqrt(delG)
b := alpha / delG
// Iterate until value is <= tolerance.
for delG > tolerance {
// Calculate next value.
for i := 0; i < n; i++ {
x[i] -= b * fi[i]
}
h /= 2
// Calculate next gradient.
fi = gradG(x, h)
// Calculate next norm.
delG = 0
for i := 0; i < n; i++ {
delG += fi[i] * fi[i]
}
delG = math.Sqrt(delG)
b = alpha / delG
// Calculate next value.
g1 := g(x)
// Adjust parameter.
if g1 > g0 {
alpha /= 2
} else {
g0 = g1
}
}
}
// Provides a rough calculation of gradient g(x). func gradG(x []float64, h float64) []float64 {
n := len(x) z := make([]float64, n) y := make([]float64, n) copy(y, x) g0 := g(x)
for i := 0; i < n; i++ {
y[i] += h
z[i] = (g(y) - g0) / h
}
return z
}
// Function for which minimum is to be found. func g(x []float64) float64 {
return (x[0]-1)*(x[0]-1)*
math.Exp(-x[1]*x[1]) + x[1]*(x[1]+2)*
math.Exp(-2*x[0]*x[0])
}
func main() {
tolerance := 0.0000006
alpha := 0.1
x := []float64{0.1, -1} // Initial guess of location of minimum.
steepestDescent(x, alpha, tolerance)
fmt.Println("Testing steepest descent method:")
fmt.Println("The minimum is at x[0] =", x[0], "\b, x[1] =", x[1])
} </lang>
- Output:
Testing steepest descent method: The minimum is at x[0] = 0.10764302056464771, x[1] = -1.223351901171944
Julia
<lang julia>using Optim, Base.MathConstants
f(x) = (x[1] - 1) * (x[1] - 1) * e^(-x[2]^2) + x[2] * (x[2] + 2) * e^(-2 * x[1]^2)
println(optimize(f, [0.1, -1.0], GradientDescent()))
</lang>
- Output:
Results of Optimization Algorithm
* Algorithm: Gradient Descent
* Starting Point: [0.1,-1.0]
* Minimizer: [0.107626844383003,-1.2232596628723371]
* Minimum: -7.500634e-01
* Iterations: 14
* Convergence: true
* |x - x'| ≤ 0.0e+00: false
|x - x'| = 2.97e-09
* |f(x) - f(x')| ≤ 0.0e+00 |f(x)|: true
|f(x) - f(x')| = 0.00e+00 |f(x)|
* |g(x)| ≤ 1.0e-08: true
|g(x)| = 2.54e-09
* Stopped by an increasing objective: false
* Reached Maximum Number of Iterations: false
* Objective Calls: 35
* Gradient Calls: 35
Perl
Calculate with bignum for numerical stability.
<lang perl>use strict; use warnings; use bignum;
sub steepestDescent {
my($alpha, $tolerance, @x) = @_; my $N = @x; my $h = $tolerance; my $g0 = g(@x) ; # Initial estimate of result.
my @fi = gradG($h, @x) ; # Calculate initial gradient
# Calculate initial norm.
my $delG = 0;
for (0..$N-1) { $delG += $fi[$_]**2 }
my $b = $alpha / sqrt($delG);
while ( $delG > $tolerance ) { # Iterate until value is <= tolerance.
# Calculate next value.
for (0..$N-1) { $x[$_] -= $b * $fi[$_] }
$h /= 2;
@fi = gradG($h, @x); # Calculate next gradient.
# Calculate next norm.
$delG = 0;
for (0..$N-1) { $delG += $fi[$_]**2 }
$b = $alpha / sqrt($delG);
my $g1 = g(@x); # Calculate next value.
$g1 > $g0 ? ($alpha /= 2) : ($g0 = $g1); # Adjust parameter. } @x
}
- Provides a rough calculation of gradient g(x).
sub gradG {
my($h, @x) = @_;
my $N = @x;
my @y = @x;
my $g0 = g(@x);
my @z;
for (0..$N-1) { $y[$_] += $h ; $z[$_] = (g(@y) - $g0) / $h }
return @z
}
- Function for which minimum is to be found.
sub g { my(@x) = @_; ($x[0]-1)**2 * exp(-$x[1]**2) + $x[1]*($x[1]+2) * exp(-2*$x[0]**2) };
my $tolerance = 0.0000001; my $alpha = 0.01; my @x = <0.1 -1>; # Initial guess of location of minimum.
printf "The minimum is at x[0] = %.6f, x[1] = %.6f", steepestDescent($alpha, $tolerance, @x);</lang>
- Output:
The minimum is at x[0] = 0.107653, x[1] = -1.223370
Phix
... and just like Go, the results don't quite match anything else. <lang Phix>-- Function for which minimum is to be found. function g(sequence x)
atom {x0,x1} = x
return (x0-1)*(x0-1)*exp(-x1*x1) +
x1*(x1+2)*exp(-2*x0*x0)
end function
-- Provides a rough calculation of gradient g(x). function gradG(sequence x, atom h)
integer n = length(x)
sequence z = repeat(0, n)
atom g0 := g(x)
for i=1 to n do
x[i] += h
z[i] = (g(x) - g0) / h
end for
return z
end function
function steepestDescent(sequence x, atom alpha, tolerance)
integer n = length(x)
atom h = tolerance,
g0 = g(x) -- Initial estimate of result.
-- Calculate initial gradient. sequence fi = gradG(x, h)
-- Calculate initial norm.
atom delG = sqrt(sum(sq_mul(fi,fi))),
b = alpha / delG
-- Iterate until value is <= tolerance.
while delG>tolerance do
-- Calculate next value.
x = sq_sub(x,sq_mul(b,fi))
h /= 2
-- Calculate next gradient.
fi = gradG(x, h)
-- Calculate next norm.
delG = sqrt(sum(sq_mul(fi,fi)))
b = alpha / delG
-- Calculate next value.
atom g1 = g(x)
-- Adjust parameter.
if g1>g0 then
alpha /= 2
else
g0 = g1
end if
end while
return x
end function
constant tolerance = 0.0000006, alpha = 0.1 sequence x = steepestDescent({0.1,-1}, alpha, tolerance) printf(1,"Testing steepest descent method:\n") printf(1,"The minimum is at x[1] = %.16f, x[1] = %.16f\n", x)</lang>
- Output:
Testing steepest descent method: The minimum is at x[1] = 0.1076572080934996, x[1] = -1.2232976080475890 -- (64 bit) The minimum is at x[1] = 0.1073980565405569, x[1] = -1.2233251778997771 -- (32 bit)
Racket
Note the different implementation of grad. I believe that the vector should be reset and only the partial derivative in a particular dimension is to be used. For this reason, I've _yet another_ result!
I could have used ∇ and Δ in the variable names, but it looked too confusing, so I've gone with grad- and del-
<lang racket>#lang racket
(define (apply-vector f v)
(apply f (vector->list v)))
- Provides a rough calculation of gradient g(v).
(define ((grad/del f) v δ #:fv (fv (apply-vector f v)))
(define dim (vector-length v))
(define tmp (vector-copy v))
(define grad (for/vector #:length dim ((i dim)
(v_i v))
(vector-set! tmp i (+ v_i δ))
(define ∂f/∂v_i (/ (- (apply-vector f tmp) fv) δ))
(vector-set! tmp i v_i)
∂f/∂v_i))
(values grad (sqrt (for/sum ((∂_i grad)) (sqr ∂_i)))))
(define (steepest-descent g x α tolerance)
(define grad/del-g (grad/del g))
(define (loop x δ α gx grad-gx del-gx b)
(cond
[(<= del-gx tolerance) x]
[else
(define δ´ (/ δ 2))
(define x´ (vector-map + (vector-map (curry * (- b)) grad-gx) x))
(define gx´ (apply-vector g x´))
(define-values (grad-gx´ del-gx´) (grad/del-g x´ δ´ #:fv gx´))
(define b´ (/ α del-gx´))
(if (> gx´ gx)
(loop x´ δ´ (/ α 2) gx grad-gx´ del-gx´ b´)
(loop x´ δ´ α gx´ grad-gx´ del-gx´ b´))]))
(define gx (apply-vector g x)) (define δ tolerance) (define-values (grad-gx del-gx) (grad/del-g x δ #:fv gx)) (loop x δ α gx grad-gx del-gx (/ α del-gx)))
(define (Gradient-descent)
(steepest-descent
(λ (x y)
(+ (* (- x 1) (- x 1) (exp (- (sqr y))))
(* y (+ y 2) (exp (- (* 2 (sqr x)))))))
#(0.1 -1.) 0.1 0.0000006))
(module+ main
(Gradient-descent))
</lang>
- Output:
'#(0.10760797905122492 -1.2232993981966753)
Raku
(formerly Perl 6)
<lang perl6>use v6.d;
sub steepestDescent(@x, $alpha is copy, $h is copy) {
my $g0 = g(@x) ; # Initial estimate of result.
my @fi = gradG(@x, $h, $g0) ; # Calculate initial gradient
# Calculate initial norm.
my $b = $alpha / sqrt(my $delG = sum(map {$_²}, @fi));
while ( $delG > $h ) { # Iterate until value is <= tolerance.
for @fi.kv -> $i, $j { @x[$i] -= $b * $j } # Calculate next value.
# Calculate next gradient and next value
@fi = gradG(@x, $h /= 2, my $g1 = g(@x));
$b = $alpha / sqrt($delG = sum(map {$_²}, @fi) ); # Calculate next norm.
$g1 > $g0 ?? ( $alpha /= 2 ) !! ( $g0 = $g1 ) # Adjust parameter. }
}
sub gradG(@x is copy, $h, $g0) { # gives a rough calculation of gradient g(x).
return map { $_ += $h ; (g(@x) - $g0) / $h }, @x
}
- Function for which minimum is to be found.
sub g(\x) { (x[0]-1)² * exp(-x[1]²) + x[1]*(x[1]+2) * exp(-2*x[0]²) }
my $tolerance = 0.0000006 ; my $alpha = 0.1;
my @x = 0.1, -1; # Initial guess of location of minimum.
steepestDescent(@x, $alpha, $tolerance);
say "Testing steepest descent method:"; say "The minimum is at x[0] = ", @x[0], ", x[1] = ", @x[1]; </lang>
- Output:
Testing steepest descent method: The minimum is at x[0] = 0.10743450794656964, x[1] = -1.2233956711774543
REXX
<lang rexx>/*REXX pgm searches for minimum values of the bi─variate function (AKA steepest descent)*/ numeric digits length( e() ) % 2 tolerance= 0.0000006
alpha= 0.1
x.0= 0.1; x.1= -1; n= 2
say center(' testing for the steepest descent method ', 79, "═") call steepestD say 'The minimum is at: x[0]=' format(x.0,,4) " x[1]=" format(x.1,,4) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gx: return (x.0-1)**2 * exp(- (x.1**2) ) + x.1 * (x.1 + 2) * exp(-2 * x.0**2) gy: return (y.0-1)**2 * exp(- (y.1**2) ) + y.1 * (y.1 + 2) * exp(-2 * y.0**2) /*──────────────────────────────────────────────────────────────────────────────────────*/ gradG: do j=0 for n; y.j= x.j /*copy X ──► Y */
end /*j*/
g0= gx()
do j=0 for n; y.j= y.j + h
z.j= (gy() - g0) / h
end /*j*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/ steepestD: h= tolerance
g0= gx()
call gradG
delG= 0
do j=0 for n; delG= delG + z.j**2
end /*j*/
delG= sqrt(delG)
b= alpha / delG
do while delG>tolerance
do j=0 for n; x.j= x.j - b*z.j
end /*j*/
h= h / 2
call gradG
delG= 0
do j=0 for n; delG= delG + z.j**2
end /*j*/
delG= sqrt(delG)
if delG=0 then return
b= alpha / delG
g1= gx()
if g1>g0 then alpha= alpha / 2
else g0= g1
end /*while*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/ e: e= 2.7182818284590452353602874713526624977572470936999595749669676277240766303 || ,
535475945713821785; return e
/*──────────────────────────────────────────────────────────────────────────────────────*/ exp: procedure; parse arg x; ix= x%1; if abs(x-ix)>.5 then ix= ix + sign(x); x= x - ix
z=1; _=1; w=z; do j=1; _= _*x/j; z= (z+_)/1; if z==w then leave; w= z; end
if z\==0 then z= z * e() ** ix; return z/1
/*──────────────────────────────────────────────────────────────────────────────────────*/ sqrt: procedure; parse arg x; if x=0 then return 0; d= digits(); numeric digits; h= d+6
numeric form; m.=9; parse value format(x,2,1,,0) 'E0' with g "E" _ .; g=g *.5'e'_ %2
do j=0 while h>9; m.j=h; h= h % 2 + 1; end /*j*/
do k=j+5 to 0 by -1; numeric digits m.k; g=(g+x/g)*.5; end /*k*/; return g</lang>
- output when using the internal default inputs:
═══════════════════ testing for the steepest descent method ═══════════════════ The minimum is at: x[0]= 0.1062 x[1]= -1.2264
Scala
<lang scala>object GradientDescent {
/** Steepest descent method modifying input values*/
def steepestDescent(x : Array[Double], learningRate : Double, tolerance : Double) = {
val n = x.size
var h = tolerance
var alpha = learningRate
var g0 = g(x) // Initial estimate of result.
// Calculate initial gradient. var fi = gradG(x,h)
// Calculate initial norm. var delG = 0.0 for (i <- 0 until n by 1) delG += fi(i) * fi(i) delG = math.sqrt(delG) var b = alpha / delG
// Iterate until value is <= tolerance.
while(delG > tolerance){
// Calculate next value.
for (i <- 0 until n by 1) x(i) -= b * fi(i)
h /= 2
// Calculate next gradient.
fi = gradG(x,h)
// Calculate next norm.
delG = 0.0
for (i <- 0 until n by 1) delG += fi(i) * fi(i)
delG = math.sqrt(delG)
b = alpha / delG
// Calculate next value.
var g1 = g(x)
// Adjust parameter.
if(g1 > g0) alpha = alpha / 2
else g0 = g1
}
}
/** Gradient of the input function given in the task*/
def gradG(x : Array[Double], h : Double) : Array[Double] = {
val n = x.size
val z : Array[Double] = Array.fill(n){0}
val y = x
val g0 = g(x)
for(i <- 0 until n by 1){
y(i) += h
z(i) = (g(y) - g0) / h
}
z
}
/** Bivariate function given in the task*/
def g( x : Array[Double]) : Double = {
( (x(0)-1) * (x(0)-1) * math.exp( -x(1)*x(1) ) + x(1) * (x(1)+2) * math.exp( -2*x(0)*x(0) ) )
}
def main(args: Array[String]): Unit = {
val tolerance = 0.0000006
val learningRate = 0.1
val x = Array(0.1, -1) // Initial guess of location of minimum.
steepestDescent(x, learningRate, tolerance)
println("Testing steepest descent method")
println("The minimum is at x : " + x(0) + ", y : " + x(1))
}
} </lang>
- Output:
Testing steepest descent method The minimum is at x : 0.10756393294495799, y : -1.2234116852966237
TypeScript
- Translation of
- [Numerical Methods, Algorithms and Tools in C# by Waldemar Dos Passos (18.2 Gradient Descent Method]
<lang Typescript> // Using the steepest-descent method to search // for minimum values of a multi-variable function export const steepestDescent = (x: number[], alpha: number, tolerance: number) => {
let n: number = x.length; // size of input array let h: number = 0.0000006; //Tolerance factor let g0: number = g(x); //Initial estimate of result
//Calculate initial gradient let fi: number[] = [n];
//Calculate initial norm
fi = GradG(x, h);
// console.log("fi:"+fi);
//Calculate initial norm let DelG: number = 0.0;
for (let i: number = 0; i < n; ++i) {
DelG += fi[i] * fi[i];
}
DelG = Math.sqrt(DelG);
let b: number = alpha / DelG;
//Iterate until value is <= tolerance limit
while (DelG > tolerance) {
//Calculate next value
for (let i = 0; i < n; ++i) {
x[i] -= b * fi[i];
}
h /= 2;
//Calculate next gradient
fi = GradG(x, h);
//Calculate next norm
DelG = 0;
for (let i: number = 0; i < n; ++i) {
DelG += fi[i] * fi[i];
}
DelG = Math.sqrt(DelG);
b = alpha / DelG;
//Calculate next value
let g1: number = g(x);
//Adjust parameter
if (g1 > g0) alpha /= 2;
else g0 = g1;
}
}
// Provides a rough calculation of gradient g(x). export const GradG = (x: number[], h: number) => {
let n: number = x.length; let z: number[] = [n]; let y: number[] = x; let g0: number = g(x);
// console.log("y:" + y);
for (let i = 0; i < n; ++i) {
y[i] += h;
z[i] = (g(y) - g0) / h;
}
// console.log("z:"+z);
return z;
}
// Method to provide function g(x). export const g = (x: number[]) => {
return (x[0] - 1) * (x[0] - 1)
* Math.exp(-x[1] * x[1]) + x[1] * (x[1] + 2)
* Math.exp(-2 * x[0] * x[0]);
}
export const gradientDescentMain = () => {
let tolerance: number = 0.0000006; let alpha: number = 0.1; let x: number[] = [2];
//Initial guesses x[0] = 0.1; //of location of minimums x[1] = -1; steepestDescent(x, alpha, tolerance);
console.log("Testing steepest descent method");
console.log("The minimum is at x[0] = " + x[0]
+ ", x[1] = " + x[1]);
// console.log("");
}
gradientDescentMain();
</lang>
- Output:
Testing steepest descent method The minimum is at x[0] = 0.10768224291553158, x[1] = -1.2233090211217854
Linear Regression
- Translation of
- [Linear Regression using Gradient Descent by Adarsh Menon]
<lang Typescript>
let data: number[][] =
[[32.5023452694530, 31.70700584656990], [53.4268040332750, 68.77759598163890], [61.5303580256364, 62.56238229794580], [47.4756396347860, 71.54663223356770], [59.8132078695123, 87.23092513368730], [55.1421884139438, 78.21151827079920], [52.2117966922140, 79.64197304980870], [39.2995666943170, 59.17148932186950], [48.1050416917682, 75.33124229706300], [52.5500144427338, 71.30087988685030], [45.4197301449737, 55.16567714595910], [54.3516348812289, 82.47884675749790], [44.1640494967733, 62.00892324572580], [58.1684707168577, 75.39287042599490], [56.7272080570966, 81.43619215887860], [48.9558885660937, 60.72360244067390], [44.6871962314809, 82.89250373145370], [60.2973268513334, 97.37989686216600], [45.6186437729558, 48.84715331735500], [38.8168175374456, 56.87721318626850], [66.1898166067526, 83.87856466460270], [65.4160517451340, 118.59121730252200], [47.4812086078678, 57.25181946226890], [41.5756426174870, 51.39174407983230], [51.8451869056394, 75.38065166531230], [59.3708220110895, 74.76556403215130], [57.3100034383480, 95.45505292257470], [63.6155612514533, 95.22936601755530], [46.7376194079769, 79.05240616956550], [50.5567601485477, 83.43207142132370], [52.2239960855530, 63.35879031749780], [35.5678300477466, 41.41288530370050], [42.4364769440556, 76.61734128007400], [58.1645401101928, 96.76956642610810], [57.5044476153417, 74.08413011660250], [45.4405307253199, 66.58814441422850], [61.8962226802912, 77.76848241779300], [33.0938317361639, 50.71958891231200], [36.4360095113868, 62.12457081807170], [37.6756548608507, 60.81024664990220], [44.5556083832753, 52.68298336638770], [43.3182826318657, 58.56982471769280], [50.0731456322890, 82.90598148507050], [43.8706126452183, 61.42470980433910], [62.9974807475530, 115.24415280079500], [32.6690437634671, 45.57058882337600], [40.1668990087037, 54.08405479622360], [53.5750775316736, 87.99445275811040], [33.8642149717782, 52.72549437590040], [64.7071386661212, 93.57611869265820], [38.1198240268228, 80.16627544737090], [44.5025380646451, 65.10171157056030], [40.5995383845523, 65.56230126040030], [41.7206763563412, 65.28088692082280], [51.0886346783367, 73.43464154632430], [55.0780959049232, 71.13972785861890], [41.3777265348952, 79.10282968354980], [62.4946974272697, 86.52053844034710], [49.2038875408260, 84.74269780782620], [41.1026851873496, 59.35885024862490], [41.1820161051698, 61.68403752483360], [50.1863894948806, 69.84760415824910], [52.3784462192362, 86.09829120577410], [50.1354854862861, 59.10883926769960], [33.6447060061917, 69.89968164362760], [39.5579012229068, 44.86249071116430], [56.1303888168754, 85.49806777884020], [57.3620521332382, 95.53668684646720], [60.2692143939979, 70.25193441977150], [35.6780938894107, 52.72173496477490], [31.5881169981328, 50.39267013507980], [53.6609322616730, 63.64239877565770], [46.6822286494719, 72.24725106866230], [43.1078202191024, 57.81251297618140], [70.3460756150493, 104.25710158543800], [44.4928558808540, 86.64202031882200], [57.5045333032684, 91.48677800011010], [36.9300766091918, 55.23166088621280], [55.8057333579427, 79.55043667850760], [38.9547690733770, 44.84712424246760], [56.9012147022470, 80.20752313968270], [56.8689006613840, 83.14274979204340], [34.3331247042160, 55.72348926054390], [59.0497412146668, 77.63418251167780], [57.7882239932306, 99.05141484174820], [54.2823287059674, 79.12064627468000], [51.0887198989791, 69.58889785111840], [50.2828363482307, 69.51050331149430], [44.2117417520901, 73.68756431831720], [38.0054880080606, 61.36690453724010], [32.9404799426182, 67.17065576899510], [53.6916395710700, 85.66820314500150], [68.7657342696216, 114.85387123391300], [46.2309664983102, 90.12357206996740], [68.3193608182553, 97.91982103524280], [50.0301743403121, 81.53699078301500], [49.2397653427537, 72.11183246961560], [50.0395759398759, 85.23200734232560], [48.1498588910288, 66.22495788805460], [25.1284846477723, 53.45439421485050]];
function lossFunction(arr0: number[], arr1: number[], arr2: number[]) {
let n: number = arr0.length; // Number of elements in X
//D_m = (-2/n) * sum(X * (Y - Y_pred)) # Derivative wrt m let a: number = (-2 / n) * (arr0.map((a, i) => a * (arr1[i] - arr2[i]))).reduce((sum, current) => sum + current); //D_c = (-2/n) * sum(Y - Y_pred) # Derivative wrt c let b: number = (-2 / n) * (arr1.map((a, i) => (a - arr2[i]))).reduce((sum, current) => sum + current); return [a, b];
}
export const gradientDescentMain = () => {
// Building the model let m: number = 0; let c: number = 0; let X_arr: number[]; let Y_arr: number[]; let Y_pred_arr: number[]; let D_m: number = 0; let D_c: number = 0;
let L: number = 0.00000001; // The learning Rate let epochs: number = 10000000; // The number of iterations to perform gradient descent
//Initial guesses
for (let i = 0; i < epochs; i++) {
X_arr = data.map(function (value, index) { return value[0]; });
Y_arr = data.map(function (value, index) { return value[1]; });
// The current predicted value of Y
Y_pred_arr = X_arr.map((a) => ((m * a) + c));
let all = lossFunction(X_arr, Y_arr, Y_pred_arr);
D_m = all[0];
D_c = all[1];
m = m - L * D_m; // Update m
c = c - L * D_c; // Update c
}
console.log("m: " + m + " c: " + c);
}
gradientDescentMain(); </lang>
zkl
with tweaked gradG
<lang zkl>fcn steepestDescent(f, x,y, alpha, h){
g0:=f(x,y); # Initial estimate of result.
fix,fiy := gradG(f,x,y,h); # Calculate initial gradient
# Calculate initial norm.
b:=alpha / (delG := (fix*fix + fiy*fiy).sqrt());
while(delG > h){ # Iterate until value is <= tolerance.
x,y = x - b*fix, y - b*fiy;
# Calculate next gradient and next value
fix,fiy = gradG(f,x,y, h/=2);
b=alpha / (delG = (fix*fix + fiy*fiy).sqrt()); # Calculate next norm.
if((g1:=f(x,y)) > g0) alpha/=2 else g0 = g1; # Adjust parameter.
}
return(x,y)
}
fcn gradG(f,x,y,h){ # gives a rough calculation of gradient f(x,y).
g0:=f(x,y); return((f(x + h, y) - g0)/h, (f(x, y + h) - g0)/h)
}</lang> <lang zkl>fcn f(x,y){ # Function for which minimum is to be found.
(x - 1).pow(2)*(-y.pow(2)).exp() + y*(y + 2)*(-2.0*x.pow(2)).exp()
}
tolerance,alpha := 0.0000006, 0.1;
x,y := 0.1, -1.0; # Initial guess of location of minimum. x,y = steepestDescent(f,x,y,alpha,tolerance);
println("Testing steepest descent method:"); println("The minimum is at (x,y) = (%f,%f). f(x,y) = %f".fmt(x,y,f(x,y)));</lang>
- Output:
Testing steepest descent method: The minimum is at (x,y) = (0.107608,-1.223299). f(x,y) = -0.750063