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Gradient descent

From Rosetta Code
Gradient descent is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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.

ALGOL 68[edit]

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32
Translation of: Go
modified to use the actual gradient function -
Translation of: Fortran

THe results agree with the Fortran sample and the Julia sample to 6 places.

PROC steepest descent = ( REF[]LONG REAL x, LONG REAL alphap, tolerance )VOID:
BEGIN
LONG REAL alpha := alphap;
LONG REAL g0 := g( x ); # Initial estimate of result. #
# Calculate initial gradient. #
[ LWB x : UPB x ]LONG REAL fi := grad g( x );
# Calculate initial norm. #
LONG REAL del g := 0.0;
FOR i FROM LWB x TO UPB x DO
del g +:= fi[ i ] * fi[ i ]
OD;
del g := long sqrt( del g );
LONG REAL b := alpha / del g;
# Iterate until value is <= tolerance. #
WHILE del g > tolerance DO
# Calculate next value. #
FOR i FROM LWB x TO UPB x DO
x[i] -:= b * fi[i]
OD;
# Calculate next gradient. #
fi := grad g( x );
# Calculate next norm. #
del g := 0;
FOR i FROM LWB x TO UPB x DO
del g +:= fi[ i ] * fi[ i ]
OD;
del g := long sqrt( del g );
IF del g > 0 THEN
b := alpha / del g;
# Calculate next value. #
LONG REAL g1 := g( x );
# Adjust parameter. #
IF g1 > g0 THEN
alpha /:= 2
ELSE
g0 := g1
FI
FI
OD
END # steepest descent # ;
# calculates the gradient of g(p). #
# The derivitives wrt x and y are (as in the Fortran sample ): #
# g' wrt x = 2( x - 1 )e^( - ( y^2 ) ) - 4xe^( -2( x^2) )y( y + 2 ) #
# g' wrt y = ( -2( x-1 )^2ye^( - (y^2) )) + e^(-2( x^2 ) )( y + 2 ) + e^( -2( x^2 ) )y #
PROC grad g = ( []LONG REAL p )[]LONG REAL:
BEGIN
[ LWB p : UPB p ]LONG REAL z;
LONG REAL x = p[ 0 ];
LONG REAL y = p[ 1 ];
z[ 0 ] := 2 * ( x - 1 ) * long exp( - ( y * y ) )
- 4 * x * long exp( -2 * ( x * x ) ) * y * ( y + 2 );
z[ 1 ] := ( -2 * ( x - 1 ) * ( x - 1 ) * y ) * long exp( - y * y )
+ long exp( -2 * x * x ) * ( y + 2 )
+ long exp( -2 * x * x ) * y;
z
END # grad g # ;
# Function for which minimum is to be found. #
# g( x, y ) = ( ( x - 1 )^2 )e^( - ( x^2 ) ) + y( y + 2 )e^( - 2(x^2)) #
PROC g = ( []LONG REAL x )LONG REAL:
( x[ 0 ] - 1 )
* ( x[ 0 ] - 1 )
* long exp( - x[ 1 ] * x[ 1 ] ) + x[ 1 ] * ( x[ 1 ] + 2 )
* long exp( - 2 * x[ 0 ] * x[ 0 ] )
;
BEGIN
LONG REAL tolerance := 0.0000006;
LONG REAL alpha := 0.1;
[ 0 : 1 ]LONG REAL x := ( []LONG REAL( 0.1, -1 ) )[ AT 0 ]; # Initial guess of location of minimum. #
steepest descent( x, alpha, tolerance );
print( ( "Testing steepest descent method:", newline ) );
print( ( "The minimum is at x[0] = ", fixed( x[ 0 ], -10, 6 ), ", x[1] = ", fixed( x[ 1 ], -10, 6 ), newline ) )
END
Output:
Testing steepest descent method:
The minimum is at x[0] =   0.107627, x[1] =  -1.223260

ALGOL W[edit]

Translation of: ALGOL 68
which is a
Translation of: Go
with the gradient function from
Translation of: Fortran

The results agree (to 6 places) with the Fortran and Julia samples.

begin
procedure steepestDescent ( long real array x ( * ); long real value alphap, tolerance ) ;
begin
long real array fi ( 0 :: 1 );
long real alpha, g0, g1, delG, b;
alpha := alphap;
g0  := g( x ); % Initial estimate of result. %
 % Calculate initial gradient. %
gradG( fi, x );
 % Calculate initial norm. %
delG  := 0.0;
for i := 0 until 1 do delG := delG + ( fi( i ) * fi( i ) );
delG  := longSqrt( delG );
b  := alpha / delG;
 % Iterate until value is <= tolerance. %
while delG > tolerance do begin
 % Calculate next value. %
for i := 0 until 1 do x(i) := x( i ) - ( b * fi(i) );
 % Calculate next gradient. %
gradG( fi, x );
 % Calculate next norm. %
delG  := 0;
for i := 0 until 1 do delG := delg + ( fi( i ) * fi( i ) );
delG  := longSqrt( delG );
if delG > 0 then begin
b  := alpha / delG;
 % Calculate next value. %
g1 := g( x );
 % Adjust parameter. %
if g1 > g0
then alpha := alpha / 2
else g0 := g1
end if_delG_gt_0
end while_delG_gt_tolerance
end steepestDescent ;
 % Provides a rough calculation of gradient g(x). %
procedure gradG ( long real array z, p ( * ) ) ;
begin
long real x, y;
x := p( 0 );
y := p( 1 );
z( 0 ) := 2 * ( x - 1 ) * longExp( - ( y * y ) )
- 4 * x * longExp( -2 * ( x * x ) ) * y * ( y + 2 );
z( 1 ) := ( -2 * ( x - 1 ) * ( x - 1 ) * y ) * longExp( - y * y )
+ longExp( -2 * x * x ) * ( y + 2 )
+ longExp( -2 * x * x ) * y
end gradG ;
 % Function for which minimum is to be found. %
long real procedure g ( long real array x ( * ) ) ;
( x( 0 ) - 1 ) * ( x( 0 ) - 1 ) * longExp( - x( 1 ) * x( 1 ) )
+ x( 1 ) * ( x( 1 ) + 2 ) * longExp( - 2 * x( 0 ) * x( 0 ) )
 ;
begin
long real alpha, tolerance;
long real array x ( 0 :: 1 );
x( 0 ) := 0.1; % Initial guess of location of minimum. %
x( 1 ) := -1;
tolerance := 0.0000006;
alpha  := 0.1;
steepestDescent( x, alpha, tolerance );
r_format := "A"; r_w := 11; r_d := 7; s_w := 0; % output formatting %
write( "Testing steepest descent method:" );
write( "The minimum is at x(0) = ", x( 0 ), ", x(1) = ", x( 1 ) )
end
end.
Output:
Testing steepest descent method:
The minimum is at x(0) =   0.1076268, x(1) =  -1.2232596

Fortran[edit]

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 gradient to be done separately.

      SUBROUTINE EVALFG (N, X, F, G)
IMPLICIT NONE
INTEGER N
DOUBLE PRECISION X(N), F, G(N)
F = (X(1) - 1.D0)**2 * EXP(-X(2)**2) +
$ X(2) * (X(2) + 2.D0) * EXP(-2.D0 * X(1)**2)
G(1) = 2.D0 * (X(1) - 1.D0) * EXP(-X(2)**2) - 4.D0 * X(1) *
$ EXP(-2.D0 * X(1)**2) * X(2) * (X(2) + 2.D0)
G(2) = (-2.D0 * (X(1) - 1.D0)**2 * X(2) * EXP(-X(2)**2)) +
$ EXP(-2.D0 * X(1)**2) * (X(2) + 2.D0) +
$ EXP(-2.D0 * 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) Double Both Variables
* G(N) Double Both Gradient
* TOL Double In Relative convergence tolerance
* IFLAG Integer Out Reverse Communication Flag
* on output: 0 done
* 1 compute G and call again
*-----------------------------------------------------------------------
SUBROUTINE GD (N, X, G, TOL, IFLAG)
IMPLICIT NONE
INTEGER N, IFLAG
DOUBLE PRECISION X(N), G(N), TOL
DOUBLE PRECISION ETA
PARAMETER (ETA = 0.3D0) ! Learning rate
INTEGER I
DOUBLE PRECISION GNORM ! norm of gradient
 
GNORM = 0.D0 ! 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
 
DO I = 1, N ! take step
X(I) = X(I) - ETA * G(I)
END DO
IFLAG = 1
RETURN ! let main program evaluate G
END ! of gd
 
PROGRAM GDDEMO
IMPLICIT NONE
INTEGER N
PARAMETER (N = 2)
INTEGER ITER, J, IFLAG
DOUBLE PRECISION X(N), F, G(N), TOL
 
X(1) = -0.1D0 ! initial values
X(2) = -1.0D0
TOL = 1.D-15
CALL EVALFG (N, X, F, G)
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, F19.15, A, F19.15, A, F19.15)',
$ 'After ', ITER, ' steps, found minimum at x=',
$ X(1), ' y=', X(2), ' of f=', F
STOP 'program complete'
END
 
Output:
After      31 steps, found minimum at x=  0.107626843548372 y= -1.223259663839920 of f= -0.750063420551493
STOP program complete

Go[edit]

This is a translation of the C# code in the book excerpt linked to above and hence also of the first Typescript example below.

However, since it was originally written, I've substituted Fortran's gradient function for the original one (see Talk page) which now gives results which agree (to 6 decimal places) with those of the Fortran, Julia, Algol 68 and Algol W solutions. As a number of other solutions are based on this one, I suggest their authors update them accordingly.

package main
 
import (
"fmt"
"math"
)
 
func steepestDescent(x []float64, alpha, tolerance float64) {
n := len(x)
g0 := g(x) // Initial estimate of result.
 
// Calculate initial gradient.
fi := gradG(x)
 
// 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]
}
 
// Calculate next gradient.
fi = gradG(x)
 
// 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(p).
func gradG(p []float64) []float64 {
z := make([]float64, len(p))
x := p[0]
y := p[1]
z[0] = 2*(x-1)*math.Exp(-y*y) - 4*x*math.Exp(-2*x*x)*y*(y+2)
z[1] = -2*(x-1)*(x-1)*y*math.Exp(-y*y) + math.Exp(-2*x*x)*(y+2) + math.Exp(-2*x*x)*y
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.Printf("The minimum is at x = %f, y = %f for which f(x, y) = %f.\n", x[0], x[1], g(x))
}
Output:
Testing steepest descent method:
The minimum is at x = 0.107627, y = -1.223260 for which f(x, y) = -0.750063.

Julia[edit]

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()))
 
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[edit]

Calculate with bignum for numerical stability.

Translation of: Raku
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);
Output:
The minimum is at x[0] = 0.107653, x[1] = -1.223370

Phix[edit]

Translation of: Go
-- 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 p)
atom {x,y} = p
p[1] = 2*(x-1)*exp(-y*y) - 4*x*exp(-2*x*x)*y*(y+2)
p[2] = -2*(x-1)*(x-1)*y*exp(-y*y) + exp(-2*x*x)*(y+2) + exp(-2*x*x)*y
return p
end function
 
function steepestDescent(sequence x, atom alpha, tolerance)
integer n = length(x)
atom g0 = g(x) -- Initial estimate of result.
 
-- Calculate initial gradient.
sequence fi = gradG(x)
 
-- 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))
 
-- Calculate next gradient.
fi = gradG(x)
 
-- 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 = 0.0000001, 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 = %.13f, y = %.13f for which f(x, y) = %.16f\n", {x[1], x[2], g(x)})
Output:

Results now match (at least) Algol 68/W, Fortran, Go, Julia, Raku, REXX, and Wren [to 6dp or better anyway].
Note that specifying a tolerance < 1e-7 causes an infinite loop on Phix, whereas REXX copes with a much smaller tolerance.
Results on 32/64 bit Phix agree to 13dp, which I therefore choose to show in full here (but otherwise would not really trust).

Testing steepest descent method:
The minimum is at x = 0.1076268243295, y = -1.2232596548816 for which f(x, y) = -0.7500634205514924

Racket[edit]

Translation of: Go

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
 
(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))
 
Output:
'#(0.10760797905122492 -1.2232993981966753)

Raku[edit]

(formerly Perl 6)

Translation of: Go
# 20200904 Updated Raku programming solution
 
sub steepestDescent(@x, $alpha is copy, $h) {
 
my $g0 = g |@x ; # Initial estimate of result.
 
my @fi = gradG |@x ; # Calculate initial gradient
 
my $b = $alpha / my $delG = sqrt ( sum @fi»² ) ; # Calculate initial norm.
 
while ( $delG > $h ) { # Iterate until value is <= tolerance.
 
@x «-»= $b «*« @fi; # Calculate next value.
 
@fi = gradG |@x ; # Calculate next gradient and next value
 
$b = $alpha / ($delG = sqrt( sum @fi»² )); # Calculate next norm.
 
my $g1 = g |@x ;
 
$g1 > $g0 ?? ( $alpha /= 2 ) !! ( $g0 = $g1 ) # Adjust parameter.
}
}
 
sub gradG(\x,\y) { # gives a rough calculation of gradient g(x).
2*(x-1)*exp(-) - 4*x*exp(-2*x²)*y*(y+2) , -2*(x-1)²*y*exp(-) + exp(-2*x²)*(2*y+2)
}
 
# Function for which minimum is to be found.
sub g(\x,\y) { (x-1)² * exp(-) + y*(y+2) * exp(-2*x²) }
 
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];
 
Output:
Testing steepest descent method:
The minimum is at x[0] = 0.10762682432947938, x[1] = -1.2232596548816097

REXX[edit]

The   tolerance   can be much smaller;   a tolerance of   1e-200   was tested.   It works, but causes the program to execute a bit slower, but still sub-second execution time.

/*REXX pgm searches for minimum values of the bi─variate function (AKA steepest descent)*/
numeric digits (length( e() ) - length(.) ) % 2 /*use half of number decimal digs in E.*/
tolerance= 1e-30 /*use a much smaller tolerance for REXX*/
alpha= 0.1
x.0= 0.1; x.1= -1
say center(' testing for the steepest descent method ', 79, "═")
call steepestD /* ┌──◄── # digs past dec. point ─►───┐*/
say 'The minimum is at: x[0]=' format(x.0,,9) " x[1]=" format(x.1,,9)
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
e: return 2.718281828459045235360287471352662497757247093699959574966967627724
g: return (x.0-1)**2 * exp(- (x.1**2) ) + x.1 * (x.1 + 2) * exp(-2 * x.0**2)
/*──────────────────────────────────────────────────────────────────────────────────────*/
gradG: x= x.0; y= x.1 /*define X and Y from the X array. */
xm= x-1; eny2= exp(-y*y); enx2= exp(-2 * x**2) /*shortcuts.*/
z.0= 2 * xm * eny2 - 4 * x * enx2 * y * (y+2)
z.1= -2 * xm**2 * y * eny2 + enx2 * (y+2) + enx2 * y
return /*a rough calculation of the gradient. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
steepestD: g0= g() /*the initial estimate of the result. */
call gradG /*calculate the initial gradient. */
delG= sqrt(z.0**2 + z.1**2) /* " " " norm. */
b= alpha / delG
do while delG>tolerance
x.0= x.0 - b * z.0; x.1= x.1 - b * z.1
call gradG
delG= sqrt(z.0**2 + z.1**2); if delG=0 then return
b= alpha / delG
g1= g() /*find minimum.*/
if g1>g0 then alpha= alpha * .5 /*adjust ALPHA.*/
else g0= g1 /* " G0. */
end /*while*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
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
output   when using the internal default inputs:
═══════════════════ testing for the steepest descent method ═══════════════════
The minimum is at:     x[0]= 0.107626824      x[1]= -1.223259655

Scala[edit]

Translation of: Go
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))
}
}
 
Output:
Testing steepest descent method
The minimum is at x : 0.10756393294495799, y : -1.2234116852966237

TypeScript[edit]

Translation of 
  •   [Numerical Methods, Algorithms and Tools in C# by Waldemar Dos Passos (18.2 Gradient Descent Method]



 
// 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();
 
 
Output:
Testing steepest descent method
The minimum is at x[0] = 0.10768224291553158, x[1] = -1.2233090211217854

Linear Regression[edit]

Translation of 
  •   [Linear Regression using Gradient Descent by Adarsh Menon]



 
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();
 

Wren[edit]

Translation of: Go
Library: Wren-math
Library: Wren-fmt
import "/math" for Math
import "/fmt" for Fmt
 
// Function for which minimum is to be found.
var g = Fn.new { |x|
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])
}
 
// Provides a rough calculation of gradient g(p).
var gradG = Fn.new { |p|
var x = p[0]
var y = p[1]
return [2*(x-1)*Math.exp(-y*y) - 4*x*Math.exp(-2*x*x)*y*(y+2),
-2*(x-1)*(x-1)*y*Math.exp(-y*y) + Math.exp(-2*x*x)*(y+2) + Math.exp(-2*x*x)*y]
}
 
var steepestDescent = Fn.new { |x, alpha, tolerance|
var n = x.count
var g0 = g.call(x) // // Initial estimate of result.
 
// Calculate initial gradient.
var fi = gradG.call(x)
 
// Calculate initial norm.
var delG = 0
for (i in 0...n) delG = delG + fi[i]*fi[i]
delG = delG.sqrt
var b = alpha/delG
 
// Iterate until value is <= tolerance.
while (delG > tolerance) {
// Calculate next value.
for (i in 0...n) x[i] = x[i] - b*fi[i]
 
// Calculate next gradient.
fi = gradG.call(x)
 
// Calculate next norm.
delG = 0
for (i in 0...n) delG = delG + fi[i]*fi[i]
delG = delG.sqrt
b = alpha/delG
 
// Calculate next value.
var g1 = g.call(x)
 
// Adjust parameter.
if (g1 > g0) {
alpha = alpha / 2
} else {
g0 = g1
}
}
}
 
var tolerance = 0.0000006
var alpha = 0.1
var x = [0.1, -1] // Initial guess of location of minimum.
 
steepestDescent.call(x, alpha, tolerance)
System.print("Testing steepest descent method:")
Fmt.print("The minimum is at x = $f, y = $f for which f(x, y) = $f.", x[0], x[1], g.call(x))
Output:
Testing steepest descent method:
The minimum is at x = 0.107627, y = -1.223260 for which f(x, y) = -0.750063.

zkl[edit]

Translation of: Go
with tweaked gradG
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)
}
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)));
Output:
Testing steepest descent method:
The minimum is at (x,y) = (0.107608,-1.223299). f(x,y) = -0.750063