Filter löschen
Filter löschen

Find minimum error function using gradient descent

11 Ansichten (letzte 30 Tage)
Elysi Cochin
Elysi Cochin am 17 Jan. 2023
Kommentiert: Elysi Cochin am 18 Jan. 2023
Find the minimum error function using gradient descent
for the function (x-1).^2 - 2
I found a code in matlabcentral, but I'm not sure how to edit it according to my requirement.
%% Code I got from matlabcentral
X = -2:0.1:2;
Y = -2:0.1:2;
[X,Y] = meshgrid(X,Y);
% Z = 2*X.^2+3*Y.^2;
Z = (x-1).^2 - 2;
surf(X,Y,Z)
hold on
x(1) = 2; % initial value of x
y(1) = 2; % initial value of y
z(1) = 2.*x(1).^2 + 3.*y(1).^2;
stepsize = 0.1;
for i = 1:30
zx = 4*x(i);
zy = 6*y(i);
x(i+1) = x(i) - stepsize*zx; %gradient descent
y(i+1) = y(i) - stepsize*zy;
z(i+1) = 2.*x(i+1).^2 + 3.*y(i+1).^2
end

Akzeptierte Antwort

Dyuman Joshi
Dyuman Joshi am 17 Jan. 2023
Bearbeitet: Dyuman Joshi am 18 Jan. 2023
I don't know about the code you found, the method of operation for the given optimization algorithm should be like this -
funcChoice=2;
%error function
switch funcChoice
case 1
fun = @(x) (x-1).^2 - 2;
fungrad = @(x) 2*(x-1);
minval = -2;
case 2
fun = @(x) exp(x) - 2*x;
fungrad = @(x) exp(x)-2;
minval = 2*(1-log(2));
case 3
fun = @(x) -log(x) + 2*x;
fungrad = @(x) -1./x+2;
minval = 1+log(2);
end
disp(fun)
@(x)exp(x)-2*x
disp(fungrad)
@(x)exp(x)-2
disp(minval)
0.6137
%starting point, can be changed as per wish
x=4;
%learning rate
rate = 0.01;
%counter
itr=0;
%tolerance, adjust accordingly
tol=1e-3;
%main algorithm
while abs(fun(x(end))-minval)>=tol
itr=itr+1;
x(end+1)=x(end)-rate*fungrad(x(end));
end
%number of iterations
itr
itr = 169
%corresponding x-value, already had a starting point, that's why +1
finalx=x(itr+1)
finalx = 0.7242
%minimum function value
finalval=fun(x(itr+1))
finalval = 0.6147
%minimum error
minError=abs(minval-finalval)
minError = 9.7524e-04
  8 Kommentare
Dyuman Joshi
Dyuman Joshi am 18 Jan. 2023
Bearbeitet: Dyuman Joshi am 18 Jan. 2023
That doesn't make any sense to me or doesn't strike any logic
Was a justification given for it? If not, check notes or ask the instructor what does it mean by a good point.
Elysi Cochin
Elysi Cochin am 18 Jan. 2023
Ok Sir. Thank you so much for your help. Thanks a lot.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Programming finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by