Golden section search algorithm

조회 수: 18 (최근 30일)
Ous Chkiri
Ous Chkiri 2021년 2월 26일
답변: Pratyush Roy 2021년 3월 31일
Errors :
with the same size and shape as the input arguments.
> In matlab.graphics.function.FunctionLine>getFunction
In matlab.graphics.function/FunctionLine/updateFunction
In matlab.graphics.function.FunctionLine.set.Function_I
In matlab.graphics.function.FunctionLine.set.Function
In matlab.graphics.function.FunctionLine
In fplot>singleFplot (line 245)
In fplot>@(f)singleFplot(cax,{f},limits,extraOpts,args) (line 200)
In fplot>vectorizeFplot (line 200)
In fplot (line 166)
In goldensection (line 30)
Matlab code :
clc; clear all; clc;
phi = double( (sqrt(5)-1) / 2 ); % golden ratio
del = .05;% small increment value
epsilon = .001; % function difference precision
max_iter = 100; % maximum number of iterations for Phase I and II.
alpha(1) = 0; % first value of alpha
alpha(2) = del; % second value of alpha is equal to the small increment value
% begin iterations for Phase I
i_iter = 1;
while i_iter <= max_iter
if ( f(alpha(i_iter)) > f(alpha(i_iter + 1)) )
alpha(i_iter + 2) = 0;
for j_iter = 1 : i_iter+1
alpha(i_iter + 2) = alpha(i_iter + 2) + del*(1.618)^(j_iter-1);
end
else
break
end
i_iter = i_iter + 1; % increment number of iterations by 1
end
% the uncertainity interval
alpha_l = 0; % set lower bound of alpha since it cannot be known how many alphas will be enough, i.e. alpha(i_iter - 2) may be null.
alpha_u = alpha(i_iter + 1);
% plot the line search function of one variable (i.e., alpha - a)
figure; hold on;
syms a;
fplot(@f,[alpha_l alpha_u],'b');
% plot title and axes labels
title( 'Golden section search on f ( alpha )' );
xlabel('alpha');
ylabel('f (alpha)');
% divide uncertainity interval using golden ratio, (i.e., set new interval
% boundaries)
a = alpha_l + (1 - phi)*(alpha_u - alpha_l);
b = alpha_l + phi*(alpha_u - alpha_l);
% calculate function values at points
f_a = f(a);
f_b = f(b);
% plot new interval boundaries on top of the function plot.
plot(a,f_a,'rd')
plot(b,f_b,'rd')
% begin iterations for Phase II
i_iter = 1;
while ( ( abs(alpha_u-alpha_l) > epsilon ) && (i_iter < max_iter) ) && (i_iter < max_iter)
if(f_a < f_b)
alpha_u = b;
b = a;
a = alpha_l + (1 - phi)*(alpha_u - alpha_l);
f_a=f(a);
f_b=f(b);
plot(a,f_a,'rd');
else
alpha_l = a;
a = b;
b = alpha_l + phi*(alpha_u - alpha_l);
f_a = f(a);
f_b = f(b);
plot(b, f_b, 'rd')
end
end
% print alpha value that minimizes the line search function
if(f_a<f_b)
fprintf('\nalpha_min = %f\n', a)
fprintf('f_min = %f\n', f_a)
plot(a,f_a,'kd', 'LineWidth', 1, 'MarkerSize', 10, 'MarkerFaceColor' , 'g')
else
fprintf('\nalpha_min = %f\n', b)
fprintf('f_min = %f\n', f_b)
plot(b,f_b,'kd', 'LineWidth', 1, 'MarkerSize', 10, 'MarkerFaceColor' , 'g')
end
hold off;
saveas(gcf,'results_plot.png')
function y = f(a)
% This function returns the value of the line search function for a given
% value of function variable.
% y = 2 - 4*a + exp(a); % Example 10.3 in the book (Arora)
% y = (11*a - 5)^2 + 16*a^2; % Homework 3 / Section B - a
% y = 0.1*(5 - a)^2 + (1 - 2*a)^2; % Homework 3 / Section B - b
y =(-19+2358*a)^2+5*(-1+588*a)^2+(-4+880*a)^4+10*(-3+2208*a)^4;
end

답변 (1개)

Pratyush Roy
Pratyush Roy 2021년 3월 31일
Hi,
fplot prefers to pass a vector to the function for efficiency, but checks whether that works and gives a warning if it doesn't work. As a workaround, you can consider changing the function f in a manner as follows:
function y = f(a)
%Here we assume that a is an array and the function works on every element of a
%to return a corresponding element in the y array.
for i=1:numel(a)
y(i) =(-19+2358.*a(i)).^2 + 5*(-1+588.*a(i)).^2 + (-4+880.*a(i)).^4 + 10.*(-3+2208.*a(i)).^4;
end
end
Hope this helps!

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