Cody

# Problem 484. Steepest Descent Method

Solution 172733

Submitted on 5 Dec 2012 by @bmtran
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### Test Suite

Test Status Code Input and Output
1   Pass
%% % Rosenbrock's banana function F=@(x) 100*(x(2)-x(1).^2).^2 + (1-x(1)).^2; gradF=@(x) [100*(4*x(1).^3-4*x(1).*x(2))+2*x(1)-2; 100*(2*x(2)-2*x(1).^2)]; x0 = [-1.9; 2.0]; x1=[ -1.4478 2.1184]; x2=[ 1.7064 2.9446]; f1=6.0419; f2=0.6068; [xmin,fmin]=SteepestDescent(F,gradF,x0,0.01,1) assert(norm(xmin-x1)<0.2||norm(xmin-x2)<0.2) assert( abs(fmin-f1)<0.5|| abs(fmin-f2)<0.5) % 2 local min

``` iter alpha f(alpha) norm(c) 0 0.000 267.6200 1270.8691 1 0.000 6.0719 14.1065 xmin = -1.4452 2.1191 fmin = 6.0719 ```

2   Pass
%% % Rosenbrock's banana function F=@(x) 100*(x(2)-x(1).^2).^2 + (1-x(1)).^2; gradF=@(x) [100*(4*x(1).^3-4*x(1).*x(2))+2*x(1)-2; 100*(2*x(2)-2*x(1).^2)]; x0 = [0; 0]; xcorrect=[1;1]; fcorrect=0; [xmin,fmin]=SteepestDescent(F,gradF,x0) % 20 iterations default assert(norm((xmin-xcorrect),inf)<1) assert(abs(fmin-fcorrect)<0.8);

``` iter alpha f(alpha) norm(c) 0 0.000 1.0000 2.0000 20 0.005 0.3983 1.2496 xmin = 0.3689 0.1360 fmin = 0.3983 ```

3   Pass
%% % Rosenbrock's banana function F=@(x) 100*(x(2)-x(1).^2).^2 + (1-x(1)).^2; gradF=@(x) [100*(4*x(1).^3-4*x(1).*x(2))+2*x(1)-2; 100*(2*x(2)-2*x(1).^2)]; x0 = [1.1; 0.9]; xcorrect=[1;1]; fcorrect=0; [xmin,fmin]=SteepestDescent(F,gradF,x0,1e-2,2000) assert(isequal(round(xmin),xcorrect)) assert(isequal(round(fmin),fcorrect))

``` iter alpha f(alpha) norm(c) 0 0.000 9.6200 150.0119 298 0.001 0.0001 0.0100 xmin = 0.9897 0.9794 fmin = 1.0730e-04 ```