Cody

Solution 172766

Submitted on 5 Dec 2012 by @bmtran
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.

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]=BFGS_Quasi_Newton(F,gradF,x0,0.01,1) % single steepest descent 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 0.0000 267.6200 1270.8691 1 0.0004 6.0204 1.6920 xmin = -1.4521814453125 2.117291015625 fmin = 6.02035111158688
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=[ 0.2927 0.0506]; fcorrect=0.63; [xmin,fmin]=BFGS_Quasi_Newton(F,gradF,x0,1e-2,2) % two iterations assert(norm(xmin-xcorrect)<0.1) assert( abs(fmin-fcorrect)<0.01)
iter alpha f(alpha) norm(c) 0 0.0000 1.0000 2.0000 1 0.0806 0.7711 5.2003 2 0.0098 0.6237 7.5488 xmin = 0.293138805215789 0.0507131696991079 fmin = 0.623677791785989
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 = [0;0]; xcorrect = [1;1;]; fcorrect = 0; [xmin,fmin]=BFGS_Quasi_Newton(F,gradF,x0) assert(norm(xmin-xcorrect)<0.01) assert(abs(fmin-fcorrect)<0.01);
iter alpha f(alpha) norm(c) 0 0.0000 1.0000 2.0000 1 0.0806 0.7711 5.2003 2 0.0098 0.6237 7.5488 3 0.6744 0.4362 2.9602 4 0.7897 0.3172 3.8160 5 1.3353 0.2743 6.6469 6 5.5758 0.0183 1.6494 7 0.0966 0.0172 0.3335 8 4.4928 0.0060 1.9915 9 2.2007 0.0012 0.7472 10 0.5005 0.0001 0.2988 11 2.1621 0.0000 0.0867 12 0.6740 0.0000 0.0012 13 1.1440 0.0000 0.0000 xmin = 0.99999963957031 0.999999356221718 fmin = 7.24057125016987e-13
4   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)]; HessF=@(x) 200*[4*x(1).^2-2*(x(2)-x(1).^2)+1/100, -2*x(1); -2*x(1), 1]; xcorrect = [1;1]; fcorrect = 0; x0 = [-1.9; 2]; [xmin,fmin]=BFGS_Quasi_Newton(F,gradF,x0,1e-4) assert(isequal(round(xmin),xcorrect)) assert(isequal(round(fmin),fcorrect))
iter alpha f(alpha) norm(c) 0 0.0000 267.6200 1270.8691 1 0.0004 6.0204 1.6920 2 0.0864 5.9008 19.6322 3 6.2485 5.3456 36.0218 4 3.5329 4.0383 28.6153 5 0.2788 3.6720 3.7520 6 1.3761 3.1057 17.4518 7 1.7003 2.7678 20.0084 8 0.9365 2.1467 4.6768 9 0.5341 1.7479 7.9198 10 1.0640 1.6516 10.5990 11 3.6971 0.9423 6.8115 12 0.2643 0.8467 1.8447 13 1.8696 0.6372 5.1444 14 2.5470 0.4593 7.2784 15 0.4177 0.3375 1.9342 16 1.0873 0.2375 4.1169 17 1.6420 0.1890 6.7381 18 1.0246 0.1027 2.8612 19 0.6739 0.0610 2.7611 20 1.5333 0.0472 4.7146 21 1.7406 0.0151 2.2041 22 0.5299 0.0053 1.1737 23 2.1143 0.0028 1.8166 24 1.0176 0.0004 0.2330 25 1.1406 0.0000 0.2085 26 1.4584 0.0000 0.0098 27 0.8172 0.0000 0.0005 28 1.0511 0.0000 0.0000 xmin = 1.00000000001576 1.00000000004913 fmin = 3.123986307874e-20