I want to determine the Steepest descent of the Rosenbruck function using Armijo steplength where x = [-1.2, 1]' (the initial column vector).
The problem is, that the code has been running for a long time. I think there will be an infinite loop created here. But I could not understand where the problem was.
Could anyone help me?
n=input('enter the number of variables n ');
% Armijo stepsize rule parameters
x = [-1.2 1]';
s = 10;
m = 0;
sigma = .1;
beta = .5;
obj=func(x);
g=grad(x);
k_max = 10^5;
k=0; % k = # iterations
nf=1; % nf = # function eval.
x_new = zeros([],1) ; % empty vector which can be filled if length is not known ;
[X,Y]=meshgrid(-2:0.5:2);
fx = 100*(X.^2 - Y).^2 + (X-1).^2;
contour(X, Y, fx, 20)
while (norm(g)>10^(-3)) && (k<k_max)
d = -g./abs(g); % steepest descent direction
s = 1;
newobj = func(x + beta.^m*s*d);
m = m+1;
if obj > newobj - (sigma*beta.^m*s*g'*d)
t = beta^m *s;
x = x + t*d;
m_new = m;
newobj = func(x + t*d);
nf = nf+1;
else
m = m+1;
end
obj=newobj;
g=grad(x);
k = k + 1;
x_new = [x_new, x];
end
% Output x and k
x_new, k, nf
fprintf('Optimal Solution x = [%f, %f]\n', x(1), x(2))
plot(x_new)
function y = func(x)
y = 100*(x(1)^2 - x(2))^2 + (x(1)-1)^2;
end
function y = grad(x)
y(1) = 100*(2*(x(1)^2-x(2))*2*x(1)) + 2*(x(1)-1);
end
Related
For my studies I had to write a PDE solver for the Poisson equation on a disc shaped domain using the finite difference method.
I already passed the Lab exercise. There is one issue in my code I couldn't fix. Function fun1 with the boundary value problem gun2 is somehow oscillating at the boundary. When I use fun2 everything seems fine...
Both functions use at the boundary gun2. What is the problem?
function z = fun1(x,y)
r = sqrt(x.^2+y.^2);
z = zeros(size(x));
if( r < 0.25)
z = -10^8*exp(1./(r.^2-1/16));
end
end
function z = fun2(x,y)
z = 100*sin(2*pi*x).*sin(2*pi*y);
end
function z = gun2(x,y)
z = x.^2+y.^2;
end
function [u,A] = poisson2(funame,guname,M)
if nargin < 3
M = 50;
end
%Mesh Grid Generation
h = 2/(M + 1);
x = -1:h:1;
y = -1:h:1;
[X,Y] = meshgrid(x,y);
CI = ((X.^2 +Y.^2) < 1);
%Boundary Elements
Sum= zeros(size(CI));
%Sum over the neighbours
for i = -1:1
Sum = Sum + circshift(CI,[i,0]) + circshift(CI,[0,i]) ;
end
%if sum of neighbours larger 3 -> inner note!
CI = (Sum > 3);
%else boundary
CB = (Sum < 3 & Sum ~= 0);
Sum= zeros(size(CI));
%Sum over the boundary neighbour nodes....
for i = -1:1
Sum = Sum + circshift(CB,[i,0]) + circshift(CB,[0,i]);
end
%If the sum is equal 2 -> Diagonal boundary
CB = CB + (Sum == 2 & CB == 0 & CI == 0);
%Converting X Y to polar coordinates
Phi = atan(Y./X);
%Converting Phi R back to cartesian coordinates, only at the boundarys
for j = 1:M+2
for i = 1:M+2
if (CB(i,j)~=0)
if j > (M+2)/2
sig = 1;
else
sig = -1;
end
X(i,j) = sig*1*cos(Phi(i,j));
Y(i,j) = sig*1*sin(Phi(i,j));
end
end
end
%Numberize the internal notes u1,u2,......,un
CI = CI.*reshape(cumsum(CI(:)),size(CI));
%Number of internal notes
Ni = nnz(CI);
f = zeros(Ni,1);
k = 1;
A = spalloc(Ni,Ni,5*Ni);
%Create matix A!
for j=2:M+1
for i =2:M+1
if(CI(i,j) ~= 0)
hN = h;hS = h; hW = h; hE = h;
f(k) = fun(X(i,j),Y(i,j));
if(CB(i+1,j) ~= 0)
hN = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i+1,j))*2/(hN^2+hN*h);
A(k,CI(i-1,j)) = -2/(h^2+h*hN);
else
if(CB(i-1,j) ~= 0) %in negative y is a boundry
hS = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j),Y(i-1,j))*2/(hS^2+h*hS);
A(k,CI(i+1,j)) = -2/(h^2+h*hS);
else
A(k,CI(i-1,j)) = -1/h^2;
A(k,CI(i+1,j)) = -1/h^2;
end
end
if(CB(i,j+1) ~= 0)
hE = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j+1),Y(i,j))*2/(hE^2+hE*h);
A(k,CI(i,j-1)) = -2/(h^2+h*hE);
else
if(CB(i,j-1) ~= 0)
hW = abs(1-sqrt(X(i,j)^2+Y(i,j)^2));
f(k) = f(k) + gun(X(i,j-1),Y(i,j))*2/(hW^2+h*hW);
A(k,CI(i,j+1)) = -2/(h^2+h*hW);
else
A(k,CI(i,j-1)) = -1/h^2;
A(k,CI(i,j+1)) = -1/h^2;
end
end
A(k,k) = (2/(hE*hW)+2/(hN*hS));
k = k + 1;
end
end
end
%Solve linear system
u = A\f;
U = zeros(M+2,M+2);
p = 1;
%re-arange u
for j = 1:M+2
for i = 1:M+2
if ( CI(i,j) ~= 0)
U(i,j) = u(p);
p = p+1;
else
if ( CB(i,j) ~= 0)
U(i,j) = gun(X(i,j),Y(i,j));
else
U(i,j) = NaN;
end
end
end
end
surf(X,Y,U);
end
I'm keeping this answer short for now, but may extend when the question contains more info.
My first guess is that what you are seeing is just numerical errors. Looking at the scales of the two graphs, the peaks in the first graph are relatively small compared to the signal in the second graph. Maybe there is a similar issue in the second that is just not visible because the signal is much bigger. You could try to increase the number of nodes and observe what happens with the result.
You should always expect to see numerical errors in such simulations. It's only a matter of trying to get their magnitude as small as possible (or as small as needed).
The code below is defined as algorithm 1 that computes the Pseudo Zernike Radial polynomials:
function R = pseudo_zernike_radial_polynomials(n,r)
if any( r>1 | r<0 | n<0)
error(':zernike_radial_polynomials either r is less than or greater thatn 1, r must be between 0 and 1 or n is less than 0.')
end
if n==0;
R =ones(n +1, length(r));
return;
end
R =ones(n +1, length(r));
rSQRT= sqrt(r);
r0 = ~logical(rSQRT.^(2*n+1)) ; % if any low r exist, and high n, then treat as 0
if any(r0)
m = n:-1:mod(n,2); ss=1:sum(r0);
R0(m +1, ss)=0;
R0(0 +1, ss)=1;
R(:,r0)=R0;
end
if any(~r0)
rSQRT= rSQRT(~r0);
R1 = zernike_radial_polynomials(2*n+1, rSQRT );
m = 2:2: 2*n+1 +1;
R1=R1(m,:);
for m=1:size(R1,1)
R1(m,:) = R1(m,:)./rSQRT';
end
R(:,~r0)=R1;
end
Then, this is algorithm 2 that calculates the moments:
and I translate into the code as follow:
clear all
%input : 2D image f, Nmax = order.
f = rgb2gray(imread('Oval_45.png'));
prompt = ('Input PZM order Nmax:');
Nmax = input(prompt);
Pzm =0;
l = size(f,1);
for x = 1:l;
for y =x;
for n = 0:Nmax;
[X,Y] = meshgrid(x,y);
R = sqrt((2.*X-l-1).^2+(2.*Y-l-1).^2)/l;
theta = atan2((l-1-2.*Y+2),(2.*X-l+1-2));
R = (R<=1).*R;
rad = pseudo_zernike_radial_polynomials(n, R);
for m = 0:n;
%find psi
if mod(m,2)==0
%m is even
newd1 = f(x,y)+f(x,y);
newd2 = f(y,x)+f(y,x);
newd3 = f(x,y)+f(x,y);
newd4 = f(y,x)+f(y,x);
x1 = newd1;
y1 = (-1)^m/2*newd2;
x2 = newd3;
y2 = (-1)^m/2*newd4;
psi = cos(m*theta)*(x1+y1+x2+y2)-(1i)*sin(m*theta)*(x1+y1-x2-y2);
else
newd1 = f(x,y)-f(x,y);
newd2 = f(y,x)-f(y,x);
newd3 = f(x,y)-f(x,y);
newd4 = f(y,x)-f(y,x);
x1 = newd1;
y1 = (-1)^m/2*newd2;
x2 = newd3;
y2 = (-1)^m/2*newd4;
psi = cos(m*theta)*(x1+x2)+sin(m*theta)*(y1-y2)+(1i)*(cos(m*theta)*(y1+y2)-sin(m*theta)*(x1-x2));
end
Pzm = Pzm+rad*psi;
end
end
end
end
However its give me error :
Error using *
Integers can only be combined with integers of the same class, or scalar doubles.
Error in main_pzm (line 44)
Pzm = Pzm+rad*psi;
The detail of the calculation can be seen here
EDIT: The code that I have pasted is too long. Basicaly I dont know how to work with the second code, If I know how calculate alpha from the second code I think my problem will be solved. I have tried a lot of input arguments for the second code but it does not work!
I have written following code to solve a convex optimization problem using Gradient descend method:
function [optimumX,optimumF,counter,gNorm,dx] = grad_descent()
x0 = [3 3]';%'//
terminationThreshold = 1e-6;
maxIterations = 100;
dxMin = 1e-6;
gNorm = inf; x = x0; counter = 0; dx = inf;
% ************************************
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
%alpha = 0.01;
% ************************************
figure(1); clf; ezcontour(f,[-5 5 -5 5]); axis equal; hold on
f2 = #(x) f(x(1),x(2));
% gradient descent algorithm:
while and(gNorm >= terminationThreshold, and(counter <= maxIterations, dx >= dxMin))
g = grad(x);
gNorm = norm(g);
alpha = linesearch_strongwolfe(f,-g, x0, 1);
xNew = x - alpha * g;
% check step
if ~isfinite(xNew)
display(['Number of iterations: ' num2str(counter)])
error('x is inf or NaN')
end
% **************************************
plot([x(1) xNew(1)],[x(2) xNew(2)],'ko-')
refresh
% **************************************
counter = counter + 1;
dx = norm(xNew-x);
x = xNew;
end
optimumX = x;
optimumF = f2(optimumX);
counter = counter - 1;
% define the gradient of the objective
function g = grad(x)
g = [(8*x(1) + 2*x(2) +10)
(2*x(1) + 16*x(2) + 1)];
end
end
As you can see, I have commented out the alpha = 0.01; part. I want to calculate alpha via an other code. Here is the code (This code is not mine)
function alphas = linesearch_strongwolfe(f,d,x0,alpham)
alpha0 = 0;
alphap = alpha0;
c1 = 1e-4;
c2 = 0.5;
alphax = alpham*rand(1);
[fx0,gx0] = feval(f,x0,d);
fxp = fx0;
gxp = gx0;
i=1;
while (1 ~= 2)
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
if (fxx > fx0 + c1*alphax*gx0) | ((i > 1) & (fxx >= fxp)),
alphas = zoom(f,x0,d,alphap,alphax);
return;
end
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx >= 0,
alphas = zoom(f,x0,d,alphax,alphap);
return;
end
alphap = alphax;
fxp = fxx;
gxp = gxx;
alphax = alphax + (alpham-alphax)*rand(1);
i = i+1;
end
function alphas = zoom(f,x0,d,alphal,alphah)
c1 = 1e-4;
c2 = 0.5;
[fx0,gx0] = feval(f,x0,d);
while (1~=2),
alphax = 1/2*(alphal+alphah);
xx = x0 + alphax*d;
[fxx,gxx] = feval(f,xx,d);
xl = x0 + alphal*d;
fxl = feval(f,xl,d);
if ((fxx > fx0 + c1*alphax*gx0) | (fxx >= fxl)),
alphah = alphax;
else
if abs(gxx) <= -c2*gx0,
alphas = alphax;
return;
end
if gxx*(alphah-alphal) >= 0,
alphah = alphal;
end
alphal = alphax;
end
end
But I get this error:
Error in linesearch_strongwolfe (line 11) [fx0,gx0] = feval(f,x0,d);
As you can see I have written the f function and its gradient manually.
linesearch_strongwolfe(f,d,x0,alpham) takes a function f, Gradient of f, a vector x0 and a constant alpham. is there anything wrong with my declaration of f? This code works just fine if I put back alpha = 0.01;
As I see it:
x0 = [3; 3]; %2-element column vector
g = grad(x0); %2-element column vector
f = #(x1,x2) 4.*x1.^2 + 2.*x1.*x2 +8.*x2.^2 + 10.*x1 + x2;
linesearch_strongwolfe(f,-g, x0, 1); %passing variables
inside the function:
[fx0,gx0] = feval(f,x0,-g); %variable names substituted with input vars
This will in effect call
[fx0,gx0] = f(x0,-g);
but f(x0,-g) is a single 2-element column vector with these inputs. Assingning the output to two variables will not work.
You either have to define f as a proper named function (just like grad) to output 2 variables (one for each component), or edit the code of linesearch_strongwolfe to return a single variable, then slice that into 2 separate variables yourself afterwards.
If you experience a very rare kind of laziness and don't want to define a named function, you can still use an anonymous function at the cost of duplicating code for the two components (at least I couldn't come up with a cleaner solution):
f = #(x1,x2) deal(4.*x1(1)^2 + 2.*x1(1)*x2(1) +8.*x2(1)^2 + 10.*x1(1) + x2(1),...
4.*x1(2)^2 + 2.*x1(2)*x2(2) +8.*x2(2)^2 + 10.*x1(2) + x2(2));
[fx0,gx0] = f(x0,-g); %now works fine
as long as you always have 2 output variables. Note that this is more like a proof of concept, since this is ugly, inefficient, and very susceptible to typos.
I'm trying to build make a code where an equation is not calculated for some certain values. I have a meshgrid with several values for x and y and I want to include a for loop that will calculate some values for most of the points in the meshgrid but I'm trying to include in that loop a condition that if the points have a specified index, the value will not be calculated. In my second group of for/if loops, I want to say that for all values of i and k (row and column), the value for z and phi are calculated with the exception of the specified i and k values (in the if loop). What I'm doing at the moment is not working...
The error I'm getting is:
The expression to the left of the equals sign is not a valid target for an assignment.
Here is my code at the moment. I'd really appreciate any advice on this! Thanks in advance
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
for i = 1:length(x)
for k = 1:length(x)
% Zeroing values where cylinder is
if sqrt(x(i,k).^2 + y(i,k).^2) < a
x(i,k) = 0;
y(i,k) = 0;
end
end
end
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
z = zeros(length(x));
phi = zeros(length(x));
for i = 1:length(x)
for k = 1:length(x)
if (i > 16 && i < 24 && k > 16 && k <= length(x))
z = 0;
phi = 0;
else
z = U_i.*r.*(1-a^2./r.^2).*sin(theta); % Stream function
phi = U_i*r.*(1+a^2./r.^2).*cos(theta); % Velocity potential
end
end
end
The original code in the question can be rewritten as seen below. Pay attention in the line with ind(17:24,:) since your edit now excludes 24 and you original question included 24.
U_i = 20;
a = 4;
c = -a*5;
b = a*10;
d = -20;
e = 20;
n = a*10;
[x,y] = meshgrid([c:(b-c)/n:b],[d:(e-d)/n:e]');
ind = find(sqrt(x.^2 + y.^2) < a);
x(ind) = 0;
y(ind) = 0;
r = sqrt(x.^2 + y.^2);
theta = atan2(y,x);
ind = true(size(x));
ind(17:24,17:length(x)) = false;
z = zeros(size(x));
phi = zeros(size(x));
z(ind) = U_i.*r(ind).*(1-a^2./r(ind).^2).*sin(theta(ind)); % Stream function
phi(ind) = U_i.*r(ind).*(1+a^2./r(ind).^2).*cos(theta(ind)); % Velocity potential
I am doing two updates for h and x given in this [paper] http://paris.cs.illinois.edu/pubs/nasser-icassp2015.pdf (You don't have to read the paper, just look for the equations 4,10 and 14 for updating h and x given on page 2 and 3).
This is the code snippet that I have tried so far. Can you tell me if its correct? Also, is there any way to optimize these for loops?
In some cases (t-tau) term was negative and MATLAB was giving an error. So, I put a condition that only implement if (t-tau)>0. Doing this is correct or is there any other way to take care of the negative indices?
%updates
Lh=10;
lambda = 0.1*(sum(reverberatedspeechspec(:))/(size(Y,1)*size(Y,2)));
S=reverberatedspeechspec;
W=basis_mel_act; W=gather(W); W = double(W); %W is the dictionary
%---initialization for H(RIR)----
H=rand(size(Y,1),Lh);
nmfIter = 50;
%---initialization for X(Activations)-----
W_trans=W';
X=W_trans*S; X=double(X);
Y = zeros(size(S,1));
for idx = 1 : size(Y,1)
Y(idx,:) = filter(S(idx,:),1,H(idx,:));
end
Stilde = zeros(size(S));
Ytilde = zeros(size(S));
for iter=1:nmfIter
% update for H
Stilde = W*X;
Ytilde = zeros(size(S));
for j=1:size(Stilde,1)
Ytilde (j,:) = filter(Stilde(j,:),1,H(j,:));
end
ratio = Y./Ytilde;
numerator = zeros(size(H));
denominator = numerator;
for k = 1 :size(Y,1)
for tau = 1:Lh
for t= 1:size(Y,2)
if gt (t-tau , 0)
numerator (k,tau) = numerator(k,tau) + ratio(k,t) * Stilde(k,t-tau);
denominator(k,tau) = denominator (k,tau) + Stilde (k,t-tau);
end
end
end
end
H = H .* numerator ./denominator ;
%updating Ytilde after getting a new value for H
for j=1:size(Stilde,1)
Ytilde (j,:) = filter(Stilde(j,:),1,H(j,:));
end
%update for X
ratio = Y./Ytilde;
ratio = [ratio zeros(size(Y,1),Lh)]; %zero padding Y and Ytilde for (t+tau) term in update of X.
Product = H.' * W; % Product of H_transpose and W in th update which is equivalent to the term ∑H(k,tau)W(k,r)
numerator = zeros(size(H));
denominator = numerator;
for r = 1:size(W,2)
for t = 1:size(Y,2)
for k = 1:size(Y,1)
for tau = 1:Lh
numerator(r,t) = numerator(r,t) + ratio(k,t+tau) * Product(tau,r);
denominator(r,t) = denominator(r,t) + Product(tau,r) + lambda;
end
end
end
end
X = X .* numerator ./denominator;
end