Getting inf in matrix?
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I wrote a code where I’m implementing an algorithm. The code works, but the values in my Z matrix are all inf and I don’t know why. I added a picture so you can see it. r in R^d should a sample drawn uniformly at random from the Fourier transformation of the Gaussian kernel. Gamma is just numbers uniformly at random in (0,2pi). cos(...) is at maximum 1, so I don’t understand why I get inf. Help is appreciated!
n=2000;
d=1000;
s=50;
Gamma=10;
S=randn(n,s);
U=qr(randn(s,d));
F=randn(n,d);
D=zeros(s,s);
for i=1:s
D(i,i)=1-(i-1)/d;
end
X = S*D*U + F/Gamma;
% This is the data matrix
-----------------------------------
function [G_2] = rnca(A,m)
[n,d]=size(A);
G=zeros(n,n);
for k=1:n
for j=1:n
G(k,j)=(1/2*pi)^(d/2)*exp(-norm(A(k,:)-A(j,:))^2/2);
end
end
Z=zeros(n,m);
gamma=0+(2*pi-0)*rand(1,m);
D=fft(G);
r=D(randsample(size(D,1),m),1:d);
for i=1:n
for l=1:m
Z(i,l)=sqrt(2/m)*cos(dot(r(l,:),A(i,:))+gamma(l));
end
end
G_2=Z*Z';
end
7 comentarios
James Tursa
el 19 de Abr. de 2020
We can't run pictures. Please post your code as text.
Walter Roberson
el 19 de Abr. de 2020
r is not in R^d because the fourier transform is complex valued. You are taking cos of a complex number and the result of that has value that is exponential growth.
Desiree
el 19 de Abr. de 2020
Walter Roberson
el 19 de Abr. de 2020
is your expectation that the fft will produce a lot of values that are exactly 0? doesn't happen much due to roundoff even for cases that theory say should be 0.
Desiree
el 19 de Abr. de 2020
Walter Roberson
el 19 de Abr. de 2020
what is mean(G)?
Desiree
el 19 de Abr. de 2020
Respuestas (0)
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