Date: Feb 27, 2013 7:40 AM
Author: Tony Kittler
Subject: Re: Alternative solution for NAN

"Torsten" wrote in message <kgku2t$t2g$1@newscl01ah.mathworks.com>...
> "Carl S." wrote in message <kgktg9$rhc$1@newscl01ah.mathworks.com>...
> > "Carl S." wrote in message <kgkpep$h9u$1@newscl01ah.mathworks.com>...
> > > "Torsten" wrote in message <kgko7b$e4d$1@newscl01ah.mathworks.com>...
> > > > "Carl S." wrote in message <kgkmon$aeo$1@newscl01ah.mathworks.com>...
> > > > > "Torsten" wrote in message <kgklbh$6ph$1@newscl01ah.mathworks.com>...
> > > > > > "Carl S." wrote in message <kgki2g$rie$1@newscl01ah.mathworks.com>...
> > > > > > > The following code gives NAN (Not a Number) values
> > > > > > > [U,D]=eig(N);
> > > > > > >
> > > > > > > To solve this problem, I wrote that
> > > > > > > while(det(N) == 0)
> > > > > > > N=(1e-10.*randi(1,size(N)))*eye(size(N));
> > > > > > > end
> > > > > > >
> > > > > > > But, the loop does not stop

> > > > > >
> > > > > > Your matrix N within the loop always has determinant (1e-10)^(size(N))
> > > > > > which may become very small if N is large.
> > > > > >
> > > > > > :( Are there any alternative solution instead of this loop to solve the NAN problem ?
> > > > > >
> > > > > > Depends on the original matrix N.
> > > > > >
> > > > > > Best wishes
> > > > > > Torsten.

> > > > >
> > > > > Dear Torsten,
> > > > > The matrix N has standard deviation values of grayscale images. So, it changes for each image. How to solve the NAN problem in this case ?

> > > >
> >
> > I have tried this,
> > u=1e-10;
> > while(det(N) == 0)
> > N=u.*eye(size(N));
> > u=u*100;
> > end
> >
> > Now, it works without giving NAN value. But, I am not sure that this algorithm correct results. Do you think that this is meaningful or I can get unexpected results ? Any suggestions ?

>
> The matrix N you get after the while loop is a scalar multiple of the identity matrix and in general has nothing in common with your original matrix N.
> You will have to find out why eig produces NaN values for your original matrix N.
> Are you sure all elements of N are finite ?
>
> Best wishes
> Torsten.


Yes, Torsten, they are finite

My goal is to fit means(mu) and standard deviations(N) to Gaussian shape. The codes that I wrote above are from the function ;

function res=MultivariateGaussianPDF(x,mu,N)
while(det(N) == 0)
N=(1e-10.*randi(1,size(N)))*eye(size(N));
end

[M,d]=size(x);
[U,D]=eig(N); % <=causes NAN problem :((

W=sqrt(inv(D))*U;
Wx=W*(x-ones(M,1)*mu)';
res=(1/sqrt((2*pi)^d*det(N)))*exp(-0.5*sum(Wx.^2,1));