```Date: Feb 27, 2013 6:08 AM
Author: Tony Kittler
Subject: Re: Alternative solution for NAN

"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 ?> > Did you check whether the matrix N already contains NaN values ?> > Best wishes> Torsten.Yes, I have checked with this codesif isnan(N)==1h=1;elseh=0;endThis code return that h=0
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