Consider A an nxn matrix. its not a special matrix and in the worst case all of its entries are non-zero. I am looking for a way to compute AA^T using matrix-vector operation. The total number of flops are (2n-1)*(n(n+1))/2 because all I have to do for a symmetric matrix like C=AA^T is to compute the diagonal entires which are C(i,i)=A(i,:)^T * A(i,:). What I want now is to compute the lower triangular part and then when I am done I just say that the upper triangular part is the same as the lower triangular part. The problem is can I do it in matrix-vector multiplication or will it force me to perform unnecessary multiplication (like multiplying elements in the upper part)? It is clear that a scalar computation would work but I am interested to know if a matrix-vector computation would work or not.
MATLAB is already smart enough to do this for you. When MATLAB encounters an expression like A*A.', it will recognize that the two operands are the same and will call a symmetric BLAS library function in the background to do the calculation. This symmetric function does exactly what you want ... only does about 1/2 the operations and generates an exact symmetric result.
I would like to put this equation into the MATLAB to calculate the global maximum of mutual information function. In order to do that, I notice that I need a gradient of the function which is described in
http://www.princeton.edu/~wbialek/rome/refs/sharpee+al_04.pdf
of equation (3.1) ( and of course we need to consider equation 3.2 too)
I do have the vector for P(x) and P(x|spike). Let's say they are called A and B respectively. Then, how do I write this down in MATLAB to calculate the gradient?
FYI, A and B are 1-by-100 vectors.
In MATLAB I need to generate a second derivative of a gaussian window to apply to a vector representing the height of a curve. I need the second derivative in order to determine the locations of the inflection points and maxima along the curve. The vector representing the curve may be quite noise hence the use of the gaussian window.
What is the best way to generate this window?
Is it best to use the gausswin function to generate the gaussian window then take the second derivative of that?
Or to generate the window manually using the equation for the second derivative of the gaussian?
Or even is it best to apply the gaussian window to the data, then take the second derivative of it all? (I know these last two are mathematically the same, however with the discrete data points I do not know which will be more accurate)
The maximum length of the height vector is going to be around 100-200 elements.
Thanks
Chris
I would create a linear filter composed of the weights generated by the second derivative of a Gaussian function and convolve this with your vector.
The weights of a second derivative of a Gaussian are given by:
Where:
Tau is the time shift for the filter. If you are generating weights for a discrete filter of length T with an odd number of samples, set tau to zero and allow t to vary from [-T/2,T/2]
sigma - varies the scale of your operator. Set sigma to a value somewhere between T/6. If you are concerned about long filter length then this can be reduced to T/4
C is the normalising factor. This can be derived algebraically but in practice I always do this numerically after calculating the filter weights. For unity gain when smoothing periodic signals, I will set C = 1 / sum(G'').
In terms of your comment on the equivalence of smoothing first and taking a derivative later, I would say it is more involved than that. As which derivative operator would you use in the second step? A simple central difference would not yield the same results.
You can get an equivalent (but approximate) response to a second derivative of a Gaussian by filtering the data with two Gaussians of different scales and then taking the point-wise differences between the two resulting vectors. See Difference of Gaussians for that approach.
I need to create a diagonal matrix containing the Fourier coefficients of the Gaussian wavelet function, but I'm unsure of what to do.
Currently I'm using this function to generate the Haar Wavelet matrix
http://www.mathworks.co.uk/matlabcentral/fileexchange/33625-haar-wavelet-transformation-matrix-implementation/content/ConstructHaarWaveletTransformationMatrix.m
and taking the rows at dyadic scales (2,4,8,16) as the transform:
M= 256
H = ConstructHaarWaveletTransformationMatrix(M);
fi = conj(dftmtx(M))/M;
H = fi*H;
H = H(4,:);
H = diag(H);
etc
How do I repeat this for Gaussian wavelets? Is there a built in Matlab function which will do this for me?
For reference I'm implementing the algorithm in section 4 of this paper:
http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=04218361
I maybe would not being answering the question, but i will try to help you advance.
As far as i know, the Matlab Wavelet Toolbox only deal with wavelet operations and coefficients, increase or decrease resolution levels, and similar operations, but do not exposes the internal matrices serving to doing the transformations from signals and coefficients.
Hence i fear the answer to this question is no. Some time ago, i did this for some of the Hart Class wavelet, and i actually build the matrix from the scratch, and then i compared the coefficients obtained with the Built-in Matlab Wavelet Toolbox, hence ensuring your matrices are good enough for your algorithm. In my case, recursive parameter estimation for time varying models.
For the function ConstructHaarWaveletTransformationMatrix it is really simple to create the matrix, because the Hart Class could be really simple expressed as Kronecker products.
The Gaussian Wavelet case as i fear should be done from the scratch too...
THe steps i suggest would be;
Although MATLAB dont include explicitely the matrices, you can use the Matlab built-in functions to recover the Gaussian Wavelets, and thus compose the matrix for your algorithm.
Build every column of the matrix with every Gaussian Wavelet, for every resolution levels you are requiring (the dyadic scales). Use the Matlab Wavelets toolbox for recover the shapes.
After this, compare the coefficients obtained by you, with the coefficients of the toolbox. This way you will correct the order of the Matrix row.
Numerically, being fj the signal projection over Vj (the PHI signals space, scaling functions) at resolution level j, and gj the signal projection over Wj (the PSI signals space, mother functions) at resolution level j, we can write:
f=fj0+sum_{j0}^{j1-1}{gj}
Hence, both fj0 and gj will induce two matrices, lets call them PHIj and PSIj matrices:
f=PHIj0*cj0+sum_{j0}^{j1-1}{PSIj*dj}
The PHIj columns contain the scaled and shifted scaling wavelet signal (one, for j0 only) for the approximation projection (the Vj0 space), and the PSIj columns contain the scaled and shifted mother wavelet signals (several, from j0 to j1-1) for the detail projection (onto the Wj0 to Wj1-1 spaces).
Hence, the Matrix you need is:
PHI=[PHIj0 PSIj0... PSIj1]
Thus you can express you original signal as:
f=PHI*C
where C is a vector of approximation and detail coefficients, for the levels:
C=[cj0' dj0'...dj1']'
The first part, for addressing the PHI build can be achieved by writing:
function PHI=MakePhi(l,str,Jmin,Jmax)
% [PHI]=MakePhi(l,str,Jmin,Jmax)
%
% Build full PHI Wavelet Matrix for obtaining wavelet coefficients
% (extract)
%FILTER
[LO_R,HI_R] = wfilters(str,'r');
lf=length(LO_R);
%PHI BUILD
PHI=[];
laux=l([end-Jmax end-Jmax:end]);
PHI=[PHI MakeWMatrix('a',str,laux)];
for j=Jmax:-1:Jmin
laux=l([end-j end-j:end]);
PHI=[PHI MakeWMatrix('d',str,laux)];
end
the wfilters is a MATLAB built in function, giving the required signal for the approximation and or detail wavelet signals.
The MakeWMatrix function is:
function M=MakeWMatrix(typestr,str,laux)
% M=MakeWMatrix(typestr,str,laux)
%
% Build Wavelet Matrix for obtaining wavelet coefficients
% for a single level vector.
% (extract)
[LO_R,HI_R] = wfilters(str,'r');
if typestr=='a'
F_R=LO_R';
else
F_R=HI_R';
end
la=length(laux);
lin=laux(2); lout=laux(3);
M=MakeCMatrix(F_R,lin,lout);
for i=3:la-1
lin=laux(i); lout=laux(i+1);
Mi=MakeCMatrix(LO_R',lin,lout);
M=Mi*M;
end
and finally the MakeCMatrix is:
function [M]=MakeCMatrix(F_R,lin,lout)
% Convolucion Matrix
% (extract)
lf=length(F_R);
M=[];
for i=1:lin
M(:,i)=[zeros(2*(i-1),1) ;F_R ;zeros(2*(lin-i),1)];
end
M=[zeros(1,lin); M ;zeros(1,lin)];
[ltot,lin]=size(M);
lmin=floor((ltot-lout)/2)+1;
lmax=floor((ltot-lout)/2)+lout;
M=M(lmin:lmax,:);
This last matrix should include some interpolation routine for having better general results in each case.
I expect this solve part of your problem.....
Hyp
There are two alternative methods to compute DCT and its inverse in MATLAB. One is dct2/idct2 and the other is the transformation matrix computed by dctmtx. Why is there an alternative way based on matrix multiplications making use of dctmtx?
"If A is square, the two-dimensional DCT of A can be computed as D*A*D'. This computation is sometimes faster than using dct2, especially if you are computing a large number of small DCTs, because D needs to be determined only once."
Where D = dctmtx(n)
Source: http://www.mathworks.com/help/toolbox/images/ref/dctmtx.html