I need to calculate the maximum of normalized cross correlation of million of particles. The size of the two parameters of normxcorr2 is 56*56. I can't parallelize the calculations. Is there any suggestion to speed up the code especially that I don't need all the results but only the maximum value of each cross correlation (to know the displacement)?
Example of the algorithm
%The choice of 170 particles is because in each time
%the code detects 170 particles, so over 10000 images it's 1 700 000 particles
particle_1=rand(54,54,170);
particle_2=rand(56,56,170);
for i=1:170
C=normxcorr2(particle_1(:,:,i),particle_2(:,:,i));
L(i)=max(C(:));
end
I don't have MATLAB so I ran the following code on this site: https://www.tutorialspoint.com/execute_matlab_online.php which is actually octave. So I implemented "naive" normalized cross correlation and indeed for these small images sizes the naive performs better:
Elapsed time is 2.62645 seconds - for normxcorr2
Elapsed time is 0.199034 seconds - for my naive_normxcorr2
The code is based on the article http://scribblethink.org/Work/nvisionInterface/nip.pdf which describes how to calculate the standard deviation needed for the normalization in an efficient way using integral image, this is the box_corr function.
Also, MATLAB's normxcorr2 returns a padded image so I took the max on the unpadded part.
pkg load image
function [N] = naive_corr(pat,img)
[n,m] = size(img);
[np,mp] = size(pat);
N = zeros(n-np+1,m-mp+1);
for i = 1:n-np+1
for j = 1:m-mp+1
N(i,j) = sum(dot(pat,img(i:i+np-1,j:j+mp-1)));
end
end
end
%w_arr the array of coefficients for the boxes
%box_arr of size [k,4] where k is the number boxes, each box represented by
%4 something ...
function [C] = box_corr2(img,box_arr,w_arr,n_p,m_p)
% construct integral image + zeros pad (for boundary problems)
I = cumsum(cumsum(img,2),1);
I = [zeros(1,size(I,2)+2); [zeros(size(I,1),1) I zeros(size(I,1),1)]; zeros(1,size(I,2)+2)];
% initialize result matrix
[n,m] = size(img);
C = zeros(n-n_p+1,m-m_p+1);
%C = zeros(n,m);
jump_x = 1;
jump_y = 1;
x_start = ceil(n_p/2);
x_end = n-x_start+mod(n_p,2);
x_span = x_start:jump_x:x_end;
y_start = ceil(m_p/2);
y_end = m-y_start+mod(m_p,2);
y_span = y_start:jump_y:y_end;
arr_a = box_arr(:,1) - x_start;
arr_b = box_arr(:,2) - x_start+1;
arr_c = box_arr(:,3) - y_start;
arr_d = box_arr(:,4) - y_start+1;
% cumulate box responses
k = size(box_arr,1); % == numel(w_arr)
for i = 1:k
a = arr_a(i);
b = arr_b(i);
c = arr_c(i);
d = arr_d(i);
C = C ...
+ w_arr(i) * ( I(x_span+b,y_span+d) ...
- I(x_span+b,y_span+c) ...
- I(x_span+a,y_span+d) ...
+ I(x_span+a,y_span+c) );
end
end
function [NCC] = naive_normxcorr2(temp,img)
[n_p,m_p]=size(temp);
M = n_p*m_p;
% compute template mean & std
temp_mean = mean(temp(:));
temp = temp - temp_mean;
temp_std = sqrt(sum(temp(:).^2)/M);
% compute windows' mean & std
wins_mean = box_corr2(img,[1,n_p,1,m_p],1/M, n_p,m_p);
wins_mean2 = box_corr2(img.^2,[1,n_p,1,m_p],1/M,n_p,m_p);
wins_std = real(sqrt(wins_mean2 - wins_mean.^2));
NCC_naive = naive_corr(temp,img);
NCC = NCC_naive ./ (M .* temp_std .* wins_std);
end
n = 170;
particle_1=rand(54,54,n);
particle_2=rand(56,56,n);
[n_p1,m_p1,c_p1]=size(particle_1);
[n_p2,m_p2,c_p2]=size(particle_2);
L1 = zeros(n,1);
L2 = zeros (n,1);
tic
for i=1:n
C1=normxcorr2(particle_1(:,:,i),particle_2(:,:,i));
C1_unpadded = C1(n_p1:n_p2 , m_p1:m_p2);
L1(i)=max(C1_unpadded(:));
end
toc
tic
for i=1:n
C2=naive_normxcorr2(particle_1(:,:,i),particle_2(:,:,i));
L2(i)=max(C2(:));
end
toc
I have a numeric dataset and I want to cluster data with a non-parametric algorithm. Basically, I would like to cluster without specifying the number of clusters for the input. I am using this code that I accessed through the MathWorks File Exchange network which implements the Mean Shift algorithm. However, I don't Know how to adapt my data to this code as my dataset has dimensions 516 x 19.
function [clustCent,data2cluster,cluster2dataCell] =MeanShiftCluster(dataPts,bandWidth,plotFlag)
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
%perform MeanShift Clustering of data using a flat kernel
%
% ---INPUT---
% dataPts - input data, (numDim x numPts)
% bandWidth - is bandwidth parameter (scalar)
% plotFlag - display output if 2 or 3 D (logical)
% ---OUTPUT---
% clustCent - is locations of cluster centers (numDim x numClust)
% data2cluster - for every data point which cluster it belongs to (numPts)
% cluster2dataCell - for every cluster which points are in it (numClust)
%
% Bryan Feldman 02/24/06
% MeanShift first appears in
% K. Funkunaga and L.D. Hosteler, "The Estimation of the Gradient of a
% Density Function, with Applications in Pattern Recognition"
%*** Check input ****
if nargin < 2
error('no bandwidth specified')
end
if nargin < 3
plotFlag = true;
plotFlag = false;
end
%**** Initialize stuff ***
%[numPts,numDim] = size(dataPts);
[numDim,numPts] = size(dataPts);
numClust = 0;
bandSq = bandWidth^2;
initPtInds = 1:numPts
maxPos = max(dataPts,[],2); %biggest size in each dimension
minPos = min(dataPts,[],2); %smallest size in each dimension
boundBox = maxPos-minPos; %bounding box size
sizeSpace = norm(boundBox); %indicator of size of data space
stopThresh = 1e-3*bandWidth; %when mean has converged
clustCent = []; %center of clust
beenVisitedFlag = zeros(1,numPts,'uint8'); %track if a points been seen already
numInitPts = numPts %number of points to posibaly use as initilization points
clusterVotes = zeros(1,numPts,'uint16'); %used to resolve conflicts on cluster membership
while numInitPts
tempInd = ceil( (numInitPts-1e-6)*rand) %pick a random seed point
stInd = initPtInds(tempInd) %use this point as start of mean
myMean = dataPts(:,stInd); % intilize mean to this points location
myMembers = []; % points that will get added to this cluster
thisClusterVotes = zeros(1,numPts,'uint16'); %used to resolve conflicts on cluster membership
while 1 %loop untill convergence
sqDistToAll = sum((repmat(myMean,1,numPts) - dataPts).^2); %dist squared from mean to all points still active
inInds = find(sqDistToAll < bandSq); %points within bandWidth
thisClusterVotes(inInds) = thisClusterVotes(inInds)+1; %add a vote for all the in points belonging to this cluster
myOldMean = myMean; %save the old mean
myMean = mean(dataPts(:,inInds),2); %compute the new mean
myMembers = [myMembers inInds]; %add any point within bandWidth to the cluster
beenVisitedFlag(myMembers) = 1; %mark that these points have been visited
%*** plot stuff ****
if plotFlag
figure(12345),clf,hold on
if numDim == 2
plot(dataPts(1,:),dataPts(2,:),'.')
plot(dataPts(1,myMembers),dataPts(2,myMembers),'ys')
plot(myMean(1),myMean(2),'go')
plot(myOldMean(1),myOldMean(2),'rd')
pause
end
end
%**** if mean doesnt move much stop this cluster ***
if norm(myMean-myOldMean) < stopThresh
%check for merge posibilities
mergeWith = 0;
for cN = 1:numClust
distToOther = norm(myMean-clustCent(:,cN)); %distance from posible new clust max to old clust max
if distToOther < bandWidth/2 %if its within bandwidth/2 merge new and old
mergeWith = cN;
break;
end
end
if mergeWith > 0 % something to merge
clustCent(:,mergeWith) = 0.5*(myMean+clustCent(:,mergeWith)); %record the max as the mean of the two merged (I know biased twoards new ones)
%clustMembsCell{mergeWith} = unique([clustMembsCell{mergeWith} myMembers]); %record which points inside
clusterVotes(mergeWith,:) = clusterVotes(mergeWith,:) + thisClusterVotes; %add these votes to the merged cluster
else %its a new cluster
numClust = numClust+1 %increment clusters
clustCent(:,numClust) = myMean; %record the mean
%clustMembsCell{numClust} = myMembers; %store my members
clusterVotes(numClust,:) = thisClusterVotes;
end
break;
end
end
initPtInds = find(beenVisitedFlag == 0); %we can initialize with any of the points not yet visited
numInitPts = length(initPtInds); %number of active points in set
end
[val,data2cluster] = max(clusterVotes,[],1); %a point belongs to the cluster with the most votes
%*** If they want the cluster2data cell find it for them
if nargout > 2
cluster2dataCell = cell(numClust,1);
for cN = 1:numClust
myMembers = find(data2cluster == cN);
cluster2dataCell{cN} = myMembers;
end
end
This is the test code I am using to try and get the Mean Shift program to work:
clear
profile on
nPtsPerClust = 250;
nClust = 3;
totalNumPts = nPtsPerClust*nClust;
m(:,1) = [1 1];
m(:,2) = [-1 -1];
m(:,3) = [1 -1];
var = .6;
bandwidth = .75;
clustMed = [];
%clustCent;
x = var*randn(2,nPtsPerClust*nClust);
%*** build the point set
for i = 1:nClust
x(:,1+(i-1)*nPtsPerClust:(i)*nPtsPerClust) = x(:,1+(i-1)*nPtsPerClust:(i)*nPtsPerClust) + repmat(m(:,i),1,nPtsPerClust);
end
tic
[clustCent,point2cluster,clustMembsCell] = MeanShiftCluster(x,bandwidth);
toc
numClust = length(clustMembsCell)
figure(10),clf,hold on
cVec = 'bgrcmykbgrcmykbgrcmykbgrcmyk';%, cVec = [cVec cVec];
for k = 1:min(numClust,length(cVec))
myMembers = clustMembsCell{k};
myClustCen = clustCent(:,k);
plot(x(1,myMembers),x(2,myMembers),[cVec(k) '.'])
plot(myClustCen(1),myClustCen(2),'o','MarkerEdgeColor','k','MarkerFaceColor',cVec(k), 'MarkerSize',10)
end
title(['no shifting, numClust:' int2str(numClust)])
The test script generates random data X. In my case. I want to use the matrix D of size 516 x 19 but I am not sure how to adapt my data to this function. The function is returning results that are not agreeing with my understanding of the algorithm.
Does anyone know how to do this?
I've found myself needing to do a least-squares (or similar matrix-based operation) for every pixel in an image. Every pixel has a set of numbers associated with it, and so it can be arranged as a 3D matrix.
(This next bit can be skipped)
Quick explanation of what I mean by least-squares estimation :
Let's say we have some quadratic system that is modeled by Y = Ax^2 + Bx + C and we're looking for those A,B,C coefficients. With a few samples (at least 3) of X and the corresponding Y, we can estimate them by:
Arrange the (lets say 10) X samples into a matrix like X = [x(:).^2 x(:) ones(10,1)];
Arrange the Y samples into a similar matrix: Y = y(:);
Estimate the coefficients A,B,C by solving: coeffs = (X'*X)^(-1)*X'*Y;
Try this on your own if you want:
A = 5; B = 2; C = 1;
x = 1:10;
y = A*x(:).^2 + B*x(:) + C + .25*randn(10,1); % added some noise here
X = [x(:).^2 x(:) ones(10,1)];
Y = y(:);
coeffs = (X'*X)^-1*X'*Y
coeffs =
5.0040
1.9818
0.9241
START PAYING ATTENTION AGAIN IF I LOST YOU THERE
*MAJOR REWRITE*I've modified to bring it as close to the real problem that I have and still make it a minimum working example.
Problem Setup
%// Setup
xdim = 500;
ydim = 500;
ncoils = 8;
nshots = 4;
%// matrix size for each pixel is ncoils x nshots (an overdetermined system)
%// each pixel has a matrix stored in the 3rd and 4rth dimensions
regressor = randn(xdim,ydim, ncoils,nshots);
regressand = randn(xdim, ydim,ncoils);
So my problem is that I have to do a (X'*X)^-1*X'*Y (least-squares or similar) operation for every pixel in an image. While that itself is vectorized/matrixized the only way that I have to do it for every pixel is in a for loop, like:
Original code style
%// Actual work
tic
estimate = zeros(xdim,ydim);
for col=1:size(regressor,2)
for row=1:size(regressor,1)
X = squeeze(regressor(row,col,:,:));
Y = squeeze(regressand(row,col,:));
B = X\Y;
% B = (X'*X)^(-1)*X'*Y; %// equivalently
estimate(row,col) = B(1);
end
end
toc
Elapsed time = 27.6 seconds
EDITS in reponse to comments and other ideas
I tried some things:
1. Reshaped into a long vector and removed the double for loop. This saved some time.
2. Removed the squeeze (and in-line transposing) by permute-ing the picture before hand: This save alot more time.
Current example:
%// Actual work
tic
estimate2 = zeros(xdim*ydim,1);
regressor_mod = permute(regressor,[3 4 1 2]);
regressor_mod = reshape(regressor_mod,[ncoils,nshots,xdim*ydim]);
regressand_mod = permute(regressand,[3 1 2]);
regressand_mod = reshape(regressand_mod,[ncoils,xdim*ydim]);
for ind=1:size(regressor_mod,3) % for every pixel
X = regressor_mod(:,:,ind);
Y = regressand_mod(:,ind);
B = X\Y;
estimate2(ind) = B(1);
end
estimate2 = reshape(estimate2,[xdim,ydim]);
toc
Elapsed time = 2.30 seconds (avg of 10)
isequal(estimate2,estimate) == 1;
Rody Oldenhuis's way
N = xdim*ydim*ncoils; %// number of columns
M = xdim*ydim*nshots; %// number of rows
ii = repmat(reshape(1:N,[ncoils,xdim*ydim]),[nshots 1]); %//column indicies
jj = repmat(1:M,[ncoils 1]); %//row indicies
X = sparse(ii(:),jj(:),regressor_mod(:));
Y = regressand_mod(:);
B = X\Y;
B = reshape(B(1:nshots:end),[xdim ydim]);
Elapsed time = 2.26 seconds (avg of 10)
or 2.18 seconds (if you don't include the definition of N,M,ii,jj)
SO THE QUESTION IS:
Is there an (even) faster way?
(I don't think so.)
You can achieve a ~factor of 2 speed up by precomputing the transposition of X. i.e.
for x=1:size(picture,2) % second dimension b/c already transposed
X = picture(:,x);
XX = X';
Y = randn(n_timepoints,1);
%B = (X'*X)^-1*X'*Y; ;
B = (XX*X)^-1*XX*Y;
est(x) = B(1);
end
Before: Elapsed time is 2.520944 seconds.
After: Elapsed time is 1.134081 seconds.
EDIT:
Your code, as it stands in your latest edit, can be replaced by the following
tic
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
% Actual work
picture = randn(xdim,ydim,n_timepoints);
picture = reshape(picture, [xdim*ydim,n_timepoints])'; % note transpose
YR = randn(n_timepoints,size(picture,2));
% (XX*X).^-1 = sum(picture.*picture).^-1;
% XX*Y = sum(picture.*YR);
est = sum(picture.*picture).^-1 .* sum(picture.*YR);
est = reshape(est,[xdim,ydim]);
toc
Elapsed time is 0.127014 seconds.
This is an order of magnitude speed up on the latest edit, and the results are all but identical to the previous method.
EDIT2:
Okay, so if X is a matrix, not a vector, things are a little more complicated. We basically want to precompute as much as possible outside of the for-loop to keep our costs down. We can also get a significant speed-up by computing XT*X manually - since the result will always be a symmetric matrix, we can cut a few corners to speed things up. First, the symmetric multiplication function:
function XTX = sym_mult(X) % X is a 3-d matrix
n = size(X,2);
XTX = zeros(n,n,size(X,3));
for i=1:n
for j=i:n
XTX(i,j,:) = sum(X(:,i,:).*X(:,j,:));
if i~=j
XTX(j,i,:) = XTX(i,j,:);
end
end
end
Now the actual computation script
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
Y = randn(10,xdim*ydim);
picture = randn(xdim,ydim,n_timepoints); % 500x500x10
% Actual work
tic % start timing
picture = reshape(picture, [xdim*ydim,n_timepoints])';
% Here we precompute the (XT*Y) calculation to speed things up later
picture_y = [sum(Y);sum(Y.*picture)];
% initialize
est = zeros(size(picture,2),1);
picture = permute(picture,[1,3,2]);
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture);
XTX = sym_mult(XTX); % precompute (XT*X) for speed
X = zeros(2,2); % preallocate for speed
XY = zeros(2,1);
for x=1:size(picture,2) % second dimension b/c already transposed
%For some reason this is a lot faster than X = XTX(:,:,x);
X(1,1) = XTX(1,1,x);
X(2,1) = XTX(2,1,x);
X(1,2) = XTX(1,2,x);
X(2,2) = XTX(2,2,x);
XY(1) = picture_y(1,x);
XY(2) = picture_y(2,x);
% Here we utilise the fact that A\B is faster than inv(A)*B
% We also use the fact that (A*B)*C = A*(B*C) to speed things up
B = X\XY;
est(x) = B(1);
end
est = reshape(est,[xdim,ydim]);
toc % end timing
Before: Elapsed time is 4.56 seconds.
After: Elapsed time is 2.24 seconds.
This is a speed up of about a factor of 2. This code should be extensible to X being any dimensions you want. For instance, in the case where X = [1 x x^2], you would change picture_y to the following
picture_y = [sum(Y);sum(Y.*picture);sum(Y.*picture.^2)];
and change XTX to
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture,picture.^2);
You would also change a lot of 2s to 3s in the code, and add XY(3) = picture_y(3,x) to the loop. It should be fairly straight-forward, I believe.
Results
I sped up your original version, since your edit 3 was actually not working (and also does something different).
So, on my PC:
Your (original) version: 8.428473 seconds.
My obfuscated one-liner given below: 0.964589 seconds.
First, for no other reason than to impress, I'll give it as I wrote it:
%%// Some example data
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
estimate = zeros(xdim,ydim); %// initialization with explicit size
picture = randn(xdim,ydim,n_timepoints);
%%// Your original solution
%// (slightly altered to make my version's results agree with yours)
tic
Y = randn(n_timepoints,xdim*ydim);
ii = 1;
for x = 1:xdim
for y = 1:ydim
X = squeeze(picture(x,y,:)); %// or similar creation of X matrix
B = (X'*X)^(-1)*X' * Y(:,ii);
ii = ii+1;
%// sometimes you keep everything and do
%// estimate(x,y,:) = B(:);
%// sometimes just the first element is important and you do
estimate(x,y) = B(1);
end
end
toc
%%// My version
tic
%// UNLEASH THE FURY!!
estimate2 = reshape(sparse(1:xdim*ydim*n_timepoints, ...
builtin('_paren', ones(n_timepoints,1)*(1:xdim*ydim),:), ...
builtin('_paren', permute(picture, [3 2 1]),:))\Y(:), ydim,xdim).'; %'
toc
%%// Check for equality
max(abs(estimate(:)-estimate2(:))) % (always less than ~1e-14)
Breakdown
First, here's the version that you should actually use:
%// Construct sparse block-diagonal matrix
%// (Type "help sparse" for more information)
N = xdim*ydim; %// number of columns
M = N*n_timepoints; %// number of rows
ii = 1:N;
jj = ones(n_timepoints,1)*(1:N);
s = permute(picture, [3 2 1]);
X = sparse(ii,jj(:), s(:));
%// Compute ALL the estimates at once
estimates = X\Y(:);
%// You loop through the *second* dimension first, so to make everything
%// agree, we have to extract elements in the "wrong" order, and transpose:
estimate2 = reshape(estimates, ydim,xdim).'; %'
Here's an example of what picture and the corresponding matrix X looks like for xdim = ydim = n_timepoints = 2:
>> clc, picture, full(X)
picture(:,:,1) =
-0.5643 -2.0504
-0.1656 0.4497
picture(:,:,2) =
0.6397 0.7782
0.5830 -0.3138
ans =
-0.5643 0 0 0
0.6397 0 0 0
0 -2.0504 0 0
0 0.7782 0 0
0 0 -0.1656 0
0 0 0.5830 0
0 0 0 0.4497
0 0 0 -0.3138
You can see why sparse is necessary -- it's mostly zeros, but will grow large quickly. The full matrix would quickly consume all your RAM, while the sparse one will not consume much more than the original picture matrix does.
With this matrix X, the new problem
X·b = Y
now contains all the problems
X1 · b1 = Y1
X2 · b2 = Y2
...
where
b = [b1; b2; b3; ...]
Y = [Y1; Y2; Y3; ...]
so, the single command
X\Y
will solve all your systems at once.
This offloads all the hard work to a set of highly specialized, compiled to machine-specific code, optimized-in-every-way algorithms, rather than the interpreted, generic, always-two-steps-away from the hardware loops in MATLAB.
It should be straightforward to convert this to a version where X is a matrix; you'll end up with something like what blkdiag does, which can also be used by mldivide in exactly the same way as above.
I had a wee play around with an idea, and I decided to stick it as a separate answer, as its a completely different approach to my other idea, and I don't actually condone what I'm about to do. I think this is the fastest approach so far:
Orignal (unoptimised): 13.507176 seconds.
Fast Cholesky-decomposition method: 0.424464 seconds
First, we've got a function to quickly do the X'*X multiplication. We can speed things up here because the result will always be symmetric.
function XX = sym_mult(X)
n = size(X,2);
XX = zeros(n,n,size(X,3));
for i=1:n
for j=i:n
XX(i,j,:) = sum(X(:,i,:).*X(:,j,:));
if i~=j
XX(j,i,:) = XX(i,j,:);
end
end
end
The we have a function to do LDL Cholesky decomposition of a 3D matrix (we can do this because the (X'*X) matrix will always be symmetric) and then do forward and backwards substitution to solve the LDL inversion equation
function Y = fast_chol(X,XY)
n=size(X,2);
L = zeros(n,n,size(X,3));
D = zeros(n,n,size(X,3));
B = zeros(n,1,size(X,3));
Y = zeros(n,1,size(X,3));
% These loops compute the LDL decomposition of the 3D matrix
for i=1:n
D(i,i,:) = X(i,i,:);
L(i,i,:) = 1;
for j=1:i-1
L(i,j,:) = X(i,j,:);
for k=1:(j-1)
L(i,j,:) = L(i,j,:) - L(i,k,:).*L(j,k,:).*D(k,k,:);
end
D(i,j,:) = L(i,j,:);
L(i,j,:) = L(i,j,:)./D(j,j,:);
if i~=j
D(i,i,:) = D(i,i,:) - L(i,j,:).^2.*D(j,j,:);
end
end
end
for i=1:n
B(i,1,:) = XY(i,:);
for j=1:(i-1)
B(i,1,:) = B(i,1,:)-D(i,j,:).*B(j,1,:);
end
B(i,1,:) = B(i,1,:)./D(i,i,:);
end
for i=n:-1:1
Y(i,1,:) = B(i,1,:);
for j=n:-1:(i+1)
Y(i,1,:) = Y(i,1,:)-L(j,i,:).*Y(j,1,:);
end
end
Finally, we have the main script which calls all of this
xdim = 500;
ydim = 500;
n_timepoints = 10; % for example
Y = randn(10,xdim*ydim);
picture = randn(xdim,ydim,n_timepoints); % 500x500x10
tic % start timing
picture = reshape(pr, [xdim*ydim,n_timepoints])';
% Here we precompute the (XT*Y) calculation
picture_y = [sum(Y);sum(Y.*picture)];
% initialize
est2 = zeros(size(picture,2),1);
picture = permute(picture,[1,3,2]);
% Now we calculate the X'*X matrix
XTX = cat(2,ones(n_timepoints,1,size(picture,3)),picture);
XTX = sym_mult(XTX);
% Call our fast Cholesky decomposition routine
B = fast_chol(XTX,picture_y);
est2 = B(1,:);
est2 = reshape(est2,[xdim,ydim]);
toc
Again, this should work equally well for a Nx3 X matrix, or however big you want.
I use octave, thus I can't say anything about the resulting performance in Matlab, but would expect this code to be slightly faster:
pictureT=picture'
est=arrayfun(#(x)( (pictureT(x,:)*picture(:,x))^-1*pictureT(x,:)*randn(n_ti
mepoints,1)),1:size(picture,2));