Using matrix structure to speed up matlab - performance

Suppose that I have an N-by-K matrix A, N-by-P matrix B. I want to do the following calculations to get my final N-by-P matrix X.
X(n,p) = B(n,p) - dot(gamma(p,:),A(n,:))
where
gamma(p,k) = dot(A(:,k),B(:,p))/sum( A(:,k).^2 )
In MATLAB, I have my code like
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
which are highly inefficient since it uses three for loops! Is there a good way to speed up this code?

Use bsxfun for the division and matrix multiplication for the loops:
gamma = bsxfun(#rdivide, B.'*A, sum(A.^2));
x = B - A*gamma.';
And here is a test script
N = 3;
K = 4;
P = 5;
A = rand(N, K);
B = rand(N, P);
for p = 1:P
for n = 1:N
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
gamma2 = bsxfun(#rdivide, B.'*A, sum(A.^2));
X2 = B - A*gamma2.';
isequal(x, X2)
isequal(gamma, gamma2)
which returns
ans =
1
ans =
1

It looks to me like you can hoist the gamma calculations out of the loop; at least, I don't see any dependencies on N in the gamma calculations.
So something like this:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
end
for p = 1:P
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end
I'm not familiar enough with your code (or matlab) to really know if you can merge the two loops, but if you can:
for p = 1:P
for k = 1:K
gamma(p,k) = dot(A(:,k),B(:,p))/sum(A(:,k).^2);
end
for n = 1:N
x(n,p) = B(n,p) - dot(gamma(p,:),A(n,:));
end
end

bxfun is slow...
How about something like the following (I might have a transpose wrong)
modA = A * (1./sum(A.^2,2)) * ones(1,k);
gamma = B' * modA;
x = B - A * gamma';

Related

The levenberg-marquardt method for solving non-linear equations

I tried implement the levenberg-marquardt method for solving non-linear equations on Julia based on Numerical Optimization using the
Levenberg-Marquardt Algorithm presentation. This my code:
function get_J(ArrOfFunc,X,delta)
N = length(ArrOfFunc)
J = zeros(Float64,N,N)
for i = 1:N
for j=1:N
Temp = copy(X);
Temp[j]=Temp[j]+delta;
J[i,j] = (ArrOfFunc[i](Temp)-ArrOfFunc[i](X))/delta;
end
end
return J
end
function get_resudial(ArrOfFunc,Arg)
return map((x)->x(Arg),ArrOfFunc)
end
function lm_solve(Funcs,Init)
X = copy(Init)
delta = 0.01;
Lambda = 0.01;
Factor = 2;
J = get_J(Funcs,X,delta)
R = get_resudial(Funcs,X)
N = 5
for t = 1:N
G = J'*J+Lambda.*eye(length(X))
dC = J'*R
C = sum(R.*R)/2;
Xnew = X-(inv(G)\dC);
Rnew = get_resudial(Funcs,Xnew)
Cnew = sum(Rnew.*Rnew)/2;
if ( Cnew < C)
X = Xnew;
R = Rnew;
Lambda = Lambda/Factor;
J = get_J(Funcs,X,delta)
else
Lambda = Lambda*Factor;
end
if(maximum(abs(Rnew)) < 0.001)
return X
end
end
return X
end
function test()
ArrOfFunc = [
(X)->X[1]+X[2]-2;
(X)->X[1]-X[2]
];
X = lm_solve(ArrOfFunc,Float64[3;3])
println(X)
return X
end
But from any starting point the step not accepted. What's I doing wrong?
Any help would be appreciated.
I have at the moment no way to test this, but one line does not make sense mathematically:
In the computation of Xnew it should be either inv(G)*dC or G\dC, but not a mix of both. Preferably the second, since the solution of a linear system does not require the computation of the inverse matrix.
With this one wrong calculation at the center of the iteration, the trajectory of the computation is almost surely going astray.

Vectorized code slower than loops? MATLAB

In the problem Im working on there is such a part of code, as shown below. The definition part is just to show you the sizes of arrays. Below I pasted vectorized version - and it is >2x slower. Why it happens so? I know that i happens if vectorization requiers large temporary variables, but (it seems) it is not true here.
And generally, what (other than parfor, with I already use) can I do to speed up this code?
maxN = 100;
levels = maxN+1;
xElements = 101;
umn = complex(zeros(levels, levels));
umn2 = umn;
bessels = ones(xElements, xElements, levels); % 1.09 GB
posMcontainer = ones(xElements, xElements, maxN);
tic
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
nn = n + 1;
mm = 1;
for m = 1 : 2 : n
umn(nn, mm) = bessels(i, j, nn) * posMcontainer(i, j, m);
mm = mm + 1;
end
end
end
end
toc % 0.520594 seconds
tic
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
umn2(nn, 1:numOfEl) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
end
end
toc % 1.275926 seconds
sum(sum(umn-umn2)) % veryfying, if all done right
Best regards,
Alex
From the profiler:
Edit:
In reply to #Jason answer, this alternative takes the same time:
for n = 1:2:maxN
nn(n) = n + 1;
numOfEl(n) = ceil(n/2);
end
for j = 1 : xElements
for i = 1 : xElements
for n = 1 : 2 : maxN
umn2(nn(n), 1:numOfEl(n)) = bessels(i, j, nn(n)) * posMcontainer(i, j, 1:2:n);
end
end
end
Edit2:
In reply to #EBH :
The point is to do the following:
parfor i = 1 : xElements
for j = 1 : xElements
umn = complex(zeros(levels, levels)); % cleaning
for n = 0:maxN
mm = 1;
for m = -n:2:n
nn = n + 1; % for indexing
if m < 0
umn(nn, mm) = bessels(i, j, nn) * negMcontainer(i, j, abs(m));
end
if m > 0
umn(nn, mm) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
if m == 0
umn(nn, mm) = bessels(i, j, nn);
end
mm = mm + 1; % for indexing
end % m
end % n
beta1 = sum(sum(Aj1.*umn));
betaSumSq1(i, j) = abs(beta1).^2;
beta2 = sum(sum(Aj2.*umn));
betaSumSq2(i, j) = abs(beta2).^2;
end % j
end % i
I speeded it up as much, as I was able to. What you have written is taking only the last bessels and posMcontainer values, so it does not produce the same result. In the real code, those two containers are filled not with 1, but with some precalculated values.
After your edit, I can see that umn is just a temporary variable for another calculation. It still can be mostly vectorizable:
betaSumSq1 = zeros(xElements); % preallocating
betaSumSq2 = zeros(xElements); % preallocating
% an index matrix to fetch the right values from negMcontainer and
% posMcontainer:
indmat = tril(repmat([0 1;1 0],ceil((maxN+1)/2),floor(levels/2)));
indmat(end,:) = [];
% an index matrix to fetch the values in correct order for umn:
b_ind = repmat([1;0],ceil((maxN+1)/2),1);
b_ind(end) = [];
tempind = logical([fliplr(indmat) b_ind indmat+triu(ones(size(indmat)))]);
% permute the arrays to prevent squeeze:
PM = permute(posMcontainer,[3 1 2]);
NM = permute(negMcontainer,[3 1 2]);
B = permute(bessels,[3 1 2]);
for k = 1 : maxN+1 % third dim
for jj = 1 : xElements % columns
b = B(:,jj,k); % get one vector of B
% perform b*NM for every row of NM*indmat, than flip the result:
neg = fliplr(bsxfun(#times,bsxfun(#times,indmat,NM(:,jj,k).'),b));
% perform b*PM for every row of PM*indmat:
pos = bsxfun(#times,bsxfun(#times,indmat,PM(:,jj,k).'),b);
temp = [neg mod(1:levels,2).'.*b pos].'; % concat neg and pos
% assign them to the right place in umn:
umn = reshape(temp(tempind.'),[levels levels]).';
beta1 = Aj1.*umn;
betaSumSq1(jj,k) = abs(sum(beta1(:))).^2;
beta2 = Aj2.*umn;
betaSumSq2(jj,k) = abs(sum(beta2(:))).^2;
end
end
This reduce running time from ~95 seconds to less 3 seconds (both without parfor), so it improves in almost 97%.
I would suspect it is memory allocation. You are re-allocating the m array in a 3 deep loop.
try rearranging the code:
tic
for n = 1 : 2 : maxN
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
for j = 1 : xElements
for i = 1 : xElements
umn2(nn, 1:numOfEl) = bessels(i, j, nn) * posMcontainer(i, j, m);
end
end
end
toc % 1.275926 seconds
I was trying this in Igor pro, which a similar language, but with different optimizations. So the direct translations don't time the same way as Matlab (vectorized was slightly faster in Igor). But reordering the loops did speed up the vectorized form.
In your second part of the code, that is setting umn2, inside the loops, you have:
nn = n + 1;
m = 1:2:n;
numOfEl = ceil(n/2);
Those 3 lines don't require any input from the i and j loops, they only use the n loop. So reordering the loops such that i and j are inside the n loop will mean that those 3 lines are done xElements^2 (100^2) times less often. I suspect it is that m = 1:2:n line that takes time, since that is allocating an array.

Low-Rank Approximation using Frobenious Norm Tolerance

I would love some help with an algorithm I've working on. My specific problem is in the while loop of the algorithm.
What I want the algorithm to do is to sequentially add a singular value in Sk(until the desired TOL is achieved) in the diagonal while making the rest of the matrix 0. The desired goal is to do this to greyscale images.
I can only get 2x2 cycles.
here is the code:
A = rand(5)*30;
[M N] = size(A);
[U S V] = svds(A);
% r = rank(A);
Sn = S(1,1);
Sk = blkdiag(Sn,zeros(N-1));
Ak = U*Sk*V;
X = A - Ak;
F = norm(X,'fro');
tol = 10;
i = 2;
while F > tol
Snk = S(i,i);
Snew = diag([Sn;Snk]);
Sk = blkdiag(Snew,zeros(N-2));
Ak = U*Sk*V';
F = norm(A-Ak,'fro');
i = i + 1;
end

Sum of outer products multiplied by a scalar in MATLAB

I would like to vectorize the sum of products hereafter in order to speed up my Matlab code. Would it be possible?
for i=1:N
A=A+hazard(i)*Z(i,:)'*Z(i,:);
end
where hazard is a vector (N x 1) and Z is a matrix (N x p).
Thanks!
With matrix multiplication only:
A = A + Z'*diag(hazard)*Z;
Note, however, that this requires more operations than Divakar's bsxfun approach, because diag(hazard) is an NxN matrix consisting mostly of zeros.
To save some time, you could define the inner matrix as sparse using spdiags, so that multiplication can be optimized:
A = A + full(Z'*spdiags(hazard, 0, zeros(N))*Z);
Benchmarking
Timing code:
Z = rand(N,p);
hazard = rand(N,1);
timeit(#() Z'*diag(hazard)*Z)
timeit(#() full(Z'*spdiags(hazard, 0, zeros(N))*Z))
timeit(#() bsxfun(#times,Z,hazard)'*Z)
With N = 1000; p = 300;
ans =
0.1423
ans =
0.0441
ans =
0.0325
With N = 2000; p = 1000;
ans =
1.8889
ans =
0.7110
ans =
0.6600
With N = 1000; p = 2000;
ans =
1.8159
ans =
1.2471
ans =
1.2264
It is seen that the bsxfun-based approach is consistently faster.
You can use bsxfun and matrix-multiplication -
A = bsxfun(#times,Z,hazard).'*Z + A

Why I got this Error The variable in a parfor cannot be classified

I'm trying to use parfor to estimate the time it takes over 96 sec and I've more than one image to treat but I got this error:
The variable B in a parfor cannot be classified
this the code I've written:
Io=im2double(imread('C:My path\0.1s.tif'));
Io=double(Io);
In=Io;
sigma=[1.8 20];
[X,Y] = meshgrid(-3:3,-3:3);
G = exp(-(X.^2+Y.^2)/(2*1.8^2));
dim = size(In);
B = zeros(dim);
c = parcluster
matlabpool(c)
parfor i = 1:dim(1)
for j = 1:dim(2)
% Extract local region.
iMin = max(i-3,1);
iMax = min(i+3,dim(1));
jMin = max(j-3,1);
jMax = min(j+3,dim(2));
I = In(iMin:iMax,jMin:jMax);
% Compute Gaussian intensity weights.
H = exp(-(I-In(i,j)).^2/(2*20^2));
% Calculate bilateral filter response.
F = H.*G((iMin:iMax)-i+3+1,(jMin:jMax)-j+3+1);
B(i,j) = sum(F(:).*I(:))/sum(F(:));
end
end
matlabpool close
any Idea?
Unfortunately, it's actually dim that is confusing MATLAB in this case. You can fix it by doing
[n, m] = size(In);
parfor i = 1:n
for j = 1:m
B(i, j) = ...
end
end

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