In the case of multiple of same matrix matA, like
matA.transpose()*matA,
You don't have to compute all result product, because the result matrix is symmetric(so only if the m>n), in my specific case is always symmetric! square.
So its enough the compute only for. ex. lower triangular part and rest only copy..... because the results of the multiple 2nd and 3rd row, resp.col, is the same like 3rd and 2nd.....And etc....
So my question is , exist way how to tell Eigen, to compute only lower part. and optionally save to only lower trinaguler part the product?
DATA = SparseMatrix<double>((SparseMatrix<double>(matA.transpose()) * matA).pruned()).toDense();
According to the documentation, you can evaluate the lower triangle of a matrix with:
m1.triangularView<Eigen::Lower>() = m2 + m3;
or in your case:
m1.triangularView<Eigen::Lower>() = matA.transpose()*matA;
(where it says "Writing to a specific triangular part: (only the referenced triangular part is evaluated)"). Otherwise, in the line you've written
Eigen will calculate the entire sparse matrix matA.transpose()*matA.
Regarding saving the resulting m1 matrix, it is the same as saving whatever type of matrix it is (Eigen::MatrixXt or Eigen::SparseMatrix<t>). If m1 is sparse, then it will be only half the size of a straightforward matA.transpose()*matA. If m1 is dense, then it will be the full square matrix.
https://eigen.tuxfamily.org/dox/classEigen_1_1SparseSelfAdjointView.html
The symmetric rank update is defined as:
B = B + alpha * A * A^T
where alpha is a scalar. In your case, you are doing A^T * A, so you should pass the transposed matrix instead. The resulting matrix will only store the upper or lower portion of the matrix, whichever you prefer. For example:
SparseMatrix<double> B;
B.selfadjointView<Lower>().rankUpdate(A.transpose());
Related
Assume that I have 2 matrix: image, filter; with size MxM and NxN.
My regular convolution looks like this and produces matrix output size (M-N+1)x(M-N+1). Basically it places the top-left corner of a filter on a pixel, convolute, then assign the sum onto that pixel:
for (int i=0; i<M-N; i++)
for (int j=0; j<M-N; j++)
{
float sum = 0;
for (int u=0; u<N; u++)
for (int v=0; v<N; v++)
sum += image[i+u][j+v] * filter[u][v];
output[i][j] = sum;
}
Next, to perform FFT:
Apply zero-padding to both image, filter to the right and bottom (that is, adding more zero columns to the right, zero rows to the bottom). Now both have size (M+N)x(M+N); the original image is at
image[0->M-1][0-M-1].
(Do the same for both matrix) Calculate the FFT of each row into a new matrix, then calculate the FFT of each column of that new matrix.
Now, I have 2 matrices imageFreq and filterFreq, both size (M+N)x(M+N), which is the FFT-ed form of the image and the filter.
But how can I get the convolution values that I need (as described in the sample code) from them?
convolution between A,B using FFT is done by per element multiplication in the frequency domain so in 1D something like this:
convert A,B by FFT
assuming the sizes are N,M of A[N],B[M] first zero pad to common size Q which is a power of 2 and at least M+N in size and then apply FFT:
Q = exp2(ceil(log2(M+N)));
zeropad(A,Q);
zeropad(B,Q);
a = FFT(A);
b = FFT(B);
convolute
in frequency domain use just element wise multiplication:
for (i=0;i<Q;i++) a[i]*=b[i];
reconstruct result
simply apply IFFT (inverse of FFT)...
AB = IFFT(a); // crop to first N (real) elements
and use only the first N element (unless algorithm used need more depends on what you are doing...)
For 2D you can either convolute directly in 2D (using 2 nested for loops) or convolve each axis separately. Beware that separating axises need also to normalize the result by some constant (which depends on dimensionality, resolution and kernel used)
So when put together (also assuming the same resolution NxN and MxM) first zero pad to (QxQ) and then:
Q = exp2(ceil(log2(M+N)));
zeropad(A,Q,Q);
zeropad(B,Q,Q);
a = FFT(A);
b = FFT(B);
for (i=0;i<Q;i++)
for (j=0;j<Q;j++) a[i][j]*=b[i][j];
AB = IFFT(a); // crop to first NxN (real) elements
And again crop to AB to NxN size (unless ...) for more info see:
How to compute Discrete Fourier Transform?
and all sublinks there... Also here at the end is 1D convolution example using NTT (its a special form of FFT) to compute bignum multiplication:
Modular arithmetics and NTT (finite field DFT) optimizations
Also if you want real result then just use only the real parts of the result (ignore imaginary part).
I'm trying to estimate a 3D rotation matrix between two sets of points, and I want to do that by computing the SVD of the covariance matrix, say C, as follows:
U,S,V = svd(C)
R = V * U^T
C in my case is 3x3 . I am using the Eigen's JacobiSVD module for this and I only recently found out that it stores matrices in column-major format. So that has had me confused.
So, when using Eigen, should I do:
V*U.transpose() or V.transpose()*U ?
Additionally, the rotation is accurate upto changing the sign of the column of U corresponding to the smallest singular value,such that determinant of R is positive. Let's say the index of the smallest singular value is minIndex .
So when the determinant is negative, because of the column major confusion, should I do:
U.col(minIndex) *= -1 or U.row(minIndex) *= -1
Thanks!
This has nothing to do with matrices being stored row-major or column major. svd(C) gives you:
U * S.asDiagonal() * V.transpose() == C
so the closest rotation R to C is:
R = U * V.transpose();
If you want to apply R to a point p (stored as column-vector), then you do:
q = R * p;
Now whether you are interested R or its inverse R.transpose()==V.transpose()*U is up to you.
The singular values scale the columns of U, so you should invert the columns to get det(U)=1. Again, nothing to do with storage layout.
Using Armadillo, how do I reshape a matrix when I only specify one dimension size?
In Matlab documentation, there is this example of such functionality:
Reshape a 6-by-6 magic square matrix into a matrix that has only 3
columns. Specify [] for the first dimension size to let reshape
automatically calculate the appropriate number of rows.
A = magic(6);
B = reshape(A,[],3);
The result is a 12-by-3 matrix, which maintains the same number of
elements (36) as the original 6-by-6 matrix. The elements in B also
maintain their columnwise order from A.
How can that be accomplished with Armadillo?
You can use .size() to get the total number of elements of your matrix and calculate the dimensions yourself.
Example:
B = reshape(A, A.size()/3, 3);
I have 2 matrices of dimension 1*280000.
I wanted to multiply one matrix with transposed second matrix using opencv.
I tried to multiply them using Multiplication operator(*).
But it is giving me error :'The total size matrix does not fit to size_t type'
As after multiplication the size will be 280000*28000 of matrix.
So,I am thinking multiplication should 32 bit.
Is there any method to do the 32bit multiplication?
Why do you want to multiply them like that? But because this is an answer, I would like to help you thinking more than just do it:
supposing that you have the two matrix: A and B (A.size() == B.size() == [1x280000]).
and A * B.t() = AB (AB is the result)
then AB = [A[0][0]*B A[0][1]*B ... A[0][279999]*B] (each column is the transposed matrix multiplied by the corresponding element of the other matrix)
AB may also be written as:
[ B[0][0]*A
B[0][1]*A
...
B[0][279999]*A]
(each row of the result will be the row matrix multiplied by the corresponding element of the column (transposed) matrix)
Hope that this will help you in what you are doing... Using a for loop you can print, or store, or what you need with the result
Just as Z boson recommended, I am using a column-major matrix format in order to avoid having to use the dot product. I don't see a feasible way to avoid it when multiplying a vector with a matrix, though. The matrix multiplication trick requires efficient extraction of rows (or columns, if we transpose the product). To multiply a vector by a matrix, we therefore transpose:
(b * A)^T = A^T * b^T
A is a matrix, b a row vector, which, after being transposed, becomes a column vector. Its rows are just single scalars and the vector * matrix product implementation becomes an inefficient implementation of dot products of columns of (non-transposed) matrix A with b. Is there a way to avoid performing these dot products? The only way I see that could do it, would involve row extraction, which is inefficient with the column-major matrix format.
This can be understood from original post on this (my first on SO)
efficient-4x4-matrix-vector-multiplication-with-sse-horizontal-add-and-dot-prod
. The rest of the discussion applies to 4x4 matrices.
Here are two methods to do do matrix times vector (v = Mu where v and u are column vectors)
method 1) v1 = dot(row1, u), v2 = dot(row2, u), v3 = dot(row3, u), v4 = dot(row4, u)
method 2) v = u1*col1 + u2*col2 + u3*col3 + u4*col4.
The first method is more familiar from math class while the second is more efficient for a SIMD computer. The second method uses vectorized math (like numpy) e.g.
u1*col1 = (u1x*col1x, u1y*col1y, u1z*col1z, u1w*col1w).
Now let's look at vector times matrix (v = uM where v and u are row vectors)
method 1) v1 = dot(col1, u), v2 = dot(col2, u), v3 = dot(col3, u), v4 = dot(col4, u)
method 2) v = u1*row1 + u2*row2 + u3*row3 + u4*row4.
Now the roles of columns and rows have swapped but method 2 is still the efficient method to use on a SIMD computer.
To do matrix times vector efficiently on a SIMD computer the matrix should be stored in column-major order. To do vector times matrix efficient on a SIMD computer the matrix should be stored in row-major order.
As far as I understand OpenGL uses column major ordering and does matrix times vector and DirectX uses row-major ordering and does vector times matrix.
If you have three matrix transformations that you do in order M1 first then M2 then M3 with matrix times vector you write it as
v = M3*M2*M1*u //u and v are column vectors - OpenGL form
With vector times matrix you write
v = u*M1*M2*M3 //u and v are row vectors - DirectX form
Neither form is better than the other in terms of efficiency. It's just a question of notation (and causing confusion which is useful when you have competition).
It's important to note that for matrix*matrix row-major versus column-major storage is irrelevant.
If you want to know why the vertical SIMD instructions are faster than the horizontal ones that's a separate question which should be asked but in short the horizontal ones really act in serial rather than parallel and are broken up into several micro-ops (which is why ironically dppd is faster than dpps).