matrix-multiplication

I have two square matrices A and B I must convert B to CSR Format and determine the product C A * B_csr = C I have found a lot of information online regarding CSR Matrix - Vector multiplication . The algorithm is: for (k = 0; k < N; k = k + 1) result[i] = 0; for (i = 0; i < N; i = i + 1) { for (k = RowPtr[i]; k < RowPtr[i+1]; k = k + 1) { result[i] = result[i] + Val[k]*d[Col[k]]; } } However, I require Matrix - Matrix multiplication. Further, it seems that most algorithms apply A_csr - vector multiplication where I require A * B_csr . My solution is to transpose the two matrices before

As per following the inital thread make efficient the copy of symmetric matrix in c-sharp from cMinor. I would be quite interesting with some inputs in how to build a symmetric square matrix multiplication with one line vector and one column vector by using an array implementation of the matrix, instead of the classical long s = 0; List<double> columnVector = new List<double>(N); List<double> lineVector = new List<double>(N); //- init. vectors and symmetric square matrix m for (int i=0; i < N; i++) { for(int j=0; j < N; j++){ s += lineVector[i] * columnVector[j] * m[i,j]; } } Thanks for your

What is the best way to calculate sum of matrices such as A^i + A^(i+1) + A^i+2........A^n for very large n? I have thought of two possible ways: 1) Use logarithmic matrix exponentiation(LME) for A^i, then calculate the subsequent matrices by multiplying by A. Problem : Doesn't really take advantage of the LME algorithm as i am using it only for the lowest power!! 2)Use LME for finding A^n and memoize the intermediate calculations. Problem: Too much space required for large n. Is there a third way? Notice that: A + A^2 = A(I + A) A + A^2 + A^3 = A(I + A) + A^3 A + A^2 + A^3 + A^4 = (A + A^2)(I

Abstract Hi, suppose you have two different independent 64-bit binary matrices A and T ( T is a transposed version of itself, using the transposed version of matrix allows during multiplication to operate on T 's rows rather than columns which is super cool for binary arithmetic) and you want to multiply these matrices the only thing is that matrix multiplication result is truncated to 64-bits and if you yield to a value greater that 1 in some specific matrix cell the resulting matrix cell will contain 1 otherwise 0 Example A T 00000001 01111101 01010100 01100101 10010111 00010100 10110000

Say we have a single channel image (5x5) A = [ 1 2 3 4 5 6 7 8 9 2 1 4 5 6 3 4 5 6 7 4 3 4 5 6 2 ] And a filter K (2x2) K = [ 1 1 1 1 ] An example of applying convolution (let us take the first 2x2 from A) would be 1*1 + 2*1 + 6*1 + 7*1 = 16 This is very straightforward. But let us introduce a depth factor to matrix A i.e., RGB image with 3 channels or even conv layers in a deep network (with depth = 512 maybe). How would the convolution operation be done with the same filter ? A similiar work out will be really helpful for an RGB case. They will be just the same as how you do with a single

I used to have Ubuntu 12.04 and recently did a fresh installation of Ubuntu 14.04. The stuff I'm working on involves multiplications of big matrices (~2000 X 2000), for which I'm using numpy. The problem I'm having is that now the calculations are taking 10-15 times longer. Going from Ubuntu 12.04 to 14.04 implied going from Python 2.7.3 to 2.7.6 and from numpy 1.6.1 to 1.8.1. However, I think that the issue might have to do with the linear algebra libraries that numpy is linked to. Instead of libblas.so.3gf and liblapack.so.3gf , I can only find libblas.so.3 and liblapack.so.3 . I also

According to MKL BLAS documentation "All matrix-matrix operations (level 3) are threaded for both dense and sparse BLAS." http://software.intel.com/en-us/articles/parallelism-in-the-intel-math-kernel-library I have built Scipy with MKL BLAS. Using the test code below, I see the expected multithreaded speedup for dense, but not sparse, matrix multiplication. Are there any changes to Scipy to enable multithreaded sparse operations? # test dense matrix multiplication from numpy import * import time x = random.random((10000,10000)) t1 = time.time() foo = dot(x.T, x) print time.time() - t1 # test