Faster way to transpose each row of a matrix and multiply the resulting vector by some other matrix? - matlab

Question

I have an input matrix X with dimensions N_rows x N_cols. I also have a sparse, tridiagonal matrix M which is square of size N_rows x N_rows. These are created as follows:
N_rows = 3;
N_cols = 6;
X = rand(N_rows,N_cols);
mm = 10*ones(N_cols,1); % Subdiagonal elements
dd = 20*ones(N_cols,1); % Main diagonal elements
pp = 30*ones(N_cols,1); % Superdiagonal elements
M = spdiags([mm dd pp],-1:1,N_cols,N_cols);
and look like the following:
>> X
X =
0.4018 0.1233 0.4173 0.9448 0.3377 0.1112
0.0760 0.1839 0.0497 0.4909 0.9001 0.7803
0.2399 0.2400 0.9027 0.4893 0.3692 0.3897
full(M)
ans =
2 3 0 0 0 0
1 2 3 0 0 0
0 1 2 3 0 0
0 0 1 2 3 0
0 0 0 1 2 3
0 0 0 0 1 2
I would like to take each row of X, and do a matrix multiplication with M, and piece the obtained rows back together to obtain an output Y. At the moment, I achieve this successfully with the following:
Y = (M*X.').';
The example above is for a 3x6 matrix for X, but in reality I need to do this for a matrix with dimensions 500 x 500, about 10000 times, and the profiler says that this operation in the bottleneck in my larger code. Is there a faster way to do this row-by-row matrix multiplication multiplication?
On my system, the following takes around 20 seconds to do this 10000 times:
N_rows = 500;
N_cols = 500;
X = rand(N_rows,N_cols);
mm = 10*ones(N_cols,1); % Subdiagonal elements
dd = 20*ones(N_cols,1); % Main diagonal elements
pp = 30*ones(N_cols,1); % Superdiagonal elements
M = spdiags([mm dd pp],-1:1,N_cols,N_cols);
tic
for k = 1:10000
Y = (M*X.').';
end
toc
Elapsed time is 18.632922 seconds.

Answer 1

You can use X*M.' instead of (M*X.').';. This saves around 35% of time on my computer.
This can be explained because transposing (or permuting dimensions) implies rearranging the elements in the internal (linear-order) representation of the matrix, which takes time.

Answer 2

Another option is using conv2:
Y = conv2(X, [30 20 10], 'same');
Explanation:
There is a tridiagonal matrix that all elements on each diagonal are identical to each other:
M =
2 3 0 0 0 0
1 2 3 0 0 0
0 1 2 3 0 0
0 0 1 2 3 0
0 0 0 1 2 3
0 0 0 0 1 2
Suppose you want to multiply the matrix by a vector:
V = [11 ;12 ;13 ;14 ;15 ;16];
R = M * V;
Each element of the vector R is computed by sum of products of each row of M by V:
R(1):
2 3 0 0 0 0
11 12 13 14 15 16
R(2):
1 2 3 0 0 0
11 12 13 14 15 16
R(3):
0 1 2 3 0 0
11 12 13 14 15 16
R(4):
0 0 1 2 3 0
11 12 13 14 15 16
R(5):
0 0 0 1 2 3
11 12 13 14 15 16
R(6):
0 0 0 0 1 2
11 12 13 14 15 16
It is the same as multiplying a sliding window of [1 2 3] by each row of M. Basically convolution applies a sliding window but first it reverses the direction of window so we need to provide the sliding window in the reversed order to get the correct result. Because of that I used Y = conv2(X, [30 20 10], 'same'); instead of Y = conv2(X, [10 20 30], 'same');.