2-D convolution as a matrix-matrix multiplication

Question

I know that, in the 1D case, the convolution between two vectors, a and b, can be computed as conv(a, b), but also as the product betw

Answer 1

If you unravel k to a m^2 vector and unroll X, you would then get:

• a m**2 vectork
• a ((n-m)**2, m**2) matrix for unrolled_X

where unrolled_X could be obtained by the following Python code:

from numpy import zeros


def unroll_matrix(X, m):
  flat_X = X.flatten()
  n = X.shape[0]
  unrolled_X = zeros(((n - m) ** 2, m**2))
  skipped = 0
  for i in range(n ** 2):
      if (i % n) < n - m and ((i / n) % n) < n - m:
          for j in range(m):
              for l in range(m):
                  unrolled_X[i - skipped, j * m + l] = flat_X[i + j * n + l]
      else:
          skipped += 1
  return unrolled_X

Unrolling X and not k allows a more compact representation (smaller matrices) than the other way around for each X - but you need to unroll each X. You could prefer unrolling k depending on what you want to do.

Here, the unrolled_X is not sparse, whereas unrolled_k would be sparse, but of size ((n-m+1)^2,n^2) as @Salvador Dali mentioned.

Unrolling k could be done like this:

from scipy.sparse import lil_matrix
from numpy import zeros
import scipy 


def unroll_kernel(kernel, n, sparse=True):

    m = kernel.shape[0]
    if sparse:
         unrolled_K = lil_matrix(((n - m)**2, n**2))
    else:
         unrolled_K = zeros(((n - m)**2, n**2))

    skipped = 0
    for i in range(n ** 2):
         if (i % n) < n - m and((i / n) % n) < n - m:
             for j in range(m):
                 for l in range(m):
                    unrolled_K[i - skipped, i + j * n + l] = kernel[j, l]
         else:
             skipped += 1
    return unrolled_K