Outer product, vectorial operation and loop - numpy

Question

I have two arrays of size 15 : A = [a_0, ... , a_14] and B = [b_0, ..., b_14]

Goal: obtain the array C of size 8 resulting from

C = [a_0] * [b_7, ..., b_14] + [a_2, a_3] * [b_3, b_4, b_5, b_6] + [a_3, a_4, a_5, a_6] * [b_2, b_3] + [a_7, ..., a_14] * [b_0]

where * is the outer product np.outer. Note that:

• each sub-array is of length 2^i for i between 0 and 3.
• from the outer product, we obtain two vectors of size (8) and two matrices of sizes (2, 4) and (4, 2). We suppose that we flatten immediately after the product, in order to be able to sum the four products and have at the end a long vector of size 8.

My implementation is the following:

inds = [0, 1, 3, 7, 15]
C = np.zeros(8)
d = 4
for i in range(d):
    left = A[inds[i]:inds[i+1]]
    right = B[inds[d-i-1]:inds[d-i]]
    C += (left[:, None]*right[None, :]).ravel()  # same as np.outer(left, right).ravel()

Question: what is the fastest way to obtain C ? i.e. is there a way to avoid having this for loop to perform the summation ?

If not: what are my options ? code in C++ ? Cython ?

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NB: this is to be generalized for loops of range(L+1) with L any integer. In the example above I have illustrated the case L=3 for better comprehension. FYI, the generalized code would look like this:

L = 3
inds = np.cumsum([2**k for k in range(0, L+1)])
inds = np.concatenate(([0], inds))
# Input arrays A and B are of size inds[-1]
C = np.zeros(2**L)
d = L+1
for i in range(d):
    left = A[inds[i]:inds[i+1]]
    right = B[inds[d-i-1]:inds[d-i]]
    C += (left[:, None]*right[None, :]).ravel()  # same as np.outer(left, right).ravel()

Answer 1

I think you can simply do:

C = np.outer(A[0], B[7:])+\
np.outer(A[[2,3]], B[[3,4,5,6]]).ravel()+\
np.outer(A[[3,4,5,6]], B[[2,3]]).ravel()+\
np.outer(A[7:], B[0]).ravel()

Am I wrong?