simple 1D example showing convolution vs correlation to test commutative/associative property - correlation

Question

So I've known this fact that convolution has commutative/associative property while correlation does not but never really written out a simple example to test this out. (proofs in equation forms exist out there)
I was just trying this simple example:
Image : [1 0 -1 0 1 0 1 1]
A kernel = [-1 0 1]
B kernel = [-1 0 1]
Using convolution, I * (A * B) should equal (I * A) * B, but when I tried this they were not equal.. which confused me a lot. would anyone be able to help out? maybe I made a human error in doing computation.. but they should be equal right? (assuming borders are treated as 0s)
And using correlation, the same should not be equal as I understand.. which they dont, but then, my convolution did not either so lol (but it should!)
Any help would be appreciated.
also,
A*B (convolution) would be [0 -2 0] right?
and AxB (cross correlation) would be [0 2 0]?

Answer 1

The convolution operator is commutative. So, you are right that I*(A*B) should be equal to (I*A)*B. Let's convert this to matrix formation first. Convolution by kernel A can be translated to multiplication by the following convolution matrix, C:
C=
[-1 0 1 0 0 0 0 0]
[ 0 -1 0 1 0 0 0 0]
[ 0 0 -1 0 1 0 0 0]
[ 0 0 0 -1 0 1 0 0]
[ 0 0 0 0 -1 0 1 0]
[ 0 0 0 0 0 -1 0 1]
[ 1 0 0 0 0 0 -1 0]
[ 0 1 0 0 0 0 0 -1]
We have taken the kernel [-1 0 1] and zero-padded it by 5 zeros on first row. Then, each row below is a cyclic shift by 1 position to the right of the preceding row.
So, if we want to compute the convolution of kernel A with I:
I = [1 0 -1 0 1 0 1 1]
We simply compute the matrix-vector multiplication:
C*I^T
(where I^T is the transposed vector).
Following the above formulation, we obtain:
I*(A*B) = (C*C)*I^T
Computing C*C you get the following matrix:
C^2=
[ 1 0 -2 0 1 0 0 0]
[ 0 1 0 -2 0 1 0 0]
[ 0 0 1 0 -2 0 1 0]
[ 0 0 0 1 0 -2 0 1]
[ 1 0 0 0 1 0 -2 0]
[ 0 1 0 0 0 1 0 -2]
[-2 0 1 0 0 0 1 0]
[ 0 -2 0 1 0 0 0 1]
and:
I*(A*B) = (C*C)*I^T =
[ 4]
[ 0]
[-2]
[ 1]
[ 0]
[-2]
[-2]
[ 1]
whereas (I*A)*B is in matrix formulation C*(C*I^T) which is equal to I*(A*B)=(C*C)*I^T from matrix associativity.
You can validate this result by running the following numpy code:
import numpy as np
C = [[-1,0,1,0,0,0,0,0]]
for k in range(7):
C.append(np.roll(C[0],k+1))
C = np.array(C)
#print(C)
I = np.transpose(np.array([[1,0,-1,0,1,0,1,1]]))
print(f'C=\n{C}\n')
print(f'C^2=\n{C#C}\n')
print(f'(C*C)*I=\n{(C#C)#I}\n')
print(f' C*(C*I)=\n{C#(C#I)}\n')
The result of running the above code is:
C=
[[-1 0 1 0 0 0 0 0]
[ 0 -1 0 1 0 0 0 0]
[ 0 0 -1 0 1 0 0 0]
[ 0 0 0 -1 0 1 0 0]
[ 0 0 0 0 -1 0 1 0]
[ 0 0 0 0 0 -1 0 1]
[ 1 0 0 0 0 0 -1 0]
[ 0 1 0 0 0 0 0 -1]]
C^2=
[[ 1 0 -2 0 1 0 0 0]
[ 0 1 0 -2 0 1 0 0]
[ 0 0 1 0 -2 0 1 0]
[ 0 0 0 1 0 -2 0 1]
[ 1 0 0 0 1 0 -2 0]
[ 0 1 0 0 0 1 0 -2]
[-2 0 1 0 0 0 1 0]
[ 0 -2 0 1 0 0 0 1]]
(C*C)*I=
[[ 4]
[ 0]
[-2]
[ 1]
[ 0]
[-2]
[-2]
[ 1]]
C*(C*I)=
[[ 4]
[ 0]
[-2]
[ 1]
[ 0]
[-2]
[-2]
[ 1]]
So you can see that the same result is obtained.
Note that A*B is not [0 -2 0] but [1 0 -2 0 1] because you take A=[-1 0 1] (zero pad it on both sides to it has 2 zeros on each side) and move it as a sliding window over B=[-1 0 1] so first you get: [-1-1, 0-1 + -10, -11 + 00 + 1-1, 10 + 01, 11] = [1 0 -2 0 1]. The resulting convolution of 2 kernels of width n will be of length 2n-1.