How is the 2D filter function used in this code?

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Daigo el 13 de En. de 2022
Comentada: Daigo el 14 de En. de 2022
In one of the cody answers (solution 712713), I found a following code:
function B = hanlelize(A)
F = flip(eye(111));
B1 = filter2(F,A);
B2 = filter2(F,A|1);
B = B1./B2;
end
This function convert an input matrix A to a Hankel matrix B by replacing each skew-diagonal of A with its mean. For example, if the input is
A =
3 7 10 2
3 5 1 2
6 3 2 7
then, the matrix B1, B2, B will be
B1 =
3 10 21 6
10 21 6 4
21 6 4 7
B2 =
1 2 3 3
2 3 3 2
3 3 2 1
B =
3 5 7 2
5 7 2 2
7 2 2 7
The code looks elegant but I'm not so familiar with the 2D filter function used in this code.
• How is the 2D filter working in the computation of B1? Why does the 2D filter by flipped identity matrix result in B1?
• How is the 2D filter working in the computation of B2? What does 'A|1' mean? I checked the documentation of filter2 but I couldn't find such an expression.
I'm looking forward to your help.
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Steve Eddins el 13 de En. de 2022
Editada: Steve Eddins el 13 de En. de 2022
For each element in A, filter2 essentially "slides" the filter F so that its center lines up with the element in A, and then it does a point-wise multiplication between the elements in A and the corresponding elements in F, and then sums all those products. With a filter that is just ones along the anti-diagonal, each element of the output matrix is found by summing up the input matrix elements along the corresponding anti-diagonal. Eh, it's a bit hard to describe in words. Let's try it with a small A and small F.
A = magic(5)
A = 5×5
17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9
F = flip(eye(3))
F = 3×3
0 0 1 0 1 0 1 0 0
B1 = filter2(F,A)
B1 = 5×5
17 47 6 15 29 47 10 21 42 36 9 23 39 55 43 16 36 57 68 5 23 37 46 5 9
Where does B1(4,2) come from?
B1(4,2)
ans = 36
It is the sum of A(4,2), the element to its upper right, A(3,3), and the element to its lower left, A(5,1).
A(4,2) + A(3,3) + A(5,1)
ans = 36
In the expression A|1, the vertical bar is the element-wise OR operator. When you OR any value with 1, you always get 1, so this expression appears to be just a tricky way to compute true(size(A)). In Cody, solutions with a smaller code size (based on some metric) are scored higher, so perhaps A|1 had a slightly smaller code size metric than true(size(A)).
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Steve Eddins el 13 de En. de 2022
I made a small correction to my answer. Where I originally wrote false(size(A)), it should have been true(size(A)).
Daigo el 14 de En. de 2022
Thank you for your detailed explanations! Now I can make a good use of filter2 function.

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