Sub-pixel Distance Transform — Acko.net

Given a 1D input array of zeroes (filled), with an area masked out with infinity (empty):

O = [·, ·, ·, 0, 0, 0, 0, 0, ·, 0, 0, 0, ·, ·, ·]

Make a matching sequence … 3 2 1 0 1 2 3 … for each element, centering the 0 at each index:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13,14] + ∞
[1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12,13] + ∞
[2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11,12] + ∞
[3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11] + 0
[4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10] + 0
[5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9] + 0
[6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8] + 0
[7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7] + 0
[8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6] + ∞
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5] + 0
[10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4] + 0
[11,10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3] + 0
[12,11,10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2] + ∞
[13,12,11,10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1] + ∞
[14,13,12,11,10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0] + ∞

You then add the value from the array to each element in the row:

[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11]
[4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10]
[5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5]
[10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4]
[11,10,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]
[∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞, ∞]

And then take the minimum of each column:

P = [3, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 3]

This sequence counts up inside the masked out area, away from the zeroes. This is the positive distance field P.

You can do the same for the inverted mask:

I = [0, 0, 0, ·, ·, ·, ·, ·, 0, ·, ·, ·, 0, 0, 0]

to get the complementary area, i.e. the negative distance field N:

N = [0, 0, 0, 1, 2, 3, 2, 1, 0, 1, 2, 1, 0, 0, 0]

That’s what the EDT does, except it uses square distance … 9 4 1 0 1 4 9 …: