Deriving the total length of segments from the glance function

Theorem:

For glance function

H = ∑ c_a*h_a

where the c_a are constants and the h_a are Heaviside functions, with impulse at a.

Then we have

(*)          ∑ c_a*a = sum of lengths of the segments

The proof is a generalization of brute force calculation, which is always of the same form. So for example, if we have:

With segments of lengths a, c, and e; separated by gaps of lengths b and d, then the left hand side of (*) is:

a - (a+b) + (a+b+c) - (a+b+c+d) + (a+b+c+d)

       b      -   (b+c)    + (b+c+d) - (b+c+d+e)

                       c      -   (c+d)   + (c+d+e)

                                       d      -    d+ e

                                                     e

Here, everything in the second row cancels most of what is above it, leaving alternating +/- a's. Being odd in number, the final sum is +a.

Now the second row was used up completely, in those cancelations. So the third row starts with c and because  of cancelations from the fourth row, result in a final value of +c. The fourth row was used up canceling things in the third row. Now the fifth row simply adds +e. In total we get a+c+e. This is the sum we wanted.

It follows that if two domains in the 2D plane have the same glance functions over all lines, then they have the same Radon transforms. So the domains are the same, up to the sorts of details usually assumed away.

Note: (sum of the gaps) = D - (sum of length of segments)