Margin with z-binning

About

Follow-up of !188 (merged). The goal is to make sure margin works with (z-)binning.

To do

• Check _build_processing_steps
• Adapt start_xyz and end_xyz when using binning
• End-to-end reconstruction test

Notes

When splitting the dataset in chunks of sub-volumes, there are a few cases to consider.

The processing is either split along the vertical direction (process projections[:, start_z:end_z, :]), or in the "angles" direction (process projections[start_a:end_a, :, :].
Either way, let's denote N the size of the dimension where the processing is split along (can be either be N_z = delta_z or n_angles.

Case 0: no binning, no margin

   C1           C3
|------|------|------|---|
          C2          C4

Simplest case, here we have

N = \displaystyle\sum_{i=1}^K C_i 

where there are K chunks C_1, \ldots, C_K.

Case 1: margin, no binning

Each chunk of size C_i is eventually truncated to C_i - M_i where M_i is the total margin (up and down).

|   C1    |    |    C3    | 
|------|--|----|--|----|--|----|
       |    C2    |    |  C4   |

N = \displaystyle\sum_{i=1}^K C_i - M_i

Case 2: binning, no margin

Each chunk of size C_i is eventually reduced to C_i // b where b denotes the binning factor, and // is the integer division.

When there is no margin, sticking together all the binned chunks yields

\displaystyle\sum_{i=1}^K C_i // b

To have this sum equal to N//b is not obvious, as generally (a + b) // c is not equal to a // c + b // c.

A sufficient condition is to have C_i \text{ mod } b = 0, i.e all the chunks sizes are multiple of the binning factor.

Case 1+2: binning and margin

We need to have

\displaystyle\sum_{i=1}^K (C_i - M_i) // b

equal to N//b.

Similarly as above, the simplest sufficient condition is C_i \text{ mod } b = 0 and M_i \text{ mod } b = 0, i.e all the chunk size and margins are multiple of b.