Confusion on total number of blocks produced after splitting image into overlapping blocks.

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So apparently i found out about the principle that states the total number of blocks produced from splitting an image into overlapping blocks is computed with the following equation.
(M-B+1)(N-B+1)
where M and N are the dimensions of the input image and B is the block size. So i suggest that, if i want to split an image of dimension 256x256 into 8x8 overlapping blocks, then according to the equation i would get:
(256-8+1)(256-8+1) = 62001 blocks in total
I'm assuming that the +1 indicates overlap by 1 pixel? Correct me if I'm wrong.
However, what would it be if i were to split into 8x8 overlapping blocks with an overlap of 4 pixels? Would the equation be:
(256-8+4)(256-8+4) = 63504 blocks in total?
I also realize that somehow the equation doesn't take into account partial blocks? Is this principle reliable, or am i interpreting it wrongly?

Accepted Answer

Guillaume
Guillaume on 14 Aug 2019
Edited: Guillaume on 14 Aug 2019
I don't think you got it right.
First, starts with non-overlapping blocks, how many blocks do you have if divide M columns into blocks of size P: M/P blocks (assuming that image size is a multiple of block size). Niow, if you slide these blocks by one column and ignoring edge effects, you get another M/P blocks. How many times can you slide these blocks before you've shifted by the full width of a block: P-1 times. Therefore, for an offest of 1, you'll have: blocks (the P-1 shifted blocks plus the orginal unshifted blocks.
Now, if you change the shift, and assuming it's a divisor of of the block size, you'll have in total P/S (S: shift) shifts before you've shifted by the full with, so you'll have in total: blocks.
Now all the above assumes that you allow the last block to shift partially out of the image. If not, you'll obviously have the last P-1 blocks you need to remove, so indeed, you'll end up with blocks. But for the case where S > 1, you'll need to remove the last blocks, so the number of blocks is .
Obviously, the same applies to the rows.

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