Replace every cell in a matrix with the average of adjcent cells - algorithm

Question

Requirement: Must be done in-place.
For example:
Given matrix
1, 2, 3
4, 5, 6
7, 8, 9
Should replace by the average of its sum of 3*3 neighbor cells and its own:
(1+2+4+5)/4, (2+1+3+4+5+6)/6 , (3+2+6+5)/4
(1+2+5+4+7+8)/6, (1+2+3+4+5+6+7+8+9)/9, (2+3+5+6+8+9)/6
(4+5+7+8)/4, (4+5+6+7+8+9)/6, (5+6+8+9)/4
which is:
All floating number convert to int
3, 3.5(3), 4 3, 3, 4
4.5(4), 5, 5.5(5) => 4, 5, 5
6, 6.5(6), 7 6, 6, 7
I tried to just iterate over the matrix and update each cell, but I found this will affect the future calculation:
Say I update the original 1 to 3, but when I when I tried to update the original 2, the original 1 becomes 3 now.
Copying the original matrix for calculating average is a workaround but it's a bad idea, Could we achieve that without using that much space?

Answer 1

In most cases, you should just create a copy of the original matrix and use that for calculating the averages. Unless creating a copy of the matrix would use more memory than you have available, the overhead should be negligible.
If you have a really large matrix, you could use a "rolling" backup (in lack of a better term). Let's say you update the cells row-by-row and you are currently in row n. You don't need a backup of row n-2, as those cells are not relevant any more, and neither of row n+1, because those are still the original values. So you can just keep a backup of the previous and the current row. Whenever you advance to the next row, discard the backup of the previous row, move the backup of the current row to previous, and create a backup of the new current row.
Some pseudo-code (not taking any edge-cases into account):
previous = [] # or whatever works for the first row
for i in len(matrix):
current = copy(matrix[i])
for k in len(matrix[i]):
matrix[i][k] = previous[k-1] + ... + current[k] + ... matrix[i+1][k+1] / 9
previous = current
(You might also keep a backup of the next row, just so you can use only the backup rows for all the values instead of having to differentiate.)

Answer 2

You must have some kind of cache for the result data so you can keep reference to the original data. I don't think there is a way around it.
If the data set is large, you could optimize by using a smaller data buffer (like looking through a keyhole) and 'scrolling' the input matrix as you update it. In your case, you could use a buffer as small as 3x3.
It is a compromise between speed and space though. The smaller your buffer, the worse the performance will be.
To visualize the problem, starting from the top-left (0,0) of the dataset:
(result values are rounded down for simplicity)
First step: update first 4 cells (prime the buffer)
// Data Set // Data Viewport // Result Set
01,02,03,04,05 01,02,03 04,04,??
06,07,08,09,10 06,07,08 06,07,??
11,12,13,14,15 11,12,13 ??,??,??
16,17,18,19,20
21,22,23,24,25
then for each iteration..
( new values indicated with [xx] )
++ update first column in Data Set from Result Set
// Data Set // Data Viewport // Result Set
[04],02,03,04,05 01,02,03 04,04,??
[06],07,08,09,10 06,07,08 06,07,??
11 ,12,13,14,15 11,12,13 ??,??,??
16 ,17,18,19,20
21 ,22,23,24,25
++ shift Data Viewport and Result Set right 1 column
// Data Set // Data Viewport // Result Set
[04],02,03,04,05 02,03,04 04,[03],??
[06],07,08,09,10 07,08,09 07,[08],??
11 ,12,13,14,15 12,13,14 ??, ?? ,??
16 ,17,18,19,20
21 ,22,23,24,25
++ update middle column of Result Set
// Data Set // Data Viewport // Result Set
[04],02,03,04,05 02,03,04 04,[05],??
[06],07,08,09,10 07,08,09 07,[08],??
11 ,12,13,14,15 12,13,14 ??, ?? ,??
16 ,17,18,19,20
21 ,22,23,24,25
At the following iteration, the data state would be:
// Data Set // Data Viewport // Result Set
04,[04],03,04,05 03,04,05 05,[06],??
06,[07],08,09,10 08,09,10 08,[09],??
11, 12 ,13,14,15 13,14,15 ??, ?? ,??
16, 17 ,18,19,20
21, 22 ,23,24,25
.. etc
Don't forget to handle the other edge cases.
*The Data Viewport representation is just for visualization. In code, the actual viewport would be the result buffer.