Donno If this post will be allowed to be published but here goes.

As I've covered before It's possible to count with 2 or more integers for every Integer. like so.
2 integers
{1,1},{1,2},{2,1},{1,3},{2,2},{3,1},{1,4}...ect
3 integers
{1,1,1},{1,1,2},{1,2,1},{2,1,1},{1,1,3},{1,2,2},{1,3,1},{2,1,2},{2,2,1},[3,1,1},{1,1,4}..ect

Anyway you can use this technique in order to reduce the numbers of a 9*9 sudoku grid and use my method to quite possibly speed the whole search up for sudoku grid problems,
First you grab your grid to solve then you go square space by square space looking horizontally vertically and by the 3*3 grid the space is in finding all the numbers that each square could be.
Then you make 2 grids one with the first integers of the integer pair count above and then the other with the second integers of the count and fill the blanks in with the lower integers of the possible numbers you've worked out. Sometimes this will mean in the first or second lower grid that your blank space can only be one number.
Then you break you 2 grids into 4 grids of smaller integers and of course yet more blank spaces will become a single number from a few possibilities.
This gives you 4 more filled in grids than the two upper grids and your two upper grids will be more filled in than the top grid(note each grid is still only 9*9).
So now for each grid you should need so many ones and so many 2's and one grid with a 3 corresponding vertically horizontally and in the 3*3 grid of your remaining black spaces, So you start with the last grid you were working on and go blank space by blank space where there should still be possible numbers in that square and you check with the 3*3 grid the space is in and vertically and horizontally for each of the bottom grids.
This should solve some more squares for the upper grids when you convert your 2 integer data back to integers for the upper grids. Then you go to the unsolved squares by unsolved square doing the same thing for the upper grid and then calculate the new solutions Into the top gird.
Now all you do is repeat the whole thing again until the grid is solved.
I predict this takes less than O 2 Log[n] n^4 or O Ln[n] n^Ln[n] time to solve n being 9 for a standard Sudoku grid. So using this method it should take a maximum of roughly 26244 steps or 162 steps. also it depends on what you call a step. If the first O notation is true then P = NP if the other is true then the issue is still unproven one way or the other.