Let's now see what we have, after the deletions:

Cells (1,2), (3,4) and (4,1) have got resolved as a result, and they take '3', '1' and '1' respectively.

Let's repeat the process of Reductions, deleting these values from their respective rows, columns and major squares respectively, till no more reductions are possible.

First, since cell (1,2) takes the value '3', we can't have any more '3' in row 1, or column 2 or the top left major square (consisting of cells (1,1), (1,2), (2,1) and (2,2)). So, delete '3' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have:

Removing the deleted numbers, the puzzle reduces to the following:

We see that Cell (1,3) now has a '1', Cell (1,4) has '4', and (4,2) has '4'.

Again, let's repeat the process of deleting these values from their respective rows, columns and major squares respectively.

Since cell (1,3) takes the value '1', we can't have any more '1' in row 1, or column 3 or the top right major square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '1' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have, again, no such '1' to be deleted, and this is what we continue to have: Since cell (1,4) takes the value '4', we can't have any more '4' in row 1, or column 4 or the top right major square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '4' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have, yet again, no such '4' to be deleted, and this is what we still continue to have:

Since cell (4,2) takes the value '4', we can't have any more '4' in row 4, or column 2 or the bottom left major square (consisting of cells (3,1), (3,2), (4,1) and (4,2)). So, delete '4' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have, once again, no such '4' to be deleted, and this is what we again continue to have: Now, we find that we have reached an impasse, being unable to resolve between cells (2,3), (2,4), (4,3) and (4,4). And we need to resort to the Tie Breaker Rule.

[Note: If we reach an impasse and it becomes clear that we have reached an impasse, we don't have to repeat the previous steps where we found no further scope for reduction is possible. We would learn to skip these steps from experience, as we solve more and more puzzles.]

In a situation like this, where we have 2 or more cells with exactly the same possibility values for different cells, and if we are unable to resolve otherwise, we break the impasse using the Tie-breaker Rule.

Let's assume one of the 2 possible values for any one of the unresolved cells. Let's start with lower value for the Lower Row No., and the Lower Column no. (You start in any order, and still you will get the same results.)

Let's assume the Value '2' in (2,3); so, let's delete '3' from Cell (2,3).

Since cell (2,3) takes the value '2', we can't have any more '2' in row 2, or column 3 or the Top Right Major Square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '2' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have:

Let's now remove the deleted nos. and see what we have:

Since cell (2,4) takes the value '3', we can't have any more '3' in row 2, or column 4 or the top right major square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '3' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have:

Let's see what we would have had in such a case:

Since cell (2,3) takes the value '3', we can't have any more '3' in row 2, or column 3 or the top right major square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '3' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have:

Let's now remove the deleted nos. and see what we have:

Since cell (2,4) takes the value '2', we can't have any more '2' in row 2, or column 4 or the top right major square (consisting of cells (1,3), (1,4), (2,3) and (2,4)). So, delete '2' from all the above cells in the Possibility Matrix where the puzzle is yet to be solved. We have:

Now, removing the deleted value, we have a Final Solution, as below:

Hey, this is ANOTHER Solution! Isn't that interesting? For all practical purposes, you should be satisfied if you get one of the final solutions. In fact, most Sudoku solvers

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