# An Irregular Colombian Sudoku

Solve the Irregular Sudoku below (by Xavier Castillo), where the usual rules apply.

The dots outside the board on top indicate how many cells in the corresponding column or row of that board contain precisely the same digit as is to be found in the same cell of the solved Sudoku on the below.

Solution:

Step by step:

1:

To start with, we can focus on thee top grid and mark which numbers are already correct/incorrect. We can then look along the rows and columns and mark any numbers which can't be placed due to numbers in the actual grid

2:

Now we can start working out what must be placed. Consider the 4th column. It has 3 numbers left, 2 which are real. However 2 of these 3 are 6s, so it can't be both 6s and hence the 2 must be placed in the actual grid. Similarly for the 8th and 9th column, the 9 and 2 must be correct, as well as the 4 and 2 respectively.

The same logic can now be done on the rows, in the 2nd row placing a 3.

3:

Going back to the first step, we can update the top grid. In particular we have the 6th row which is now complete. in the first step, only rows and columns were focused on, but we can now consider boxes as well, leading to a few more red circles.

Putting the 3 in the actual grid, the top right box is now only missing an 8 which can be placed. This means the 3rd column is now complete in the top grid, and the remaining 3 correct numbers can be placed in the real grid. Finally, updating the top grid gives:

4:

Consider the bottom right box. Because of the 9th column, the 2x2 box jutting out to the left must contain 1,6 and 4. The 8th column must have a 6 due to 1 and 4 being in the same column.

The 3rd row has a 2 and 5 remaining. This means that the 2 in the same box in the top grid is incorrect, and this leaves 2 5s in the 5th column, with 1 being correct, so 5 is the correct number in this column.

However this also means in the 3rd row, the 5 can't be in the 5th column, so we can complete this row. This leaves one place for a 2 in a box on the right.

5:

We know in the bottom right box, it must be 1 and 4 in the left bit. So 1 and 4 are taken in that column for the top grid. This rules out a 1 in row 8, meaning this row can then be completed up top, placing a 4 on the left.

We can use the initial logic on column 1, there are 4 remaining correct numbers, 5 remaining numbers, but a duplicate 9. So the 6, 5 and 2 are correct and can be placed. Updating the grid, we see the 4th row is now complete up top, which in turn completes column 5 and we can place another 5.

6:

Looking up top, the 1 in the second row must be in the middle top box. This means the box top left has a 1 that can be placed in the first column. This completes the first column up top and we can place the 9, whilst also completing the second row.

In the top left box, there is only one place for a 2. This rules out an 8 up top, which allows us to complete the top row and place a 4. This completes the 3rd column. This then completes the bottom row, placing an 8.

7:

The first column can be completed, also completing the top left box. 5 in the second row can be placed. 3 in the top row can be placed, and a 6 in the bottom left box.

This 6 means the 1 up top is wrong, and the entire top grid can now be completed, placing several numbers down below:

8:

Now we can focus on the final grid.

We can fill in the 1 in the bottom right box, the 7 can then be placed in the final column. The 6th row can then be completed, allowing the second column to also be completed. The bottom left box can then be finished. The 8th column can now also be completed.

The 5th row can be completed, as can the middle right box. Nearly there, we are at this position:

9:

From here, the grid solves with simple sudoku logic. The 4th column completes, followed by the second row, then first row, 5th column, and 4th column. This leaves just 10 remaining cells, all of which solve simply, giving the final solution: