Colombian Sudoku, Cows and Bulls again…

Question

Fill the sudoku on the left so that each column, row, box, and colored box of the same color contain the numbers 1 to 9. The dots show how many numbers in the row or column of the grid on the right match the solved sudoku on the left.

Answer 1

Completed grid:

As usual, we begin by using our given digits to mark cells which match and cells which cannot match. We then:

Look in column 3. We need 5 matches, but three cells (R1, R3 and R7) cannot match, and only one of R4C3 or R8C3 can be a match, so all of R2C3, R5C3, R6C3 and R9C3 are matches, and we can fill digits. Now note that the only candidates for R1C3 are 3/4/7: 7 is blocked by the region, and 3 is blocked because one of R4C3 or R8C3 must be 3, so this digit is 4, and we can mark a 3/7 pair in the column. The new digits also give us some known non-matches in the grid. Progress thus far:

As usual, I forget that the lack of a clue in column 8 actually means there are no matches in that column. So let's mark that. Now look at row 8:

We need four matches, and while there are a lot of possible cells that can match, there are only 4 digits: 3/4/6/7, thus each digit must be in a matching cell. In particular, this resolves the 3/7 pair in column 3. Moreover, R8C4 must be 6. Now we can use purple colored region to force R7C5 to be 6. This forces R5C5 to not be a match, and now we can look at matches in C5. We need 3 more, but only one of R3C5 or R6C5 can be a match, so R1C5 and R4C5 must be matches, allowing us to fill digits. Marking the now forced non-matches, we have the following grid:

Now in column 4:

We need three more matches in four candidate cells; one of R2C4 and R9C4 must be 2, so R3C4 and R4C4 must be matches. Also, using the purple box we know R6C4 must be 5 (actually had that a bit ago). By sudoku, R8C9 is 5, R3C6 is 5 (giving R3 the two matches it needs), as is R4C1 and R2C2. All 5s are placed. Now in row 2 we need 4 matches, which forces R2C1 to be 6. Sudoku then forces R9C2 to be 6. Now the second match in row 9 must be R9C4, a 1. Now a bit more sudoku: R1C4 and R2C4 must be a 3/8 pair, which resolves, leaving a 4/7 pair below and a 1/9 pair in the region. Progress:

Now in row 8:

We need four matches, which forces R8C7 to be 4. We have all of our matches in C1, so the rest are non-matches. In column 5, we must have 4 matches, forcing R6C5 to be 4, which resolves the 4/7 pair in C4. Now in column 1, irregular sudoku forces R1C1, R2C1 and R3C1 to be 3/6/9, giving us a 3/9 pair. We can pencilmark column 2 as well. But now some small things I missed: the 3 in R5C2 looks right, resolving R5C5, and R9C6 must be 4 by sudoku, and this forces R5C1 to be 4. Looking at the blue region, we see the 2 in R3C4 forces R1C2 to be 2. Progress:

A little more Columbian sudoku:

Column 2 needs 3 matches, resolving the 1/4 pair, and forcing R6C2 to not be 7, since the 7 must be a match in R7 or R8. Now in column 6, the 7 must also be in R7 or R8. However, it cannot be in R7C6, for if it were, neither R7C6 nor R8C6 would be a match, leaving only 3 possible matches in the column. So R8C6 is 7, forcing R7C2 to be 7. Now sudoku in the column forces R7C6 to be 8. In R7, this leaves 1/3/9 to be placed, but the 3/9 pair in C1 forces R7C1 to be 1, leaving a 3/9 pair in R7. This also forces R8C8 to be 1, which looks up to resolve the 1/4/8 triple in row 3. Progress:

The green region:

The 7/8 in R7C6/R8C6 must appear in the lower right region in R9C8/R9C9, and are resovled by the 8 in row 3. Sudoku places the 2 in R9C1. Row 8 and the lower left corner region now resolve. We need three matches in row 6, forcing R6C7 to be 2. Sudoku places the last 7 in row 1, which gives a match, and there are no more matches in R1. R2C7 can only be 1 or 9, giving a 1/9 pair, and a 2/4 pair in the remaining cells in the row. Sudoku forces R5C7 to be 8. And we can pencilmark everything at this point, giving:

Hopefully finishing up:

The green region comes to the rescue. R6C6 cannot equal either R6C8 or R7C8, forcing it to equal R7C9. This forces both cells to be 3. This gets us most of the way home, but we are left with a small tangle which is resolved by noting that Columbian sudoku in R4 forces R4C9 to be 2. The rest resolves trivially.