Here is one more Colombia Sudoku devised by Xavier Castillo. The usual sudoku rules apply.

The dots outside the board above 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 below.

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  • 1
    $\begingroup$ Just to clarify for myself, the one dot above column six indicates that exactly one cell in column six of the top grid will exactly match in the solution in the bottom grid. $\endgroup$
    – LeppyR64
    Jan 29 at 12:12
  • 1
    $\begingroup$ @LeppyR64: That is indeed the case. $\endgroup$ Jan 29 at 12:14

1 Answer 1


Breaking in:

Row 8 is a good place to start, because it has 4 matches and that target grid only has 4 different digits in the row. This immediately places R8C2 as 3, and R8C1 is forced to be 2 by the already existing 2 in column 9. We'll also go through the grid and shade known matches with the target grid green, and cells that cannot match red. By doing this coloring, we see that in column 8 there are only three possible matching cells (R2/5/8), and of these the target digits in R2/8 are both 8. This forces R5C8 to be 7. The grid thus far:

Initial Grid

More similar logic:

Row 3 also gives us a corner to pick at. We must have four matches: one is in C7, and C3/4/6/8 cannot be matches, so three of C1/2/5/9 must match. But the target digits for C1/2 are both 6, forcing R3C5 to be 3 and R3C9 to be 9. Now in column 9, we must have four matches: one is in R3, and of the remaining five possible locations, for two the target digit is 8, and for two the target digit is 5. So the remaining possibility must be a match, forcing R5C9 to be 1. This gives us three matches in row 5, forcing all of the remaining cells in that row to not match. At this point, I realized I could have made the same leap in column 1 earlier, but now's not too bad, since it forces our fourth match in row 3, in R3C2. The grid thus far:


Moving on:

In column 4, we need three matches in five possible locations. There are two 8s and two 7s in the target digit set, forcing R7C4 to be 3. That seems to be all we can force from matching logic, so let's try some sudoku. The 2 in column 2 must be in row 1. The 3 in column 3 must be in either R1 or R2, but some interesting match logic resolves it. If the 3 is in R1C3, then neither R1 nor R2 is a match in this column, which means all of the rest of the cells must match. However, this would force two 9s in the column. Thus the 3 must go in R2C3. Continuing with 3s, we see the only place in box 5 for the 3 is R5C6. We're close on row 3, so let's pencilmark candidates. Note that beyond basic sudoku, we can remove 8 as a candidate from R3C8 by matching logic in column 8. The grid thus far:


When the going gets tough...

There's some weird, but interesting logic in column 3. Column 3 has six matches, and thus three misses: one is in row 3, and two of the remaining candidates have target digit 9, so at least one of those is a miss. For each of the digits 1,5,6,7, they cannot go in a position where any of the others is in the matching grid, since that would force two additional misses. Moreover, 9 cannot go into any of these places either, since that would force three additional misses. This doesn't break anything open yet, but it does give us some restricted candidate sets in this column.
We can finish pencilmarking in box 1, noting that the 9 in box 1 must be in column 1. This blocks 9 from being in R5C1, and the miss in that cell means it cannot be 4 either, leaving only 5 and 6 as candidates. We then pencilmark across row 5. This doesn't seem to be getting us far...need a big idea. The grid thus far:


Hopefully a big idea:

Let's look at row 1. There are five possible match candidates to give three matches. Two of the target digits are 8s, and two 5s. This does force the remaining digit to be a match, forcing R1C7 to be 4. Not only does this remove a pencilmark from R1C3, but it forces it to be 5...if it were not, then it must be 8, which would force three more misses in row 1. And THIS is the big break we're looking for, because box 1 resolves completely now. We can also resolve all of 3s in the grid, since the 3 in row 1 can only go in C9. Now we are left with only one candidate for the remaining needed match in row 1, which forces R1C4 to be 8, which completes the row. The grid thus far:


Doing some clean up:

Now we need to go clean up our potential matches, and this yields two new deductions. First, in row 8 we're forced to have a 7 in C3, as the other two target digits are both 8s, and we must have two more matches. Second, in column 9, we must have an 8 in R6C9, as the other two target digits are 5, and again we must have two more matches. Now back to sudoku, the 8 in R1C4 looks down to R5, forcing a 9. This leaves a 5/6 pair in row 5, which then forces a 4/8 pair. Sudoku forces the 9 in box 6 to be in R4C7, which is now a miss. This forces R7C7 to be a 2, since it must be a match to get three matches in the column. Sudoku places the 6 in box 6, which resolves the 5/6 pair in its row. Pencilmarking in box 3 shows that the only candidate for R2C9 is 5; sudoku, combined with match logic on row 2 resolves the box, as well as column 7, and placing the 8 in box 9, which is a match. The grid thus far:


Hopefully finishing up:

In column 6, we must have a single match, and the target digit for both remaining candidates is a 4, forcing R8C6 to be 9. This looks up and resolves box 2. Pencilmarking in column 6, with the just-noted restriction on 4s in column 6 forces R7C6 to be 1. The resulting 4/8 pair in box 5 forces R6C5 not to match. Good thing I looked there, because now I see the 6 in box 5 looking left to resolve column 1. Looking in row 4, we see we have only one place for the 6 in C3, which resolves column 3 as well. The remainder resolves with sudoku. The final grid:

Final Grid


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