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This Colombian Sudoku is by Xavier Castillo. Usual Sudoku rules apply. Additionally, digits on each of the two diagonals must all be different.

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.

enter image description here

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    $\begingroup$ Sudoku Pad link for solving grid, for interested solvers: sudokupad.app/f52n5a5g2s $\endgroup$ Apr 18 at 18:09
  • $\begingroup$ @JeremyDover do you have the sudokupad app installed or did you make that puzzle on the website? $\endgroup$
    – ACB
    May 10 at 10:52
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    $\begingroup$ @ACB Probably the easiest way to do this is to create the puzzle at sudokumaker.app, which has a button to automatically export to sudokupad. You can do the puzzle at Sudokumaker as well, but I like the Sudokupad play engine a bit better. $\endgroup$ May 10 at 12:15
  • $\begingroup$ @JeremyDover ah, thanks. I didn't know about sudokumaker.app. $\endgroup$
    – ACB
    May 10 at 13:57

2 Answers 2

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Here you go:

Solution to the sudoku

Now some precisions:

The most interesting techniques used in the process was thanks to the added constraint of correspondance with the underlying graph. Obviously, you can sometimes infer directly by counting the red and green remaining. But the trick you can use when this is not enough to pinpoint colors is to instead look at groups of cells that have the same number in the underlying grid. If you have as many remaining greens as groups on a particular line, you can infer that there is a green in each one of them, thus the numbers corresponding to these groups are not somewhere else on the line. It can let you fill a number in single cell groups. The steps taken were a back and forth between the classical sudoku rules and these new one to pinpoint numbers.

I am editing this in to make up for my unknowing of the logical path rule (at the time I took on this puzzle the grid-deduction tag was not present and so I was oblivious to the logical path rule) and the fact that I can't easily go back and detail every step of my solution, without spending a considerable amount of time, comparable to the time it took me to solve it. It was definitely more difficult than I thought, thanks for sharing it, that was really fun to solve. I've put in green the cells that correspond with the underlying grid and in red those who differ for the sake of readability.

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    $\begingroup$ It is a generally accepted rule on PSE that when it comes to grid deduction puzzles, some kind of solution path is provided. $\endgroup$
    – PDT
    Apr 19 at 6:09
  • $\begingroup$ Oh. I did not know that. Now it feels super bad knowing that I had it but because I did not write down every deduction it is basically worthless. $\endgroup$
    – Fluorine
    Apr 19 at 8:53
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    $\begingroup$ @PDT This was a guaranteed bait for new contributors. I had no way of knowing that rule before hand. Maybe there should be some indication somewhere either in the question or in the tag description to avoid such awkward cases. Let me tell you it is discouraging to know only an experienced user could've gotten it no matter what because of a non written rule. And it makes me wonder for any new solution I post on the site will I get topped for a similar reason. $\endgroup$
    – Fluorine
    Apr 19 at 9:08
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    $\begingroup$ Tell me about it I had a similar experience a few years ago it’s all part of the process 😃 $\endgroup$
    – PDT
    Apr 19 at 9:36
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    $\begingroup$ Truthfully, it doesn't have to be. I include a blurb explaining that I what a solution path whenever I post a [grid-deduction] (e.g.) $\endgroup$
    – bobble
    Apr 19 at 13:49
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We begin my identifying whether the given digits match the given grid, and also locate cells that cannot be matches in other locations based on these digit placements...matches are in blue, known cells that cannot match are in red:

Initial match assessment

Now let's look at basic counts:

Column 6 provides the break-in. We need six matches in the column, and there are only six possibilities, so we can fill in the match candidates, and mark any new forced non-matches. We'll pencilmark the remaining cell candidates, since we get some sudoku and diagonal reductions. In column 7, we are permitted only one match which occurs in row 7, so we can pencilmark the remaining cells in this column...no immediate payoff.

Looking in row 6, we see we must have two matches with four possible locations. However, three of those locations have match digit 8, which can contribute only one match. First, this forces R6C4 to be 7; but also it forces R6C6 to be 4, since it cannot be 1 by diagonal rules, nor 8 since that would yield just one match in the row. The grid thus far:

Progress 1

Some more column counts:

Column 4 now has its two matches, so we can pencilmark. Column 9 needs four matches, with six possible locations. Two of these need an 8, two need a 6, so the 9 and 1 must be matches, and one of the 8s and one of the 6s will be the other two matches. Now row 3 has its two matches, and we can pencilmark accordingly. This puts pressure on column 5, where we have six locations for three matches, but three are 8s, two are 2s, and one is a 7, so the 7 is placed in R1C5, which by sudoku forces R1C5 to be 3. We can pencilmark column 5, noting that the 8 must be in a matching location, as well as the 2s. The grid thus far:

Progress 2

Some pencilmarking with deeper logic:

In row 8, we need 5 total matches. We've matched the 5 in column 6, so we need four matches from the digits 2/6/7/8/9. We need to be careful how we fail to match one of them, because if (for example) we place a 2 in the only place a 6 can match, we've violated the rules, because now the 2 won't match AND the 6 won't match. We can use this to pencilmark row 8 a little bit more granularly. In particular, this removes 2/6/8/9 as a pencilmark from R8C1, 2/6/7/9 as a pencilmark in R8C, 9 from R8C5 (many other digits were already removed), and 2/7/8/9 from R8C9.

And I'm dumb:

Because the lack of dots in row 2 indicates that there are no matches, rather than that the clue is not given. BEtter go back and fix that...now what? Well, we can pencilmark R2C8. It can't be 3/9 by sudoku, 6/7 by diagonal rules, and through sudoku/diagonal it sees both legs of the 1/8 pair in column 6. So this cell is 2/4/5. The grid thus far:

Progress 3

Some small ball:

R5C1 cannot be 9, and thus cannot match. If it were, then R5C5 would be 8 and thus also match, creating too many matches in row 5.

And a bigger break:

R4C6 cannot be 8. If it were, sudoku in box 5 would force R4C4 to be 2, which would leave only two places for a match in row 4. Hence R4C6 must be 1, resolving the 1/8 pair in C6.

An inference chain:

R5C5 cannot be 9. If it were, focus on R5C3 with pencilmark 4/5/8/9. It cannot be 9 by sudoku. It cannot be 4, since if it were, we would only have two matches in row 5. It cannot be 8, since the 9 in R5C5 prevents either 9 in row 8 from being a match by diagonal sudoku, forcing R8C3 to be 8. Hence R5C3 must be 5, but now we have four matches in column 3, a contradiction. Hence R5C5 is 8, which is a match and gives us all matches in row 5, allowing us some pencilmarking in the row. Moreover, the 9 in on the off diagonal must be in box 7, removing 9 as a candidate off the diagonal. The grid thus far:

Progress 4

An interesting deduction:

Look in box 7. Regardless of position, the 7 and 9 in this box must be a match. Moreover, row 7 can provide at most one match in these columns, and box 4 can only provide at most one match in these columns. Since there must be four matches in these two columns, these figures must be exact. So either the 2 or 4 in columns 1/2 of row 7 must match. Hence by counting, there's no match in the rest of row 7, which lets us remove 2 as a candidate from R7C8, and 8 as a candidate for R7C9. This latter is most important, since it leaves 2 and 4 as the only candidates for this cell. But the other 2/4 must appear in box 7 as one of the matches. Hence we can remove 2 and 4 as a candidate from any other cell in the row.

But now we have pressure in column 9. We need two more matches, and the three available locations have two 6s and an 8, so the 8 must match. This means R6C2 cannot match, which by the above forces R4C1 to match. The grid thus far:

Progress 5

Another (shorter) inference chain:

R9C3 cannot be 5. If it were, then we would have 3 matches in C3, forcing R8C3 to be a non-match. This forces R1C3 to be an 8, which restricts R3C2 to 4/5. It also forces R8C1 to be a match, namely 7, restricting R3C1 to 4/5. But this gives three cells in R3 restricted to 4/5, a contradiction. Hence R9C3 is not 5, and not a match, and we conclude R8C3 IS a match, being 8. This gives us a 1/2/4 triple in row 7, which places the 8 and 9. Sudoku lets us place all the 8s in the grid, and we now have a 1/4 pair in box 8, forcing R8C5 to be 2, a match. After some sudoku reductions, we get the following grid:

Progress

Finishing up, I hope:

The 5 in column 3 must be in box 1, resolving row 3. The diagonal then looks into box 7, placing the 5 and 9. In row 9, Colombian rules force R9C2 to be 7, resolving row 9 and thus box 9. Colombian rules place the 1 and 2 in row 7. With the final Colombian deduction that R4C8 must be 2, the rest finishes with basic sudoku. The finished grid:

Finished

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    $\begingroup$ D'oh! Realized I got sniped after posting. But I'll leave mine here as it provides a logical path. $\endgroup$ Apr 19 at 3:47

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