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An entry in Fortnightly Topic Challenge #47: "Wacky Sudokus"


This puzzle consists of three overlapping Sudokus, coloured red, blue and green, so each cell will contain three digits, one of each colour. Taken individually, the digits of each colour will form a Sudoku in the grid. In addition, the coloured Sudokus are mutually orthogonal: for each pair of distinct colours A and B and each pair of (possibly equal) digits x and y, there is a unique cell containing an A-coloured x and a B-coloured y. An accepted solution will provide the answer as well as the key logical steps needed to reach it. I hope you enjoy!

Grid

TEXT/COLORBLIND VERSION

RED CLUES

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|8| | | | | | | |3|
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| |3| | | | | |2| |
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|4| |2| | | |8| |6|
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| |8| |2| |7| |3| |
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| | |1| |5| |7| | |
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| | | | | | | | | |
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| |5| |7| |4| |1| |
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|9| |3| |8| |2| |4|
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| |4| | | | | |6| |
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BLUE CLUES

-------------------
| | | |1| |2| | | |
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|1| | | |9| | | |6|
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| |9| | | | | |2| |
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| | |7| | | |9| | |
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| | | |5| |9| | | |
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| | | | |7| | | | |
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| | | | | | | | | |
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| |2| |9| |8| |7| |
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|9| |8| |3| |4| |2|
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GREEN CLUES

-------------------
| |7| | | | | |9| |
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| | |6| | | |7| | |
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| | | |5| |7| | | |
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|8| | | |5| | | |4|
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|7|5| | | | | |3|9|
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| |2|4|3| |9|1|5| |
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|4| |2| |3| |5| |1|
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| | | | | | | | | |
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| | | |4| |2| | | |
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SOLVER NOTE

As clued, the individually coloured Sudokus do not have unique solutions; the mutual orthogonality condition is vital to creating a unique solution to the entire puzzle.

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  • $\begingroup$ Slightly confused about the mutually orthogonal part, to clarify, say there is a red 8 and a blue 6, the blue 8 and the red 6 are in the same place on the grid? $\endgroup$ – Beastly Gerbil Jan 21 at 15:28
  • $\begingroup$ Not quite. If there is a red 8 and blue 6 in one cell, there will be no other cell with this combination. However, there will be a cell with a red 6 and blue 8, necessarily elsewhere, in the grid. $\endgroup$ – Jeremy Dover Jan 21 at 15:52
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Should've known what I'd gotten myself into when starting, whew. Great puzzle! Really hard, but I hope I got everything right.


0th Step: Try to squeeze out everything we can by normal Sudoku first:

enter image description here

This blocks g4b9, g6b2, g5b8 and r1g1

1st Step:

G R5C5 and G R2C2 must be 4 because g4b9 is blocked. G R9C1 must be 5 because g5b8 is blocked. This gives us R1C6 = 4, G R1C3 = 5, G R2C9 = 5, G R2C6 = 3, G R4C6 = 1 by normal Sudoku. New pairs: r5g4, r3g4, g5b9, g4b2, g5b6, r7g1
With g4b2, G R3C7 must also be 4, which gives us G R8C8 = 4 by normal Sudoku. These add the pairs r8g4 and g4b7 to the collection.

enter image description here

2nd Step:

B R9C4 can only be 6 or 7 by Sudoku, but since g4b7 is blocked, it must be 6. R R8C8 is either 5 or 7 by Sudoku, but must be 7 because of r5g4. This gives us R R2C9 = 7 New pairs: g4b6, r7g4, r7b7, r7g5, r7b6, r7b8

Now R R9C3 must be 7 because of the new r7g5. Now some normal Sudoku logic for box 4, 7, 9 and 5. R R4C9 must be 1, it can't be 5 since r5g4. R R6C9 in turn is 5. R R9C4 must be 9. Added r8g2, r9g1, r3g5, r5b4, r8b2, r2g9, r1g4, r3g7, r7g6, r2g2
R R9C4 must actually be 9, because 1 is blocked by r1g4 and 3 is blocked by r3g4! Of course this adds r9g4 and r9b6 to the blacklist.

enter image description here

3rd step:

G R4C4 cannot be 2 because of r2g2, the only place where 2 can go in G box 5 is G R5C4, giving an additional pair g2b5. G R5C7 must be 8 because of r7g6. Now some Sudoku logic for green, giving us additional pairs of r2g7, r6b9, g8b7, r4b8, r1g7

G R4C7 must be 2 since 6 is blocked by the newly added r6b9. G R7C4 is 9 because 6 is blocked by r7g6. Now Sudoku rules on G R7C2, G R4C8 and G R9C8. New pairs: g2b9, r7g9, r5g6, r3g6, r6g8.

enter image description here

Last step:

At this point I'm going to leave out the pair listing and reasons, since it gets repetitive. The basic idea is to keep a list of all used pairs and remove possible candidates from a cell if it appears in one of them, after that you can basically apply normal Sudoku on it.
The rest of the green grid can be solved now (but not entirely by Sudoku logic):
enter image description here
And then also the red one:
enter image description here
Obviously we now already have a pairing for each blue cell, so we can also finish the blue grid if we just have the patience :)
enter image description here

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    $\begingroup$ Damn I was about halfway through, good job! :P This was very hard so great answer! $\endgroup$ – Beastly Gerbil Jan 21 at 20:32
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    $\begingroup$ More or less correct, but there is a typo in the red grid, upper middle square...you have two 9s. I checked everything else though, and it is just a typo. Great job! I hope you enjoyed. $\endgroup$ – Jeremy Dover Jan 21 at 20:46
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    $\begingroup$ @JeremyDover Whoops, fixed that. Must've happened in the rush when I realized that now is not the best time to get sniped :P Yes, it was very enjoyable! $\endgroup$ – Lukas Rotter Jan 21 at 20:56

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