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

Other puzzles in this series

Welcome to the fifth puzzle in this series! For more information about the series, see the first puzzle and the introduction. Enjoy!

(Reminder that clicking on the puzzle will give a grid without the background, to save you from having to screenshot!)

            enter image description here

Who said the sudoku can't involve parquetry!?


  • Normal Sudoku rules apply
  • Some cells are included in multiple rows and columns
  • The following must also contain the digits 1-9 with no repeats:
    • The 2x2 squares in the middle of the boxes
    • The L shaped 'cell' in the top left of each box
    • The L shaped 'cell' in the bottom right of each box

I really like this type of sudoku, as there is some really interesting logic that can be used.

So far, the sudokus themselves have been rather straightforward, but for this one I would like a bit more of an explanation, of both the logic used and the solution path if that's do-able!

Good luck!

  • $\begingroup$ Also, yes I have made a mistake and posted puzzle 6 as puzzle 5. Puzzle 5 will replace puzzle 6. $\endgroup$ Commented Jan 14, 2021 at 21:11

1 Answer 1


Completed Grid

Finished grid


A quick note: in each row and column of 4x4-cell squares, each number must appear in a multi-cell box exactly once. Proof left to reader.

In the bottom right square, the 6 needs to span both of the middle rows, meaning the 2x2 must be 6. Similarly, the 2 in this square must be in the upper right corner, since it can only be in the far right column; this fills the "only far column" boxes in this square, so the 1 is forced as well. With this 1 placed, we force the 1 in the lower left square to be in the upper left corner, and the 1 in the middle right square to be in the middle right box.

Using similar expanded Sudoku logic, the 3 in the middle left square must be in the lower right corner, since it has to cover both of the bottom two rows. Moreover, the 3 in the bottom right square must be in the upper left corner, since this is the only multi-cell block in the right column of squares that is not already barred from being a 3. This forces the 3 in the upper right square, and then the 3 in the upper left as well. The grid thus far:

Progress 1

Continuing in the right column of squares:

The 7 must be in the lower right corner of the lower right square, since it must cover the rightmost 2 columns, forcing the 7 in the bottom middle square. Now consider 4s: at least one of the multi-cell boxes in the right column of squares must be a 4. It cannot be either lower-right L, and it cannot be the upper left L in the middle right square, since it would block any 4 in the upper right square. So the middle 2x2 in the middle right square must be 4. We also see the 6 in this square must be in the middle left cell, forcing the 9 to be the upper left L, and also forces the 9 in the bottom right square. Continuing in the middle right square, we see the remaining boxes are 5 and 8. But the 8 cannot be in the single cell box, because then there would be no multi-cell box in the middle row of squares that could contain 8. This lets us finish the middle right and bottom right squares off. Progress so far:

Progress 2

Some additional deductions:

The only multi-cell box in the middle row of squares that can accommodate 1 is the lower right L of the middle box. All lower left Ls must contain each number once, so the upper left one must be 6, which lets us complete all the 6s. Additional Sudoku deductions let us fill more of the grid, in particular filling a 9 in the lower right L of the upper middle square. This leaves only the lower left 2x2 to accommodate a 9, and lets us finish all of the 9s and 1s. Additional deductions with standard Sudoku logic ensue, yielding:


Finishing up:

4, 5, 7 and 8 are the remaining digits. We cannot absolutely place them in the upper row of squares, but we can use color to associate which squares must be the same:

Note particularly that blue must be one of 4 or 8, while either pink or purple could be 7. Looking in the first column, we see that 7 cannot be in the upper left of the middle left square, since that would leave no place for the 7 in the upper left square, and there is already a 7 in the lower right square. So pink must be 7. The grid fills in at this point.

  • $\begingroup$ Fantastic answer! Very thoroughly explained, and nicely layed out. Hope this was enjoyable! (Also, I think that at the end of the 14 I'll award a bounty to the answer that I think is 'best', and I'd say right now this would be in the lead!) $\endgroup$ Commented Jan 14, 2021 at 20:37
  • $\begingroup$ Thanks @BeastlyGerbil! I really, really liked it! Definitely felt like there was some spatial, non-sudoku logic mixed in there, and it made for a nice variation. I've been enjoying your other puzzles in the series, just more slowly than others are :-) $\endgroup$ Commented Jan 14, 2021 at 21:16
  • $\begingroup$ Yeah this is one of my favourite variations, it has some really nice logic. There's some other really nice logic involving the single cells: for every number in a row or column, it must appear in two single cells (and one multi as you found), however, the two single cells cant be in, for instance columns 1 and 3, or columns 2 and 3 as then the multi cell needed doesn't exist. Therefore, the single cells must be either together on one side, or separated by a 2x2 box. Also, plenty more on the way still, not even halfway through! :P $\endgroup$ Commented Jan 14, 2021 at 21:20

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