Lately I've been enjoying a little game called Flow Fit: Sudoku.


It's basically sudoku meets jigsaw. You need to place the pieces into predetermined slots, and numbers cannot repeat in the same rows/columns. If you make a mistake, the game alerts you immediately:


And I've been wondering - is it possible to make a mistake, but not realize it until the very last piece? I cannot come up with such an example, but neither can I come up with a reason why it's impossible.

So, here's my challenge - make such a puzzle! Or prove that it's impossible. The rules:

  • The grid can be of any size
  • Pieces cannot be rotated/flipped by the player (but you can present them with pre-rotated pieces like in the screenshots above)
  • The allowed pieces are: rectangles 1x2, 1x3, 2x2; "Z" (4 squares); "T" (4 squares), "L" (3 or 4 squares). [Clarification: you can rotate/flip these shapes every possible way when creating the puzzle, but the user who will be solving your puzzle won't be able to rotate/flip them]
  • The grid must have one and only one correct solution
  • It must be possible to place the pieces in such a way, that all but one piece are placed in the grid without conflicts, but putting in the last piece would result in a conflict.
  • Optional rule: no two pieces are the same. At least, so far I haven't seen a board in the game that would include two identical pieces. However I'll also accept answers that have identical pieces.

The game also has boards with additional tweaks like holes or hints or wildcard pieces. I'm not looking for a board with them, but if that's all you can come up with, you'll get an upvote from me too (no checkmark though).

  • $\begingroup$ A couple of clarifications: (1) The image shows "S" pieces but not their mirror-image "Z" pieces; I assume both are possible? (2) What about the vertically-oriented versions of the "T" piece? (3) I take it the goal is for it to be possible to put all but one of the pieces in so that the remaining piece would physically fit but the numbers on it would violate the constraints? (So being able to fit in the other pieces but then not have the last one fit into the remaining space won't do.) $\endgroup$
    – Gareth McCaughan
    Commented Aug 25, 2021 at 11:21
  • $\begingroup$ @GarethMcCaughan - "Yes" on everything. All rotations and mirrorings of the mentioned pieces are allowed. And "Yes" about the end result too. Remember - the puzzle MUST have a unique correct solution too, otherwise this would be trivial. $\endgroup$
    – Vilx-
    Commented Aug 25, 2021 at 11:52
  • $\begingroup$ +1 from me just because I hadn't seen Flow Fit: Sudoku before. Thanks for the recommendation! $\endgroup$
    – livingtech
    Commented Aug 26, 2021 at 15:19

1 Answer 1


There is a simple, almost trivial way in which the conditions can be satisfied. The smallest example of this works on a 3x3 grid.

The board looks like this:


with the pieces

1  123  31  23


 Solution    Dead end
 231         31.  1
 312         23.  2
 123         123

You can use a similar domino arrangement as part of a larger board. It is of course very easy to avoid the dead end solution, cause the vertical domino would normally be placed before you do the others.

  • $\begingroup$ Yesss! I was hoping for a trivial solution like this! :) $\endgroup$
    – Vilx-
    Commented Aug 25, 2021 at 15:53
  • $\begingroup$ Is it simple to extrapolate onto a 9x9 board? What if the pieces must include all shapes shown in the question? (pure curiosity - great answer! +1) $\endgroup$
    – TCooper
    Commented Aug 25, 2021 at 21:42
  • $\begingroup$ @TCooper - there are too many shapes to include them ALL in a single board. Remember - each rotation/flip counts as a separate shape. So that's 25 different shapes if I'm not miscounting. I don't think you can even make a 9x9 board like that. But, yeah, it's possible to make various interesting challenges from this. :) $\endgroup$
    – Vilx-
    Commented Aug 25, 2021 at 23:11
  • $\begingroup$ It would be trivial to make the dead end more likely in a larger board with any "L, S, T or Z" piece for the B segment. And the relative positions of the 1,2,3 arrangement make it pretty simple to extrapolate without giving the trick away while placing other pieces. $\endgroup$ Commented Aug 26, 2021 at 3:25
  • 1
    $\begingroup$ @TCooper - go for it! 🙂 I'm sure there's a lot of fun to be had by trying to make various boards under various constraints. 🙂 $\endgroup$
    – Vilx-
    Commented Aug 26, 2021 at 15:29

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