Suppose that a 5x5 square has been tiled with five (not necessarily distinct) pentominoes. Is it true that there will necessarily exist at least one Latin square of size 5x5 (using the numbers 1,2,3,4,5) that can overlay the 5x5 square of pentominoes such that each pentomino contains all the numbers 1,2,3,4,5?

  • $\begingroup$ Incidentally, there is a name for this type of puzzle (when it is possible). It would be called a "Jigsaw Sudoku". $\endgroup$ Jul 11, 2022 at 15:24
  • $\begingroup$ @DarrelHoffman In fact, I have made a number of these Jigsaw Sudoku puzzles (always size 5x5). I call them Pentadoku puzzles. $\endgroup$ Jul 11, 2022 at 19:09

1 Answer 1


It's not always possible; one counter-example is:


Let's try to find a Latin square orthogonal mate. By symmetry, we begin with


We are forced to place a 1 and a 5 as follows (considering the pentomino labelled "c"):


Next, we are forced to place a 1 and a 5 as follows (considering the pentomino labelled "b"):


And now the Latin square property implies the cells labelled x above contain only 5s and 1s, but this violates orthogonality with the pentomino labelled "e".

As an aside, a similar argument works for 4x4 squares:

aaab   1234
acbb   41..
ccdb   .4.1
cddd   ..x.

(can't fill in the cell marked x with both a 1 and a 4) and 6x6 squares

aaaaab   123456
acbbbb   61....
ccdddb   .6...1
cddeff   ..xxx.
cdeeff   ..xxx.
ceeeff   ..xxx.

(at least two 1s or two 6s will intersect the e's) and 7x7 squares

aaaaaab   1234567
acbbbbb   71.....
ccddggb   .7....1
cddeegg   ..xxxx.
cdeeeeg   ..xxxx.
cdefffg   ..xxxx.
cdffffg   ..xxxx.

(at least two 1s or two 7s will intersect the f's).


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