You are asked to dissect an $N \times N$ square into polyomino pieces such that each piece shares a portion of its boundary with exactly $D$ other pieces, and no piece has area exceeding $N$. This can be achieved for $D \le 5$.

For $D=2, 3, 4$ the smallest such squares are of size $2 \times 2$, $3 \times 3$, $4 \times 4$, respectively:
enter image description here

Find the smallest square for $D=5$.

Credit: inspired by this puzzle.

EDIT / Bounty offer by Will Octagon Gibson:

Two answers have already been posted using 8x8 squares and 16 polyominoes.

I plan to award a bounty of 100 if a correct answer is posted using a smaller square or fewer polyominoes.

I plan to increase the bounty to 200 if a correct answer is posted using a smaller square AND fewer polyominoes.

These offers will expire on November 30, 2023 at approximately 11:59 p.m. (my time).

Good luck!

  • 1
    $\begingroup$ Wasn't this proven by Gerhard in that other puzzle? $\endgroup$
    – Quark
    Jan 26, 2015 at 7:31
  • 2
    $\begingroup$ That answer is not a square, and it can't be trivially extended to a square, because you will have pieces with an area larger than N. $\endgroup$
    – dmg
    Jan 26, 2015 at 8:39
  • $\begingroup$ @Quark - Gerhard has proven that for $D=5$ you need to dissect into at least 12 pieces. $\endgroup$
    – Johannes
    Jan 26, 2015 at 9:38
  • $\begingroup$ "and no piece has area exceeding N". Is this condition necessary? $\endgroup$ Sep 17, 2019 at 2:32

3 Answers 3


I think I've found a solution for an 8x8 square. I do not know if it is the minimum solution or how to prove that:


It was definitely fun to try and find this! Took me a while. Excellent puzzle.

Some comments on how I got to the solution (Rather a chronology than a full deduction):

- It has been proven in the other puzzle that at least 12 tiles would be needed.
- It was clear, that each piece has to have at least 2 squares.
- I was intuitively convinced that the solution would have a 4-fold rotational symmetry. (No proof for that whatsoever. More a 'feeling'.) So I set out with trying patterns in this symmetry. If I get 1/4th of the pieces in place, the other 3/4th would be correct automatically.
- I figured the rim would be the difficult part, as long-stretched tiles are needed, and their length is limited by the square-size.
- So I first tried to create a 7x7 square with fitting tiles to the border and leaving room for 5 connections.
- This did not work out (because of the symmetry centre being 'left out'). So I then retried the same with a 8x8 grid, ending up with:
- Trying to extend the colours towards the centre (always using 4-fold rotational symmetry on the extensions) I soon ran into obstacles, which forced me to 'shift' the outer rim to:
- And then it was just a matter of extending inwards and realizing that the number of tiles is not yet enough. Adding 4 colours did the trick.

  • $\begingroup$ Well done. This 16 pieces solution is almost certainly the minimum. There are alternative solutions with fewer pieces, but they typically have a 3-fold symmetry and require 3 pieces to cover the full boundary, thereby violating the requirement of the area not exceeding the linear dimension. Not a proof, but makes it pretty plausible smaller squares won't allow for a solution. $\endgroup$
    – Johannes
    Jan 27, 2015 at 15:05

It is also possible with rectangles.

square dissection or square dissection

  • $\begingroup$ Nice. Didn't see those. $\endgroup$
    – BmyGuest
    Jan 27, 2015 at 22:52

A near miss for $D=5$ and $N=7$, with $12$ polyominoes and largest area $8$ instead of $\le 7$:

enter image description here

The underlying $5$-regular connected planar graph is the icosahedral graph.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.