Jigsaw puzzles are awesome and making them is an easy way to get rich. So I bought a cutting machine to make puzzles. Unfortunately I could only afford the cheapest machine available. All pieces are arranged in the standard rectangular pattern, no fancy shapes or arrangements.

Additionally the machine can only do a single shape for the tap and the blank. That means any two pieces with a tap and a blank fit together.

I still want to have my puzzles a unique correct solution. So in spite of the individual pieces fitting together in lots of ways there should be only one way to arrange all pieces into a rectangle.

Question: What is the biggest puzzle in terms of number of pieces I can make?

Here are two examples of puzzles I could make with the machine. Any individual piece can be described by listing the four sides in clockwise order. Each can be a (s)ide, a (t)ap or a (b)lank.

Puzzle A consists of the four pieces (s,s,t,t), (s,s,t,b), (s,s,b,t) and (s,s,b,b). This can be assembled as a 2 by 2 puzzle in a unique way (up to rotation). This is a good puzzle.

Puzzle B consists of the four pieces (s,s,t,t), (s,s,t,b), (s,s,t,b) and (s,s,b,b). This can be assembled as a 2 by 2 puzzle in two distinct ways. So there is no unique solution and I don't want that.

PS: This can be solved with pen and paper, no computer required. If someone wants to make helpful pictures feel free to edit.

Edit in response to comments: If a piece can be rotated in place the picture would still be off so that would count as distinct ways to assemble. Sides are only on the boundary, not inside the puzzle.

  • $\begingroup$ I suppose it is not OK if a piece can be rotated ? So (btbt) cannot be used? $\endgroup$
    – Florian F
    Feb 3 at 12:23
  • $\begingroup$ Can flat sides appear between 2 pieces? $\endgroup$
    – Florian F
    Feb 3 at 12:24

2 Answers 2


There are only so many types of pieces, and we can only use one of each to have a unique solution. Starting with some (lol) edge cases:

ssss is possible, but would be a 1-piece puzzle.

sssb, ssst, stst, sbsb, and stsb could make a 1x5 puzzle, which improves on the 2x2 in the original post. (sbst is equivalent to stsb by rotation.)

Now that that's out of the way, we have:

corner pieces: sstt, sstb, ssbb, ssbt. There are four, which is good, since four is all we need.

edge pieces: sttt, sttb, stbt, stbb, sbtt, sbtb, sbbt, sbbb. There are eight, which is enough to make a 4x4, 3x5, or 2x6 puzzle, giving us a theoretical maximum of 16 pieces.

internal pieces: tttb, ttbb, tbbb. (There is also tttt, tbtb, and bbbb, but as pointed out in comments, those are all rotationally symmetrical, so would not provide a unique solution.) There are only three, so we can't fill in the 4x4, but we can fill in the 3x5.

With these pieces, the largest puzzle we can make is a

3x5, consisting of 15 pieces:
completed puzzle of 15 pieces

  • $\begingroup$ bbbb and tttt are both rotationally symmetrical, so cannot be used. $\endgroup$
    – fljx
    Feb 3 at 15:35
  • $\begingroup$ @fljx d'oh. of course they are. updated accordingly. $\endgroup$
    – juicifer
    Feb 3 at 15:57
  • 1
    $\begingroup$ The question asks for the largest set of pieces that have a unique solution. That set of fifteen has multiple solutions - e.g. rotate the centre three pieces 180 degrees as a group. $\endgroup$
    – fljx
    Feb 3 at 16:07
  • $\begingroup$ This looks pretty good but @fljx comment is correct. This jigsaw still admits multiple solutions. Nice picture by the way. $\endgroup$
    – quarague
    Feb 3 at 19:00
  • $\begingroup$ so it does. back to the drawing board, I guess $\endgroup$
    – juicifer
    Feb 3 at 20:16

Partial answer, providing an upper bound

There are 4 possible corner pieces and 4 corners to fill, so all possible corner pieces must be used. There are 8 possible edge pieces, so the perimeter of our completed puzzle cannot exceed 16, which limits the possible sizes to $4×4,3×5,2×6,3×4,2×5,3×3,2×4,2×3,2×2$. (We can easily exclude $1×n$ puzzles.)

We will now show that the rectangle cannot have perimeter 16.

It is easy to show that the subpattern formed by corner pieces is always distinct up to rotation from its reflection except in square puzzles where the only exceptions place the two-tap and two-blank pieces at opposite corners. If the perimeter is 16 all 8 possible edge pieces must be used, but reflection forms an automorphism of this group; all interior pieces are also reflection-symmetric. A perimeter-16 rectangle thus cannot possibly have a unique solution from this reflection argument, except perhaps in the 4×4 diagonal-symmetric case. That exception does not lead to a valid puzzle either, since the two interior pieces not on the line of symmetry would have to be identical. Hence we eliminate the $4×4,3×5,2×6$ cases.

  • $\begingroup$ Which reflection do you mean? For rectangles you can reflect along a horizontal or along a vertical, for squares you reflect along the diagonals as well. $\endgroup$
    – quarague
    Feb 3 at 18:59
  • $\begingroup$ @quarague All of them. $\endgroup$ Feb 3 at 19:04

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