This is an offshoot of this question.
It is clear that there can be $3^4=81$ different types of pieces. I was wondering, given one of each type (and with no rotations or flips allowed), whether it was possible to create a standard jigsaw puzzle, using just those $81$ pieces. By "standard jigsaw puzzle" I mean one of dimensions $m \times n$ where all perimeter sides are straight.
A $1\times 81$ puzzle is clearly not possible as it would require all $81$ pieces to be straight on both the left and the right side. A similar argument holds for a $81\times 1$ puzzle.
A $3\times 27$ puzzle would require $27$ pieces with a left straight side and $27$ pieces with a right straight edge, and while it is true that such two sets exist, they have an overlap of $9$ pieces. This is therefore not possible. As before, a similar argument holds for a $27\times 3$ puzzle.
This leaves the $9\times 9$ possibility. A priori, I can see no reason this shouldn't be possible. And I think I have a proof that it is possible, which is what I would like your opinion on.
Each black side of a piece is fixed, but each green side of a piece represents a possible connection type. A connection type could be a "straight - straight", "concave - convex" or "convex - concave" type. It seems to me that if we run through every possible connection type for each piece, one of those scenarios must give a puzzle where each of the $81$ piece types is used exactly once.
Am I right?