If you happen to own two compatible sets of heptominoes (and hey, who doesn't these days?), then by discarding the duplicate mirror-symmetric pieces and turning a bunch of the remaining bits over so you've got distinct mirror image pairs, you have yourself a set of the one-sided heptominoes. Difficulty-wise they seem to slot in somewhere between heptominoes and octominoes - the main challenge now being resisting the urge to flip over pieces. The total area they cover is 7x196 = 1372 but as one of the pieces has a hole an extra unit cell must be accounted for when deciding on shapes to fill.

Anyway, I found this 32x43 rectangle, just since it was my first time doing anything with the one-sided heptominoes and I didn't want to take on anything too elaborate straight out of the gate. The first two-thirds of the solution process feel pretty much the same as solving with the standard turn-over-able heptominoes, but when you get right down to the last few bits there are a lot more near misses - times where the remaining hole could be filled with the mirror images of the remaining pieces but not the pieces I was holding.

Fig. 1: A rectangle with the one-sided heptominoes. As far as I know there are no pieces which have been accidentally turned over, but I can't say for definite and I can't be bothered to check either.

If you remove the harbour heptomino from this set then you are left with 195 pieces and a total area of 1365 units. This can make (among other things) a 37x37 square with the four corner pieces removed:

**[ Home > Polyominoes > One-sided Heptominoes ]**

Lewis Patterson. Last updated 08/05/21.