If you happen to own two compatible sets of heptominoes, 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. (I suppose you could just discard the holey heptomino and see what you can do with an area of 1365. Maybe a 37x37 with 4 corner squares removed?)
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.
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| 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. |
In other news, I now have a website over at polyominoes.co.uk. Right now it's a complete mess and still very much under construction, but eventually it will have a lot of the information from this blog on it, just tidied up and organised in a slightly more logical way. Posts sorted chronologically make sense for some things but it also makes looking for specific things a complete nightmare, even when I manage to remember to tag them correctly.
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