It turns out there are a family of shapes that approximate hexagons with angles of 90 and 135 degrees that are just as versatile as rectangles, if a little less pleasing to the eye (especially if you're not too fussy about them having a few holes in 'em.) These also have an added bonus: the diagonal edges make these a slightly more challenging solution than squares and rectangles.
Again, odd x odd bounding boxes seem to work best with this, because an even width or height would mean a line of symmetry bisecting the entire shape into two congruent 110-cell sub-shapes, which would mean the overall shape has balanced checkerboard parity. So with the aid of a hideous and confusing spreadsheet I knocked together, I found the following set of solvable problems:
3x73 - 9 holes
Yeah no. I think we can generally rule out the possibility of any purely 3xn constructions with hexominoes. I roughly outlined my reasoning here (but that's not a rigorous mathematical proof so it could well be wildly incorrect...)
5x54 - 3 holes
Pros: At this width solving these shapes feels not much different from solving a normal rectangle.
Cons: Solving a width 5 rectangle isn't exactly a walk in the park.
7x35 - 11 holes
At this point there's a sort of 'choose your own difficulty' option built in - with the difficulty levels being hard, harder and hardest, naturally. Two immediate options for arranging the 11 internal holes are having them as one big 11-omino, or as a line of 11 individual monominoes. The long 11-omino option creates two stretches of treacherous 3xn space which takes a bit of doing:
Whereas the individual holes introduce all manner of feisty challenges.
This one's left as an exercise for the reader. I spent a little while trying to solve it and got fairly close, but not close enough. I think I gave it like 45 minutes or so and wound up with like 6 pieces remaining a few times. But they were never nice pieces.
9x29 - 11 holes
This one was tough, although in hindsight this could be because of a complete lack of solving strategy. The obvious (looking back now) thing to do would have been to solve the ends first, then either the top half or the bottom half of the middle (if we think of the row of holes as a dividing line), trying to leave a fairly convex-looking endgame to be filled in last with the 2x2-block-y pieces. Instead I just went at it the way a nine-year old would his first all-you-can-eat buffet, and I paid for it dearly by spending maybe close to two hours (!) getting those last few bits in.
You know you've utterly stuffed it when you're down to the last five pieces and one of them is the stair-step hexomino (as it's known in Conway's Game of Life terminology).
11x25 - 5 holes
This is the start of the sweet spot, so to speak. Wide enough so that the central holes don't get in the way too much, but not so wide that there's miles of diagonal edge to contend with.
Of course, the fact that there's only 5 holes maybe makes this a little easier too. You could just as easily do this with the holes arranged vertically. Or with one bigger hole shaped like a pentomino of your choice.
13x23 - 5 holes
Can be done with the five holes the other way on. (So can the 11-height one but I didn't think of it at the time.)
I think there is a 15xn possible but it's going to have like 17 holes or something, and I think there's a point where unless you can arrange them in a particularly pleasing way it just gets silly. So I'll skip over this and go straight to...
17x21 - 3 holes and 19x21 - 9 holes
No solution images here, for the simple reason that I just plain can't be arsed. If you want them that badly, acquire a set of hexominoes and have a go at finding them yourself. More fun than looking at a picture of someone else's solution, guaranteed. Or feed them into a solver program, they'll eat these for breakfast.
And finally, 21x21 - 11 holes
This is just a rotated square at this point. And I've seen a solution for this on the internet somewhere, with the 11 holes as the straight 11-omino dead centre. As far as I can remember it was illustrating something to do with parity imbalance, and this was the example where the difference between black and white squares is exactly ±22, the maximum permissible. I assume that the restrictions that places on parity-imbalanced pieces means it's a head-bendingly difficult solution.
[EDIT: The solution I saw somewhere was here, on page 9 of Chessics, issue 28]
(The above solution was found using a computer search. Because it sounded like it might be difficult to do, and I'm always one to back down from a difficult challenge when one presents itself.)
Perhaps this was the URL:
ReplyDelete[http://www.mattbusche.org/blog/article/polycube]
...if not, it still might contain some ideas of interest ... ?
Yes, that was the site. Thanks for finding it! It looks like the diamond in their example is a slightly different design, I must have mis-remembered it. Both shapes have the ±22 parity though.
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