yo can I bum one a those Evil Renees
Josh "cortex" Millard
@joshmillard
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Makes art and stuff, used to run http://www.metafilter.com. he/him, Portland, OR, USA. email at [email protected]
The implication of the uniqueness of this one specimen is that *every other solution* for n >=3 is simply the concatenation of two or more previous solutions for smaller values of n. Which makes sense: other than this one configuration that constantly overlaps its layers of I shapes, every other solution ends in a nice flat wall on both sides. They can be mashed together arbitrarily to create a longer solution made up of distinct unrelated parts.
And that just keeps going. For every n >= 3, there is a new specimen that follows this pattern of 2L + (n-2)I, just sticking in a new I piece each time. You could think of it as flip-flopping a bit for each n, or as extending two related odd-n and even-n configurations in a more straightforward "add to I's" way.
Coming back to this to look at some patterns and make a couple more notes.
First off: for odd n, there's never going to be a solution that uses ONLY L or ONLY I shapes. Which makes sense: for a 2x2n rectangle for odd n, you'd need to arrange n-1 I's or L's into most of the space (and the only way we know how to do that is 2x4 blocks) and then fit one extra one into...a 2x2 space. Oops! So for odd n, those specimens just will never exist.
(as always, I manage to goof when doing this stuff on the fly; there are in fact [at least] eleven specimens for n = 4, missed a I + L variant in my haste!)
Once we hit n = 4, we're in full swing: there's finally specimens for all seven combinations of pieces, and ten total unique solutions controlling for symmetry.
I have to stop and pay attention to work for a bit, but one of the things I'm excited to fiddle with more is looking at where genuinely new non-decomposable patterns show up (like the sixth one down on the left in this image) where it's not just a concatenation of previously attested smaller specimens.
(If we don't control for symmetry on the n=3 specimens, we get the nine total that we see in Mark's set, the results of flipping some of these horizontally and or vertically.)
Does this mean we don't ever have to care about the mirroring of L pieces, tho? Nope! We just had to yet. That changes immediately with...
n = 3
For which we have four solutions, controlling for symmetry (several more that I've ignored if you don't, my set will keep diverging more and more from Mark's because of this). The bottom-most of which is a case where we have BOTH versions of L/J in play -- there are times when that WILL matter!
There's also no single-piece solutions except 3 O's.
The first couple of values of n are very trivial:
n = 1 (so 2x2n = 2x2) the only solution is a single O.
n = 2 (so 2x4) has four solutions: a pair of O's, a pair of I's, and two different orientations of a pair of L's.
But me being me, I want to avoid repetitions across mirror/rotational symmetry, so I'm going to decide that second pair-of-L's isn't interestingly different from the first one. I'm only going to count that once. So really we have three n=2 solutions, controlling for symmetry.
