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]
@scruss wait till you see the galactic distribution of hoary marmot-resembling hoary marmot maps
@ben would that we could all pull that off
I think that's as far as my brain is going with it at this point. My one lingering thought to play with is whether it'd bear fruit to treat L and J pieces as distinct after all in laying out specimens -- I might be prematurely simplifying things treating them as indistinct and keeping myself from noticing some other interesting patterns.
And once we're talking concatenation it starts to feel interesting and tingly to remember that this all started with an observation about the Fibonacci sequence, which is all ABOUT concatenating smaller values.
People capable of crunchy analysis can tackle the actual formal combinatorics here, I know better than to try and put myself and others through that. But it's neat!
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.
The other interesting thing is where totally new specimens show up -- that is, solutions that aren't a combination of two smaller solutions.
The answer seems to be "literally only in the I+L row", and only and exactly one for each successive value of n >= 3.
I've highlighted all brand-new specimens in green.
n=1 introduces the O as a 2x2 solution
n=2 has two I's and two L's respectively as new 2x4 solutions
n=3 has a new L+L+I 2x6 solution
n=4 has a new L+L+I+I
n=5 has a new L+L+I+I+I...
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.
@AmyZenunim @pronoiac without a profit motive, the shittiest dudes you ever met wouldn’t do anything. Shitty dude culture would collapse.
(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!)
