Post #2672369
2026-03-10 15:44 UTC
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.
Replies (1)
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@joshmillard@mastodon.social 2026-03-10 15:51
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.