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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 18, 2016 at 18:16 comment added Matthew Kahle I think the only problem with the quoted construction is that the algorithm sometimes gets stuck. But as someone else pointed out, this does not mean that it is not sampling correctly––you just have to start over whenever you get stuck. All one needs to check is that every $n$-omino gets counted exactly $n$ times, which seems straightforward unless I am missing something. Still, this does not answer your questions about the expected number of $2 \times 2$ squares as $ n \to \infty$.
Jul 18, 2016 at 16:04 history asked Wolfgang CC BY-SA 3.0