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Mar 9, 2023 at 20:53 answer added Peter Taylor timeline score: 6
Mar 9, 2023 at 9:13 comment added Peter Taylor Ah, that was my stupidity. I saw that in the condition of Theorem 0.2, made a note of it, and then forgot about it when writing the code. The other thing which is missing is a gcd filter to remove e.g. 3 3 3 3 3 3 for $q=9$. I've also optimised the checking and it can do up to 300 well within the 1 minute restriction of tio.run. On a desktop it takes a bit over a quarter of an hour to run up to 1000.
Mar 9, 2023 at 7:15 comment added Ethan Dlugie @PeterTaylor thank you very much for the code implementation! And to answer your question, the parts of the partition should be strictly between $0$ and $q$. A cone point of curvature $2\pi \cdot 0/q$ is actually a smooth point, and a cone point of curvature $2\pi \cdot q/q$ is actually a cylindrical end rather than a cone point. So your example is disallowed.
Mar 8, 2023 at 23:40 comment added Peter Taylor So you want to count partitions $\lambda \vdash 2q$ such that $\lambda_i + \lambda_j < q \implies (q - \lambda_i - \lambda_j) | q \vee \lambda_i = \lambda_j \wedge (q - \lambda_i - \lambda_j) | 2q$? If so then the additional thought gets you to $q = 100$ in seconds with fairly unoptimised code. However, I'm confused as to why the appendix of [3] doesn't list 4 1 1 1 1 for $q=4$ so it may be more complicated than I've understood.
Mar 8, 2023 at 17:55 comment added Ethan Dlugie @TimothyChow in fact for a given denominator $q$, the task would be to run through partitions of $2q$. So much worse!
Mar 8, 2023 at 2:01 comment added Timothy Chow The number of partitions of 999 is 23127843459154899464880444632250 so whatever Thurston was doing, he certainly wasn't running through all these partitions one by one.
Mar 8, 2023 at 1:38 history edited Ethan Dlugie CC BY-SA 4.0
Adding an extra comment with an approach to address one question
Mar 7, 2023 at 20:10 history edited LSpice CC BY-SA 4.0
Typo
Mar 7, 2023 at 20:07 history asked Ethan Dlugie CC BY-SA 4.0