buster

@ohad@mathstodon.xyz @MartinEscardo@mathstodon.xyz Definitely, yes, with h=2 in that paragraph. Not a big deal at all, but I stumbled on it. I would certainly get more seriously lost a bit later on, but that's hardly the author's fault 🙂

@MartinEscardo@mathstodon.xyz He lost me at if |A|=k, then \(2k-1\leq|A|\leq\binom{k+1}2\)

@peterluschny@mathstodon.xyz @leonardom@mathstodon.xyz This paper ([1]) discusses algorithms for generating all labelled graphs on 𝑛 vertices where the vertices have given degrees 𝑑₁,…,𝑑ₙ. The 'no isolated vertices' and 'friendship paradox' conditions depend only on the degrees, so we can generate all lists of degrees, check the conditions, and if they are satisfied, generate and count the graphs and add the count to an accumulator. We can exploit symmetry a bit by considering only lists 𝑑₁≤…≤𝑑ₙ and multiplying the count by the number of distinct permutations of the list. [1] https://www.sciencedirect.com/science/article/pii/0377042791901173

@gregeganSF@mathstodon.xyz I'm lazy, so I put a huge amount of effort into montecarloing it, was surprised, and only then thought (just) hard enough to see that it was (probably) ok to use affine transformations to reduce to the case of the equilateral triangle.