@J.A Presumably you would be more interested in the flat torus (obtained by abstractly identifying opposite edges of a parallelogram, or from the Clifford embedding in $\mathbb{R}^4$) rather than the (non-uniform) surface of a doughnut?

I suppose it's subjective what you mean by 'shape'. Whilst the open and closed unit discs are distinct as sets, I would say that they are the same 'shape'.

(Upvoted because the FFT example is a good one.) Archimedes did not discover integral calculus! He had a 'method of exhaustion' to determine the area of a region bounded by a curve, which is similar to Riemann-Darboux integration. However, this is not 'integration' until coupled with Descartes' idea of representing algebraic functions as curves.

Yes, we can think of it as choosing a random quaternion q according to a spherically-symmetric (Gaussian) distribution. Then the formula for Uniform[-1, 1] is just one coordinate of the image of q/|q| under the Hopf map, and we're done by Archimedes' theorem that the orthogonal projection of a uniform distribution over S^2 is a U[-1, 1] distribution.

I can confirm that your `smallest examples' are the only ones with orders below $10^{10}$.

This reminds me of one of Professor Imre Leader's example sheets, which features the question "what can you infer from the previous question about the lecturer's ability to typeset matrices?". Another one, interspersed with serious questions asking for proofs of various equivalences involving the well-ordering principle, was "what's yellow and equivalent to the axiom of choice?".

Actually, arc-transitivity isn't enough, either; the Dyck graph lacks your property (and has just 32 vertices).

Well, the counter-example shows that vertex-transitivity is not enough. My friend and colleague Gabriel Gendler proved that Kneser graphs and cycle graphs have your property, so maybe arc-transitivity suffices. en.wikipedia.org/wiki/Arc-transitive_graph

"finite periodic series of steps up and down" -- should that be `up and right'?