I hope that whoever conjectured the heptagon was the worst possible convex shape for dense packing would have considered small perturbations like this.

]]>I have a combinatorial question: is it known how do the number of different paths joining the finer and the coarser partition grows with n? Each path correspond to a different sequence of “splittings” of the whole….

Thanks!

]]>(Every edge in a normally colored graph is normal, i.e. it uses together with its four neighbours either only three colors or all five colors.)

]]>$$d^4 – 2 u d^2 + q \ge 0$$

is easily converted to an inequality with no false positives at all, just by explicitly choosing the lower root of the quadratic in $d^2$:

$$d^2 \le u – \sqrt{u^2 – q}$$

Here, $q$ is entirely independent of the displacement vector $d$, while $u$ depends on a unit vector $\hat{d}$ in the direction of the displacement.

My instinct up until now has been to try to clear as many square roots as possible to make the calculations simpler, but now that I’ve realised the downside of that I should try the opposite approach!

]]>The Milne quartic could also give “false positives”, but only when the centre of the ellipsoid lay outside the sphere. But with this degree-16 polynomial, the false positives appear inside the sphere.

I suspect this means that it’s impossible to find a single polynomial expression whose sign being positive is necessary and sufficient for the existence of a nested simplex, for an ellipsoid at an arbitrary location inside the sphere, in four or more dimensions.

That’s disappointing, but there’s still the simple formula for the axis-aligned case in $n$ dimensions, so maybe I can think of another way to prove that that’s necessary as well as sufficient.

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