(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.

]]>There are 12,870 terms in the polynomial, so it’s not very enlightening as it stands, but I’m hoping to find a way to rewrite it in terms of traces, determinants, etc.

]]>$$(d_1^2 – (R \pm s_1 \pm s_2 \pm s_3 \pm s_4)(R \pm s_1 \pm s_2 \pm s_3 \pm s_4))$$

I’ve now found a 1,423-parameter family of degree 16, and it *does* permit the axially aligned case to be written as a product of eight factors like this.

But this requirement doesn’t fix all the parameters, so I’m hoping to pin down the rest by imposing some further symmetries: requiring both the planar displacement and 3D specialisations to factor neatly into four quartics that differ only in choices of signs for the $s_i$.

]]>I found a 151-parameter family of degree 12 polynomials that specialise to give the correct 3D case and the 4D planar displacement case, and also the axially-aligned case which is a further specialisation of the planar case. But when specialised for the axially-aligned case, the degree 12 polynomials factored into four terms of the form:

$$(d_1^2 – (R \pm s_1 \pm s_2 \pm s_3 \pm s_4))$$

along with a quartic that depended on some of the parameters, but which could **not** be turned into a product of terms of the above form, for any choice of the parameters.

So I’m now looking for a family of degree 16, to see if I can get that kind of extra symmetry. I have a huge expression that I know must contain the 4D form as a factor, and though it’s too big to actually factorise I can at least *check* candidate factors.

$$ (d_1^2-(R+s_1+s_2-s_R) (R+s_1-s_2+s_R)) (d_1^2-(R-s_1-s_2-s_R) (R-s_1+s_2+s_R))+\\

(d_2^2-(R-s_1-s_2-s_R) (R+s_1-s_2+s_R)) (d_2^2-(R+s_1+s_2-s_R) (R-s_1+s_2+s_R))+\\

2 d_1^2 d_2^2- (R-s_1-s_2-s_R)(R+s_1+s_2-s_R) (R+s_1-s_2+s_R) (R-s_1+s_2+s_R) \ge 0$$

Here $s_R$ is the sum of all the semi-axes other than $s_1$ and $s_2$. In 2D, we set $s_R=0$ and this expression becomes the 2D Milne quartic.

]]>