(sqrt(x^2 + y^2) - R1)^2 + z^2 - R2^2 = 0

into

(x^2+y^2+z^2+R1^2-R2^2)^2 - 4R1^2(x^2+y^2) = 0

And, what I was thinking about, was if we solved this polynomial for xy or xz. It seems like it would become two circles, displaced or concentric. That is, we would derive its plane intercepts into exact solutions of two circles. They would be 2 discrete quadratic polynomials, acting as roots to the quartic. It's funny how I'm learning all of this backwards. Explore hypershapes first, then learn the math.

(x^2+y^2+z^2+R1^2-R2^2)^2 - 4R1^2(x^2+y^2) = xy

sets z=0 , and should make

(x^2+y^2+x+y-(R1+R2))*(x^2+y^2+x+y-(R1-R2)) = z

Where their diameters R1 are equidistant to the cut open R2

and,

(x^2+y^2+z^2+R1^2-R2^2)^2 - 4R1^2(x^2+y^2) = xz

sets y=0, and should make

((x+R1)^2+z^2+(x+R1)+z-R2)*((x-R1)^2+z^2+(x-R1)+z-R2) = y

Where their centers are equidistant to the cut open R1 along X

There might be some errors in that. I also have no clue how to do those calculations, but I'd like to learn. It makes me think about higher toratopes. If the ditorus can make 8 x-intercepts, it needs an 8 degree polynomial. Same for a tiger, which comes out as all complex, or half real and complex, for x. It seems like complex roots are another way of saying 'empty cut'. The whole shape is still there, just not intercepting the plane that was set to zero. Like a parabola that's in the +x+y quadrant. It has complex roots, because it's translated away from the x-plane. It lies outside the 1st dimension. So, if we cut to empty, we have all complex roots. But, translating out makes real roots and complex, as we see only half the shape.

So, if solving for some 2-plane, the 8 degree poly for a ditorus should have four exact roots of four circles. These four would be spaced apart in an array corresponding exactly to the diameters that were cut open. A tiger would have four circle intercepts as well, one in each quadrant. A tiger also seems it would have a more complex polynomial, as it introduces an implicit where the coefficients are polynomials.

Being that we're working with discrete surfaces, we can also solve for discrete and exact roots of lower polynomials. Crazy thing is, the cut algorithm has already pre-derived every solution ahead of time. Like the intercept array of a 6D toratope. We get at most 8 tori, in an array spaced apart from cut diameters. Which is 8 quartic polynomials as the 3D intercept 'points' of a hypershape. That 5D surface has a 32 degree polynomial, and we can derive every single one of its exact solutions for comparison. That sounds kind of interesting, and perhaps an avenue of research. Maybe it'll shed some light on root finding algorithms. It also might be a valuable proof system, to show how the discrete polynomials have several discrete roots, in various hyperplanes.

This also means, for base-species toratopes,

dimension - degree of poly

----------------------------------

1D - 1 degree

2D - 2

3D - 4

4D - 8

5D - 16

6D - 32

7D - 64

8D - 128

9D - 256

10D - 512

degree = 2^(1-dim)

or, as a better generalization, consider the amount of distinct diameters, and

degree = 2^(R

_{sum})

So, if the implicits for 4 of the five 4D toratopes are:

((III)I) = (sqrt(x^2 + y^2 + z^2) - R1)^2 + w^2 - R2^2 = 0

(((II)I)I) = (sqrt((sqrt(x^2 + y^2) - R1)^2 + z^2) - R2)^2 + w^2 - R3^2 = 0

((II)II) = (sqrt(x^2 + y^2) - R1)^2 + z^2 + w^2 - R2^2 = 0

((II)(II)) = (sqrt(x^2 + y^2) - R1a)^2 + (sqrt(z^2 + w^2) - R1b)^2 - R2^2 = 0

how does one go about making them polynomials? Or, is there an easier way using the torus quartic? Maybe, actually.

EDIT: made some minor corrections to the torus quartic intercepts