Coloring polynomials

We had another bi proof, Quanta is on it.

What does is mean to color a functions?

Functions (z) =y
Take a complete sequence of the function (1,y,z) pairs.  Pick a large number of  y and z pairs.

Then pick a spot on the 1My,Mz that matches you expected N.  Divide the Mz into x and the My into you y.

Huffman encode both sequences. You have a self sampled solution.

Let us do this for some polynomial with all its zeros on the unit circle. Polynomial with zeros on the unit circle should be identifiable.  In those polynomials, the Z value at y=1 is repeated.

The polynomial is p(z) = y.  I  do this because I want to count z in to set resolution. But I want to match a Markov triple with z the large number.

Color the ball. It is as white as you can get, now look at the point y=1 (this is as close as we get to Null).  The number of z colors around that spot in a diameter is the resolution of your unit circle.

Pick a spot where y is large. The number of z colors is the resolution of large values, a lower.  The resolutions values will hold for any lower degree polynomial  Resolution goes with the degree of the polynomial.

Jumping down Markov should be equivalent to coloring with lowered resolution.

The complex plane as the maths like it, is a bit curved. The farther out from the center, the fewer bins of resolution.

Any set of polynomials with roots on the unit circle fall fill the plan if they are coprime in degree. Any set of these, in counting primes, will optimally divide the complex plan. Then take some arbitrary polynomial and encode if Z values against the same X values.   hen compare the y values, after quantization, of the two pairs.  Try to ge both their ilog(i) to match by re-arrnagment, forget z. You want to match them in resolution, then you can identify where the match on your colored ball.  Look at the spot where y=1, as close to a zero as you can get, then read off x, the closest fractional match to your zero.  But that should match the equivalent zero in terms of angle in the unit circle, and you can project that to get the solution on the complex plane.

Unitary polynomials, they call them.  All polynomials will color the same  such that tops and bottoms have lower resolution. So any polynomial can have its 'resolution' circle on the complex plane with position identifiable. The roots of the base can translate the unit circle to match.  Maximum resolution around the y = 1 spot. So I am moving plane used by the unitaries and moving it to match the resolution pattern of the zeros for another polynomial.   I can read of the zero values.

When my Markov x is 1 I am just marking between operator solutions.