I need to color a beach ball surface with M colors, I want to know the relative counts of each color that makes whitest. My first color has an arrival rate of one, set by standard. Now my second color must measure white relative to the first. The first has a half bit of precision. But my update rate is exactly the same as the first color. Why not make my second color arrive at twice the rate? Because i will always have the same solution, make it which within one.quarter. If it is postive definite, the my best choice is to always add a quarter and get the same distance closer to white, the color is not needed. So the next color is one third. And by induction the next color is one fifth.
Now mu buttons to be colored are evenly spaced, my arrival rates need to match button size. The surface is thus a fixed integer. But my white has precision one fifth, so my button count is very close to a penatbit count.
So, you see, this is why here is a maximum liquidity match between a pentabit and a sphere surface. And in the color generator there are intsances when some color combinations o not increase precision, and these are omitted. The various three colors of error update must be equally probable in value. But leaving no color is a prohibit. So there are five paths for one bit, three for the other and two he first. The error generator rank is three, in the three dimensional case. So the trimmed factorial tree is defined.
There is a surface resolution that fits better than any other, and he rates constant, the coloring algorithm finite.
The actual N count must be the sum of a pentabit tribit and binary digit sequence, as all three error terms are independent. The most liquid round off error is the Phi series, so match that power series against the surface area. Get the count and set the rate. At that point the trhees bi mus have counted to the constant, and the two and fives via rollover.
The goal is to pack the error band for closed ball then transform to flat surface.
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