Counting shades of color, a work in progress

I have a beach ball with buttons equally spaced on its surface.  I can set a button one of three colors or give is a color tint. Then I move to the next button, and walk the beach ball.

This description is not complete, it is a first pass, read with knowledge of errors.

My circular paths are the various shade of a three color system.   I keep three counters that count the color deficits. Thus when the blue counter counts, it tells me there is not enough blue and soon I have to color a button blue.

When ever one of the paths rolls over to the top, that means I have placed that color on the current chip. on a button in my beach ball. Othewrise I just shade the current button.

Every clock tick I will either place a color or increment one of the shade counters.    Now my color shades are relatively prime,  any two colors that have rolled over in sequence is illegal as the 2,3,5 counters never are multiple of one another. Also every clock tick I move to a different adjacent button, and set a tint.

The game is finite, there area finite set of shadings.  If I just placed a yellow, then the next chip cannot be yellow, and the same for all three colors.

So straight away, I can rule out some states of the counters, which I have done below, then with the remaining states I control a coloring tree which tells me which is the whitest move I can make.

Consider, for example the state 0,2,4.  This state is as Yellow as can be.  I cannot place another yellow on an adjacent button and must set a tint.  This count say I have 0 yellows ready to place and 2 blue and 4 red; the last to colors are as off white as I can be.  104 is as blue as possible and 110 as red as possible.

From these boundary conditions I can construct a coloring tree.  By convention I will start my tree immediately after placing a yellow. Thus, after each yellow button I trurn to the tree top. Below are the legal starting conditions, each taking a path through the tree, each node of the tree is either a tint setting  or a color setting. Each step in the tree is an adjacent button from the last.

Consider 011, I have just set a button to yellow and just tint the adjacent buttons.. My red and blue counters are minimal. I will set two adjacent buttons with either a slight reddish blue or a slight bluish red. Then the next button gets a yellow tint. The red blue and blue red tints merge to a single shade. Then I set the next button Yellow. The two blue-red buttons both lead to the same button with the same tint.

Note. Once I set a button to absolute colomn on the beach ball I will never change the color as that inplies the one path length a multiple of the other, but the path lengths have to be relatively prime. So the steady state is that I end up traversing the tinted buttons and adjust their shade.

When do I win? When the color algorithm wants to recolor a button. If we originally set all butons to black then their is an algorithm that sets each button once, though you may visit tinted buttons repeatedly  until all buttons are set to one of the colors. So backtracking on tinted colors is the name of the game.


011 -> 021 -> 121
            012 ->112  -> 122 -> 022 ->  023 -> 123 -> 124 -> 024

013 -> 014 -> 114  -> 124 -> 024
                                    110 -> 010

            023 ->                             
022 -> 023 -> 123-> 124 -> 024

024

I wan to retire from the science and tech biz.

But here is my point.  Take a flat surface, evenly square and start coloring. Some squares go missing and causes the flat surface to roll into a ball with the correct number to close the sphere, that number is huge. If you cannot find enough squares to color you cycle, like the proof says. All the applications of this theory will likely not have the large enough number of buttons, and the story then becomes periodically updating those color counts adiabatically to adapt to the N you have.