I can handle one step only, among the loops. The smaller loop counts around back to zero the fastest, the red loop is the lowest and their steps are 2,3,5.
How often will on loop have a pending step? Count the smallest three times and bump the next largest by on. The points on the loop have label, let me tell you the labels.
On the smallest, they are no update pending, one update pending. The count the same on the larger loops, none,one,tow,...
Now ask, ham many times will the red have two updates pending? Twice (I broke out all thecombinations. The middle loop will have two pending once, among all the possible paths, and the smallest none.
Now how often all have one item pending? Once, and this is what Markov specified in the Markov tree. This is the basic idea on graph spectra, lowering and raising the resolution of a graph. The idea is to remove the loops, and leave the thinned spanning tree which counts out the various loops as paths through the tree.
The paths that are either all pending once, or one pending twice are the update All the other paths are merge and thinned, there are more than one path leading to the outcome on account of commutative property. So we end up with five paths through rather then the factorial 30.
This is self sampling, the different colors are sampling each other, as if each button on our beach ball are trying to self adjust simultaneously. The ambiguous paths are merged until the next step when they will resolve. The case where all have oine pending is unrexolvable, and are shoved down a layer on the beach ball and identify the number of buttons in the lower layer. This is why we get a tree, we are making a sphere in layers. The case where each has one pending mark a location where the local is white, and can be matched by the layer underneath.
There should be a solution for four colors, and five, I think. Still looking at this.
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