Markov jumping

I put this section on the math page, my donation to the book.

For the modeler I am talking about resolution, this is the method of taking your deviations of colors and changing the resolution in surface coloring, a great tool, by the way. Add the detail you need, but if you want a broad look, take your coloring graph to the compiler, even though it may have ten thousand paths. The compiler can thin the tree, keeping redundancy minimized.  You will train it with the typical sequence, that is your Pea Tree coloring algorithm has resolution, as long as those node jumps include you most detailed equations. So we go directly from your existing charts, right into surface coloring, one of the greatest tools around, be we need a complete, representative sample, an event sequence that would trigger the complete traversal of your Pea tree. The must have tool of the century, the holy grail. This is the fast transform for AI, a really neat training trick.

x,y,z -> 3xyz

Hypothesis
x,y,-z -> -3xyz   True by use of symmetry about the z axis
x,y,b-z -> 3xy(b-z)     Simply the uniqueness of integers.

Degree b on left mean 3xy must satisfy on right 
But 3xy is an integer in a integer system. Note I am using connected set theory, no operation of the type used will generate a non integer, operations closed under integers.

x,y,3xy-z is a solution. This must be true for any permutation of x,y,z; the system is commutative withn the solution triples.  So from any point on the tree, swap z with any axis of jump.

Proof:

x^2 + y^2 + (b-z)^2  ->    3xy(b-z) = 3xyb-3xyz

x*2 + y^2 _+z^2 =  -b^2 +2bz +3xyb - 3xyz

let b = 3xyz then the right becomes
b^2 - b^2 + bz = 3xyz