AZ simple test for 5D stability

Treat the system as a continuous model, sand prune the factorial tree to maintain out five partitions.  As you step down the tree, the most probable update combination is rejected, or rather emitted to maintain a partition. This is a correct assumption in the continuous case. So all the paths belowe the most probable are eliminated. The next step down the rmaining graph, throw out the most probable and so on.

In the end, look a those most probable number, what is the significance of them? How many digits needed to separate out the continuous queues.  Then, find an N that converts those probabilities all to the nearest, different, absolute count, in ascending order.  If that N cannot be found, then 5D is unstable.  There must be one N performing the task for their to be N stability.

If one finds a stable N then one works backwards to write the Dophantine equation. Like a miracle come true. This is all the miracle of Poisson. There needs a recursive tree. If I know N to M qubits of precision, the I need to know M+1 for M qubits of precision.  The M digits count N. I need an expansion tree to stabilize on the correct N.

I haven't done it, it is an I dunno. But why should there be a solution? It is a prime problem and they are all the same, we assume. I can have Quanta Magazine find out whop is working this and get an answer. Can I just make a request?

This most likely thing.

We are asking which of the update queues most likely rolled over.  And at the top of the 3D tree is the spin, so those paths dropped and in the node down, spin is 0 or 1, so the next rollover if charge, and the next will be magnetism.  So the tree gets pruned by 1/3 then 1/3 then 1/5. The spin combinations are 11,01 or 10, 00.  There is commutivity in spin at this point.   Droppin a node we expect chae to show a rollover of 3, but commutivity applies here.  011 and 110 and 101 combine. As do 001,010,011. In other words, these are equally probable, none a rollover.  So this is how branches collapse, commutivity works because we want the most probable rollover and can combine thos with equal probability since there is no maximum.

In the 5D system we have the most accurate bit the sevens qubit.  The previous qubits are biased toward zero adjustment, there is no negative adjustment, the number is always less than one  unit digit. It is white in the sense of all those channels will be set once per light step.

Now, at the bottom of the error generator we get the least likely things, the stuff we could not compute at the constant light steps, this is Planck's uncertainty. All the residual paths get packed not the error term. Spin will st the spin bit and looks like: 0,0,1,0,1,0,0,1 and so one. It cunts more even than odd, there are actually an qual and odd N value. The charge bit will set at 0,1,o 2 of a third. To count the integers, you have to count the firings. Hence the connection between fermions and bosons.

So we should be able to continue this in 5D with no problem, I dunno?