You can pick one of three colors, and make the light get brighter. The light cannot go above a limit, else you lose one round. But if you make six moves and keep the light below max, you win one round. After six safe rounds, the three colors will be at their error minimum count. The light changes brigthness after two moves.
We know some things. The light bulb must be quantized under uncertainty because my guesses will not be consistent. I think this is it. I need to know the surfaces the follows minimum uncertainty.
From there we can make some quickies. I want to stabilize my error counts so the brightness levels become clear.
So, I want my selections to be equally likely. I need a relative count and probability close to log relative count, it must look as Gaussian as any of the other counters.
The counters will be modulo to six, in some fashion. So we give them two angles of separation and one of radius, like and good college kid. These too will be quantized to a surface, zero point at the bottom, as will the Lie graph. There will be a surface path, geodesic, having the complete minimum sequence of brightness. Markov is about finding the perpendicular of that surface, and centering it then quantizing it.
It will be characterized as a separation of three error surfaces. And there is some Lie group stuff that applies.
One complete sequence is xyz, in integers. Each of them characterized by binomials, matched by matching error value and relative count. In essence there are selections from the binomial
At stability, each of my color error distribution must be a horizontal on this tree.Relative likelihood and relative error matched to look like a fair coin flip. There will be an optimum integer solution to the number of fair coin flips per color.
Simple solution. The minimum brightness is 111. From here I have discrete Poisson solution to match my discrete binomial. I want to count up to the maximum amount, six.
(One of my angles separates the colors in sequence, c^2 + s^2 guides the options. The other measures the difference in a near perpendicular, c^2-s^2 = 1 I will have a salad bowl.
The discrete Poisson is the relative arrival times, as the two angles are independent even though the flat angle count is almost integer. The relative arrival times expanded to accommodate the independence. The relative arrival count is really a unit of frequency, bandwidth. It is delta hyperbolic angle times delta circular angle.
My feeble mind has trouble sorting probability and frequency and finite counts and combined uncertainty. But I think this one does it better than my previous.
Count the tree, those coefficients on the left flank need to be sum to multiple of multiple of three if my binomials are balanced. That is because I know I have a multiple of three needed to adjust the center position, I need to adjust my unfair coin by some 1/3, as that will be the closest match.
They all have to roll over to the minimum count, the left flank. So we can see three fixation points which match the three moments. The center, and the two left flanks all have a algebraic relationship.
This little discovery:
Moreover, he pointed out that , an approximation of the original Diophantine equation, is equivalent to with f(t) = arcosh(3t/2).[6] The conjecture was proved[disputed ] by Greg McShane and Igor Rivin in 1995 using techniques from hyperbolic geometry.[7]The nth Lagrange number can be calculated from the nth Markov number with the formula
He is just accommodating the extra uncertainty of the circular component, giving it 1/3 of the bandwidth. Then he says the bandwidth must obey the golden ration, if the binomials are optimumly aligned and there is no better solution, he must be haviig the most irrational round off error.
Nothing here is restricted to 3D, but the total N goes way up. The adjustment is removing the Mth moment in the distribution.
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