Dimensional Equation Tested

I previously posted an equation on the Dimensionism group that perhaps I should attribute to my brother Brian.

In my two- to three- year history of intermittent searching for better expressions of the data, only recently was I able to re-orient myself towards something close to a correct formulation for multiple types of sets. This was due in large part to a hint (or perhaps, if I don't remember, much more than a hint...) which was embarrassingly "2 X 2" "2 X 2"... It's unclear whether what he meant at the time was that he had solved the problem (probably he had, since he's such a smart person), or whether 2 X 2 was simply a notion of how the two categories crossed in the simplest version of a quadra arrangement.

At any rate, the resulting equation was:  2 * {2^ (M root of C) - 1}.

Some initial testing early this morning suggests that this equation is highly effective at pruning out trivial cases, something that might be noted may be characteristic of my brother's intelligence rather than mine.

Here is the equation originally posted at:  http://www.facebook.com/dimensionism):

This equation works for odd numbers of categories:
2 * {2^ (M root of C) - 1}

And, this equation works for even numbers:

2 ^ (M root of C).

An approximation can be reached for quadra sets by taking the square root of the number of categories.

[Pending new results with penta, some of the figures are being revised. Some of the following is theoretical, as the methodology may be under dispute. The very viability of categorical deductions has been scrutinized in the past].


QUADRA TEST [even number]
2 ^ (4 root of 4)
= 2, correct.

2 ^ (4 root of 16)
= 4, correct, in the non-unity method.

2 ^ (4 root of 64)
= 8, correct so far as I know.


DUALISTIC TEST

2 ^ {2 root of 2}
= 2, correct.

2 ^ {2 root of 4}
= 4, correct.

2 ^ {2 root of 8}
= 8, correct.


TRINITARY TEST

2 * {2 ^ (3 root of 3) - 1}
= 2, correct! (forwards and backwards, instead of combinations of two).

2 * {2 ^ (3 root of 9) - 1}
= 6, correct! (3 * 2 * 1 combinations).

2 * {2 ^ (3 root of 27) - 1}
= 14. ? I predict greater efficiency.

The more general estimate gives 2 ^ (3 root of 27) = 8 deductions for 27 categories.


PENTA TEST

2 * {2 ^ (5 root of 5) - 1}
= 2, correct! (2 rotations of the diagram, no symmetric altercations).

2 * (2 ^ {5 root of 25} - 1)
= 6, (3 opposites * 2 opposites * 1 opposite, and no more opposites!)

The Dimensional Equation --- Simple and Complex Solution

Via the Dimensionism Group on Facebook (a group I run myself):

"If it's exponential, then my guess is that it relates to the square root"---Brian Coppedge [my younger brother by 2 years and 2 days]

Referring to the number of categorical deductions for a given category set, this proved to be true, at least for two dimensions of modular 4.

The simplest equation is to take the square root of the number of categories. Another equation is 2 * ( 2 ^ (M root of C) - 1), where C is the number of categories, and M is the modular value, such as 4 for quadra.

This equation simply gives the number of categorical deductions, not how to tabulate them.

In my formulation, the tabulations are equal to all legal cyclical orders (made complex by modularity), in which the opposite positions are occupied by opposite categories. There are also simpler ways to reach the combinations. The result is essentially binary for every set level, but the exact order counts, and some combinations are superfluous, which is not always obvious. Set-equivalence must be defined in a formalized way.

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