Re: Quantum Gravity 297.4: p^2 = (x1 - y1 - 1)^2 + (x2 - y2 - 1)^2 is Maximum iff (x1-->y1) and (x2-->y2) are Maximum

On Nov 19, 1:00 am, OsherD <mdocto...@xxxxxxxxx> wrote:
From Osher Doctorow

Readers who are familiar with the Euclidean distance squared (taken in
2 dimensions for simplicity here):

1) d^2 = (x1 - y1)^2 + (x2 - y2)^2

may be in for a surprise when we consider the "proximity squared",
defined as:

2) p^2 = (x1 - y1 - 1)^2 + (x2 - y2 - 1)^2  for 0 < = y1 < = x1 < = 1,
0 < = y2 < = x2 < = 1

It is rather easy to prove that:

3) p^2 is maximum (= 2) iff 1 + y1 - x1 = 1 and 1 + y2 - x2 = 1, which
holds iff x1 = x2 = 1 and y1 = y2 = 0, that is to say at the points
(x1, y1) = (1, 0), (x2, y2) = (1, 0).

But recall that:

4) (x1 --> y1) = (definition) 1 + y1 - x1 = P ' (A --> B) where x1 = P
(A), y1 = P(B), P(B) < = P(A)

and similarly for (x2 --> y2), so (3) is equivalent to:

5) p^2 is maximum (= 2) iff (x1-->y1) = 1 and (x2 --> y2) = 1 (or if
we call the latter two respectively p1, p2, then p1 = 1 and p2 = 1.

This is a considerable advantage arguably of Probable Causation/
Influence P(A-->B) or P ' (A-->B) over Euclidean distance because the
maximum of the former two is so easily identifiable and corresponds to
componentwise invariance of Probable Causation/Influence.

If one desires to keep the scale [0, 1] for p^2 or a generalization of
p^2, then one can simply divide p^2 by the number of dimensions (2 in
this example) to get p^2 = 1 at maximum.

To avoid Reader confusion, I point out that p in the past has been
used in my posts for (x-->y), but here it is used as defined
implicitly by (2) above.  However, p1 and p2 correspond to my usages
in prior posts.

Osher Doctorow


Brilliant, Radical, and Right.

I like the "proximity squared" approach. Excellent post Osher.


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