Arithmetic to ensure positives

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Amit il 26 Dic 2015
Commentato: Amit il 4 Gen 2016
Hello all:
I am looking for a simple logic to ensure 'positive' for multiple variables. In pseudo code terms want to replace
if {(a-b>=0).and.(c-d>=0).and.(e-f>=0) then...}
without the use of boolean 'and' or special functions like max. Pure arithmetic will be much helpful.
My inspiration is
(a-b)^2 + (c-d)^2 + (e-f)^2 = 0
which uniquely ensures/enforces eqalities, a=b, c=d, e=f.
Any parallels.
Much appreciated.

Risposte (2)

Walter Roberson
Walter Roberson il 27 Dic 2015
This is not possible to do without at least one comparison, and comparisons are not pure arithmetic. Your inspiration (a-b)^2 + (c-d)^2 + (e-f)^2 = 0 involves a comparison and so is not pure arithmetic.
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Walter Roberson
Walter Roberson il 27 Dic 2015
When you say that your optimization scheme is very fragile, are you talking about attempting to code constraints in a manner that is differentiable?
Amit il 3 Gen 2016
Yes actually Walter. In deed, hoping for that. Thanks in deed for bringing me closer to asking proper question here.
In the last few days, I tried in built GRG Non Linear Scheme with 'min' function, as a work around. Hoping to do something better.
Thanks for your attention.

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Walter Roberson
Walter Roberson il 4 Gen 2016
(a-b)^2 + (c-d)^2 + (e-f)^2 = 0 is differentiable only because it is smoothly invertible, that it can be translated into a series of variable reductions. Inequalities cannot be inverted that way. You cannot even code a > 0 invertibly -- if you could then c>=d could be coded as (c-d)^2 - delta_c = 0 together with however you coded delta_c > 0.
Unless, that is, you are okay with coding Heaviside functions, in which delta_c > 0 translates to Heaviside(delta_c) - 1 = 0 after having defined Heaviside(0) as 0 (Heaviside(0) does not have a fixed value, not really; one of the common conventions says Heaviside(0) = 1/2).
But diff(Heaviside(delta_c),delta_c) is Dirac(delta_c) and that is considered a distribution rather than a particular value, definitely not continuously differentiable. I would not consider it suitable for the use in this situation, but perhaps the theory of GRG is more flexible than I am.
  1 Commento
Amit il 4 Gen 2016
Dear Walter:
Thanks much. I learnt a few things from you answer. I understood what you are saying. Though this is pushing my problem in 'high mathematics' rather than come out of it.
Perhaps what I asked is not available, but let me seek a little bit more by keeping it still as an open question.
Meanwhile, the GRG scheme in deed appears to be accommodating.
Look forward to more thoughts, if any.
Regards, Amit

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