# 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.
Regards.
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### Risposte (2)

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 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.

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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.
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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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