MATLAB Answers


Arithmetic to ensure positives

Asked by Amit
on 26 Dec 2015
Latest activity Commented on by Amit
on 4 Jan 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.


Sign in to comment.


2 Answers

Answer by Walter Roberson
on 27 Dec 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.


John D'Errico
on 27 Dec 2015
Note that your "target" idea,
(a-b)^2 + (c-d)^2 + (e-f)^2 = 0
is actually a terribly written thing in terms of numerical analysis. Sorry, but that is a bad thing to do when you might be working with floating point numbers.
As Image Analyst says, just write some basic code that does what you want instead of looking for some elegant (to you) code fragment that looks nice, but does bad things.
When you say that your optimization scheme is very fragile, are you talking about attempting to code constraints in a manner that is differentiable?
on 3 Jan 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.

Sign in to comment.

Answer by Walter Roberson
on 4 Jan 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 Comment

on 4 Jan 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

Sign in to comment.