[Ifeffit] max number of restraints

Matt Newville newville at cars.uchicago.edu
Sat Apr 14 14:56:24 CDT 2007

Hi Victor,

> I am confused.

I think Bruce had the same idea as you:  Because each restraint is
appended to the vector to be minimized in the least square sense, the
important quantity is
       Sum_restraints ( restraint_i)^2
and so you could just do that ahead of time, and then say that
sqrt(Sum_restraints[restraint_i^2]) is a good substitute for the set
of restraints.  It does have the appealing feature that it separates
there interdependence compared to a simple addition (where you could
easily get competition between restraints).

I think adding in quadrature is not quite right, as not only is the
minimum of chi-square important, but also the ability to find the
minimum and explore the parameter space.  By adding in quadrature, you
prevent the restraints from being negative.    Well, the sign *is*
arbitrary (is it data-model or model-data??), but the ability to
switch signs is important.

I  am not sure whether adding in quadrature is worse than a simple
addition, but neither is ideal. For sure, having enough individual
restraints is the best approach.


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