[Matplotlib-users] "Olympic rings" optimization problems with SpaceFuncs

[Dmitrey <dmi...@uk...>]

Hi all, recently I have implemented some new functionality for the free 
geometric software SpaceFuncs <http://openopt.org/SpaceFuncs> and would 
like to introduce a couple of optimization examples with Olympic rings, 
devoted to the Olympic games - 2012, that are going on now. They are 
valid for upcoming <http://openopt.org/coming> version of SpaceFuncs 
<http://openopt.org/SpaceFuncs> only (already available in subversion 
repository). In similar way you can solve systems of linear/nonlinear 
geometrical (in any space dimension) equations (see OpenOpt SLE 
<http://openopt.org/SLE> and SNLE <http://openopt.org/SNLE>).

Graphical output is powered by matplotlib 
<http://matplotlib.sourceforge.net/>.

 1.

    Let's search 5 points (1 from each ring) that form polygon with
    maximum area. Code is here
    <http://openopt.org/SpaceFuncsDoc#An_example_with_olympic_circles>
    (or here
    <http://trac.openopt.org/openopt/browser/PythonPackages/SpaceFuncs/examples/maxArea.py>).
    max area

 2.

    Well, (1) was quite easy problem that could be done by many ways
    with many different software. Let's consider more difficult example.
    Suppose we have several points with different positive or negative
    weights and would like to place them in the olympic circles so that
    their centroid (barycenter) would be as close to a target as
    possible; let's also demand distance between any 2 points has to be
    no less than 0.15. It is difficult combinatorial optimization
    problem - each point may belong to one of 5 circles, so number of
    possibilities is 5^m, where m is number of points; moreover, we have
    m * (m-1) / 2 difficult nonlinear nonconvex equality constraints
    (requirements on distances). Fortunately, OpenOpt has awesome solver
    interalg <http://openopt.org/interalg>, that easily solves this
    problem (for quite small m, although), moreover, with specifiable
    accuracy fTol: abs(f-f*) <= fTol; we will consider 3 points of
    weights [15, -0.15, 3]. Code is here
    <http://openopt.org/SpaceFuncsDoc#Another_example_with_olympic_circles>
    (or here
    <http://trac.openopt.org/openopt/browser/PythonPackages/SpaceFuncs/examples/centroid.py>)
    centroid

Best wishes for you country in the Olympic games,
D.
------------------------------------
http://openopt.org/Dmitrey