combinatorical problem !?

Hi there,

up to now I used Eclipse as a Prolog system, but now I guess I have a 'real' 
Eclipse problem, but I don't know how to set it up :-(

I have a set of balls lying on a table (red, green and blue ones) represented 
as facts like

ball( ID, COLOR ):

ball( 1, red ).
ball( 2, blue ).
ball( 3, blue ).
ball( 4, green ).
ball( 5, red ).
ball( 6, green ).
ball( 7, red ).
ball( 8, green ).
ball( 9, blue ).
ball( 10, red ).

Further I have a set of costs (distances between balls on the table)

cost( BallID1, BallID2, Cost ):

cost( 1, 2, 5.0 ).
cost( 1, 3, 3.0 ).
cost( 1, 4, 3.0 ).
cost( 2, 1, 5.0 ).
cost( 2, 4, 2.0 ).
cost( 2, 8, 3.0 ).
cost( 2, 10, 4.0 ).
cost( 3, 1, 3.0 ).
cost( 3, 5, 4.0 ).
cost( 3, 6, 6.0 ).
cost( 3, 7, 2.0 ).
cost( 3, 4, 5.0 ).
cost( 4, 1, 3.0 ).
cost( 4, 3, 5.0 ).
cost( 4, 7, 4.0 ).
cost( 4, 2, 2.0 ).
cost( 5, 3, 4.0 ).
cost( 5, 6, 4.0 ).
cost( 6, 3, 6.0 ).
cost( 6, 5, 4.0 ).
cost( 6, 7, 3.0 ).
cost( 7, 4, 4.0 ).
cost( 7, 3, 2.0 ).
cost( 7, 6, 3.0 ).
cost( 7, 8, 3.0 ).
cost( 7, 9, 3.0 ).
cost( 8, 2, 3.0 ).
cost( 8, 7, 3.0 ).
cost( 8, 9, 4.0 ).
cost( 8, 10, 2.0 ).
cost( 9, 8, 4.0 ).
cost( 9, 7, 3.0 ).
cost( 9, 10, 5.0 ).
cost( 10, 2, 4.0 ).
cost( 10, 8, 2.0 ).
cost( 10, 9, 5.0 ).

Now I want to group the balls into triples of three different colored balls.
But obviously there are many solutions, so I want the solution with the 
minimal sum of costs.
Because the number of green, blue and red balls is not equal, there will be 
some balls left (not belonging to a group) in the end (In this example one 
red ball will be left).

This seems to me like a typical Eclipse problem, maybe there's somebody out 
there to give me a hint how to set this up ;-)

Many thanks in advance

Manuel

-- 
Manuel Weindorf (weindorf@ipf.uni-karlsruhe.de)
Institut fuer Photogrammetrie und Fernerkundung, Universitaet Karlsruhe
Postfach 6980 D-76128 Karlsruhe, Tel. +49 721 608-6010 Fax +49 721 608-8450
http://www-ipf.bau-verm.uni-karlsruhe.de