Herwig 7.3.0
|
At the moment it does nothing in the case that also par3 is present.
The method assumes par3 == PDPtr() (otherwise we don't know how to proceed: a possible, trivial way would be to randomly select two of the three (anti-)quarks and treat them as a (anti-)diquark, reducing the problem to two components as treated below. In the normal (two components) situation, the strategy is the following: treat in the same way the two possibilities: (d dbar) (i=0) and (u ubar) (i=1) as the pair quark-antiquark necessary to form a pair of hadrons containing the input flavour par1 and par2; finally, select the one that produces the lightest pair of hadrons, compatible with the charge conservation constraint.