Hi Leva.
Yes, you can do this with a *weighted* max-sat problem (instance).
if you have Y soft clauses, you can weight each hard clause with the
value of Y+1, and all soft clauses with a weight of 1.
Then, if/when the weight of all unsatisfied clauses is <= Y, you have
satisfied all of the hard clauses.
A less elegant solution to this that uses regular (unweighted) max-sat
is as follows: for each hard clause, add Y duplicates of it to the
instance. This achieves the same goal but will change the behaviour
of many algorithms (perhaps in good ways, perhaps in bad ways). One
disadvantage of this approach is that increases the memory
requirements and can slow down algorithms considerably as there is
more "bookkeeping" to keep track of.
-Dave
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