/** Allocate relations for SubsetSig top-down. */
private Expression allocateSubsetSig(SubsetSig sig) throws Err {
// We must not visit the same SubsetSig more than once, so if we've been here already, then
// return the old value right away
Expression sum = sol.a2k(sig);
if (sum != null && sum != Expression.NONE) return sum;
// Recursively form the union of all parent expressions
TupleSet ts = factory.noneOf(1);
for (Sig parent : sig.parents) {
Expression p =
(parent instanceof PrimSig) ? sol.a2k(parent) : allocateSubsetSig((SubsetSig) parent);
ts.addAll(sol.query(true, p, false));
if (sum == null) sum = p;
else sum = sum.union(p);
}
// If subset is exact, then just use the "sum" as is
if (sig.exact) {
sol.addSig(sig, sum);
return sum;
}
// Allocate a relation for this subset sig, then bound it
rep.bound("Sig " + sig + " in " + ts + "\n");
Relation r = sol.addRel(sig.label, null, ts);
sol.addSig(sig, r);
// Add a constraint that it is INDEED a subset of the union of its parents
sol.addFormula(r.in(sum), sig.isSubset);
return r;
}
/** Allocate relations for nonbuiltin PrimSigs bottom-up. */
private Expression allocatePrimSig(PrimSig sig) throws Err {
// Recursively allocate all children expressions, and form the union of them
Expression sum = null;
for (PrimSig child : sig.children()) {
Expression childexpr = allocatePrimSig(child);
if (sum == null) {
sum = childexpr;
continue;
}
// subsigs are disjoint
sol.addFormula(sum.intersection(childexpr).no(), child.isSubsig);
sum = sum.union(childexpr);
}
TupleSet lower = lb.get(sig).clone(), upper = ub.get(sig).clone();
if (sum == null) {
// If sig doesn't have children, then sig should make a fresh relation for itself
sum = sol.addRel(sig.label, lower, upper);
} else if (sig.isAbstract == null) {
// If sig has children, and sig is not abstract, then create a new relation to act as the
// remainder.
for (PrimSig child : sig.children()) {
// Remove atoms that are KNOWN to be in a subsig;
// it's okay to mistakenly leave some atoms in, since we will never solve for the
// "remainder" relation directly;
// instead, we union the remainder with the children, then solve for the combined solution.
// (Thus, the more we can remove, the more efficient it gets, but it is not crucial for
// correctness)
TupleSet childTS = sol.query(false, sol.a2k(child), false);
lower.removeAll(childTS);
upper.removeAll(childTS);
}
sum = sum.union(sol.addRel(sig.label + " remainder", lower, upper));
}
sol.addSig(sig, sum);
return sum;
}
