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Add simplify_clause::try_remove_equations

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This commit is contained in:
Jakob Rath 2022-12-22 14:53:42 +01:00
parent 5bd63ab7c5
commit b5af2164f4
5 changed files with 113 additions and 15 deletions
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74
src/math/polysat/simplify_clause.cpp
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@ -11,21 +11,6 @@ Author:
Notes:
TODO: inspired from bench23, trying to strength weak FI lemma.
If we have both:
p <= q
p - q == 0
Then remove the equality.
If we have both:
p < q
p - q == 0
Then merge into p <= q.
TODO: from test_ineq_basic5: (mod 2^4)
Lemma: -0 \/ -1 \/ 2 \/ 3
-0: -4 > v1 + v0 [ bvalue=l_false @0 pwatched=1 ]
@ -93,6 +78,8 @@ namespace polysat {
bool simplify_clause::apply(clause& cl) {
LOG_H1("Simplifying clause: " << cl);
bool simplified = false;
if (try_remove_equations(cl))
simplified = true;
#if 0
if (try_recognize_bailout(cl))
simplified = true;
@ -102,6 +89,63 @@ namespace polysat {
return simplified;
}
/**
* If we have:
* p <= q
* p - q == 0
* Then remove the equality.
*
* If we have:
* p < q
* p - q == 0
* Then merge into p <= q.
*/
bool simplify_clause::try_remove_equations(clause& cl) {
LOG_H2("Try removing equations");
sat::literal_set eqns;
for (sat::literal lit : cl)
if (s.lit2cnstr(lit).is_eq())
eqns.insert(lit);
sat::literal_set to_remove;
for (unsigned i = cl.size(); i-- > 0; ) {
sat::literal const lit = cl[i];
signed_constraint const c = s.lit2cnstr(lit);
if (!c->is_ule())
continue;
if (c->is_eq())
continue;
LOG_V(10, "Examine: " << lit_pp(s, lit));
pdd const p = c->to_ule().lhs();
pdd const q = c->to_ule().rhs();
signed_constraint const eq = s.m_constraints.find_eq(p - q);
if (!eq)
continue;
if (!eqns.contains(eq.blit()))
continue;
to_remove.insert(eq.blit());
if (c.is_positive()) {
// c: p <= q
// eq: p == q
LOG("Removing " << eq.blit() << ": " << eq << " because it subsumes " << cl[i] << ": " << s.lit2cnstr(cl[i]));
}
else {
// c: p > q
// eq: p == q
cl[i] = s.ule(q, p).blit();
LOG("Merge " << eq.blit() << ": " << eq << " and " << lit << ": " << c << " to obtain " << cl[i] << ": " << s.lit2cnstr(cl[i]));
}
}
// Remove superfluous equations
if (to_remove.empty())
return false;
unsigned j = 0;
for (unsigned i = 0; i < cl.size(); ++i)
if (!to_remove.contains(cl[i]))
cl[j++] = cl[i];
cl.m_literals.shrink(j);
return true;
}
// If x != k appears among the new literals, all others are superfluous.
// TODO: this seems to work for lemmas coming from forbidden intervals, but in general it's too naive (esp. for side lemmas).
bool simplify_clause::try_recognize_bailout(clause& cl) {