void resolve(vec<Lit>& main, vec<Lit>& other, Var x)
{
Lit p;
bool ok1 = false, ok2 = false;
for (int i = 0; i < main.size(); i++){
if (var(main[i]) == x){
ok1 = true, p = main[i];
main[i] = main.last();
main.pop();
break;
}
}
for (int i = 0; i < other.size(); i++){
if (var(other[i]) != x)
main.push(other[i]);
else{
if (p != ~other[i])
printf("PROOF ERROR! Resolved on variable with SAME polarity in both clauses: %d\n", x+1);
ok2 = true;
}
}
if (!ok1 || !ok2)
printf("PROOF ERROR! Resolved on missing variable: %d\n", x+1);
sortUnique(main);
}
ClauseId Proof::addRoot(vec<Lit>& cl)
{
cl.copyTo(clause);
sortUnique(clause);
if (trav != NULL)
trav->root(clause);
if (!fp.null()){
putUInt(fp, index(clause[0]) << 1);
for (int i = 1; i < clause.size(); i++)
putUInt(fp, index(clause[i]) - index(clause[i-1]));
putUInt(fp, 0); // (0 is safe terminator since we removed duplicates)
}
return id_counter++;
}
/*_________________________________________________________________________________________________
|
| newClause : (ps : const vec<Lit>&) (learnt : bool) -> [void]
|
| Description:
| Allocate and add a new clause to the SAT solvers clause database. If a conflict is detected,
| the 'ok' flag is cleared and the solver is in an unusable state (must be disposed).
|
| Input:
| ps - The new clause as a vector of literals.
| learnt - Is the clause a learnt clause? For learnt clauses, 'ps[0]' is assumed to be the
| asserting literal. An appropriate 'enqueue()' operation will be performed on this
| literal. One of the watches will always be on this literal, the other will be set to
| the literal with the highest decision level.
|
| Effect:
| Activity heuristics are updated.
|________________________________________________________________________________________________@*/
void Solver::newClause(const vec<Lit>& ps_, bool learnt)
{
if (!ok) return;
vec<Lit> qs;
if (!learnt){
assert(decisionLevel() == 0);
ps_.copyTo(qs); // Make a copy of the input vector.
// Remove duplicates:
sortUnique(qs);
// Check if clause is satisfied:
for (int i = 0; i < qs.size()-1; i++){
if (qs[i] == ~qs[i+1])
return; }
for (int i = 0; i < qs.size(); i++){
if (value(qs[i]) == l_True)
return; }
// Remove false literals:
int i, j;
for (i = j = 0; i < qs.size(); i++)
if (value(qs[i]) != l_False)
qs[j++] = qs[i];
qs.shrink(i - j);
}
const vec<Lit>& ps = learnt ? ps_ : qs; // 'ps' is now the (possibly) reduced vector of literals.
if (ps.size() == 0){
ok = false;
}else if (ps.size() == 1){
// NOTE: If enqueue takes place at root level, the assignment will be lost in incremental use (it doesn't seem to hurt much though).
if (!enqueue(ps[0]))
ok = false;
}else if (ps.size() == 2){
// Create special binary clause watch:
watches[index(~ps[0])].push(GClause_new(ps[1]));
watches[index(~ps[1])].push(GClause_new(ps[0]));
if (learnt){
check(enqueue(ps[0], GClause_new(~ps[1])));
stats.learnts_literals += ps.size();
}else
stats.clauses_literals += ps.size();
n_bin_clauses++;
}else{
// Allocate clause:
Clause* c = Clause_new(learnt, ps);
if (learnt){
// Put the second watch on the literal with highest decision level:
int max_i = 1;
int max = level[var(ps[1])];
for (int i = 2; i < ps.size(); i++)
if (level[var(ps[i])] > max)
max = level[var(ps[i])],
max_i = i;
(*c)[1] = ps[max_i];
(*c)[max_i] = ps[1];
// Bump, enqueue, store clause:
claBumpActivity(c); // (newly learnt clauses should be considered active)
check(enqueue((*c)[0], GClause_new(c)));
learnts.push(c);
stats.learnts_literals += c->size();
}else{
// Store clause:
clauses.push(c);
stats.clauses_literals += c->size();
}
// Watch clause:
watches[index(~(*c)[0])].push(GClause_new(c));
watches[index(~(*c)[1])].push(GClause_new(c));
}
}