/** If ex is a simple combination of Relations, then return that combination, else return null. */
private Expression sim(Expr ex) {
while (ex instanceof ExprUnary) {
ExprUnary u = (ExprUnary) ex;
if (u.op != ExprUnary.Op.NOOP && u.op != ExprUnary.Op.EXACTLYOF) break;
ex = u.sub;
}
if (ex instanceof ExprBinary) {
ExprBinary b = (ExprBinary) ex;
if (b.op == ExprBinary.Op.ARROW || b.op == ExprBinary.Op.PLUS || b.op == ExprBinary.Op.JOIN) {
Expression left = sim(b.left);
if (left == null) return null;
Expression right = sim(b.right);
if (right == null) return null;
if (b.op == ExprBinary.Op.ARROW) return left.product(right);
if (b.op == ExprBinary.Op.PLUS) return left.union(right);
else return left.join(right);
}
}
if (ex instanceof ExprConstant) {
switch (((ExprConstant) ex).op) {
case EMPTYNESS:
return Expression.NONE;
}
}
if (ex == Sig.NONE) return Expression.NONE;
if (ex == Sig.SIGINT) return Expression.INTS;
if (ex instanceof Sig) return sol.a2k((Sig) ex);
if (ex instanceof Field) return sol.a2k((Field) ex);
return null;
}
/** 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;
}
/**
* Returns a relational encoding of the problem.
*
* @return a relational encoding of the problem.
*/
public Formula rules() {
final List<Formula> rules = new ArrayList<Formula>();
rules.add(x.function(queen, num));
rules.add(y.function(queen, num));
final Variable i = Variable.unary("n");
final Variable q1 = Variable.unary("q1"), q2 = Variable.unary("q2");
// at most one queen in each row: all i: num | lone x.i
rules.add(x.join(i).lone().forAll(i.oneOf(num)));
// at most one queen in each column: all i: num | lone y.i
rules.add(y.join(i).lone().forAll(i.oneOf(num)));
// no queen in a blocked position: all q: Queen | q.x->q.y !in blocked
rules.add(q1.join(x).product(q1.join(y)).intersection(blocked).no().forAll(q1.oneOf(queen)));
// at most one queen on each diagonal
// all q1: Queen, q2: Queen - q1 |
// let xu = prevs[q2.x] + prevs[q1.x],
// xi = prevs[q2.x] & prevs[q1.x],
// yu = prevs[q2.y] + prevs[q1.y],
// yi = prevs[q2.y] & prevs[q1.y] |
// #(xu - xi) != #(yu - yi)
final Expression ordClosure = ord.closure();
final Expression q2xPrevs = ordClosure.join(q2.join(x)), q1xPrevs = ordClosure.join(q1.join(x));
final Expression q2yPrevs = ordClosure.join(q2.join(y)), q1yPrevs = ordClosure.join(q1.join(y));
final IntExpression xDiff =
(q2xPrevs.union(q1xPrevs)).difference(q2xPrevs.intersection(q1xPrevs)).count();
final IntExpression yDiff =
(q2yPrevs.union(q1yPrevs)).difference(q2yPrevs.intersection(q1yPrevs)).count();
rules.add(xDiff.eq(yDiff).not().forAll(q1.oneOf(queen).and(q2.oneOf(queen.difference(q1)))));
return Formula.and(rules);
}
/**
* Returns the conjecture theorem_3_8_5.
*
* @return theorem_3_8_5
*/
public final Formula theorem385() {
// all c: curve, p, q, r: point |
// c->p->q->r in between =>
// incident.c - q in q.(p.(c.between)) + ((c.between).r).q
final Variable c = Variable.unary("C");
final Variable p = Variable.unary("P");
final Variable q = Variable.unary("Q");
final Variable r = Variable.unary("R");
final Formula f0 = c.product(p).product(q).product(r).in(between);
final Expression e0 = q.join(p.join(c.join(between)));
final Expression e1 = c.join(between).join(r).join(q);
final Formula f1 = incident.join(c).difference(q).in(e0.union(e1));
return f0.implies(f1)
.forAll(p.oneOf(point).and(q.oneOf(point)).and(r.oneOf(point)).and(c.oneOf(curve)));
}
/** 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;
}