我试图在c ++中找到线性系统的所有整数解。但它并没有找到所有解决方案(只有一些)。问题在哪里?
expr mk_and(expr_vector args) {
vector<Z3_ast> array;
for (int i = 0; i < args.size(); i++)
array.push_back(args[i]);
return to_expr(args.ctx(), Z3_mk_and(args.ctx(), array.size(), &(array[0])));
}
expr mk_or(expr_vector args) {
vector<Z3_ast> array;
for (int i = 0; i < args.size(); i++)
array.push_back(args[i]);
return to_expr(args.ctx(), Z3_mk_or(args.ctx(), array.size(), &(array[0])));
}
expr mk_add(expr_vector args) {
vector<Z3_ast> array;
for (int i = 0; i < args.size(); i++)
array.push_back(args[i]);
return to_expr(args.ctx(), Z3_mk_add(args.ctx(), array.size(), &(array[0])));
}
vector<vector<int>> solveLinearEquations(vector<vector<int>> linEquations, vector<int> vect, int from, int to){
context c;
const unsigned N = linEquations[0].size();
expr_vector x(c);
for (unsigned i = 0; i < N; i++) {
std::stringstream x_name;
x_name << "x_" << i;
x.push_back(c.int_const(x_name.str().c_str()));
}
solver s(c);
for (unsigned i = 0; i < N; i++) {
s.add(x[i] >= from);
s.add(x[i] <= to);
}
for (int i = 0; i < linEquations.size(); i++) {
expr_vector linVector(c);
for (int j = 0; j< linEquations[i].size(); j++) {
if (linEquations[i][j] != 0) {
linVector.push_back(linEquations[i][j] * x[j]);
}
}
s.add(mk_add(linVector) == vect[i]);
}
vector<vector<int> > solutions;
while(true) {
if (s.check() == sat) {
model m = s.get_model();
expr_vector ve(c);
vector<int> sol; sol.clear();
for (unsigned i = 0; i < N; i++) {
ve.push_back(x[i] != m.eval(x[i]));
int val;
Z3_get_numeral_int(c, m.eval(x[i]), &val);
sol.push_back(val);
}
solutions.push_back(sol);
expr_vector ve2(c);
ve2.push_back(mk_and(ve));
s.add(mk_or(ve2));
}
else {
break;
}
}
return solutions;
}
例如:linearSystem {{1,1,0,0,0,0},{0,0,1,1,1,1},{0,1,0,1,2,3} ,{0,3,0,1,4,9}},vect {3,4,6,12},从= 0到= 3只找到2个解,但有5个。任何想法?
答案 0 :(得分:1)
s.add(mk_or(ve2));
做什么?删除它。此外,您需要将s.add(mk_or(ve));
更改为x0 != 1 || x1 != 7 || ...
。
不确定你想在那里建立什么样的表达。您需要Array of STD Objects
。