从z3模型读取z3数组的func interp

Question

假设我在公式中有2个数组,我想用z3检查它的可满足性.如果z3返回sat,我想读入z3模型中的第一个数组,并将其打印为键,值对和默认值.后来我想将它转换为地图并对其进行进一步分析.这是我运行的示例:

void find_model_example_arr() {
  std::cout << "find_model_example_involving_array\n";
  context c;
  sort arr_sort = c.array_sort(c.int_sort(), c.int_sort());
  expr some_array_1 = c.constant("some_array_1", arr_sort);
  expr some_array_2 = c.constant("some_array_2", arr_sort);
  solver s(c);

  s.add(select(some_array_1, 0) > 0);
  s.add(select(some_array_2, 5) < -4);
  std::cout << s.check() << "\n";

  model m = s.get_model();
  std::cout << m << "\n";

  expr some_array_1_eval = m.eval(some_array_1);

  std::cout << "\nsome_array_1_eval = " << some_array_1_eval << "\n";

  func_decl some_array_1_eval_func_decl = some_array_1_eval.decl();
  std::cout << "\nThe Z3 expr(fun_decl) for some_array_1_eval is : " << some_array_1_eval_func_decl << "\n";

  // ERROR here
  func_interp fun_interp = m.get_func_interp(some_array_1_eval_func_decl);
  // This works well 
  //func_interp fun_interp = m.get_func_interp(m.get_func_decl(0)); 

  std::cout << "\nThe Z3 expr(fun_interp) for the array is : " << fun_interp << "\n";

  unsigned num_entries = fun_interp.num_entries();
  for(unsigned i = 0; i < num_entries; i++) 
  {
    z3::func_entry entry = fun_interp.entry(i);
    z3::expr k = entry.arg(0);

    z3::expr v = entry.value();

    std::cout << "\n(key,value): (" << k << "," << v << ")";
  }

  z3::expr default_value = fun_interp.else_value();
  std::cout << "\nDefault value:" << default_value;
}

我得到以下输出:

find_model_example_involving_array 
sat 
(define-fun some_array_1 () (Array Int Int)   
  (_ as-array k!0)) 
(define-fun some_array_2 () (Array Int Int)  
  (_ as-array k!1))
(define-fun k!0 ((x!1 Int)) Int
  (ite (= x!1 0) 1
    1)) 
(define-fun k!1 ((x!1 Int)) Int
  (ite (= x!1 5) (- 5)
    (- 5)))

some_array_1_eval = (_ as-array k!0)

The Z3 expr(fun_decl) for some_array_1_eval is : 
(declare-fun as-array () (Array  Int Int)) 
unexpected error: invalid argument

相反,如果我注释掉第一行并使用第二行,即使用以下代码块:

// ERROR here
// func_interp fun_interp = m.get_func_interp(some_array_1_eval_func_decl);
// This works well 
func_interp fun_interp = m.get_func_interp(m.get_func_decl(0)); 

我得到了我正在寻找的输出:

(key,value): (0,1)
Default value:1

这是问题所在？我怎么知道m.get_func_decl(0)是与some_array_1对应的？例如,如果我使用m.get_func_decl(1),我会得到错误的(键,值)对.换句话说,如何从模型中获取数组的func_interp(定义为z3 expr)？

Answer 1

数组模型的表示确实有点令人困惑.的含义

(define-fun some_array_1 () (Array Int Int)   
  (_ as-array k!0)) 

是数组的模型是要被解释为数组some_array_1的函数k!0(通过调用来表示as-array.后者是参数函数,因此没有参数,以获得some_array_1的模型函数的实际定义我们必须查找哪些函数as-array调用.在给定的示例中,我们可以按如下方式执行此操作,首先通过检查一些断言来确保我们实际上具有预期格式的数组模型:

assert(Z3_get_decl_kind(c, some_array_1_eval_func_decl) == Z3_OP_AS_ARRAY); 
assert(Z3_is_app(c, some_array_1_eval));
assert(Z3_get_decl_num_parameters(c, some_array_1_eval_func_decl) == 1);
assert(Z3_get_decl_parameter_kind(c, some_array_1_eval_func_decl, 0) == 
       Z3_PARAMETER_FUNC_DECL);
func_decl model_fd = func_decl(c, 
                   Z3_get_decl_func_decl_parameter(c, some_array_1_eval_func_decl, 0));

然后函数声明model_fd保存model(k!0)指定的实际函数,我们可以通过函数解释

  func_interp fun_interp = m.get_func_interp(model_fd);