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我正在使用带有 c api 的 Z3。是否可以找出给定Z3_ast变量是否对应于or_b1_b2下面的子句?

Z3_ast or_b1_b2 = mk_binary_or(c,mk_bool_var(c,"b1"),mk_bool_var(c,"b2"));

谢谢

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1 回答 1

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是的,Z3 API 提供了几个用于检查/遍历 AST 的函数。Z3 API 是极简的,但它具有编写函数所需的所有要素,例如:is_propositional_variableis_literalis_clause. 这是一个例子:

// Return true if the given ast is an application of the given kind
int is_app_of(Z3_context c, Z3_ast a, Z3_decl_kind k) {
   if (Z3_get_ast_kind(c, a) != Z3_APP_AST) 
       return 0;
   return Z3_get_decl_kind(c, Z3_get_app_decl(c, Z3_to_app(c, a))) == k;
}

// Return true if the given ast is an OR.
int is_or(Z3_context c, Z3_ast a) {
    return is_app_of(c, a, Z3_OP_OR);
}

// Return true if the given ast is a NOT.
int is_not(Z3_context c, Z3_ast a) {
    return is_app_of(c, a, Z3_OP_NOT);
}

// Return true if the given ast is an application of an unintepreted symbol.
int is_uninterp(Z3_context c, Z3_ast a) {
    return is_app_of(c, a, Z3_OP_UNINTERPRETED);
}

// Return true if the given ast is a uninterpreted constant.
// That is, it is application (with zero arguments) of an uninterpreted symbol.
int is_uninterp_const(Z3_context c, Z3_ast a) {
    return is_uninterp(c, a) && Z3_get_app_num_args(c, Z3_to_app(c, a)) == 0;
}

// Return true if the given ast has Boolean sort (aka type).
int has_bool_sort(Z3_context c, Z3_ast a) {
    return Z3_get_sort_kind(c, Z3_get_sort(c, a)) == Z3_BOOL_SORT;
}

// Return true if the given ast is a "propositional variable".
// That is, it has Boolean sort and it is uninterpreted.
int is_propositional_var(Z3_context c, Z3_ast a) {
    return is_uninterp_const(c, a) && has_bool_sort(c, a);
}

// Return true if the given ast is a "literal".
// That is, it is a "propositional variable" or the negation of a propositional variable.
int is_literal(Z3_context c, Z3_ast a) {
    if (is_propositional_var(c, a))
        return 1;
    if (is_not(c, a))
        return is_propositional_var(c, Z3_get_app_arg(c, Z3_to_app(c, a), 0));
    return 0;
}

// Return true if the given ast is a "clause".
// That is, it is a literal, or a disjuction (OR) of literals.
int is_clause(Z3_context c, Z3_ast a) {
    if (is_literal(c, a)) {
        return 1; // unit clause
    }
    else if (is_or(c, a)) {
        unsigned num;
        unsigned i;
        num = Z3_get_app_num_args(c, Z3_to_app(c, a));
        for (i = 0; i < num; i++) {
            if (!is_literal(c, Z3_get_app_arg(c, Z3_to_app(c, a), i)))
                return 0;
        }
        return 1;
    }
    else {
        return 0;
    }
}
于 2012-01-14T14:47:10.703 回答