python - 在 Z3Py 中编码可接受的集合

Question

基于论证框架理论，我尝试使用 Z3Py 证明者对可接受的集合进行编码。然而，我遇到了一些问题，希望能得到任何关于如何改进它的指示。

基于 Wikipedia 的定义，一组参数 E 是可接受的，当且仅当它是无冲突的，并且其所有参数对于 E 都是可接受的。

基于以下带参数的有向图:a、b、c、d、e 和关系:(a, d)、(a, b)、(b, c)、(c, b)、(c , d), (e, a) 容许集合的正确解是:[], [e], [c], [c, e], [b, e], [b, d, e]

我开始使用 Z3Py，但在编码时遇到问题。

到目前为止我有这个:

from z3 import *

a, b, c, d, e = Bools('a b c d e')
arguments = [a, b, c, d, e]
solver = Solver()
relations = [(a, d), (a, b), (b, c), (c, b), (c, d), (e, a)]

# ensure there are no conflicting arguments
for relation in relations:
    solver.append(Implies(relation[0], Not(relation[1])))

for argument in arguments:
    # if not attacked then removed any arguments attacked by this argument
    if len([item for item in relations if item[1] == argument]) == 0:
        solver.append([Not(attacked[1]) for attacked in relations if attacked[0] == argument])

    # if self attacking remove the argument
    if len([item for item in relations if item[1] == argument and item[0] == argument]) > 0:
        solver.append(Not(argument))

    # get all attackers
    attackers = [item[0] for item in relations if item[1] == argument]
    for attacker in attackers:
        # get defenders of the initial argument (arguments attacking the attacker)
        defenders = [item[0] for item in relations if item[1] == attacker]
        defending_z3 = []
        if len(defenders) > 0:
            for defender in defenders:
                if defender not in attackers:
                    defending_z3.append(defender)
            if len(defending_z3) > 0:
                solver.append(Implies(Or([defend for defend in defending_z3]), argument))
        else:
            solver.append(Not(argument))

# get solutions
solutions = []
while solver.check() == sat:
    model = solver.model()
    block = []
    solution = []
    for var in arguments:
        v = model.eval(var, model_completion=True)
        block.append(var != v)
        if is_true(v):
            solution.append(var)
    solver.add(Or(block))
    solutions.append(solution)

for solution in solutions:
    print(solution)

执行时，它给出以下解决方案:[]、[c]、[d]、[b, d]、[b, d, e]，其中只有 3 个是正确的:

1. [b, d] 不正确，因为 d 无法保卫自己(e 是 d 的防守者)
2. [d] 又不正确，因为 d 无法保护自己
3. 缺少[e]
4. 缺少[c, e]
5. 缺少[b, e]

如有任何帮助，我们将不胜感激。

Answer 1

对于 SMT 解决来说，这确实是一个很大的问题，即使您有非常大的集合，z3 也能够相对轻松地处理此类问题。

您的编码想法是正确的，但您混淆了相等性:在测试一个变量是否与另一个变量相同时应该小心。为此，请使用 Python 的 eq 方法。 (如果您使用 ==，那么您将获得符号相等测试，这不是您在这里寻找的。)

鉴于此，我倾向于将您的问题编码如下:

from z3 import *

a, b, c, d, e = Bools('a b c d e')
arguments = [a, b, c, d, e]
solver = Solver()
relations = [(a, d), (a, b), (b, c), (c, b), (c, d), (e, a)]

# No conflicting arguments
for relation in relations:
    solver.add(Not(And(relation[0], relation[1])))

# For each element, if it is attacked, then another element that
# attacks the attacker must be present:
for argument in arguments:
    # Find the attackers for this argument:
    attackers = [relation[0] for relation in relations if argument.eq(relation[1])]

    # We must defend against each attacker:
    for attacker in attackers:
        defenders = [relation[0] for relation in relations if relation[1].eq(attacker)]

        # One of the defenders must be included:
        solver.add(Implies(argument, Or(defenders)))

# Collect the solutions:
while solver.check() == sat:
    model = solver.model()
    block = []
    solution = []

    for var in arguments:
        v = model.eval(var, model_completion=True)
        block.append(var != v)
        if is_true(v):
            solution.append(str(var))

    solver.add(Or(block))
    solution.sort()
    print(solution)

希望代码应该易于理解。如果没有，请随时提出具体问题。

当我运行它时，我得到:

[]
['c']
['c', 'e']
['e']
['b', 'e']
['b', 'd', 'e']

这似乎与您预测的解决方案相符。