在Z3Py中编码可允许的集合

Question

基于论证框架理论，我正在尝试使用Z3Py证明者对可允许集合进行编码。但是，我遇到了一些问题，希望对如何改进它有所指点。

基于Wikipedia的定义，当且仅当它没有冲突并且其所有参数对于E都是可接受的时，才允许使用一组参数E。

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

我开始使用Z3Py，但是在编码时遇到问题。

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，它为我提供以下解决方案：[]，[c]，[d]，[b，d]，[b，d，e]，其中只有3个是正确的：

1. [[b，d]是不正确的，因为d无法自卫（e是d的捍卫者）
2. [[d]再次不正确，因为d无法自卫]
3. [e]丢失
4. [c，e]丢失
5. [[b，e]丢失

任何帮助将不胜感激。

Answer 1

这对于解决SMT来说确实是一个很大的问题，即使您拥有非常多的set，z3也将能够相对轻松地处理此类问题。

您的编码思路是正确的，但是您正在混淆相等性：在测试变量是否与另一个变量相同时，请务必小心。为此，请使用Python的eq方法。 （如果您使用==，则将获得符号相等性测试，这不是您在此处查找的内容。）

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)

希望这些代码应易于遵循。如果没有，请随时提出具体问题。

运行时，我得到：

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

似乎与您预期的解决方案相符。