Question

使用python，将一串AND和OR转换为disjunctive normal form（也称为＆＃34;产品总和＆＃34;）的最佳方法是什么？

b AND (c OR (a AND d))

变为

(b AND c) OR (b AND a AND d)

我想按字典顺序对这些进行排序

(a AND b AND d) OR (b AND c)

Answer 1

也许这个图书馆可以提供帮助：pyeda

这是一种将表达式转换为DNF的方法：to_dnf()

当然你必须把你的字符串变成一个有效的pyeda表达式......

Answer 2

以下是部分回答问题的示例（即不是解析/排序）

REPR_FORM_AS_TEXT = False


class Or(frozenset):
    def __repr__(self):
        if REPR_FORM_AS_TEXT:
            return " OR ".join(repr(e) for e in self)
        else:
            return super().__repr__()


class And(frozenset):
    def __repr__(self):
        if REPR_FORM_AS_TEXT:
            if len(self) > 1:
                return "(" + " AND ".join(repr(e) for e in self) + ")"
            else:
                return "".join(repr(e) for e in self)
        else:
            return super().__repr__()


def dnf(expr, ):
    """Normalize a boolean expression to its DNF.

    Expr can be an element, it this case it returns Or({And({element})}).
    Expr can be an Or(...) / And(...) expressions, in which cases it returns also a disjunctive normalised form (removing identical elements)

    assert dnf(And((Or((4, 5, 1)), Or((1, 2)), 7, 7))) == Or(
        (
            And((2, 5, 7)),
            And((1, 5, 7)),
            And((1, 2, 7)),
            And((1, 7)),
            And((2, 4, 7)),
            And((1, 4, 7)),
        )
    )
    """

    if not isinstance(expr, (Or, And)):
        result = Or((And((expr,)),))
    elif isinstance(expr, Or):
        result = Or(se for e in expr for se in dnf(e))
    elif isinstance(expr, And):
        total = []
        for c in itertools.product(*[dnf(e) for e in expr]):
            total.append(And(se for e in c for se in e))
        result = Or(total)
    return result

Python中的析取范式

2 个答案: