有没有一种已知的算法可以通过数字比较来简化布尔表达式？

这是一个慢速动态规划算法，与你所想到的一样。

from collections import defaultdict, namedtuple
from heapq import heappop, heappush
from itertools import product
from math import inf

# Constructors for Boolean expressions. False and True are also accepted.
Lt = namedtuple("Lt", ["lhs", "rhs"])
Eq = namedtuple("Eq", ["lhs", "rhs"])
Gt = namedtuple("Gt", ["lhs", "rhs"])
And = namedtuple("And", ["lhs", "rhs"])
Or = namedtuple("Or", ["lhs", "rhs"])

# Variable names. Arbitrary strings are accepted.
A = "A"
B = "B"
C = "C"

# Example formulas.
first_example = And(Gt(A, 5), Eq(A, 6))
second_example = Or(And(Gt(B, 2), Eq(C, 2)), And(Gt(B, 2), Lt(C, 2)))
third_example = Or(And(Gt(A, 1), Gt(B, 1)), And(Gt(A, 0), Gt(B, 2)))
fourth_example = Or(Lt(A, 6), Gt(A, 5))
fifth_example = Or(And(Eq(A, 2), Gt(C, 2)), And(Eq(B, 2), Lt(C, 2)))

# Returns a map from each variable to the set of values such that the formula
# might evaluate differently for variable = value-1 versus variable = value.
def get_critical_value_sets(formula, result=None):
    if result is None:
        result = defaultdict(set)
    if isinstance(formula, bool):
        pass
    elif isinstance(formula, Lt):
        result[formula.lhs].add(formula.rhs)
    elif isinstance(formula, Eq):
        result[formula.lhs].add(formula.rhs)
        result[formula.lhs].add(formula.rhs + 1)
    elif isinstance(formula, Gt):
        result[formula.lhs].add(formula.rhs + 1)
    elif isinstance(formula, (And, Or)):
        get_critical_value_sets(formula.lhs, result)
        get_critical_value_sets(formula.rhs, result)
    else:
        assert False, str(formula)
    return result


# Returns a list of inputs sufficient to compare Boolean combinations of the
# primitives returned by enumerate_useful_primitives.
def enumerate_truth_table_inputs(critical_value_sets):
    variables, value_sets = zip(*critical_value_sets.items())
    return [
        dict(zip(variables, values))
        for values in product(*({-inf} | value_set for value_set in value_sets))
    ]


# Returns both constants and all single comparisons whose critical value set is
# a subset of the given ones.
def enumerate_useful_primitives(critical_value_sets):
    yield False
    yield True
    for variable, value_set in critical_value_sets.items():
        for value in value_set:
            yield Lt(variable, value)
            if value + 1 in value_set:
                yield Eq(variable, value)
            yield Gt(variable, value - 1)


# Evaluates the formula recursively on the given input.
def evaluate(formula, input):
    if isinstance(formula, bool):
        return formula
    elif isinstance(formula, Lt):
        return input[formula.lhs] < formula.rhs
    elif isinstance(formula, Eq):
        return input[formula.lhs] == formula.rhs
    elif isinstance(formula, Gt):
        return input[formula.lhs] > formula.rhs
    elif isinstance(formula, And):
        return evaluate(formula.lhs, input) and evaluate(formula.rhs, input)
    elif isinstance(formula, Or):
        return evaluate(formula.lhs, input) or evaluate(formula.rhs, input)
    else:
        assert False, str(formula)


# Evaluates the formula on the many inputs, packing the values into an integer.
def get_truth_table(formula, inputs):
    truth_table = 0
    for input in inputs:
        truth_table = (truth_table << 1) + evaluate(formula, input)
    return truth_table


# Returns (the number of operations in the formula, the number of Ands).
def get_complexity(formula):
    if isinstance(formula, bool):
        return (0, 0)
    elif isinstance(formula, (Lt, Eq, Gt)):
        return (1, 0)
    elif isinstance(formula, And):
        ops_lhs, ands_lhs = get_complexity(formula.lhs)
        ops_rhs, ands_rhs = get_complexity(formula.rhs)
        return (ops_lhs + 1 + ops_rhs, ands_lhs + 1 + ands_rhs)
    elif isinstance(formula, Or):
        ops_lhs, ands_lhs = get_complexity(formula.lhs)
        ops_rhs, ands_rhs = get_complexity(formula.rhs)
        return (ops_lhs + 1 + ops_rhs, ands_lhs + ands_rhs)
    else:
        assert False, str(formula)


# Formula compared by complexity.
class HeapItem:
    __slots__ = ["_complexity", "formula"]

    def __init__(self, formula):
        self._complexity = get_complexity(formula)
        self.formula = formula

    def __lt__(self, other):
        return self._complexity < other._complexity

    def __le__(self, other):
        return self._complexity <= other._complexity

    def __eq__(self, other):
        return self._complexity == other._complexity

    def __ne__(self, other):
        return self._complexity != other._complexity

    def __ge__(self, other):
        return self._complexity >= other._complexity

    def __gt__(self, other):
        return self._complexity > other._complexity


# Like heapq.merge except we can add iterables dynamically.
class Merge:
    __slots__ = ["_heap", "_iterable_count"]

    def __init__(self):
        self._heap = []
        self._iterable_count = 0

    def update(self, iterable):
        iterable = iter(iterable)
        try:
            value = next(iterable)
        except StopIteration:
            return
        heappush(self._heap, (value, self._iterable_count, iterable))
        self._iterable_count += 1

    def __iter__(self):
        return self

    def __next__(self):
        if not self._heap:
            raise StopIteration
        value, index, iterable = heappop(self._heap)
        try:
            next_value = next(iterable)
        except StopIteration:
            return value
        heappush(self._heap, (next_value, index, iterable))
        return value


class Combinations:
    __slots__ = ["_op", "_formula", "_best_formulas", "_i", "_n"]

    def __init__(self, op, formula, best_formulas):
        self._op = op
        self._formula = formula
        self._best_formulas = best_formulas
        self._i = 0
        self._n = len(best_formulas)

    def __iter__(self):
        return self

    def __next__(self):
        if self._i >= self._n:
            raise StopIteration
        formula = self._op(self._formula, self._best_formulas[self._i])
        self._i += 1
        return HeapItem(formula)


# Returns the simplest equivalent formula, breaking ties in favor of fewer Ands.
def simplify(target_formula):
    critical_value_sets = get_critical_value_sets(target_formula)
    inputs = enumerate_truth_table_inputs(critical_value_sets)
    target_truth_table = get_truth_table(target_formula, inputs)
    best = {}
    merge = Merge()
    for formula in enumerate_useful_primitives(critical_value_sets):
        merge.update([HeapItem(formula)])
    best_formulas = []
    for item in merge:
        if target_truth_table in best:
            return best[target_truth_table]
        formula = item.formula
        truth_table = get_truth_table(formula, inputs)
        if truth_table in best:
            continue
        n = len(best_formulas)
        for op in [And, Or]:
            merge.update(Combinations(op, formula, best_formulas))
        best[truth_table] = formula
        best_formulas.append(formula)


print(simplify(first_example))
print(simplify(second_example))
print(simplify(third_example))
print(simplify(fourth_example))
print(simplify(fifth_example))

输出：

Eq(lhs='A', rhs=6)
And(lhs=Lt(lhs='C', rhs=3), rhs=Gt(lhs='B', rhs=2))
And(lhs=And(lhs=Gt(lhs='B', rhs=1), rhs=Gt(lhs='A', rhs=0)), rhs=Or(lhs=Gt(lhs='B', rhs=2), rhs=Gt(lhs='A', rhs=1)))
True
Or(lhs=And(lhs=Eq(lhs='B', rhs=2), rhs=Lt(lhs='C', rhs=2)), rhs=And(lhs=Gt(lhs='C', rhs=2), rhs=Eq(lhs='A', rhs=2)))

也许您可以考虑变量的间隔，例如：

(A > 5) && (A == 6)

假设你有一个变量，为它设置一个初始间隔：。AA: [-∞, ∞]

您阅读的每个条件，您都可以缩短间隔：

(A > 5)  sets the interval for A: [6, ∞]
(A == 6) sets the interval for A: [6, 6]

对于间隔的每次更新，请检查新条件是否可行，例如：

(A > 5)  sets the interval for A: [6, ∞]
(A == 5) out of the interval, impossible condition.

再举一个例子：

((B > 2) && (C == 2)) || ((B > 2) && (C < 2))

最初： 和 .B: [-∞, ∞]C: [-∞, ∞]

((B > 2) && (C == 2))

(B > 2)  sets the interval for B: [3, ∞]
(C == 2) sets the interval for C: [2, 2]

下一个条件附加了 ，因此您可以添加间隔：||

((B > 2) && (C < 2)) 

(B > 2) sets the interval for B: [3, ∞]
(C < 2) sets the interval for C: [2, 2] U [-∞, 1] = [-∞, 2]