Python 如何简化这些布尔语句

我被告知用python将其压缩为更简化的形式。是否可以有秩序地这样做

(not ((b or not c) and (not a or not c))) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))

(not (a and not b) or (not c and b)) and (not b) or (not a and b and not c) or (a and not b))

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可以使用Symphy计算和简化布尔表达式：

注意：本例中的解析器相当幼稚：

from sympy.logic.boolalg import to_dnf
from sympy.abc import a, b, c

def translate(expr):
    e = list(expr)
    res = [' ' for _ in range(len(e))]
    for start in range(len(e)):
        if expr[start: start+3] == 'not':
            res[start] = '~'
            res[start+1] = ''
            res[start+2] = ''
        elif expr[start: start+3] == 'and':
            res[start] = '&'
            res[start+1] = ''
            res[start+2] = ''
        else:
            if res[start] == ' ':
                res[start] = e[start]

    expr = ''.join(res)         
    e = list(expr)
    res = [' ' for _ in range(len(e))]
    for start in range(len(e)):
        if expr[start: start+2] == 'or':
            res[start] = '|'
            res[start+1] = ''
        else:
            if res[start] == ' ':
                res[start] = e[start]

    res = [elt for elt in res if elt != ' ' or elt != '']
    return ''.join(res)


exp1 = '(not ((b or not c) and (not a or not c))) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))'
exp2 = '(not (a and not b) or (not c and b)) and (not b) or (not a and b and not c) or (a and not b)'


print('exp1:', '\n', eval(translate(exp1)), '\n', to_dnf(eval(translate(exp1)), simplify=True))
print('exp2:', '\n', eval(translate(exp2)), '\n', to_dnf(eval(translate(exp2)), simplify=True))

输出简化表达式 正确性验证： 对于每个输入，计算原始表达式和简化表达式的真值表

def evaluate(a, b, c):
    exp1 = (not ((b or not c) and (not a or not c))) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))
    exp2 = (not (a and not b) or (not c and b)) and (not b) or (not a and b and not c) or (a and not b)
    return exp1, exp2

def evaluate2(a, b, c):
    exp1 = a or (c and not b)
    exp2 = not b or (not a and not c)
    return exp1, exp2

values = [(1, 1, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)]
for vals in values:
    print(f'{vals}: {evaluate(*vals)}, {evaluate2(*vals)}')

真值表输出：

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@User2357112Karnaugh映射很好，但是源代码不是太复杂，无法填充单元格吗？不太确定这个问题与Python有什么关系。在任何使用布尔运算符的语言中都需要执行相同的过程。而且

def evaluate(a, b, c):
    exp1 = (not ((b or not c) and (not a or not c))) or (not (c or not (b and c))) or (a and not c) and (not a or (a and b and c) or (a and ((b and not c) or (not b))))
    exp2 = (not (a and not b) or (not c and b)) and (not b) or (not a and b and not c) or (a and not b)
    return exp1, exp2

def evaluate2(a, b, c):
    exp1 = a or (c and not b)
    exp2 = not b or (not a and not c)
    return exp1, exp2

values = [(1, 1, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)]
for vals in values:
    print(f'{vals}: {evaluate(*vals)}, {evaluate2(*vals)}')

(1, 1, 1): (True, False), (1, False)
(1, 1, 0): (True, False), (1, False)
(1, 0, 1): (True, True), (1, True)
(0, 1, 1): (0, 0), (False, False)
(1, 0, 0): (True, True), (1, True)
(0, 1, 0): (0, True), (0, True)
(0, 0, 1): (True, True), (True, True)
(0, 0, 0): (0, True), (0, True)