如何简化这些布尔语句

Question

有人告诉我在python中以更简化的形式压缩它。有可能这样做吗？

(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))

Answer 1

您可以使用sympy评估和简化布尔表达式：

（注意：此示例中的解析器非常幼稚）：

(void *) 1

输出简化表达式

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))

正确性验证：

对于每个输入，请评估原始表达式和简化表达式的真值表。

exp1: 
 ~(c | ~(b & c)) | ~((b | ~c) & (~a | ~c)) | (a & ~c & (~a | (a & b & c) | (a & (~b | (b & ~c))))) 
 a | (c & ~b)
exp2: 
 (a & ~b) | (b & ~a & ~c) | (~b & ((b & ~c) | ~(a & ~b))) 
 ~b | (~a & ~c)

真值表输出：

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)}')