我有代表 SAT 论坛的嵌套真值列表,如下所示:
I have nested lists of truth values representing SAT forumlas, like this:
[[[0, True, False], [0, True, False], [0, True, 1]], [[0, True, True], [2, True, True], [3, False, True]], [[1, False, False], [1, False, False], [3, False, True]]]
代表
([x0=0] + [x0=0] + [x0=1]) * ([x0=1] + [x1=1] + [-x2=1]) * ([-x3=0] + [-x3=0] + [-x2=1])
我想计算整个公式的真值.第一步是将每个子句中文字的真值相加.
I would like to calculate the truth value of the whole formula. First step would be adding up the truth values of the literals in each clause.
像这样:
clause_truth_value = None
for literal in clause:
# multiply polarity of literal with its value
# sum over all literals
clause_truth_value += literal[1]*literal[2]
如果clause_truth_value
求和后为True
,则子句整体为真.
if clause_truth_value
is True
after the summation, the clause is true as a whole.
但我没有得到预期的结果:
But I am not getting what I expected:
True + True = 2
不符合预期
True * True = 1
符合预期
False + False = 0
符合预期
False * False = 0
符合预期
所以... True 只是 1 而 False 是 0... 这太糟糕了,我预计算术运算符会为布尔代数重载.有没有一种优雅的方法可以用布尔变量进行布尔运算?
so... True is simply 1 and False is 0... that sucks, I expected the arithmetic operators to be overloaded for the boolean algebra. Is there an elegant way to do do boolean arithmetic with boolean variables?
在 Python 中,True == 1
和 False == 0
,为 True
和 False
是 bool
类型,它是 int
的子类型.当您使用运算符 +
时,它会隐式添加 True
和 False
的整数值.
In Python, True == 1
and False == 0
, as True
and False
are type bool
, which is a subtype of int
. When you use the operator +
, it is implicitly adding the integer values of True
and False
.
int(True)
# 1
int(False)
# 0
您真正想要的是将 True
和 False
视为二进制数.
What you really want is to treat True
and False
as binary numbers.
int(False & False)
# 0
int(True & False)
# 0
int(True & True)
# 1
来自 Python 中的位运算符:
x &是的
执行按位与".输出的每一位为 1,如果x与y的对应位为1,否则为0.
Does a "bitwise and". Each bit of the output is 1 if the corresponding bit of x AND of y is 1, otherwise it's 0.
x |是的
执行按位或".如果对应的位为 0,则输出的每一位为 0x AND of y 为 0,否则为 1.
Does a "bitwise or". Each bit of the output is 0 if the corresponding bit of x AND of y is 0, otherwise it's 1.
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