True + True =2.优雅地执行布尔算术? [英] True + True = 2. Elegantly perform boolean arithmetic?

问题描述

我有嵌套的真值列表，它们代表SAT论坛，例如:

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是True，则该子句在整体上为真.

if clause_truth_value is True after the summation, the clause is true as a whole.

但是我没有得到我所期望的:

But I am not getting what I expected:

True + True = 2与预期不符

True * True = 1符合预期

False + False = 0符合预期

False * False = 0符合预期

所以... True仅是1而False是0 ...很烂，我希望算术运算符会因布尔代数而重载.有没有一种优雅的方法可以对布尔变量进行布尔算术?

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?

推荐答案

在Python中，True == 1和False == 0的类型分别为bool和False，而bool是int的子类型. .使用运算符+时，它隐式地添加了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中的按位运算符:

x& y

x & y

执行按位与".如果 y的x AND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 | y

执行按位或".如果输出的相应位为0，则输出的每个位均为0 y的x AND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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