问题
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]]]
representing
([x0=0] + [x0=0] + [x0=1]) * ([x0=1] + [x1=1] + [-x2=1]) * ([-x3=0] + [-x3=0] + [-x2=1])
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.
like this:
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]
if clause_truth_value is True after the summation, the clause is true as a whole.
But I am not getting what I expected:
True + True = 2 that's not as expected
True * True = 1 that's as expected
False + False = 0 that's as expected
False * False = 0 that's as expected
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?
回答1:
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
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
From Bitwise Operators in Python:
x & y
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
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.
回答2:
There is a way to do boolean arithmetic. That way is to use boolean operators.
- and, as you call "*", is
and. - or, as you call "+", is
or.
来源：https://stackoverflow.com/questions/38276843/true-true-2-elegantly-perform-boolean-arithmetic