lo tolmencre - 3 months ago 6x

Python Question

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`

`True`

But I am not getting what I expected:

`True + True = 2`

`True * True = 1`

`False + False = 0`

`False * False = 0`

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?

Answer

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.

Source (Stackoverflow)

Comments