python build a dynamic growing truth table

user_43679 0
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My question is simple: "how to build a dynamic growing truth table in python in an elegant way?"

for n=3

for p in False, True:
    for q in False, True:
        for r in False, True:
            print '|{0} | {1} | {2} |'.format(int(p),int(q), int(r))

for n=4

for p in False, True:
    for q in False, True:
        for r in False, True:
            for s in False, True:
                print '|{0} | {1} | {2} | {3}'.format(int(p),int(q), int(r), int(s))

I would like to have a function which takes n as a parameter and builds up the table, it is not necessary to print the table, returning a data structure representing the table is fine also.

truthtable logic python
8 Answers

Easy math aproach:

a = lambda x: [x//4%2,x//2%2,x%2]

for i in range(8):
    print(a(i))

[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]

EDIT:

A more general format wold be:

def truth_table(n):
    for i in range(2**n):
        line = [i//2**j%2 for j in reversed(range(n))]
        print(line)

This will only print the result as:

>>> truth_table(3)
[0, 0, 0]
[0, 0, 1]
[0, 1, 0]
[0, 1, 1]
[1, 0, 0]
[1, 0, 1]
[1, 1, 0]
[1, 1, 1]
user_26534860

Here's an implementation that uses functional programming, doesn't use loops, and doesn't import any external libraries:

get_bits = lambda n: [*map(lambda x:[*map(int,bin(x)[2:].zfill(n))],range(2**n))]

Then call get_bits with the desired number of parameters:

>>> get_bits(3)
[[0, 0, 0],
 [0, 0, 1],
 [0, 1, 0],
 [0, 1, 1],
 [1, 0, 0],
 [1, 0, 1],
 [1, 1, 0],
 [1, 1, 1]]

Check out my GitHub repo to see my fleshed-out project for generating Truth Tables from boolean expression strings.

user_131619030
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