简体   繁体   中英

python build a dynamic growing truth table

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.

Useitertools.product() :

table = list(itertools.product([False, True], repeat=n))

Result for n = 3 :

[(False, False, False),
 (False, False, True),
 (False, True, False),
 (False, True, True),
 (True, False, False),
 (True, False, True),
 (True, True, False),
 (True, True, True)]

itertools really is the way to go as has been pointed out by everyone. But if you really want to see the nuts and bolts of the algorithm required for this, you should look up recursive descent . Here's how it would work in your case:

def tablize(n, truths=[]):
    if not n:
        print truths
    else:
        for i in [True, False]:
            tablize(n-1, truths+[i])

Tested, working

Hope this helps

List comprehensions are, of course, more Pythonic.

def truthtable (n):
  if n < 1:
    return [[]]
  subtable = truthtable(n-1)
  return [ row + [v] for row in subtable for v in [0,1] ]

Results, indented for clairity:

truthtable(1)
[ [0],
  [1] ]

truthtable(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] ]

As a generator function with yield :

def truthtable (n): 
  if n < 1:
    yield []
    return
  subtable = truthtable(n-1)
  for row in subtable:
    for v in [0,1]:
      yield row + [v]

Also simply changing the return from an array comprehension to a generator expression makes the return type equivalent to the yield version's generator function:

def truthtable (n):
  if n < 1:
    return [[]]
  subtable = truthtable(n-1)
  return ( row + [v] for row in subtable for v in [0,1] )

who here likes raw 1-liners?

>>> truthtable = lambda n: [[(v>>i)&1 for i in range(n-1,-1,-1)] for v in range(1<<n)] if n>0 else [[]]

100% tested and working.
(can't copy/paste result, or above code, cause I'm on a phone for Internet)

Have a look at the itertools module

In [7]: [i for i in itertools.product([0,1], repeat=3)]
Out[7]: 
[(0, 0, 0),
 (0, 0, 1),
 (0, 1, 0),
 (0, 1, 1),
 (1, 0, 0),
 (1, 0, 1),
 (1, 1, 0),
 (1, 1, 1)]

returning a datastructure representing the table is fine

...in that case range(2 ** n) is all you need. Each number in the range represents a row in the truth table. The i th bit of the binary representation of the number k is 1 if and only if the i th variable is true in the k th row of the table.

If you want an actual table you can use:

[ [ ((row >> bit_index) & 1) == 1 for bit_index in range(n)] 
  for bit_index in range(2 ** n) ]

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]

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.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM