For n variables, there exists 2^(2^n) distinct boolean functions. For example, if n=2, then there exists 16 possible boolean functions which can be written in sum of product form, or product of sum forms. The number of possible functions increases exponentially with n.
I am looking for an algorithm which can generate all these possible boolean rules for n variables. I have tried to search at various places, but have not found anything suitable till now. Most of the algorithms are related to simplifying or reducing boolean functions to standard forms.
I know even for the number of rules become too large even for n=8 or 9, but can somebody please help me out with the relevant algorithm if it exists?
A boolean function of n variables has 2^n possible inputs. These can be enumerated by printing out the binary representation of values in the range 0 <= x < 2^n.
For each one of the those possible inputs, a boolean function can output 0 or 1. To enumerate all the possibilities (i.e. every possible truth table). List the binary values in range 0 <= x < 2^(2^n).
Here's the algorithm in Python:
from __future__ import print_function
from itertools import product # forms cartesian products
n = 3 # number of variables
print('All possible truth tables for n =', n)
inputs = list(product([0, 1], repeat=n))
for output in product([0, 1], repeat=len(inputs)):
print()
print('Truth table')
print('-----------')
for row, result in zip(inputs, output):
print(row, '-->', result)
The output looks like this:
All possible truth tables for n = 3
Truth table
-----------
(0, 0, 0) --> 0
(0, 0, 1) --> 0
(0, 1, 0) --> 0
(0, 1, 1) --> 0
(1, 0, 0) --> 0
(1, 0, 1) --> 0
(1, 1, 0) --> 0
(1, 1, 1) --> 0
Truth table
-----------
(0, 0, 0) --> 0
(0, 0, 1) --> 0
(0, 1, 0) --> 0
(0, 1, 1) --> 0
(1, 0, 0) --> 0
(1, 0, 1) --> 0
(1, 1, 0) --> 0
(1, 1, 1) --> 1
Truth table
-----------
(0, 0, 0) --> 0
(0, 0, 1) --> 0
(0, 1, 0) --> 0
(0, 1, 1) --> 0
(1, 0, 0) --> 0
(1, 0, 1) --> 0
(1, 1, 0) --> 1
(1, 1, 1) --> 0
Truth table
-----------
(0, 0, 0) --> 0
(0, 0, 1) --> 0
(0, 1, 0) --> 0
(0, 1, 1) --> 0
(1, 0, 0) --> 0
(1, 0, 1) --> 0
(1, 1, 0) --> 1
(1, 1, 1) --> 1
... and so on
If you want the output in algebraic form rather than truth tables, the algorithm is the same:
from __future__ import print_function
from itertools import product # forms cartesian products
n = 3 # number of variables
variables = 'abcdefghijklmnopqrstuvwxyz'[:n]
pairs = [('~'+var, var) for var in variables]
print('All possible algebraic expressions for n =', n)
inputs = list(product(*pairs))
for i, outputs in enumerate(product([0, 1], repeat=len(inputs))):
terms = [''.join(row) for row, output in zip(inputs, outputs) if output]
if not terms:
terms = ['False']
print('Function %d:' % i, ' or '.join(terms))
The output looks like this:
All possible algebraic expressions for n = 3
Function 0: False
Function 1: abc
Function 2: ab~c
Function 3: ab~c or abc
Function 4: a~bc
Function 5: a~bc or abc
Function 6: a~bc or ab~c
Function 7: a~bc or ab~c or abc
Function 8: a~b~c
Function 9: a~b~c or abc
Function 10: a~b~c or ab~c
Function 11: a~b~c or ab~c or abc
Function 12: a~b~c or a~bc
Function 13: a~b~c or a~bc or abc
Function 14: a~b~c or a~bc or ab~c
Function 15: a~b~c or a~bc or ab~c or abc
Function 16: ~abc
Function 17: ~abc or abc
Function 18: ~abc or ab~c
Function 19: ~abc or ab~c or abc
Function 20: ~abc or a~bc
Function 21: ~abc or a~bc or abc
Function 22: ~abc or a~bc or ab~c
Function 23: ~abc or a~bc or ab~c or abc
Function 24: ~abc or a~b~c
Function 25: ~abc or a~b~c or abc
Function 26: ~abc or a~b~c or ab~c
Function 27: ~abc or a~b~c or ab~c or abc
Function 28: ~abc or a~b~c or a~bc
Function 29: ~abc or a~b~c or a~bc or abc
Function 30: ~abc or a~b~c or a~bc or ab~c
Function 31: ~abc or a~b~c or a~bc or ab~c or abc
Function 32: ~ab~c
Function 33: ~ab~c or abc
... and so on
As mentioned in comments, there's a one-to-one relation between numbers and truth tables. For example, we can represent the truth table
0 0 0 | 1
0 0 1 | 1
0 1 0 | 0
0 1 1 | 0
1 0 0 | 1
1 0 1 | 0
1 1 0 | 1
1 1 1 | 0
by the binary number 01010011 (the topmost row is represented by the least-significant bit).
It is obviously just a matter of looping over numbers to generate these representations:
for f := 0 to 2^(2^n) - 1:
# do something with f
What can we do with f? We can evaluate it, for example. Say we want to know f(0,1,0). It's as simple as interpreting the argument as the binary number x = 010 and doing some bit-magic:
def evaluate(f, x):
return (f & (1<<x)) != 0
We can also find its disjunctive normal form by just checking which bits are 0:
def dnf(f):
for x := 0 to 2^n - 1:
if f & (1<<x) != 0:
print binary(x) + " OR "
Giving a result like 000 OR 001 OR 100 OR 110 (OR) for the function above.
I've changed the code posted by Raymond Hettinger. Now you can:
- generate all Boolean expressions up to n_ary,
- pick a particular function,
- perform evaluation, and
- get answers in binary or True/False manner.
Code:
from itertools import product
def allBooleanFunctions(kk):
"""Finds all Boolean functions for indegree kk"""
inputs1 = list(product([0, 1], repeat=kk))
variables = 'abcdefghijklmnopqrstuvwxyz'[:kk]
pairs = [('( not '+var+' )', var) for var in variables]
inputs = list(product(*pairs))
bool_func=[]
for i, outputs in enumerate(product([0, 1], repeat=len(inputs))):
terms = [' and '.join(row) for row, output in zip(inputs, outputs) if output]
if not terms:
terms = ['False']
bool_func.append(('('+(') or ('.join(terms))+')'))
return bool_func
n_ary=2 # number of inputs; keep it n_ary<=5
boolean_function_analytical=allBooleanFunctions(n_ary)
print('All Boolean functions of indegree:'+str(n_ary)+'\n')
print(boolean_function_analytical)
print
print('A Boolean function:'+'\n')
print(boolean_function_analytical[2])
# Evaluate third boolean function:
a=1 # first input
b=0 # second input
c=0 # third input
d=0 # fourth input
print('a='+str(a)+'; b='+str(b)+'; c='+str(c)+'; d='+str(d)+'\n')
print('Evaluation in 0/1 manner:')
print(int(eval((boolean_function_analytical[2]))))
print('Evaluation in True/False manner:')
print(bool(eval((boolean_function_analytical[2]))))
Result:
All Boolean functions of indegree:2
['(False)', '(a and b)', '(a and ( not b ))', '(a and ( not b )) or (a and b)', '(( not a ) and b)', '(( not a ) and b) or (a and b)', '(( not a ) and b) or (a and ( not b ))', '(( not a ) and b) or (a and ( not b )) or (a and b)', '(( not a ) and ( not b ))', '(( not a ) and ( not b )) or (a and b)', '(( not a ) and ( not b )) or (a and ( not b ))', '(( not a ) and ( not b )) or (a and ( not b )) or (a and b)', '(( not a ) and ( not b )) or (( not a ) and b)', '(( not a ) and ( not b )) or (( not a ) and b) or (a and b)', '(( not a ) and ( not b )) or (( not a ) and b) or (a and ( not b ))', '(( not a ) and ( not b )) or (( not a ) and b) or (a and ( not b )) or (a and b)']
A Boolean function:
(a and ( not b ))
a=1; b=0; c=0; d=0
Evaluation in 0/1 manner:
1
Evaluation in True/False manner:
True
Have fun!!!
来源：https://stackoverflow.com/questions/22600552/algorithm-for-generating-all-possible-boolean-functions-of-n-variables