Find all 3^4 permutation using lists
Solution 1:
There is simple and bruteforce, but working, example:
import itertools
import pprint
balls = set('1234')
splits = []
for i in range(0,len(balls)+1): # python ranges are upper-bound exclusive
bucket_1_possibilities = (set(x) for x in itertools.combinations(balls,i))
for bucket1 in bucket_1_possibilities:
for j in range(0,len(balls)-i+1): # in the second cell there may be up to len(balls)-i balls
bucket_2_possibilities = list(set(x) for x in itertools.combinations(balls-bucket1,j))
for bucket_2 in bucket_2_possibilities:
splits.append([list(bucket1), list(bucket_2),list(balls-bucket1-bucket_2)])
pprint.pprint(splits)
print(len(splits))
Itertools.combinations (https://docs.python.org/3/library/itertools.html#itertools.combinations) allow you to get possible contents of first cell. From remaining balls you get possible contents of second cells, and so on.
Solution 2:
Instead of lists in list, a simpler interpretation is to think of a list of the balls position. See code below. The code is written in a lengthy way so that the logic is easy to follow.
combinations = []
for ball1 in range(3): # for all possible buckets we can put ball #1
for ball2 in range(3): # for all possible buckets we can put ball #2
for ball3 in range(3): # for all possible buckets we can put ball #3
for ball4 in range(3): # for all possible buckets we can put ball #4
ball_positions = [ball1, ball2, ball3, ball4]
buckets = [[], [], []]
for ball in range(4): # For each ball
bucket = ball_positions[ball] # This is the ball's position
buckets[bucket].append(ball) # So let's put the ball in this bucket
combinations.append(buckets)
for i in range(len(combinations)):
print(i, combinations[i])
Solution 3:
You can use product from itertools to iterate over all combinations of several iterators. I think you are looking for something like this:
from itertools import product
main_list = []
for i,j,k,l in product(range(3),range(3),range(3),range(3)):
sub_list = [[] for _ in range(3)]
sub_list[i].append(1)
sub_list[j].append(2)
sub_list[k].append(3)
sub_list[l].append(4)
main_list.append(sub_list)
print(main_list)
or even more pythonic - replace product(range(3),range(3),range(3),range(3))
with product(*[range(3) for _ in range(4)])
and for better printing instead of print(main_list)
use
for combination in main_list:
print(*combination)