How to find all groups of subsets of the set A? Set sections in Python

Question

I want to find a set defining algorithm Ato find all groups of subsets that satisfy the following condition:

x ∪ y ∪ .... z = A, where x, y, ... z ∈ Group

and

∀ x, y ∈ Group: x ⊆ A, y ⊆ A, x ∩ y = ∅ = {}

and

∀ x ∈ Group: x! = ∅

Note. I hope to define it correctly, I do not like mathematical symbols.

I made the following approach only to search for groups of two subsets:

from itertools import product, combinations

def my_combos(A):
  subsets = []
  for i in xrange(1, len(A)):
    subsets.append(list(combinations(A,i)))
  combos = []
  for i in xrange(1, 1+len(subsets)/2):
    combos.extend(list(product(subsets[i-1], subsets[-i])))
  if not len(A) % 2:
    combos.extend(list(combinations(subsets[len(A)/2-1], 2)))
  return [combo for combo in combos if not set(combo[0]) & set(combo[1])]

my_combos({1,2,3,4})

I get the following conclusion: these are all groups consisting of two subsets

[
  ((1,), (2, 3, 4)), 
  ((2,), (1, 3, 4)), 
  ((3,), (1, 2, 4)), 
  ((4,), (1, 2, 3)), 
  ((1, 2), (3, 4)), 
  ((1, 3), (2, 4)), 
  ((1, 4), (2, 3))
]

..... but groups of one, three, four subsets ....

Question:

How can I find a general solution?

For example, the following expected result:

my_combos({1,2,3,4})

[
  ((1,2,3,4)),
  ((1,2,3),(4,)),
  ((1,2,4),(3,)),
  ((1,3,4),(2,)),
  ((2,3,4),(1,)),
  ((1,2),(3,4)),
  ((1,3),(2,4)),
  ((1,4),(2,3)),
  ((1,2),(3,),(4,)),
  ((1,3),(2,),(4,)),
  ((1,4),(2,),(3,)),
  ((1,),(2,),(3,4)),
  ((1,),(3,),(2,4)),
  ((1,),(4,),(2,3)),
  ((1,),(4,),(2,),(3,))
]

Answer 1

:

def partitions(A):
    if not A:
        yield []
    else:
        a, *R = A
        for partition in partitions(R):
            yield partition + [[a]]
            for i, subset in enumerate(partition):
                yield partition[:i] + [subset + [a]] + partition[i+1:]

:

• .
• , .
• , - . , , frozensets, . , , .

:

for partition in partitions({1, 2, 3, 4}):
    print(partition)

[[4], [3], [2], [1]]
[[4, 1], [3], [2]]
[[4], [3, 1], [2]]
[[4], [3], [2, 1]]
[[4, 2], [3], [1]]
[[4, 2, 1], [3]]
[[4, 2], [3, 1]]
[[4], [3, 2], [1]]
[[4, 1], [3, 2]]
[[4], [3, 2, 1]]
[[4, 3], [2], [1]]
[[4, 3, 1], [2]]
[[4, 3], [2, 1]]
[[4, 3, 2], [1]]
[[4, 3, 2, 1]]

( ):

from time import time
print('elements partitions seconds')
for n in range(14):
    t0 = time()
    number = sum(1 for partition in partitions(range(n)))
    print('{:5}{:10}{:11.2f}'.format(n, number, time() - t0))

elements partitions seconds
    0         1       0.00
    1         1       0.00
    2         2       0.00
    3         5       0.00
    4        15       0.00
    5        52       0.00
    6       203       0.00
    7       877       0.00
    8      4140       0.06
    9     21147       0.07
   10    115975       0.36
   11    678570       2.20
   12   4213597      13.56
   13  27644437      87.59

OEIS.