Generating the powerset of a multiset - algorithm

Question

Suppose I have a multiset
{a,a,a,b,c}
from which I can make the following selections:
{}
{a}
{a,a}
{a,a,a}
{a,a,a,b}
{a,a,a,b,c}
{a,a,a,c}
{a,a,b}
{a,a,b,c}
{a,a,c}
{a,b}
{a,b,c}
{a,c}
{b}
{b,c}
{c}
Notice that the number of selections equals 16. The cardinality of a powerset of a multiset, card(P(M)), is defined on OEIS as
card(P(M)) = prod(mult(x) + 1) for all x in M
where mult(x) is the multiplicity of x in M and prod is the product of the terms. So for our example, this would amount to 4 x 2 x 2 = 16.
Let's say, for example, that the multiplicity of these elements is very high:
m(a) = 21
m(b) = 36
m(c) = 44
Then
card(P(M)) = 22 * 37 * 45 = 36630.
But if we were to treat all those elements as distinct - as a set - the cardinality of the powerset would be
card(P(S)) = 2^(21+36+44) = 2535301200456458802993406410752.
The "standard" solution for this problem suggests to just compute the powerset of the set where all of the elements are treated as distinct, and then prune the results to remove the duplicates. That's a solution with O(2^n) complexity.
Does a general algorithm for generating a powerset of a multiset with complexity on the order of card(P(M)) exist?

Answer 1

powerset recipe with itertools
What you are asking is usually called the powerset and is available as an itertools recipe, as well as a function in the module more_itertools. See the documentation:
itertools recipe;
more_itertools.powerset.
multiset = ['a', 'a', 'a', 'b', 'c']
#
# USING ITERTOOLS
#
import itertools
def powerset(iterable):
"powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
s = list(iterable)
return itertools.chain.from_iterable(itertools.combinations(s, r) for r in range(len(s)+1))
print(list(powerset(multiset)))
# [(), ('a',), ('a',), ('a',), ('b',), ('c',), ('a', 'a'), ('a', 'a'), ('a', 'b'), ('a', 'c'), ('a', 'a'), ('a', 'b'), ('a', 'c'), ('a', 'b'), ('a', 'c'), ('b', 'c'), ('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'c'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'c'), ('a', 'b', 'c'), ('a', 'a', 'a', 'b'), ('a', 'a', 'a', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'a', 'b', 'c')]
#
# USING MORE_ITERTOOLS
#
import more_itertools
print(list(more_itertools.powerset(multiset)))
# [(), ('a',), ('a',), ('a',), ('b',), ('c',), ('a', 'a'), ('a', 'a'), ('a', 'b'), ('a', 'c'), ('a', 'a'), ('a', 'b'), ('a', 'c'), ('a', 'b'), ('a', 'c'), ('b', 'c'), ('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'c'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'c'), ('a', 'b', 'c'), ('a', 'a', 'a', 'b'), ('a', 'a', 'a', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'b', 'c'), ('a', 'a', 'a', 'b', 'c')]
Powerset of a collections.Counter object
In Python, multisets are usually represented with a collections.Counter rather than with a list. The class collections.Counter is a subclass of dict; it implements dictionaries that map elements to counts, as well as several useful methods such as building a Counter by counting occurrences in a sequence.
Taking the powerset of a Counter is the topic of another question on stackoverflow:
How to generate all the subsets of a Counter?
Although I am not aware of an already-implemented method doing this in standard modules, the answer to that question presents one solution using itertools:
import collections
import itertools
multiset = collections.Counter(['a', 'a', 'a', 'b', 'c'])
# Counter({'a': 3, 'b': 1, 'c': 1})
def powerset(multiset):
range_items = [[(x, z) for z in range(y + 1)] for x,y in multiset.items()]
products = itertools.product(*range_items)
return [{k: v for k, v in pairs if v > 0} for pairs in products]
print(powerset(multiset))
# [{}, {'c': 1}, {'b': 1}, {'b': 1, 'c': 1}, {'a': 1}, {'a': 1, 'c': 1}, {'a': 1, 'b': 1}, {'a': 1, 'b': 1, 'c': 1}, {'a': 2}, {'a': 2, 'c': 1}, {'a': 2, 'b': 1}, {'a': 2, 'b': 1, 'c': 1}, {'a': 3}, {'a': 3, 'c': 1}, {'a': 3, 'b': 1}, {'a': 3, 'b': 1, 'c': 1}]

Answer 2

This will give you all the combinations of lst as tuples. Hope this answers your question.
from itertools import combinations
lst = ['a', 'a', 'a', 'b', 'c']
combs = set()
for i in range(len(lst)+1):
els = [tuple(x) for x in combinations(lst, i)]
for item in tuple(els):
combs.add(item)
print(combs)

Answer 3

The best way I like to think of it is that you start out with the empty set and then for each character you are making a choice of either adding it to the current existing sets or not adding it. Since you have 2 choices at each step, the number of total elements in a powerset is 2^n. Implementing this can be done through this Java code:
public List<List<Integer>> subsets(int[] nums) {
// list variable contains all of the subsets
List<List<Integer>> list = new ArrayList<>();
// add the empty set to start with
list.add(new ArrayList<Integer>());
for (int i = 0; i < nums.length; i++) {
//find current list size
int length = list.size();
// Loop through and add the current element to all existing
subsets
//Represents making the choice of adding the element
for (int j = 0; j < length; j++) {
// making a copy of current subset list
ArrayList<Integer> temp = new ArrayList<>(list.get(j));
temp.add(nums[i]);
list.add(temp);
}
}
return list;
}