No general solution to k-combination of multiset?

Question

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The problem to solve

Solve k-combination whose input may contain repeated items, and return only unique combinations.

e.g input is [0,1,2,2,4], k = 2, the result is:

{(0, 1), (2, 4), (1, 2), (0, 4), (1, 4), (0, 2), (2, 2)}

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The solution I can think of

The only general solution I can think is to use a set or map to extract unique results.

e.g with python

from itertools import combinations
print (set(combinations([0, 1, 2, 2, 4], 2)))

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Questions

• Is there a general solution to do this, without using a set/map for filtering.

I can generate combinations of unique items, in order (refer: https://stackoverflow.com/a/76048486), but if there are repeated items, then there will be repeated combinations.

I've read this: https://math.stackexchange.com/a/554237 , seems it says there is no general solution to get all unique combinations, though there do have a formula to get count of unique combinations.

But, I'm not sure.

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@Update - Summary & tips

• According to the answers, there are ways both via iteration & recursion to do this.
• When translating from python to go:
  • yield in python, need chan in go, or u need return all combinations (which can be huge for large input).
  • python's else for for and while is omitted when break.
  • list slicing in python will make a copy (of reference ?), while slicing a go slice will reuse the same underlying array, (so in go, if the slice being copied will be modified later, and u don't want that change in the copy, then u should copy/append it into a new slice).

Answer 1

Here’s a non-recursive enumerator for combinations in lexicographic order, optimized for simplicity over speed. Knuth’s The Art of Computer Programming almost certainly has a speed-optimized version.

def combinations(a, k):
    a = sorted(a)
    while True:
        yield a[:k]
        for i in range(k - 1, -1, -1):
            # a[i + 1 :] is sorted. Sort a[i:].
            x = a[i]
            j = i + 1
            while j < len(a):
                if x < a[j]:
                    break
                a[j - 1] = a[j]
                j += 1
            a[j - 1] = x
            # a[j] is the next greatest element after x. If there are enough
            # elements after it, rotate them into position.
            if j + (k - i) <= len(a):
                rotate(a, i, j, j + (k - i))
                break
        else:
            break


def rotate(a, i, j, k):
    reverse(a, i, j)
    reverse(a, j, k)
    reverse(a, i, k)


def reverse(a, i, k):
    for j in range((k - i) // 2):
        a[i + j], a[(k - 1) - j] = a[(k - 1) - j], a[i + j]


print(*combinations([0, 1, 2, 2, 4], 2))
print(*combinations([0, 1, 2, 2, 4], 3))
print(*combinations([0, 0, 0, 1, 1, 1], 3))

Output:

[0, 1] [0, 2] [0, 4] [1, 2] [1, 4] [2, 2] [2, 4]
[0, 1, 2] [0, 1, 4] [0, 2, 2] [0, 2, 4] [1, 2, 2] [1, 2, 4] [2, 2, 4]
[0, 0, 0] [0, 0, 1] [0, 1, 1] [1, 1, 1]

Answer 2

Below is an algorithm I developed some time back for this task:

def multiset_combinations(v, m):
    
    v = sorted(v)
    f = [0]
    j = 0

    # Populating f, a list of expanded indices. E.g. For
    #
    #    v = [2, 2, 33, 33, 33, 40]
    #
    # f would be:
    #
    #    [0, 0, 1, 1, 1, 2]
    
    for i in range(1, len(v)):
        if v[i] != v[i - 1]:
            j += 1

        f.append(j)

    v = list(set(v))
    r = [0] * m
    n = len(v)
    
    z   = f[:m]
    idx = [0] * n
    p   = len(f) - m
    
    last = f[-m:]
    last[-1] += 1
    
    # Find the first occurence of each index. E.g. For the
    # same example above, idx would be:
    #
    #   [0, 2, 5]
    
    for i in range(n):
        idx[i] = f.index(i)
    
    # The main algorithm

    while True:
        while z[-1] < n:
            for j in range(m):
                r[j] = v[z[j]]

            z[-1] += 1
            yield r.copy()

        if z == last:
            break
        
        for i in range(m - 2, -1, -1):
            if z[i] != f[p + i]:
                z[i] += 1
                
                for j, k in zip(range(i + 1, m),
                                range(idx[z[i]] + 1, idx[z[i]] + m)):
                    z[j] = f[k]
                
                break

Calling it:

print(*multiset_combination([0, 1, 2, 2, 4], 2))
#> [0, 1] [0, 2] [0, 4] [1, 2] [1, 4] [2, 2] [2, 4]

print(*multiset_combination([0, 1, 2, 2, 4], 3))
#> [0, 1, 2] [0, 1, 4] [0, 2, 2] [0, 2, 4] [1, 2, 2] [1, 2, 4] [2, 2, 4]

print(*multiset_combination([0, 0, 0, 1, 1, 1], 3))
#> [0, 0, 0] [0, 0, 1] [0, 1, 1] [1, 1, 1]

It is pretty efficient as well. It can generate 753564 in just under a second:

v1 = [1,
      2, 2,
      3, 3, 3,
      4, 4, 4, 4,
      5, 5, 5, 5, 5,
      6,
      7, 7,
      8, 8, 8,
      9, 9, 9, 9,
      10, 10, 10, 10, 10,
      11,
      12, 12,
      13, 13, 13,
      14, 14, 14, 14,
      15, 15, 15, 15, 15]
      
def get_time(v, m):
    t1 = time.time()
    list(multiset_combinations(v, m))
    t2 = time.time()
    return t2 - t1
    
get_time(v1, 10)
#> 0.8702290058135986