Finding all subsets of specified size - algorithm

Question

I've been scratching my head about this for two days now and I cannot come up with a solution. What I'm looking for is a function f(s, n) such that it returns a set containing all subsets of s where the length of each subset is n.
Demo:
s={a, b, c, d}
f(s, 4)
{{a, b, c, d}}
f(s, 3)
{{a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}}
f(s, 2)
{{a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}}
f(s, 1)
{{a}, {b}, {c}, {d}}
I have a feeling that recursion is the way to go here. I've been fiddling with something like
f(S, n):
for s in S:
t = f( S-{s}, n-1 )
...
But this does not seem to do the trick. I did notice that len(f(s,n)) seems to be the binomial coefficient bin(len(s), n). I guess this could be utilized somehow.
Can you help me please?

Answer 1

Let us call n the size of the array and k the number of elements to be out in a subarray.
Let us consider the first element A[0] of the array A.
If this element is put in the subset, the problem becomes a (n-1, k-1) similar problem.
If not, it becomes a (n-1, k) problem.
This can be simply implemented in a recursive function.
We just have to pay attention to deal with the extreme cases k == 0 or k > n.
During the process, we also have to keep trace of:
n: the number of remaining elements of A to consider
k: the number of elements that remain to be put in the current subset
index: the index of the next element of A to consider
The current_subset array that memorizes the elements already selected.
Here is a simple code in c++ to illustrate the algorithm
Output
For 5 elements and subsets of size 3:
3 4 5
2 4 5
2 3 5
2 3 4
1 4 5
1 3 5
1 3 4
1 2 5
1 2 4
1 2 3
#include <iostream>
#include <vector>
void print (const std::vector<std::vector<int>>& subsets) {
for (auto &v: subsets) {
for (auto &x: v) {
std::cout << x << " ";
}
std::cout << "\n";
}
}
// n: number of remaining elements of A to consider
// k: number of elements that remain to be put in the current subset
// index: index of next element of A to consider
void Get_subset_rec (std::vector<std::vector<int>>& subsets, int n, int k, int index, std::vector<int>& A, std::vector<int>& current_subset) {
if (n < k) return;
if (k == 0) {
subsets.push_back (current_subset);
return;
}
Get_subset_rec (subsets, n-1, k, index+1, A, current_subset);
current_subset.push_back(A[index]);
Get_subset_rec (subsets, n-1, k-1, index+1, A, current_subset);
current_subset.pop_back(); // remove last element
return;
}
void Get_subset (std::vector<std::vector<int>>& subsets, int subset_length, std::vector<int>& A) {
std::vector<int> current_subset;
Get_subset_rec (subsets, A.size(), subset_length, 0, A, current_subset);
}
int main () {
int subset_length = 3; // subset size
std::vector A = {1, 2, 3, 4, 5};
int size = A.size();
std::vector<std::vector<int>> subsets;
Get_subset (subsets, subset_length, A);
std::cout << subsets.size() << "\n";
print (subsets);
}
Live demo

Answer 2

One way to solve this is by backtracking. Here's a possible algorithm in pseudo code:
def backtrack(input_set, idx, partial_res, res, n):
if len(partial_res == n):
res.append(partial_res[:])
return
for i in range(idx, len(input_set)):
partial_res.append(input_set[i])
backtrack(input_set, idx+1, partial_res, res, n) # path with input_set[i]
partial_res.pop()
backtrack(input_set, idx+1, partial_res, res, n) # path without input_set[i]
Time complexity of this approach is O(2^len(input_set)) since we make 2 branches at each element of input_set, regardless of whether the path leads to a valid result or not. The space complexity is O(len(input_set) choose n) since this is the number of valid subsets you get, as you correctly pointed out in your question.
Now, there is a way to optimize the above algorithm to reduce the time complexity to O(len(input_set) choose n) by pruning the recursive tree to paths that can lead to valid results only.
If n - len(partial_res) < len(input_set) - idx + 1, we are sure that even if we took every remaining element in input_set[idx:] we are still short at least one to reach n. So we can employ this as a base case and return and prune.
Also, if n - len(partial_res) == len(input_set) - idx + 1, this means that we need each and every element in input_set[idx:] to get the required n length result. Thus, we can't skip any elements and so the second branch of our recursive call becomes redundant.
backtrack(input_set, idx+1, partial_res, res, n) # path without input_set[i]
We can skip this branch with a conditional check.
Implementing these base cases correctly, reduces the time complexity of the algorithm to O(len(input_set) choose k), which is a hard limit because that's the number of subsets that there are.

Answer 3

subseqs 0 _ = [[]]
subseqs k [] = []
subseqs k (x:xs) = map (x:) (subseqs (k-1) xs) ++ subseqs k xs
Live demo
The function looks for subsequences of (non-negative) length k in a given sequence. There are three cases:
If the length is 0: there is a single empty subsequence in any sequence.
Otherwise, if the sequence is empty: there are no subsequences of any (positive) length k.
Otherwise, there is a non-empty sequence that starts with x and continues with xs, and a positive length k. All our subsequences are of two kinds: those that contain x (they are subsequences of xs of length k-1, with x stuck at the front of each one), and those that do not contain x (they are just subsequences of xs of length k).
The algorithm is a more or less literal translation of these notes to Haskell. Notation cheat sheet:
[] an empty list
[w] a list with a single element w
x:xs a list with a head of x and a tail of xs
(x:) a function that sticks an x in front of any list
++ list concatenation
f a b c a function f applied to arguments a b and c

Answer 4

Here is a non-recursive python function that takes a list superset and returns a generator that produces all subsets of size k.
def subsets_k(superset, k):
if k > len(superset):
return
if k == 0:
yield []
return
indices = list(range(k))
while True:
yield [superset[i] for i in indices]
i = k - 1
while indices[i] == len(superset) - k + i:
i -= 1
if i == -1:
return
indices[i] += 1
for j in range(i + 1, k):
indices[j] = indices[i] + j - i
Testing it:
for s in subsets_k(['a', 'b', 'c', 'd', 'e'], 3):
print(s)
Output:
['a', 'b', 'c']
['a', 'b', 'd']
['a', 'b', 'e']
['a', 'c', 'd']
['a', 'c', 'e']
['a', 'd', 'e']
['b', 'c', 'd']
['b', 'c', 'e']
['b', 'd', 'e']
['c', 'd', 'e']