def sub(s, out, i):

    if i >= len(s):

        print(out)

        return

    sub(s, out + [s[i]], i + 1)  # Concatenate out with [s[i]]

    sub(s, out, i + 1)

​

sub([1, 2, 3], [], 0)

/*  C++ code to generate all possible subsequences.

    Time Complexity O(n * 2^n) */

#include<bits/stdc++.h>

using namespace std;

void printSubsequences(int arr[], int n)

{

    /* Number of subsequences is (2**n -1)*/

    unsigned int opsize = pow(2, n);

    /* Run from counter 000..1 to 111..1*/

    for (int counter = 1; counter < opsize; counter++)

    {

        for (int j = 0; j < n; j++)

        {

            /* Check if jth bit in the counter is set

                If set then print jth element from arr[] */

            if (counter & (1<<j))

                cout << arr[j] << " ";

        }

        cout << endl;

    }

}

// Driver program

int main()

{

    int arr[] = {1, 2, 3, 4};

    int n = sizeof(arr)/sizeof(arr[0]);

    cout << "All Non-empty Subsequences\n";

    printSubsequences(arr, n);

    return 0;

}