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I have to calculate power set of set which may have more elements upto 10^5. I tried an algo and the code below but it failed (I think cause large value of pow(2, size)).

void printPowerSet(int *set, int set_size)
{
unsigned int pow_set_size = pow(2, set_size);
int counter, j,sum=0;
 for(counter = 0; counter < pow_set_size; counter++)
{
  for(j = 0; j < set_size; j++)
   {
      if(counter & (1<<j))
     std::cout<<set[i]<<" ";
   }
   std::cout<<sum;
   sum=0;
   printf("\n");
}
}

Is there any other algorithm or how can I fix this one (if it is possible)??
OR
Can you suggest me how to do it i.e. finding subset of large set.

As pointed out in an answer it seems I'm stuck in X-Y problem. Basically, I need sum of all subsets of any set. Now if you can suggest me any other approach to solve the problem.
Thank you.

Community
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DKP
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  • "I think cause large value of `pow(2, size)`" - Well, does it work correctly with a small value of `size`??? Don't **conjecture** what the problem is, debug your code, find out **exactly** what the problem is, and if you're still having problems resolving it on your own, then publish it here along with your intermediate conclusions. – barak manos Apr 04 '16 at 09:36
  • The number of all subsets is ~10^30102. Are you sure you have to calculate all subsets? Seems like an [X-Y problem](http://meta.stackexchange.com/questions/66377/what-is-the-xy-problem) for me. – awesoon Apr 04 '16 at 09:36
  • @barak yes, it works correctly for small `size`. – DKP Apr 04 '16 at 09:42
  • And what **did** you see when you **used** a large value of `size` and **debugged** your code step-by-step? – barak manos Apr 04 '16 at 09:43
  • At what line of code exactly? What were the values of each of the (relevant) variables? – barak manos Apr 04 '16 at 09:46
  • well `size` was approx `10^4` and say I have stored set as-- for(int i=1;i<=t;++i) { cin>>i; } where `t==10^5` – DKP Apr 04 '16 at 09:49
  • Try for values between `n = 15 to n = 30` you will get you point of failure, by that i mean the size is growing exponentially, hence it would become impossible to compute. – uSeemSurprised Apr 04 '16 at 09:52
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    Of course it is possible to compute this set. You have to be careful with the implementation, which _will_take a long time---I am willing to bet it won't terminate in your lifetime---,but it will correctly compute the power set eventually. – blazs Apr 04 '16 at 09:54
  • As said by @blazs and in my answer it is `not possible` to print the entire power set not in this lifetime not in the next million lifetimes. – uSeemSurprised Apr 04 '16 at 10:03
  • @uSeemSurprised no, I'm not getting the result. – DKP Apr 04 '16 at 10:04
  • even though the power set of a set of numbers of size 10^5 is a trillion billion gazillion times bigger than the number of atoms in the universe, i don't that means we shouldn't try to come up with an algorithm to compute it. I just means we can't store the set and we need to be careful about buffer sizes and recursion depth. – Richard Hodges Apr 04 '16 at 10:27

3 Answers3

1

Here is an algorithm which will print out the power set of any set that will fit in your computer's memory.

Given enough time, it will print the power set of a set of length 10^5.

However, "enough time" will be something like several trillion billion gazillion years.

c++14

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

#include <iostream>

void printPowerset (const vector<int>& original_set)
{
    auto print_set = [&original_set](auto first, auto last) -> ostream&
    {
        cout << '(';
        auto sep = "";
        for ( ; first != last  ; ++first, sep = ",")
        {
            cout << sep << original_set[(*first) - 1];
        }
        return cout << ')';
    };

    const int n = original_set.size();
    std::vector<int> index_stack(n + 1, 0);
    int k = 0;


    while(1){
        if (index_stack[k]<n){
            index_stack[k+1] = index_stack[k] + 1;
            k++;
        }

        else{
            index_stack[k-1]++;
            k--;
        }

        if (k==0)
            break;

        print_set(begin(index_stack) + 1, begin(index_stack) + 1 + k);
    }
    print_set(begin(index_stack), begin(index_stack)) << endl;
}

int main(){

    auto nums = vector<int> { 2, 4, 6, 8 };
    printPowerset(nums);

    nums = vector<int> { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 };
    printPowerset(nums);

    return 0;
}

expected results:

first power set (4 items):

(2)(2,4)(2,4,6)(2,4,6,8)(2,4,8)(2,6)(2,6,8)(2,8)(4)(4,6)(4,6,8)(4,8)(6)(6,8)(8)()

second power set (10 items)

(2)(2,4)(2,4,6)(2,4,6,8)(2,4,6,8,10)(2,4,6,8,10,12)(2,4,6,8,10,12,14)(2,4,6,8,10,12,14,16)(2,4,6,8,10,12,14,16,18)(2,4,6,8,10,12,14,16,18,20)(2,4,6,8,10,12,14,16,20)(2,4,6,8,10,12,14,18)(2,4,6,8,10,12,14,18,20)(2,4,6,8,10,12,14,20)(2,4,6,8,10,12,16)(2,4,6,8,10,12,16,18)(2,4,6,8,10,12,16,18,20)(2,4,6,8,10,12,16,20)(2,4,6,8,10,12,18)(2,4,6,8,10,12,18,20)(2,4,6,8,10,12,20)(2,4,6,8,10,14)(2,4,6,8,10,14,16)(2,4,6,8,10,14,16,18)(2,4,6,8,10,14,16,18,20)(2,4,6,8,10,14,16,20)(2,4,6,8,10,14,18)(2,4,6,8,10,14,18,20)(2,4,6,8,10,14,20)(2,4,6,8,10,16)(2,4,6,8,10,16,18)(2,4,6,8,10,16,18,20)(2,4,6,8,10,16,20)(2,4,6,8,10,18)(2,4,6,8,10,18,20)(2,4,6,8,10,20)(2,4,6,8,12)(2,4,6,8,12,14)(2,4,6,8,12,14,16)(2,4,6,8,12,14,16,18)(2,4,6,8,12,14,16,18,20)(2,4,6,8,12,14,16,20)(2,4,6,8,12,14,18)(2,4,6,8,12,14,18,20)(2,4,6,8,12,14,20)(2,4,6,8,12,16)(2,4,6,8,12,16,18)(2,4,6,8,12,16,18,20)(2,4,6,8,12,16,20)(2,4,6,8,12,18)(2,4,6,8,12,18,20)(2,4,6,8,12,20)(2,4,6,8,14)(2,4,6,8,14,16)(2,4,6,8,14,16,18)(2,4,6,8,14,16,18,20)(2,4,6,8,14,16,20)(2,4,6,8,14,18)(2,4,6,8,14,18,20)(2,4,6,8,14,20)(2,4,6,8,16)(2,4,6,8,16,18)(2,4,6,8,16,18,20)(2,4,6,8,16,20)(2,4,6,8,18)(2,4,6,8,18,20)(2,4,6,8,20)(2,4,6,10)(2,4,6,10,12)(2,4,6,10,12,14)(2,4,6,10,12,14,16)(2,4,6,10,12,14,16,18)(2,4,6,10,12,14,16,18,20)(2,4,6,10,12,14,16,20)(2,4,6,10,12,14,18)(2,4,6,10,12,14,18,20)(2,4,6,10,12,14,20)(2,4,6,10,12,16)(2,4,6,10,12,16,18)(2,4,6,10,12,16,18,20)(2,4,6,10,12,16,20)(2,4,6,10,12,18)(2,4,6,10,12,18,20)(2,4,6,10,12,20)(2,4,6,10,14)(2,4,6,10,14,16)(2,4,6,10,14,16,18)(2,4,6,10,14,16,18,20)(2,4,6,10,14,16,20)(2,4,6,10,14,18)(2,4,6,10,14,18,20)(2,4,6,10,14,20)(2,4,6,10,16)(2,4,6,10,16,18)(2,4,6,10,16,18,20)(2,4,6,10,16,20)(2,4,6,10,18)(2,4,6,10,18,20)(2,4,6,10,20)(2,4,6,12)(2,4,6,12,14)(2,4,6,12,14,16)(2,4,6,12,14,16,18)(2,4,6,12,14,16,18,20)(2,4,6,12,14,16,20)(2,4,6,12,14,18)(2,4,6,12,14,18,20)(2,4,6,12,14,20)(2,4,6,12,16)(2,4,6,12,16,18)(2,4,6,12,16,18,20)(2,4,6,12,16,20)(2,4,6,12,18)(2,4,6,12,18,20)(2,4,6,12,20)(2,4,6,14)(2,4,6,14,16)(2,4,6,14,16,18)(2,4,6,14,16,18,20)(2,4,6,14,16,20)(2,4,6,14,18)(2,4,6,14,18,20)(2,4,6,14,20)(2,4,6,16)(2,4,6,16,18)(2,4,6,16,18,20)(2,4,6,16,20)(2,4,6,18)(2,4,6,18,20)(2,4,6,20)(2,4,8)(2,4,8,10)(2,4,8,10,12)(2,4,8,10,12,14)(2,4,8,10,12,14,16)(2,4,8,10,12,14,16,18)(2,4,8,10,12,14,16,18,20)(2,4,8,10,12,14,16,20)(2,4,8,10,12,14,18)(2,4,8,10,12,14,18,20)(2,4,8,10,12,14,20)(2,4,8,10,12,16)(2,4,8,10,12,16,18)(2,4,8,10,12,16,18,20)(2,4,8,10,12,16,20)(2,4,8,10,12,18)(2,4,8,10,12,18,20)(2,4,8,10,12,20)(2,4,8,10,14)(2,4,8,10,14,16)(2,4,8,10,14,16,18)(2,4,8,10,14,16,18,20)(2,4,8,10,14,16,20)(2,4,8,10,14,18)(2,4,8,10,14,18,20)(2,4,8,10,14,20)(2,4,8,10,16)(2,4,8,10,16,18)(2,4,8,10,16,18,20)(2,4,8,10,16,20)(2,4,8,10,18)(2,4,8,10,18,20)(2,4,8,10,20)(2,4,8,12)(2,4,8,12,14)(2,4,8,12,14,16)(2,4,8,12,14,16,18)(2,4,8,12,14,16,18,20)(2,4,8,12,14,16,20)(2,4,8,12,14,18)(2,4,8,12,14,18,20)(2,4,8,12,14,20)(2,4,8,12,16)(2,4,8,12,16,18)(2,4,8,12,16,18,20)(2,4,8,12,16,20)(2,4,8,12,18)(2,4,8,12,18,20)(2,4,8,12,20)(2,4,8,14)(2,4,8,14,16)(2,4,8,14,16,18)(2,4,8,14,16,18,20)(2,4,8,14,16,20)(2,4,8,14,18)(2,4,8,14,18,20)(2,4,8,14,20)(2,4,8,16)(2,4,8,16,18)(2,4,8,16,18,20)(2,4,8,16,20)(2,4,8,18)(2,4,8,18,20)(2,4,8,20)(2,4,10)(2,4,10,12)(2,4,10,12,14)(2,4,10,12,14,16)(2,4,10,12,14,16,18)(2,4,10,12,14,16,18,20)(2,4,10,12,14,16,20)(2,4,10,12,14,18)(2,4,10,12,14,18,20)(2,4,10,12,14,20)(2,4,10,12,16)(2,4,10,12,16,18)(2,4,10,12,16,18,20)(2,4,10,12,16,20)(2,4,10,12,18)(2,4,10,12,18,20)(2,4,10,12,20)(2,4,10,14)(2,4,10,14,16)(2,4,10,14,16,18)(2,4,10,14,16,18,20)(2,4,10,14,16,20)(2,4,10,14,18)(2,4,10,14,18,20)(2,4,10,14,20)(2,4,10,16)(2,4,10,16,18)(2,4,10,16,18,20)(2,4,10,16,20)(2,4,10,18)(2,4,10,18,20)(2,4,10,20)(2,4,12)(2,4,12,14)(2,4,12,14,16)(2,4,12,14,16,18)(2,4,12,14,16,18,20)(2,4,12,14,16,20)(2,4,12,14,18)(2,4,12,14,18,20)(2,4,12,14,20)(2,4,12,16)(2,4,12,16,18)(2,4,12,16,18,20)(2,4,12,16,20)(2,4,12,18)(2,4,12,18,20)(2,4,12,20)(2,4,14)(2,4,14,16)(2,4,14,16,18)(2,4,14,16,18,20)(2,4,14,16,20)(2,4,14,18)(2,4,14,18,20)(2,4,14,20)(2,4,16)(2,4,16,18)(2,4,16,18,20)(2,4,16,20)(2,4,18)(2,4,18,20)(2,4,20)(2,6)(2,6,8)(2,6,8,10)(2,6,8,10,12)(2,6,8,10,12,14)(2,6,8,10,12,14,16)(2,6,8,10,12,14,16,18)(2,6,8,10,12,14,16,18,20)(2,6,8,10,12,14,16,20)(2,6,8,10,12,14,18)(2,6,8,10,12,14,18,20)(2,6,8,10,12,14,20)(2,6,8,10,12,16)(2,6,8,10,12,16,18)(2,6,8,10,12,16,18,20)(2,6,8,10,12,16,20)(2,6,8,10,12,18)(2,6,8,10,12,18,20)(2,6,8,10,12,20)(2,6,8,10,14)(2,6,8,10,14,16)(2,6,8,10,14,16,18)(2,6,8,10,14,16,18,20)(2,6,8,10,14,16,20)(2,6,8,10,14,18)(2,6,8,10,14,18,20)(2,6,8,10,14,20)(2,6,8,10,16)(2,6,8,10,16,18)(2,6,8,10,16,18,20)(2,6,8,10,16,20)(2,6,8,10,18)(2,6,8,10,18,20)(2,6,8,10,20)(2,6,8,12)(2,6,8,12,14)(2,6,8,12,14,16)(2,6,8,12,14,16,18)(2,6,8,12,14,16,18,20)(2,6,8,12,14,16,20)(2,6,8,12,14,18)(2,6,8,12,14,18,20)(2,6,8,12,14,20)(2,6,8,12,16)(2,6,8,12,16,18)(2,6,8,12,16,18,20)(2,6,8,12,16,20)(2,6,8,12,18)(2,6,8,12,18,20)(2,6,8,12,20)(2,6,8,14)(2,6,8,14,16)(2,6,8,14,16,18)(2,6,8,14,16,18,20)(2,6,8,14,16,20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Richard Hodges
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0

It is not possible, as you pointed out yourself power set contains pow(2, size) no of elements. You need to print the entire set that can only be done by generating the set using backtracking, there is nothing better.

To find sum of all subsets of a given set is however a simpler problem.

If the no of elements is n, then each element in the array occurs 2^(n-1) times in the power set.

Pseudo code : 

int sum = 0;

for(int i = 0;i < n;i++){
    sum += ( set[i] * ( 1 << (n-1) ) );    //pow(2, n-1) == 1 << (n-1)
}

cout << sum << endl;

You would need a Big Integer Library for larger values of n.

uSeemSurprised
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0

Since number of elements in the set can go upto 10^5 and hence the size of the set will go upto 2^(10^5) which is huge. You can just either print it or if you want to store it , store it in a file.

Priyansh Goel
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