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Power Set

Power Set Power set P(S) of a set S is the set of all subsets of S. For example S = {a, b, c} then P(s) = {{}, {a}, {b}, {c}, {a,b}, {a, c}, {b, c}, {a, b, c}}.

If S has n elements in it then P(s) will have 2^n elements


Algorithm:

Input: Set[], set_size
1. Get the size of power set
    powet_set_size = pow(2, set_size)
2  Loop for counter from 0 to pow_set_size
     (a) Loop for i = 0 to set_size
          (i) If ith bit in counter is set
               Print ith element from set for this subset
     (b) Print seperator for subsets i.e., newline

Example:

Set  = [a,b,c]
power_set_size = pow(2, 3) = 8
Run for binary counter = 000 to 111

Value of Counter            Subset
    000                    -> Empty set
    001                    -> a
    011                    -> ab
   100                     -> c
   101                     -> ac
   110                     -> bc
   111                     -> abc

Program:

#include <stdio.h>
#include <math.h>
 
void printPowerSet(char *set, int set_size)
{
    /*set_size of power set of a set with set_size
      n is (2**n -1)*/
    unsigned int pow_set_size = pow(2, set_size);
    int counter, j;
 
    /*Run from counter 000..0 to 111..1*/
    for(counter = 0; counter < pow_set_size; counter++)
    {
      for(j = 0; j < set_size; j++)
       {
          /* Check if jth bit in the counter is set
             If set then pront jth element from set */
          if(counter & (1<<j))
            printf("%c", set[j]);
       }
       printf("\n");
    }
}
 
/*Driver program to test printPowerSet*/
int main()
{
    char set[] = {'a','b','c'};
    printPowerSet(set, 3);
 
    getchar();
    return 0;
}

Time Complexity: O(n2^n)

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