Printing all subsets of {1,2,3,…n} without using array or loop

Given a natural number n, print all the subsets of the set \{1, 2, 3, ..., n\} without using any array or loop (only the use of recursion is allowed).

Examples:

Input : n = 4
Output : { 1 2 3 4 }
         { 1 2 3 }
         { 1 2 4 }
         { 1 2 }
         { 1 3 4 }
         { 1 3 }
         { 1 4 }
         { 1 }
         { 2 3 4 }
         { 2 3 }
         { 2 4 }
         { 2 }
         { 3 4 }
         { 3 }
         { 4 }
         { }

Input : n = 2
Output : { 1 2 }
         { 1 }
         { 2 }
         { }


Approach:

  • Start from num = 2^n - 1 upto 0.
  • Consider the binary representation of num with n bits.
  • Start from the leftmost bit which represents 1, the second bit represents 2 and so on until nth bit which represents n.
  • Print the number corresponding to the bit if it is set.
  • Perform the above steps for all values of num until it is equal to 0.

Let’s understand the above approach through an example:

Considering input n = 4, start from num = 2^n - 1 = 15.

and so on … until num = 0.

Below is the implementation of the above approach:

Java

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// Java code to print all subsets 
// of {1, 2, 3, n} without using
// array or loop, just recursion.
class GfG 
{
  
    // This recursive function calls subset
    // function to print the subsets one by one. 
    // numBits --> number of bits needed to 
    // represent the number (simply input value n).
    // num --> Initially equal to 2 ^ n - 1 and 
    // decreases by 1 every recursion until 0.
    static void printSubSets(int numOfBits, int num) 
    {
        if (num >= 0
        {
            System.out.print("{ ");
              
            // Print the subset corresponding to 
            // binary representation of num.
            subset(numOfBits - 1, num, numOfBits);
            System.out.println("}");
              
            // Call the function recursively to 
            // print the next subset.
            printSubSets(numOfBits, num - 1);
  
        } else
            return;
    }
  
    // This function recursively prints the 
    // subset corresponding to the binary 
    // representation of num.
    // nthBit --> nth bit from right side 
    // starting from n and decreases until 0.
    static void subset(int nthBit, int num, int numOfBits) 
    {
        if (nthBit >= 0
        {
            // Print number in given subset only
            // if the bit corresponding to it 
            // is set in num.
            if ((num & (1 << nthBit)) != 0)
            {
                System.out.print(numOfBits - nthBit + " ");
  
            }
              
            // Check for the next bit 
            subset(nthBit - 1, num, numOfBits);
        } else
            return;
    }
      
    // Driver code
    public static void main(String[] args) 
    {
        int n = 4;
        printSubSets(n, (int) (Math.pow(2, n)) -1);
    }
}
  
// This code is contributed by laststringx

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Python3

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# Python3 code to print all subsets 
# of {1, 2, 3, …n} without using
# array or loop, just recursion.
  
# This recursive function calls subset
# function to print the subsets one by one. 
# numBits --> number of bits needed to 
# represent the number (simply input value n).
# num --> Initially equal to 2 ^ n - 1 and 
# decreases by 1 every recursion until 0.
def printSubsets(numOfBits, num):
      
    if num >= 0:
        print("{", end = " ")
  
        # Print the subset corresponding to 
        # binary representation of num.
        subset(numOfBits-1, num, numOfBits)
        print("}")
  
        # Call the function recursively to 
        # print the next subset.
        printSubsets(numOfBits, num-1)
          
    else:
        return
  
# This function recursively prints the 
# subset corresponding to the binary 
# representation of num.
# nthBit --> nth bit from right side 
# starting from n and decreases until 0.
def subset(nthBit, num, numOfBits):
      
    if nthBit >= 0:
          
        # Print number in given subset only
        # if the bit corresponding to it 
        # is set in num.
        if num & (1 << nthBit) != 0:
            print(numOfBits - nthBit, end = " ")
          
        # Check for the next bit 
        subset(nthBit-1, num, numOfBits)
          
    else:
        return
  
# Driver Code    
n = 4
printSubsets(n, 2**n - 1)

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Output:

{ 1 2 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 2 }
{ 1 3 4 }
{ 1 3 }
{ 1 4 }
{ 1 }
{ 2 3 4 }
{ 2 3 }
{ 2 4 }
{ 2 }
{ 3 4 }
{ 3 }
{ 4 }
{ }

Time Complexity: O(n*2^n)



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Improved By : laststringx