A backtracking approach to generate n bit Gray Codes ?

In this section we will see how we can generate the gray codes of n bits using backtracking approach? The n bit gray code is basically bit patterns from 0 to 2^n – 1 such that successive patterns differ by one bit. So for n = 2, the gray codes are (00, 01, 11, 10) and decimal equivalent is (0, 1, 3, 2). The program will generate the decimal equivalent of the gray code values.

Algorithm

generateGray(arr, n, num)

begin
   if n = 0, then
      insert num into arr
      return
   end if
   generateGray(arr, n-1, num)
   num := num XOR (1 bit left shift of n-1)
   generateGray(arr, n-1, num)
end

Example

 Live Demo

#include<iostream>
#include<vector>
using namespace std;
void generateGray(vector<int>&arr, int n, int &num){
   if(n==0){
      arr.push_back(num);
      return;
   }
   generateGray(arr, n-1, num);
   num = num ^ (1 << (n-1));
   generateGray(arr, n-1, num);
}
vector<int> gray(int n){
   vector<int> arr;
   int num = 0;
   generateGray(arr, n, num);
   return arr;
}
main() {
   int n;
   cout << "Enter number of bits: ";
   cin >> n;
   vector<int> grayCode = gray(n);
   for(int i = 0; i<grayCode.size(); i++){
      cout << grayCode[i] << endl;
   }
}

Output

Enter number of bits: 3
0
1
3
2
6
7
5
4