Generate n-bit Gray Codes | Set 2

Given a number n, generate bit patterns from 0 to 2^n-1 such that successive patterns differ by one bit.

Examples:

Input: n=2
Output: 00 01 11 10
Every adjacent element of gray code differs 
only by one bit.
So the n bit grey codes are: 00 01 11 10

Input: n=3
Output: 000 001 011 010 110 111 101 100
Every adjacent element of gray code differs  
only by one bit.
So the n bit gray codes are:
 000 001 011 010 110 111 101 100

Another approach of Generate n-bit Gray Codes has already been discussed.

Approach:
The idea is to get gray code of binary number using XOR and Right shift operation.

  1. The first bit(MSB) of the gray code is same as the first bit(MSB) of binary number.
  2. The second bit(from left side) of the gray code equals to XOR of first bit(MSB) and second bit(2nd MSB) of the binary number.
  3. The third bit(from left side) of the gray code equals to XOR of the second bit(2nd MSB) and third bit(3rd MSB) and so on..

In this way, the gray code can be calculated for the corresponding binary number. So, it can be observed that the ith element can be formed by bitwise XOR of i and floor(i/2) which is equal to the bitwise XOR of i and (i >> 1) i.e., i right-shifted by 1. By performing this the MSB of the binary number is kept intact and all the other bits are performed bitwise XOR with its adjacent higher bit.

C++

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// C++ program to generate n-bit
// gray codes
#include <bits/stdc++.h>
using namespace std;
  
// Function to convert decimal to binary
void decimalToBinaryNumber(int x, int n)
{
    int* binaryNumber = new int(x);
    int i = 0;
    while (x > 0) {
        binaryNumber[i] = x % 2;
        x = x / 2;
        i++;
    }
  
    // leftmost digits are filled with 0
    for (int j = 0; j < n - i; j++)
        cout << '0';
  
    for (int j = i - 1; j >= 0; j--)
        cout << binaryNumber[j];
}
  
// Function to generate gray code
void generateGrayarr(int n)
{
    int N = 1 << n;
    for (int i = 0; i < N; i++) {
  
        // generate gray code of corresponding
        // binary number of integer i.
        int x = i ^ (i >> 1);
  
        // printing gray code
        decimalToBinaryNumber(x, n);
  
        cout << endl;
    }
}
  
// Drivers code
int main()
{
    int n = 3;
    generateGrayarr(n);
    return 0;
}

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Java

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// Java program to generate
// n-bit gray codes
import java.io.*;
  
class GFG {
  
    // Function to convert
    // decimal to binary
    static void decimalToBinaryNumber(int x,
                                      int n)
    {
        int[] binaryNumber = new int[x];
        int i = 0;
        while (x > 0) {
            binaryNumber[i] = x % 2;
            x = x / 2;
            i++;
        }
  
        // leftmost digits are
        // filled with 0
        for (int j = 0; j < n - i; j++)
            System.out.print('0');
  
        for (int j = i - 1;
             j >= 0; j--)
            System.out.print(binaryNumber[j]);
    }
  
    // Function to generate
    // gray code
    static void generateGrayarr(int n)
    {
        int N = 1 << n;
        for (int i = 0; i < N; i++) {
  
            // generate gray code of
            // corresponding binary
            // number of integer i.
            int x = i ^ (i >> 1);
  
            // printing gray code
            decimalToBinaryNumber(x, n);
  
            System.out.println();
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int n = 3;
        generateGrayarr(n);
    }
}
  
// This code is contributed
// by anuj_67.

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Python3

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# Python program to generate
# n-bit gray codes
  
# Function to convert
# decimal to binary
def decimalToBinaryNumber(x, n):
    binaryNumber = [0]*x;
    i = 0;
    while (x > 0):
        binaryNumber[i] = x % 2;
        x = x // 2;
        i += 1;
  
    # leftmost digits are
    # filled with 0
    for j in range(0, n - i):
        print('0', end ="");
  
    for j in range(i - 1, -1, -1):
        print(binaryNumber[j], end ="");
  
# Function to generate
# gray code
def generateGrayarr(n):
    N = 1 << n;
    for i in range(N):
          
        # generate gray code of
        # corresponding binary
        # number of integer i.
        x = i ^ (i >> 1);
  
        # printing gray code
        decimalToBinaryNumber(x, n);
  
        print();
  
# Driver code
if __name__ == '__main__':
    n = 3;
    generateGrayarr(n);
  
# This code is contributed by 29AjayKumar

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C#

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// C# program to generate
// n-bit gray codes
using System;
  
class GFG {
  
    // Function to convert
    // decimal to binary
    static void decimalToBinaryNumber(int x,
                                      int n)
    {
        int[] binaryNumber = new int[x];
        int i = 0;
        while (x > 0) {
            binaryNumber[i] = x % 2;
            x = x / 2;
            i++;
        }
  
        // leftmost digits are
        // filled with 0
        for (int j = 0; j < n - i; j++)
            Console.Write('0');
  
        for (int j = i - 1;
             j >= 0; j--)
            Console.Write(binaryNumber[j]);
    }
  
    // Function to generate
    // gray code
    static void generateGrayarr(int n)
    {
        int N = 1 << n;
        for (int i = 0; i < N; i++) {
  
            // Generate gray code of
            // corresponding binary
            // number of integer i.
            int x = i ^ (i >> 1);
  
            // printing gray code
            decimalToBinaryNumber(x, n);
  
            Console.WriteLine();
        }
    }
  
    // Driver code
    public static void Main()
    {
        int n = 3;
        generateGrayarr(n);
    }
}
  
// This code is contributed
// by anuj_67.

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Output:

000
001
011
010
110
111
101
100

Complexity Anslysis:

  • Time Complexity: O(n).
    Only one traversal from 0 to (n-1) is needed.
  • Auxiliary Space: O(log x).
    A space of (log x) is required for binary representation of (x).

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