Given the Binary code of a number as a decimal number, we need to convert this into its equivalent Gray Code. Assume that the binary number is in the range of integers. For the larger value, we can take a binary number as string.
In gray code, only one bit is changed in 2 consecutive numbers.
Examples:
Input: 1001
Output: 1101
Explanation: 1001 -> 1101 -> 1101 -> 1101Input: 11
Output: 10
Explanation: 11 -> 10
Approach:
The idea is to check whether the last bit and second last bit are same or not, if it is same then move ahead otherwise add 1.
Follow the steps to solve the given problem:
- binary_to_grey(n)
- if n == 0
- grey = 0;
- else if last two bits are opposite to each other
- grey = 1 + 10 * binary_to_gray(n/10))
- else if last two bits are same
- grey = 10 * binary_to_gray(n/10))
- grey = 10 * binary_to_gray(n/10))
Below is the implementation of the above approach :
CPP
// CPP program to convert Binary to// Gray code using recursion#include <bits/stdc++.h>using namespace std;// Function to change Binary to// Gray using recursionint binary_to_gray(int n){ if (!n) return 0; // Taking last digit int a = n % 10; // Taking second last digit int b = (n / 10) % 10; // If last digit are opposite bits if ((a && !b) || (!a && b)) return (1 + 10 * binary_to_gray(n / 10)); // If last two bits are same return (10 * binary_to_gray(n / 10));}// Driver Functionint main(){ int binary_number = 1011101; printf("%d", binary_to_gray(binary_number)); return 0;} |
Java
// Java program to convert// Binary code to Gray codeimport static java.lang.StrictMath.pow;import java.util.Scanner;class bin_gray { // Function to change Binary to // Gray using recursion int binary_to_gray(int n, int i) { int a, b; int result = 0; if (n != 0) { // Taking last digit a = n % 10; n = n / 10; // Taking second last digit b = n % 10; if ((a & ~b) == 1 || (~a & b) == 1) { result = (int)(result + pow(10, i)); } return binary_to_gray(n, ++i) + result; } return 0; } // Driver Function public static void main(String[] args) { int binary_number; int result = 0; binary_number = 1011101; bin_gray obj = new bin_gray(); result = obj.binary_to_gray(binary_number, 0); System.out.print(result); }}// This article is contributed by Anshika Goyal. |
Python3
# Python3 code to convert Binary# to Gray code using recursion# Function to change Binary to Gray using recursiondef binary_to_gray(n): if not(n): return 0 # Taking last digit a = n % 10 # Taking second last digit b = int(n / 10) % 10 # If last digit are opposite bits if (a and not(b)) or (not(a) and b): return (1 + 10 * binary_to_gray(int(n / 10))) # If last two bits are same return (10 * binary_to_gray(int(n / 10)))# Driver Codebinary_number = 1011101print(binary_to_gray(binary_number), end='')# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to convert// Binary code to Gray codeusing System;class GFG { // Function to change Binary to // Gray using recursion static int binary_to_gray(int n, int i) { int a, b; int result = 0; if (n != 0) { // Taking last digit a = n % 10; n = n / 10; // Taking second last digit b = n % 10; if ((a & ~b) == 1 || (~a & b) == 1) { result = (int)(result + Math.Pow(10, i)); } return binary_to_gray(n, ++i) + result; } return 0; } // Driver Function public static void Main() { int binary_number; binary_number = 1011101; Console.WriteLine(binary_to_gray(binary_number, 0)); }}// This article is contributed by vt_m. |
PHP
<?php// PHP program to convert Binary to // Gray code using recursion// Function to change Binary to// Gray using recursionfunction binary_to_gray($n){ if (!$n) return 0; // Taking last digit $a = $n % 10; // Taking second // last digit $b = ($n / 10) % 10; // If last digit are // opposite bits if (($a && !$b) || (!$a && $b)) return (1 + 10 * binary_to_gray($n / 10)); // If last two // bits are same return (10 * binary_to_gray($n / 10));} // Driver Code $binary_number = 1011101; echo binary_to_gray($binary_number);// This code is contributed by Ajit?> |
Javascript
<script>// JavaScript program to convert// Binary code to Gray code // Function to change Binary to // Gray using recursion function binary_to_gray(n, i) { let a, b; let result = 0; if (n != 0) { // Taking last digit a = n % 10; n = n / 10; // Taking second last digit b = n % 10; if ((a & ~ b) == 1 || (~ a & b) == 1) { result = (result + Math.pow(10,i)); } return binary_to_gray(n, ++i) + result; } return 0; } // Driver code let binary_number; let result = 0; binary_number = 1011101; result = binary_to_gray(binary_number,0); document.write(result); // This code is contributed by code_hunt.</script> |
Output
1110011
Time Complexity: O(logN), Traverse through all the digits, as there are logN bits.
Auxiliary Space: O(1)
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