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# Gray to Binary and Binary to Gray conversion

Binary Number is the default way to store numbers, but in many applications, binary numbers are difficult to use and a variety of binary numbers is needed. This is where Gray codes are very useful.

Gray code has a property that two successive numbers differ in only one bit because of this property gray code does the cycling through various states with minimal effort and is used in K-maps, error correction, communication, etc.

How to generate n bit Gray Codes?

Following is 2-bit sequence (n = 2)
00 01 11 10
Following is 3-bit sequence (n = 3)
000 001 011 010 110 111 101 100
And Following is 4-bit sequence (n = 4)
0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111
1110 1010 1011 1001 1000

n-bit Gray Codes can be generated from a list of (n-1)-bit Gray codes using the following steps.

1. Let the list of (n-1)-bit Gray codes be L1. Create another list L2 which is the reverse of L1.
2. Modify the list L1 by prefixing a ‘0’ in all codes of L1.
3. Modify the list L2 by prefixing a ‘1’ in all codes of L2.
4. Concatenate L1 and L2. The concatenated list is the required list of n-bit Gray codes.

Please refer Generate n-bit Gray Codes for a detailed program.

How to Convert Binary To Gray and Vice Versa?

Binary : 0011
Gray   : 0010

Binary : 01001
Gray   : 01101

In computer science many a time we need to convert binary code to gray code and vice versa. This conversion can be done by applying the following rules :

Binary to Gray conversion :

1. The Most Significant Bit (MSB) of the gray code is always equal to the MSB of the given binary code.
2. Other bits of the output gray code can be obtained by XORing binary code bit at that index and previous index.

Binary code to gray code conversion

Gray to binary conversion :

1. The Most Significant Bit (MSB) of the binary code is always equal to the MSB of the given gray code.
2. Other bits of the output binary code can be obtained by checking the gray code bit at that index. If the current gray code bit is 0, then copy the previous binary code bit, else copy the invert of the previous binary code bit.

Gray code to binary code conversion

Below is the implementation of the above steps.

## C++

 // C++ program for Binary To Gray// and Gray to Binary conversion#include using namespace std; // Helper function to xor two characterschar xor_c(char a, char b) { return (a == b) ? '0' : '1'; } // Helper function to flip the bitchar flip(char c) { return (c == '0') ? '1' : '0'; } // function to convert binary string// to gray stringstring binarytoGray(string binary){    string gray = "";     // MSB of gray code is same as binary code    gray += binary[0];     // Compute remaining bits, next bit is computed by    // doing XOR of previous and current in Binary    for (int i = 1; i < binary.length(); i++) {        // Concatenate XOR of previous bit        // with current bit        gray += xor_c(binary[i - 1], binary[i]);    }     return gray;} // function to convert gray code string// to binary stringstring graytoBinary(string gray){    string binary = "";     // MSB of binary code is same as gray code    binary += gray[0];     // Compute remaining bits    for (int i = 1; i < gray.length(); i++) {        // If current bit is 0, concatenate        // previous bit        if (gray[i] == '0')            binary += binary[i - 1];         // Else, concatenate invert of        // previous bit        else            binary += flip(binary[i - 1]);    }     return binary;} // Driver program to test above functionsint main(){    string binary = "01001";    cout << "Gray code of " << binary << " is "         << binarytoGray(binary) << endl;     string gray = "01101";    cout << "Binary code of " << gray << " is "         << graytoBinary(gray) << endl;    return 0;}

## Java

 // Java program for Binary To Gray// and Gray to Binary conversionimport java.io.*;class code_conversion{       // Helper function to xor    // two characters    char xor_c(char a, char b)    {        return (a == b) ? '0' : '1';    }     // Helper function to flip the bit    char flip(char c)    {        return (c == '0') ? '1' : '0';    }     // function to convert binary  // string to gray string    String binarytoGray(String binary)    {        String gray = "";         // MSB of gray code is same        // as binary code        gray += binary.charAt(0);         // Compute remaining bits, next bit is        // computed by doing XOR of previous        // and current in Binary        for (int i = 1; i < binary.length(); i++)        {             // Concatenate XOR of previous bit            // with current bit            gray += xor_c(binary.charAt(i - 1),                          binary.charAt(i));        }       return gray;    }       // function to convert gray code    // string to binary string    String graytoBinary(String gray)    {        String binary = "";         // MSB of binary code is same        // as gray code        binary += gray.charAt(0);         // Compute remaining bits        for (int i = 1; i < gray.length(); i++)        {                       // If current bit is 0,            // concatenate previous bit            if (gray.charAt(i) == '0')                binary += binary.charAt(i - 1);                     // Else, concatenate invert of            // previous bit            else                binary += flip(binary.charAt(i - 1));        }         return binary;    }     // Driver program to test above functions    public static void main(String args[])        throws IOException    {        code_conversion ob = new code_conversion();        String binary = "01001";        System.out.println("Gray code of " +                           binary + " is " +                           ob.binarytoGray(binary));         String gray = "01101";        System.out.println("Binary code of " +                           gray + " is " +                           ob.graytoBinary(gray));    }  } // This code is contributed by Anshika Goyal.

## Python3

 # Python3 program for Binary To Gray# and Gray to Binary conversion # Helper function to xor two charactersdef xor_c(a, b):    return '0' if(a == b) else '1'; # Helper function to flip the bitdef flip(c):    return '1' if(c == '0') else '0'; # function to convert binary string# to gray stringdef binarytoGray(binary):    gray = "";     # MSB of gray code is same as    # binary code    gray += binary[0];     # Compute remaining bits, next bit    # is computed by doing XOR of previous    # and current in Binary    for i in range(1, len(binary)):                 # Concatenate XOR of previous        # bit with current bit        gray += xor_c(binary[i - 1],                      binary[i]);     return gray; # function to convert gray code# string to binary stringdef graytoBinary(gray):     binary = "";     # MSB of binary code is same    # as gray code    binary += gray[0];     # Compute remaining bits    for i in range(1, len(gray)):                 # If current bit is 0,        # concatenate previous bit        if (gray[i] == '0'):            binary += binary[i - 1];         # Else, concatenate invert        # of previous bit        else:            binary += flip(binary[i - 1]);     return binary; # Driver Codebinary = "01001";print("Gray code of", binary, "is",             binarytoGray(binary)); gray = "01101";print("Binary code of", gray, "is",               graytoBinary(gray));     # This code is contributed by mits

## C#

 // C# program for Binary To Gray// and Gray to Binary conversion.using System; class GFG {     // Helper function to xor    // two characters    static char xor_c(char a, char b)    {        return (a == b) ? '0' : '1';    }     // Helper function to flip the bit    static char flip(char c)    {        return (c == '0') ? '1' : '0';    }     // function to convert binary    // string to gray string    static String binarytoGray(String binary)    {        String gray = "";         // MSB of gray code is same        // as binary code        gray += binary[0];         // Compute remaining bits, next        // bit is computed by doing XOR        // of previous and current in        // Binary        for (int i = 1; i < binary.Length; i++) {             // Concatenate XOR of previous            // bit with current bit            gray += xor_c(binary[i - 1],                          binary[i]);        }         return gray;    }     // function to convert gray code    // string to binary string    static String graytoBinary(String gray)    {         String binary = "";         // MSB of binary code is same        // as gray code        binary += gray[0];         // Compute remaining bits        for (int i = 1; i < gray.Length; i++) {             // If current bit is 0,            // concatenate previous bit            if (gray[i] == '0')                binary += binary[i - 1];             // Else, concatenate invert of            // previous bit            else                binary += flip(binary[i - 1]);        }         return binary;    }     // Driver program to test above    // functions    public static void Main()     {        String binary = "01001";        Console.WriteLine("Gray code of "                          + binary + " is "                          + binarytoGray(binary));         String gray = "01101";        Console.Write("Binary code of "                      + gray + " is "                      + graytoBinary(gray));    }} // This code is contributed by nitin mittal.



## Javascript



Output

Gray code of 01001 is 01101
Binary code of 01101 is 01001

Time Complexity: O(n)

Auxiliary Space: O(n)

Here, n is length of the binary string.

Binary to Gray using Bitwise Operators

## C++

 // C++ program for above approach#include using namespace std; int greyConverter(int n) { return n ^ (n >> 1); } int main(){    int n = 3;    cout << greyConverter(n) << endl;     n = 9;    cout << greyConverter(n) << endl;     return 0;}

## Java

 // Java Program for above approachpublic class Main {    public static int greyConverter(int n)    {        return n ^ (n >> 1);    }     // Driver code    public static void main(String[] args)    {        int n = 3;        System.out.println(greyConverter(n));         n = 9;        System.out.println(greyConverter(n));    }} // This code is contributed by divyesh072019

## Python3

 # Python3 code for above approachdef greyConverter(n):     return n ^ (n >> 1)  n = 3print(greyConverter(n)) n = 9print(greyConverter(n)) # This code is contributed by divyeshrabadiya07

## C#

 // C# program for above approachusing System; class GFG {     static int greyConverter(int n) { return n ^ (n >> 1); }     // Driver Code    public static void Main(string[] args)    {        int n = 3;        Console.WriteLine(greyConverter(n));         n = 9;        Console.WriteLine(greyConverter(n));    }} // This code is contributed by rutvik_56

## Javascript



Output

2
13

Time Complexity: O(1)

Auxiliary Space: O(1)

Gray to Binary using Bitwise Operators

## C++

 // C++ program for above approach#include using namespace std; int binaryConverter(int n){    int res = n;    while (n > 0)    {        n >>= 1;        res ^= n;    }    return res;} // Driver Codeint main(){    int n = 4;       // Function call    cout << binaryConverter(n) << endl;     return 0;} // This code is contributed by sshrey47

## Java

 // Java Program for above approachpublic class Main {    public static int binaryConverter(int n)    {        int res = n;         while (n > 0) {            n >>= 1;            res ^= n;        }         return res;    }     // Driver code    public static void main(String[] args)    {        int n = 4;        System.out.println(binaryConverter(n));    }} // This code is contributed by sshrey47

## Python3

 # Python3 code for above approachdef binaryConverter(n):    res = n     while n > 0:        n >>= 1        res ^= n     return res  n = 4print(binaryConverter(n)) # This code is contributed by sshrey47

## C#

 using System; public class GFG {     static int binaryConverter(int n)    {        int res = n;         while (n > 0) {            n >>= 1;            res ^= n;        }         return res;    }     static public void Main()    {        int n = 4;        Console.WriteLine(binaryConverter(n));    }} // This code is contributed by sshrey47

## Javascript



Output

7

Time Complexity: O(log(n))

Auxiliary Space: O(1)