# Binary to Gray code using recursion

Given Binary code of a number as a decimal number, we need to convert this into its equivalent Gray Code.

Examples :

```Input : 1001
Output : 1101

Input : 11
Output : 10```

In gray code, only one bit is changed in 2 consecutive numbers.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm :

```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))
```

Below is implementation of above approach :

## CPP

 `// CPP program to convert Binary to  ` `// Gray code using recursion ` `#include ` `using` `namespace` `std; ` ` `  `// Function to change Binary to ` `// Gray using recursion ` `int` `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 Function ` `int` `main() ` `{ ` `    ``int` `binary_number = 1011101; ` `     `  `    ``printf``(``"%d"``, binary_to_gray(binary_number)); ` `    ``return` `0; ` `} `

## Java

 `// Java program to convert ` `// Binary code to Gray code ` `import` `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 recursion ` `def` `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 Code ` `binary_number ``=` `1011101` `print``( binary_to_gray(binary_number), end``=``'') ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# program to convert ` `// Binary code to Gray code ` `using` `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

 ` `

Output :

`1110011`

Assume that the binary number is in the range of integer. For larger value we can take binary number as string.

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