Given a number n, the task is to generate n bit Gray codes (generate bit patterns from 0 to 2^n-1 such that successive patterns differ by one bit)
Examples:
Input : 2
Output : 0 1 3 2
Explanation :
00 - 0
01 - 1
11 - 3
10 - 2
Input : 3
Output : 0 1 3 2 6 7 5 4
We have discussed an approach in Generate n-bit Gray Codes
This article provides a backtracking approach to the same problem. Idea is that for each bit out of n bit we have a choice either we can ignore it or we can invert the bit so this means our gray sequence goes upto 2 ^ n for n bits. So we make two recursive calls for either inverting the bit or leaving the bit as it is.
// CPP program to find the gray sequence of n bits. #include <iostream> #include <vector> using namespace std;
/* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the
bit i.e we can leave the number as it is. */
void grayCodeUtil(vector< int >& res, int n, int & num)
{ // base case when we run out bits to process
// we simply include it in gray code sequence.
if (n == 0) {
res.push_back(num);
return ;
}
// ignore the bit.
grayCodeUtil(res, n - 1, num);
// invert the bit.
num = num ^ (1 << (n - 1));
grayCodeUtil(res, n - 1, num);
} // returns the vector containing the gray // code sequence of n bits. vector< int > grayCodes( int n)
{ vector< int > res;
// num is passed by reference to keep
// track of current code.
int num = 0;
grayCodeUtil(res, n, num);
return res;
} // Driver function. int main()
{ int n = 3;
vector< int > code = grayCodes(n);
for ( int i = 0; i < code.size(); i++)
cout << code[i] << endl;
return 0;
} |
// JAVA program to find the gray sequence of n bits. import java.util.*;
class GFG
{ static int num;
/* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ static void grayCodeUtil(Vector<Integer> res, int n)
{ // base case when we run out bits to process
// we simply include it in gray code sequence.
if (n == 0 )
{
res.add(num);
return ;
}
// ignore the bit.
grayCodeUtil(res, n - 1 );
// invert the bit.
num = num ^ ( 1 << (n - 1 ));
grayCodeUtil(res, n - 1 );
} // returns the vector containing the gray // code sequence of n bits. static Vector<Integer> grayCodes( int n)
{ Vector<Integer> res = new Vector<Integer>();
// num is passed by reference to keep
// track of current code.
num = 0 ;
grayCodeUtil(res, n);
return res;
} // Driver function. public static void main(String[] args)
{ int n = 3 ;
Vector<Integer> code = grayCodes(n);
for ( int i = 0 ; i < code.size(); i++)
System.out.print(code.get(i) + "\n" );
} } // This code is contributed by Rajput-Ji |
# Python3 program to find the # gray sequence of n bits. """ we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. """ def grayCodeUtil(res, n, num):
# base case when we run out bits to process
# we simply include it in gray code sequence.
if (n = = 0 ):
res.append(num[ 0 ])
return
# ignore the bit.
grayCodeUtil(res, n - 1 , num)
# invert the bit.
num[ 0 ] = num[ 0 ] ^ ( 1 << (n - 1 ))
grayCodeUtil(res, n - 1 , num)
# returns the vector containing the gray # code sequence of n bits. def grayCodes(n):
res = []
# num is passed by reference to keep
# track of current code.
num = [ 0 ]
grayCodeUtil(res, n, num)
return res
# Driver Code n = 3
code = grayCodes(n)
for i in range ( len (code)):
print (code[i])
# This code is contributed by SHUBHAMSINGH10 |
// C# program to find the gray sequence of n bits. using System;
using System.Collections.Generic;
class GFG
{ static int num;
/* we have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ static void grayCodeUtil(List< int > res, int n)
{ // base case when we run out bits to process
// we simply include it in gray code sequence.
if (n == 0)
{
res.Add(num);
return ;
}
// ignore the bit.
grayCodeUtil(res, n - 1);
// invert the bit.
num = num ^ (1 << (n - 1));
grayCodeUtil(res, n - 1);
} // returns the vector containing the gray // code sequence of n bits. static List< int > grayCodes( int n)
{ List< int > res = new List< int >();
// num is passed by reference to keep
// track of current code.
num = 0;
grayCodeUtil(res, n);
return res;
} // Driver function. public static void Main(String[] args)
{ int n = 3;
List< int > code = grayCodes(n);
for ( int i = 0; i < code.Count; i++)
Console.Write(code[i] + "\n" );
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to find the gray sequence of n bits. /* We have 2 choices for each of the n bits either we can include i.e invert the bit or we can exclude the bit i.e we can leave the number as it is. */ function grayCodeUtil(res, n, num)
{ // Base case when we run out bits to process
// we simply include it in gray code sequence.
if (n == 0)
{
res.push(num[0]);
return ;
}
// Ignore the bit.
grayCodeUtil(res, n - 1, num);
// Invert the bit.
num[0] = num[0] ^ (1 << (n - 1));
grayCodeUtil(res, n - 1, num);
} // Returns the vector containing the gray // code sequence of n bits. function grayCodes(n)
{ let res = [];
// num is passed by reference to keep
// track of current code.
let num = [0];
grayCodeUtil(res, n, num);
return res;
} // Driver code let n = 3; let code = grayCodes(n); for (let i = 0; i < code.length; i++)
document.write(code[i] + "<br>" );
// This code is contributed by gfgking </script> |
Output:
0
1
3
2
6
7
5
4
Time Complexity: O(2n)
Auxiliary Space: O(n)