Generate n-bit Gray Codes

3.4

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

Examples:

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

The above sequences are Gray Codes of different widths. Following is an interesting pattern in Gray Codes.

n-bit Gray Codes can be generated from list of (n-1)-bit Gray codes using following steps.
1) Let the list of (n-1)-bit Gray codes be L1. Create another list L2 which is 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 required list of n-bit Gray codes.

For example, following are steps for generating the 3-bit Gray code list from the list of 2-bit Gray code list.
L1 = {00, 01, 11, 10} (List of 2-bit Gray Codes)
L2 = {10, 11, 01, 00} (Reverse of L1)
Prefix all entries of L1 with ‘0’, L1 becomes {000, 001, 011, 010}
Prefix all entries of L2 with ‘1’, L2 becomes {110, 111, 101, 100}
Concatenate L1 and L2, we get {000, 001, 011, 010, 110, 111, 101, 100}

To generate n-bit Gray codes, we start from list of 1 bit Gray codes. The list of 1 bit Gray code is {0, 1}. We repeat above steps to generate 2 bit Gray codes from 1 bit Gray codes, then 3-bit Gray codes from 2-bit Gray codes till the number of bits becomes equal to n. Following is C++ implementation of this approach.

// C++ program to generate n-bit Gray codes
#include <iostream>
#include <string>
#include <vector>
using namespace std;

// This function generates all n bit Gray codes and prints the
// generated codes
void generateGrayarr(int n)
{
    // base case
    if (n <= 0)
        return;

    // 'arr' will store all generated codes
    vector<string> arr;

    // start with one-bit pattern
    arr.push_back("0");
    arr.push_back("1");

    // Every iteration of this loop generates 2*i codes from previously
    // generated i codes.
    int i, j;
    for (i = 2; i < (1<<n); i = i<<1)
    {
        // Enter the prviously generated codes again in arr[] in reverse
        // order. Nor arr[] has double number of codes.
        for (j = i-1 ; j >= 0 ; j--)
            arr.push_back(arr[j]);

        // append 0 to the first half
        for (j = 0 ; j < i ; j++)
            arr[j] = "0" + arr[j];

        // append 1 to the second half
        for (j = i ; j < 2*i ; j++)
            arr[j] = "1" + arr[j];
    }

    // print contents of arr[]
    for (i = 0 ; i < arr.size() ; i++ )
        cout << arr[i] << endl;
}

// Driver program to test above function
int main()
{
    generateGrayarr(4);
    return 0;
}

Output

000
001
011
010
110
111
101
100

Asked in: AmazonMicrosoft

This article is compiled by Ravi Chandra Enaganti. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



3.4 Average Difficulty : 3.4/5.0
Based on 39 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.