Question

Write a program go chech the numbers are 4placed successively in a graycode sequence

Answer 1

Generate n-bit Gray Codes

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 the implementation of this approach.

C++