Hamming code Implementation in C/C++
Pre-requisite: Hamming Code
Given a message bit in the form of an array msgBit[], the task is to find the Hamming Code of the given message bit.
Examples:
Input: S = “0101”
Output:
Generated codeword:
r1 r2 m1 r4 m2 m3 m4
0 1 0 0 1 0 1
Explanation:
Initially r1, r2, r4 is set to ‘0’.
r1 = Bitwise XOR of all bits position that has ‘1’ in its 0th-bit position.
r2 = Bitwise XOR of all bits that has ‘1’ in its 1st-bit position.
r3 = Bitwise XOR of all bits that has ‘1’ in its 2nd-bit position.Input: S = “0111”
Output:
Generated codeword:
r1 r2 m1 r4 m2 m3 m4
0 0 0 1 1 1 1
Approach: The idea is to first find the number of redundant bits which can be found by initializing r with 1 and then incrementing it by 1 each time while 2r is smaller than (m + r + 1) where m is the number of bits in the input message. Follow the below steps to solve the problem:
- Initialize r by 1 and increment it by 1 until 2r is smaller than m+r+1.
- Initialize a vector hammingCode of size r + m which will be the length of the output message.
- Initialize all the positions of redundant bits with -1 by traversing from i = 0 to r – 1 and setting hammingCode [2i – 1] = -1. Then place the input message bits in all the positions where hammingCode[j] is not -1 in order where 0 <= j < (r + m).
- Initialize a variable one_count with 0 to store the number of ones and then traverse from i = 0 to (r + m – 1).
- If the current bit i.e., hammingCode[i] is not -1 then find the message bit containing set bit at log2(i+1)th position by traversing from j = i+2 to r+m by incrementing one_count by 1 if (j & (1<<x)) is not 0 and hammingCode[j – 1] is 1.
- If for index i, one_count is even, set hammingCode[i] = 0 otherwise set hammingCode[i] = 1.
- After traversing, print the hammingCode[] vector as the output message.
Below is the implementation of the above approach:
C
// C program for the above approach#include <math.h>#include <stdio.h>// Store input bitsint input[32];// Store hamming codeint code[32];int ham_calc(int, int);void solve(int input[], int);// Function to calculate bit for// ith positionint ham_calc(int position, int c_l){ int count = 0, i, j; i = position - 1; // Traverse to store Hamming Code while (i < c_l) { for (j = i; j < i + position; j++) { // If current boit is 1 if (code[j] == 1) count++; } // Update i i = i + 2 * position; } if (count % 2 == 0) return 0; else return 1;}// Function to calculate hamming codevoid solve(int input[], int n){ int i, p_n = 0, c_l, j, k; i = 0; // Find msg bits having set bit // at x'th position of number while (n > (int)pow(2, i) - (i + 1)) { p_n++; i++; } c_l = p_n + n; j = k = 0; // Traverse the msgBits for (i = 0; i < c_l; i++) { // Update the code if (i == ((int)pow(2, k) - 1)) { code[i] = 0; k++; } // Update the code[i] to the // input character at index j else { code[i] = input[j]; j++; } } // Traverse and update the // hamming code for (i = 0; i < p_n; i++) { // Find current position int position = (int)pow(2, i); // Find value at current position int value = ham_calc(position, c_l); // Update the code code[position - 1] = value; } // Print the Hamming Code printf("\nThe generated Code Word is: "); for (i = 0; i < c_l; i++) { printf("%d", code[i]); }}// Driver Codevoid main(){ // Given input message Bit input[0] = 0; input[1] = 1; input[2] = 1; input[3] = 1; int N = 4; // Function Call solve(input, N);} |
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to generate hamming codevector<int> generateHammingCode( vector<int> msgBits, int m, int r){ // Stores the Hamming Code vector<int> hammingCode(r + m); // Find positions of redundant bits for (int i = 0; i < r; ++i) { // Placing -1 at redundant bits // place to identify it later hammingCode[pow(2, i) - 1] = -1; } int j = 0; // Iterate to update the code for (int i = 0; i < (r + m); i++) { // Placing msgBits where -1 is // absent i.e., except redundant // bits all positions are msgBits if (hammingCode[i] != -1) { hammingCode[i] = msgBits[j]; j++; } } for (int i = 0; i < (r + m); i++) { // If current bit is not redundant // bit then continue if (hammingCode[i] != -1) continue; int x = log2(i + 1); int one_count = 0; // Find msg bits containing // set bit at x'th position for (int j = i + 2; j <= (r + m); ++j) { if (j & (1 << x)) { if (hammingCode[j - 1] == 1) { one_count++; } } } // Generating hamming code for // even parity if (one_count % 2 == 0) { hammingCode[i] = 0; } else { hammingCode[i] = 1; } } // Return the generated code return hammingCode;}// Function to find the hamming code// of the given message bit msgBit[]void findHammingCode(vector<int>& msgBit){ // Message bit size int m = msgBit.size(); // r is the number of redundant bits int r = 1; // Find no. of redundant bits while (pow(2, r) < (m + r + 1)) { r++; } // Generating Code vector<int> ans = generateHammingCode(msgBit, m, r); // Print the code cout << "Message bits are: "; for (int i = 0; i < msgBit.size(); i++) cout << msgBit[i] << " "; cout << "\nHamming code is: "; for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";}// Driver Codeint main(){ // Given message bits vector<int> msgBit = { 0, 1, 0, 1 }; // Function Call findHammingCode(msgBit); return 0;} |
Output:
The generated Code Word is: 0001111
Time Complexity: O((M + R)2) where M is the number of bits in the input message and R is the number of redundant bits
Auxiliary Space: O(M + R)
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