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Hamming code Implementation in C++ with receiver side verification of code

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  • Difficulty Level : Hard
  • Last Updated : 20 Jan, 2023
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C++




#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
class hamming{
    public:
        string data;   //it is the raw data received
        int m , r = 0; // n is the length of raw data and r is the number of redundant bits
        char * msg; // it will store the all bits (data + redundant). We made it dynamic because at compile time we dont know how much redundant bits will be there, we will initialize memory to it once we know the number of redundant bits.
        hamming(string data){
              this->data = data;
            //reversing the data received
            reverse(data.begin(),data.end());
            m = data.size();
            int power = 1;
 
            //finding the number of redundant bits and storing them in r
            while(power < (m + r + 1)){
                r++;
                power*=2;
            }
            //Allocating memory to our dynamic msg array(Note we are using one based indexing).
            msg = new char[m+r+1];
            int curr = 0;
 
            //initializing the msg with data bits and for redundant bits, an initial value of n
            for(int i = 1 ; i <= m+r ; i++){
                if(i & (i-1)){
                    msg[i] = data[curr++];
                }
                else msg[i] = 'n';
            }
            //function call to set the redundant bits
            setRedundantBits();
        }
        //function to show the whole msg
        void showmsg(){
            cout << "the data packet to be sent is :   ";
            for(int i = m+r ; i >= 1 ; i--){
                cout << msg[i] << " ";
            }
            cout << endl;
        }
        void setRedundantBits(){
            //for first redundant bit, check all those data bits at index where the first bit of index is set(1) similarly for second redundant bit, check all those data bits at index where the second bit of index is set(1), similarly for third redundant bit check all those data bits at index where the third bit of index is set to 1 and so on.
            int bit = 0;
            //outer loop runs for redundant bits (1 ,2 ,4 ,8 ....)
            for(int i = 1 ; i  <= m+r ; i*=2){
                int count = 0;
                //inner loop runs for data bits
                for(int j = i+1 ; j<=m+r ; j++){
                    // checking if the data bit corresponds to our redundant bit or not using bit manipulation
                    if(j & (1 << bit)){
                        if(msg[j] == '1') count++; // counting the number of ones in corresponding data bits
                    }
                }
                //setting up redundant bits
                if(count & 1) msg[i] = '1';
                else msg[i] = '0';
                //increasing the bit position.
                bit++;
            }
            //showing up the message to be sent(data + redundant)
            showmsg();
        }
        void receiver(){
            //this ans will store the redundant bits, if they were right then according to even parity they will store 0 else if some error was made in a bit it will store 1
            string ans = "";
            int bit = 0;
            //this loop corresponds to the logic used in set redundant bits function
            for(int i = 1 ; i  <= m+r ; i*=2){
                int count = 0;
                for(int j = i+1 ; j<=m+r ; j++){
                    if(j & (1 << bit)){
                        if(msg[j] == '1') count++;
                    }
                }
                //incrementing the ans variable with the parity of redundant bit
                // if it was right then add 0 else 1
                if(count & 1){
                    if(msg[i] == '1') ans.push_back('0');
                    else ans.push_back('1');
                }
                else{
                    if(msg[i]=='0') ans.push_back('0');
                    else ans.push_back('1');
                }
                bit++;
            }
            // if the ans had any occurrence of 1 then there is some fault
            if(ans.find('1') != string::npos){
                int power = 1;
                int wrongbit = 0;
                //evaluating the binary expression of ans variable
                for(int i = 0 ; i < ans.size() ; i++){
                    if(ans[i]=='1') wrongbit+=power;
                    power*=2;
                }
                cout << "bit number " << wrongbit << " is wrong and having error " << endl;
            }
            // if the ans dont have any occurrence of 1 then it is correct
            else{
                cout << "correct data packet received " << endl;
            }
        }
};
int main(){
      string data = "1011001";
    hamming h(data);
    // manipulating any ith data bit to check if receiver is detecting a error in that bit. If you eliminate the following line then correct code will be sent to receiver following that no error is received
     
    //h.msg[i] == '0' ? h.msg[i] = '1' : h.msg[i] = '0';
 
    h.receiver();
    return 0;
}
//time complexity = O(nlogn) where , n = databits + redundant bits
//this code is contributed by Mayank Sharma.


Output

the data packet to be sent is :   1 0 1 0 1 0 0 1 1 1 0 
correct data packet received 

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 bits
int input[32];
 
// Store hamming code
int code[32];
 
int ham_calc(int, int);
void solve(int input[], int);
 
// Function to calculate bit for
// ith position
int 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 bit 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 code
void 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 Code
void 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 code
vector<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 Code
int main()
{
    // Given message bits
    vector<int> msgBit = { 0, 1, 0, 1 };
 
    // Function Call
    findHammingCode(msgBit);
 
    return 0;
}


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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