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Cyclic Redundancy Check and Modulo-2 Division
  • Difficulty Level : Hard
  • Last Updated : 22 Dec, 2020

CRC or Cyclic Redundancy Check is a method of detecting accidental changes/errors in the communication channel. 
CRC uses Generator Polynomial which is available on both sender and receiver side. An example generator polynomial is of the form like x3 + x + 1. This generator polynomial represents key 1011. Another example is x2 + 1 that represents key 101. 

n : Number of bits in data to be sent 
    from sender side.  
k : Number of bits in the key obtained 
    from generator polynomial.

Sender Side (Generation of Encoded Data from Data and Generator Polynomial (or Key)): 

  1. The binary data is first augmented by adding k-1 zeros in the end of the data
  2. Use modulo-2 binary division to divide binary data by the key and store remainder of division.
  3. Append the remainder at the end of the data to form the encoded data and send the same

 Receiver Side (Check if there are errors introduced in transmission)
Perform modulo-2 division again and if the remainder is 0, then there are no errors. 

In this article we will focus only on finding the remainder i.e. check word and the code word.

Modulo 2 Division:
The process of modulo-2 binary division is the same as the familiar division process we use for decimal numbers. Just that instead of subtraction, we use XOR here.



  • In each step, a copy of the divisor (or data) is XORed with the k bits of the dividend (or key).
  • The result of the XOR operation (remainder) is (n-1) bits, which is used for the next step after 1 extra bit is pulled down to make it n bits long.
  • When there are no bits left to pull down, we have a result. The (n-1)-bit remainder which is appended at the sender side.

Illustration:
Example 1 (No error in transmission): 

Data word to be sent - 100100
Key - 1101 [ Or generator polynomial x3 + x2 + 1]

Sender Side:

sender

Therefore, the remainder is 001 and hence the encoded 
data sent is 100100001.

Receiver Side:
Code word received at the receiver side  100100001

receiver y

Therefore, the remainder is all zeros. Hence, the
data received has no error.

 
Example 2: (Error in transmission) 

Data word to be sent - 100100
Key - 1101

Sender Side:

sender

Therefore, the remainder is 001 and hence the 
code word sent is 100100001.

Receiver Side
Let there be an error in transmission media
Code word received at the receiver side - 100000001

receiver n

Since the remainder is not all zeroes, the error
is detected at the receiver side.
 

 
Implementation
Below implementation for generating code word from given binary data and key.



C++




#include<bits/stdc++.h>
using namespace std;
  
// Returns XOR of 'a' and 'b'
// (both of same length)
string xor1(string a, string b)
{
     
    // Initialize result
    string result = "";
     
    int n = b.length();
     
    // Traverse all bits, if bits are
    // same, then XOR is 0, else 1
    for(int i = 1; i < n; i++)
    {
        if (a[i] == b[i])
            result += "0";
        else
            result += "1";
    }
    return result;
}
 
// Performs Modulo-2 division
string mod2div(string divident, string divisor)
{
     
    // Number of bits to be XORed at a time.
    int pick = divisor.length();
     
    // Slicing the divident to appropriate
    // length for particular step
    string tmp = divident.substr(0, pick);
     
    int n = divident.length();
     
    while (pick < n)
    {
        if (tmp[0] == '1')
         
            // Replace the divident by the result
            // of XOR and pull 1 bit down
            tmp = xor1(divisor, tmp) + divident[pick];
        else
         
            // If leftmost bit is '0'.
            // If the leftmost bit of the dividend (or the
            // part used in each step) is 0, the step cannot
            // use the regular divisor; we need to use an
            // all-0s divisor.
            tmp = xor1(std::string(pick, '0'), tmp) +
                  divident[pick];
                   
        // Increment pick to move further
        pick += 1;
    }
     
    // For the last n bits, we have to carry it out
    // normally as increased value of pick will cause
    // Index Out of Bounds.
    if (tmp[0] == '1')
        tmp = xor1(divisor, tmp);
    else
        tmp = xor1(std::string(pick, '0'), tmp);
         
    return tmp;
}
 
// Function used at the sender side to encode
// data by appending remainder of modular division
// at the end of data.
void encodeData(string data, string key)
{
    int l_key = key.length();
     
    // Appends n-1 zeroes at end of data
    string appended_data = (data +
                            std::string(
                                l_key - 1, '0'));
     
    string remainder = mod2div(appended_data, key);
     
    // Append remainder in the original data
    string codeword = data + remainder;
    cout << "Remainder : "
         << remainder << "\n";
    cout << "Encoded Data (Data + Remainder) :"
         << codeword << "\n";
}
  
// Driver code
int main()
{
    string data = "100100";
    string key = "1101";
     
    encodeData(data, key);
     
    return 0;
}
 
// This code is contributed by MuskanKalra1


Python3




# Returns XOR of 'a' and 'b'
# (both of same length)
def xor(a, b):
 
    # initialize result
    result = []
 
    # Traverse all bits, if bits are
    # same, then XOR is 0, else 1
    for i in range(1, len(b)):
        if a[i] == b[i]:
            result.append('0')
        else:
            result.append('1')
 
    return ''.join(result)
 
 
# Performs Modulo-2 division
def mod2div(divident, divisor):
 
    # Number of bits to be XORed at a time.
    pick = len(divisor)
 
    # Slicing the divident to appropriate
    # length for particular step
    tmp = divident[0 : pick]
 
    while pick < len(divident):
 
        if tmp[0] == '1':
 
            # replace the divident by the result
            # of XOR and pull 1 bit down
            tmp = xor(divisor, tmp) + divident[pick]
 
        else:   # If leftmost bit is '0'
            # If the leftmost bit of the dividend (or the
            # part used in each step) is 0, the step cannot
            # use the regular divisor; we need to use an
            # all-0s divisor.
            tmp = xor('0'*pick, tmp) + divident[pick]
 
        # increment pick to move further
        pick += 1
 
    # For the last n bits, we have to carry it out
    # normally as increased value of pick will cause
    # Index Out of Bounds.
    if tmp[0] == '1':
        tmp = xor(divisor, tmp)
    else:
        tmp = xor('0'*pick, tmp)
 
    checkword = tmp
    return checkword
 
# Function used at the sender side to encode
# data by appending remainder of modular division
# at the end of data.
def encodeData(data, key):
 
    l_key = len(key)
 
    # Appends n-1 zeroes at end of data
    appended_data = data + '0'*(l_key-1)
    remainder = mod2div(appended_data, key)
 
    # Append remainder in the original data
    codeword = data + remainder
    print("Remainder : ", remainder)
    print("Encoded Data (Data + Remainder) : ",
          codeword)
 
# Driver code
data = "100100"
key = "1101"
encodeData(data, key)


Output:

Remainder :  001
Encoded Data (Data + Remainder) :  100100001

Note that CRC is mainly designed and used to protect against common of errors on communication channels and NOT suitable protection against intentional alteration of data (See reasons here)

Implementation using Bit Manipulation:
CRC codeword generation can also be done using bit manipulation methods as follows:

C++




// C++ Program to generate CRC codeword
#include<stdio.h>
#include<iostream>
#include<math.h>
 
using namespace std;
 
// function to convert integer to binary string
string toBin(long long int num){
    string bin = "";
    while (num){
        if (num & 1)
            bin = "1" + bin;
        else
            bin = "0" + bin;
        num = num>>1;
    }
    return bin;
}
 
// function to convert binary string to decimal
long long int toDec(string bin){
    long long int num = 0;
    for (int i=0; i<bin.length(); i++){
        if (bin.at(i)=='1')
            num += 1 << (bin.length() - i - 1);
    }
    return num;
}
 
// function to compute CRC and codeword
void CRC(string dataword, string generator){
    int l_gen = generator.length();
    long long int gen = toDec(generator);
 
    long long int dword = toDec(dataword);
 
     // append 0s to dividend
    long long int dividend = dword << (l_gen-1);      
 
    // shft specifies the no. of least
    // significant bits not being XORed
    int shft = (int) ceill(log2l(dividend+1)) - l_gen; 
    long long int rem;
 
    while ((dividend >= gen) || (shft >= 0)){
 
        // bitwise XOR the MSBs of dividend with generator
        // replace the operated MSBs from the dividend with
        // remainder generated
        rem = (dividend >> shft) ^ gen;               
        dividend = (dividend & ((1 << shft) - 1)) | (rem << shft);
 
        // change shft variable
        shft = (int) ceill(log2l(dividend + 1)) - l_gen;
    }
 
    // finally, AND the initial dividend with the remainder (=dividend)
    long long int codeword = (dword << (l_gen - 1)) | dividend;
    cout << "Remainder: " << toBin(dividend) << endl;
    cout << "Codeword : " << toBin(codeword) << endl;
}
 
int main(){
    string dataword, generator;
    dataword = "10011101";
    generator = "1001";
    CRC(dataword, generator);
    return 0;
}


Python3




# Python3 program to generate CRC codeword
from math import log, ceil
 
def CRC(dataword, generator):
    dword = int(dataword, 2)
    l_gen = len(generator)
 
    # append 0s to dividend
    dividend = dword << (l_gen - 1)   
 
    # shft specifies the no. of least significant
    # bits not being XORed
    shft = ceil(log(dividend + 1, 2)) - l_gen    
 
    # ceil(log(dividend+1 , 2)) is the no. of binary
    # digits in dividend
    generator = int(generator, 2)
 
    while dividend >= generator or shft >= 0:
 
        # bitwise XOR the MSBs of dividend with generator
        # replace the operated MSBs from the dividend with
        # remainder generated
        rem = (dividend >> shft) ^ generator   
        dividend = (dividend & ((1 << shft) - 1)) | (rem << shft)
         
        # change shft variable
        shft = ceil(log(dividend+1, 2)) - l_gen
 
    # finally, AND the initial dividend with the remainder (=dividend)
    codeword = dword << (l_gen-1)|dividend
    print("Remainder:", bin(dividend).lstrip("-0b"))
    print("Codeword :", bin(codeword).lstrip("-0b"))
 
# Driver code
dataword = "10011101"
generator = "1001"
CRC(dataword, generator)


Output:

Remainder: 100
Codeword : 10011101100

 

References:
https://en.wikipedia.org/wiki/Cyclic_redundancy_check

This article is contributed by Jay Patel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

 

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