本文概述
- Python3
- C ++
CRC使用生成多项式在发送方和接收方均可用。示例生成器多项式的形式如x3+ x +1。此生成多项式表示键1011。另一个示例是x2+1代表键101。
n : Number of bits in data to be sent from sender side.k : Number of bits in the key obtained from generator polynomial.
发送方(从数据和生成多项式(或密钥)生成编码数据):
首先通过在数据末尾添加k-1个零来扩充二进制数据
使用modulo-2二进制除法可通过键对二进制数据进行除法并存储除法的余数。
在数据末尾追加其余部分以形成编码数据并发送
.
接收方(检查传输中是否引入了错误)
再次执行模2除法, 如果余数为0, 则没有错误。
在本文中, 我们将只专注于查找余数, 即校验字和代码字。
模2部门:
模2二进制除法的过程与我们用于十进制数的常见除法过程相同。只是, 我们在这里使用XOR而不是减法。
- 在每个步骤中, 将除数(或数据)的副本与被除数(或键)的k位进行异或。
- XOR操作的结果(余数)为(n-1)位, 将其拉低1位以使其为n位长后, 将用于下一步。
- 当没有剩余可下拉的位时, 我们得到结果。 (n-1)位余数附加在发送方。
示例1(传输无错误):
Data word to be sent - 100100Key - 1101 [ Or generator polynomial x3 + x2 + 1]Sender Side:Therefore, the remainder is 001 and hence the encoded data sent is 100100001.Receiver Side:Code word received at the receiver side100100001Therefore, the remainder is all zeros. Hence, thedata received has no error.
示例2 :(传输错误)
Data word to be sent - 100100Key - 1101Sender Side:Therefore, the remainder is 001 and hence the code word sent is 100100001.Receiver SideLet there be an error in transmission mediaCode word received at the receiver side - 100000001 Since the remainder is not all zeroes, the erroris detected at the receiver side.
实现
以下是从给定的二进制数据和密钥生成代码字的Python实现。
# 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 = https://www.lsbin.com/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 = "https://www.lsbin.com/100100"
key = "1101"
encodeData(data, key)
输出如下:
('Remainder : ', '001')('Encoded Data (Data + Remainder) : ', '100100001')
输出如下:
Remainder :001Encoded Data (Data + Remainder) :100100001
请注意, CRC主要设计用于防止通信信道上常见的错误, 而不适用于防止数据有意更改(请参阅原因)。这里)
使用位操作的实现:
CRC码字的生成也可以使用以下位处理方法来完成:
Python3
# Python program to generate CRC codeword
from math import log, ceildef 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)
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;
}
输出如下:
Remainder: 100Codeword : 10011101100
参考文献:
https://en.wikipedia.org/wiki/Cyclic_redundancy_check
本文作者:杰伊·帕特尔(Jay Patel)。如果你喜欢lsbin并希望做出贡献, 那么你也可以写一篇文章并将你的文章邮寄到contribution@lsbin.org。查看你的文章出现在lsbin主页上, 并帮助其他Geeks。
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