I'm doing some CRC tests in Matlab. I have a CRC-8 function and I would like to know which combination of bit flips leads to the exact same CRC.
Lets consider an array of bits like
Input = [1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,1];
How would I now go through every possible combination of bit flips? I have done some research but didnt find a good answer, maybe you can help.
CRC-8
has eight bits to represent the signature of the input bit-string:
Your experiment could be to generate all possible 2^n
different input bit-strings of length n
and register all strings which lead to a given signature.
Note that you have to start with the same initial 8-bit status word for every experiment.
Your example input is 16 bit long. There are 2^16 = 65536
different bit-strings possible with 16 bits. Try each of them to find out, which strings lead to the same signature as your input.
To get to know the operation of a CRC-8
circuitry, look at one of the animated CRC calculators
available online. You could also have a look at the Wikipedia article on CRC computation .
Using a C#
program (see below), I have done this experiment. Regardless of the ordering of the input bit-string (from-left-to-right or vice versa), 256 of the 65536 bit-strings have the same signature. What could we expect more from a 8-bit hashing method, which has just 256 different signatures to offer?
My code:
using System;
namespace akCrc8Eval
{
class Crc8Evaluator
{
// CRC 8 lookup table
// https://github.com/WasatchPhotonics/CRC8_Example
private static byte[] CRC_8_TABLE =
{
0, 94,188,226, 97, 63,221,131,194,156,126, 32,163,253, 31, 65,
157,195, 33,127,252,162, 64, 30, 95, 1,227,189, 62, 96,130,220,
35,125,159,193, 66, 28,254,160,225,191, 93, 3,128,222, 60, 98,
190,224, 2, 92,223,129, 99, 61,124, 34,192,158, 29, 67,161,255,
70, 24,250,164, 39,121,155,197,132,218, 56,102,229,187, 89, 7,
219,133,103, 57,186,228, 6, 88, 25, 71,165,251,120, 38,196,154,
101, 59,217,135, 4, 90,184,230,167,249, 27, 69,198,152,122, 36,
248,166, 68, 26,153,199, 37,123, 58,100,134,216, 91, 5,231,185,
140,210, 48,110,237,179, 81, 15, 78, 16,242,172, 47,113,147,205,
17, 79,173,243,112, 46,204,146,211,141,111, 49,178,236, 14, 80,
175,241, 19, 77,206,144,114, 44,109, 51,209,143, 12, 82,176,238,
50,108,142,208, 83, 13,239,177,240,174, 76, 18,145,207, 45,115,
202,148,118, 40,171,245, 23, 73, 8, 86,180,234,105, 55,213,139,
87, 9,235,181, 54,104,138,212,149,203, 41,119,244,170, 72, 22,
233,183, 85, 11,136,214, 52,106, 43,117,151,201, 74, 20,246,168,
116, 42,200,150, 21, 75,169,247,182,232, 10, 84,215,137,107, 53
};
// Calculate 8-bit signature for byte array
// inspired by: https://github.com/WasatchPhotonics/CRC8_Example
public byte Calc_CRC_8(byte[] DataArray, int Length)
{
byte CRC = 0; // initial value always 0
for (int i = 0; i < Length; i++)
{
CRC = CRC_8_TABLE[CRC ^ DataArray[i]];
}
return CRC;
}
// Calculate 8-bit signature for 16-bit unsigned int
private byte Calc_CRC_8(UInt16 n)
{
const int DIM = 16;
byte CRC = 0; // initial value always 0
for (int i = 0; i < DIM; i++)
{
byte b = (byte)(((n & (1 << i)) == 0) ? 0 : 1);
CRC = CRC_8_TABLE[CRC ^ b];
}
return CRC;
}
public void getBitStringsWithSameSignature(byte[] bytes)
{
byte sig = Calc_CRC_8(bytes, bytes.Length);
int count = 0;
for (int n = 0; n < UInt16.MaxValue; n++)
{
if (sig == Calc_CRC_8((UInt16)n))
{
count++;
}
}
Console.WriteLine($"Found {count} bit-strings with signature {sig}");
}
}
}
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