How do I calculate CRC32 mathematically?

Question

I want to calcuate the CRC32 algorithm using polynomials directly but I don't know how. I found the generating polynomial listed here https://en.wikipedia.org/wiki/Cyclic_redundancy_check. This corresponds to 0xedb88320 in the example code.

Can someone please define a mathematical specification that calculates the same result as this algorithm?

Answer 1

The most common CRC32 is borrowed from the 32-bit Frame Check Sequence in the 1988 edition of CCITT V.42, section 8.1.1.6.2, available here, which gives a mathematical definition (note: remove the obviously spurious $1$ after $x^{30}$ in the English edition).

I prefer this alternate definition with some of the math on polynomial replaced by equivalent operations on bits:

1. Consider the message as a sequence of $n$ bits in chronological order (if the message is structured in words or bytes: with low-order bit first unless otherwise specified).
2. Append $32$ one bits, forming a sequence of $n+32$ bits.
3. Complement the first $32$ bits of that sequence.
4. Form the binary polynomial of degree (at most) $n+31$, with term $x^{n+32-j}$ present (resp. absent) when the $j$^th bit in the result of the previous step is one (resp. zero), for $j$ with $0<j\le n+32$.
5. Compute the remainder of the polynomial division of that polynomial by the binary polynomial $x^{32}+x^{26}+x^{23}+x^{22}+x^{16}+x^{12}+x^{11}+x^{10}+x^8+x^7+x^5+x^4+x^2+x+1$, forming a binary polynomial of degree (at most) $31$.
6. Form the $32$-bit sequence with the $j$^th bit one (resp. zero) when the term $x^{32-j}$ is present (resp. absent) in that polynomial, for $j$ with $0<j\le 32$.
7. Append that $32$-bit sequence to the ORIGINAL message (if it is to be converted to bytes: the first bit of that sequence shall be the low-order bit of the first byte unless otherwise specified).

Note: Inserting $32$ bits at step 2 allows a receiver to process bits of the message and $32$-bit sequence uniformly as they are being received, without knowing the frontier between the message and the final 32-bit sequence until after the end of that sequence. Step 3 makes it likely that suppression of bits in the message is detected including for zero bits at the beginning of the message.

Note: In term of binary polynomials (according to the conventions in steps 4 and 6), the combination of steps 2 and 3 changes $M(x)$ to $M(x)\cdot x^{32}+\displaystyle\sum_{i=n}^{n+31}{x^i}+\sum_{i=0}^{31}{x^i}$.

Note: Steps 4, 5 and 6 can be replaced by:

• repeat until the sequence has exactly 32 bits:
  • if the first bit is one:
    • complement the 7^th, 10^th, 11^th, 17^th, 21^th, 22^th, 23^th, 25^th, 26^th, 28^th, 29^th, 31^th, 32^th and 33^th bit.
  • remove the first bit.