The Cyclic Redundancy Check

1. The other aspect of the CRC that makes it valuable is the surprising fact that it is actually quite simple to calculate.

   • In a transmission system, the ideal is to be able to calculate the checksum bit by bit as the data arrives (rather than having to wait for the whole frame to arrive before we can start the process). This is how the calculation of the CRC is performed.

   • As a result, the best way to undertand the computation is to assume you have received a message and computed its CRC when to your surprise an extra bit of the message arrives. How could you calculate the CRC of the message with the added bit included based on the CRC you had computed before you knew about the extra bit?

   • Well, let's call the partial message T(X) and assume that the partial CRC is R(x) such that for some Q(X)

     T(X) = Q(x)G(x) + R(x)

     and lets call the extra bit "b".

   • By multiplying both sides of the equation above by x, adding b to both sides and then regrouping a bit we obtain

     (T(X)x + b) = (x Q(x))G(x) + (R(x)x + b)

   • If R(x)x is of degree less than G(x) then we are done. In this case, by the definition of polynomial division, R(x)x + b is the remainder of dividing the polynomial corresponding to the complete message by G(x). Since multiplying a polynomial by x just raises the power of all its terms, this means we can obtain the CRC for the polynomial with b added by throwing away the first bit of the old CRC (which must have been 0) and adding b as the last bit.

   • If the first bit of the old CRC was 1, then R(x)x will have degree k. Luckily, in this case

     R(x)x + b - Q(x)

     is guaranteed to have degree less than k since the coefficients of x^k in both Q(x) and R(x)x must be 1 and 1+1 = 0.

     In this case, regrouping the terms of the equation above to yield

     (T(X)x + b) = (x Q(x) + 1)G(x) + (R(x)x + b - G(x))

     allows us to conclude that this time

     R(x)x + b - Q(x)

     must be the new remainder.

     Accordingly, when the first bit of the CRC is 1, we can compute the new CRC by first shifting the old CRC left by one, then appending the new bit b and then adding G(x). Since addition mod 2 is just the exclusive or operation, we can just exclusive or the bits of the shifted CRC with the bits corresponding to the coefficients of G(x).

   • As the text shows, these rules can be expressed by cute little diagrams that can ultimately be turned into simple and highly efficient circuits for computing and verifying CRCs.