Question

I am taught that given:

message M = 101001
polynomial C = x^3 + x^2 + 1 = 1101

I should add k bits to the end of the message such that the result P is divisible by C (where k is the degree of the polynomial, 3 in this case).

I can find no 3 bit combination (XYZ) that when appended to M satisfies this criteria.

Does anyone know what is wrong with my understanding?

Answer 1

I’m 5 months late to this, but here goes :

Perhaps, thinking about this by integer (or binary) division is counterproductive. Better to work it out by the continuous XOR method – which gives a checksum of 001, rather than the expected 100. This, when appended to the source generates the check value 101001001.

Try this C code to see a somewhat descriptive view.

I’m no expert, but I got most of my CRC fundamentals from here. Hope that helps.