Polynomial generator required to detect single bit error in Cyclic Redundancy Check codes

Question

I was reading about CRC coding from two books:

• Data Communication and Networking by Forouzan Page 294
• Computer Network by Tanenbaum Page 188

They use following notations:

• $d(x)$: dataword to be sent (as a polynomial)
• $c(x)$: codeword sent (as a polynomial)
• $e(x)$: error (as a polynomial)
• $c(x)+e(x)$: codeword sent with error introduced (if any)
• $g(x)$: generator polynomial to be used at CRC encoder (for creating c(x)) and decoder (for checking if error is introduced or not during transmission)

Forouzan then states following:

A single-bit error is $e(x)=x^i$, where i is the position of the bit. If a single-bit error is caught, then $e(x)=x^i$ is not divisible by $g(x)$. (Note that when we say not divisible, we mean that there is a remainder.) If $g(x)$ has at least two terms and the coefficient of $x^0$ is not zero (the rightmost bit is 1), then $e(x)$ cannot be divided by $g(x)$ and all single bits errors can be caught.

He then gives example of generator:

• $g(x)=x+1$ saying that it can catch all single bit error (with which I have some doubts)
• $g(x)=x^3$ saying that all single-bit errors in positions 1 to 3 are caught rest are left uncaught (with which I dont have any confusion).

For example, consider the below example for generator $g(x)=x+1$:

Tanenbaum says:

If $g(x)$ contains two or more terms, $e(x)$ will never divide into $g(x)$, so all single-bit errors will be detected.

My doubts:

Q1. Forouzan states additional requirement to Tanenbaum that "the coefficient of $x^0$ is not zero (the rightmost bit is 1)". Whats correct? and why? Will generator $g(x)=x^3+x^2$ (which satisfies Tanenbaum's statement) capture all single bit errors? Do we always need "$+1$" in $g(x)$ for capturing any single bit error?

Q2. I feel I dont understand the reason/logic behind statements made by both authors, thats why I am not able to decide on myself whether "$+1$" should be there in $g(x)$ or not. Whats the logic then behind both of statements (whichever is correct)? More precisely: why $e(x)$ with single bit error will not be dividable by $g(x)$ if

$g(x)$ has at least two terms and the coefficient of $x^0$ is not zero

as Forouzan says or if

$g(x)$ contains two or more terms

as Tanenbaum says (whichever is correct).

I must be missing something very very basic stuff here.

Answer 1

You just need to concentrate on the polynomial division here. First consider the case where we need to detect one - bit error . For this case

$ e(x) = x^k $

we can choose any polynomial with terms >=2 , since it will not divide the error polynomial completely .

Next let us see the case of odd number of errors. Notice how this is different from one bit error. This is a burst error. Consider the polynomial

$ g(x) = (x+1) $

If we multiply this polynomial with any other polynomial the number of terms we get as a result is always even. This result in general is true you can try this for assurance.

Consider any error polynomial with odd number of errors . This polynomial will have odd number of terms in the polynomial. example :

$e(X) = x^3 + x^2 + x $

Now

$\nexists p(x) \quad such\quad that \quad p(x) * g(x) = e(x) $

In other words no polynomial could be fund which gives odd number of terms when multiplied by x+1 . This observation suggests x+1 should be a factor of our generator function in order to catch odd- number of errors. As 1 - bit error is an odd-bit error x+1 also detects one - bit errors. Both books are correct in their reasoning.