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+24 votes

Let $G(x)$ be the generator polynomial used for CRC checking. What is the condition that should be satisfied by $G(x)$ to detect odd number of bits in error?

- $G(x)$ contains more than two terms
- $G(x)$ does not divide $1+x^k$, for any $k$ not exceeding the frame length
- $1+x$ is a factor of $G(x)$
- $G(x)$ has an odd number of terms.

+35 votes

Best answer

Let me first explain building blocks to this problem. Before answering this, we shouldknow the relationship between Sent codeword, Received codeword, CRC generator and error polynomial.

let's take an example:

Sent codeword $=10010\ (=x^4+x)$

Received codeword $= 10\color{blue}{1}10$ (error at $2nd$ bit ) $(=x^4+x^2+x)$

Now, i can write **Sent codeword = Received codeword + error**

$(10010 =10\color{blue}{1}10 + 00100,$ here we do modulo $2$ arithmetic

i.e $1+1 =0$ without carry )

in polynomial also we can see $x^4+x = x^4+x^2+x+x^2 = x^4+\bf{2}x^2+x=\bf{x^4+x}$

**(here multiplying with 2 means 0 **because it corresponds to binary modulo 2 arithmetic **which is 1+1 = 0 **(not 2) **)**^{ }

^{OR }

We can also write,

**Received codeword = Sent codeword + error **(Check it using same method as above)

Sent codeword $C(x),$ Received codeword $R(x)$ and error $E(x)$.

Now we have R**(x) = C(x) + E(x)**. and let CRC polynomial be G**(x)**.

$G(x)$ always divides $C(x),$ and if there is an error then $G(x)$ should not divide $R(x)$.

**Lets check - **

$R(x) \bmod G(x) =\left(C(x) + E(x)\right) \bmod G(x)$ (for simplicity i am writing mod as division)

$\dfrac{R(x)}{G(x)}=\dfrac{C(x)}{G(x)}+\dfrac{E(x)}{G(x)}$

$G(x)$ always divides $C(x)$

$\Rightarrow \dfrac{R(x)}{G(x)}=0+\dfrac{E(x)}{G(x)}$

If $G(x)$ divides $E(x)$ also this would mean $G(x)$ divides $R(x)$. We know that, if $G(x)$ does not properly divide $R(x)$ then there is an error but we are never sure if there is error or not when $G(x)$ divides $R(X)$.

As we saw, $G(x)$ divides $R(x)$ or not totally depends on $G(x)$ divides $E(x)$ or not.

Whole strength of $G(x)$ lies if it does not divide any possible $E(x)$.

Lets see again $E(x)$, if there is an error in $3^{rd}$ and $4^{th}$ bit from left $\text{(LSB is 0th bit )}$ then $E(X) = x^4+x^3$.( it does not matter error is from toggling $1$ to $0$ or $0$ to $1$ ) **Check with above example.**

Now come to question. it says $G(x)$ should detect odd number of bits in error?.

If number of bits are odd then terms in $E(x)$ would be odd.

for instance if $1^{st}, 2nd$ and $5^{th}$ bit got corrupted then $E(x) = x^5+x^2+x.$

It is clear that if any function $f(x)$ has a factor of $x-k,$ then at x=k, $f(x)$ would be zero. I.e. $f(x) = 0$ at $x=k$.

- We want to detect odd number of bits that means received message $R(x)$ contains an odd number of inverted bits, then $E(x)$ must contain an odd number of terms with coefficients equal to $1$.
- As a result, $E(1)$ must equal to $1$
**(remember 1+1 = 0, 1+1+1 = 1.**$E(1)$ is not zero, this means $x+1$ is not a factor of $E(x)$.

Any Odd number of times sum of one's is = 1). - Now I want $G(x)$ not to be a factor of $E(x),$ So that $G(x)$ wont divide $E(x)$ and i would happily detect odd number of bits.
- So, if we make sure that $G(1) = 0,$ we can conclude that $G(x)$ does not divide any $E(x)$ corresponding to an odd number of error bits. In this case, a CRC based on $G(x)$ will detect any odd number of errors.
- As long as $1+x$ is a factor of $G(x), G(x)$ can never divide $E(x).$ Because we know $E(x)$ dont have factor of $1+x$.

**Option C.**

(**Option B **might confuse you, If $G(x)$ has some factor of the form $x^k+1$ then also $G(x)$ would detect all odd number of errors, **But in Option B,** language is changed, and that too we should not have any upper bound on $k$)

+1

@Sachin Mittal1 ,

- As a result, E(1) must equal to 1 (remeber 1+1 = 0, 1+1+1 = 1. Any Odd number of times sum of one's is = 1). E(1) is not zero, this means x+1 is not a factor of E(x).

How does E(1) not equal to zero means x+1 is not a factor of E(x)?? Shouldn't it be x-1 instead of x+1 here?

0

Yes I also agree with shraddha priya

If at X=1 E(X)!=0 it means X-1 is not a factor of E(X) .

also

if at X=1 we want E(1)=1 and G(1)=0

then E(1) mod G(1)=1 mod 0= NaN

+2

@ shraddha priya @akb1115 this is not proper polynomial arithmetic its different here x+1 if its generator shouldn't divide the error polynomial hence as error polynomial contains odd no of digits when we do E(1) we get a 1 as we should do x-or to odd no of 1s and if the polynomial x+1 we substitute 1 we get 0 hence 0 doesnt divide 1 hence our polynomial doesn't divide the error and thats what we want... if our polynomial contains odd no of ones it divides the E(x) as we get 1

+4 votes

+1 vote

+1 vote

If 1+x is factor of generator polynomial then we can find all odd number of error. So** B** is true.

To detect 1 bit error we need x^{k} +1. where k is any constant.

0 votes

if we have any odd number of errors the arithmetic modulo sum will be 1.

example: $x^{3}+x^{2}+x^{1}$

Now if generator polynomial g(x) has x+1 as factor, substituting x as 1 should give zero. Here we are substituting x as 1 and not -1 because it is modulo 2. In $x^{3}+x^{2}+x^{1}$ put x as 1, 1+1+1=3 mod2=1. Hence x+1 does not divide this and can detect the error. same is true for all odd number of errors but not for even number of errors.

example: $x^{3}+x^{2}+x^{1}$

Now if generator polynomial g(x) has x+1 as factor, substituting x as 1 should give zero. Here we are substituting x as 1 and not -1 because it is modulo 2. In $x^{3}+x^{2}+x^{1}$ put x as 1, 1+1+1=3 mod2=1. Hence x+1 does not divide this and can detect the error. same is true for all odd number of errors but not for even number of errors.

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