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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?

1. $G(x)$ contains more than two terms
2. $G(x)$ does not divide $1+x^k$, for any $k$ not exceeding the frame length
3. $1+x$ is a factor of $G(x)$
4. $G(x)$ has an odd number of terms.

$Ans: C$

$*$ Error should not be divisible by your generator

$a)$ $\dfrac{E(x)}{G(x)}=\dfrac{1+x+x^2}{1+x+x^2}: Wrong$

$b)$ $G(x)$ does not divide $1+x^k,$ so let's take $G(x)=x$

$\dfrac{E(x)}{G(x)}=\dfrac{x+x^2+x^3}{x}: Wrong$

$d)$ $\dfrac{E(x)}{G(x)}=\dfrac{1+x+x^2}{1+x+x^2}: Wrong$

$c)$ $\dfrac{E(x)}{G(x)}=\dfrac{E(x)}{(1+x)G(x)}=\dfrac{odd \#\ of\ terms}{even \#\ of\ terms}:Not\ possible: \checkmark$

Reference :-Tanenbaum

answer to question is marked by red.

Rest marked by yellow are also properties of G(x) to catch different type of errors.  edited

@KUSHAGRA गुप्ता Why did you take ${E(x)}$$={x+x^2+x^3}$ when it says $G(x)$ does not divide $1+x^k$ ? How is $x+x^{2}+x^{3}$  the expansion of $1+x^{k}$ ?

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.
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 $E(x)$.
• 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 $k$)

For even number of bit errors -

It's impossible to tell if there are exactly $0$ error bits, when we're using CRC, because there's always a chance that the error goes undetected.

However, if we exclude $0$ from even numbers, then the helpful/important things are-

1. $G(x)$ should not divide $E(x)$.

2. Since there are even no. of error bits, the polynomial $E(1) = 0$

3. $G(x)$ will divide $E(x)$ if $G(1) = 0$, and by point $1$, we want to avoid that.

4. So, if we have $G(x) = 1$, then $G(x)$ won't divide $E(x)$.

5. What does it mean for $G(x)$ to be equal to $1$? It means the generator polynomial $G(x)$ should have odd number of terms.

6. So, if I do $G(x) = 1 + x^m + x^{n}$, then $G(1)$ which will always be an odd number, which are represented by $1$ in modular $2$ arithmetic, whereas even numbers are represented by $0$.

7. I don't know if it's okay to say that $1 + x + x^2$ is enough to detect even no. of errors.

Great explanation! Thank you Sachin!
edited by

I think for even no of error bits statement is not correct

please see below example-

E(x)=$x^{3}+x^{4}+x^{5}+x^{6}+x^{7}+x^{8}$ which can be written as ($1+x+x^{2}$)($x^{3}+x^{6}$)

E(x) has even no of errors but it is divisible by G(x) $1+x+x^{2}$

Please correct me if wrong...

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.
by

Thanks for explanation.
too short..

The better the generator polynomial, the more likelihood of detecting the errors.

There's no universal CRC polynomial that can detect all errors, but we can establish some guidelines of a good CRC generator polynomial. Let's denote the generator polynomial function by g(x).

• To detect single bit errors, g(x) must have at least two terms.
»For eg: g(x) = x + 1.

• To detect all odd number of errors, g(x) must have an even number of terms.
This means (1 + x) should be a factor of g(x). [Option C is correct]
»For eg: g(x) = x + 1 or,
»g(x) =  $x^{4} + x^{3} + x + 1$

• To detect burst errors, ie, errors of the form "1 <any combo of 0's and 1's> 1", we must pick g(x) such that it has a degree b, where b is the length of the burst error.
Moreover, If g(x) does not have x as a factor, it'll detect any error except a specific error of length b+1.

Hence, Option C

### 1 comment

@Jashan sir, If  $G(x) = x^4 + x + 1$ then $1+x$ is not factor of $G(x)$ right?

Even if $G(x)$ contains the term $1+x$ it’s NOT necessary that $1+x$ is factor of $G(x)$. $1+x$ is factor of $G(x)$ if and only if $G(x)$ contains even number of terms.

Is my understanding correct?

Let G(x) be the generator polynomial.

C(x) be the sent codeword, R(x) be the received code and E(x) be the error polynomial.

We can write: R(x) = C(x) + E(x)                             { This can be written by mod 2 addition and skipping the carry}

As we know that: "To detect error, G(x) should not divide R(x)".

=> G(x) should not divide E(x). As, $\frac{R(x)}{G(x)} = \frac{C(x)}{G(x)} + \frac{E(x)}{G(x)}$

Further, $\frac{R(x)}{G(x)} = 0 + \frac{E(x)}{G(x)}$

This means if E(x) is not divisible by G(x) then R(x) will not also be divisible by G(x).

Now, To detect odd number of bits in error, E(x) will contain odd number of terms.

Example: let sent codeword= 101010 and Received codeword = 100100 i.e three bits (1st, 2nd and 3rd bit) are in error.

Therefore, $E(x) = x^{3}+x^{2}+x$.

A) G(x) contain more than 2 terms.

let, $G(x) = x^{3}+x^{2}+x$ then $\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{x^{3}+x^{2}+x}$

So, E(x) is divisible by G(x). So, A is not the right option.

B) G(x) does not divide by $1+x^{k}$, for any k not exceeding frame length.

here, k=6 [ frame length or code length ]

let $G(x) = x$   &   $E(x) = x^{3}+x^{2}+x$.

$\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{x}$. here, E(x) can be divisible by G(x). So, B is not the right option.

C) $(1+x)$ is a factor of G(x).

$\frac{E(x)}{G(x)} = \frac{x^{3}+x^{2}+x}{(1+x)G(x)} = \frac{odd no. of terms}{even no. of terms}$

here, E(x) wil never be divisible by G(x). So, (C) is correct option.

D) G(x) has odd number of terms

let $G(x) = x^{3}+x^{2}+x$.

$\frac{E(x)}{G(x)} = \frac{odd No of terms}{Odd number of terms}$

here, G(x) may divide E(x). So, this canot be the option.

Hence, Correct Ans: (C)