Proof that a language involving $gcd$ is not context-free

Question

How would you prove that the following language is not context-free?

$$ L= \{a^n b^m |\, gcd(n,m)=1 \}$$

I suspect the solution uses the pumping lemma, but I'm not sure how to apply it.

Answer 1

Let $p$ be the pumping length guaranteed by the pumping lemma (for context free languages). Then we choose $m \neq n$ such that $m,n \geq p$ and are both prime. Then clearly $s = a^{n}b^{m} \in L$.

By the pumping lemma we can divide $s$ such that $s=uvxyz$ and

1. $|vxy| \leq p$
2. $|vy| \geq 0$
3. $s' = uv^{i}xy^{i}z \in L$ for all $i \in \mathbb{N}$

For this language we get three similar cases, and one trivial case. The trivial case is where either $v$ or $y$ contains both $a$'s and $b$'s, in which case $s'$ doesn't have the correct ordering, and thus $s'\notin L$.

The nontrivial cases:

1. $v$ and $y$ are both strings of $a$'s, then when we pump $s$ we get $s' = a^{n+ik}b^{n}$ where $k\geq 1$. Then $s' \in L$ if $gcd(n+ik,m)=1$, however the modular equation $n+ik\equiv 0 (m)$ has the solution $i \equiv nk^{-1} (m)$, and as $m$ is prime we are guaranteed that $k^{-1}$ exists. Therefore any element of the residue class $i \equiv nk^{-1} (m)$ would give us $gcd(n+ik,m) > 1$. Ergo $s' \notin L$.
   
   Short version: we can pump the string so that $n+ik$ is a multiple of $m$ for any $k$ with $1 \leq k \leq p < m,n$ (which is what we set up at the start).
2. $v$ and $y$ are both strings of $b$'s, but this case is just the symmetric case to case 1, so we derive a contradiction in this case too.
3. $v = a^{k}$ and $y = b^{h}$ for some $k$ and $h$ with $1 \leq k+h \leq p$. Then when we pump we get $s' = a^{n+ik}b^{m+ih}$. Now $s' \in L$ if $n+ik \equiv m+ih (m)$ has no solution, but as I'm sure you can see from here, rearranging just gives $n+i(k+h)\equiv 0(m)$, and as before this has solution $i \equiv n(k+h)^{-1}(m)$. So again we derive a contradiction.

Thus there is at least one string in $L$ that cannot be divided as per the pumping lemma and still have all pumping results remain in $L$. Therefore $L$ is not context free. $Q.E.D$

Some additional notes: of course there is an infinite number of strings which can't be pumped, and it's not actually necessary for $m$ and $n$ to be prime, it just simplifies that argument, as we don't have to fiddle around with a prime decomposition - though if you did want to, you just pick one of the prime factors and do the same (or even all of them if you want).

Answer 2

Trick: A language is not Context free language if it has same Comaprision more then one times For Instance :L = {a^nb^c^n|n>0} as you know this is not context free why:? as you can see number of b shuld be same as the number of a "This is first Comarision" and after that again number of c shuld be same of the number of b or number of previous a.Here 2 stack is needed to remember for equal number of occurance of a and b.so this is not "pda" because you know pda has only one stack...

Another Beautifull Example: L={a^nb^na^mb^m|m,n >0} Tell me weather this language is CFL or not. 

Again Here Look for number of Comparision between same Entity: Here Number of b shuld be same as number of a...In case of a^nb^n...Here only one comarision between "n" now move to the a^mb^m Again here number of b depend up on number of a, where number of doesn't depend on any one so between a and b for power m only one comparision is going on that's why it is L={a^nb^na^mb^m|m,n >0} is CFL

Ex: L = {a^nb^ma^nb^m | m>n>0} Tell me weather It's CFL or not By seeing number of comaprision Answer: Yes This is CFL

Now Lets move to your Questions:

L= {a^n,b^m | gcd(n,m)=1}

Here we are restricted to choose value of n and m...so that it's gcd always come 1.... Find Comarision Here:

 What Will be the value of m,n ???
 It's depend upon condition that whenever you will choose value of m, or n it shuld be either > and equal to m in case of value selection for m or same for value selection for m (This is first comarision between m and n or reverse for selecting value)  Now think 

Another comaprision : GCD of m, n shuld be 1 (This is another comarision for m, n) so here two comarision is present show it is not CFL.

 And Acrding to pumping Lemma Theorem:
    1)Select any string belongs to L
    2)devide it in xYz part make sure that Y shuld not have o length.
    3)Take any value of "i" acrding  to your wish.
    4)make x(Y)^iz and check weather it's belong to L if not then it will prove that it is not regular. i>and equal to 1.

Think: Can you construct a PDA For this, Condition you have one one Stack...Defenately You will come to conclusion ...If you still didn't get my point I will write more for you....