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I am trying to prove that $L=${$a^mb^n | n=m^2$} is not a CFL with the help of the pumping lemma for CFL's.

I chose $w=a^mb^{m^2}$ = $a^{m-S}a^Sb^Tb^{m^2-T}$ $\in L$

And now in order to contradict the pumping lemma assumption I am looking for $i$ such that $a^{m-S}a^{S^i}b^{T^i}b^{m^2-T}$ $\notin L$ for $i>=0$.

Is it possible to find such value of $i$ ? How do i prove that $b^{m^2-T+Ti}\neq b^{(m+1)^2}$ or $a^{m-S+Si}\neq a^{m+1}$ ?

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up vote 3 down vote accepted

You’re not using the pumping lemma correctly: you’ve not mentioned the pumping length, you’re not allowed to specify a particular breakdown of $w$ into five (not four) pieces, and you only have to show that there is at least one $i\ge 0$ such that $uv^ixy^iz\notin L$ (where $w=uvxyz$ with appropriate restrictions). I’ll do most of a correct argument and leave just the last bit for you.

Assume that $L$ is a CFL, and let $p$ be a pumping number for $L$. Let $w=a^pb^{p^2}$. By the pumping lemma we can decompose $w$ as $uvxyz$ so that $|vxy|\le p$, $|vy|\ge 1$, and $uv^ixy^iz\in L$ for all $i\ge 0$. There are three cases that have to be considered:

  1. $vxy=a^n$ for some positive $n\le p$.
  2. $vxy=a^mb^n$ for some positive $m$ and $n$ such that $m+n\le p$.
  3. $vxy=b^n$ for some positive $n\le p$.

In cases (1) and (3) you should have no trouble seeing that if $i\ne 1$, then $uv^ixy^iz\notin L$; only case (2) requires any real work.

Suppose, then, that $vxy=a^mb^n$ for some positive $m$ and $n$ such that $m+n\le p$. If $v$ contains a $b$, it’s easy to see that $uv^ixy^iz\notin L$ when $i>1$, since $v$ contains $ba$. Similarly, $y$ cannot contain an $a$. Thus, there are $r,s\ge 0$ such that $v=a^r$, $y=b^s$, and $r+s\ge 1$; of course $x=a^{m-r}b^{n-s}$. Then for $i\ge 0$ we have $$uv^ixy^iz=\underbrace{a^{p-m}}_u\underbrace{a^{ir}}_{v^i}\underbrace{a^{m-r}b^{n-s}}_x\underbrace{b^{is}}_{y^i}\underbrace{b^{p^2-n}}_z\;,$$ which is in $L$ iff

$$(p-m+ir+m-r)^2=n-s+is+p^2-n\;,$$ i.e., iff


Let $k=i-1$ and multiply out and simplify to get $2kpr+k^2r^2=ks$. Thus, either $k=0$, or $2pr+kr^2=s$. Use this to show that there is at most one value of $i$ different from $1$ such that $uv^ixy^iz\in L$, and conclude that $L$ is not context-free; you’ll want to recall that $r$ and $s$ are not both $0$.

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Thank you again..! – Yoni Hassin Jan 13 '13 at 17:26
@Yoni: You’re welcome! – Brian M. Scott Jan 13 '13 at 17:30

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