# Lecture 04

## A Non-regular Language

Thesis I Decision problems = languages

Thesis II Solving = recognizable by a finite automata

$$L=\{o^n1^n:n\geq 0\}$$

Theorem $$L_{0^n1^n}$$ is not regular

Pigeonhole Principle If I have $$n$$ pigeons and I put them in $$m$$ boxes for any $$m<n$$, at least one box contains more than one pigeon.

Assume there is a DFA $$M$$ with $$m$$ states, that recognize $$L_{0^n1^n}$$. Consider the state in $$M$$ $$r_0,...,r_m$$ reached after reading $$\epsilon, 0, 00,...,0^m$$.

By the pigeonhole principle, there exists $$0\leq k < l \leq m$$ such $$r_k=r_l$$.

So $$M$$ is in the same state after reading $$0^k$$ and $$0^l$$, so it mush also be in the same state after reading $$0^kw$$ and $$0^lw$$ for any string $$w\in\{0,1\}^*$$.

But then consider $$w=1^k$$. Then $$M$$ either accepts or rejects both $$0^k1^k$$ and $$0^l1^k$$. In either case, $$M$$ does not recognized $$L_{0^n1^n}$$ $$\Rightarrow\Leftarrow$$.

## Pumping Lemma

For any regular language $$L$$, there exists $$p\geq 1$$ (pumping length), such that every string $$s\in L$$ of length at least greater than $$p$$ can be expressed as $$s=xyz$$ such that

1. $$\forall i\geq 0$$, $$xy^iz\in L$$
2. $$|y|>0$$
3. $$|xy|\leq p$$

Proof Let $$M$$ be the smallest DFA that recognizes $$L$$. Let $$m$$ be the number of states in $$M$$.

Set $$p=m$$, fix any $$s\in L$$ of length $$|s|>p$$ by the pigeonhole principle, if $$r_0,...,r_p$$ are the states reached in $$M$$ after reading $$i=0,1,...,p$$ symbols of $$s$$, there exists $$0\leq k < l \leq p$$ such that $$r_k=r_l$$.

Let $$x=s_1,...,s_k$$ $$y=s_{k+1},...,s_l$$, $$z=s_{l+1},...,s_{|s|}$$ we have that

1. $$|xy|=l\leq p$$
2. $$|y|\geq 1 > 0$$ because $$l>k$$
3. $$r_k=r_l$$ mean that $$M$$ reaches the same state after reading $$xw$$ and $$xyw$$ for any $$m\in\Sigma^*$$.

In particular, with $$w=z$$ we see that $$M$$ reaches the same state on $$xz$$ and $$xyz=s$$. Since $$M$$ accepts $$s$$, it accepts $$xz$$ as well.

When $$w=yz$$, we see that $$M$$ reaches same state on both $$xyz=s$$ and $$xyyz$$, so it accepts $$xyyz$$.

Rest: Induction on $$i\geq 2$$

I step: Induction Hypothesis states that we have already shown $$xy^{i-1}z\in L$$.

Now consider $$xy^iz$$. Set $$w=y^{i-1}z$$. Then $$M$$ reaches the same state on both $$xy^{i-1}z\in L$$ and $$xyy^{i-1}z=xy^iz$$. So $$xy^iz\in L$$ also.

Definition $$L$$ has the pumping property if it satisfies the condition of the pumping lemma.

## Using the Pumping Lemma

To show $$L$$ is not regular, it suffices to show that it does not have the pumping property.

(Show that $$\forall p\geq 1$$, there is a string $$s\in L$$ of length $$|s|\geq p$$ such that for any decomposition $$s=xyz$$ with $$|y| > 0$$ and $$|xy|\leq p$$, there exists $$i\geq 0$$ such that $$xy^iz\not\in L$$)

E.g. $$L_{0^n1^n}$$ is not regular

Proof For any $$p\geq 1$$, we can choose $$s=0^p1^p\in L$$.

Fix any decomposition $$s=xyz$$ such that $$|y| > 0$$ and $$|xy|\leq p$$. In any such decomposition, $$y=0^j$$ for some $$1\leq j\leq p$$.

So when $$i=2$$, $$xy^2z=0^{p+j}1^p\not\in L_{0^n1^n}$$

$$E.g.$$

1. $$L=\{w\in\{0,1\}^*:w$$ has more $$1$$s that $$0$$s $$\}$$
• $$0^p1^{p+1}$$
2. $$L=\{ww: w\in\{0,1\}^*\}$$
• $$0^p10^p1$$
3. $$L=\{1^{n^2}: n\geq 0\}$$
• $$s=1^{p^2}$$

## Glimpse Beyond

• CS 462
• Myhill-Nerode Theorem
• Theorem: If $$L$$ is a language over $$\Sigma=\{1\}$$ that is recognized by an NFA with $$n$$ states, it can be recognized by a DFA with $$e^{\sqrt{n}\log{n}}$$ states.