Regular expressions are everywhere. We try to solve almost all of the computing problems on string with a regular expression. But the primary understanding of what a regular expression can and cannot do is missing in most of the solutions. We would try to build some set of thumb rules to find if regular expression is the right solution to a problem. Inorder to do that, we need to understand some specific properties of regular languages. Languages are set of strings accepted by a regular expression in this case. We try to model computing as decision problem of checking whether a particular element belongs to a set or not. The expressibility of regular expression is nothing but what sets can and cannot be defined through regular expression.

## Structure of regular expression

We need to define our notion of regular expression properly to infer some of the properties that it could carry. We define regular expression inductively as follows

$E = x \shortmid E.E \shortmid E+E \shortmid E \ast$

where $E$ is the regular expression and $x$ is any character.

One of the first property we can say about a regular language is any finite set of strings is regular. Since we can just add $+$ between each of the strings in the finite set and call it a regular expression. So the only case we need to care about is the expressibility of infinite sets.

## Ultimately periodic

If we take a infinite regular set and sort it by the length of the string, the length would eventually be periodic. What ultimately periodic means is that, the length of a string will eventually be regularly spaced by a constant or a finite set of constants from other.

For example, consider the regular expression $a(bbb)\ast c$ if you consider the string $abbbbbbbbbbbba$ has length 3 more than $abbbbbbbbba$ which has length 3 more than $abbba$ and those are the only strings in the language. There need not a be a single constant, consider the regular expression $a(bb\shortmid ccc)\ast d$ there are 2 constants here and every string should have length either 2 or 3 more than another if you start taking considerably long strings. With finite sets this aint true, thats exactly why its “ultimately” periodic.

So any set of string which doesnt have this property is regular. For example, $b^k$ where $k$ has to be a perfect square, is not a regular language. Since its infinite and not ultimately periodic. This method is not very useful to distinguish all non-regular languages to be not regular. Since length of context free language is also ultimately

## Pumping lemma

This is the fool proof way to say whether a particular language is regular or not. Formally, pumping lemma states that if there is always a way to partition any string in a regular language of length more than some $p$ as $xyz$ such that $xy^nz$ also belongs to the language, where $len(y) \ge 1$ and $len(xy) \le p$ and $n \ge 0$.

Some examples.

• Set of all strings on $\{a\}$ with length multiples of $4$ is definitely regular. Where $p=4$ $x=\epsilon$ $z=\epsilon$
• Set of all strings of form $ab*a$ is also regular (actually definition itself is a regular expression) where $p=3$ $x=a$ $z=a$

Some non examples

• Set of all strings of the form $a^nb^n$, has no division for any finite $p$. If $y$ is just $a$’s or just $b$’s then the pumped value will have more $a$’s than $b$’s and more $b$’s than $a$’s respectively. If it contained both $a$ and $b$ the pumped string wont have the form all $a$’s followed by all $b$’s
• Set of all well bracketed expression, ie., bracket matched perfectly $\{(,)\}$. If $p \le 6$ consider the string $(())()$ which has no partition, if $p \le 8$ $((()))()$ and so on so forth. We can give a counter example for every choice of $p$

The intuition is simple, only case where a regular language is not a finite set is when it has a Kleene star in its regular expression. Pumping is a way to clearly express what Kleene star does.

## Myhill-Nerode

One other good way to find if a language is regular, is by finding prefixes of a language that could be distinguished. Distinguished here mean, if $u$ and $w$ are prefixes of strings in a regular language, they are indistinguishable if for all suffix $x$ either both $ux$ and $wx$ are part of the language or both aren’t. We put such indistinguishable strings in a partition. If a language has finite number of such partitions then its regular otherwise it isn’t

Some examples

• Set of all strings on $\{a\}$ with length multiples of $4$, have 4 partitions where any string of length $l$, $l \mod 4 = 1$, $l \mod 4 = 2$, $l \mod 4 = 3$, $l \mod 4 = 0$
• Set of all strings of form $ab*a$, have 3 partitions $\epsilon$, $ab*$, $ab*a$

Some non examples

• Set of all strings of the form $a^nb^n$, has infinite such partitions since for each $k$ $ab^k$ suffix is accepted only for prefix $a^{k-1}$.
• Set of well bracketed expressions, also has infinite such partitions, since for each $k$ $)^k$ suffix is accepted by only prefix(es) that needs $k$ closing braces to match.

The intituition is, each such partition can be modelled as a state in the DFA, since any two prefixes become indistinguishable when it reaches the same state. This theorem is usually used to find minimal DFA, but can also be used to find whether a language is regular or not.