Properties of Regular Languages

Chapter 4

Regular Languages

A language is regular iff it is accepted by some DFA.

Example: L = {w {a, b}* :

every a is immediately followed by a b}.

Languages: Regular or Not?

a*b* is regular. {anbn: n 0} is not.

{w {a, b}* : every a is immediately followed by b} is regular.

{w {a, b}* : every a has a matching b somewhere} is not

Note: We may define a grammar for a non-regular language, but not a regular grammar (left- or right-linear)

Closure Properties of Regular Languages

● Union

● Concatenation

● Kleene star

● Complement

● Intersection

● Difference

● Reverse

● Letter substitution

Showing that a Language is Regular

Theorem: Every finite language is regular.

Proof: If L is the empty set, then it is defined by the regular expression and so is regular. If it is any finite language composed of the strings s1, s2, … sn for some positive integer n, then it is defined by the regular expression:

s1 + s2 + … + sn

So it too is regular.

Showing that a Language is Regular

Example:

Let L = L1 ∩ L2, where:L1 = {anbn, n 0}, and L2 = {bnan, n 0}

L1 and L2 are not regular, but L = λ, which is regular

More Examples

● L1 = {w {0 - 9}*: w is the social security number of the current US president}.

● L2 = {1 if Santa Claus exists and 0 otherwise}

● L3 = {1 if God exists and 0 otherwise}

Examples of finite languages. Even though they may be unknown (or debatable), the languages are finite; therefore regular.

Showing That L is Regular

1. Show that L is finite.

2. Exhibit an FSM for L.

3. Exhibit a regular expression for L.

4. Show that the number of equivalence classes of L is finite.

5. Exhibit a regular grammar for L.

6. Exploit the closure theorems.

Identifying Nonregular Languages

Finite languages are regular, but what about infinite languages???

Some are regular: a*b*

Some are not: L = {anbn: n 0}

Pigeonhole Principle

If n objects are put into m boxes, and if n > m, then at least one box must have more than 1 object in it.

Indicates repetitionis needed

Showing that a Language is Not Regular

Every regular language can be accepted by some FSM.

It can only use a finite amount of memory to record essential properties.

The only way to generate/accept an infinite language with a finite description is to use:

• Kleene star (in regular expressions), or • cycles (in automata).

This forces some kind of simple repetitive cycle within the strings.

Example:ab*a generates aba, abba, abbba, abbbba, etc.

How Long a String Can Be Accepted?

What is the longest string that a 3-state FA can accept?

Theorem – Long Strings

Theorem: Let M = (Q, , , q0, F) be any DFSM. If M accepts any string of length |Q| or greater, then that string will force M to visit some state more than once (thus traversing at least one loop).

Proof: M must start in one of its states. Each time it reads an input character, it visits some state. So, in processing a string of length n, M creates a total of n + 1 state visits. If n+1 > |Q|, then, by the pigeonhole principle, some state must get more than one visit. So, if n |Q|, then M must visit at least one state more than once.

The Pumping Lemma for Regular Languages

Let L be an infinite regular language.

So, m 1 : strings w L, where |w| m

x, y, z w = xyz,

|xy| m,

|y| 1, and

wi = xyi z is also in L, for all i = 0, 1, 2, …

Note:

m is the number of states in an FA that accepts L

y represents a repetition in string w (a loop in the FA)

i represents the number of times through the loop

The Pumping Lemma for Regular Languages

Paraphrased:

Every sufficiently long string in L can be broken into three parts in such a way that an arbitrary number of repetitions of the middle part yields another string in L. We say that the middle string is “pumped”.

Example: {anbn: n 0} is not Regular Choose w to be ambm

(We get to choose any w; m is from pumping lemma).

1 2a a a a a … a a a a a b b b b … b b b b b b

x y z

We show that there is no x, y, z with the required properties:|xy| m, |y| 1, andwi = xyi z is also in L, for all i = 0, 1, 2, …

Since |xy| m, y must be in region 1. So y = ap for some p 1. Assume that i = 2 (we pump one extra copy of y), this means:

am+pbm

which L, since it has more a’s than b’s.

Using the Pumping Theorem Effectively

To choose w (a string in the language L):• Choose a w with as homogeneous as possible in

initial region of length at least m.• Choose a w that is only barely in L.

To choose i:• Try letting i be either 0 or 2.• If that doesn’t work, analyze L to see if there is

some other specific value that will work.

PalEven = {ssR : s {a, b}*}

1. Let w = ambmbmam Therefore, |w| = 4m, |w| m.

2. w = xyz, where |xy| m and and |y| 1; therefore y must occur in the first m characters, so y = ap, p > 0.

3. Let i = 2 (make an extra copy of y). This means that the first half of the string has more a’s than the second half of the string and is not in PalEven.

Therefore, by contradiction of the pumping lemma, PalEven is not a regular language.

{axby: x < y}

L = {axby: x < y}

Assume L is regular and m = # of states in the FA that accepts L:

1. Let w = ambm+1 Therefore, |w| = 2m + 1, |w| m.

2. w = xyz, where |xy| m and and |y| 1; therefore y must occur in the first m characters, so y = ap, p > 0.

3. Let i = 2 (make an extra copy of y). This means that the resulting string is am+p bm+1

Therefore, by contradiction of the pumping lemma, PalEven is not a regular language.

{axby: x y}

{axby: x y}

Assume L is regular and m = # of states in the FA that accepts L:

1. Let w = ambm Therefore, |w| = 2m, |w| m.

2. w = xyz, where |xy| m and and |y| 1; therefore y must occur in the first m characters, so y = ap, p > 0.

3. Let i = 0 (pump out y). This means that the resulting string is am-p bm. Since p > 0, there are now fewer a’s than b’s.

Therefore, by contradiction of the pumping lemma, PalEven is not a regular language.

PoetryThe Pumping Lemma

Any regular language L has a magic number pAnd any long-enough word in L has the following property:Amongst its first p symbols is a segment you can findWhose repetition or omission leaves x amongst its kind.

So if you find a language L which fails this acid test,And some long word you pump becomes distinct from all the

rest,By contradiction you have shown that language L is notA regular guy, resiliant to the damage you have wrought.

But if, upon the other hand, x stays within its L,Then either L is regular, or else you chose not well.For w is xyz, and y cannot be null,And y must come before p symbols have been read in full.

As mathematical postscript, an addendum to the wise:The basic proof we outlined here does certainly generalize.So there is a pumping lemma for all languages context-free,Although we do not have the same for those that are r.e.

-- Martin Cohn