Non Regular Language

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Nonregular Languages

The language that cannot be expressed byany regular expression is called aNonregular language.

The languages PALINDROME is theexamples of nonregular languages.

Note: It is to be noted that a nonregular

language, by Kleenes theorem, cant beaccepted by any FA or TG.

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Example

Consider the language L = {, ab, aabb,aaabbb, } i.e. {anbn: n=0,1,2,3,}

Suppose, it is required to prove that thislanguage is nonregular. Let, contrary, L bea regular language then by Kleenestheorem it must be accepted by an FA,say, F.

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Example contd

Since every FA has finite number of states

then the language L (being infinite)accepted by Fmust have words of lengthmore than the number of states. Which

shows that, Fmust contain a circuit.

For the sake of convenience suppose thatF has 10 states. Consider the word a9 b9from the language L and let the pathtraced by this word be shown as under

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Example Contd

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Example Contd

But, looping the circuit generated by the

states 3,4,6,5,3 with a-edges once more, F

also accepts the word a9+4 b9, while a13b9 is

not a word in L. It may also be observed that,

because of the circuit discussed above, F

also accepts the words a9(a4 )m b9, m = 1,2,3,

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Example Contd

Moreover, there is another circuitgenerated by the states 9,10,9. Includingthe possibility of looping this circuit, F

accepts the words a9(a4)m b9(b2 )n wherem, n=0,1,2,3,(m and n not being 0simultaneously).Which shows that F

accepts words that are not belonging to L.

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Proof Contd

Thus there is no FA which accepts thelanguage L. which shows, by Kleenestheorem, that the language L cant be

expressed by any regular expression. Itmay be noted that apparently anbn seemsto be a regular expression of L, but in fact itis not. The observations made from this

example, generalize the theorem (alsocalled the Pumping lemma) regarding theinfinite regular language as follows:

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Pumping Lemma

Statement

Let L be any infinite regular language (thathas infinite many words), defined over analphabet then there exist three strings x,

y and z belonging to * (where y is not the

null string) such that all the strings of theform xynz for n = 1,2,3, are the words in

L.

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Pumping Lemma Contd

Proof

If L is a regular language, then according toKleenes theorem, there exists an FA, say, Fthat accepts this language.

Now F, by definition, must have finite numberof states while the language has infinitelymany words, which shows that there is no

restriction on the length of words in L, becauseif there were such restriction then the languagewould have finite many words.

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Pumping Lemma Contd

Let w be a word in the language L, so that

the length of word is greater than thenumber of states in F. In this case the pathgenerated by the word w, is such that it

cannot visit a new state for each letter i.e.there is a circuitin this path.

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Pumping Lemma Contd

The word w, in this case, may be divided

into three parts.

1. The substring which generates the pathfrom initial state to the state which is

revisited first while reading the word w.This part can be called X and X can be anull string.

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Pumping Lemma Contd2. The substring which generates the circuit

starting from the state which was lead by X.This part can be called as Y which cannot benull string.

3. The substring which is the remaining part ofthe word after Y, call this part as Z. It may benoted that this part may be null string as the

word may end after Y or Z part may itself bea circuit.

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Pumping Lemma Contd

Thus the word may be written as w = xyzwhere x, y and z are the strings, also ycant be a null string. Now this is obvious

that, looping the circuit successively, thewords xyyz, xyyyz, xyyyyz, will also beaccepted by this FA i.e. xynz, n=1,2,3, will be words in L.

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Pumping Lemma Contd

Remark: In the above theorem, it is notaffected if the z-part has circuit. To provethe theorem it is only to find a circuit and

then looping that circuit, is all that isneeded. While looping the circuit thevolume of the string y (or z) is pumped, so

the theorem is also called the Pumpinglemma. Following are the examples

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Example

Consider the following 6 states FA, say, Fwhich accepts an infinite language

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Example Contd

Let the word w = bbbababa, belonging tothe language L, so that the length of word isgreater than 6 (the number of states in F).

In this case the path generated by this wordis such that it cannot visit a new state foreach letter i.e. there is acircuit in this path.

The word w, in this case, may be dividedinto three parts.

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Example Contd

The substring which generates the pathfrom initial state to the state which isrevisited first while reading the word w.

This can be called as part xand this maybe null string.

The substring which generates the circuit

starting from the start state which was leadby x, this part can be called as yand thiscannot be null string.

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Example Contd The substring which is the remaining part of the

word after y, this part can be called as z. It maybe noted that this part may be null string as theword may end after y or z-part may itself be acircuit.

Thus the word w may be written as w = xyz,where x, y, z are strings belonging to * and ycannot be null string.

The state 2 is such that it is revisited first whilereading the word w. So the word w can bedecomposed, according to pumping lemma, asw = xyz = (b)(bba)(baba)

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Example Contd

If y-part of w is continuously pumped, theresulting strings will be accepted by F andhence will be words in the language acceptedby F. Thus, by pumping lemma, the languageaccepted by F is regular.

Remark: If the pumping lemma is applieddirectly on the language L = {an bn :n=0,1,2,3,}, it can be observed that for the

word w = (aaa)(aaaabbbb)(bbb) where x = aaa, y = aaaabbbb and z = bbb

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Remarks Contd

xyyz will contain as many number of as asthere are bs but this string will not belong to Lbecause the substring ab can occur at themost once in the words of L, while the string

xyyz contains the substring ab twice.

On the other hand if y-part consisting of onlyas or bs, then xyyz will contain number of as

different from number of bs. This shows thatpumping lemma does not hold and hence thelanguage is not regular.

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Example

Consider the language EQUAL, of strings,

defined over = {a, b}, with number of asequal to number of bs, i.e. EQUAL = {,ab,aabb,abab,baba,abba,}

From the definition of EQUAL, it is clear that{an bn } = a* b* EQUAL Obviously a* b*defines a regular language while {an bn } hasbeen proved nonregular.

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Example Contd

Using the theorem that intersection of two

regular languages is, regular; it can beproved that the EQUAL is not regular.

Because if it is considered regular then thelanguage {an bn } will, being intersection ofregular languages, be regular language,

which is impossible.

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Example contd

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Following are the remarks regarding these examplesRemarks

In the previous examples, languages are proved to

be regular or nonregular using pumping lemma. Infact to prove a certain language to be regular, it is notneeded to use the full force of pumping lemma i.e. fora wordwith length greater than the number of states

of the machine, decomposing the word into xyz andfor a language to be regular it is sufficient that xyyz isin L. The condition that xynz is in L for n>2, providesthat the language is infinite.


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Example contd

Consider the language PALINDROME and a

word w = aba belonging to PALINDROME.Decomposing w = xyzwhere x=a, y=b, z=a. Itcan be observed that the strings of the formxynz for n=1,2,3, , belong toPALINDROME. Which shows that thepumping lemma holds for the languagePALINDROME (which is non regularlanguage). To overcome this drawback ofpumping lemma, a revised version ofpumping lemma is to be introduced.

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Assignment

Describe the revised version of pumpinglemma with example

Due date 11.12.2011

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Non Regular Language