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Recap Lecture-2

Kleene Star Closure, Plus operation, recursivedefinition of languages, INTEGER, EVEN,

factorial, PALINDROME, {a

n

b

n

}, languages ofstrings (i) ending in a, (ii) beginning and endingin same letters, (iii) containing aa or bb(iv)containing exactly aa,

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Task

Q)

1) Let S={ab, bb} and T={ab, bb, bbbb} Show

that S*= T*[Hint S*T* and T*S*]

2) Let S={ab, bb} and T={ab, bb, bbb} Show thatS* T* But S*T*

Solution: Since ST , so every string belongingto S* , also belongs to T* but bbb is a stringbelongs to T*but does not belong to S*.

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3) Let S={a, bb, bab, abaab} be a set of strings. Areabbabaabab and baabbbabbaabb in S*? Does any wordin S*have odd number of bs?

Solution: since abbabaabab can be grouped as(a)(bb)(abaab)ab , which shows that the last member ofthe group does not belong to S, so abbabaabab is not inS*, while baabbbabbaabb can not be grouped as

members of S, hence baabbbabbaabb is not in S*

. Sinceeach string in S has even number of bs so there is nopossiblity of any string with odd number of bs to be inS*.

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Task

Q1)Is there any case when S+ contains? Ifyes then justify your answer.

Solution: consider S={,a} thenS+ ={, a, aa, aaa, }Hereis in S+ as member of S. Thuswillbe in S+ , in this case.

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Q2) Prove that for any set of strings S

i. (S+)*=(S*)*

Solution: In generalis not in S+ , whiledoes belong to S*. Obviouslywill now be

in (S+

)*

, while (S*

)*

and S*

generate thesame set of strings. Hence (S+)*=(S*)*.

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Q2) continued

ii) (S+)+=S+

Solution: since S+ generates all possible

strings that can be obtained byconcatenating the strings of S, so (S+)+generates all possible strings that can beobtained by concatenating the strings ofS+ , will not generate any new string.

Hence (S+)+=S+

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Q2) continued

iii) Is (S*)+=(S+)*

Solution: sincebelongs to S* ,sowill

belong to (S*

)+

as member of S*

.Moreovermay not belong to S+, in general, whilewill automatically belong to (S+)*.

Hence (S

*

)

+

=(S

+

)

*

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Regular Expression

As discussed earlier that a*generates

, a, aa, aaa,

and a+

generates a, aa, aaa, aaaa, , so thelanguage L1= {, a, aa, aaa, } andL2= {a, aa, aaa, aaaa, } can simply beexpressed by a*and a+, respectively.

a*and a+ are called the regular expressions(RE) for L1 and L2respectively.

Note:a+, aa* and a*a generate L2.

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Recursive definition of Regular

Expression RE)

Step 1: Every letter of includingis aregular expression.

Step 2: If r1and r2 are regular expressions then

1.(r1)

2.r1r2

3.r1+ r2and

4.r1*

are also regular expressions.

Step 3: Nothing else is a regular expression.

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Defining Languages (continued)

Method 3 (Regular Expressions)

Consider the language L={, x, xx, xxx,}of strings, defined over = {x}.

We can write this language as the Kleene starclosure of alphabet or L=*={x}*

this language can also be expressed by the

regular expression x

*

.Similarly the language L={x, xx, xxx,},

defined over = {x}, can be expressed bythe regular expression x+.

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Now consider another language L, consistingof all possible strings, defined over= {a, b}. This language can also be

expressed by the regular expression(a + b)*.

Now consider another language L, of stringshaving exactly double a, defined over= {a, b}, then its regular expression maybe

b*aab*

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Now consider another language L, of evenlength, defined over = {a, b}, then itsregular expression may be

((a+b)(a+b))*

Now consider another language L, of oddlength, defined over = {a, b}, then itsregular expression may be

(a+b)((a+b)(a+b))* or

((a+b)(a+b))*(a+b)

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Remark

It may be noted that a language may beexpressed by more than one regularexpressions, while given a regular expressionthere exist a unique language generated by thatregular expression.

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Consider the language, defined over

={a, b}, of words starting with double aand ending in double b then its regular

expression may be aa(a+b)*bbConsider the language, defined over

={a, b} of words starting with a and

ending in b OR starting with b and endingin a, then its regular expression may bea(a+b)*b+b(a+b)*a

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TASK

Consider the language, defined over

={a, b} of words beginning with a, then

its regular expression may be a(a+b)*

Consider the language, defined over

={a, b} of words beginning and endingin same letter, then its regular expressionmay be (a+b)+a(a+b)*a+b(a+b)*b

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TASK

Consider the language, defined over

={a, b} of words ending in b, then its

regular expression may be (a+b)

*

b.Consider the language, defined over

={a, b} of words not ending in a, then itsregular expression may be (a+b)*b +. It is to

be noted that this language may also beexpressed by ((a+b)*b)*.

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SummingUP Lecture 3

RE, Recursive definition of RE, defininglanguages by RE, { x}*, { x}+, {a+b}*,Language of strings having exactly one aa,

Language of strings of even length, Languageof strings of odd length, RE defines uniquelanguage (as Remark), Language of stringshaving at least one a, Language of strings

havgin at least one a and one b, Language ofstrings starting with aa and ending in bb,Language of stringsstarting with and endingin different letters.