ME test series

Question

Given that L1=(bba*baa*) andL2=(ab*).
the regular expression corresponding to language L3=L1/L2 is given by
 a) bba*b
b)bba*baa*
c)bba*ba*
d)none

Comments

Option b) ?

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opttion c) ??

Answer 1

Here L1 ends with 'a' and L2 is ab^* , means simply if L2 contains at least one 'b' then there is no string of L1 that has that suffix that can ends with 'b' ..

Now go to case when L2 is just 'a' , now see the pattern of strings of L1 that can end with 'a'

we can see , those can be bba^*b , bba*ba , bba*baa , bba*baaa , bba*baaaa ,,,,,so the pattern is bba*ba^*

So in that way C becomes answer...

Comments

given made easy  ans is b

 

L1/L2=bba*baa*/ab*

         =bba*baa*/a

          =bba*baa*

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Hello mohit , bba^*b is one possible right quotient string ....

read it more carefully

the right quotient (or simply quotient) of a formal language L 1 with a formal language L 2 is the language consisting of strings w such that wx is in L 1 for some string x in L 2.