33 views
L={w|the number of occurences of '011'  in w is equal to the number of occurences of '111'in w}

Is the above language regular?
0

No It is not regular. with number of occurrences of '011' we cannot generate number of occurrences of  '111'

011011011011011011

A good Question https://gateoverflow.in/1995/gate2014-2-36

The problem involves comparisons between the number of occurrences of the two strings. Whenever such kind of problem occurs one should look whether we need extra storage or not for comparing.If extra storage(Stack) is required then the language is definitely not regular.

In this problem,

1. first we have to count the occurrence of the string '011'  and save the count
2. after that we can count the occurrence of  string '111'.
3. compare both the counts.

It requires a storage to save the count of string '011' for comparing later with '111' , hence it is not a regular language.

Another approach :

If a Finite Automata could be created for a given language then the language is said to be Regular.

We cannot create a finite automata for the given language hence the language is not regular.

1