564 views

1 Answer

0 0 votes

In Exercise 9 problem the automata was deterministic because we knew when we need to start popping out from the stack but here we don't know what is the midpoint of the given string and from which step we need to start popping elements from the stack.

Here we have to find the center using brute force kind of approach , whenever the top of stack element and the input element are same then that input element could be either the center of string or could be the just another element so we could either push that element in stack or we could then go to another state and start popping elements, this is a non deterministic move as we can't say which move we should  take and hence the given language is non-deterministic language.

Position:
Show:

Related questions

0 0 votes
0 0 answers
1.0k
1.0k views
Naveen Kumar 3 asked Jun 23, 2019
1,047 views
Give an example of a deterministic context-free language whose reverse is not deterministic.
0 0 votes
0 0 answers
360
360 views
Naveen Kumar 3 asked Jun 23, 2019
360 views
Show that under the conditions of Exercise 16, $L_1 ∩ L_2$ is a deterministic context-free language.
0 0 votes
1 1 answer
627
627 views
Naveen Kumar 3 asked Jun 23, 2019
627 views
Show that if $L_1$ is deterministic context-free and $L_2$ is regular, then the language $L_1 ∪ L_2$ isdeterministic context-free.
0 0 votes
1 1 answer
471
471 views
Naveen Kumar 3 asked Jun 23, 2019
471 views
Show that every regular language is a deterministic context-free language.