# Peter Linz Edition 4 Exercise 7.3 Question 7 (Page No. 200)

87 views
Give reasons why one might conjecture that the following language is not deterministic.
$L =$ { $a^nb^mc^k : n = m$ or $m = k$}.
0
Not deterministic
0
I think it can also be written as L={a^n b^n c^n} and then it is CSL it can solved by Linear bounder automata
2
@ravijha thats not correct you changed the question,now the strings in language accpeted going to be less if you say that way.

As one can see there are two conditions which can make the language accepted by pda. So in first condition we would be pushing a and popping b to check equivalence for n=m and just do nothing when you see c , similarly for second condition we goona skip a and push b and pop c to check equivalence of number of b'c and c's. So this way its gonna make it non deterministic.
0

@Shubham Shukla 6 okay got it

## Related questions

1
91 views
Give an example of a deterministic context-free language whose reverse is not deterministic.
Show that under the conditions of Exercise 16, $L_1 ∩ L_2$ is a deterministic context-free language.
Show that if $L_1$ is deterministic context-free and $L_2$ is regular, then the language $L_1 ∪ L_2$ is deterministic context-free.