643 views
0 votes
0 votes
How a^i b^j c^k |  i= j+k is CFL?

Please explain

Thanks

2 Answers

0 votes
0 votes
From $i$=$j$+$k$ , we can say no.of $a's$ =no. of $b's$+ no.of $c's$.

With the help of stack, whenever we see $a's$ push them and pop $b's$ and $c's$. If the stack is empty string is accepted otherwise rejected.

So,with the help of this we can say given language is CFL. Particularly it's DCFL($Deterministic$  $  CFL$)

Related questions

242
views
0 answers
0 votes
Vedantthakkar asked Feb 24
242 views
Consider a regular language R and a context free language C. Let the PDA that recognizes C be called P=(QP,∑,Γ,δP,q0P,FP), and the DFA that reconginzes R be (QR,∑,δR,q0R,FR).
212
views
1 answers
1 votes
vedantk asked Jan 10
212 views
If it’s DCFL then also construct the DPDA ?
280
views
1 answers
1 votes
823
views
2 answers
1 votes
ggwon asked Dec 29, 2022
823 views
L = {$a^{n+m}b^{n}a^{m} | n,m \geq 0$}Is the above language DCFL or CFL ?