retagged by
3,400 views

3 Answers

Best answer
9 votes
9 votes

S1: Every DCFL has an unambiguous grammar. True [DCFL has at least one grammar which is unambiguous But  DCFG can be ambiguous]

S2: Every language accepted by DPDA with the final state is also accepted by DPDA with the empty stack. False [The set of languages accepted by a DPDA by the empty stack is a subset of the set of languages accepted by a DPDA by final state. Example 0*, 1* which is accepted by final state but not with empty stack since we can only do push but don't know when we start poping]

selected by
4 votes
4 votes
Regarding S2: Conversion from Final state to Empty state PDA & vice-versa involves adding epsilon transition. It does not leave DCFL to DCFL but makes it CFL. Thus S2 is false.

Related questions

0 votes
0 votes
0 answers
1
sripo asked Oct 13, 2018
601 views
L1={an bn | n>=0}L2={bn cn | n>=0}What is L1.L2 ?Is it an b2n cn ?
2 votes
2 votes
0 answers
2
Tuhin Dutta asked Jan 18, 2018
745 views
$ L_1\ =\ \{ \ a^mb^mb^na^n | m,n\ >\ 0\}\\ L_2\ =\ \{ \ a^mb^na^nb^m | m,n\ >\ 0\}$Find $L_1 \cap L_2$
1 votes
1 votes
1 answer
3
0 votes
0 votes
0 answers
4
raj_uddeshya157 asked Dec 27, 2023
79 views
Can $\Sigma^{*}$ be called DCFL? If yes, what would the state transition diagram of its PDA look like?