Regularity of CFG and DCFL - Computer Science Stack Exchange
2021年10月13日 · DCFL are not closed under union, but closed under complement. The undecidability of regularity for CFL is usually obtained from two properties of the context-free languages: (1) they are closed under union, and (2) universality, i.e., equality to Σ∗, is undecidable. To the contrary, DCFL (1) are not closed under union, (2) have decidable ...
formal languages - Why are DCFL not closed under concatenation …
2017年10月12日 · I understand that DCFL they are not closed under concatenation or Union. As without non determinism, PDA cannot decide when to jump to the next one in case of concatenation and without epsilon moves Union is not possible. However, DCFL is a proper subset of CFL (unambiguous) and CFL is closed under union and concatenation.
Union of a Deterministic Context-free language and a Regular …
2016年5月29日 · In formal language theory, deterministic context-free languages (DCFL) are a proper subset of context-free languages. They are the context-free languages that can be accepted by a deterministic pushdown automaton.
nondeterminism - How to determine if a language is deterministic ...
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Intersection of Deterministic Context-Free Languages
2021年4月22日 · 1 Answer. 2. The statement itself doesn't imply that. It just means that the intersection of two DCFLs might or might not be a DCFL; and thus presumably could be a DCFL; could be a CFL but not a DCFL; or could be not context-free. However, we know more actually. The intersection of two DCFLs could be, of course, a DCFL.
Are DCFLs closed under concatenation with a regular language?
2015年6月21日 · ($\{a^ib^ic^i\}$ is not DCFL and the class is closed under complement.) Is there a reason why the presented disproof wouldn't apply to concatenation with regular languages and if so, is there a more understandable proof of the closure then the one referenced?
Techniques to prove a language is not DCFL
2015年2月11日 · I know that DCFL is closed under complementation and intersection with regular languages. By using these we can prove that a language is not DCFL. Are there any other techniques that will help me to
every DCFL has an LR(1), an LALR(1) and even an SLR(1) grammar
2017年1月10日 · Here is one discussion that says, "every DCFL has an LR(1), an LALR(1) and even an SLR(1) grammar." And wiki says, "a language can be generated by an LR(k) grammar if and only if it is deterministic [and context-free], if and only if it can be generated by an LR(1) grammar" Is the first statement in that discussion forum is correct?
Are DCFLs closed under reversal? - Computer Science Stack …
8. According to this chart, DCFLs are closed under reversal. However, I am not convinced as the intuitive proof (reversing the arrows of the controlling finite state machine and switching the pushes and pops) for this seems to depend on non-determinism in choosing the null transition to take from the initial state (since the new initial state ...
Is the language $L=\\{a^nb^n\\} \\cup \\{aa*\\}$ DCFL?
This will infinitely interrupt the simulation of the parallel finite automaton. Hence we need the requirement that the DPDA we start with has no such $\varepsilon$-loops. That is non-trivial, but as a normal-form it has been obtained in the construction for closure under complement for DCFL.