L=L2-L1 = { a^n b^n|n>=**1**} and hence it is DCFL

The Gateway to Computer Science Excellence

+3 votes

$L_1 =\{a^n b^m c^n \mid m,n \geq 0\}$ and $L_2=\{ a^n b^n\mid n\geq 0\}$. If $L=L_2-L_1$ then $L$ is

- finite language
- regular language
- DCFL
- not DCFL

+6 votes

Best answer

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.3k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.1k
- Non GATE 1.5k
- Others 1.5k
- Admissions 595
- Exam Queries 576
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 17

50,645 questions

56,567 answers

195,749 comments

101,708 users