- Type
- Quiz
- Course
- CS402 Theory of Automata
- Size
- PDF 838 KB
- Updated
- 23 Oct 2026
You get 3 free files a day on the website. Add the Chrome extensionSign in with the extension for 30 a day.
Preview text
Text from pages 2, 3, 4 of 7. Download the file for all 7 pages.
- Let FA1 accepts many strings and FA2 accepts no string, then FA1 + FA2 will be equal to: o (A) (FA2)* o (B) FA2 o (C) FA2 − FA1 o (D) FA1 Correct
- Every NFA can be considered to be a -------- as well, but the converse may not be true: o (A) TG o (B) FA Correct o (C) PDA o (D) GTG
Q3. Which of the following is used to delay the transmission of signal along the wire by one step (clock pulse)?
o (A) Holdup box o (B) Slow box o (C) Reduce box o (D) Delay box Correct
- Strings x, y, z belong to Σ such that xz ∈ L but yz ∉ L where L ⊆ Σ are:** o (A) Indistinguishable o (B) Distinguishable Correct o (C) Undetermined o (D) Both distinguishable and indistinguishable
Q5. How many new states are introduced while developing NFA for the closure of an FA?
o (A) Four o (B) One o (C) Three o (D) Two Correct
- Let FA3 be an FA corresponding to FA1FA2, then the final state of FA3 must correspond to the final state of: o (A) FA1 or FA2 o (B) FA1 and FA2 o (C) FA1 only o (D) FA2 only Correct
Q7. If new A = 1 NAND (1 AND 1), then what will be the value of new A?
o (A) 0 Correct o (B) 1 o (C) 10 o (D) 01
- In Moore machine the output depends on:
Related CS402 files
All CS402 files →Get more out of VUBuddy, for free
- Unlimited downloads
- The explanation behind every MCQ answer
- Study offline on slow data
- 30 downloads a day instead of 3
- Sign in and buy Pro Semester here
One Pro Semester membership works in both the extension and the app.