NPTEL Theory of Computation Week 1

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Questions and Answers

Any NFA having n states can be converted to a DFA having at most (select the smallest possible value)

  • 2n states (correct)
  • n2 states
  • 3n states
  • 2n states (correct)

What is the minimum number of states required for a DFA that accepts the language L = {w | the number of 1’s in w is divisible by 3}?

  • 1
  • 4
  • 2
  • 3 (correct)

What is the ϵ-closure of state q1 in the given NFA?

  • {q0, q1} (correct)
  • {q1}
  • {q0, q1, q2, q3}
  • {q0, q1, q2}

Which of the following strings is accepted by the DFA below?

<p>01011101 (B)</p> Signup and view all the answers

Let L1 and L2 be two regular languages. What can we say about the language L = {w1 w2 | w1 ∈ L1, w2 ∈ L2}?

<p>L is regular (D)</p> Signup and view all the answers

What is the language accepted by the given NFA?

<p>{w ∈ {0, 1}* | w begins with 0 or ends with 1} (B)</p> Signup and view all the answers

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Study Notes

NFA to DFA Conversion

  • An NFA with n states can be converted to a DFA with at most 2^n states, since the states of the DFA correspond to subsets of the NFA's states.

Counting States in DFA

  • For the language L = {w | the number of 1’s in w is divisible by 3}, a DFA requires a minimum of 3 states to track the count of 1's modulo 3.

ε-Closure in NFA

  • The ε-closure of a state q includes all states reachable from q through ε-transitions only. For state q1, the ε-closure is {q0, q1}.

Accepted Strings by DFA

  • The DFA accepts the string "01011101". Other provided strings do not lead to the accept state.

Regular Languages Concatenation

  • The concatenation of two regular languages L1 and L2 results in a language L that is guaranteed to also be regular.

NFA Accepted Language

  • The language accepted by a specific NFA includes strings that either begin with 0 or end with 1. Verification involves checking for computation paths leading to accept states based on input string characteristics.

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