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Section 2.3 Deterministic and Nondeterministic Finite Acceptors $PRACTICE QUESTION$ Question 3: ** Convert the nfa defined by $\delta(q_0, a) = \{q_0, q_1\}$ $\delta(q_1, b) = \{q_1, q_2\}$ $\delta(q_2, a) = \{q_2\}$ with initial state $q_0$ and final state $q_2$ into an equivalent dfa. $PRACTICE QUESTION$ Convert the nfa defined by $\delta(q_0, a) = \{q_0, q_1\}$ $\delta(q_1, b) = \{q_1, q_2\}$ $\delta(q_2, a) = \{q_2\}$ $\delta(q_1, \lambda) = \{q_1, q_2\}$ with initial state $q_0$ and final state $q_2$ into an equivalent dfa. 5. (GRADED 3 points) On $\Sigma = \{0, 1\}$ Draw the NFA figure defined by: $\delta(q_0, 1) = \{q_1\}$ $\delta(q_0, \lambda) = \{q_2\}$ $\delta(q_1, 1) = \{q_2\}$ $\delta(q_2, 1) = \{q_1\}$ $\delta(q_2, 0) = \{q_3, q_4\}$ $\delta(q_4, 0) = \{q_2\}$ q0 is the initial state and q3 is the final(accepting) state Convert the NFA to a DFA (show all your work)

          Section 2.3 Deterministic and Nondeterministic Finite Acceptors
\(PRACTICE QUESTION\) Question 3:
** Convert the nfa defined by
$\delta(q_0, a) = \{q_0, q_1\}$
$\delta(q_1, b) = \{q_1, q_2\}$
$\delta(q_2, a) = \{q_2\}$
with initial state $q_0$ and final state $q_2$ into an equivalent dfa.
\(PRACTICE QUESTION\)
Convert the nfa defined by
$\delta(q_0, a) = \{q_0, q_1\}$
$\delta(q_1, b) = \{q_1, q_2\}$
$\delta(q_2, a) = \{q_2\}$
$\delta(q_1, \lambda) = \{q_1, q_2\}$
with initial state $q_0$ and final state $q_2$ into an equivalent dfa.
5. (GRADED 3 points) On $\Sigma = \{0, 1\}$
Draw the NFA figure defined by:
$\delta(q_0, 1) = \{q_1\}$
$\delta(q_0, \lambda) = \{q_2\}$
$\delta(q_1, 1) = \{q_2\}$
$\delta(q_2, 1) = \{q_1\}$
$\delta(q_2, 0) = \{q_3, q_4\}$
$\delta(q_4, 0) = \{q_2\}$
q0 is the initial state and q3 is the final(accepting) state
Convert the NFA to a DFA (show all your work)

Section 2.3 Deterministic and Nondeterministic Finite Acceptors
PRACTICE QUESTION Question 3:
** Convert the nfa defined by
δ(q0, a) = {q0, q1}
δ(q1, b) = {q1, q2}
δ(q2, a) = {q2}
with initial state q0 and final state q2 into an equivalent dfa.
PRACTICE QUESTION
Convert the nfa defined by
δ(q0, a) = {q0, q1}
δ(q1, b) = {q1, q2}
δ(q2, a) = {q2}
δ(q1, λ) = {q1, q2}
with initial state q0 and final state q2 into an equivalent dfa.
5. (GRADED 3 points) On Σ = {0, 1}
Draw the NFA figure defined by:
δ(q0, 1) = {q1}
δ(q0, λ) = {q2}
δ(q1, 1) = {q2}
δ(q2, 1) = {q1}
δ(q2, 0) = {q3, q4}
δ(q4, 0) = {q2}
q0 is the initial state and q3 is the final(accepting) state
Convert the NFA to a DFA (show all your work)

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