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)
