UGCNET CSE December 2022: 32

Question

 

In the $\varepsilon$-NFA, $M=\left(\left\{\mathrm{q}_{0}, \mathrm{q}_{1}, \mathrm{q}_{2}, \mathrm{q}_{3}\right\},\{\mathrm{a}\}, \delta, \mathrm{q}_{0},\left\{\mathrm{q}_{3}\right\}\right)$ where ' $\delta$ ' is given in the transition table below, what is the minimum length of string to reach to the final state?

	$\varepsilon$	$\mathrm{a}$
$\mathrm{q}_{0}$	$\left\{\mathrm{q}_{1}\right\}$	{}
$\mathrm{q}_{1}$	$\left\{\mathrm{q}_{2}\right\}$	{}
$\mathrm{q}_{2}$	{}	$\left\{\mathrm{q}_{2}, \mathrm{q}_{3}\right\}$
$\mathrm{q}_{3}$	{}	{}

 

• A. $0$
• B. $1$
• C. $2$
• D. $3$

(Option $1[39425]) 1$
(Option $2[39426]) 2$
(Option $3[39427]) 3$
(Option $4[39428]) 4$

Answer Given by Candidate : $2$

Answer 1

from the given $\epsilon$ NFA it is clear that only 1 length string is enough to reach the final state.

$q_0 \xrightarrow \epsilon q_1\xrightarrow \epsilon q_2\xrightarrow a q_3$

Note: $\epsilon.L=L,\epsilon$ is zero length string.

Option (1) is correct.