Automata to Regular Expression

Question

Ans. C

Comments

automata is accepting all the string starting with 0 including the null string.....compliment should be starting with 1 excluding the null string....C is correct

Answer 1

Find the smallest length string and then proceed in lexicographic ordering...find out first 3 or 4 string of the language....then compare with the options....if any of the strings is present in the option, it will be discarded

Answer 2

If we try to write the regular expression of the language accepted by above NFA,it looks something like :

$\epsilon + 01^{*} + (01)^{*}$

From the above expression we can say that language L(M) accepts all strings starting with zero or the null string.Or, equivalently in set-builder notation:

$L(M)=\left \{ w:w\epsilon (\left \{ 0,1 \right \}^{*}\wedge {w} \ starts\ with\ zero \vee it's\ a\ null\ string) \right \}$

$\sim L(M)=\left \{ w: w \in \left \{ 0,1 \right \}^{*}\wedge w\ starts\ with\ 1 \right \}$

$So,regex(\sim L(M))=1(0+1)^{*}$

Answer 3

option c is right.

Answer 4

The NFA represents the null string, starts with 0 and followed by anything so compliments is nothing but 1 followed by anything.

Comments

B not be answer because there is union