CS 390 Solutions to Homework 8




4.33

             





4.35 (a)

             





2. $R(1,3,3)$ is a regular expression for the language accepted by the NFA.

$R(1,3,3) = R(1,3,2) + R(1,3,2)R(3,3,2)^{*}R(3,3,2)$
$R(1,3,2) = R(1,3,1) + R(1,2,1)R(2,2,1)^{*}R(2,3,1)$
Now $R(1,3,1) = b$, $R(1,2,1) = a$, $R(2,2,1) = \Lambda$ and $R(2,3,1) = a$.
Hence $R(1,3,2) = b + aa$.
Also $R(3,3,2) = \Lambda + ab + aaa$
Hence $R(1,3,3) = b + aa + (b + aa)(\Lambda + ab + aaa)^{*}(\Lambda + ab + aaa)$
$ = b + aa + (b + aa)(\Lambda + ab + aaa)^{+}$
$ = (b + aa)(\Lambda + ab + aaa)^{*}$.
$ = (b + aa)(ab + aaa)^{*}$.
Thus the language accepted by the NFA is $ (b + aa)(ab + aaa)^{*}$.