CS 390 Solutions to Homework 6

Textbook pp. 160 - 162:

4.18 (a)

         



4.20 (c) Left: (a + b )*
              Right: a* + b*
So they accept different languages. The right one does not accept ab for example.

(d) Convert the left one to FA. Then it is the same as the right one. Hence they accept the same language.



4.40 (b)

We are going to prove it by proving $\Lambda(S) \subseteq \Lambda (\Lambda(S))$ and $\Lambda (\Lambda(S)) \subseteq \Lambda(S)$.
(Recall that for sets A and B, A = B if and only if A $\subseteq$ B and B $\subseteq$ A.)

First by the definition of $\Lambda$-closure, $\Lambda(S) \subseteq \Lambda (\Lambda(S))$.
To prove $\Lambda (\Lambda(S)) \subseteq \Lambda(S)$ we use induction on $\Lambda (\Lambda(S))$.
Basis Step: $\Lambda(S) \subseteq \Lambda(S)$ since $\Lambda(S) = \Lambda(S)$
Induction Step: Let $x$ be an arbitrary element in $\Lambda (\Lambda(S))$ that has the property of being in $\Lambda(S)$.
Then by the definition of $\Lambda(S)$, $\delta (x, \Lambda ) \subseteq \Lambda(S)$.
Hence $\Lambda (\Lambda(S)) \subseteq \Lambda(S)$.
Hence $\Lambda(S) = \Lambda (\Lambda(S))$.



4.28 (b)

State	Input	Next State	|	State	Input	Next State
1	a	{2}	|	3	a	{2, 4}
2	a	{1, 5}	|	4	b	{1, 3}
2	b	{1, 3}	|	5	a	{2}

The accepting states are 1, 3 and 5.

4.29 (a)