CS 390 Solutions to Test II

November 9, 2004

1(a). Find an $DFA$ that recognizes the same language as the following $NFA$:

State $q$	$a$	$b$	State $q$	$a$	$b$
$1$	{$2, 3$}	$\emptyset$	$4$	$\emptyset$	$\emptyset$
$2$	{$2, 5$}	{$4$}	$5$	{$4$}	$\emptyset$
$3$	$\emptyset$	{$5$}

The initial state is $1$ and the accepting states are $4$ and $5$.

Answer:

2. Construct an $NFA-\Lambda$ that accepts the language corresponding to
$(ab^{*} + b)^{*}$ from the simplest $NFA$'s. DO NOT SIMPLIFY.

Answer:

3. For the following $NFA-\Lambda$, answer the questions below:

State $q$	$a$	$b$	$\Lambda$
$1$	{$3$}	$\emptyset$	{$2$}
$2$	$\emptyset$	{$2$}	{$4$}
$3$	$\emptyset$	{$4$}	{$1$}
$4$	$\emptyset$	$\emptyset$	$\emptyset$

(a) Find $\Lambda (\{1, 3\})$

Answer:

$\{1, 2, 3, 4\}$

(b) Find $\delta^{*} (1, ab)$

Answer:

$\{2, 4\}$


4. Prove by structural(general) induction that for DFA, $\delta^{*} (q, xy) = \delta^{*} ( \delta^{*}(q,x), y)$.

Note that the set of strings $\Sigma$ over the alphabet $\{a, b\}$ and $\delta^{*}$ for DFA are defined recursively as follows:

Basis Clause: $a \in \Sigma$ and $b \in \Sigma$
Inductive Clause: If any string $x$ in $\Sigma$, then $xa$ and $xb$ are in $\Sigma$.
Extremal Clause: Nothing is in $\Sigma$ unless it is obtained by Basis and Inductive Clauses .

Basis Clause: $\delta^{*} (q, \Lambda )$ = $q$
Inductive Clause: For any state $q$ and any string $y$ in $\Sigma$ and any symbol $a$ of $\Sigma$,
$\delta^{*} (q, ya) = \delta (\delta^{*} (q, y), a)$.

Proof:
Basis Step: $y = \Lambda$.
If $y = \Lambda$, then $\delta^{*} (q, xy)$ = $\delta^{*} (q, x)$.
Also $\delta^{*} (\delta^{*} (q,x), y) = \delta^{*} (\delta^{*} (q,x), \Lambda) = \delta^{*} (q, x)$ by the definition of $\delta^{*}$.
Hence $\delta^{*} (q, xy) = \delta^{*} ( \delta^{*}(q,x), y)$ if $y = \Lambda$.
Inductive Step: Assume that for any string $y$ $\delta^{*} (q, xy) = \delta^{*} ( \delta^{*}(q,x), y)$.
We are going to show that $\delta^{*} (q, xya) = \delta^{*} ( \delta^{*}(q,x), ya)$ for any symbol a of the alphabet.
$\delta^{*} ( \delta^{*}(q,x), ya) = \delta (\delta^{*}( \delta^{*}(q,x), y), a)$ by the definition of $\delta^{*}$.
= $\delta ( \delta^{*}(q,xy), a)$ by the induction hypothesis.
= $\delta^{*} (q, xya)$ by the definition of $\delta^{*}$.