Exam 12: Regular Expressions and Finite State Automata

Correct Answer:

Verified

Suppose not. That is suppose there is a finite-state automaton $A$ that accepts the language $L$ consisting of all strings of $x$ 's and $y$ 's of the form $x^{n} y^{n}$ where $n$ is a positive integer. [A contradiction will be derived.] Since $A$ has only a finite number of states, these states can be denoted $s_{1}, s_{2}, s_{3}, \ldots, s_{k}$ , where $k$ is a positive integer.Consider all input strings that consist entirely of $x$ 's: $x, x^{2}, x^{3}, \ldots$ . Now there are infinitely many such strings and only finitely many states. Thus, by the pigeonhole principle, there must be a state $s_{m}$ and two input strings $x^{p}$ and $x^{q}$ with $p \neq q$ such that when either $x^{p}$ or $x^{q}$ are input to $A, A$ goes to state $s_{m}$ . Now, by supposition, $A$ accepts the string $x^{p} y^{p}$ . This means that after $p x^{\prime}$ 's have been input, at which point $A$ is in state $s_{m}$ , inputting $p$ additional b's sends $A$ into an accepting state, say $s_{a}$ . But that implies that inputting $x^{q} y^{p}$ also sends $A$ to $s_{a}$ , which means that $x^{q} y^{p}$ is also accepted by $A$ . The reason is that after $q x$ 's have been input, $A$ is also in state $s_{m}$ , and from that point, we know that inputting $p$ additional b's sends $A$ to state $s_{a}$ , which is an accepting state. Hence
$x ^ { q } y ^ { p }$ is in $L$ where $p \neq q$
But, by supposition, all the strings in $L$ have the form $x ^ { n } y ^ { n }$ , which have the same number of that there is no finite-state automaton that accepts the language $L$ consisting of all strings of $x ^ { \prime }$ 's and $y$ 's of the form $x ^ { n } y ^ { n }$ where $n$ is a positive integer.

Correct Answer:

Verified

a. The language accepted by the automaton is the set of all strings of 0 's and 1 's that contain exactly one 1 or that contain exactly two 1 's.
b.

Correct Answer:

Verified

a. $N \left( s _ { 2 } , a \right) = s _ { 2 }$
b. $s _ { 2 }$
c. The strings $a b b b a a$ and $a$ are accepted by the automaton. The strings $a b a b$ and $b b a b$ are not accepted by the automaton.
d. The language accepted by the automaton is the set of all strings of $a$ 's and $b$ 's that end in an $a$ .
e. $( a \mid b ) ^ { * } a$