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\title{Noncomputability lecture summary}
\author{M. George}
\date{October 20, 2014}
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\section{Why are automata useful?}
\begin{itemize}
\item They are a nice abstraction to have in mind when building certain kinds of
programs \begin{itemize}
\item Network protocols (alphabet is send and receive messages)
\item Games (alphabet is player actions, state is character)
\item Thread state in an operating system (started, running, waiting for input \dots)
\item For one way to use these ideas in your programs, see the ``State design pattern''.
\end{itemize}
\item They are closely related to regular expressions, which are a useful tool
for text processing (including compilers).
\item Having a formal model of computation allows us to prove things about
\emph{all possible programs}. For example, we can prove that there are languages
that cannot be recognized by any program.
\end{itemize}
\section{Pumping lemma}
We expanded on Sid's proof that a machine with $n$ states cannot count past $n$.
This is formalized in the \emph{pumping lemma}:
\paragraph{Pumping lemma:} Given a machine $M$ having $m$ states, and a string
$x \in L(M)$ with $|x| > m$, there exists strings $u$, $v$, and $w$, with $|v|
\geq 1$, such that $x = uvw$ (the concatenation of $u$, $v$ and $w$), and for
all $c \geq 0$, $uv^cw \in L$.
The proof is similar to the proof given by Sid; since $|x| \geq m$, we must repeat
some state while processing $x$. Let $v$ be the substring of $x$ that is
processed within the loop, and let $u$ be the string that takes the machine from
$q_0$ to $q_i$ and $w$ the string that takes $q_j$ to a final state $q_f \in F$.
The fact that we can ``pump'' $v$ corresponds to the fact that we can go around
the loop any number of times.
Note: once we introduce a bit more formal notation, it would be a good exercise
to write out this proof more formally.
\section{Using the pumping lemma}
We proved that the language $L = \{0^n1^n \mid n \in \mathbb{N}\}$ is not
recognizable.
\paragraph{Proof:} Suppose (for contradiction) that there is a machine $M$ with
$L(M) = L$. Suppose $M$ has $m$ states. We know $0^m1^m \in L$. But
$|0^m1^m| > m$ (which is the number of states of $M$), so we can apply the
pumping lemma to split $x$ into $uvw$. $v$ can occur in one of three places in
the string $x$; it can contain only zeros, only ones, or some number of zeros
followed by some number of ones. The pumping lemma tells us that $uvvw \in L(M)$,
but this gives a contradiction in each of the three cases. In the first case, $uvvw$
has more zeros than ones, so it is not in $L$. In the second case, $uvvw$ has
more ones than zeros, and is thus not in $L$. In the third case, $uvvw$ contains
some ones followed by some zeros, and is thus not in $L$. In all three possible
cases we have shown that $uvvw \notin L$ which contradicts the fact that $uvvw
\in L(M)$. Thus there can be no machine $M$.
\section{Non-computability}
The above shows that not all languages are DFA-recognizable. This raises the
question: are there more powerful models that can recognize any language?
Choose your favorite language (say Java). We have a function $L$ from Java
programs to sets of strings: on a given program $P$,
\begin{align*}
L & : & \text{The set of all programs} \rightarrow 2^{\Sigma^*} \\
L & : & P \mapsto \{x \in \Sigma^* \mid P(x)\textrm{ returns true}\}
\end{align*}
Asking whether there are any uncomputable sets is the same as asking if this
function is surjective. Answer: it cannot be, because the set of all programs
is a subset of the set of all strings, and thus is countable (you proved this
on the homework), but the set of all languages is uncouontable (you also proved
this on the homework).
Thus there are sets that Java programs cannot recognize.
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