\documentstyle[12pt]{article}
\pagestyle{empty}
\include{psfig}
\begin{document}
For a one sided discrete event system where without loss of generality
$T \subseteq [0,\infty)$ define {\em cantor metric}
\[ d(s_1,s_2) = \frac{1}{2^t} \]
where $t$ is the smallest tag in $T$ such that $s_1\!\!\mid^t\, \neq
s_2\!\!\mid^t$, or if $s_1=s_2$ then $d(s_1,s_2)=0$.

An {\em open neighbourhood} in this metric space is the set of all the
signals which have the same prefix. More formally, for $a \in S$, $r
\in \Re$, $0 < r \leq 1$
\begin{eqnarray}
N(a,r) & = & \{\underline{s}:d(a,\underline{s}) < r\} \nonumber \\
       & = & \{\underline{s}:\underline{s}\!\!\mid^{\tau} =
            a\!\!\mid^{\tau}\} \nonumber
\end{eqnarray}
where $\tau = \log_2\left(\frac{1}{r}\right)$.

\vspace{10pt}
Problem: Let $S_d$ be the set of one sided discrete event signals and
the cantor metric is defined. Show that $S_d$ is Huasdorff, i.e.,
given $s,s'\in S_d$, $s\neq s'$ show that there exist open sets $U_1$
and $U_2$ such that $s\in U_1$, $s'\in U_2$ and $U_1\cap U_2=\phi$
\vspace{10pt}

We can also describe causality in the cantor metric space
\begin{itemize}
\item {\bf Causality} $d(F(s),F(s'))\leq d(s,s')$
\item {\bf Strict Causality} $d(F(s),F(s'))< d(s,s')$
\item {\bf Delta Causality} $\exists k < 1, k = 2^{-\Delta}$ such that
$d(F(s),F(s')) \leq kd(s,s')$ ($F$ is a {\em contraction mapping})
\end{itemize}
\vspace{5pt}
 
We now investigate whether a causal system placed in feedback configuration
deterministic. We will show that if the system is strictly causal then
the feedback configuration is guaranteed to have at most one behaviour,
i.e., it can have either a unique behaviour or no behaviour at all. We
will also show that if the system is delta causal then the feedback
configuration has a unique behaviour and we can systematically find
it. 

We will first show that the cantor metric is indeed a metric. We need
to show that it satisfies the four conditions that any metric must
satisfy. For $s, s', s'' \in S_d$,
\begin{enumerate}
\item $d(s,s') = d(s',s)$ by definition
\item $d(s,s') \geq 0$ by definition
\item $d(s,s') = 0$ iff $s=s'$ by definition
\item Triangle inequality $d(s,s')+d(s',s'') \geq d(s,s'')$
\end{enumerate}
We will prove a stronger condition than the last condition by showing
that $\max(d(s,s'), d(s',s'') \geq d(s,s''))$, i.e., the cantor metric
is an {\em ultra metric}.\vspace{5pt}
{\bf Proof} Without loss of generality let $d(s,s') \geq
d(s',s'')$. $\exists\, \tau_1 \tau_2$ such that $s\!\!\mid^{\tau_1} =
s'\!\!\mid^{\tau_1}$ and  $s'\!\!\mid^{\tau_2} =
s''\!\!\mid^{\tau_2}$. Since $d(s,s') \geq d(s',s'')$, $\tau_2 \geq
\tau_1$. Hence $\exists\, \tau_3 \geq \tau_1$ such that
$s\!\!\mid^{\tau_3} = s''\!\!\mid^{\tau_3}$. Hence $d(s,s') \geq
d(s,s'')$.
\vspace{5pt}

{\large \bf Composing Functional Processes}
\begin{itemize}
\item {\bf Parallel Composition} The parallel composition of functional
(causal, strictly causal or delta causal) processes is functional
(causal, strictly causal or delta causal).
\item {\bf Cascade Composition} The cascade composition of functional
(causal, strictly causal or delta causal) processes is functional
(causal, strictly causal or delta causal).
\item {\bf Source Composition} The parallel composition of functional
(causal, strictly causal or delta causal) processes and source
processes is functional (causal, strictly causal or delta causal) if
all the source processes are determinate (determinate, strictly
causal, delta causal).
\item {\bf Feedback Composition} We will modify the general feedback
configuration slightly. We replace the input with equivalent
determinate source process. We also make all the output only signals
as inputs. For $f:S^N \rightarrow S^N$, define the {\em semantics} to
be a fixed point of $f$, i.e., $s$ such that $f(s)=s$. If no or one
signal satisfies this semantic, then the system is determinate
otherwise it is indeterminate. 
\begin{itemize}
\item If $f$ is strictly causal, then it has at most one fixed
point. Hence the feedback composition is determinate.
\item {\em (Banach fixed point theorem)} If the metric space is
complete and $f$ is delta causal, then it has exactly one fixed point
and that fixed point can be found by starting with any signal tuple
$s_0$ and finding the limit of $s_1 = f(s_0), s_2 = f(s_1) \ldots$
\item If the metric space is compact (for instance if $V$ is finite and
time is discrete), then $f$ only needs to be strictly causal to apply
the Banach fixed point theorem.
\end{itemize}
\end{itemize}
{\bf Lemma} Let $(S_d, d)$ be some metric space and let the function
$f:S^d \rightarrow S^d$ be strictly causal, i.e., $d(f(s),f(s') <
d(s,s') \forall s,s' \in S_d$. Then if $f(s_1)=s_1$ and $f(s_2)=s_2$
then $s_1=s_2$.

{\bf Proof} Assume $s_1 \neq s_2$ then $d(f(s),f(s') < d(s,s')
\forall s,s' \in S_d$ implies $d(s,s') < d(s,s')$ which is a
contradiction. Hence $s_1=s_2$.

Why do we have to restrict ourself to strict causality? Consider the
following system with a delta causal process $f$. Define an input output
process pair such that output is the same as input. Now this composite
process is no longer delta causal but it is causal
nevertheless. Even though $f$ is delta causal but the overall system is
causal and in the given feedback configuration, it is indeterminate.

Now consider the following example. The process has two inputs $s_1$
and $s_2$ and an output $s_3$. $T=[0,\infty), V=\{1,2,3\ldots\}$ \\
foreach $e=(\tau, v) \in s_1$, let $e'=(\tau+1,v+1) \in s_3$ \\
foreach $e=(\tau, v) \in s_2$, let $e'=(\tau+1/v^2,v+1) \in s_3$ \\
if collision, choose $s_1$.\\ Now consider this process in a feedback
configuration with the output $s_3$ fed back to $s_2$ and the input at
$s_1=\{(i,1), i=0,1,2,\ldots\}$. Clearly the system is a discrete event
system and obeys strict causality but in the feedback configuration,
the behaviour of the system is not discrete event. Hence strict
causality need not preclude absence of behaviour. 
\end{document}