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%%tth:\special{html:<title> An introduction to information theory and entropy</title>}

%Slide 1
\title{{\LARGE\bf An introduction to information theory and entropy}\newline \newline \newline}
\author{Tom Carter
\newline
\newline
\newline
\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS}{http://astarte.csustan.edu/\~\ tom/SFI-CSSS}}
%%tth:\href{http://astarte.csustan.edu/~tom/SFI-CSSS}{http://astarte.csustan.edu/\~tom/SFI-CSSS}
\vfill
Complex Systems Summer School
\newline
}
\date{June, 2006}
\maketitle

%Slide 2
\sectionhead{Our general topics:}
%%tth:\begin{itemize}
%%tth:\item
\tthdump{\hyperlink{Measuring complexity}
               {\ $\circledcirc$ Measuring complexity\newline}}
%%tth:\makehyperlink{Measuring complexity}
%%tth:\item
\tthdump{\hyperlink{Some probability background}
	        {$\circledcirc$ Some probability background\newline}} 
%%tth:\makehyperlink{Some probability background}
%%tth:\item
\tthdump{\hyperlink{Basics of information theory}
	        {$\circledcirc$ Basics of information theory\newline}} 
%%tth:\makehyperlink{Basics of information theory}
%%tth:\item
\tthdump{\hyperlink{Some entropy theory}
	        {$\circledcirc$ Some entropy theory\newline}} 
%%tth:\makehyperlink{Some entropy theory}
%%tth:\item
 \tthdump{\hyperlink{The Gibbs inequality}
	         {$\circledcirc$ The Gibbs inequality\newline}} 
%%tth:\makehyperlink{The Gibbs inequality}
%%tth:\item
\tthdump{\hyperlink{A simple physical example (gases)}
	         {$\circledcirc$ A simple physical example (gases)\newline}} 
%%tth:\makehyperlink{A simple physical example (gases)}
%%tth:\item
\tthdump{\hyperlink{Shannon's communication theory}
	         {$\circledcirc$ Shannon's communication theory\newline}} 
%%tth:\makehyperlink{Shannon's communication theory}
%%tth:\item
\tthdump{\hyperlink{Application to Biology (analyzing genomes)}
	         {$\circledcirc$ Application to Biology (analyzing\\\ \ \ \ \ genomes)\newline}} 
%%tth:\makehyperlink{Application to Biology (analyzing genomes)}
%%tth:\item
\tthdump{\hyperlink{Some other measures}
	        {$\circledcirc$ Some other measures\newline}} 
%%tth:\makehyperlink{Some other measures}
%%tth:\item
\tthdump{\hyperlink{Some additional material}
	        {$\circledcirc$ Some additional material\newline}} 
%%tth:\makehyperlink{Some additional material}
%%tth:\item
	 \tthdump{\hyperlink{Examples using Bayes' Theorem}
	         {$\circledcirc$ Examples using Bayes' Theorem\newline}}
%%tth:\makehyperlink{Examples using Bayes' Theorem}
%%tth:\item
 \tthdump{\hyperlink{Analog channels}
	         {$\circledcirc$ Analog channels\newline}} 
%%tth:\makehyperlink{Analog channels}
%%tth:\item
\tthdump{\hyperlink{A Maximum Entropy Principle}
	         {$\circledcirc$ A Maximum Entropy Principle\newline}} 
%%tth:\makehyperlink{A Maximum Entropy Principle}
%%tth:\item
\tthdump{\hyperlink{Application: Economics I (a Boltzmann Economy)}
	         {$\circledcirc$ Application: Economics I (a Boltzmann Economy)\newline}} 
%%tth:\makehyperlink{Application: Economics I (a Boltzmann Economy)}
%%tth:\item
\tthdump{\hyperlink{Application: Economics II (a power law)}
	         {$\circledcirc$ Application: Economics II (a power law)\newline}} 
%%tth:\makehyperlink{Application: Economics II (a power law)}
%%tth:\item
\tthdump{\hyperlink{Application to Physics (lasers)}
	         {$\circledcirc$ Application to Physics (lasers)\newline}} 
%%tth:\makehyperlink{Application to Physics (lasers)}
%%tth:\item
\tthdump{\hyperlink{References}
	        {$\circledcirc$ References\newline}} 
%%tth:\makehyperlink{References}
%%tth:\end{itemize}

\pagedone

\quotesection{The quotes}
%%tth:\begin{itemize}

%%tth:\item
\tthdump{\hyperlink{Science, wisdom, and counting}
               {\ $\circledcirc$ Science, wisdom, and counting\newline}}
%%tth:\makehyperlink{Science, wisdom, and counting}
%%tth:\item
	 \tthdump{\hyperlink{Being different -- or random}
	        {$\circledcirc$ Being different -- or random\newline}}
%%tth:\makehyperlink{Being different -- or random}
%%tth:\item
	 \tthdump{\hyperlink{Surprise, information, and miracles}
	        {$\circledcirc$ Surprise, information, and miracles\newline}}
%%tth:\makehyperlink{Surprise, information, and miracles}
%%tth:\item
	 \tthdump{\hyperlink{Information (and hope)}
	        {$\circledcirc$ Information (and hope)\newline}}
%%tth:\makehyperlink{Information (and hope)}
%%tth:\item
	 \tthdump{\hyperlink{H (or S) for Entropy}
	         {$\circledcirc$ H (or S) for Entropy\newline}}
%%tth:\makehyperlink{H (or S) for Entropy}
%%tth:\item
 	 \tthdump{\hyperlink{Thermodynamics}
	         {$\circledcirc$ Thermodynamics\newline}}
%%tth:\makehyperlink{Thermodynamics}
%%tth:\item
	 \tthdump{\hyperlink{Language, and putting things together}
	         {$\circledcirc$ Language, and putting things together\newline}}
%%tth:\makehyperlink{Language, and putting things together}
%%tth:\item
	 \tthdump{\hyperlink{Tools}
	         {$\circledcirc$ Tools}}
%%tth:\makehyperlink{Tools}
%%tth:\end{itemize}
%\thepage

\tthdump{\hyperlink{Our general topics:}{\hfil To topics $\leftarrow$}}
%%tth:{\special{html: <a href="\#Top of file">       Back to top of file</a>}}
\pagedone

%Slide 3

\quotesection{Science, wisdom, and counting}
%%tth:\begin{quote}
``Science is organized knowledge. Wisdom is organized life.''

- Immanuel Kant 

``My own suspicion is that the universe is not only stranger than we suppose,
but stranger than we can suppose.''

- John Haldane

 ``Not everything that can be counted counts, and not everything that counts can be counted.'' 
 
      - Albert Einstein (1879-1955) 
      

``The laws of probability, so true in general, so fallacious in particular .''

- Edward Gibbon 

%%tth:\end{quote}

\pagedone

\sectionhead{Measuring complexity}

\begin{itemize}
       \item Workers in the field of complexity face a classic problem:  how can we tell that
            the system we are looking at is actually a complex system?  (i.e., should we even
            be studying this system? :-)
            
            Of course, in practice, we will study the systems that interest us, for whatever reasons,
            so the problem identified above tends not to be a real problem.  On the other hand,
            having chosen a system to study, we might well ask ``How complex is this system?''
            
            In this more general context,  we probably want at least to be able to compare two
            systems, and be able to say that system A is more complex than system B.  Eventually,
            we probably would like to have some sort of numerical rating scale.
            
\pagedone
          
          \item Various approaches to this task have been
            proposed, among them:
          \begin{enumerate}
              \item  Human observation and (subjective) rating
 %                  (i.e., use
  %                 the human perceptual system as a measurement instrument)
              \item  Number of parts or distinct elements (what counts as a distinct part?)
              \item  Dimension (measured how?)
              \item  Number of parameters controlling the system
              \item  Minimal description (in which language?)
              \item  Information content (how do we define/measure information?)
              \item  Minimal generator/constructor (what machines/methods can we use?)
              \item  Minimum energy/time to construct (how would evolution count?)
          \end{enumerate}

\pagedone
          
          \item  Most (if not all) of these measures will actually be measures associated
               with a {\em model} of a phenomenon.  Two observers (of the same phenomenon?)
               may develop or use very different models, and thus disagree in their assessments
               of the complexity.  For example, in a very simple case, counting the number of
               parts is likely to depend on the scale at which the phenomenon is viewed (counting
               atoms is different from counting molecules, cells, organs, etc.).
               
               We shouldn't expect to be able to come up with a single universal measure
               of complexity.  The best we are likely to have is a measuring system useful
               by a particular observer, in a particular context, for a particular purpose.
               
               My first focus will be on measures related to how surprising or unexpected
               an observation or event is.  This approach has been described as {\em
               information theory}.
            
\end{itemize}
\pagedone

\quotesection{Being different -- or random}
%%tth:\begin{quote}
``The man who follows the crowd will usually get no further than the crowd. The man who walks alone is likely to find himself in places no one has ever been before. Creativity in living is not without its attendant difficulties, for peculiarity breeds contempt. And the unfortunate thing about being ahead of your time is that when people finally realize you were right, they'll say it was obvious all along. You have two choices in life: You can dissolve into the mainstream, or you can be distinct. To be distinct is to be different. To be different, you must strive to be what no one else but you can be. ''

 -Alan Ashley-Pitt
 

 ``Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin.''
  
      - John von Neumann (1903-1957) 
%%tth:\end{quote}
 
 \pagedone
  
\sectionhead{Some probability background}

\begin{itemize}
	\item At various times in what follows, I may float between two notions of the {\em probability}
	      of an event happening.  The two general notions are:
	\begin{enumerate}
		\item A {\em frequentist} version of probability:
		
		       In this version, we assume we have a set of possible events, each of which
		       we assume occurs some number of times.  Thus, if there are $N$ distinct possible
		       events $(x_1, x_2, \ldots , x_N),$ no two of which can occur simultaneously,
		       and the events occur with frequencies $(n_1, n_2, \ldots , n_N)$,
		       we say that the probability of event $x_i$ is given by
		       $$P(x_i) = \frac{n_i}{\sum_{j=1}^{N}{n_j}}$$
		       This definition has the nice property that
		       $$\sum_{i=1}^{N}P(x_i) = 1$$
		       
		\item An {\em observer relative} version of probability:
		
		       In this version, we take a statement of {\em probability} to be an
		       assertion about the belief that a specific observer has of the occurrence of
		       a specific event.
		       
		       Note that in this version of {\em probability}, it is possible that two different
		       observers may assign different probabilities to the same event.
		       
		       Furthermore, the {\em probability} of an event, for me, is likely to change as
		       I learn more about the event, or the context of the event.
		       
\pagedone
		       
		\item In some (possibly many) cases, we may be able to find a reasonable correspondence
		       between these two views of probability.  In particular, we may sometimes be able
		       to understand the {\em observer relative} version of the probability of an event
		       to be an approximation to the {\em frequentist} version, and to view new knowledge
		       as providing us a better estimate of the relative frequencies.
		
	\end{enumerate}
	
\end{itemize}

\pagedone

%Slide 4
\begin{itemize}
	\item I won't go through much, but some probability basics, where $a$ and $b$ are events: \newline
		$ P(not\ a) = 1 - P(a).$\newline
		$ P(a\ or\ b) = P(a) + P(b) - P(a\ \mathrm{and}\ b).$ \newline
		We will often denote $P(a\ \mathrm{and}\ b)$ by $P(a, b)$.
		If $P(a, b) = 0$, we say $a$ and $b$ are mutually exclusive.\newline
	\item Conditional probability: \newline\newline
	       $ P(a \vert b) $ is the probability of $a$, given that we know $b$.
	       The joint probability of both $a$ and $b$ is given by:
	       	$$P(a, b) = P(a \vert b) P(b).$$
	        Since $P(a, b) = P(b, a)$, we have Bayes' Theorem:
	        $$P(a \vert b)P(b)  = P(b \vert a) P(a),$$
                 or
                $$P(a \vert b) = \frac{P(b \vert a) P(a)}{P(b)}.$$ 
	        
\end{itemize}

\pagedone
%Slide 5

\begin{itemize}
	 \item If two events $a$ and $b$ are such that
	        $$P(a \vert b) = P(a),$$
	        we say that the events $a$ and $b$ are {\em independent}.  Note that from
	        Bayes' Theorem, we will also have that
	        $$P(b \vert a) = P(b),$$
	        and furthermore,
	        $$P(a, b) = P(a \vert b)P(b) = P(a)P(b).$$
	        This last equation is often taken as the definition of {\em independence}.
    
	\item We have in essence begun here the development of a mathematized methodology for drawing
	       inferences about the world from uncertain knowledge.  We could say that our observation
	       of the coin showing heads gives us {\em information} about the world.  We will develop a
	       formal mathematical definition of the {\em information} content of an event which occurs
	       with a certain probability.
\end{itemize}
\pagedone

\quotesection{Surprise, information, and miracles}
%%tth:\begin{quote}
``The opposite of a correct statement is a false statement. The opposite of a profound truth may well be another profound truth.''
 
      - Niels Bohr (1885-1962)
      
``I heard someone tried the monkeys-on-typewriters bit trying for the plays of W. Shakespeare, but all they got was the collected works of Francis Bacon.'' 

- Bill Hirst 

``There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.''
 
      - Albert Einstein (1879-1955) 
%%tth:\end{quote}
      
\pagedone
\sectionhead{Basics of information theory}
\begin{itemize}
	\item We would like to develop a usable measure of the {\em information} we get
	       from observing the occurrence of an event having probability $p$ .  Our first reduction
                will be to ignore any particular features of the event, and only observe whether or
                not it happened.  In essence this means that we can think of the event as the 
                observance of a symbol whose probability of occurring is $p$.

               We will thus be defining the {\em information} in terms of the probability $p$.

\pagedone
	       
	 \item      We will want our {\em information} measure $I(p)$ to have several properties:
	       \begin{enumerate}
	            \item Information is a non-negative quantity:  $I(p) \ge 0$.
		     \item If an event has probability 1, we get no information from the occurrence
		          of the event:  $I(1) = 0$.
		     \item If two independent events occur (whose joint probability is the product of
		          their individual probabilities), then the information we get from observing
		          the events is the sum of the two informations:
		          $I(p_1*p_2) = I(p_1) + I(p_2).$
		          (This is the critical property \ldots)
		     \item We will want our {\em information} measure to be a continuous (and, in fact,
		          monotonic) function of the probability (slight changes in probability should result
		          in slight changes in {\em information}).
             \end{enumerate}
\pagedone

     \item  We can therefore derive the following:
          \begin{enumerate}
               \item $I(p^2) = I(p*p) = I(p) + I(p) = 2*I(p)$
               \item Thus, further, $I(p^n) = n*I(p)$ \newline
                    (by induction \ldots)
               \item  $I(p) = I((p^{1/m})^m) = m * I(p^{1/m})$, so $I(p^{1/m}) = \frac{1}{m}*I(P)$
                      and thus in general
                      $$I(p^{n/m}) = \frac{n}{m}*I(p)$$
               \item  And thus, by continuity, we get, for $0 < p \le 1$, and $a>0$ a real number:
                      $$I(p^a) = a*I(p)$$
           \end{enumerate}
           
      \item  From this, we can derive the nice property:
           $$I(p) = -\log_b(p) = \log_b(1/p)$$
           for some base $b$.


\end{itemize}
\pagedone

%Slide 9
\begin{itemize}
       \item Summarizing: from the four properties,
       \begin{enumerate}
            \item $I(p) \ge 0$
            \item $I(p_1*p_2) = I(p_1) + I(p_2)$
            \item $I(p)$ is monotonic and continuous in $p$
            \item $I(1) = 0$
       \end{enumerate}
       we can derive that
       $$I(p) = \log_b(1/p) = - \log_b(p),$$
       for some positive constant $b$.  The base $b$ determines the units we
       are using.

       We can change the units by changing the base, using the formulas, for $b_1, b_2, x > 0$,
            $$x = b_1^{\log_{b_1}(x)}$$
       and therefore
       $$\log_{b_2}(x) = \log_{b_2}(b_1^{\log_{b_1}(x)}) = (\log_{b_2}(b_1))(\log_{b_1}(x)).$$

\end{itemize}
\pagedone

\begin{itemize}
       \item  Thus, using different bases for the logarithm results in {\em information} measures
            which are just constant multiples of each other, corresponding with measurements
            in different units:
            \begin{enumerate}
            \item $\log_2$ units are {\em bits} (from 'binary')
            \item $\log_3$ units are {\em trits}(from 'trinary')
            \item $\log_e$ units are {\em nats} (from 'natural logarithm') (We'll use $\ln(x)$
                 for $\log_e(x)$)
            \item $\log_{10}$ units are {\em Hartleys}, after an early worker in the field.
            \end{enumerate}
      
      \item Unless we want to emphasize the units, we need not bother to specifiy
            the base for the logarithm, and will write $\log(p)$.  Typically, we will think
            in terms of $\log_2(p)$.
            
\end{itemize}
\pagedone

\begin{itemize}            
       \item  For example, flipping a fair coin once will give us events $h$ and $t$ each
            with probability $1/2$, and thus a single flip of a coin gives us $-\log_2(1/2) = 1$ bit
            of information (whether it comes up $h$ or $t$).
            
            Flipping a fair coin $n$ times (or, equivalently, flipping $n$ fair coins) gives us
            $-\log_2((1/2)^n) = \log_2(2^n) = n*\log_2(2) = n$ bits of information.
            
            We could enumerate a sequence of 25 flips as, for example:
            $$hthhtththhhthttththhhthtt$$
            or, using $1$ for $h$ and $0$ for $t$, the 25 bits
            $$1011001011101000101110100.$$
            
            We thus get the nice fact that $n$ flips of a fair coin gives us $n$ bits of information,
            and takes $n$ binary digits to specify.  That these two are the same reassures us that
            we have done a good job in our definition of our {\em information} measure \ldots
    
\end{itemize}

\pagedone

\quotesection{Information (and hope)}
%%tth:\begin{quote}
``In Cyberspace, the First Amendment is a local ordinance.''

- John Perry Barlow


``Groundless hope, like unconditional love, is the only kind worth having.''

- John Perry Barlow

``The most interesting facts are those which can be used several times, those which have a chance of recurring.  \ldots  Which, then, are the facts that have a chance of recurring?  In the first place, simple facts.''

H. Poincare, 1908
%%tth:\end{quote}

\pagedone
%Slide 10
\sectionhead{Some entropy theory}
\begin{itemize}
     \item Suppose now that we have $n$ symbols $\{a_1, a_2, \ldots, a_n\}$, and some
          source is providing us with a stream of these symbols.  Suppose further that
          the source emits the symbols with probabilities $\{p_1, p_2, \ldots, p_n\}$, respectively.
          For now, we also assume that the symbols are emitted independently
          (successive symbols do not depend in any way on past symbols).

          What is the average amount of {\em information} we get from each symbol we see
           in the stream?
\end{itemize}          
\pagedone
 
 \begin{itemize}        
     \item What we really want here is a weighted average.  If we observe the symbol $a_i$,
          we will get be getting $\log(1/p_i)$ {\em information} from that particular observation.
          In a long run (say $N$) of observations, we will see (approximately)
          $N*p_i$ occurrences of symbol $a_i$ (in the frequentist sense, that's what it means to say
          that the probability of seeing $a_i$ is $p_i$).  Thus, in the $N$ (independent) observations, we
          will get total information $I$ of
          $$I = \sum_{i=1}^{n}(N*p_i)*\log(1/p_i).$$
          But then, the average information we get per symbol observed will be
          \begin{eqnarray*}
             I/N & = & (1/N)\sum_{i=1}^{n}(N*p_i)*\log(1/p_i)\\
             & = & \sum_{i=1}^{n}p_i*\log(1/p_i)
          \end{eqnarray*}
          
          Note that $\lim_{x\rightarrow0} x*\log(1/x) = 0$, so we can, for our purposes,
          define $p_i*\log(1/p_i)$ to be $0$ when $p_i = 0$.


\end{itemize}          
\pagedone
%Slide 11
 \begin{itemize} 

     \item  This brings us to a fundamental definition.  This definition is essentially due to
          Shannon in 1948, in the seminal papers in the field of information theory.
          
          As we have observed, we have defined {\em information} strictly in terms of
          the probabilities of events.  Therefore, let us suppose that we have a set of
          probabilities (a probability distribution) $P = \{p_1, p_2, \ldots, p_n\}$.
          We define the {\em entropy} of the distribution $P$ by:
          $$H(P) = \sum_{i=1}^{n}p_i*\log(1/p_i).$$
          
          I'll mention here the obvious generalization, if we have a continuous rather
          than discrete probability distribution $P(x)$:
          $$H(P) = \int P(x)*\log(1/P(x))dx.$$
          
	 
\end{itemize}

\pagedone

%Slide 12

     \begin{itemize}
          \item  Another worthwhile way to think about this is in terms of
               {\em expected value}.  Given a discrete probability distribution
               $P = \{p_1, p_2, \ldots, p_n\}$, with $p_i \ge 0$ and $\sum_{i=1}^n p_i = 1$,
               or a continuous distribution $P(x)$ with $P(x) \ge 0$ and $\int P(x)dx = 1$,
               we can define the {\em expected value} of an associated discrete set
               $F = \{f_1, f_2, \ldots, f_n\}$ or function $F(x)$ by:
               $$ <F> = \sum_{i=1}^n f_i p_i$$
               or
               $$ <F(x)> = \int F(x) P(x) dx.$$
               
               With these definitions, we have that:
               $$H(P) = <I(p)>.$$
               
               In other words, the {\em entropy} of a probability distribution is just the
               {\em expected value} of the {\em information} of the distribution.

\end{itemize}

\pagedone

%Slide 12

     Several questions probably come to mind at this point:
     \begin{itemize}
          \item  What properties does the function $H(P)$ have?  For example,
               does it have a maximum, and if so where?
          \item  Is {\em entropy} a reasonable name for this?  In particular, the
               name {\em entropy} is already in use in thermodynamics.  How are
               these uses of the term related to each other?
          \item  What can we do with this new tool?
          \item  Let me start with an easy one.  Why use the letter $H$ for entropy?
               What follows is a slight variation of a footnote, p. 105, in the book
          {\em Spikes} by Rieke, et al. :-)
\end{itemize}

\pagedone


\quotesection{H (or S) for Entropy}
%%tth:\begin{quote}         
          ``The enthalpy is [often] written U.  V is the volume, and Z is the partition function.  P and
          Q are the position and momentum of a particle.  R is the gas constant, and of course T is
          temperature.  W is the number of ways of configuring our system (the number of states),
          and we have to keep X and Y in case we need more variables.  Going back to the first half
          of the alphabet, A, F, and G are all different kinds of free energies (the last named for
          Gibbs).  B is a virial coefficient or a magnetic field.  I will be used as a symbol for
          information; J and L are angular momenta.  K is Kelvin, which is the proper unit of T.
          M is magnetization, and N is a number, possibly Avogadro's, and O is too easily
          confused with 0.  This leaves S . . .'' and H.  In {\em Spikes} they also eliminate H
          (e.g., as the Hamiltonian).  I, on the other hand, along with Shannon and others, prefer
          to honor Hartley.  Thus, H for entropy . . .
%%tth:\end{quote}

\pagedone

%Slide 14

\sectionhead{The Gibbs inequality}

\begin{itemize}

     \item First, note that the function $\ln(x)$ has derivative $1/x$.  From this, we find that
          the tangent to $\ln(x)$ at $x = 1$ is the line $y = x - 1$.  Further, since $\ln(x)$
          is concave down, we have, for $x > 0$, that
          $$\ln(x) \le x - 1,$$
          with equality only when $x = 1$.
          
          Now, given two probability distributions, $ P = \{p_1, p_2, \ldots, p_n\}$
          and $Q = \{q_1, q_2, \ldots, q_n\}$, where $p_i, q_i \ge 0$ and $\sum_i p_i = \sum_i q_i = 1$,
          we have
          \begin{eqnarray*}
               \sum_{i=1}^n p_i \ln\left(\frac{q_i}{p_i}\right) & \le & \sum_{i=1}^n p_i
                              \left(\frac{q_i}{p_i} - 1\right)
                   =  \sum_{i=1}^n(q_i - p_i)\\
              & = & \sum_{i=1}^n q_i - \sum_{i=1}^n p_i  =  1 - 1 = 0,
          \end{eqnarray*}
          with equality only when $p_i = q_i$ for all $i$.  It is easy to see that the inequality
          actually holds for any base, not just $e$.

\end{itemize}

\pagedone

% \sectionhead{Magic}
% \begin{quote}
% ``The Universe is full of magical things patiently waiting for our wits to
% grow sharper.''
% 
% -Eden Phillpotts
% \end{quote}
% 
% \begin{quote}
% ``Any sufficiently advanced technology is indistinguishable from magic.''
% 
% -Arthur C. Clarke
% \end{quote}
% 
% \pagedone
%Slide 15

\begin{itemize}
	\item We can use the Gibbs inequality to find the probability distribution which
	     maximizes the entropy function.  Suppose $ P = \{p_1, p_2, \ldots, p_n\} $ is a
	     probability distribution.  We have
      \begin{eqnarray*}
	     H(P) - \log(n) & = & \sum_{i=1}^n p_i \log(1/p_i) - \log(n)\\
	      & = & \sum_{i=1}^n p_i \log(1/p_i) - \log(n) \sum_{i=1}^n p_i \\
	      & = & \sum_{i=1}^n p_i \log(1/p_i) - \sum_{i=1}^n p_i \log(n)\\
	      & = & \sum_{i=1}^n p_i(\log(1/p_i) - \log(n))\\
	      & = & \sum_{i=1}^n p_i(\log(1/p_i) + \log(1/n))\\
	      & = & \sum_{i=1}^n p_i\log\left(\frac{1/n}{p_i}\right)\\
	      \\
	      & \le & 0,
	\end{eqnarray*}
	with equality only when $p_i = \frac{1}{n}$ for all $i$.  The last step is the application
	of the Gibbs inequality.

\end{itemize}
\pagedone


%Slide 16
\begin{itemize}
       \item What this means is that
               $$0 \le H(P) \le \log(n).$$
               We have $H(P) = 0$ when exactly one of the $p_i$'s is one and all the rest
               are zero.  We have $H(P) = \log(n)$ only when all of the events have the same
               probability $\frac{1}{n}$.
       
              That is, the maximum of the entropy function is the $\log()$ of the number of possible
              events, and occurs when all the events are equally likely.
       \item  An example illustrating this result:
                How much information can a student get from a single grade?  First, the
                maximum information occurs if all grades have equal probability (e.g.,
                in a pass/fail class, on average half should pass if we want to maximize
                the information given by the grade).
   
   \pagedone
               
                The maximum information the
                student gets from a grade will be:
                
                Pass/Fail : 1 bit.
                
                A, B, C, D, F : 2.3 bits.
                
                A, A-, B+, . . ., D-, F : 3.6 bits.
                
                Thus, using +/- grading gives the students about 1.3 more bits of information
                per grade than without +/-, and about 2.6 bits per grade more than pass/fail.
 
      \item  If a source provides us with a sequence chosen from 4 symbols (say
      A, C, G, T), then the maximum average information per symbol is 2 bits.
      If the source provides blocks of 3 of these symbols, then the maximum
      average information is 6 bits per block (or, to use different units, 4.159 nats
      per block).
                     
\end{itemize}
\pagedone
%Slide 17

We ought to note several things. 
\begin{itemize}
  
      \item  First, these definitions of {\em information}
          and {\em entropy} may not match with some other uses of the terms.

          For example, if we know that a source will, with equal probability, transmit either the
          complete text of Hamlet or the complete text of Macbeth (and nothing else), then receiving
          the complete text of Hamlet provides us with precisely 1 bit of information.

          Suppose a book contains ascii characters.  If the book is to provide us with {\em information}
          at the maximum rate, then each ascii character will occur with equal probability -- it will be
          a random sequence of characters.

\end{itemize}
\pagedone

\begin{itemize}

     \item Second, it is important to recognize that our definitions of {\em information} and
          {\em entropy} depend only on the probability distribution.  In general, it won't make
          sense for us to talk about the {\em information} or the {\em entropy} of a source
          without specifying the probability  distribution.
          
          Beyond that, it can certainly happen that two different observers of the same
          data stream have different models of the source, and thus associate different
          probability distributions to the source.  The two observers will then assign
          different values to the {\em information} and {\em entropy} associated with
          the source.
          
          This observation (almost :-) accords with our intuition:  two people listening to the same
          lecture can get very different information from the lecture.  For example, without
          appropriate background, one person might not understand anything at all, and therefore
          have as probability model a completely random source, and therefore get much
          more information than the listener who understands quite a bit, and can therefore
          anticipate much of what goes on, and
          therefore assigns non-equal probabilities to successive words . . .

\end{itemize}
\pagedone

\quotesection{Thermodynamics}
%%tth:\begin{quote}
``A theory is the more impressive the greater the simplicity of its premises is, the more different kinds of things it relates, and the more extended its area of applicability.  Therefore the deep impression which classical thermodynamics made upon me.  It is the only physical theory of universal content which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown (for the special attention of those who are skeptics on principle).''

- A. Einstein, 1946

``Thermodynamics would hardly exist as a profitable discipline if it were not that the natural limit to the size of so many types of instruments which we now make in the laboratory falls in the region in which the measurements are still smooth.''

- P. W. Bridgman, 1941
%%tth:\end{quote}

 \pagedone
    
\sectionhead{A simple physical example (gases)}

\begin{itemize}

       \item  Let us work briefly with a simple model for an idealized gas.  Let us assume
            that the gas is made up of $N$ point particles, and that at some time $t_0$ all
            the particles are contained within a (cubical) volume $V$.  Assume that through
            some mechanism, we can determine the location of each particle sufficiently
            well as to be able to locate it within a box with sides $1/100$ of the sides
            of the containing volume $V$.  There are $10^6$ of these small boxes within
            $V$.
            
       \item  We can now develop a (frequentist) probability model for this system.  For
            each of the $10^6$ small boxes, we can assign a probability $p_i$ of finding a gas
            particle in the small box by counting the number of particles $n_i$ in the box,
            and dividing by $N$.  That is, $p_i = \frac{n_i}{N}$.  From this probability
            distribution, we can calculate an entropy:
            \begin{eqnarray*}
             H(P) & = & \sum_{i=1}^{10^6}p_i*\log(1/p_i)\\
                     & = & \sum_{i=1}^{10^6}\frac{n_i}{N} * \log(N/n_i)
            \end{eqnarray*}
            
            If the particles are evenly distributed among the $10^6$ boxes, then we will
            have that each $n_i = N/10^6$, and in this case the entropy will be:
            \begin{eqnarray*}
             H(evenly) & = & \sum_{i=1}^{10^6}\frac{N/10^6}{N} * \log\left(\frac{N}{N/10^6}\right)\\
                            & = & \sum_{i=1}^{10^6}\frac{1}{10^6} * \log(10^6)\\
                            & = & \log(10^6).
            \end{eqnarray*}
            
           
\end{itemize}

\pagedone

     There are several ways to think about this example.  
     
\begin{itemize}

       \item  First, notice that the
            calculated entropy of the system depends in a strong way on the relative
            scale of measurement.  For example, if the particles are evenly distributed, and
            we increase our accuracy of measurement by a factor of $10$ (i.e., if each
            small box is $1/1000$ of the side of $V$), then the calculated maximum entropy will be
            $\log(10^9)$ instead of $\log(10^6)$.
            
            For physical systems, we know that quantum limits (e.g., Heisenberg uncertainty
            relations) will give us a bound on the accuracy of our measurements, and
            thus a more or less natural scale for doing entropy calculations.  On the other
            hand, for macroscopic systems, we are likely to find that we can only make
            relative rather than absolute entropy calculations.
            
       \item  Second, we have simplified our model of the gas particles to the extent that
            they have only one property, their position.  If we want to talk about the {\em state}
            of a particle, all we can do is specify the small box the particle is in at time
            $t_0$.  There are thus $Q=10^6$ possible {\em states} for a particle, and the
            maximum entropy for the system is $\log(Q)$.  This may look familiar for
            equilibrium statistical mechanics \ldots
            
       \item  Third, suppose we generalize our model slightly, and allow the particles to
            move about within $V$.  A {\em configuration} of the system is then simply a list of
            $10^6$ numbers $b_i$ with $1 \le b_i \le N$ (i.e., a list of the numbers of particles
            in each of the boxes).  Suppose that the motions of the
            particles are such that for each particle, there is an equal probability that it
            will move into any given new small box during one (macroscopic) time step.
            How likely is it that at some later time we will find the system
            in a ``high'' entropy configuration?  How likely is it that if we start the system
            in a ``low'' entropy configuration, it will stay in a ``low'' entropy configuration
            for an appreciable length of time?  If the system is not currently in a ``maximum''
            entropy configuration, how likely is it that the entropy will increase in succeeding
            time steps (rather than stay the same or decrease)?
            
            Let's do a few computations using combinations:
                $$\binom{n}{m} = \frac{n!}{m! * (n - m)!},$$
            and Stirling's approximation:
                $$n! \approx   \sqrt{2\pi}\ n^ne^{-n}\sqrt{n}.$$
 
  \pagedone
          
            Let us start here:
            
            There are $10^6$ configurations with all the particles sitting in exactly
            one small box, and the entropy of each of those configurations is:
            $$H(all\ in\ one) = \sum_{i=1}^{10^6}p_i * \log(1/p_i) = 0,$$
            since exactly one $p_i$ is $1$ and the rest are $0$.
            These are obviously minimum entropy configurations.
           
            Now consider pairs of small boxes.  The number of configurations with all the particles
            evenly distributed between two boxes is:
            \begin{eqnarray*}
            \binom{10^6}{2} & = & \frac{10^6!}{(2)!(10^6 - 2)!}\\
                                  & = & \frac{10^6 * (10^6 - 1)}{2}\\
                                  & = & 5 * 10^{11},
            \end{eqnarray*}

            which is a (comparatively :-) large number.  The entropy of each of these configurations is:
            $$H(two\ boxes) = 1/2 * \log(2) + 1/2 * \log(2) = \log(2).$$
            We thus know that there are at least $5 * 10^{11} + 10^6$ configurations.
            If we start the system in a configuration with entropy $0$,
            then the probability that at some later time it will be in a configuration with
            entropy $\ge \log(2)$ will be
            \begin{eqnarray*}
                  & \ge & \frac{5*10^{11}}{5*10^{11} + 10^6} = (1 - \frac{10^6}{5*10^{11} + 10^6})\\
                  & \ge &(1 - 10^{-5}).
            \end{eqnarray*}
           
            As an example at the other end, consider the number of configurations with
            the particles distributed almost equally, except that half the boxes are short
            by one particle, and the rest have an extra.  The number of such configurations
            is:
            \begin{eqnarray*}
            \binom{10^6}{10^6/2} & = & \frac{10^6!}{(10^6/2)!(10^6 - 10^6/2)!}\\
                       & = & \frac{10^6!}{((10^6/2)!)^2}
            \end{eqnarray*}
            \begin{eqnarray*}
                       & \approx & \frac{\sqrt{2\pi}{(10^6)}^{10^6}e^{-10^6}\sqrt{10^6}}
                                                  {(\sqrt{2\pi}(10^6/2)^{10^6/2}e^{-(10^6/2)}\sqrt{10^6/2})^2}\\
                       & = & \frac{\sqrt{2\pi}{(10^6)}^{10^6}e^{-10^6}\sqrt{10^6}}
                                                  {{2\pi}(10^6/2)^{10^6}e^{-(10^6)}{10^6/2}}\\
                       & = & \frac{2^{10^6+1}\sqrt{10^6}}
                                                  {\sqrt{2\pi}\sqrt{10^6}}\\
                      & \approx & 2^{10^6}\\
                      & = &          (2^{10})^{10^5}\\
                      & \approx & 10^{3*10^5}.
            \end{eqnarray*}            
            
            Each of these configurations has entropy essentially equal to $\log(10^6)$.
            
            From this, we can conclude that if we start the system in a configuration with
            entropy $0$ (i.e., all particles in one box), the probability that later it will be
            in a higher entropy configuration will be $> (1 - 10^{-3*10^5})$.

  %          \begin{eqnarray*}
  %               & \ge & \frac{10^{3*10^5}}{10^3*10^5 + 10^6} = (1 - \frac{10^6}{10^{3*10^5} + 10^6})\\
  %               & > &(1 - 10^{-3*10^5 + 6}).
  %          \end{eqnarray*}
          
            Similar arguments (with similar results in terms of probabilities) can be made
            for starting in any configuration with entropy appreciably less than $\log(10^6)$
            (the maximum).  In other words, it is overwhelmingly probable that as time passes,
            macroscopically, the system will increase in entropy until it reaches the maximum.

            In many respects, these general arguments can be thought of as a ``proof''
            (or at least an explanation) of
            a version of the second law of thermodynamics:
            Given any macroscopic system which is free to change configurations, and given any
            configuration with entropy less than the maximum, there will be overwhelmingly many
            more accessible configurations with higher entropy than lower entropy, and thus, with
            probability indistinguishable from $1$, the system will (in macroscopic time steps) 
            successively change to configurations with higher entropy until it reaches the maximum.

\end{itemize}

\pagedone


\pagedone

\quotesection{Language, and putting things together}
%%tth:\begin{quote}
``An essential distinction between language and experience is that language separates out from the living matrix little bundles and freezes them; in doing this it produces something totally unlike experience, but nevertheless useful.''

- P. W. Bridgman, 1936

 ``One is led to a new notion of unbroken wholeness which denies the classical
 analyzability of the world into separately and independently existing parts.
 The inseparable quantum interconnectedness of the whole universe is the
 fundamental reality.''
 
 - David Bohm
%%tth:\end{quote}
\pagedone

% \sectionhead{Shocking}
% \begin{quote}
% ``Anyone who is not shocked by quantum theory has not understood it.'' --Neils Bohr
% \end{quote}
% 
% \begin{quote}
% ``One is led to a new notion of unbroken wholeness which denies the classical
% analyzability of the world into separately and independently existing parts.
% The inseparable quantum interconnectedness of the whole universe is the
% fundamental reality.'' --David Bohm
% \end{quote}
% 
% \begin{quote}
% ``I don't like it, and I'm sorry I ever had anything to do with it.'' --Erwin Schrodinger
% \end{quote}
% 
% \pagedone
% %Slide 19

\sectionhead{Shannon's communication theory}
\begin{itemize}

	\item In his classic 1948 papers, Claude Shannon laid the foundations
	     for contemporary {\em information}, {\em coding}, and {\em communication}
	     theory.  He developed a general model for communication systems, and a set
	     of theoretical tools for analyzing such systems.
	     
	     His basic model consists of three parts:  a sender (or source), a
	     channel, and a receiver (or sink).  His general model also includes
	     encoding and decoding elements, and noise within the channel.
	     

\end{itemize}

\vfil
%\centerline{\epsfxsize=5in \epsfbox{Communication_model.epsf}}
\centerline{\includegraphics[width = 5in]{Communication_Model}}
\centerline{Shannon's communication model}

\pagedone

\begin{itemize}

       \item  In Shannon's discrete model, it is assumed that the source provides a stream
       of symbols selected from a finite alphabet $A = \{a_1, a_2, \ldots, a_n\}$, which
       are then encoded.  The code is sent through the channel (and possibly disturbed
       by noise).  At the other end of the channel, the receiver will decode, and derive
       information from the sequence of symbols.
       
       Let me mention at this point that sending information from {\em now} to
       {\em then} is equivalent to sending information from {\em here} to {\em there},
       and thus Shannon's theory applies equally as well to information storage questions
       as to information transmission questions.
             
\end{itemize}

\pagedone

\begin{itemize}

       \item One important question we can ask is, how efficiently can we encode
            information that we wish to send through the channel?  For the moment,
            let's assume that the channel is noise-free, and, of course, that
            the receiver can accurately recover everything that is transmitted through
            the channel.  What we need, then, is a way to encode the stream of source
            symbols for transmission through the channel in such a way that the encoded
            stream can be uniquely decoded at the receiving end.
            
            If the alphabet of the channel (i.e., the symbols that can actually be
            carried by the channel) is $C = \{c_1, c_2, \ldots, c_r\}$, then an encoding
            of the source alphabet $A$ is just a function $f: A \to C^*$ (where $C^*$ is
            the set of all possible finite strings of symbols from $C$).  For future
            calculations, let $l_i = |f(a_i)|, i = 1, 2, \ldots, n$ (i.e., $l_i$ is the
            length of the string encoding the symbol $a_i \in A$).
            
\pagedone
 
       \item There is a nice inequality concerning the lengths of code strings for
            uniquely decodable (and/or instantaneous) codes, called the
            McMillan/Kraft inequality.  There is a uniquely decodable code with lengths
            $l_1, l_2, \ldots, l_n$ if and only if
            $$\sum_{i=1}^n \frac{1}{r^{l_i}} \le 1.$$
            
            The necessity of this inequality can be seen from looking at
            $$K^n = \left[\sum_{i=1}^n \frac{1}{r^{l_i}}\right]^n.$$
            
            We can rewrite this as
            $$K^n = \sum_{k=n}^{nl} \frac{N_k}{r^k}$$
            where $l$ is the length of the longest code and $N_k$ is the number of
            encodings of strings having encoded length $k$.
            
\pagedone

            Note that $N_k$ cannot be greater than
            $r^k$ (the total number of strings of length $k$, whether they encode
            anything or not).  From this we can see that
            $$K^n \le \sum_{k=n}^{nl} \frac{r^k}{r^k} = nl - n + 1 \le nl.$$
            
            From this we can conclude that $K \le 1$ (as desired), since otherwise
            $K^n$ would exceed $nl$ for some (possibly large) $n$.
            
            We can now prove a very important property of the entropy -- the entropy
            gives a lower bound for the efficiency of an encoding scheme (in other
            words, a lower bound on the possible compression of a data stream).
            
            With $K$ defined as above, we can define a set of numbers $Q_i$
            (pseudo-probabilities) by
            $$Q_i = \frac{r^{-l_i}}{K}.$$
            We call these pseudo-probabilities because we have $0 < Q_i \le 1$ for
            all $i$, and
            $$\sum_{i = 1}^n Q_i = 1.$$
            
            If $p_i$ is the probabilities of observing $a_i$ in the data stream,
            then we can apply the Gibbs inequality to get
            $$\sum_{i = 1}^n p_i \log \left(\frac{Q_i}{p_i}\right) \le 0,$$
            
            or
            $$\sum_{i = 1}^n p_i \log \left(\frac{1}{p_i}\right) \le 
                \sum_{i = 1}^n p_i \log \left(\frac{1}{Q_i}\right).$$
                
            The left hand side is the entropy of the source, say $H(S)$.
            Recalling the definition of $Q_i$, and that $K \le 1$, we find
            $$H(S) \le \sum_{i = 1}^n p_i \left( \log(K) - log\left(r^{-l_i}\right)\right)$$
            $$ = \log(K) + \sum_{i = 1}^n p_i l_i \log(r) \le \log(r)
                \sum_{i = 1}^n p_i l_i.$$
                
\pagedone
                
       \item From this, we can draw an important conclusion.  If we let
            $L = \sum_{i = 1}^n p_i l_i$, then $L$ is just the average length of code words
            in the encoding.  What we have, then, is that
            $$H(S) \le L \log(r).$$
            In other words, the entropy gives us a lower bound on average code length
            for any uniquely decodable symbol-by-symbol encoding of our data stream.
            Note that, for example, if we calculate entropy in bits and use binary
            ($r = 2$) encoding, then we have simply
            $$H(S) \le L.$$
            Shannon went beyond this, and showed that the bound (appropriately recast)
            holds even if we use extended coding systems where we group symbols together
            (into ``words'') before doing our encoding.  The generalized form of this
            inequality is called {\em Shannon's noiseless coding theorem}.
            
\pagedone

       \item In building encoding schemes for data streams (or, alternatively, in
            building data compression schemes), we will want to use our best
            understandings of the structure of the data stream -- in other words,
            we will want to use our best probability model of the data stream.
            Shannon's theorem tells us that, since the entropy gives us a lower
            bound on our encoding efficiency, if we want to improve our schemes,
            we will have to develop successively better probability models.
            
            One way to think about a scientific theory is that a theory is just
            an efficient way of encoding (i.e., structuring) our knowledge about
            (some aspect of) the world.  A good theory is one which reduces the
            (relative) entropy of our
            (probabilistic) understanding of the system (i.e., that decreases
            our average lack of knowledge about the system) \ldots
            

\end{itemize}

\pagedone

\begin{itemize}

       \item  Shannon went on to generalize to the (more realistic) situation in
            which the channel itself is noisy.  In other words, not only are we
            unsure about the data stream we will be transmitting through the
            channel, but the channel itself adds an additional layer of probability
            to our transmissions.
            
            Given a source of symbols and a channel with noise (in particular, given
            probability models for the source and the channel noise), we can talk about
            the {\em capacity} of the channel.
            The general model Shannon worked with involved two sets of symbols, the input
            symbols and the output symbols.  Let us say the two sets of symbols are
            $A = \{a_1, a_2, \ldots, a_n\}$ and $B = \{b_1, b_2, \ldots, b_m\}$.  Note
            that we do not necessarily assume the same number of symbols in the two sets.
            Given the noise in the channel, when symbol $b_j$ comes out of the channel,
            we can not be sure which $a_i$ was put in.  The channel is characterized by
            the set of probabilities $\{P(a_i|b_j)\}$.
            
       \item We can then consider various
            related information and entropy measures.  First, we can consider the
            information we get from observing a symbol $b_j$.  Given a probability model
            of the source, we have an {\em a priori} estimate $P(a_i)$ that symbol $a_i$
            will be sent next.  Upon observing $b_j$, we can revise our estimate to $P(a_i|b_j)$.
            The change in our information (the {\em mutual information}) will be given by:
       \begin{eqnarray*}
            I(a_i; b_j) & = & \log\left(\frac{1}{P(a_i)}\right) - \log\left(\frac{1}{P(a_i|b_j)}\right)\\
                           & = & \log\left(\frac{P(a_i|b_j)}{P(a_i)}\right)
       \end{eqnarray*}
       
            We have the properties:
       \begin{eqnarray*}
            I(a_i; b_j) & = & I(b_j; a_i)\\
            I(a_i; b_j) & = & \log(P(a_i|b_j)) + I(a_i)\\
            I(a_i; b_j) & \le & I(a_i)
       \end{eqnarray*}
       
       If $a_i$ and $b_j$ are independent (i.e., if $P(a_i, b_j) = P(a_i) * P(b_j)$), then
       $I(a_i; b_j) = 0.$
  
       \item What we actually want is to average the {\em mutual information} over all the
       symbols:
        \begin{eqnarray*}
            I(A; b_j) & = & \sum_iP(a_i|b_j) * I(a_i;b_j)\\
                          & = & \sum_iP(a_i|b_j) * \log\left(\frac{P(a_i|b_j)}{P(a_i)}\right)\\
            I(a_i;B) & = & \sum_jP(a_i|b_j) * \log\left(\frac{P(b_j|a_i)}{P(b_j)}\right),
       \end{eqnarray*}
       and from these,
       \begin{eqnarray*}
            I(A; B) & = & \sum_iP(a_i) * I(a_i;B)\\
                       & = & \sum_i\sum_jP(a_i, b_j) * \log\left(\frac{P(a_i, b_j)}{P(a_i)P(b_j)}\right)\\
                       & = & I(B; A).
       \end{eqnarray*}
       
       We have the properties:
       $ I(A; B) \ge 0,$
       and
       $ I(A; B) = 0$
       if and only if $A$ and $B$ are independent.
       
\pagedone

       \item  We then have the definitions and properties:
            
       \begin{eqnarray*}
            H(A) & = & \sum_{i = 1}^{n} P(a_i) * \log (1/P(a_i)) \\
            H(B) & = & \sum_{j = 1}^{m} P(b_j) * \log (1/P(b_j)) \\
            H(A|B) & = & \sum_{i = 1}^{n}\sum_{j = 1}^mP(a_i|b_j) * \log (1/P(a_i|b_j)) \\
            H(A, B) & = & \sum_{i = 1}^{n}\sum_{j = 1}^mP(a_i, b_j) * \log (1/P(a_i, b_j)) \\
            H(A, B) & = & H(A) + H(B|A)\\
                         &= & H(B) + H(A|B),
       \end{eqnarray*}
       and furthermore:
       \begin{eqnarray*}
            I(A; B) & = & H(A) + H(B) - H(A, B)\\
                        & = & H(A) - H(A|B)\\
                        & = & H(B) - H(B|A)\\
                        & \ge & 0
       \end{eqnarray*}

\pagedone

       \item  If we are given a channel, we could ask what is the maximum possible information
            that can be transmitted through the channel.  We could also ask what mix of the symbols
            $\{a_i\}$ we should use to achieve the maximum.  In particular, using the definitions
            above, we can define the {\em Channel Capacity} $C$ to be:
            $$C = \max_{P(a)}\ I(A; B).$$
            
       \item  We have the nice property that if we are using the channel at its capacity, then
            for each of the $a_i$,
            $$I(a_i;B) = C,$$
            and thus, we can maximize channel use by maximizing the use for each symbol
            independently.

\pagedone
            
       \item  We also have Shannon's main theorem:
       
            For any channel, there exist ways of encoding input symbols such that we can
            simultaneously utilize the channel as closely as we wish to the capacity, and
            at the same time have an error rate as close to zero as we wish.
            
       \item  This is actually quite a remarkable theorem.  We might naively guess that
            in order to minimize the error rate, we would have to use more of the channel
            capacity for error detection/correction, and less for actual transmission of
            information.  Shannon showed that it is possible to keep error rates low and
            still use the channel for information transmission at (or near) its capacity.

\pagedone
            
       \item  Unfortunately, Shannon's proof has a a couple of downsides.  The first is that
            the proof is non-constructive.  It doesn't tell us how to construct the coding system
            to optimize channel use, but only tells us that such a code exists.  The second is that
            in order to use the capacity with a low error rate, we may have to encode very large
            blocks of data.  This means that if we are attempting to use the channel in real-time,
            there may be time lags while we are filling buffers.  There is thus still much work
            possible in the search for efficient coding schemes.
            
            Among the things we can do is look at natural coding systems (such as, for example,
            the DNA coding system, or neural systems) and see how they use the capacity of their
            channel.  It is not unreasonable to assume that evolution will have done a pretty good
            job of optimizing channel use \ldots       
           
\end{itemize}

\pagedone

\quotesection{Tools}
%%tth:\begin{quote}
``It is a recurring experience of scientific progress that what was yesterday an object of study, of interest in its own right, becomes today something to be taken for granted, something understood and reliable, something known and familiar -- a tool for further research and discovery.''

-J. R. Oppenheimer, 1953

 ``Nature uses only the longest threads to weave her patterns, so that each
small piece of her fabric reveals the organization of the entire tapestry.''

 - Richard Feynman
%%tth:\end{quote}
\pagedone

\sectionhead{Application to Biology (analyzing genomes)}

\begin{itemize}
       \item  Let us apply some of these ideas to the (general) problem of analyzing genomes.
       
       We can start with an example such as the comparatively small genome of Escherichia coli,
       strain K-12, substrain MG1655, version M52.  This example has the convenient features:
       \begin{enumerate}
            \item  It has been completely sequenced.
            \item  The sequence is available for downloading
                 \href{http://www.genome.wisc.edu/}{(http://www.genome.wisc.edu/)}.
            \item  Annotated versions are available for further work.
            \item  It is large enough to be interesting (somewhat over 4 mega-bases, or  4 million
                 nucleotides), but not so huge as to be completely unwieldy.
            \item  The labels on the printouts tend to make other people using the printer
                 a little nervous :-)
            \item Here's the beginning of the file:
            \begin{verbatim}
>gb|U00096|U00096 Escherichia coli
   K-12 MG1655 complete genome
AGCTTTTCATTCTGACTGCAACGGGCAATATGTCT
CTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC
TTCTGAACTGGTTACCTGCCGTGAGTAAATTAAAA
TTTTATTGACTTAGGTCACTAAATACTTTAACCAA
TATAGGCATAGCGCACAGACAGATAAAAATTACAG
AGTACACAACATCCATGAAACGCATTAGCACCACC
ATTACCACCACCATCACCATTACCACAGGTAACGG
TGCGGGCTGACGCGTACAGGAAACACAGAAAAAAG
CCCGCACCTGACAGTGCGGGCTTTTTTTTTCGACC
AAAGGTAACGAGGTAACAACCATGCGAGTGTTGAA
            \end{verbatim}
       \end{enumerate}
\end{itemize}

\pagedone

\begin{itemize}

       \item  In this exploratory project, my goal has been to apply the {\em information}
            and {\em entropy} ideas outlined above to genome analysis.  Some of the results I have
            so far are tantalizing.  For a while, I'll just walk you through some preliminary work.
            While I am not an expert in genomes/DNA, I am hoping that
            some of what I am doing can bring fresh eyes to the problems
            of analyzing genome sequences, without too many preconceptions.  It is
            at least conceivable that my naivet\'e will be an advantage \ldots
            
\pagedone
            
       \item  My first step was to generate for myself a ``random genome'' of comparable
            size to compare things with.  In this case, I simply used the Unix `random'
            function to generate a file containing a random sequence of about 4 million
            A, C, G, T.  In the actual genome, these letters stand for the nucleotides
            adenine, cytosine, guanine, and thymine.
            
            Other people working in this area have taken some other approaches to this
            process, such as randomly shuffling an actual genome (thus maintaining the
            relative proportions of A, C, G, and T).  Part of the justification for this
            methodology is that actual (identified) coding sections of DNA tend to have
            a ratio of C+G to A+T different from one.  I didn't worry about this issue
            (for various reasons).

\pagedone
            
       \item  My next step was to start developing a (variety of) probability model(s) for
            the genome.  The general idea that I am working on is to build
            some automated tools to locate ``interesting'' sections of a genome.  Thinking
            of DNA as a coding system, we can hope that ``important'' stretches
            of DNA will have entropy different from other stretches.  Of course,
            as noted above, the entropy measure depends in an essential way on the
            probability model attributed to the source.  We will want to try to build a
            model that catches important aspects of what we find interesting or significant.
            We will want to use our knowledge of the systems in which DNA is embedded
            to guide the development of our models.  On the other hand, we probably
            don't want to constrain the model too much.  Remember that information
            and entropy are measures of {\em unexpectedness}.  If we constrain our model
            too much, we won't leave any room for the unexpected!
            
  \pagedone
            
            \item We know, for example, that simple repetitions have low entropy.  But if
            the code being used is redundant (sometimes called {\em degenerate}), with multiple
            encodings for the same symbol (as is the case for DNA codons), what looks
            to one observer to be a random stream may be recognized by another observer
            (who knows the code) to be a simple repetition.
            
            \item The first element of my probability model(s) involves the observation that
            coding sequences for peptides and proteins are encoded via {\em codons},
            that is, by sequences of blocks of triples of nucleotides.  Thus, for example,
            the codon AGC on mRNA (messenger RNA) codes for the amino acid serine
            (or, if we happen to be reading in the reverse direction, it might code for alanine).
            On DNA, AGC codes for UCG  or CGA on the mRNA, and thus could code for
            cysteine or arginine.

\pagedone
            
%            \centerline{\includegraphics[bb = 0 0 450 562, width = 5in]{genetic_code}}
           \centerline{\includegraphics[width = 5in]{genetic_code}}
%            \includegraphics[bb = 0 0 450 562, width = 5in]{genetic_code}
            Amino acids specified by each codon sequence on mRNA. \\
            A = adenine G = guanine C = cytosine T = thymine U = uracil\\
            Table from \href{http://www.accessexcellence.org/AB/GG/}
                 {http://www.accessexcellence.org}
            
  \pagedone
            
            Key for the above table:

   Ala: Alanine  \\  Arg: Arginine  \\  Asn: Asparagine  \\  Asp: Aspartic acid \\
   Cys: Cysteine  \\   Gln: Glutamine \\   Glu: Glutamic acid   \\   Gly: Glycine  \\                                              
   His: Histidine  \\  Ile: Isoleucine \\    Leu: Leucine \\ Lys: Lysine   \\   
   Met: Methionine \\  Phe:  Phenylalanine  \\   Pro: Proline   \\   Ser: Serine   \\
   Thr: Threonine  \\   Trp: Tryptophane  \\  Tyr: Tyrosine  \\  Val: Valine   \\


   
\end{itemize}
\pagedone

\begin{itemize}

     \item  For our first model, we will consider each three-nucleotide codon to be a
          distinct symbol.  We can then take a chunk of genome and estimate the probability
          of occurence of each codon by simply counting and dividing by the length.
          At this level, we are assuming we have no knowledge of where codons start,
          and so in this model, we assume that ``readout'' could begin at any nucleotide.
          We thus use each three adjacent nucleotides.
          
          For example, given the DNA chunk:
 \begin{verbatim}
AGCTTTTCATTCTGACTGCAACGGGCAATATGTC
\end{verbatim}

          we would count:
          
 \begin{verbatim}
AAT  1  AAC  1  ACG  1  ACT  1  AGC  1
ATA  1  ATG  1  ATT  1  CAA  2  CAT  1 
CGG  1  CTG  2  CTT  1  GAC  1  GCA  2
GCT  1  GGC  1  GGG  1  GTC  1  TAT  1     
TCA  1  TCT  1  TGA  1  TGC  1  TGT  1
TTC  2  TTT  2
\end{verbatim}
\pagedone

     \item  We can then estimate the entropy of the chunk as:
             $$\sum p_i * \log_2(1/p_i) = 4.7\ bits.$$
             The maximum possible entropy for this chunk would be:
             $$\log_2(27) = 4.755\ bits.$$
             
     \item  We want to find ``interesting'' sections (and features) of
          a genome.  As a starting place, we can slide a ``window'' over the genome,
          and estimate the entropy within the window.  The plot below shows the
          entropy estimates for the E. coli genome, within a window of size 6561
          ($= 3^8$).   The window
          is slid in steps of size 81 ($= 3^4$).  This results in 57,194 values, one for
          each placement of the window.  For comparison, the values for a ``random''
          genome are also shown.
     
\end{itemize}
%\pagedone

\vfill
\centerline{\includegraphics[width = 5.5in]{ecoli+random_6561_81}}
\centerline{Entropy of E. coli and random}
%\centerline{and random,}
\centerline{window 6561, slide-step 81}

\pagedone

\begin{itemize}

     \item At this level, we can make the simple observation that the actual genome
          values are quite different from the comparative random string.  The values
          for E. coli range from about 5.8 to about 5.96, while the random values are clustered
          quite closely above 5.99 (the maximum possible is $\log_2(64) = 6$).
          
     \item From here, there are various directions we could go.  With a given window
          size and step size (e.g., 6561:81, as in the given plot),
          we can look at interesting features of the {\em entropy} estimates.
          For example, we could look at regions with high entropy, or low entropy.
          We could look at regions where there are abrupt changes in entropy, or
          regions where entropy stays relatively stable.

\pagedone
          
     \item  We could change the window size, and/or step size.  We could work
          to develop adaptive algorithms which zoom in on interesting regions, where
          ``interesting'' is determined by criteria such as the ones listed above.
          
     \item  We could take known coding regions of genomes, and develop entropy
          ``fingerprints'' which we could then try to match.
          
     \item  There are various ``data massage'' techniques we could use.  For example,
          we could take the fourier transform of the entropy estimates, and explore that.
          Below is an example of such a fourier transform.  Notice that it has some
          interesting ``periodic'' features which might be worth exploring.  It is also
          interesting to note that the fourier transform of the entropy of a ``random''
          genome has the shape of approximately $1/f = 1/f^1$ (not unexpected \ldots), whereas
          the E. coli data are closer to $1/f^{1.5}$.
          
     \item The discrete Fourier transform of a sequence $(a_j)_{j=0}^{q-1}$
        is the sequence
        $(A_k)_{k=0}^{q-1}$
        where
        $$A_k=\frac{1}{\sqrt{q}}\sum_{j=0}^{q-1}a_je^{\frac{2\pi ijk}{q}}$$
        One way to think about this is that
        $(A_k) = F ((a_j))$
        where the linear transformation $F$ is given by:
        $$[F]_{j,k}=\frac{1}{\sqrt{q}}e^{\frac{2\pi ijk}{q}}$$
        Note that the inverse of $F$ is its conjugate transpose $F^\dag$ -- that is,
        $$[F^{-1}]_{k,j}=\frac{1}{\sqrt{q}}e^{-\frac{2\pi ijk}{q}}.$$
        The plots that follow are log-log plots of the norms
        $|A_k| = (A_k\bar{A_k})^{1/2}$ (power spectra).

\end{itemize}

\pagedone
%\vfill
%\centerline{\includegraphics[width = 6in]{ecoli_fft}}
\centerline{\includegraphics[width = 5.5in]{ecoli_6561_81_fft}}
\centerline{Fourier transform of E. coli}
  \centerline{window 6561, slide-step 81}
  
\pagedone
%\vfill
\centerline{\includegraphics[width = 5.5in]{random_6561_81_fft}}
\centerline{Fourier transform of random}
  \centerline{window 6561, slide-step 81}

\pagedone

\pagedone

\sectionhead{Some other measures}

\begin{itemize}
       \item  There have been various approaches to expanding on the idea of entropy
            as a measure of complexity.  One useful generalization of entropy was developed
            by the Hungarian mathematician A. R\'enyi.  His method involves looking
            at the moments of order $q$ of a probability distribution $\{p_i\}$:
            $$S_q = \frac{1}{q-1}\log\sum_i p_i^q$$
            If we take the limit as $q\rightarrow 1$, we get:
            $$S_1 = \sum_i p_i\log(1/p_i),$$
            the entropy we have previously defined.  We can then think of $S_q$ as
            a generalized entropy for any real number $q$.

\pagedone 
          
       \item  Expanding on these generalized entropies, we can then define a
            generalized {\em dimension} associated with a data set.  If we imagine
            the data set to be distributed among bins of diameter $r$, we can let $p_i$ be the
            probability that a data item falls in the $i$'th bin (estimated by counting the
            data elements in the bin, and dividing by the total number of items).  We can
            then, for each $q$, define a dimension:
            $$D_q = \lim_{r\rightarrow 0}\frac{1}{q-1}\frac{\log\sum_ip_i^q}
                  { \log(r)}.$$
       \item Why do we call this a generalized {\em dimension}?
       
            Consider $D_0$.
            First, we will adopt the (analyst's?) convention that $p_i^0 = 0$ when $p_i = 0$.  Also,
            let $N_r$ be the number of non-empty bins (i.e., the number of bins of diameter
            $r$ it takes to cover the data set).
            
            Then we have:
            $$D_0 = \lim_{r\rightarrow 0}\frac{\log\sum_ip_i^0}{\log(1/r)}
                      = \lim_{r\rightarrow 0}\frac{\log(N_r)}{\log(1/r)}$$
            Thus, $D_0$ is the Hausdorff dimension $D$, which is frequently in the
            literature called the {\em fractal dimension} of the set.
            
            Three examples:
            \begin{enumerate}
               \item  Consider the unit interval $[0,1]$.  Let $r_k = 1/2^k$.  Then
                    $N_{r_k} = 2^k$, and
                    $$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^k)}{\log(2^k)} = 1.$$
               \item  Consider the unit square $[0,1]X[0,1]$.  Again, let $r_k = 1/2^k$.  Then
                    $N_{r_k} = 2^{2k}$, and
                    $$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^{2k})}{\log(2^k)} = 2.$$
\pagedone
               \item  Consider the {\em Cantor set}:
               
\vspace{1cm}
%\centerline{
\includegraphics[width = 5.5in]{Cantor_set}
%}

                      The construction of the Cantor set is suggested by the diagram.  The
                      Cantor set is what remains from the interval after we have removed
                      middle thirds countably many times.  It is an
                      uncountable set, with measure (``length'') 0.  For this set we will
                      let $r_k = 1/3^k$.  Then $N_{r_k} = 2^k$, and
                      $$D_0 = \lim_{k\rightarrow\infty}\frac{\log(2^k)}{\log(3^k)}
                           = \frac{\log(2)}{\log(3)} \approx 0.631.$$
                      The Cantor set is a traditional example of a {\em fractal}.  It is
                      self similar, and has $D_0 \approx 0.631$, which is strictly greater
                      than its topological dimension $(=\ 0)$.
\pagedone
                      It is an important example since many nonlinear dynamical systems have
                      trajectories which are locally the product of a Cantor set with a manifold
                      (i.e., Poincar\'e sections are generalized Cantor sets).
                      
                      An interesting example of this phenomenon occurs with the {\em logistics}
                      equation:
                      $$x_{i+1} = k * x_i * (1 - x_i)$$
                      with $k > 4$.  In this case (of which you rarely see pictures \ldots), most
                      starting points run off rapidly to $- \infty$, but there is a {\em strange
                      repellor}(!) which is a Cantor set.  It is a {\em repellor} since arbitrarily
                      close to any point on the trajectory are points which run off to $- \infty$.
                      One thing this means is that any finite precision simulation will not capture
                      the {\em repellor} \ldots
            \end{enumerate}

\pagedone

       \item  We can make several observations about $D_q$:
       
       \begin{enumerate}
            \item  If $q_1 \le q_2$, then $D_{q_1} \le D_{q_2}$.
            
            \item  If the set is strictly self-similar with equal probabilities $p_i = 1/N$, then
                 we do not need to take the limit as $r\rightarrow 0$, and
                 \begin{eqnarray*}
                      D_q & = & \frac{1}{q-1}\frac{\log(N * (1/N)^q)}{\log(r)}\\
                             & = & \frac{\log(N)}{\log(1/r)}\\
                             & = & D_0
                  \end{eqnarray*}
                 for all $q$.  This is the case, for example, for the Cantor set.
                 
            \item  $D_1$ is usually called the {\em information dimension}:
                 $$D_1 = \lim_{r\rightarrow 0}\frac{\sum_ip_i * \log(1/p_i)}{\log(r)}$$
                 The numerator is just the entropy of the probability distribution.
                 
            \item  $D_2$ is usually called the {\em correlation dimension}:
                 $$D_2 = \lim_{r\rightarrow 0}\frac{\log \sum_ip_i^2}{\log(r)}$$
                 This dimension is related to the probability of finding two elements of
                 the set within a distance $r$ of each other.
       \end{enumerate}
            
\end{itemize}

\pagedone

\sectionhead{Some additional material}

       What follows are some additional examples, and expanded discussion of
       some topics  \ldots

\pagedone

\sectionhead{Examples using Bayes' Theorem}
\begin{itemize}		
	\item A quick example: \newline
	       Suppose that you are asked by a friend to help them understand the results of
	       a genetic screening test they have taken.  They have been told that they
	       have tested positive, and that the test is 99\% accurate.  What is the
	       probability that they actually have the anomaly?
	       
%%	       (Hint:  We don't yet have enough information \ldots)
%% \pagedone
               You do some research, and find out that the test screens for a genetic anomaly
	       that is believed to occur in one person out of 100,000 on average.  The lab that
	       does the tests guarantees that the test is 99\% accurate.  You push the question,
	       and find that the lab says that one percent of the time, the test falsely reports
	       the absence of the anomaly when it is there, and one percent of the time the test
	       falsely reports the presence of the anomaly when it is not there.
	       The test has come
	       back positive for your friend.  How worried should they be?  Given this much
	       information, what can you calculate as the probability they actually have the anomaly?

	       In general, there are four possible situations for an individual being tested:
	       \begin{enumerate}
	         \item Test positive (Tp), and have the anomaly (Ha).
	         \item Test negative (Tn), and don't have the anomaly (Na).
	         \item Test positive (Tp), and don't have the anomaly (Na).
	         \item Test negative (Tn), and have the anomaly (Ha).
	       \end{enumerate}
\pagedone
	       We would like to calculate for our friend the probability they actually have
	       the anomaly (Ha), given that they have tested positive (Tp):
	       $$P(Ha \vert Tp).$$
	       We can do this using Bayes' Theorem.
	       
	       We can calculate:
	       \begin{eqnarray*}
	         P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}.
	       \end{eqnarray*}

	       We need to figure out the three items on the right side of the equation.  We can
	       do this by using the information given.
\pagedone
	       Suppose the screening test was done on
	       10,000,000 people.  Out of these $10^7$ people, we expect there to be
	       $10^7/10^5 = 100$ people with the anomaly, and 9,999,900 people without the
	       anomaly.  According to the lab, we would expect the test results to be:
	       \begin{itemize}
	         \item Test positive (Tp), and have the anomaly (Ha):
	           $$ 0.99 * 100 = 99\ \mathrm{people}.$$
	         \item Test negative (Tn), and don't have the anomaly (Na):
	           $$ 0.99 * 9,999,900 = 9,899,901\ \mathrm{people}.$$
	         \item Test positive (Tp), and don't have the anomaly (Na):
	           $$ 0.01 * 9,999,900 = 99,999\ \mathrm{people}.$$
	         \item Test negative (Tn), and have the anomaly (Ha):
	           $$ 0.01 * 100 = 1\ \mathrm{person}.$$
	       \end{itemize}
\pagedone
           Now let's put the the pieces together:
           \begin{eqnarray*}
             P(Ha) & = & \frac{1}{100,000}\\
               \\
                   & = & 10^{-5}\\
               \\
             P(Tp) & = & \frac{99 + 99,999}{10^7}\\
               \\
                  & = & \frac{100,098}{10^7}\\
               \\
                  & = & 0.0100098\\
               \\
             P(Tp \vert Ha) & = & 0.99
             \end{eqnarray*}
\pagedone 
           Thus, our calculated probability that our friend actually has the anomaly is:
           \begin{eqnarray*}
	         P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}\\
	              \\
	                        & = & \frac{0.99 * 10^{-5}}{0.0100098}\\
	              \\
	                        & = & \frac{9.9 * 10^{-6}}{1.00098 * 10^{-2}}\\
	              \\
	                        & = & 9.890307 * 10^{-4}\\
	              \\
	                        & < & 10^{-3}
	       \end{eqnarray*}
	       
	       In other words, our friend, who has tested {\em positive}, with a test that
	       is 99\% correct, has less that one chance in 1000 of actually having the anomaly!
\pagedone
     \item There are a variety of questions we could ask now, such as, ``For this anomaly,
           how accurate would the test have to be for there to be a greater than 50\%
           probability that someone who tests positive actually has the anomaly?''
           
           For this, we need fewer false positives than true positives.  Thus, in
           the example, we would need fewer than 100 false positives out of the
           9,999,900 people who do not have the anomaly.  In other words, the proportion
           of those without the anomaly for whom the test would have to be correct
           would need to be greater than:
           $$\frac{9,999,800}{9,999,900} = 99.999\%$$
\pagedone
    \item  Another question we could ask is, ``How prevalent would an anomaly have to be
           in order for a 99\% accurate test (1\% false positive and 1\% false negative)
           to give a greater than 50\% probability of actually having the anomaly when
           testing positive?''
           
           Again, we need fewer false positives than true positives.  We would therefore
           need the actual occurrence to be greater than 1 in 100 (each false positive
           would be matched by at least one true positive, on average).
\pagedone
    \item  Note that the current population of the US is about 280,000,000 and the
           current population of the world is about 6,200,000,000.   Thus, we could
           expect an anomaly that affects 1 person in 100,000 to affect about 2,800
           people in the US, and about 62,000 people worldwide, and one affecting
           one person in 100 would affect 2,800,000 people in the US, and 62,000,000
           people worldwide \ldots
%%\pagedone
    \item  Another example:  suppose the test were not so accurate?  Suppose the test
           were 80\% accurate (20\% false positive and 20\% false negative).  Suppose
           that we are testing for a condition expected to affect 1 person in 100.
           What would be the probability that a person testing positive actually has
           the condition?
\pagedone           
           We can do the same sort of calculations.

           Let's use 1000
           people this time.  Out of this sample, we would expect 10 to have the
           condition.
           \begin{itemize}
	         \item Test positive (Tp), and have the condition (Ha):
	           $$ 0.80 * 10 = 8\ \mathrm{people}.$$
	         \item Test negative (Tn), and don't have the condition (Na):
	           $$ 0.80 * 990 = 792\ \mathrm{people}.$$
	         \item Test positive (Tp), and don't have the condition (Na):
	           $$ 0.20 * 990 = 198\ \mathrm{people}.$$
	         \item Test negative (Tn), and have the condition (Ha):
	           $$ 0.20 * 10 = 2\ \mathrm{people}.$$
	       \end{itemize}
\pagedone
           Now let's put the the pieces together:
           \begin{eqnarray*}
             P(Ha) & = & \frac{1}{100}\\
               \\
                   & = & 10^{-2}\\
               \\
             P(Tp) & = & \frac{8 + 198}{10^3}\\
               \\
                  & = & \frac{206}{10^3}\\
               \\
                  & = & 0.206\\
               \\
             P(Tp \vert Ha) & = & 0.80
             \end{eqnarray*}
\pagedone 
           Thus, our calculated probability that our friend actually has the anomaly is:
           \begin{eqnarray*}
	         P(Ha \vert Tp) & = & \frac{P(Tp \vert Ha) * P(Ha)}{P(Tp)}\\
	              \\
	                        & = & \frac{0.80 * 10^{-2}}{0.206}\\
	              \\
	                        & = & \frac{8 * 10^{-3}}{2.06 * 10^{-1}}\\
	              \\
	                        & = & 3.883495 * 10^{-2}\\
	              \\
	                        & < & .04
	       \end{eqnarray*}
	       
	       In other words, one who has tested {\em positive}, with a test that
	       is 80\% correct, has less that one chance in 25 of actually having this
	       condition.  (Imagine for a moment, for example, that this is a drug test
	       being used on employees of some corporation \ldots)
\pagedone
    \item  We could ask the same kinds of questions we asked before:
           \begin{enumerate}
             \item  How accurate would the test have to be to get a better than 50\%
                    chance of actually having the condition when testing positive?
                    
                    (99\%)
             \item  For an 80\% accurate test, how frequent would the condition
                    have to be to get a better than 50\% chance?
                    
                    (1 in 5)
           \end{enumerate}
\pagedone
    \item  Some questions:
           \begin{enumerate}
             \item  Are these examples realistic?  If not, why not?
             \item  What sorts of things could we do to improve our results?
             \item  Would it help to repeat the test?  For example, if the
               probability of a false positive is 1 in 100, would that mean
               that the probability of two false positives on the same
               person would be 1 in 10,000 ($\frac{1}{100} * \frac{1}{100}$)?
               If not, why not?
             \item  In the case of a medical condition such as a genetic anomaly,
               it is likely that the test would not be applied randomly, but would
               only be ordered if there were other symptoms suggesting the anomaly.
               How would this affect the results?
           \end{enumerate}
   \item  Another example: \newline
	       Suppose that Tom, having had too much time on his hands while an undergraduate
	       Philosophy major, through much practice at prestidigitation, got to the point where if
	       he flipped a coin, his flips would have the probabilities:
	       $$P(h) = 0.7,\ P(t) = 0.3.$$
	       Now suppose further that you are brought into a room with 10 people in it, including
	       Tom, and on a table is a coin showing heads.  You are told further that one of the 10
	       people was chosen at random, that the chosen person flipped the coin and put it on the
	       table, and that research shows that the overall average for the 10 people each flipping
	       coins many times is:
	       $$P(h) = 0.52,\ P(t) = 0.48.$$
	       What is the probability that it was Tom who flipped the coin?
	       
	       By Bayes' Theorem, we can calculate:
	       \begin{eqnarray*}
	         P(\mathrm{Tom} \vert h) & = & \frac{P(h \vert \mathrm{Tom}) P(\mathrm{Tom})}{P(h)}
                           = \frac{0.7 * 0.1}{0.52}\\
	              & = & 0.1346.
	       \end{eqnarray*}
	       Note that this estimate revises our {\em a priori} estimate of the probability of Tom being
	       the flipper up from 0.10.
	       
	       This process (revising estimated probability) of course depends in a critical way on
	       having {\em a priori} estimates in the first place \ldots
	       
\end{itemize}

\pagedone

\sectionhead{Analog channels}

\begin{itemize}

       \item  The part of Shannon's work we have looked at so far deals with discrete
            (or digital) signalling systems.  There are related ideas for continuous
            (or analog) systems.  What follows gives a brief hint of some of
            the ideas, without much detail.
            
       \item  Suppose we have a signalling system using band-limited signals (i.e.,
            the frequencies of the transmissions are restricted to lie within some
            specified range).  Let us call the bandwidth $W$.  Let us further assume
            we are transmitting signals of duration $T$.  In order to reconstruct a given
            signal, we will need $2WT$ samples of the signal.  Thus, if we are sending
            continuous signals, each signal can be represented by $2WT$ numbers $x_i$,
            taken at equal intervals.  We can associate with each signal an {\em energy},
            given by:
            $$E = \frac{1}{2W}\sum_{i = 1}^{2WT}x_i^2.$$
            The distance of the signal (from the origin) will be
            $$r = \left(\sum x_i^2\right)^{1/2} = (2WE)^{1/2}$$
            We can define the {\em signal power} to be the average energy:
            $$S = \frac{E}{T}.$$
            Then the radius of the sphere of transmitted signals will be:
            $$r = (2WST)^{1/2}.$$
            Each signal will be disturbed by the noise in the channel.  If we measure the power
            of the noise $N$ added by the channel, the disturbed signal will lie in a sphere
            around the original signal of radius $(2WNT)^{1/2}$.  Thus the original
            sphere must be enlarged to a larger radius to enclose the disturbed signals.
            The new radius will be:
            $$r = \left(2WT(S + N)\right)^{1/2}.$$
            In order to use the channel effectively and minimize error (misreading of
            signals), we will want to put the signals in the sphere, and separate them
            as much as possible (and have the distance between the signals at least twice
            what the noise contributes \ldots).  We thus want to divide the sphere up
            into sub-spheres of radius $= (2WNT)^{1/2}.$  From this, we can get an
            upper bound on the number $M$ of possible messages that we can reliably
            distinguish.  We can use the formula for the volume of an $n$-dimensional
            sphere:
            $$V(r, n) = \frac{\pi^{n/2}r^n}{\Gamma(n/2 + 1)}.$$
            We have the bound:
         \begin{eqnarray*}
             M & \le & \frac{\pi^{WT}\left(2WT(S + N)\right)^{WT}}{\Gamma(WT + 1)}
                             \frac{\Gamma(WT + 1)}{\pi^{WT}(2WTN)^{WT}}\\
                  & = & \left(1 + \frac{S}{N}\right)^{WT}
         \end{eqnarray*}
            The information sent is the $\log$ of the number of messages sent (assuming
            they are equally likely), and hence:
            $$I = \log(M) = WT * \log\left(1 + \frac{S}{N}\right),$$
            and the rate at which information is sent will be:
            $$W * \log\left(1 + \frac{S}{N}\right).$$
            We thus have the usual {\em signal/noise} formula for channel capacity \ldots
 
 \pagedone
            
            \item  An amusing little side light:  ``Random'' band-limited natural phenoma
                 typically display a power spectrum that obeys a power law of the general                 form $\frac{1}{f^\alpha}$.  On the other
                 hand, from what we have seen, if we want to use a channel optimally,
                 we should have essentially equal power at all frequencies in the band.
                 This means that a possible way to engage in SETI (the search for
                 extra-terrestrial intelligence) will be to look for bands in which there
                 is white noise!  White noise is likely to be the signature of (intelligent)
                 optimal use of a channel \ldots
\end{itemize}

\pagedone

\sectionhead{A Maximum Entropy Principle}

    \begin{itemize}
       \item  Suppose we have a system for which we can measure certain macroscopic
            characteristics.  Suppose further that the system is made up of many microscopic
            elements, and that the system is free to vary among various states.  Given
            the discussion above, let us assume that with probability essentially equal to
            1, the system will be observed in states with maximum entropy.
            
            We will then sometimes be able to gain understanding of the system by
            applying a {\em maximum information entropy} principle (MEP), and, using
            Lagrange multipliers, derive formulae for aspects of the system.
\pagedone            
      \item  Suppose we have a set of macroscopic measurable characteristics $f_k$,
           $k = 1, 2, \ldots, M$
           (which we can think of as constraints on the system), which we assume are
           related to microscopic characteristics via:
           $$\sum_ip_i * f_i^{(k)} = f_k.$$
           Of course, we also have the constraints:
           $$p_i \ge 0,\ \mathrm{and}$$
           $$\sum_ip_i = 1.$$
           We want to maximize the entropy, $\sum_ip_i\log(1/p_i)$, subject to these
           constraints.  Using Lagrange multipliers $\lambda_k$ (one for each constraint),
           we have the general solution:
           $$p_i = \exp\left(- \lambda - \sum_k\lambda_kf_i^{(k)}\right).$$
           If we define $Z$, called the partition function, by
           $$Z(\lambda_1, \ldots, \lambda_M) = \sum_i\exp\left(-\sum_k\lambda_kf_i^{(k)}\right),$$
           then we have $e^\lambda = Z$, or $\lambda = \ln(Z)$.
           
    \end{itemize}
\pagedone

\sectionhead{Application: Economics I (a Boltzmann Economy)}

    \begin{itemize}
        \item Our first example here is a very simple economy.  Suppose there is
            a fixed amount of money ($M$ dollars), and a fixed number of agents ($N$)
            in the economy.
            Suppose that during each time step, each agent randomly selects another
            agent and transfers one dollar to the selected agent.  An agent having
            no money doesn't go in debt.  What
            will the long term (stable) distribution of money be?
            
            This is not a very realistic economy -- there is no growth, only a redistribution
            of money (by a random process).  For the sake of argument, we can imagine
            that every agent starts with approximately the same amount of money, although
            in the long run, the starting distribution shouldn't matter.
            
\pagedone

        \item For this example, we are interested in looking at the distribution of money
            in the economy, so we are looking at the 
            probabilities $\{p_i\}$ that an agent has the amount of money $i$.  We are
            hoping to develop a model for the collection $\{p_i\}$.
            
            If we let $n_i$ be the number of agents who have $i$ dollars, we have two
            constraints:
            $$\sum_in_i * i = M$$
            and
            $$\sum_in_i = N.$$
            Phrased differently (using $p_i = \frac{n_i}{N}$), this says
            $$\sum_ip_i * i = \frac{M}{N}$$
            and
            $$\sum_ip_i = 1.$$
            
\pagedone

        \item  We now apply Lagrange multipliers:
            \begin{eqnarray*}
               L = \sum_ip_i\ln(1/p_i)
                & - &\lambda\left[\sum_ip_i*i - \frac{M}{N}\right] \\
                & - & \mu\left[\sum_ip_i - 1\right],
             \end{eqnarray*}
            from which we get
            $$\frac{\partial L}{\partial p_i} = -[1 + \ln(p_i)] - \lambda i - \mu = 0.$$
            
            We can solve this for $p_i$:
            $$\ln(p_i) = - \lambda i - (1 + \mu)$$
            and so
            $$p_i = e^{-\lambda_0}e^{-\lambda i}$$
            (where we have set $1 + \mu \equiv \lambda_0)$.
            
\pagedone
            
        \item Putting in constraints, we have
            \begin{eqnarray*}
                1 & = & \sum_i p_i \\
                  & = & \sum_i e^{-\lambda_0}e^{-\lambda i}\\
                  & = & e^{-\lambda_0} \sum_{i = 0}^M e^{-\lambda i},
            \end{eqnarray*}
            and
            \begin{eqnarray*}
                \frac{M}{N} & = & \sum_i p_i * i \\
                  & = & \sum_i e^{-\lambda_0}e^{-\lambda i} * i \\
                  & = & e^{-\lambda_0} \sum_{i = 0}^M e^{-\lambda i} * i.
            \end{eqnarray*}
            We can approximate (for large $M$)
            $$ \sum_{i = 0}^M e^{-\lambda i}  \approx \int_0^Me^{-\lambda x}dx
                    \approx \frac{1}{\lambda},$$
            and
            $$\sum_{i = 0}^M e^{-\lambda i} * i \approx \int_0^Mxe^{-\lambda x}dx
                \approx \frac{1}{\lambda^2}.$$
                
\pagedone

            From these we have (approximately)
            $$e^{\lambda_0} = \frac{1}{\lambda}$$
            and 
            $$e^{\lambda_0}\frac{M}{N} = \frac{1}{\lambda^2}.$$
            From this, we get
            $$\lambda = \frac{N}{M} = e^{-\lambda_0},$$
            and thus (letting $T = \frac{M}{N}$) we have:
            \begin{eqnarray*}
              p_i & = & e^{-\lambda_0}e^{-\lambda i} \\
                  & = & \frac{1}{T}e^{-\frac{i}{T}}.
            \end{eqnarray*}
            This is a Boltzmann-Gibbs distribution, where we can think of $T$ (the
            average amount of money per agent) as the
            ``temperature,'' and thus we have a ``Boltzmann economy'' \ldots
            
            Note: this distribution also solves the functional equation
            $$p(m_1)p(m_2) = p(m_1 + m_2).$$
            
\pagedone
            
        \item This example, and related topics, are discussed in
            
            {\em Statistical mechanics of money}
            
            by Adrian Dragulescu and Victor M. Yakovenko,
            
            \tthdump{\href{http://arxiv.org/abs/cond-mat/0001432}
                {http://arxiv.org/abs/cond-mat/0001432}}
%%tth:\href{http://arxiv.org/abs/cond-mat/0001432}
%%tth:{http://arxiv.org/abs/cond-mat/0001432}
            
            and
            
            {\em Statistical mechanics of money: How saving propensity affects its distribution}
            
            by Anirban Chakraborti and Bikas K. Chakrabarti
            
            \tthdump{\href{http://arxiv.org/abs/cond-mat/0004256}
                {http://arxiv.org/abs/cond-mat/0004256}}
%%tth:\href{http://arxiv.org/abs/cond-mat/0004256}
%%tth:{http://arxiv.org/abs/cond-mat/0004256}

    \end{itemize}
    
\pagedone

\sectionhead{Application: Economics II (a power law)}

    \begin{itemize}
        \item Suppose that a (simple) economy is made up of many agents $a$, each with
            wealth at time $t$ in the amount of $w(a, t)$.  (I'll leave it to you to
            come up with a reasonable definition of ``wealth'' -- of course we will
            want to make sure that the definition of ``wealth'' is applied consistently
            across all the agents.)  We can also look at the total wealth in the economy
            $W(t) = \sum_aw(a,t)$.
            
            For this example, we are interested in looking at the distribution of wealth
            in the economy, so we will assume there is some collection $\{w_i\}$ of
            possible values for the wealth an agent can have, and associated probabilities
            $\{p_i\}$ that an agent has wealth $w_i$.  We are hoping to develop a
            model for the collection $\{p_i\}$.
            
\pagedone

        \item In order to apply the maximum entropy principle, we want to look at
            global (aggregate/macro) observables of the system that reflect (or are made
            up of) characteristics of (micro) elements of the system.
            
            For this example, we can look at the growth rate of the economy.  A reasonable
            way to think about this is to let $R_i = w_i(t_1) / w_i(t_0)$ and
            $R = W(t_1)/W(t_0)$ (where $t_0$ and $t_1$ represent time steps of the economy).
            The growth rate will then be $\ln(R)$.  We then have the two constraints on the
            $p_i$:
            $$\sum_ip_i * \ln(R_i) = \ln(R)$$
            and
            $$\sum_ip_i = 1.$$
            
\pagedone

        \item  We now apply Lagrange multipliers:
            \begin{eqnarray*}
               L = \sum_ip_i\ln(1/p_i)
                & - &\lambda\left[\sum_ip_i\ln(R_i) - \ln(R)\right] \\
                & - & \mu\left[\sum_ip_i - 1\right],
             \end{eqnarray*}
            from which we get
            $$\frac{\partial L}{\partial p_i} = -[1 + \ln(p_i)] - \lambda \ln(R_i) - \mu = 0.$$
            
            We can solve this for $p_i$:
            $$p_i = e^{-\lambda_0}e^{-\lambda\ln(R_i)} = e^{-\lambda_0}R_i^{-\lambda}$$
            (where we have set $1 + \mu \equiv \lambda_0)$.
            
            Solving, we get $\lambda_0 = \ln(Z(\lambda))$, where $Z(\lambda) \equiv
            \sum_i R_i^{-\lambda}$ (the partition function) normalizes the probability
            distribution to sum to 1.  From this we see the power law (for $\lambda > 1$):
            $$p_i = \frac{R_i^{-\lambda}}{Z(\lambda)}.$$
            
\pagedone

        \item We might actually like to calculate specific values of $\lambda$, so we
            will do the process again in a continuous version.  In this version, we
            will let $R = w(T)/w(0)$ be the relative wealth at time T.  We want to
            find the probability density function $f(R)$, that is:
            $$\max_{\{f\}} H(f) = - \int_1^\infty f(R) \ln(f(R))dR,$$
            subject to
            \begin{eqnarray*}
                 \int_1^\infty f(R)dR & = & 1, \\
                 \int_1^\infty f(R)\ln(R)dR & = & C\ln(R),
            \end{eqnarray*}
            where $C$ is the average number of transactions per time step.
            
            We need to apply the calculus of variations to maximize over a class
            of functions.
\pagedone            
            When we are solving an extremal problem of the form
            $$\int F[x, f(x), f'(x)]dx,$$
            we work to solve
            $$\frac{\partial F}{\partial f(x)} - \frac{d}{dx}
                \left(\frac{\partial F}{\partial f'(x)}\right) = 0.$$
                
            Our Lagrangian is of the form
            \begin{eqnarray*}
                L \equiv & - & \int_1^\infty f(R) \ln(f(R)) dr -
                    \mu \left(\int_1^\infty f(R) dR - 1\right) \\
                    & - & \lambda\left(\int_1^\infty f(R)\ln(R)dR - C*\ln(R)\right).
            \end{eqnarray*}
            Since this does not depend on $f'(x)$, we look at:
            $$\frac{\partial[-f(R)\ln f(R) - \mu(f(R) - 1) - \lambda(f(R)\ln R - R)]}
                {\partial f(R)}$$
                $$= 0$$
            from which we get
            $$f(R) = e^{-(\lambda_0 - \lambda \ln(R))} = R^{-\lambda}e^{-\lambda_0},$$
            where again $\lambda_0 \equiv 1 + \mu$.
            
\pagedone

            We can use the first constraint to solve for $e^{\lambda_0}$:
            $$e^{\lambda_0} = \int_1^\infty R^{-\lambda}dR
                = \left[\frac{R^{-\lambda + 1}}{1 - \lambda}\right]_1^\infty
                = \frac{1}{\lambda - 1},
            $$
            assuming $\lambda > 1$.  We therefore have a power law distribution for
            wealth of the form:
            $$f(R) = (\lambda - 1)R^{-\lambda}.$$
            
            To solve for $\lambda$, we use:
            $$C * \ln(R) = (\lambda - 1)\int_1^\infty R^{-\lambda}\ln(R)dR.$$
            Using integration by parts, we get
            \begin{eqnarray*}
               C*\ln(R)
                 & = & (\lambda - 1)\left[\ln(R)\frac{R^{1 - \lambda}}{1 - \lambda}
                        \right]_1^\infty \\
                 &\ \ & - (\lambda - 1)\int_1^\infty\frac{R^{-\lambda}}{1 - \lambda}dR \\
                 & = & (\lambda - 1)\left[\ln(R)\frac{R^{1 - \lambda}}{1 - \lambda}
                        \right]_1^\infty
                        + \left[\frac{R^{1 - \lambda}}{1 - \lambda}\right]_1^\infty.
            \end{eqnarray*}
            
\pagedone

            By L'H\^opital's rule, the first term goes to zero as $R \to \infty$,
            so we are left with
            $$C * \ln(R) = \left[\frac{R^{1 - \lambda}}{1 - \lambda}\right]_1^\infty
                = \frac{1}{\lambda - 1},$$
            or, in other terms,
            $$\lambda - 1 = C * \ln(R^{-1}).$$
            
            For much more discussion of this example, see the paper {\em A Statistical
            Equilibrium Model of Wealth Distribution} by Mishael Milakovic, February, 2001,
            available on the web at:
            
% \tthdump{\href{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&paper\_id=214}
%   {http://www.econometricsociety.org/
%       cgi-bin/conference/download.cgi
%       ?db\_name=SCE2001\&paper\_id=214}}
% %%tth:\href{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&pape\r_id=214}
% %%tth:{http://www.econometricsociety.org/cgi-bin/conference/download.cgi?db\_name=SCE2001\&paper\_id=214}

\tthdump{\href{http://astarte.csustan.edu/\~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
  {http://astarte.csustan.edu/\~\ tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}}
%%tth:\href{http://astarte.csustan.edu/~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
%%tth:{http://astarte.csustan.edu/~tom/SFI-CSSS/Wealth/wealth-Milakovic.pdf}
                       
            
    \end{itemize}
    
\pagedone
\sectionhead{Application to Physics (lasers)}


    \begin{itemize}           
      \item  We can also apply this maximum entropy principle to physics examples.  Here is
           how it looks applied to a single mode laser.  For a laser,
           we will be interested in the intensity of the light emitted, and the coherence property
           of the light will be observed in the second moment of the intensity.  The
           electric field strength of such a laser will have the form
           $$E(x, t) = E(t)\sin(kx),$$
           and E(t) can be decomposed in the form
           $$E(t) = Be^{-i\omega t} + B^*e^{i\omega t}.$$
           If we measure the intensity of the light over time intervals long compared to the
           frequency, but small compared to fluctuations of $B(t)$, the output will be
           proportional to $BB^*$ and to the loss rate, $2\kappa$, of the laser:
           $$I = 2\kappa B B^*.$$
           The intensity squared will be
           $$I^2 = 4\kappa^2B^2B^{*2}.$$
\pagedone 
           \item If we assume that $B$ and $B^*$ are continuous random variables associated with
           a stationary process, then the information entropy of the system will be:
           $$H = \int p(B,B^*)\log\left(\frac{1}{p(B,B^*)}\right)d^2B.$$
           The two constraints on the system will be the averages of the intensity and the
           square of the intensity:
           \begin{eqnarray*}
                f_1 & = & <2\kappa B B^*>,\\
                f_2 & = & <4\kappa^2B^2B^{*2}>.
           \end{eqnarray*}
           Then, of course, we will let
           \begin{eqnarray*}
                f_{B,B^*}^{(1)} & = & 2\kappa B B^*,\\
                f_{B,B^*}^{(2)} & = & 4\kappa^2B^2B^{*2}.
           \end{eqnarray*}
           We can now use the method outlined above, finding the maximum entropy 
           general solution derived via Lagrange multipliers for this system.
\pagedone
           \item  Applying the general solution, we get:
           $$p(B, B^*) = \exp\left[- \lambda - \lambda_12\kappa BB^*
                                             - \lambda_24\kappa^2(BB^*)^2\right],$$
            or, in other notation:
            $$p(B, B^*) = N * \exp(- \alpha|B|^2 - \beta|B|^4).$$
            This function in laser physics is typically derived by solving the
            Fokker-Planck equation belonging to the Langevin equation for the system.
            
            \item  For quick reference, the typical generic Langevin equation looks like:
            $$\dot{\mathbf{q}} = K(\mathbf{q}) + \mathbf{F}(t)$$
            where $\mathbf{q}$ is a state vector, and the fluctuating forces $F_j(t)$ are typically
            assumed to have
            \begin{eqnarray*}
                 <F_j(t)> & = & 0\\
                 <F_j(t)F_{j'}(t')> & = & Q_j\delta_{jj'}\delta(t - t').
            \end{eqnarray*}
\pagedone
            \item  The associated generic Fokker-Planck equation for the distribution function
            $f(q, t)$ then looks like:
            $$\frac{\partial f}{\partial t} = - \sum_j\frac{\partial}{\partial q_j}(K_jf)
                                             + \frac{1}{2}\sum_{jk}Q_{jk}\frac{\partial^2}
                                                   {\partial q_j\partial q_k}f.$$
            The first term is called the drift term, and the second the diffusion term.
            This can typically be solved only for special cases \ldots
            
            \item  For much more discussion of these topics, I can recommend the
                 book {\em Information and Self-organization, A Macroscopic Approach
                 to Complex Systems} by Hermann Haken, Springer-Verlag Berlin, New
                 York, 1988.
\end{itemize}

\pagedone
 \footnotesize
 \bibliographystyle{plain}
\tthdump{\hypertarget{References}{}\hyperlink{Our general topics:}{\hfil \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ To top $\leftarrow$}}
%%tth:{\special{html: <A NAME="References"></A><a href="\#Top of file">       Top</a>}}

\vspace{-1.0 in}

\begin{thebibliography}{12}
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Bar-Yam, Yaneer,
{\em Dynamics of Complex Systems (Studies in Nonlinearity) },
Westview Press, Boulder, 1997.

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% {\bf 21}, No. 12, p. 905, 1982.
%  
% \bibitem{bennett3}
% Bennett, C. H. and Landauer, R., 
% The fundamental physical limits of computation,
% {\em Scientific American,} July 38--46, 1985.
 
%  \bibitem{bennett4}
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%  {\em Scientific American} {\bf 257} no. 5 (November) pp 88--96,
%  1987.

 \bibitem{brillouin}
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 {\em Science and information theory}
 Academic Press, New York, 1956.

\bibitem{brooks}
Brooks, Daniel R., and Wiley, E. O.,
{\em Evolution as Entropy}, Toward a Unified Theory of Biology,
Second Edition,
University of Chicago Press, Chicago, 1988.

\bibitem{campbell}
Campbell, Jeremy,
{\em Grammatical Man}, Information, Entropy, Language, and Life,
Simon and Schuster, New York, 1982.

% \bibitem{church}
% Church, Alonzo, 
% An unsolvable problem of elementary number theory,
% {\em Amer. J. Math.} {\bf 58} 345--363, 1936.

% \bibitem{clausen}
% Clausen, M., 
% Fast Generalized Fourier transforms, 
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\end{thebibliography}

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% 
% \sectionhead{On-line references}
% 
% 
% Some of the references listed above are available on line.  They are listed again here for easy access:
% 
% %\bibitem{abrams2}
% Abrams D S and Lloyd S, 
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% solution for NP-complete and $\#$P problems,
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% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9801041}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9801041}
% {http://xxx.lanl.gov/abs/quant-ph/9801041}
% 
% 
% %\bibitem{aharonov5}
% Aharonov D, Beckman D, Chuang I and  Nielsen M,
% What Makes Quantum Computers Powerful? 
% \hyperref{http://wwwcas.phys.unm.edu/\~mnielsen/science.html}{}{}
% %\hyperURL{http}{wwwcas.phys.unm.edu/~mnielsen}{science.html}
% {http://wwwcas.phys.unm.edu/\~mnielsen/science.html}
% % 
% % 
% 
% %\bibitem{decoherence2}
%  Chuang I L, Laflamme R and Paz J P, 
% Effects of Loss and Decoherence on a Simple Quantum Computer,
% %in {\it LANL e-print} quant-ph/9602018,  http://xxx.lanl.gov (1996)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9602018}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9602018}
% {http://xxx.lanl.gov/abs/quant-ph/9602018}
% 
% %\bibitem{grover2}
%  Grover L K, 
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% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9711043}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9711043}
% {http://xxx.lanl.gov/abs/quant-ph/9711043}
% 
% %\bibitem{grover4}
%  Grover L K, 
% A fast quantum mechanical algorithm for estimating the median,
% %in {\it LANL e-print} quant-ph/9607024,  http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9607024}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9607024}
% {http://xxx.lanl.gov/abs/quant-ph/9607024}
% 
% 
% %\bibitem{knill4}
% Knill E, Laflamme R and Zurek W H 1997
% Resilient quantum computation: error models and thresholds
% %in {\it LANL e-print} quant-ph/9702058,  http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9702058}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9702058}
% {http://xxx.lanl.gov/abs/quant-ph/9702058}
% 
% \pagedone
% 
% 
% %\bibitem{preskill2}
% Preskill J 1997
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% %in {\it LANL e-print} quant-ph/9712048,  http://xxx.lanl.gov (1997), 
% to appear in {\it Introduction to Quantum
% Computation},  edited by H.-K. Lo, S. Popescu, and T. P. Spiller
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9712048}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9712048}
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% 
% %\bibitem{preskill3}
% Preskill J, Kitaev A, Course notes for Physics 229, Fall 1998, Caltech Univ.,
% \hyperref{http://www.theory.caltech.edu/people/preskill/ph229}{}{}
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% {http://www.theory.caltech.edu/people/preskill/ph229}
% 
% 
% %\bibitem{rieffel}
% Rieffel E, Polak W
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% %{\it LANL e-print} quant-ph/9809016,  http://xxx.lanl.gov (1998),
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9809016}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9809016}
% {http://xxx.lanl.gov/abs/quant-ph/9809016}
% 
% %\bibitem{Steane-97}
% Steane A,
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% %(preprint in {\it LANL e-print} quant-ph/9708022, http://xxx.lanl.gov)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9708022}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9708022}
% {http://xxx.lanl.gov/abs/quant-ph/9708022}
% 
% %\bibitem{zalka2}
% Zalka C, 
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% %in {\it LANL e-print} quant-ph/9711070http://xxx.lanl.gov (1997)
% \hyperref{http://xxx.lanl.gov/abs/quant-ph/9711070}{}{}
% %\hyperURL{http}{xxx.lanl.gov/abs/quant-ph}{9711070}
% {http://xxx.lanl.gov/abs/quant-ph/9711070}
% 

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