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\mainlylecture

\lecture{notes4}{Induction, Recursive Definitions, and Recursive
  Algorithms}{February 15, 2001} 





\section{Recursive Definitions}
A {\em recursive definition} says how to build something from
a simpler version of the same thing. \\

They have two parts:
\begin{itemize}
\item Base case(s) that don't depend on anything else
\item Combination cases that depend on simpler cases
\end{itemize}
Note the relationship to a {\em recursive algorithm.}  We will also
see a close relation to induction.

In our definitions, we will frequently use the symbol ``$=$'' in a
special sense that means what appears on the left is defined to be
what appears on the right\.  We expect you to recognize from context
which use we intend and, in the definitional case, only to use the
statement as a definition and not as an algebraic statement\.  Here
are several examples:

\begin{description}

\item[The Factorial function:]\mbox{}
  \begin{itemize}
  \item $0!=1$
  \item $(n+1)! = (n+1)n!$ for $n \ge 1$.
  \end{itemize}

\item[The Fibonacci numbers:]

  These are interesting numbers that arise, e.g., in biology, where
  they model some types of growth processes (plants, cells, rabbit
  populations, etc.)\.  The Fibonacci numbers are written as $F_i$, $i
  = 0, 1, 2, \ldots$\.  They are defined recursively by: \\ 
  \begin{align*}
    F_0 &= 0\\ 
    F_1 &= 1\\ 
    F_i &= F_{i-1} + F_{i-2} \mbox{ for $i \geq 2$.} 
  \end{align*}
  Note there 2 base cases, since each combination relies on previous two
  values.

  What is $F_4$?  Well, $F_2 = F_0+F_1 = 1$, $F_3 = F_1+F_2 = 2$, so
  $F_4 = 3$.  The sequence starts out $0, 1, 1, 2, 3, 5, 8, 13,
  21,\ldots$.

\item[ $\Sigma$ notation (the traditional style):] (We've been using
  this notation informally already)
  \begin{itemize}
  \item $\Sigma_{i=1}^0 f(i) = 0$ \\
  \item $\Sigma_{i=1}^{n+1} f(i) = \Sigma_{i=1}^n f(i) + f(n+1)$. 
  \end{itemize}

\item[Simultaneous recursive definitions.]

  You can define several things at the same time, in terms of each
  other.  For example, we may define two functions $f$ and $g$ from
  $\naturals$ to $\naturals$, recursively, by:
  \begin{itemize}
  \item
    $f(0) = 1$
  \item
    $g(0) = 1$
  \item
    $f(n+1) = f(n) + g(n)$, $n \geq 0$
  \item
    $g(n+1) = f(n) \times g(n)$, $n \geq 0$
  \end{itemize}

\end{description}


We must take care that concepts defined by recursive definitions are
well-defined\.  Below are some examples that aren't well-defined\.  In
these examples, the notion being defined always appears on the
left-hand side\.
\begin{description}
\item[$f(n)=2+f(n-1)$.]
  This definition has no base case.
\item[\rm $f(1)=1$ and $f(n) = 1+f(n+1)$.]

  This still has no sufficient base case, because $f(n)$ is defined in
  terms of $f(n+1)$\.  So for positive integers $n$ other than 1,
  $f(n)$ is defined in terms of ``more complex'' cases instead of
  simpler ones\.  Note that if we had written $f(n+1) = f(n)-1$, we
  would have meant something different (we would be defining $f(n+1)$
  in terms of $f(n)$) and $f$ would have been well-defined for all
  positive integers\.  Also, note that the first way makes $f$
  well-defined for all integers {\em less than} or equal to 1.

\item[\rm $f(1)=1$ and $f(n) = 3f(n/2)$.]  This doesn't give a definition
  for odd values of $n$, since $n/2$ is not a natural number in this
  case.

\item[\rm A mystery.]  Mathematicians have been wondering about this one
  for a while:
  \begin{itemize}
  \item $f(1) = 1$
  \item If $n$ is even, then $f(n) = f(n/2)$.
  \item If $n$ is odd, then $f(n) = f(3n+1)$.
  \end{itemize}
  If $f$ is well defined, then $f(n)=1$ everywhere.  But nobody knows if
  the recursion always terminates (though it's been checked for many numbers).
\end{description}

\section{Inductive Proofs on Recursive Definitions} 

A recursive definition usually leads to inductive
proofs---the induction follows the definition down to the base case\.
The examples in this section involve functions that are defined
recursively\.  

\subsection{Fibonacci Numbers}

The Fibonacci numbers (defined earlier) satisfy many cute identities,
so they provide lots of fun for mathematicians\.  For example:
\begin{theorem}
  $\forall n \geq 0 (\Sigma_{i=0}^n F_i^2 = F_n F_{n+1})$
\end{theorem}

Example: $n = 4$: \\
$0^2 + 1^2 + 1^2 + 2^2 + 3^2 = 15 = 3 \cdot 5$\.
\\
Let's try a proof by (standard, not strong) induction\.  The theorem
statement suggests trying it with $P(n)$ defined as:
\[\sum_{i=0}^n F_i^2 = F_n F_{n+1}
\]

\structuredproof{
  Prove: $\forall n \geq 0 (\Sigma_{i=0}^n F_i^2 = F_n F_{n+1})$ \\
  1. (Base) $\Sigma_{i=0}^0 F_i^2 = F_0 F_1$ \\
  2. (Inductive step) $\forall n \geq 0
  (\Sigma_{i=0}^n F_i^2 = F_n F_{n+1} \Rightarrow
  \Sigma_{i=0}^{n+1} F_i^2 = F_{n+1} F_{n+2})$ \reason{???}\\
  3. QED               \reason{induction}\\
}

\noindent Now we stare at the gap between $P(n)$ and $P(n+1)$\.
$P(n+1)$ is given by a summation that's obtained from that for $P(n)$
by adding one term; this suggests that, once again, we subtract.  The
difference is just the term $F_{n+1}^2$\.  Now, we are assuming that
the original $P(n)$ summation totals $F_n F_{n+1}$ and want to show
that the new $P(n+1)$ summation totals $F_{n+1} F_{n+2}$\.  So we
would {\em like\/} the difference to be
\[
F_{n+1} F_{n+2} - F_n F_{n+1}
\]

So, the actual difference is $F_{n+1}^2$ and the difference we want is
$F_{n+1} F_{n+2} - F_n F_{n+1}$\.  Are these the same\?  We want to
check that:
\[
F_{n+1}^2 = F_{n+1} F_{n+2} - F_n F_{n+1}
\]
But this is true, because it is really the Fibonacci definition in
disguise: to see this, divide by $F_{n+1}$\.  \forstudents{Here's the complete
proof:

\structuredproof{
  Prove: $\forall n \geq 0 (\Sigma_{i=0}^n F_i^2 = F_n F_{n+1})$ \\
  1. (Base) $\Sigma_{i=0}^0 F_i^2 = F_0 F_1$ 
  \reason{Both sides are $0$.} \\
  2. (Inductive step) $\forall n \geq 0 
  (\Sigma_{i=0}^n F_i^2 = F_n F_{n+1} \Rightarrow
  \Sigma_{i=0}^{n+1} F_i^2 = F_{n+1} F_{n+2})$ \noreason \\
  \> 1. Fix $n \geq 0$. \\
  \> 2. Assume $\Sigma_{i=0}^n F_i^2 = F_n F_{n+1}$ \\
  \> 3. $\Sigma_{i=0}^{n+1} F_i^2 = F_{n+1} F_{n+2}$ \\
  \> \> 1. $\Sigma_{i=0}^{n+1} F_i^2 = \Sigma_{i=0}^n F_i^2 + F_{n+1}^2$
  \reason{Summation has one extra term} \\
  \> \> 2. $\Sigma_{i=0}^n F_i^2 + F_{n+1}^2 = F_n F_{n+1} + F_{n+1}^2$
  \reason{Inductive hypothesis (2.2)} \\
  \> \> 3. $F_n F_{n+1} + F_{n+1}^2 = F_{n+1}(F_n + F_{n+1})$
  \reason{Algebra} \\
  \> \> 4. $F_{n+1}(F_n + F_{n+1}) = F_{n+1} F_{n+2}$ 
  \reason{Fibonacci definition} \\
  \> \> 5. QED                         
  \reason{Combining the equations} \\
  \> 4. QED 
  \reason{Implication, UG} \\
  3. QED                           
  \reason{Induction}\\
}
}

\subsubsection{Another false theorem}

Here is another false ``theorem''\.  The fallacy is like the blocks
example---we don't have enough base cases.

\begin{falsethm}
  All Fibonacci numbers are even.
\end{falsethm}

We try strong induction, where $P(n)$ is defined to be ``$F_n$ is
even.'' 
\structuredproof{
  Prove: $\forall n \geq 0 [P(n)]$ \\
  1. (Base) $P(0)$
  \reason{$F_0 = 0$, which is even.} \\
  2. (Inductive step) 
  $\forall n \geq 0 [(\forall i, 0 \leq i \leq n (P(i))) \Rightarrow
  P(n+1)]$
  \reason{???} \\
  3. QED      \reason{Strong induction}\\
}

The inductive step is supposed to prove that $F_{n+1}$ is even, using
the fact that all smaller Fibonacci numbers are even\.
The idea is to express $F_{n+1}$ as usual, as $F_{n-1} + F_n$\.
Then use the strong inductive hypothesis to say that $F_{n-1}$ and $F_n$
are both even\.
The sum is therefore also even\.
\\
What's the error?  We proved:
$P(0)$ \\
$P(0) \wedge P(1) \Rightarrow P(2)$ \\
$P(1) \wedge P(2) \Rightarrow P(3)$, etc. \\
But we never proved $P(1)$.

\subsubsection{A lower bound}

Here's a correct strong-induction proof about Fibonacci numbers\.  It
shows a lower bound on their values.
\begin{lemma}
  $\forall n \geq 2 [F_n \geq (\frac{3}{2})^{n-2}]$
\end{lemma}
\begin{proof}
  Use strong induction\.
  Let $P(n) = F_n \geq (\frac{3}{2})^{n-2}$
  \begin{structproof}
    Prove: $\forall n \geq 2 [P(n)]$\\
    1. (Base) $P(2)$
    \> \> \> \> \>  $F_2 = 1 = (\frac{3}{2})^0$ \\
    2. (Base) $P(3)$ 
    \> \> \> \> \>  $F_3 = 2 \geq (\frac{3}{2})^1$ \\
    3. (Inductive step) $\forall n \geq 3 [(\forall  i, 2 \leq i \leq n (P(i))) \Rightarrow P(n+1)]$\\
    \> 1. Fix $n \geq 3$ \\
    \> 2. Assume $\forall  i, 2 \leq i \leq n (P(i))$ \\
    \> 3. $P(n+1)$
    \> \> \> \> ??? \\
    \> 4. QED
    \> \> \> \> Implication, UG \\
    4. QED
    \> \> \> \> \> Strong induction
  \end{structproof}

  \noindent
  Now in proving the ??? we are only worrying about $n \geq 3$ and we may
  assume the inequality for all $i$ from $2$ up to $n$, inclusive\.
  Because of the way Fibonacci numbers are defined, it turns out that the
  ones we need are just $n$ and $n-1$\.
  Those are in the range for which we can assume the property. 
  \begin{structproof}
    \> 3. $P(n+1)$ \\
    \> \> 1. $F_n \geq (\frac{3}{2})^{n-2}$
    \> \> \> Inductive hypothesis 3.2 \\
    \> \> 2. $F_{n-1} \geq (\frac{3}{2})^{n-3}$
    \> \> \> Inductive hypothesis 3.2 \\
    \> \> 3. $F_{n+1} = F_n + F_{n-1}$
    \> \> \> Definition \\
    \> \> 4. $F_{n+1}\geq (\frac{3}{2})^{n-2} + (\frac{3}{2})^{n-3}$ 
    \> \> \> Plug the inequalities into the equation \\
    \> \> 5. QED
    \> \> \> Algebra \\
    \> \> \> 1. $\frac{3}{2} + 1 \geq (\frac{3}{2})^2$
    \> \> Arithmetic \\
    \> \> \> 2. $(\frac{3}{2})^{n-2} + (\frac{3}{2})^{n-3} \geq (\frac{3}{2})^{n-1}$
    \> \> Algebra \\
    \> \> \> 3. $F_{n+1} \geq (\frac{3}{2})^{n-1}$
    \> \> Algebra (3.3.4 and 3.3.5.2) \\
    \> \> \> 4. QED
    \> \> Definition of $P(n+1)$ 
  \end{structproof}
\end{proof}


\subsection{Double Induction}

Here's a trickier use of recursive definitions and induction, used
twice, in a nested way, in the same proof\.  Define a function of two
natural number variables recursively as follows: \\ 
$f(0,k) = 1$ for all $k \geq 0$. \\ 
$f(n,0) = 1$ for all $n \geq 0$.\\ 
For $n, k \geq 0,$ $f(n+1,k+1) = f(n,k+1) + f(n+1,k)$. \\ 
\\
Why does this define a function\?  We can see by looking at a table\.
The first two rules fill in the first row and column, respectively\.
For each other entry, you obtain the value by adding the upper and
left neighbors\.  It is left to the reader to see that this defines a
unique function, that is, fills in all the spaces, in exactly one way.
\[
\begin{array}{r|rrrrr}
   & 0   & 1  &  2  &  3  &  4 \\
\hline
0  & 1   & 1  &  1  &  1  &  1 \\
1  & 1   & 2  &  3  &  4  &  5 \\
2  & 1   & 3  &  6  & 10  & 15 \\
3  & 1   & 4  & 10  & 20  & 35 \\
4 & 1 & 5 & 15 & 35 & 70\\
\end{array}
\]
If you turn this rectangle on its corner, you get the famous
``Pascal's triangle.''
We will see these numbers later in the term (around lecture 16)\.  
For now, here is a formula (which we pull out of thin air), giving the 
exact value of the function on any pair of arguments:
\[
f(n,k) = \frac{(n+k)!}{n!\;
  k!}.
\]

Here we are using the factorial function defined earlier.  As an
example, 
\[
f(3,4) = \frac{7!}{3!4!} =
\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot
  1\cdot 4\cdot 3\cdot 2\cdot 1} = 35.
\]

\begin{lemma}
  $\forall n, k \in N [f(n,k) = \frac{(n+k)!}{n! k!}]$.
\end{lemma}

\noindent We can prove this by induction\.  But there are two
variables, $n$ and $k$\.  So, we will do the proof in 2 levels\.  The
main strategy is by induction on $n$\.  And for each particular $n$,
we work by induction on $k$\.  We are doing induction inside an
induction\.  Let's break up the statement to make the two levels
clearer\.  The specific statement we want to prove is:
\begin{lemma}
  $\forall n [\forall k [f(n,k) = \frac{(n+k)!}{n! k!}]]$
\end{lemma}

\noindent Define $Q(n,k)$ to be the predicate $f(n,k) =
\frac{(n+k)!}{n! k!}$\.  $Q(n,k)$ says that the value in position
$(n,k)$ is correct (satisfies the formula)\.  Define $P(n)$ to be the
statement $\forall k [Q(n,k)]$, that is, $\forall k [f(n,k) =
\frac{(n+k)!}{n!  k!}]$\.  In English, $P(n)$ means that all the
values in row $n$ are correct\.

\forstudents{
Now we use the standard induction pattern: 
\begin{structproof}
  Prove: $\forall n \geq 0 (P(n))$ \\
  1. (Base) $P(0)$ \\
  2. (Inductive step) $\forall n \geq 0 [P(n) \Rightarrow P(n+1)]$ \\
  3. QED 
  \> \> \> \> \> Induction
\end{structproof}


Expanding this pattern:
\begin{structproof}
  Prove: $\forall n \geq 0 [P(n)]$ \\
  1. (Base) $P(0)$ \\
  \> 1. $\forall k [f(0,k) = \frac{(0+k)!}{0!k!}]$
  \> \> \> \> Both sides are 1 (for any $k$) \\
  \> 2. QED 
  \> \> \> \> Definition of $P(0)$ \\
  2. (Inductive step) $\forall n \geq 0 [P(n) \Rightarrow P(n+1)]$ \\
  \> 1. Fix $n \geq 0$. \\
  \> 2. Assume $P(n)$, that is, row $n$ is correct. \\
  \> 3. $P(n+1)$, that is, row $n+1$ is correct.      
  \> \> \> \> ??? \\
  \> 4. QED    
  \> \> \> \> Implication, UG \\
  3. QED                           
  \> \> \> \> \> Induction 
\end{structproof}

Now, how to prove 3\?  It involves saying something about the whole
row $n+1$\.  
We use induction to handle all the elements in this row one at a
time\.  
Our goal 3 can be rewritten, using the definition of $P$ in terms of
$Q$\.  
Note that this is all talking just about the particular row $n+1$.

\begin{structproof}
  \> 3. $P(n+1)$ \\
  \> \> 1. $\forall k [Q(n+1,k)]$ \\
  \> \> \> 1. (Base) $Q(n+1,0)$ \\
  \> \> \> 2. (Inductive step) $\forall k \geq 0 [Q(n+1,k) \Rightarrow Q(n+1,k+1)]$. \\
  \> \> \> 3. QED  
  \> \> Induction \\
  \> \> 2. QED
  \> \> \> Definition of $P(n+1)$
\end{structproof}

\noindent
The base case for the inner induction follows by expanding the
definition of $Q(n+1,0)$:
\begin{structproof}
  \> \> \> 1. (Base) $Q(n+1,0)$ \\
  \> \> \> \> 1. $f(n+1,0) = \frac{(n+1+0)!}{(n+1)! 0!}$ 
  \> Both sides are 1 \\
  \> \> \> \> 2. QED
  \> Definition of $Q(n+1)$
\end{structproof}

\noindent
The inductive step involves a little algebra:
\begin{structproof}
  XXX\=XXX\=XXX\=XXX\=XXX\=XXXXXXXXXXXXXXXXXXXXX\=    \kill
  \> \> \> 2. (Inductive step) $\forall k \geq 0 [Q(n+1,k) \Rightarrow Q(n+1,k+1)]$. \\
  \> \> \> \> 1. Fix $k \geq 0$. \\
  \> \> \> \> 2. Assume $Q(n+1,k)$. \\
  \> \> \> \> 3. $Q(n+1,k+1)$ \\
  \> \> \> \> \> 1. $f(n+1,k+1)= f(n,k+1) + f(n+1,k)$ \\
  \> \> \> \> \> \> Definition of $f(n+1,k+1)$ \\
  \> \> \> \> \> 2. $f(n+1,k) = \frac{(n+1+k)!}{(n+1)!k!}$ 
  \> Inductive hypothesis 2.3.1.2.2, \\
  \> \> \> \> \> \> definition of $Q(n+1,k)$. \\
  \> \> \> \> \> 3. $f(n,k+1) = \frac{(n+k+1)!}{n!(k+1)!}$ 
  \> Inductive hypothesis 2.2, \\
  \> \> \> \> \> \> definition of $P(n,k+1)$. \\
  \> \> \> \> \> 4. $f(n+1,k+1)=\frac{(n+k+1)!}{n!(k+1)!}+\frac{(n+1+k)!}{(n+1)!k!}
  =\frac{(n+1+k+1)!}{(n+1)!(k+1)!}$ \\
  \> \> \> \> \> \> Combining equations, algebra \\
  \> \> \> \> \> 5. QED                   
  \> Definition of $Q(n+1,k+1)$ \\
  \> \> \> \> 4. QED
  \> \> Induction
\end{structproof}
}

The ``inner'' inductive hypothesis (the one involving $Q(n+1,k)$)
is invoked to get information about the
previous element in the same row, $n+1$\. 
The ``outer'' inductive hypothesis (the one involving $P(n,*)$),
is invoked to get information about element $k+1$ of the previous row\.
(The proof uses UI in the invocation of the outer inductive hypothesis.)

\section{Recursive structure definitions and structural induction}

So far we've been looking at recursively defined functions of the
natural numbers\.  Recursive definitions are also a useful way to
describe other types of objects (structures) that are built up from
smaller pieces\.  Structural induction is an elegant variant of
induction for proving things about these recursively defined objects\.
Instead of using a number $n$, the induction is based explicitly on
the recursive structure of the definition.

\subsection{Some recursive definitions for data types}

\begin{definition}
  The set $F$ of fully parenthesized Boolean algebra expressions satisfies: 
  \begin{enumerate}
  \item The symbols $0$ and $1$ are in $F$. 
  \item If $e$ and $e'$ are in $F$ then the expression $[e \wedge e']$ is
    in $F$. 
  \item
    If $e$ and $e'$ are in $F$ then the expression $[e \vee e']$ is
    in $F$. 
  \item
    If $e$ is in $F$ then the expression $[\neg e]$ is in $F$. 
  \item
    $F$ contains nothing else. 
  \end{enumerate}
\end{definition}


\begin{definition}
  The set $T$ of binary trees satisfies: 
  \begin{enumerate}
  \item
    A single node (just a dot) is in $T$.
  \item 
    If $t$ is in $T$, then the structure formed from $t$, a single
    new node, and an arrow labeled ``left'' from the new node to
    $t$, is also in $T$. 
  \item
    If $t$ is in $T$, then the structure formed from $t$, a single
    new node, and an arrow labeled ``right'' from the new node to
    $t$, is also in $T$. 
  \item
    If $t$ and $t'$ are in $T$, then the structure formed from $t$, $t'$,
    and a single new node, with an arrow labeled ``left'' from the new
    node to $t$ and another arrow labeled ``right'' from the new node to
    $t'$, is also in $T$. 
  \end{enumerate}
\end{definition}
Since the descriptions above are wordy, it's nice to introduce some
notation, like makeleft$(t)$ to denote the thing built up from $t$
according to the first recursive rule\.  Likewise, makeright$(t)$,
makeboth$(t,t')$.

\forlecture{ (diagram)\\ }

Here is a definition that we already used informally:
\begin{definition}
  The set $S$ of (finite length) strings over alphabet $A$ satisfies: 
  \begin{enumerate}
  \item  $\lambda$, the empty string, is in $S$. 
  \item  If $s$ is in the set $S$ and $a \in A$, then $sa$ is in $S$. 
  \item $S$ contains nothing else. 
  \end{enumerate}
\end{definition}
This is a recursive description (definition) of the set of strings\. 
Notice that it includes every (finite length) string\. 
Moreover, every string has a unique derivation using these rules (this
isn't strictly necessary, but is often convenient in proofs).  

Finally, here is a definition of lists for 6.001:
\begin{definition}
  The set $L$ of lists over atom set $A$ satisfies:
  \begin{enumerate}
  \item $()$, the empty list, is in $L$
  \item If $\ell \in L$ and $a \in A$, then $(a\ . \ \ell)$ is in $L$
  \item $L$ contains nothing else
  \end{enumerate}
\end{definition}
Note the close similarity to the definition of strings.

\subsection{Structural induction on recursive data type definitions}

Structural induction is used to prove a property $P$ of all the elements
of some recursively-defined data type\. 

\subsubsection{The Method}

The proof consists of two steps: 
\begin{itemize}
\item Prove for the ``base cases'' of the definition. 
\item Prove for the result of any combination rule, assuming that it
  is true for all the parts.  
\end{itemize}

Structural induction looks like a new inference rule, but it isn't: it
is standard (strong) induction on the \emph{number of formation steps}
it takes to build the object at hand out of the basic objects of the
recursive definition.


For Boolean expressions:
                                %[[[Victor: slide]]]
\begin{structproof}
  Prove: $\forall e \in F (P(e))$ \\
  1. (Base) $P(0)$ \\
  2. (Base) $P(1)$ \\
  3. (Inductive step) 
  $\forall e, e' \in F (P(e) \wedge P(e') \Rightarrow P([e \wedge e']))$\\
  4. (Inductive step) 
  $\forall e, e' \in F (P(e) \wedge P(e') \Rightarrow P([e \vee e']))$ \\
  5. (Inductive step) 
  $\forall e \in F (P(e) \Rightarrow P([\neg e]))$\\
  6. QED
  \> \> \> \> \> Structural induction on Boolean expressions
\end{structproof}
(Note that we are being careful to distinguish parentheses that are
art of the boolean expression (represented by brackets) from
parenthesis used in our proof formulas (represented by standard
parentheses).

For binary trees:
                                %[[[Victor: slide]]]
\begin{structproof}
  Prove: $\forall t \in T (P(t))$ \\
  1. (Base) $P(n)$, where $n$ is the single-node tree. \\
  2. (Inductive step) 
  $\forall t \in T (P(t) \Rightarrow P(\ms{makeleft}(t)))$ \\
  3. (Inductive step) 
  $\forall t \in T (P(t) \Rightarrow P(\ms{makeright}(t)))$ \\
  4. (Inductive step) 
  $\forall t,t' \in T (P(t) \wedge P(t') \Rightarrow P(\ms{makeboth}(t,t')))$ \\
  5. QED
  \> \> \> \> \> Structural induction on binary trees
\end{structproof}

For strings over alphabet $A = \{ 0,1 \}$:
                                %[[[Victor: slide]]]
\begin{structproof}
  Prove: $\forall s \in S (P(s))$ \\
  1. (Base) $P(\lambda)$ \\
  2. (Inductive step) $\forall s \in S (P(s) \Rightarrow P(s 0))$ \\
  3. (Inductive step) $\forall s \in S (P(s) \Rightarrow P(s 1))$ \\
  4. QED
  \> \> \> \> \> Structural induction on strings.
\end{structproof}
You can see how this would extend to different alphabets.

\subsubsection{Examples}

Here's example for each of the data types: 

\begin{theorem}
  Every Boolean expression has the same number of left and right parentheses. 
\end{theorem}
\begin{proof}
  Define $P(e)$: Expression $e$ has the same number of left and right
  parentheses.
                                %[[[Victor: slide]]]
  \begin{structproof}
    Prove: $\forall e \in F (P(e))$ \\
    1. (Base) $P(0)$ 
    \> \> \> \> \> The expression $0$ has no parens. \\
    2. (Base) $P(1)$ 
    \> \> \> \> \> No parens \\
    3. (Inductive step) 
    $\forall e, e' \in F (P(e) \wedge P(e') \Rightarrow P([e \wedge e'])$\\
    \> 1. Fix $e, e' \in F$. \\
    \> 2. Assume $P(e) \wedge P(e')$, that is, each of $e$ and $e'$ has the
    same number \\
    \> \quad of left and right parens. \\
    \> 3. $P([e \wedge e'])$, that is, the expression $e \wedge e'$ has the
    same number of \\
    \> \quad left and right parens. \\
    \> \> \> \> \> By inductive hypothesis (3.2), and the fact \\ 
    \> \> \> \> \> that we're adding one left and one right \\
    \> \> \> \> \> paren. \\
    \> 4. QED 
    \> \> \> \> Implication, UG \\
    4. (Inductive step) 
    $\forall e, e' \in F (P(e) \wedge P(e') \Rightarrow P([e \vee e'])$
    \\
    \> \> \> \> \> Similar to the previous case. \\
    5. (Inductive step) 
    $\forall e \in F (P(e) \Rightarrow P([\neg e])$
    \> \> \> \> \> Similar to the previous case. \\
    6. QED
    \> \> \> \> \> Structural induction on Boolean expressions
  \end{structproof}
\end{proof}

\begin{theorem}
  The number of edges in any binary tree is exactly one fewer than the
  number of nodes. 
\end{theorem}
\begin{proof}
  Define $P(t)$: Binary tree $t$ has exactly $1$ fewer edges than nodes\.
  We are using the notation $\ms{nodes}(t)$, $\ms{edges}(t)$\. 
  The key is that the inductive steps all preserve the ``deficit'' of
  1\. 
  For instance, combining two trees involves adding two edges and one
  node; since each subtree had a ``deficit'' of 1, this maintains the
  deficit of 1:
                                %[[[Victor: slide]]]
  \begin{structproof}
    Prove: $\forall t \in T (P(t))$ \\
    1. (Base) $P(n)$, where $n$ is the single-node tree. 
    \> \> \> \> \> $0$ edges and $1$ node \\
    2. (Inductive step) 
    $\forall t \in T (P(t) \Rightarrow P(\ms{makeleft}(t))$ \\
    \> 1. Fix $t$. \\
    \> 2. Assume $P(t)$, that is, $|\ms{edges}(t)|+1 = |\ms{nodes}(t)|$. \\
    \> 3. $P(\ms{makeleft}(t))$, that is,
    $|\ms{edges}(\ms{makeleft}(t))|+1 = |\ms{nodes}(\ms{makeleft}(t))|$. \\
    \> \> 1. $|\ms{edges}(\ms{makeleft}(t))| = |\ms{edges}(t)| + 1$ \\
    \> \> 2. $|\ms{nodes}(\ms{makeleft}(t))| = |\ms{nodes}(t)| + 1$. \\
    \> \> 3. QED
    \> \> \> Algebra, using 2.2, 2.3.1, 2.3.2 \\
    \> 4. QED
    \> \> \> \> Implication, UG \\
    3. (Inductive step) 
    $\forall t \in T (P(t) \Rightarrow P(\ms{makeright}(t))$ 
    \> \> \> \> \> Similar to the previous case. \\
    4. (Inductive step) 
    $\forall t, t' \in T (P(t) \wedge P(t') \Rightarrow P(\ms{makeboth}(t,t'))$ \\
    \> 1. Fix $t$. \\
    \> 2. Assume $P(t) \wedge P(t')$, that is, $|\ms{edges}(t)|+1 =
    |\ms{nodes}(t)|$ and $|\ms{edges}(t')|+1 =|\ms{nodes}(t')|$. \\
    \> 3. $P(\ms{makeboth}(t,t'))$, that is,
    $|\ms{edges}(\ms{makeboth}(t,t'))|+1 = |\ms{nodes}(\ms{makeboth}(t))|$. \\
    \> \> 1. $|\ms{edges}(\ms{makeboth}(t,t'))| = |\ms{edges}(t)| +
    |\ms{edges}(t')| + 2$ \\
    \> \> 2. $|\ms{nodes}(\ms{makeboth}(t,t'))| = |\ms{nodes}(t)| +
    |\ms{nodes}(t')| + 1$. \\
    \> \> 3. QED
    \> \> \> Algebra, using 4.2, 4.3.1, and 4.3.2 \\
    \> 4. QED
    \> \> \> \> Implication, UG \\
    5. QED
    \> \> \> \> \> Structural induction on binary trees
  \end{structproof}
\end{proof}


\subsubsection{A string Theorem}

Here is a more complex proof, illustrating a combination of structural
induction and strengthening the hypothesis.

\forstudents{
\begin{theorem}
  In a string of $0$s and $1$s, the number of occurrences of the pattern
  $01$ is less than or equal to the number of occurrences of $10$, plus
  one.
\end{theorem}

Let's try to prove this by structural induction\.  First we must
define $P(s)$\.  Let's write $\ms{num}(pat,s)$ as the number of
occurrences of the pattern string pat in s\.  Now our inductive
hypothesis is
\[
P(s): \ms{num}(01,s) \leq \ms{num}(10,s) + 1. 
\]
If you try to prove this by structural induction, you will get
stuck\.
Why\? 
Consider what happens when you add $1$ at the end\.  
This could increase the number of $01$s without increasing the number of
$10$s\. 

So, to prove by structural induction on strings, let's strengthen the
hypothesis by adding another clause\.  If a string ends in $0$ then
the number of $01$s is less than or equal to the number of $10$s\.
That solves the problem by weakening what we have to show when the
string ends in $1$\.  But maybe it causes another problem somewhere
else\.  Let's give it a try: \\ 
\\
Redefine $P(s)$: \\
$\ms{num}(01,s) \leq \ms{num}(10,s) + 1$, \\
and \\
If $s$ ends in $0$ then $\ms{num}(01,s) \leq \ms{num}(10,s)$. \\
\\
This means that, for each inductive step, we have two things to show:
                                %[[[Victor: slide]]]
\structuredproof{
  Prove: $\forall s \in S (P(s))$ \\
  1. (Base) $P(\lambda)$ 
  \> \> \> \> \> \reason{No patterns of either kind.} \\
  2. (Inductive step) $\forall s \in S (P(s) \Rightarrow P(s 0))$ \\
  \> 1. Fix $s$. \\
  \> 2. Assume $P(s)$. \\
  \> 3. $P(s 0)$ \\
  \> \> 1. $\ms{num}(01,s0) \leq \ms{num}(10,s0) + 1$. 
  \> \> \> \reason{???} \\
  \> \> 2. If $s 0$ ends in $0$ then $\ms{num}(01,s0) \leq
  \ms{num}(10,s0)$. \noreason \\
  \> \> \> \> \> \reason{???} \\
  \> \> 3. QED
  \> \> \> \reason{Conjunction} \\
  \> 4. QED 
  \> \> \> \> \reason{Implication, UG} \\
  3. (Inductive step) $\forall s \in S (P(s) \Rightarrow P(s 1))$ \\
  \> 1. Fix $s$. \\
  \> 2. Assume $P(s)$. \\
  \> 3. $P(s 1)$ \\
  \> \> 1. $\ms{num}(01,s1) \leq \ms{num}(10,s1) + 1$. 
  \> \> \> \reason{???} \\
  \> \> 2. If $s 1$ ends in $0$ then $\ms{num}(01,s1) \leq
  \ms{num}(10,s1)$.\noreason \\
  \> \> \> \> \> \reason{???} \\
  \> \> 3. QED
  \> \> \> \reason{Conjunction} \\
  \> 4. QED 
  \> \> \> \> \reason{Implication, UG} \\
  4. QED
  \> \> \> \> \> \reason{Structural induction on strings.}\\
}

First let's consider $s1$\. 
This is the case that looks dangerous, because it might increase the
number of $01$s\. 
We have to prove two statements\. 
The second is easy, because the new string doesn't end in $0$\. 
We say it's ``vacuously true''\. 

The first statement now takes some work\. 
We might be adding to the number of $01$s\.  
However, if we do, the previous string must have ended with $0$\. 
Then the inductive hypothesis says that the previous string had to
satisfy the stronger inequality in the second statement\. 
Adding one to the LHS of the stronger inequality yields the weaker
inequality we want\.

The following proof fragment considers cases based on whether $s$ ends
in $0$ or not\. 
If not, it might end in $1$, or might be empty (don't forget this
possibility). 

\begin{structproof}
  \> 3. $P(s 1)$ \\
  \> \> 1. $\ms{num}(01,s1) \leq \ms{num}(10,s1) + 1$. \\
  \> \> \> 1. If $s$ ends in $0$ then $\ms{num}(01,s1) \leq\ms{num}(10,s1) + 1$. \\
  \> \> \> \> 1. Assume $s$ ends in $0$. \\
  \> \> \> \> 2. $\ms{num}(01,s) \leq \ms{num}(10,s)$ 
  \> Inductive hypothesis (3.2), part 2. \\
  \> \> \> \> 3. $\ms{num}(01,s1) = \ms{num}(01,s) + 1$ 
  \> Adding one more $01$. \\
  \> \> \> \> 4. $\ms{num}(10,s1) = \ms{num}(10,s)$ \\
  \> \> \> \> 5. $\ms{num}(01,s1) \leq\ms{num}(10,s1) + 1$. 
  \> Algebra (combining 3.3.1.1.2, ...3, and ...4) \\
  \> \> \> \> QED 
  \> Implication \\
  \> \> \> 2. If $s$ ends in $1$ then $\ms{num}(01,s1)\leq\ms{num}(10,s1) + 1$. \\
  \> \> \> \> \> Inductive hypothesis, part 1; no new $01$s. \\
  \> \> \> 3. If $s = \lambda$ then $\ms{num}(01,s1) \leq\ms{num}(10,s1)+ 1$. \\
  \> \> \> \> \> $s1$ is just $1$, which has no $01$s. \\
  \> \> \> 4. QED 
  \> \> Cases \\
  \> \> 2. If $s 1$ ends in $0$ then $\ms{num}(01,s1) \leq \ms{num}(10,s1)$. \\
  \> \> \> \> \> Vacuously true, because $s1$ doesn't end in $0$. \\
  \> \> 3. QED
  \> \> \> Conjunction \\
\end{structproof}
Of course, you could also expand the step for $s$ ending in $1$ into a
careful series of inequalities. \\
\\
Now consider $s0$\.
We hope that what we did to make the $s1$ case work doesn't mess up
the $s0$ case\. 
But we have to check\. 

The first statement is easy\. 
It follows from the first statement of the inductive
hypothesis for $s$, because we are not increasing the number of $01$s\. 
But now the second statement takes more work\.
The difficulty is that the new string ends in $0$, which means that we
have to show the stronger inequality in the second statement\.
But to do this, we might only have the weaker inequality for the previous
string\.
The argument again depends on what the previous string $s$ ended with\.
So again, we consider cases, based on whether $s$ ends in $0$ or $1$,
or is empty\.
If $s$ ends in $0$ we rely on the second statement of the inductive
hypothesis for $s$ (with the stronger inequality), whereas if $s$ ends
in $1$ we rely on the first statement (with the weaker inequality)\.
In this case, we have to ``turn the weaker inequality into the
stronger inequality''.

\structuredproof[38ex]{
  \> 3. $P(s 0)$ \\
  \> \> 1. $\ms{num}(01,s0) \leq \ms{num}(10,s0) + 1$.
  \reason{Induct. hypothesis, part 1; no new $01$s.} \\
  \> \> 2. If $s 0$ ends in $0$ then $\ms{num}(01,s0) \leq\ms{num}(10,s0)$. \\
  \> \> \> 1. $\ms{num}(01,s0) \leq \ms{num}(10,s0)$. \\
  \> \> \> \> 1. If $s$ ends in $0$ then $\ms{num}(01,s0)
  \leq\ms{num}(10,s0)$. \noreason \\
  \reason{Induct. hypothesis, part 2; no new $01$s} \\
  \> \> \> \> 2. If $s$ ends in $1$ then $\ms{num}(01,s0)
  \leq\ms{num}(10,s0)$. \noreason \\ 
  \> \> \> \> \> 1. Assume $s$ ends in $1$. \\
  \> \> \> \> \> 2. $\ms{num}(01,s) \leq \ms{num}(10,s) + 1$. 
  \reason{Inductive hypothesis, part 1} \\
  \> \> \> \> \> 3. $\ms{num}(01,s0) = \ms{num}(01,s)$ \\
  \> \> \> \> \> 4. $\ms{num}(10,s0) = \ms{num}(10,s) + 1$ 
  \reason{Exactly one new occurrence of $10$.} \\
  \> \> \> \> \> 5. $\ms{num}(01,s0) \leq \ms{num}(10,s0)$ 
  \reason{Algebra} \\
  \> \> \> \> \> 6. QED
  \reason{Implication} \\
  \> \> \> \> 3. If $s = \lambda$ then $\ms{num}(01,s0) \leq
  \ms{num}(10,s0)$. \noreason \\ 
  \> \> \> \> \> \> 
\reason{$s0$ is just $1$, which has no $01$s or $10$s.} \\
  \> \> \> \> 4. QED 
  \> \> \reason{Cases} \\
  \> \> \> 2. QED
  \> \> \reason{Propositional reasoning (truth table)} \\
  \> \> 3. QED 
  \> \> \> \reason{Conjunction}\\
}

If you actually write out all these cases in the proof, you will
notice that some facts are stated repeatedly, e.g., that when you add
a $0$ to the end of a string you are not increasing the number of $01$s. \\
To avoid having to state these facts several times, you can move them
earlier in the proof\.
}

\subsection{Recursively-defined functions on recursively-defined data
  types}

Recursive definitions provide a natural way to define functions whose
domains are recursively-defined data types\.
\\
\begin{example}(Boolean Expressions)
  Define $\ms{eval}$, a function from Boolean expressions to $\{ 0, 1
  \}$, by: 
\begin{quote}
  $\ms{eval}(0) = 0$ \\
  $\ms{eval}(1) = 1$ \\
  $\ms{eval}([e \wedge e']) = 1$ if $\ms{eval}(e) = \ms{eval}(e') = 1$, $0$ otherwise. \\
  $\ms{eval}([e \vee e']) = 0$ if $\ms{eval}(e) = \ms{eval}(e') = 0$, $1$ otherwise. \\
  $\ms{eval}([\neg e]) = 0$ if $\ms{eval}(e) = 1$, $1$ otherwise. 
\end{quote}
\end{example}

\begin{example} (Strings)
For strings over alphabet $A$, define the function $\ms{length}(s)$ as
follows:
\begin{quote}
$\ms{length(\lambda)}=0$ (the empty string)\\
$\ms{length}(sa) = \ms{length}(s)+1$ for string $s$ and $a\in A$.\\
\end{quote}
\end{example}

\begin{example} (Binary Trees)
  Define the function $\ms{numnodes}(t)$ (a formal definition of what
  we previously used intuitively to be the number of nodes in the tree
  $t$), by: 
\begin{quote}
  $\ms{numnodes}(\mbox{single node}) = 1$ \\
  $\ms{numnodes}(\ms{makeleft}(t)) = \ms{numnodes}(t) + 1$ \\
  $\ms{numnodes}(\ms{makeright}(t)) = \ms{numnodes}(t) + 1$ \\
  $\ms{numnodes}(\ms{makeboth}(t,t')) = \ms{numnodes}(t)
  +\ms{numnodes}(t') + 1$ 
\end{quote}

Similarly,  define $\ms{numedges}(t)$: 
\begin{quote}
  $\ms{numedges}(\mbox{single node}) = 0$ \\
  $\ms{numedges}(\ms{makeleft}(t)) = \ms{numedges}(t) + 1$ \\
  $\ms{numedges}(\ms{makeright}(t)) = \ms{numedges}(t) + 1$ \\
  $\ms{numedges}(\ms{makeboth}(t)) = \ms{numedges}(t) +
  \ms{numedges}(t') + 2$ 
\end{quote}
\end{example}

We can use structural induction to prove properties of functions
defined in this way\.
For example, we can reprove the relationship between numbers of nodes
and edges in terms of the formal functions $\ms{numnodes}$ and
$\ms{numedges}$\.
(The argument is pretty much the same as before, just using the formal
function definitions instead of relying on our understanding of the
sizes of sets.) \\

\forstudents{
Define $P(t)$: 
$\ms{numnodes}(t) = \ms{numedges}(t) + 1$. 
                                %[[[Victor: slide]]]
\begin{structproof}
  Prove: $\forall t \in T (P(t))$ \\
  1. (Base) $P(t_0)$, where $t_0$ is the single-node tree. 
  \> \> \> \> \> By definition, $\ms{numnodes}(t_0) = 1$ \\
  \> \> \> \> \> and $\ms{numedges}(t_0) = 0$ \\
  2. (Inductive step) 
  $\forall t \in T (P(t) \Rightarrow P(\ms{makeleft}(t))$ \\
  \> 1. Fix $t$. \\
  \> 2. Assume $P(t)$, that is, $\ms{numedges}(t)+1 = \ms{numnodes}(t)$. \\
  \> 3. $P(\ms{makeleft}(t))$, that is,
  $\ms{numedges}(\ms{makeleft}(t))+1 = \ms{numnodes}(\ms{makeleft}(t))$. \\
  \> \> 1. $\ms{numedges}(\ms{makeleft}(t)) = \ms{numedges}(t) + 1$ \\
  \> \> \> \> \> Definition of $\ms{numedges}$ \\
  \> \> 2. $\ms{numnodes}(\ms{makeleft}(t)) = \ms{numnodes}(t) + 1$ \\
  \> \> \> \> \> Definition of $\ms{numnodes}$ \\
  \> \> 3. QED
  \> \> \> Algebra \\
  \> 4. QED \\
  3. (Inductive step) 
  $\forall t \in T (P(t) \Rightarrow P(\ms{makeright}(t))$ 
  \> \> \> \> \> Similar to the previous case. \\
  4. (Inductive step) 
  $\forall t, t' \in T (P(t) \wedge P(t') \Rightarrow P(\ms{makeboth}(t,t'))$ \\
  \> 1. Fix $t$. \\
  \> 2. Assume $P(t) \wedge P(t')$, that is, \\
  \> \quad $\ms{numedges}(t)+1 = \ms{numnodes}(t)$ and $\ms{numedges}(t')+1 =\ms{numnodes}(t')$. \\
  \> 3. $P(\ms{makeboth}(t,t'))$, that is,
  $\ms{numedges}(\ms{makeboth}(t,t'))+1 = \ms{numnodes}(\ms{makeboth}(t))$. \\
  \> \> 1. $\ms{numedges}(\ms{makeboth}(t,t')) = \ms{numedges}(t) +
  \ms{numedges}(t') + 2$ \\
  \> \> \> \> \> Definition of $\ms{numedges}$ \\
  \> \> 2. $\ms{numnodes}(\ms{makeboth}(t,t')) = \ms{numnodes}(t) +
  \ms{numnodes}(t') + 1$ \\
  \> \> \> \> \> Definition of $\ms{numnodes}$ \\
  \> \> 3. QED
  \> \> \> Algebra \\
  \> 4. QED \\
  5. QED
  \> \> \> \> \> Structural induction on binary trees
\end{structproof}
}

\forstudents{ We've connected structural induction to recursive
  definitions.  In a sense, this connection also holds for the natural
  numbers.  One can think of the natural numbers as a recursively
  defined set ($0\in\naturals$, and if $n\in\naturals$ then $n+1 \in
  \naturals$), and of normal induction as a structural induction on
  this recursive definition. }

\section{Induction and Recursive Algorithms}

We've spent a lot of time studying induction\.
This is because induction comes up all the time in analyzing
computation\.
Why?  Well, induction is a ``one step at a time'' proof method.
Computations also evolve ``one step at a time.''

In fact, many of our inductive proofs can be seen as \emph{algorithms}
for solving a problem\.  And many of our recursive definitions can be
seen as algorithms for computing a value.

We proved ``any $2^n$ by $2^n$ board can be tiled with Ls''\.  If you
look at the inductive proof, we actually gave a \emph{procedure} for
carrying out that tiling\.  We break the board in four, tile each
piece, and merge the tilings\.  What kind of algorithm is this? It's
\emph{recursive:} it calls itself on smaller subproblems.
Similarly the inductive proof for Nim gave a game-winning algorithm\.

The recursive definitions of the Fibonacci numbers is also expressing
a recursive algorithm for computing them\.  The same holds for the
factorial function, multiplication, etc.

Just about any inductive proof that a problem has a solution is in
fact providing a recursive algorithm for solving the problem\.

Similarly, when you have a recursive algorithm that you think solves
some problem, induction is the usual way to prove that it works.

\subsection{RSA}

RSA is at the core of many of the secure communication protocols we
use today (SSL, etc)\.  RSA works by treating any given message
(sequence of bits) as a big number $x$ and applying a mathematical
function that ``hides'' the number $x$ unless you know a secret key.

More precisely, it computes the remainder $r$ of $x^a$ on division by
$b$, where $a$ and $b$ are two specially constructed \emph{large}
numbers\.  It turns out that if someone is given $r$ and knows the
right secret about the origin of $a$ and $b$, they can ``invert'' the
exponentiation and get $x$\.  But if you don't have the secret, the
inversion seems to be very hard.

So, how can we compute the exponentiation?\\
The most obvious way is to work from the \emph{recursive definition}:
\begin{itemize}
\item $x^0=1$
\item $x^{n}=x\cdot x^{n-1}$
\end{itemize}
This is a definition of $x^n$, but also suggests a recursive algorithm
for computing it:
\begin{structproof}
  Procedure $\ms{Exp}(x,n)$\\
  \\
  {\bf if} $n=0$\\
  \> return 1\\
  {\bf else}\\
  \> return $x*\ms{Exp}(x,n-1)$
\end{structproof}
Now it's pretty ``obvious'', but how would you \emph{prove} that
algorithm $\ms{Exp}$ returns the right answer?\\
Well, it doesn't always: what happens if you call $\ms{Exp}(5,-2)$?
$\ms{Exp}(8,2.5)$?\\ 
(Trying to prove correctness can reveal implicit assumptions.  We
assume $n$ is a natural number).

\begin{theorem}
  For all $x,n \in \nats$, a call to $\ms{Exp}(x,n)$ returns $x^n$
\end{theorem}
Note: the procedure $\ms{Exp}$ above can be thought of as (i) a
computational process or (ii) a recursive function definition (defined
as the output of the computational process)\.  We need to prove that it
defines the \emph{same} function as $x^n$.

We are actually claiming two distinct things in the theorem:
\begin{description}
\item[Termination.] The algorithm returns something (i.e., doesn't run
  forever). 
\item[Correctness.] The algorithm returns the \emph{right} something.
\end{description}
\begin{proof}
  By induction on $n$\.  The base case, that $\ms{Exp}(x,0)$ returns
  $x^0=1$, follows by inspection: on that call, the ``if'' condition is
  satisfied, so we return 1\.  For the inductive step, assume
  $\ms{Exp}(x,n)$ returns $x^n$ and prove correctness for $n+1$\.
  Consider the call 
  to $\ms{Exp}(x,n+1)$\.  $n+1>0$, so the ``else'' clause is invoked\.  We
  call $\ms{Exp}(x,n)$ which, by assumption, returns $x^n$\.  We multiply
  that quantity by $x$, yielding $x^{n+1}$, and return it\.  This proves
  that $\ms{Exp}(x,n+1)$ terminates and returns the right value.
\end{proof}

OK, we've shown the procedure returns the right answer\.
Is it a good implementation?
How many multiplications are done to compute $x^n$?  By induction,
there are $n$ multiplications\.
Remember those ``128 bit keys'' we weren't allowed to export to
foreign countries?  That
means the number $n$ has about 128 bits\.
So we will be doing about $2^{128}$ multiplications\.
We'll still be working on it when 6.042 finishes...

\subsection{Fast Exponentiation}

Let's cut down on the number of multiplies\.  Suppose we want to
compute $x^a$ and $a$ is even, say $2c$\.  So we want to compute
$x^a=x^{2c}$\.  Is there any obvious way to beat $a$ multiplies?
Yes\. Note
$x^{2c}=(x^c)^2$\.  So let's compute $x^c$ ($c$ multiplies) and then
square it (1 multiply)\. This totals $c+1 =1+ a/2$ multiplies.

What if $a$ is odd? 

Let's generalize this idea to an algorithm:

\begin{proofoverhead}
  Procedure $\ms{FastExp}(x,n)$\\
  \\
  {\bf if} $n=0$\\
  \> return 1\\
  {\bf else}\\
  \> {\bf if} $n$ is even\\
  \> \> let $y=\ms{FastExp}(x,n/2)$\\
  \> \> return $y^2$\\
  \> {\bf else}\\
  \> \> let $y=\ms{FastExp}(x,(n-1)/2)$\\
  \> \> return $x*y^2$
\end{proofoverhead}

\begin{theorem}
  For all $x,n \in \nats$, a call to $\ms{FastExp}(x,n)$ returns $x^n$
\end{theorem}
\begin{proof}
  We'll use induction again.  The base case ($n=0$) follows by
  inspection as before\.  So let's prove the inductive step.

  What kinds of subproblems are we going to use to prove it works for
  $n$?  {\bf Not} just $n-1$, so let's use strong induction.

  Since it keeps the notation simpler, we will prove $P(n)$ from
  smaller cases instead of $P(n+1)$\.  Suppose the inductive
  hypothesis is true for every $k < n$, let's prove it for $n$\.
  Suppose we call $\ms{FastExp}(x,n)$\.  We can assume $n\ne 0$ (we
  already did a proof for 0) so the {\bf else} clause is selected in
  the algorithm\.  We consider two cases.

  If $n$ is even (say $n=2m$) then we  call
  $\ms{FastExp}(x,m)$\.   By strong induction this returns $x^m$, which
  we assign to $y$\.
  Then we return $y=(x^m)^2=x^{n}$ as required, so the proof works for
  this case.

  If $n$ is odd (say $n=2m+1$) then we call   $\ms{FastExp}(x,m)$ and
  set $y$ to the result\.  
  Then we return $x*y^2=x*(x^m)^2=x^{2m+1}=x^n$\.
  So the proof works in this case too\.
  These are all the cases.  QED.
\end{proof}

How many multiplications does {\bf this} algorithm do?

\begin{theorem}
  If $n$ has $b$ bits, then $\ms{FastExp}(x,n)$ performs at most $2b$
  multiplications.  
\end{theorem}
\begin{proof}
  Induction on $b$.\\
  Base case:  If $b=0$, then $n=0$ and we return without doing
  any multiplications.  So the base case is true.
  Inductive step: assuming it true for $b$ bits, consider a $b+1$-bit
  number $n$.\\
  We execute a recursive call to   $\ms{FastExp}$ on either $n/2$
  or $(n-1)/2$, depending whether $n$ is odd or even.\\
  Both of these numbers have at most $b$ bits.\\
  So by induction, the recursion does at most $2b$ multiplications.\\
  Then we do either one more (even case) or two more (odd case). So at
  most two more.\\
  Overall, at most $2b+2 = 2(b+1)$ multiplications.  QED.
\end{proof}

Using this algorithm, a 128-bit key means only 256 multiplications:
quite an improvement!


\subsection{Greatest Common Divisor}

\begin{definition}
Given 2 positive natural numbers, $a, b \in N^+$, their \emph{greatest
  common divisor (GCD)} is the largest number that divides both
without remainder.
\end{definition}

(Recall the notation: $a \mid b$ means $a$ divides $b$.)

Finding the  GCD is a problem that must be solved to generate the
``secret'' used to decrypt RSA messages.

One of the first recorded algorithms---Euclid's algorithm---finds
greatest common divisors\.  Here is a key lemma for ``simplifying''
the problem:

\begin{lemma}
  Given $x>y$, let $r=x-y$.  Then any common divisor of $x$ and $y$ is
  also a common divisor of $y$ and $r$, and vice versa.
\end{lemma}
\begin{corollary}
  GCD$(x,y)=$GCD$(y,r)$
\end{corollary}
\begin{proof} (of the corollary).  The pair s$(x,y)$ and $(y,r)$ have exactly
  the same set of divisors (by lemma).  Thus, the biggest number in
  each set, namely the greatest common divisor, is the same.
\end{proof}
\begin{proof} (of the lemma)\.  We prove one direction: that any
  divisor of $x$ and $y$ also divides $r$\.  Suppose $d \mid x$ and $d
  \mid y$\.  So $x=zd$ and $y=wd$\.  Thus $r=x-y=d(z-w)$ so $d \mid r$,
  meaning $d$ is a common divisor of $y$ (by assumption) and $r$ (by
  deduction)\.  The other direction of proof is similar.
\end{proof}

From this lemma we get Euclid's algorithm:
\begin{structproof}
  Procedure $\ms{Euclid}(x,y)$\\
  \\
  {\bf if} $x=0$ return $y$\\
  {\bf else if} $y=0$ return $x$\\
  {\bf else if} $x > y$ return $\ms{Euclid}(y,x-y)$\\
  {\bf else} return $\ms{Euclid}(x,y-x))$
\end{structproof}

Example: \\
Find $\ms{GCD}(112, 84)$ = ? \\
Step 1: $112 - 84 = 28$, reduce to $(28,84)$. \\
Step 2: $84 - 28 = 56$, reduce to $(28,56)$. \\
Step 3: $56 - 28 = 28$, reduce to $(28,28)$. \\
Step 4: $28 - 28 = 0$, reduce to $28,0$\\
Now done, answer is $28$. \\
\\

Let's prove this algorithm does the right thing\.
First faint worry: maybe we've got an infinite recursion?
The first thing to prove for any algorithm: it \emph{terminates}

\begin{lemma}
  For any natural number inputs $x$ and $y$, $\ms{Euclid}(x,y)$ returns
  a value.
\end{lemma}
Why do we think this is true intuitively?  Because the arguments
``shrink'' on each recursive call and eventually reach 0.  Induction
lets us formalize this.
\begin{proof}
  Induction on $x+y$\.  More formally, we prove by strong induction on
  $n$ that for all inputs $x$ and $y$ such that $x+y=n$, the algorithm
  terminates.

  Base case: if $x+y=0$, then $x$ is 0 so we return  immediately.

  Inductive step.  Assume the claim for $x+y<n$ and prove it for
  $x+y=n$\.
  Now consider two cases\.  Either $x \ge y$ or $y \ge x$.  

  Suppose first that $x \ge y$. So consider two subcases\.  If $y=0$,
  then we return immediately and the inductive step is proved\.  If $y
  > 0$, then we make a recursive call with arguments $y$ and $x-y <
  x-0 =x$\. So the sum of the two new arguments is less than
  $x+y=n$\. So the inductive hypothesis holds: the recursive call
  returns a value\. So our original call returns (the same) value.

  What about the second case ($y \ge x$)?  Well, I would have to write
  the same proof all over again, swapping $x$ and $y$ everywhere\.  To
  avoid such tedium, we say ``without loss of generality, assume $x
  \ge y$\.  This is code for ``the other case works exactly the same
  way'' and completes the proof.
\end{proof}

\begin{lemma}
  For any natural number inputs $x$ and $y$, $\ms{Euclid}(x,y)$
  returns GCD$(x,y)$.
\end{lemma}
\begin{proof}
  We'll use induction on $x+y$ as before.  The base case of $x+y=0$ is
  trivial (assuming we define GCD$(0,0)$ to be 0, which seems fair).

  For the inductive step where $x+y= n > 0$, assume without loss of
  generality that $x \ge y$\.  If $y=0$, we return $x$, which is the
  correct answer\.  If $y>0$, then we return the result of a recursive
  call on $\ms{Euclid}(y,x-y)$\.  This means (by strong induction) that
  we return GCD$(y,x-y)$\.  But we proved in a lemma above that
  GCD$(x,y)=$GCD$(y,x-y)$\. So we are returning the right value, namely
  GCD$(x,y)$.
\end{proof}

One might be concerned about the GCD$(0,0)$ definition\.  Technically
every number is a common divisor of 0 and 0, so it should be infinite?

But this turns out not to matter\.  We never actually make a recursive
call to $(0,0)$---as soon as one argument is 0, we terminate.  

So even though we still need a base case for the induction pattern, we
never actually apply that base case\.  Instead our real base case is
implicit in the ``subcase'' of the inductive step where one argument
is 0.

This is an example of a hack to the definition or algorithm that makes
the proof cleaner\.  If we wanted to avoid considering GCD$(0,0)$, we
could prove by induction that any call to $\ms{Euclid}(x,y)$ that
\emph{does not have both arguments equal to 0} returns the correct
value\.  However, we would then have to waste time on our induction
step checking that this extra condition was still satisfied in the
recursion\.  By making an arbitrary, unimportant definition of
GCD$(0,0)$, we save ourselves some work.

As described, Euclid's algorithm may perform as many as $x$
substractions (when $y=1$)\.  A small variation on our
algorithm---using remainder on division by $y$ instead of subtracting
$y$---can dramatically speed up the algorithm\.  It turns out that
with this change, only $2\log x$ divisions are needed to solve the
problem.

\end{document}
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