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\lecture{notes5}{Notes 5}{February 15, 2000}

\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\?
Because 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.  
\begin{itemize}
\item 
We proved any $2^n$ by $2^n$ board can be tiled with L shapes\.
If you look at the inductive proof, we actually gave a
  \emph{procedure} for carrying out that tiling\.
Break the board in four, tile each piece, and merge the tilings\.
What kind of algorithm is this? \emph{recursive:} it calls
 itself on smaller subproblems.
\item The induction for existence of a fair ordering in a tournament
  gives a recursive algorithm for finding one.
\item 
The definition of the Fibonacci numbers is also expressing
a recursive algorithm for computing them.  
\item 
And the inductive proof for Nim gave a game-winning algorithm.
\end{itemize}

Just about any inductive proof that a problem has a solution is in
fact providing a recursive algorithm for solving the problem\.
Conversely, when you have a recursive algorithm that you think solves
some problem, induction is the usual way to prove that it works\.  In
fact, an inductive proof can be thought of as a \emph{recursive
  algorithm} for writing down a proof:
\begin{structproof}
Procedure Write-Proof-For($n$)\\
\\
{\bf if} $n$ is 0\\
\> Write-the-base-case\\
{\bf else}\\
\> Write-Proof-For($n-1$)\\
\> Write-Inductive-Step($n$)\\
\end{structproof}
Assuming we have rule for writing down the base case $P(0)$, and
writing down the inductive step $P(n-1) \implies P(n)$ for any $n$, we
can write down a proof of $P(k)$ recursively: we recursively prove
$P(k-1)$, then we use the inductive step proof to prove
$P(k-1)\implies P(k)$.

\subsection{RSA}

RSA is at the core of many of the secure communication protocols we
use today (SSH, 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 $x$.

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 gives 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{itemize}
\item The algorithm returns something (i.e., doesn't run forever)
\item The algorithm returns the \emph{right} something.
\end{itemize}
\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 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 $\ms{Exp}(x,n+1)$ 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
bits keys'' we aren't allowed to export because cryptography is
considered a weapon\?  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 $x^a=x^{2c}$\\
Any obvious way to beat $a$ multiplies?\\
Note $x^{2c}=(x^c)^2$\\
So compute $x^c$ ($c$ multiplies) and then square it (1 multiply)\.
Total: $c+1 =1+ a/2$ multiplies\.
What if $a$ is odd?\\

Generalize this:

\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}
  Induction again.  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\.
Suppose the theorem 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\.
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}


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

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

Finding 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.  Key lemma for ``simplifying'' 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).  $(x,y)$ 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).  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?\\
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 both $x$ and $y$ are 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}
Induction on $x+y$ as before.  The base case 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 $x \ge y$.  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 a lemma above that
GCD$(x,y)=$GCD$(y,x-y)$ so we are returning the right value.  
\end{proof}

Wait a minute, what's with the GCD$(0,0)$ definition?  Well,
technically every number is a common divisor of 0 and 0, so it should
be infinite?  

Why doesn't this matter?  Because we never actually make a recursive
call to $(0,0)$---once one number 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 base case is
implicit in the ``subcase'' of the inductive step where on argument is 0.

\section{Sets}

Today we give an overview of set theory, including relations on sets. \\
Probably much of the first part of this will be review from high
school math. \\
\\
We will define an interesting kind of relation called an {\em
equivalence relation\/}. \\
\\
An important example of equivalence relation is 
number equivalence mod $m$, where $m$ is a positive integer. 
We will define this relation and give some of its properties. \\
We will conclude with an application of modular equivalence 
ideas to get fast RSA encryption and decryption.


\subsection{Definitions}

A {\em set} is a collection of objects. \\
(Collection, object are undefined terms.) \\
\\
The order of elements is not important. I.e., the sets 
$\{1,2\}$ and $\{2,1\}$ are considered the same set\.
Each element is included only once. So, if we write
$\{a,b\}$ or $\{a,a,b\}$ they both denote the same set
and the element $a$ is considered to be present in the set 
only once. (repeating an element in the description of 
a set may be useful sometimes)\.
\\
Examples of sets: 

\begin{itemize}
\item $\{1,2,3\}$. 

A simple finite set, obtained by enumerating all the elements.

\item $\emptyset$, the {\em empty set}.

A special case of a finite set, with {\em no\/} elements.

\item $\{\emptyset\}$.

This funny set is different from $\emptyset$. \\
It has one element, which is a set.

\item $\{1,\{1,2\}\}$.

This one contains two elements, one of which is a set itself. \\
The set happens to contain the other element $1$, but that does not
contradict the condition that an element is included only once. 

\item $\{n\in \naturals  \given n \hbox{ is even }\}$.

An infinite set, specified by giving a universal set (domain of
discourse) and a property. 
In general, given a domain of discourse $D$ and a predicate $P$
over $D$, $\{x\in D \mid P(x)\}$ denotes the set of all 
$x$ in $D$ that satisfy predicate $P$.
\end{itemize} 

We write that $a\in A$ (``$a$ is in $A$'' or ``$a$ is an element of
$A$'') if $a$ is an object in the set $A$. \\
\\
The empty set satisfies the proposition $(\forall x) x \notin \emptyset$. \\
\\
Sets can be defined by axioms and deduction rules, giving a very
formal {\em axiomatic set theory\/}. \\
But usually, people reason about sets less formally, using {\em naive
set theory\/}. \\
Either way, set theory can be used as the basis for defining almost
everything else in mathematics and is therefore
regarded as the most fundamental part of math. \\
\\
E.g., natural numbers can be defined as certain kinds of sets (!).
For example we can represent the number $0$ by the empty set $\emptyset$,  
number $1$ by the set $\{\emptyset\}$, number $2$
by $\{\emptyset,\{\emptyset\}\}$, and so on.
In general, if $S_n$ is the set representing the number $n$,
the successor of $n$ can be represented by the set
$S_{n+1} = \{\emptyset\}\cup \{\{x\}\mid x\in S_n\}$.
Then the properties natural numbers can be {\em proved as
theorems\/} of set theory, 
rather than being assumed. \\
\\
Some typical axioms of axiomatic set theory: \\
(Seem ridiculously obvious but aren't)

\begin{itemize}
\item
The {\em axiom of extensionality} says that two sets are equal iff they have
the same elements. \\
\\
E.g., the two sets $\{n\in \naturals  \mid 2 \mbox{ divides } n\}$ and
$\{2n \mid n \in \naturals\}$ \\
are defined in different ways. \\
But they are ``equal'' as sets because they have exactly the same
elements. \\

\item
The {\em axiom of comprehension\/} says that we can form a new set by including
all objects of a given set that share a common property. \\
\\
For example, 
$\{x \in {\Bbb R} \given P(x)\} \hbox{  where  } P(x)=\hbox{ ``x is
rational''}$ denotes the set of all rational numbers.
\\
The axioms says that this is a set. \\
\end{itemize}

The set building notation that we used above has given the universe of
discourse explicitly.  \\
Sometimes, when this seems obvious, we will leave this out. \\
Just like what we did for predicate quantifiers. \\
E.g., we might write:
$\{n \mid n \hbox{ is even }\}$
if the universe $\naturals$ seems clear. \\
(Could be confusing -- the universe could alternatively be the
integers, positive and negative.)

\begin{defin}
We say that $A\subseteq B$ (``$A$ is a subset of $B$'' or 
``$A$ is contained in $B$'') if every
element of $A$ is also an element of $B$ (that is, $(\forall x)(x\in
A \implies x \in B)$).
\end{defin}

\begin{defin}
 We say that $A$ is a {\em proper} subset of $B$
($A\subset B$) if $A\subseteq B$ and $A\neq B$.
\end{defin}

Note $(\forall X) \emptyset \subseteq X$, i.e. the empty set 
is contained in any other set. \\
\\
Q: Can we have both $A\in B$ and $A\subseteq B$? \\
\\
A: Yes, consider $\{ 1, \{ 1 \} \}$.

\begin{theorem}
$A=B$ if and only if $A \subseteq B$ and $B \subseteq A$.
\end{theorem}

Proof by axiom of extensionality.  Ignore here. \\
\\
This observation is very important as it provides a simple way to
prove that 2 sets are equal. \\
\begin{example}
Prove that the two sets $\{n\in \naturals  \mid 2 \mbox{ divides } n\}$ and
$\{2n \mid n \in \naturals\}$ are equal. \\
\\
Must show containment in both ways. \\
$\subseteq$: \\
We assume $x \in \naturals$ and $2 \mbox{ divides } n$, and show that
there is some $y \in \naturals$ such that $x = 2y$. \\
This is true by definition of divisibility. \\
$\supseteq$: \\
Assume $x = 2y$.  Then $x$ is divisible by $2$, again by definition
of divisibility. 
\end{example}
\begin{example}
Slightly more interestingly: \\
\\
The set of evens $\geq 2$, i.e.,
$\{ 2n \mid n \in \naturals, n \geq 1 \}$, \\
is equal to 
$\{ x \in \naturals \mid \exists y, z \mbox{ odd }, x = y + z \}$.\\
$\subseteq$: \\
If $x = 2n$, then write $x = 1 + (x - 1)$, sum of two odds. \\
$\supseteq$: \\
If $x = y + z$, where both $y$ and $z$ are odd, then write $y = 2a +
1$, $z = 2b + 1$. \\
Then $x = 2(a+b+1)$, which is even.
\end{example}

The size of a set $A$ (the number of elements it contains) is written
$|A|$.  \\
For example, the set $\{a,b,c\}$ has size 
$|\{a,b,c\}|=3$. The empty set has size 
$|\emptyset|=0$.\\
The size of a  finite set is a natural number. \\
If $A$ is an infinite set, we write $|A|=\infty$.

Notice that for any two finite sets $A,B$, we have 
$|A\cup B| \leq |A|+|B|$ and equality holds if and only if 
$A\cap B=\emptyset$.
If we allow $A$ and $B$ to be infinite sets,
we still have $|A\cup B| \leq |A|+|B|$
if we define $x+\infty = \infty+x=\infty$ and 
$x\leq \infty$ for all $x\in \naturals$ and $x=\infty$.

\subsection{Russell's paradox}

We might think that set theory is obvious, but this is not the case. \\
Consider the set $S = \{x \mid x \mbox{ is a set}\}$.  \\
This seems like a perfectly reasonable set (well, maybe a bit
strange---note $S \in S$). \\
If our set theory allows for such an entity, we get into trouble as
follows. \\
\\
Consider the set $Q=\{x\in S \given x \hbox{ is a set and }x\not\in x\}$. \\
\\
If $S$ is a set, then so is $Q$ by standard axioms like
comprehension. \\
Note $Q \subseteq S$.  \\
\\
Then some sets are in $Q$ and some aren't. \\
\\
Example of something in $Q$: \\
The set $A = \{1,2,3\}$ is not an element of itself, so $A \in Q$. \\
\\
Example of something that is not in $Q$: \\
The set 
$P = \{\hbox{ all infinite sets}\}$ is a set (by comprehension from $S$). \\
Therefore, $P \in P$ as the set $P$ itself is infinite. \\
So $P \not \in Q$. \\
\\
The troublesome question is: is $Q\in Q$? 
\begin{enumerate}
\item If $Q \in Q$ then $Q \not \in Q$, a contradiction.
\item If $Q \not \in Q$ then $Q \in Q$ also a contradiction.
\end{enumerate} 

This is clearly a paradox. \\
In fact, this paradox, at the time it was discovered by Russell
($\sim1900$), was a blow to the then current attempts to formalize all
of mathematics axiomatically. \\
\\
Mathematicians had to work very hard to find a consistent set of axioms
(note that a consistent set of axioms does not allow this paradox to
arise). \\
\\
The problem with naive set theory is that you can form a set
by describing its members, as we did for $Q$ above.  \\
Axiomatic set theory has a specific set of rules for how you can form
sets---rules that don't allow the set $Q$ above to be specified, and
therefore avoid the paradox. \\
\\
Fortunately, we'll be working naively...

\subsection{Basic set operations}

There are several standard operations on sets. \\
In axiomatic set theory, the claims that these operations yield
new sets are axioms.

\begin{enumerate}
\item Union

$A\cup B=\{x\given x\in A \hbox{  or  } x\in B\}$. 

\item Intersection 

$A\cap B=\{x\given x\in A \hbox{  and  } x\in B\}$. 

\item Difference

$A-B=\{x\given x\in A \hbox{  and  } x\not\in B\}$. 

Also written as $A\setminus B$.

\item Complement

$A^c=\{x\given x\not\in A\}$. 

Also written as $\bar A$.

\item Symmetric difference

$A \oplus B= A \cup B - A\cap B$.
\end{enumerate} 

Sets and operations on sets can be represented using Venn diagrams
(see textbook, sections 1.4, 1.5). \\
\\
Union and intersection can be extended to apply to more than two sets.  \\
Given a finite collection of sets $A_i$, $i=1,\ldots,n$,
\[
\bigcup_{i=1}^n A_i
\]
is the set of elements in at least one of the $A_i$, and  \\
\\
\[
\bigcap_{i=1}^n A_i
\]
is the set of elements in all of the $A_i$.
Both of these make sense if $i=1$ -- reduce to just $A_1$. \\
In fact, for $i = 0$, default says union of no sets is $\emptyset$ and
intersection of no sets is universal set. \\
This definition of union and intersection of no sets might seem
arbitrary, but in fact it can be motivated in several ways.
For example, if we want to define the union of a finite 
family of sets $A_1,A_2,\ldots, A_n$, we can define
$\bigcup_{i=1}^{1} A_1 = A_1$ and 
$\bigcup_{i=1}^{n+1}A_i = (\bigcup_{i=1}^{n} A_i)\cup A_{n+1}$.
If we want this recursive definition to work also for base case 
$n=0$ we must set $\bigcup \emptyset = \bigcup_{i=1}^0 A_i = \emptyset$.
Analogously, for intersection. The recursive definition 
of  $\bigcap_{i=1}^{n+1}A_i = (\bigcap_{i=1}^{n} A_i)\cap A_{n+1}$
must have base case 
$\bigcap \emptyset = \bigcap_{i=1}^0 A_i = U$
if we want $\bigcap \{A_1\} = (\bigcap \emptyset)\cap A_{1} = A_1$.
\\

Instead of using explicit index, can just consider a set of sets:
If ${\cal S}$ is a {\em set of sets}, then 
\[
\bigcup {\cal S}
\] 
denotes the union of all the sets in ${\cal S}$, in any order. \\
(Can prove that the order doesn't matter, will do below.) \\
Likewise, 
\[
\bigcap {\cal S} 
\]
denotes the intersection of all the sets in ${\cal S}$. \\
\\
Complement is different from all the other operations, because it
assumes some universe $U$ from which we extract the things not in $A$.  \\
All sets in question are supposed to be subsets of $U$. \\
Given a universal set $U$, the complement operation 
can be expressed as $A^c = U - A$. \\
The universe $U$ is often implicit, where there seems to be little
chance of confusion. \\
\\
There is a close correspondence between set operations and logical operations:
consider sets $A$, $B$ and $C$ defined by 
$A=\{x\mid P(x)\}$, $B=\{x\mid Q(x)\}$, $C=\{x\mid R(x)\}$, 
where $P,Q,R$ are predicates. \\
\begin{itemize}
\item set union vs. predicate or: 
 $A\cup B =  C$ if and only if 
 $P(x)\vee  Q(x) \equiv  R(x)$ \\
\item set intersection vs. predicate and: 
 $A\cap B =  C$ if and only if 
 $P(x) \wedge Q(x)\equiv R(x)$ \\
\item set difference vs. predicate ``non implication'':
 $A - B = C$ if and only if 
 $\neg (P(x)\rightarrow Q(x)) \equiv R(x)$\\ 
\item set complement vs. predicate not:
 $A^c=B$ if and only if $\neg P(x) \equiv Q(x)$\\
\item set symmetric difference vs. predicate exclusive or:
 $A\oplus B = C$ if and only if $P(x)\oplus Q(x)\equiv R(x)$.
\end{itemize}

Predicates and sets often form alternative ways of looking at the same
notions: \\
Given predicates $p$ and $q$, define $P = \{ x \mid p(x) \}$, \\
$Q = \{ x \mid q(x) \}$. \\
Then $P \cup Q = \{ x \mid p(x) \vee q(x) \}$. \\
Etc.
\\
\\
\subsection{Power set (set of all subsets)} 

Another very important operation on sets is that of the {\em power
set} of a given set $A$. \\
It is denoted by ${\cal P}(A)$ and consists of the set of all subsets
of $A$. \\
In other words, ${\cal P}(A) = \{x\given x\subseteq A\}$. \\
\\
In particular, for any set $A$,
$\emptyset\in{\cal P}(A)$ and $A\in {\cal P}(A)$.
\\
Ex: \\
If $A=\{1,2\}$ then ${\cal P}(A)=\{\emptyset,\{1\},\{2\},\{1,2\}\}$.

\begin{theorem}
If $A$ has $n$ elements, then $P(A)$ has $2^n$ elements.
\end{theorem}

\begin{proof}
Ordinary Induction on $n$. 
BASE:  $n = 0$.  Then $P(A)$ has one element, $\emptyset$. \\
INDUCTIVE: Assume for $n$, prove for $n+1$. \\
Pick one element, $a$. \\
The subsets can be divided into two groups -- those containing $a$ and
those not containing $a$. \\
Moreover, the set of subsets of $A$ not containing $a$ is 
${\cal P}(A-\{a\})$, and the set of subsets of $A$ containing $a$
is the set $\{\{a\}\cup B \mid B\in {\cal P}(A-\{a\})$ which has 
the same number of elements as ${\cal P}(A-\{a\})$.
By induction hypothesis ${\cal P}(A-\{a\})$ has $2^{n-1}$
elements,
so the total is $2^{n-1} + 2^{n-1} = 2^n$.
\end{proof}

Q: I did a base case of $0$ here.  Does the inductive step from $0$ to
$1$ work correctly? \\
\\
A: Yes, it corresponds to the sets $\{\emptyset\}$ and $\{ a \}$. \\
\\
Because of this cardinality relationship, 
${\cal P}(A)$ is sometimes denoted by $2^A$.

\subsection{Set identities}

There are a large number of set identities involving relationships
among sets built using various operations. \\
See p. 50 of book. \\
\\
Here we just list some sample properties, and show how they can be
proved.

\begin{theorem}
(Distributivity:  Union distributes over intersection) \\
$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
\end{theorem}

\begin{proof}
We show the two sets are equal by proving that each is contained in the
other. \\
\begin{enumerate}
\item
$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$. 

Suppose $x \in A \cup (B \cap C)$.  \\
Then by definition, $x \in A$ or $x \in (B \cap C)$.  \\
We argue by cases.  \\
\\
If $x \in A$, then by definition $x \in A \cup B$ and similarly $x \in
A \cup C$.  \\
It follows by definition of intersection that $x$ is in the intersection
of these two sets, which is the RHS. \\
\\
On the other hand, if $x \in (B \cap C)$, then $x \in B$ and $x \in C$. \\
Then by definition of union, $x \in A \cup B$ and $x \in A \cup C$. \\
It again follows that $x$ is in the intersection. 
\\
\item
$A \cup (B \cap C) \supseteq (A \cup B) \cap (A \cup C)$. \\
\\
Suppose $x \in (A \cup B) \cap (A \cup C)$. \\
It follows that $x \in A\cup B$ and $x \in A \cup C$. \\
We argue by cases. \\
\\
If $x \in A$, then certainly $x \in A \cup (B \cap C)$. \\
\\
If $x \notin A$, then since $x \in A \cup B$, we know $x \in B$. \\
Similarly, $x \in C$. \\
Therefore, $x \in B \cap C$. \\
It follows that $x \in A \cup (B \cap C)$. 
\end{enumerate}
\end{proof}


\begin{theorem}
De Morgan's rule: $(A\cup B)^c=A^c\cap B^c$.
\end{theorem} 
\begin{proof}
We have to show that $(A\cup B)^c \subseteq A^c \cap B^c$ 
and that $A^c \cap B^c \subseteq (A\cup B)^c$. 

\begin{enumerate}
\item
$(A\cup B)^c\subseteq A^c\cap B^c$.

Let $x\in (A\cup B)^c$. 
Then by definition of complement, $x \notin A \cup B$.  
It follows (proof by contradiction) that $x \notin A$ and $x \notin B$.  
Therefore, again by definition of complement, $x \in A^c$ and $x \in B^c$.  \\
Then by definition of intersection, $x \in A^c \cap B^c$.

\item
$A^c\cap B^c\subseteq (A\cup B)^c$.

Let $x \in A^c \cap B^c$. \\
Then $x$ is in $A^c$ and in $B^c$. \\
Therefore, $x$ is in neither $A$ nor $B$ and therefore not in $A \cup B$. \\
We conclude that $x \in (A \cup B)^c$.
\end{enumerate} 
\end{proof} 

The above two theorems have {\em dual theorems} produced by swapping
$\cap$ and $\cup$. \\
This is a general phenomenon -- look at p. 50, arranged in pairs. \\
Duality can be applied to any theorem that is true for {\em all} sets.
Each theorem in a pair can be obtained from the other by 
swapping the operations $\cap$ and $\cup$, 
and the sets $\emptyset$ and $U$.
Notice that complement operations are not changed.

\begin{theorem}
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
\end{theorem}

\begin{theorem}
 $(A\cap B)^c=A^c\cup B^c.$
\end{theorem}

\subsection{Cartesian products}

Another important set operation is the {\em Cartesian product}. \\
For sets $A$ and $B$ define $A\times B=\{(a,b)\given a\in A\hbox{  and
}b\in B\}$. \\
Here, $(a,b)$ is considered an {\em ordered} pair so that $(a,b)\neq (b,a)$.

\begin{example}
If $A=\{0,1\}$ and $B=\{x,y\}$ then $A\times
B=\{(0,x),(0,y),(1,x),(1,y)\}$.
\end{example}

Order is important!  \\
In the above example, $(a,0) \notin A \times B$, though $(a,0) \in B
\times A$. \\
\\
We can generalize Cartesian products to apply to more than 2 sets and define 
$A_1 \times \cdots \times A_n 
= \prod_{i=1}^n A_i 
= \{(a_1,\cdots,a_n) \given a_i\in A_i \hbox{ for   }1\le i \le n\}$.
Also, we write $A^n$ for $\prod_{i=1}^n A$.

Examples: 
\begin{itemize}
\item $\{0,1\}^n$ is the set of all $n$-tuples of 0's and 1's. 
Or, length $n$ strings of $0$'s and $1$'s.
\item $\Re^2 = \Re\times \Re$ is the real plane.
\end{itemize} 

\section{Relations}

A ``relation'' is a fundamental mathematical notion expressing a
relationship between elements of sets. \\
It's an abstract notion useful in practice for modeling many
different sorts of relationships. \\
It's the basis of the {\em relational data base\/} model, the standard
data model for practical data processing systems.

\subsection{Definitions}

A {\em binary relation} {\em from} 
a set $A$ {\em to} a set $B$ is a subset 
$R\subseteq A\times B$. \\
A binary relation {\em on} a set $A$ is a subset $R\subseteq A^2$
(I.e., a relation from $A$ to $A$)\.
We often write $a\sim_R b$ or $aRb$ instead of  $(a,b)\in R$. \\
\\
We can also define a {\em ternary} relation on $A$ as a subset
$R\subseteq A^3$ or, in general, an $n$-ary relation as a subset
$R\subseteq A^n$, or $R\subseteq A_1\times A_2 \times\cdots\times A_n$
if the sets $A_i$ are different. \\
\\
Relations are used to model many different things:

\begin{enumerate}
\item \label{classes} 
The relation ``is taking class'' as a subset of
$\{$students at MIT$\}\times\{$classes at MIT$\}$. 
A relation from students to classes.
\item \label{roommates} 
The relation ``is living in the same room'' as a subset of
$\{$students at MIT$\}\times \{$students at MIT$\}$.
A relation on students.
\item \label{reachable} 
The relation ``can drive from first to second city''.
(Not necessarily directly -- just some way, on some roads.)
\item \label{wire} 
Relation on computers, ``are connected (directly) by a wire''
\item \label{meet} ``meet one another on a given day''
\item \label{likes}  ``likes''
\item \label{le} 
Let $A=\naturals$ and define $a\sim_R b$ iff $a\le b$.
\item \label{finite} 
Let $A={\cal P}(\naturals)$ and define $a\sim_R b$ iff $|a \cap b|$ is finite.
\item \label{dist} Let $A=\Re^2$ and define $a\sim_R b$ iff $d(a,b)=1$.
\item \label{contain} Let $A={\cal P}(\{1,\cdots,n\})$ and let $a\sim_R b$ if
$a\subseteq b$.
\item In a tournament, the ``who beats who'' relation.
\item ``Implies'' is a relation on the set of boolean formulae.
\end{enumerate} 

\subsection{Properties of Relations}

There are several standard properties that are satisfied by (some)
relations\.
These occur commonly enough that they are worth identifying\.
Sometimes interesting things can be deduced just from the abstract
properties.

A binary relation $R$ on $A$ is:
\begin{enumerate}
\item {\em reflexive} if for every $a\in A$, $a\sim_R a$.
\item {\em irreflexive} if for every $a \in A$, $a\not\sim_R a$.
\item {\em symmetric} if for every $a,b\in A$, $a\sim_R b$ implies $b\sim_R a$.
\item {\em antisymmetric} if for every $a,b\in A$, $a\sim_R b$ and
$b\sim_R a$ implies $a=b$.
\item {\em transitive} if for every $a,b,c\in A$, $a\sim_R b$ and $b\sim_R c$
implies $a\sim_R c$. 
\item {\em asymmetric} if for every $a,b\in A$, $a\sim_R b$ implies 
$\neg(b\sim_R a)$.
\end{enumerate} 

There's no sense to the specific negative prefixes; they just need to
be remembered\.  The difference between antisymmetric and asymmetric
relations, is that antisymmetric relations may contain pairs $(a,a)$,
i.e., elements can be in relations with themselves, while in an
asymmetric relation this is not allowed\.  Clearly, any asymmetric
relation is also antisymmetric, but not vice versa.

Among our examples:
\begin{itemize}
\item Relation~\ref{roommates} is reflexive, symmetric, transitive. 
\item Relation~\ref{reachable} is reflexive, transitive\.  Not
  necessarily symmetric, since roads could be one-way (Cambridge
  story), but in actuality...\.  But definitely not antisymmetric.
\item Relation~\ref{wire} is symmetric but not transitive\.
Whether it is reflexive is open to interpretation.
\item Relation~\ref{meet} likewise. 
\item Relation~\ref{likes} is (unfortunately) not symmetric.  Not
antisymmetric.  Not transitive.  Not even reflexive!
\item Relation~\ref{le} is reflexive, antisymmetric, transitive.
\item Relation~\ref{finite} is not reflexive.  It is symmetric.  Not
transitive\.  Evens$\cap$Odds is finite (empty), but not Evens$\cap$Evens.
\item Relation~\ref{dist} is only symmetric.
\item Relation~\ref{contain} is reflexive, antisymmetric and transitive.
\end{itemize} 

\subsection{Representation}

There are different ways of representing relations, e.g., for handling
in a computer program. \\
\\
We can describe by properties, as we did above. \\
That's about all we can do, for infinite sets. \\
But for finite sets, usually we use some method that explicitly
enumerates all the elements of the relation. \\
\\
Some alternatives: Lists, Matrices, Graphs. 

\subsubsection{Lists}

To represent a finite relation from set $A$ to set $B$, just
list all the pairs. \\
\\
Ex: Consider for example the relation from 
$A = \{ 0, 1, 2, 3 \}$ to $\{ a,b,c \}$ defined by the list 
$\{ (0,a), (0, c), (1, c), (2, b), (1,a) \}$. \\

Ex: The divisibility relation on natural numbers $\{ 1, \cdots, 12 \}$
is represented by the list: \\
$\{ (1,1), (1,2), \ldots, (1,12), (2,2), (2,4), \ldots, (2,12), $\\
$ (3,3), (3,6), (3,9), (3,12), (4,4), (4,8), (4,12), (5,5), (6,6), $\\
$(6,12), (7,7), (8,8), (9,9), (10,10), (11,11), (12,12) \}$. \\
\\

A more compact representations lists the ``partners'' for each item:\\
$1: 1,2,3,4,\ldots$\\
$2: 2,4,6,8,\ldots$\\
$3: 3,6,9,12$\\
etc.\\

Reflexivity in this representation: Contains all pairs $(a,a)$. \\
Symmetry: If contains $(a,b)$ then contains $(b,a)$. \\
Transitivity: If contains $(a,b)$ and $(b,c)$ then contains $(a,c)$. \\
\\

\subsubsection{Boolean matrices}

Used for finite sets $A$, $B$ \\
Rows for elements of $A$, columns for $B$. \\
$1$ in position if the pair is in the relation, $0$ otherwise.

Ex: The relation from
$\{0,1,2,3\}$ to $\{a,b,c\}$ of our first example is 
represented by the matrix
\[\begin{array} {r|ccc}
 &a& b&  c \\ 
\hline
0&1& 0&  1\\
1&1& 0&  1\\
2&0& 1&  0\\
3&0& 0&  0\\
\end{array}\]

Ex: The divisibility relation over $\{1,2,\ldots,12\}$ is 
represented by the matrix

\[\begin{array}{r|cccccccccccc}
  &1  &2  &3  &4  &5  &6  &7  &8  &9 &10 &11 &12\\
\hline
1 &1  &1  &1  &1  &1  &1  &1  &1  &1  &1  &1  &1\\
2 &0  &1  &0  &1  &0  &1  &0  &1  &0  &1  &0  &1\\
3 &0  &0  &1  &0  &0  &1  &0  &0  &1  &0  &0  &1\\
4 &0  &0  &0  &1  &0  &0  &0  &1  &0  &0  &0  &1\\
5 &0  &0  &0  &0  &1  &0  &0  &0  &0  &1  &0  &0\\
6 &0  &0  &0  &0  &0  &1  &0  &0  &0  &0  &0  &1\\
7 &0  &0  &0  &0  &0  &0  &1  &0  &0  &0  &0  &0\\
8 &0  &0  &0  &0  &0  &0  &0  &1  &0  &0  &0  &0\\
9 &0  &0  &0  &0  &0  &0  &0  &0  &1  &0  &0  &0\\
10&0  &0  &0  &0  &0  &0  &0  &0  &0  &1  &0  &0\\
11&0  &0  &0  &0  &0  &0  &0  &0  &0  &0  &1  &0\\
12&0  &0  &0  &0  &0  &0  &0  &0  &0  &0  &0  &1\\
\end{array}\]

Reflexivity in this representation:  major diagonal is all 1 \\
Symmetry: the matrix is symmetric around major diagonal \\
Transitivity: ... \\


\subsubsection{Digraphs} 

Digraphs are ways to make a \emph{picture} of a relation\.  The
picture can reveal lots of ``shape'' (topological) properties of the
relation that are not clear from the other representations (the matrix
is good for \emph{algebraic} properties).

The real power of graphs is for relations on a single set $A$.  To
represent a relation on $A$ as a digraph we draw a dot (vertex) for
each element of $A$, and draw an arrow from first element to second
element of each pair in the relation\.  The digraph may contain
self-loops, i.e.  arrows from a dot to itself, associated to the
elements $a$ such that $(a,a)$ is in the relation.

For relations from $A$ to $B \ne A$, you use two ``columns'' of
vertices: $A$ on left, $B$ on right, and draw arrows from $A$ vertices
to $B$ vertices\.  But we will focus on relations on $A$.

The divisibility relation over $\{1,2,\ldots,12\}$
is represented by the digraph
\begin{center}
\mbox{\psfig{figure=dividehasse.ps}}
\end{center}

Reflexivity: All nodes have self-loops. \\
Symmetry: all edges are bidirectional (in which case, instead of
drawing lots of bidirectional arrows, we just draw a line with no arrowhead). \\
Transitivity: Short-circuits, for any sequence of consecutive arrows,
there is a single arrow from the first to the last node. 

We will devote time later to the topological properties of graphs\.
For now, just a way to represent relations.





\end{document}

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