Class notes
Introduction
Document also available in PDF, Postscript, DVI, Text, LaTeX, LyX.
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^{1}
 1
 Those notes grew up from the class MACS 358 that I taught during
three semesters at the Colorado School of Mines, using the textbook
Mathematical Structures for Computer Sciences (4th edition)
by Judith L. Gersting[1].
Copyright Nicolas M. Thiéry 19992000
They are released, with the approbation of the publisher of the textbook,
under the GNU Free Documentation License (see http://www.gnu.org/copyleft/fdl.html):
in short, copying, redistribution in any form, modification or reuse
is permitted, and even encouraged, as long as I get credit for the
original version, and don't get blamed for other's modifications.
Moreover, any suggestions for improvements are warmly welcomed.
I use LyX http://www.lyx.org to typeset them, as well as to
display them in class on a LCDprojector plugged in my laptop. The
dvi, PDF, ps, HTML and pure text versions are then created using L^{A}T_{E}X,
Hevea http://para.inria.fr/~maranget/hevea/, Htmlpp http://htmlpp.sourceforge.net,
and some Makefile magic. I highly recommend the PDF of ps versions,
since the HTML and pure text versions are far from perfect due to
the limitations of those formats.
Table of Contents
1.1 Course Objectives
Mathematical tools for solving problems arising from computer science.
Using a computer to solve problems.
1.2 Pólya's hints for solving a problem
There are two main kinds of problems:

Problems to solve
 Problems to prove
Here is a list of general questions that will guide you when you try
to solve a problem.
Most of them are borrowed from ``How to solve it'' by Pólya[3, p. XVI].
(*subliminal hint* you definitely should read this book!)
Whenever you get stuck and don't know what to do, browse through them,
and most likely one of them will give you something to try out.
1.2.1 Phase 1: Understanding the problem
You have to understand the problem.

What is the goal ?

What is the unknown?
 What are the data ?
 Do you need all data ? Can you simplify the data ?
 What is the condition ?

Is it possible to satisfy the condition?
 Is the condition sufficient to determine the unknown?
 Or is it insufficient? or redundant? or contradictory?
 Separate the various parts of the condition. Can you write them down?
 Look at examples. Draw a figure. Introduce suitable notations.
 How would you store the data on a computer ?
1.2.2 Phase 2: Devising a plan
Find the connection between the data and the unknown.
You may be obliged to consider auxiliary problems if an immediate
connection cannot be found.
You should obtain eventually a plan of the solution.

Have you seen it before?

Have you seen the same problem in a slightly different form?
 Do you know a related problem? Do you know a theorem that could be
useful?
 Look at the unknown! Try to think of a familiar problem having the
same or similar unknown.
 Here is a problem related to yours and solved before. Could you use
it?

Could you use its result? Could you use its method?
 Should you introduce some auxiliary element in order to make its use
possible?
 Could you restate the problem? Could you restate it still differently?
 Go back to definitions.
 If you cannot solve the proposed problem, try to solve first some
related problem. Could you imagine a more accessible related problem?
A more general problem? A more special problem? An analogous problem?

Could you solve a part of the problem? Keep only a part of the condition,
drop the other part. How far is the unknown then determined, how can
it vary?
 Could you derive something useful from the data? Could you think of
other data appropriate to determine the unknown? Could you change
the unknown or the data, or both if necessary, so that the new unknown
and the new data are nearer to each other?
 Did you use all the data? Did you use the whole condition? Have you
taken into account all essential notions involved in the problems?
1.2.3 Phase 3: Carrying out the plan
Carry out your plan.

Carrying out your plan of the solution, check each step. Can
you see clearly that the step is correct? Can you prove it is correct?
1.2.4 Phase 4: Looking back
Examine the solution obtained.

Can you check the result? Can you check the argument?
 Can you derive the result differently? Can you see it at a glance?
 What's the structure behind?
 Can you use the result, or the method for some other problem?
 Can you make it an algorithm ?

Is your algorithm correct ?
 Can you prove it is correct ?
1.3 Some ``research'' about Mazes
1.4 Conclusion
The purpose of this exercise was to browse quickly through what we
are going to do this semester:
1.4.1 General problem solving techniques:
We have used Pólya's list of questions as a guide to solve our problems.
We will do this over and over. Usually the main difficulty with a
problem is to get started. Whenever you get stuck on a problem, and
don't know what to do, you should refer to this list, and see if some
of the questions could give you something to try out.
At some time, we had a solution for exiting from a maze. However,
we were not sure if it worked all the time or not. Well, the only
way to be sure that something is correct is to PROVE it.
We will need to learn how to prove things before anything else. In
particular we will want to prove that certain algorithms are correct.
That will be the core of our first month of class.
So what's a proof ? Basically, it's a message written by a human A
to convince another human B that some fact is true (possibly A and
B are the same person). When B reads through the message, he should
not have any choice at the end but to say "I agree".
To achieve this goal, our primary tool will be Formal Logic.
It's not easy to write good proofs. For example, a proof should be
short enough that you don't get lost in the details, and detailed
enough that you are sure not to have left a hole somewhere. Learning
to write proofs is like learning to write. The only way is to write
a lot of proofs yourself, and to take model on other's proofs. We
will learn this progressively.
1.4.3 Discrete structures:
We have seen that the very structure of a maze (once we have removed
all extraneous information like color, shape and so on) can be formalized
with a graph, that is a set of nodes which are connected or not by
edges.
A graph is a good example of discrete object, or structure (in opposition
to a continuous object like a curve). We are going to see other discrete
structures, and learn to recognize them when the arise at the very
heart of problems. We are also going to see how to deal with such
structures (algorithms and such).
1.4.4 Counting objects of a certain kind (Combinatorics)
How many mazes? how many ordered trees ?
1.4.5 Algebraic structures
That's another kind of structure that can arise in our problems. Addition,
multiplication and other algebraic operations are very powerful tools.
We will see that such operations can often be defined for other objects
that the usual integers or real number.
1.4.6 Making a solution into an algorithm, and implementing it
Part I
Formal Logic
Chapter 2 Propositional Logic
2.1 Introduction
2.1.1 What is a proof ?
Définition 1
A proof is a piece of text written by a human to convince another
human that some fact is true.
Problem 1
[1, p. 2]
You have been selected to serve on jury duty for a criminal case.
The attorney for the defense argues as follow:
``If my client is guilty, then the knife was in the drawer. Either
the knife was not in the drawer, or Jason Pritchard saw the knife.
If the knife was not there on October 10, it follows that Jason Pritchard
did not see the knife. Furthermore, if the knife was there on October
10, then the knife was in the drawer and also the hammer was in the
barn. But we all know that the hammer was not in the barn. Therefore,
ladies and gentlemen of the jury, my client is innocent.''
Are you convinced ?
Problems:

What is the goal ?
 What are the hypothesis ?
 What is the logical link between the statements ?
 There is plenty of unrelevant information.
A good proof consists of:

A description of the goal
 Hypothesis
 Statements, one after the other.
Each one derives logically from the previous ones.
 Conclusion
Let's look at a very simple example:
Théorème 1
Let x and y be two integers.
If x is odd and y is odd, then xy is odd.
Proof.
Assume x is odd and y is odd.
Let k be an integer such that x=2k+1.
Let l be an integer such that y=2l+1.
Then, xy=(2k+1)(2l+1)=2(2kl+k+l)+1.
Note that 2kl+k+l is an integer.
Conclusion: xy is odd.
2.1.2 Formal and Informal proofs
What do we mean by ``derives logically from the previous ones''
?
What operations shall we consider as legal ?
To learn more about this, we need a model of what we intuitively
consider as true, or logic.
This model we are going to construct is called formal logic.
Formal systems
Définition 2
A formal system
is build by providing:

a set of strings: the axioms
 a set of rewriting rules
A string that can be obtained from the axioms by applying rewriting
rules is called a theorem
of the formal system.
The proof
of a theorem is a description of the axioms and rules
used to get the theorem.
The set of all theorems is called the language
defined by the
formal system.
Exemple 1
The MIU formal system[2]
A formal system does not need to have any meaning or semantic.
It's pure syntax.
We will build formal logic as a formal system, with a meaning!

The strings will be logic formulas
 The axioms will be simple logic formulas that we interpret intuitively
true
 The rewriting rules will be transformations we interpret intuitively
as valid
Then, any theorem of this formal system shall be interpreted as true.
We will have to be very careful with the choice of the axioms and
rules!
Proofs and programming
Writing a proof and a writing a program are very similar:
Proofs 
Programming 
Purpose 
Theorem 
Function prototype 
How to use it 
Proof 
Function definition 
How it works 
Intuitive proof 
Comments 
Explanation for the reader 
Informal proof 
Program in a high level language 
Human readable 
Formal proof 
Program in assembly language 
Machine readable 
Checking an informal proof 
Compiling 
Human readable > Machine readable 
This comparison is not gratuitous. There are deep links between proofs
and programming which have been studied a lot in the last decades.
This is called the CurryHoward correspondence.
2.2 Statements, Symbolic Representation, and Tautologies
As a first step, let's define logic formulas and their interpretation.
2.2.1 Statements and Propositions
Définition 3
A statement (or proposition) is a sentence which is
either true or false.
Exemple 2
Which sentences below are statements ?

The sky is cloudy
 It's raining
 Are you happy ?
 He is happy
 Three divides five
 There are life forms on other planets in the universe
Notation 1
A, B, C: symbols representing a statement.
Each of those symbols can be given a truth value, either True (T for
short) or False (F for short).
1 is a statement if truth value True.
0 is a statement if truth value False.
2.2.2 Connectives
How can we combine two statements ?
And (conjunction), or (disjunction)
Exemple 3
``The sky is cloudy'': A. ``It's raining'': B.

``The sky is cloudy AND it's raining'': A/\ B
 ``The sky is cloudy OR it's raining'': A\/ B
Définition 4
The truth table
of a formula describes when it's true or not,
given the truth value of the symbols.
A 
B 
A/\ B 
A\/ B 
T 
T 


T 
F 


F 
T 


F 
F 


Exclusive or
Exemple 4
``Cheese or desert''
You get cheese: A
You get desert: B
Définition 5
A exclusive or B: A xor B.
Truth table:
A 
B 
A xor B 
T 
T 

T 
F 

F 
T 

F 
F 

Implication
Exemple 5
``If I get my driver license, I will open a bottle of champagne''.
``I get my driver license'': A.
``I open a bottle of champagne'': B.
Définition 6
A implies
B: A> B.
Truth table:
A 
B 
A> B 
T 
T 

T 
F 

F 
T 

F 
F 

Why this should be true:

no real meaning but we have to pickup one to make it easy
 Most reasonable is true
Why this should be false:
This one can be tricky. To help you make up your mind, here are some
examples:

If I don't get my driver license, but still I open a bottle of champagne,
am I a liar?
 ``If I open a bottle of champagne, I will get my driver license''
 ``When pig flies, I'll give you 2 billions dollars''
Well, as strange as it can sound, F> T is a true formula.
If you start from an utterly false hypothesis, you can prove anything.
``Avec des si, on pourrait mettre Paris en bouteille''
``With ifs, Paris could be put in a bottle''
Remarque 1
A> B and B> A are not the same!
Sounds stupid ?
Look carefully around you! This is the most common logic error in
peoples argumentations.
And it's not always an involuntary error.
Exercice 1
Give the antecedent and the consequent in the following sentences

If the rain continues, then the river will flood
 A sufficient condition for network failure is that the central switch
goes down
 It's raining only if there are clouds in the sky
 It's raining if there are clouds in the sky
 It's raining if and only if there are clouds in the sky
 The avocados are ripe only if they are dark and soft
 A good diet is a necessary condition for a healthy cat
Equivalence
Exemple 6
The knight will win if and only if his armor is strong
The knight will win: A.
His armor is strong: B.
This sentence contains two informations:

The knight will win if his armor is strong: A< B.
 The knight will win only if his armor is strong: A> B.
Définition 7
A is equivalent to
B: A<> B.
Truth table:
A 
B 
A<> B 
T 
T 

T 
F 

F 
T 

F 
F 

Not
So far we defined binary connectives, that is connectives which took
two arguments.
Here is now a unary connective, taking just one argument.
Exemple 7
It will not rain tomorrow
The knight will win: A.
His armor is strong: B.
This sentence contains two informations:

The knight will win if his armor is strong: A< B.
 The knight will win only if his armor is strong: A> B.
Définition 8
not A: A' (more usual notation: ¬ A)
Truth table:
2.2.3 Complex Well Formed Formulas
Définition 9
A formula
is a string of

symbols: A, B, ...
 connectives: \/, /\, >, < , '
, ...
 parenthesis: (, ), [, ]
A well formed formula
(or wff) is a formula which satisfies
certain syntax rules.
Notation 2
Symbols representing wff: P, Q, R.
Order of precedence
The formula A/\ B> C' is ambiguous. It can be read
as:

(A/\ B)>(C'),
 A/\[B>(C')],
 [(A/\ B)> C]'
Solution 1: always put all parenthesis (not very readable).
Solution 2: put an order of precedence between the operators:

Connectives within parentheses, innermost parenthesis first
 '
 /\, \/
 >, <
 <>
With this order of preference, A/\ B> C' unambiguously
refer to (A/\ B)>(C)'.
Evaluation of a wff
From the truth values of the basic statements, and the elementary
truth tables, one can evaluate the truth value of a wff.
Exemple 8
A/\ B> C', with A true, B false and C true:
A/\ B: false
C': false
A/\ B> C': true
There is a recursive algorithm to compute the evaluation of a wff.
Truth table of a wff
This yields an algorithm for computing the truth table of a wff.
A 
B 
C 
A/\ B 
C' 
(A/\ B)>(C') 
T 
T 
T 



T 
T 
F 



T 
F 
T 



T 
F 
F 



F 
T 
T 



F 
T 
F 



F 
F 
T 



F 
F 
F 



Problem 3
What's the size of a truth table?
Problem 4
Can any truth table be obtained by a wff?
2.2.4 Tautologies and Contradictions
Définition 10
A tautology is a statement that is always true.
A contradiction is a statement that is always false.
Exemple 9
A\/ A'
``Today the sun shine or today the sun does not shine''.
Exercice 2
Are the following wff tautologies or contradictions ?

A/\ A'
 A<> A
 (A/\ B)<>(B/\ A)
 (A/\ B)> A
Algorithm to check for a tautology?
Compute the truth table!
It's slow (2^{n}, where n is the number of propositions), but
it works.
For certain simple forms of propositions, there is a much faster algorithm
(see [1, Tautology test, p. 13]).
Exemple 10
Is [(A/\ B> C')/\(E< D)]<>[(E< D)/\(A/\ B> C')]
a tautology ?
Let P:=(A/\ B> C') and Q:=(E< D).
The formula above is [P/\ Q]<>[Q/\ P].
Let's prove it's a tautology.
Take some truth values for A, B, C, D, E.
P and Q will then have some truth value.
But we know that (A/\ B)<>(B/\ A) is a tautology.
So [P/\ Q]<>[Q/\ P] will be true!
We don't actually need to compute the truth table of P and Q.
Remarque 2
Whenever you have some tautology, if you replace ALL occurrences of
some basic statement (say A) by a more complicated wff, you still
get a tautology.
2.2.5 Equivalence of wff
Let P and Q be two wff, and suppose P<> Q is
a tautology.
Then P and Q have the same truth table.
Définition 11
P and Q are equivalent (noted P <=> Q) if they have
the same truth table.
Remarque 3
Writing P<> Q or P <=> Q is different.
In the first case, you just represent a formula, which can be true
or false.
In the second case, you claim that the wff P<> Q is
a tautology.
Basic equivalent wff
What is the negation of ``Peter is tall and thin'' ?
Identity (id) 
A\/0 <=> A 

A/\1 <=> A 
Complement (comp) 
A\/ A' <=> 1 

A/\ A' <=> 0 
Commutativity (comm) 
A\/ B <=> B\/ A 

A/\ B <=> B/\ A 
Associativity (ass) 
(A\/ B)\/ C <=> A\/(B\/ C) <=> A\/ B\/ C 

(A/\ B)/\ C <=> A/\(B/\ C) <=> A/\ B/\ C 
Distributivity (dist) 
A\/(B/\ C) <=> (A\/ B)/\(A\/ C) 

A/\(B\/ C) <=> (A/\ B)\/(A/\ C) 
Double negation (dn) 
A'' <=> A 
De Morgan's Law (dm) 
(A\/ B)' <=> A'/\ B' 

(A/\ B)' <=> A'\/ B' 
Implication (imp) 
A'\/ B <=> A> B 
Remarque 4
Notice the duality between \/ and /\.
Unlike + and · they play a perfectly symmetric role in all
formulas.
More equivalent wff using substitution
Exemple 11
(A/\ B)/\1 <=> (A/\ B).
Remarque 5
Let P and Q be equivalent wff.
If you replace all occurrences of some basic statement by another
wff, you still get two equivalent wff's.
Exemple 12
A/\(B\/ C) <=> A/\(C\/ B)
Remarque 6
Let P and Q be equivalent wff.
If P appears exactly as a subformula of a bigger wff, then
you can substitute P by Q and get an equivalent wff.
What's the point ?
Rewriting a formula to simplify it:
Exercice 3
[1, Example 7 p.12]

If ((outflow > inflow) and not ((outflow>inflow) and (pressure < 1000)))
Do something;
Else
Do something else;
EndIf
Write the wwf corresponding to the test.
Rewrite this wwf using equivalence rules.
Electronic circuits and logical gates:
Exercice 4
Build an electronic circuit for (A/\ B)> C'.

Case 1: you have electronic gates TRUE, FALSE, AND, OR, NAND.
 Case 2: you only have electronic gates NAND.
2.3 Propositional logic
We have seen propositions, well formed formula for formalizing statements.
We now need to formalize arguments.
Exemple 13
[1, Example 17 p. 28]
Let's formalize the following argument:
``If interest rates drop, the housing market will improve.
Either the federal discount rate will drop, or the housing market
will not improve.
Interest rates will drop.
Therefore the federal discount rate will drop.''
Basic statements:

I: Interest rates drop
 H: The housing market will improve
 F: The federal discount rate will drop
Argument: (I> H)/\(F\/ H')/\ I > F
Définition 12
An argument is a wff of the form P_{1}/\ P_{2}/\···/\ P_{n} > Q.
P_{1},···,P_{n} are the hypothesis.
Q is the conclusion.
The argument is valid iff P_{1}/\ P_{2}/\···/\ P_{n} > Q
is a tautology.
Exercice 5
Prove that the following arguments are valid:

[A>(B> C)] > [(A/\ B)> C]
 [(A/\ B)> C] > [A>(B> C)]
Why is this reasonable ?
``If the sun shine then if you don't put sun cream, you will get
burned'': A > (B> C)
``If the sun shine and you don't put sun cream, then you will get
burned'': (A/\ B) > C
Exemple 14
``George Washington was the first president of the United States.
Therefore everyday has 24 hours.''
We put some more restrictions to only keep ``meaningful'' arguments.

Q has some relation with the P_{i}.
 the P_{i} are not in contradiction
2.3.2 Proof sequences
So now, how to prove that an argument is a tautology ?

Use the truth table
 Use formal logic!
Exemple 15
Let's prove that the argument A\/(A\/ B)' > (A\/ B') is
valid.
We are lazy to compute the truth table.
Instead, we assume that the hypothesis is valid, and use a series
of equivalences to deduce that the conclusion is valid.

1. 
A\/(A\/ B)' 
(Hypothesis) 
2. 
A\/(A'/\ B') 
(De Morgan's, 1) 
3. 
(A\/ A')/\(A\/ B') 
(Distributivity, 2) 
4. 
1/\(A\/ B') 
(Complement, 3) 
5. 
(A\/ B') 
(Identity, 4) 

What we just did is called a proof sequence.
This is our first proof using formal logic.
Exercice 6
Prove that the following argument is valid:
(A\/ B)'\/(A'/\ B) > A'
Définition 13
Formal Logic: wff + set of derivation rules (formal system).
What derivation rules shall we accept?
We want our formal logic to be:

Correct
 Complete
 Minimal
 Powerful
2.3.3 Looking for derivation Rules
Exemple 16
A/\ B > A
Equivalences rules are not enough.
We can't prove the previous statement, because A/\ B is strictly
stronger than A.
We have build equivalence rules from equivalences P<> Q.
We can build Inference rules from any valid argument P> Q.
Rule name 
From 
Can derive 
modus ponens (mp) 
P, (P> Q) 
Q 
modus tollens (mt) 
(P> Q), Q' 
P' 
conjunction (con) 
P, Q 
P/\ Q 
simplification (sim) 
P/\ Q 
P, Q 
addition (add) 
P 
P\/ Q 
Exemple 17
A/\(A> B)/\(B> C) > C

1. 
A/\(A> B)/\(B> C) 
(Hypothesis) 
2. 
A 
(Simplification, 1) 
3. 
A> B 
(Simplification, 1) 
4. 
B 
(Modus ponens, 2, 3) 
5. 
B> C 
(Simplification, 1) 
6. 
C 
(Modus ponens, 4, 5) 

Exemple 18
(A>(B\/ C))/\ B'/\ C' > A'

1. 
(A>(B\/ C))/\ B'/\ C' 
(Hypothesis) 
2. 
(A>(B\/ C)) 
(Simplification 1) 
3. 
(B'/\ C') 
(Simplification 1) 
4. 
(B\/ C)' 
(De Morgan's 3) 
5. 
A' 
(Modus tollens 4, 5) 

Exercice 7
Prove the validity of the following arguments using formal logic:

A/\(B> C)/\[(A/\ B)>(D\/ C')]/\ B> D
 A'/\(A\/ B)> B
 A'/\ B/\[B>(A\/ C)]> C
 A/\ A'> B
Theorem 1
This formal logic is correct and complete.
The correctness of the formal logic comes from the way we built the
rules from valid arguments.
Proving it's complete is far more difficult.
2.3.4 Methods to make it easier
Our logic is complete, but still not very practical to use.
Here are some tricks to make it easier.
Most of the time, the proofs start like this:

1. 
(A>(B\/ C))/\ B'/\ C' 
(hyp) 
2. 
(A>(B\/ C)) 
(Simp 1) 
3. 
(B'/\ C') 
(Simp 1) 
··· 
··· 
··· 

The steps 2 and 3 are pretty trivial, and it's reasonable to start
directly with:

1. 
(A>(B\/ C)) 
(hyp) 
2. 
(B'/\ C') 
(hyp) 
··· 
··· 
··· 

The deduction method
Exemple 19
(A> B)/\(B> C) > (A> C)
The obvious start is:

1. 
(A> B) 
(hyp) 
2. 
(B> C) 
(hyp) 

But then it's pretty painful:

3. 
A'\/ B 
(imp 1) 
4. 
A'\/(B> C) 
(add 2) 
5. 
(A'\/ B)/\(A'\/(B> C)) 
(con 3, 4) 
6. 
A'\/(B/\(B'> C)) 
(dist 5) 
7. 
A'\/ C 
(mp 6) 
8. 
A> C 
(imp 7) 

Here it's easier to first rewrite the argument into an equivalent
argument.
We have seen that P>(Q> R) and (P/\ Q)> R
are equivalent.
So we can prove A/\(A> B)/\(B> C) > C
instead, which is straightforward:

1. 
A 
(hyp) 
1. 
(A> B) 
(hyp) 
2. 
(B> C) 
(hyp) 
3. 
B 
(mt 1, 3) 
4. 
C 
(mt 2, 4) 

This trick is called the deduction method.
It's powerful because we have:

a simpler goal
 more hypothesis
Exercice 8
(A> B)/\[B>(C> D)]/\[A>(B> C)] > (A> D)
Additional deduction rules
As we have seen, any valid argument could be transformed into a new
rule.
Exemple 20
Here are two useful rules.
Rule name 
From 
Can derive 
hypothetical syllogism (hs) 
P> Q, Q> R 
P> R 
contraposition 
P> Q 
Q'> P' 
Adding those new rules won't change the power of the formal logic,
but can make it more convenient. On the other hand, a minimal formal
logic is nice since it's less error prone, and it's easier to remember
all the rules.
It's up to you to choose which rules you want to use (and learn).
Exemple 21
(P> Q)/\(P'> Q)> Q.
2.4 Verbal arguments
We can now start checking verbal arguments.
Exemple 22
Let's prove the following verbal argument[1, Example 17 p. 28]:
``If interest rates drop, the housing market will improve.
Either the federal discount rate will drop, or the housing market
will not improve.
Interest rates will drop.
Therefore the federal discount rate will drop.''
Exercice 9
Prove the following verbal argument[1, Ex. 30 p. 32]:
``If the program is efficient, it executes quickly.
Either the program is efficient or it has a bug.
However, the program does not execute quickly.
Therefore it has a bug.''
2.5 Summary
2.5.1 Different levels of logic
Three levels for arguments:

formal argument as a string of symbols, proof sequence (syntactic
level).
 formal argument as a truth table (semantic level).
 verbal argument (interpretation level).
Any verbal argument based on propositions can be translated into a
formal argument.
The propositional logic system is complete and correct:

Any formal argument proved syntactically is valid semantically.
 Any semantically valid formal argument can be proved syntactically
(it can be hard, though!).
2.5.2 Syntactic proof versus semantic proof
Testing an argument by computing it's truth table is purely mechanical.
So, what's the point of formal logic ?

It's often faster and shorter to write
 You have more chance to grab the meaning of an argument
 In predicate logic we won't be able to rely on truth tables
Chapter 3 Predicate Logic (sections 1.3, 1.4)
3.1 Introduction
Exemple 23
``Every man has a brain; Socrate is a man; therefore Socrate has
a brain.''
Is propositional logic expressive enough for this sentence ?
What is missing ?
3.1.1 Quantifiers, Predicates, Validity
Exemple 24
``Every man has a brain''
``For all man, this man has a brain''
New features: constants / variables / quantifier / predicates
Définition 14
Quantifiers:

For all
 For every
 There exists
 For at least one
 For each
 For any
 For some
Universal quantifier
: (for all x)
Existential quantifier
: (there exists
x)
Always place the quantifiers inside ( ).
3.1.2 Predicates:
Définition 15
Predicate: a statement that describes a property of a variable
Exemple 25
P(x) : x has a brain
``For all man, this man has a brain''
Translates into (for allx ) P(x)
As in propositional logic, we can build well formed formulas
from predicates and logic connectives.
Exemple 26
(for allx ) B(x)> R(x)': if x is blue, then x is not red.
3.1.3 Domain of interpretation:
If x is a car, the sentence above (for allx ) P(x) is false.
The truth of the sentence depends on where you take x from!
Définition 16
Domain of interpretation:
The collection of objects from which x may be selected
Exemple 27
(for allx ) (x>0)
DOI: all integers
DOI: all integers > 10
DOI: all humans
Exemple 28
For any thing, if it's a man then it has a brain
P(x): x is a man
Q(x): x has a brain
(for allx ) P(x)> Q(x)
Exercice 10
Write a formula for the following sentence:
``There exists a man which is called Socrate''
3.1.4 NAry predicates:
Définition 17
Binary predicate: predicate which takes two variables
NAry predicate : predicate which takes several variables

P(x,y)=x>y (DOI: integers)
 P(Q,B): Q is the author of the book B (DOI: all authors /
books)
 Everyone has exactly one best friend:
B(X,Y): Y is the best friend of X
(for allx ) (there existsy ) B(X,Y)/\[ (for all Z) B(X,Z)>(Z=Y) ]
For any person X, there exists a person Y, such that:

Y is the best friend of X
 For any person Z, if Z is the best friend of X, then Z
is actually Y.
3.1.5 Truth value
Can we define the Truth table of a wff ?
Problem 1
The domain of interpretation may be infinite.
To assess the truth value, we need to know:

What is the domain of interpretation.
 What property of elements of the DOI does P(x) represent.
 An assignment for each constant.
Définition 18
This is called an interpretation.
Exemple 29
[1, Exercise 1 p. 41]
What is the truth value of the following wffs for this interpretation:
DOI: integers; O(x): x is odd; L(x): x<10; G(x): x>9

(there existsx ) O(x)
There exists an odd number
 (for allx ) [L(x)> O(x)]
For any integer x, if x is strictly lower than 10, then x is odd;
Any integer strictly lower than 10 is odd;
 (there existsx ) [L(x)/\ G(x)]
There exists an integer x such that x is strictly lower than 10 and
x is strictly larger than 9
 (for allx ) [L(x)\/ G(x)]
For any integer, either x is strictly lower than 10 or x is strictly
bigger that 9
Exercice 11
[1, Exercise 3 p. 41]
Are the following wffs true for this interpretation:
DOI: states of the US.
Q(x,y): x is north of y; P(x): x starts with the letter
M; a is "Mississippi".

(for allx ) P(x)
 (for allx ) (for ally ) (for allz ) [ (Q(x,y)/\ Q(y,z)) > Q(x,z) ]
 (there existsy ) (there existsx ) Q(x,y)
 (for allx ) (there existsy ) [P(y)/\ Q(x,y)]
For any state x, there exists a state y such that y starts
with M and x is north of y.
 (there existsy ) Q(a,y)
Exercice 12
Translate the following sentences into wffs, with:
DOI: world; D(x): x is a day; S(x): x is sunny; R(x): x is
rainy;
M is "Monday"; T is "Tuesday".

All days are sunny:
 Some days are not rainy:
 Every day that is sunny is not rainy:
 Some days are sunny and rainy:
 No day is both sunny and rainy:
 It is always a sunny day only if it is a rainy day:
 No day is sunny:
 Monday was sunny; therefore every day will be sunny:
 It rained both Monday and Tuesday:
 If some day is rainy, then every day will be sunny:
3.1.6 Dummy variables / free variables:
Définition 19
A dummy variable is a variable which is linked to a quantifier.
The name of the variable is irrelevant.
Same thing as a local variable in a program.
Exemple 30
(there existsx ) Q(x) is the same wff as (there existsz ) Q(z)
(there existsx ) (for ally ) Q(x,y) is the same wff as (there existsz ) (for allt ) Q(z,t) or (there existsy ) (for allx ) Q(y,x)
Définition 20
A variable is free if it is not linked to a quantifier.
Same thing as a global variable in a program.
Résumé 2
Where are we?

We can build all the wff of predicate logic.
 Given a wff P, and an interpretation, we can decide the truth value
of P.
Définition 21
Argument: P_{1}/\ P_{2}/\···/\ P_{k} > Q

In propositional logic, an argument is valid if it's a tautology.
I.e. if it's true whatever truth value are assigned to each basic
proposition.
 In predicate logic, an argument is valid if it's intrinsically
true.
I.e. if it's true for ANY interpretation.
Exercice 13
[1, Exercise 16 p. 35]
Give interpretations to prove that the following wffs are not valid:

(there existsx ) A(x)/\(there existsx ) B(x) > (there existsx ) [ A(x)/\ B(x) ]
 (for allx ) (there existsy ) P(x,y) > (there existsx ) (for ally ) P(x,y)
 (for allx ) [P(x)> Q(x)] > (for allx ) [P(x)\/ Q(x)]
 (for allx ) [ A(x)' ] <> [(for allx ) A(x) ]'
3.2 Predicate logic
There is an infinity of interpretations, so there is no algorithm
to check the validity of a predicate.
We will have to use REASON:

Reuse the rules from propositional logic
 Accept a few basic new rules as intuitively valid
 Use formal logic with those rules to prove arguments
3.2.1 Universal instantiation:
Exemple 31
We want to be able to prove the following argument:
Every human is mortal; Socrate is a man; Therefore Socrate is mortal.
H(x): x is a human; M(x): x is a mortal; s: Socrate
(for allx ) [H(x) > M(x)]/\ H(s) > M(s)
Définition 22
Rule of universal instantiation (ui):
From: (for allx ) P(x)
Can derive: P(s)
Note: s can be any constant.
We decide that this rule is valid, because it is intuitively
valid.
Proof sequence for: (for allx ) [H(x)> M(x)]/\ H(s) > M(s)

1. 
(for allx ) [H(x)> M(x)] 
(hyp) 
2. 
H(s) 
(hyp) 
3. 
H(s)> M(s) 
(ui 1) 
4. 
M(s) 
(mp 2, 3) 

Exercice 14
Prove (for allx ) [ P(x)> R(x) ]/\[ R(y)' ] > [ P(y)' ]
3.2.2 Universal generalization:
Exemple 32
``Every human is a living being. Every living being is mortal.Therefore
every human is mortal.''
H(x): is a human
L(x): is a living being
M(x) is mortal
(for allx ) [H(x)> L(x)] /\ (for allx ) [L(x)> M(x)] > (for allx ) [H(x)> M(x)]
This is clearly something we want to be able to prove, but we cannot
use hypothetical syllogism directly!
Définition 23
Rule of universal generalization (ug)
From: P(s)
Can derive: (for allx ) P(x)
s must be an arbitrary element of the domain.
Exemple 33
(for allx ) [H(x)> L(x)] /\ (for allx ) [L(x)> M(x)] > (for allx ) [H(x)> M(x)]

1. 
(for allx ) H(x)> L(x) 
(hyp) 
2. 
(for allx ) L(x)> M(x) 
(hyp) 
3. 
H(s)> M(s) 
(ui 1) 
4. 
L(s)> M(s) 
(ui 2) 
5. 
H(s)> M(s) 
(hs 3,4) 
6. 
(for allx ) [H(x)> M(x)] 
(ug 5) 

Exemple 34
s: Socrate; H(x): x is a man; M(x): x is mortal:

1. 
M(s) 
(hyp) 
2. 
H(s)'\/ M(s) 
(add 1) 
3. 
H(s)> M(s) 
(imp 2) 
4. 
(for allx ) [H(x)> M(x)] 
(ug 3) 

Socrate is a mortal, therefore every man is a mortal.
This proof sequence is incorrect: you cannot apply ug at step 4.
Indeed s is Socrate, and not an arbitrary element of the domain.
Exercice 15
Prove the following arguments:

(for allx ) [P(x)/\ Q(x)] > (for allx ) P(x)/\(for allx ) Q(x)
 (for allx ) P(x)/\(for allx ) Q(x) > (for allx ) (P(x)/\ Q(x))
 (for allx ) [P(x)/\(for ally ) Q(x,y)] > (for allx ) (for ally ) Q(x,y)
3.2.3 Existential instantiation:
Exemple 35
DOI: contents of the fridge.
There exists a fruit; therefore, I can take a fruit.
F(x): x is a fruit
(there existsx ) F(x) > F(s)
Définition 24
Rule of existential instantiation (ei):
From: (there existsx ) P(x)
Can derive: P(s)
s must be a newly created variable
Exemple 36
(for allx ) [P(x)> Q(x)] /\(there existsy ) P(y) > Q(s)

1. 
(for allx ) [P(x)> Q(x)] 
(hyp) 
2. 
(there existsy ) P(y) 
(hyp) 
3. 
P(s) 
(ei 2) 
4. 
P(s)> Q(s) 
(ui 1) 
5. 
Q(s) 
(mp 3,4) 

3.2.4 Existential generalization:
Exemple 37
DOI: contents of the fridge
I see a fruit, therefore there exists a fruit.
F(s) > (there existsx ) F(x)
Définition 25
Rule of existential generalization (eg)
From: F(s)
Can derive:(there existsx ) F(x)
Exemple 38
(for allx ) [P(x)> Q(x)] /\ (there existsy ) P(y) > (there existsy ) Q(y)

1. 
(for allx ) [P(x)> Q(x)] 
(hyp) 
2. 
(there existsy ) P(y) 
(hyp) 
3. 
P(s) 
(ei 2) 
4. 
P(s)> Q(s) 
(ui 1) 
5. 
Q(s) 
(mp 3,4) 
6. 
(there existsy ) Q(y) 
(eg 5) 

Exemple 39
(there existsx ) P(x)/\(there existsx ) Q(x) > (there existsx ) [P(x)/\ Q(x)]

1. 
(there existsx ) P(x) 
(hyp) 
2. 
(there existsx ) Q(x) 
(hyp) 
3. 
P(s) 
(ei 1) 
4. 
Q(s) 
(ei 2) 
5. 
P(s)/\ Q(s) 
(add 3,4) 
6. 
(there existsx ) [P(x)/\ Q(x)] 
(eg 5) 

We have seen that the previous argument is incorrect:
So, where is the flaw in the following proof ?
For example, s cannot have been created by existential instantiation
Exercice 16
Prove or disprove the following arguments

(there existsx ) (for ally ) Q(x,y) > (for ally ) (there existsx ) Q(x,y)
 (for allx ) (there existsy ) Q(x,y) > (there existsy ) (for allx ) Q(x,y)
3.2.5 Deduction Method and Temporary Hypothesis:
Exemple 40
P(s)>(for ally ) Q(x,y) > (for ally ) [P(s)> Q(x,y)]
3.3 Conclusion
Goal: formalization of arguments and proofs.

Propositional logic:

propositions A, B, ...
connectives, well formed formula
truth table
argument valid iff wff is a tautology
proofs:

compute the truth table (algorithm)
 formal logic, deduction rules
Ex 24, 37
 Predicate logic

variables, predicates
domain of interpretation
connectives, wff
interpretation
truth table: all possible interpretations (infinite)
argument valid iff wff is true for all possible interpretations
proofs:

no algorithm
 formal logic, deduction rules
Those two logic are correct and complete.
They are still not powerful enough to represent all real life argument.
For this we would need of more powerful logics (2nd order), that allows
for quantifying other sets.
Problem: it's often difficult, if not impossible to prove that those
logics are complete and correct!
We won't need to go into those details.
We have seen enough lowlevel logic to help us write safely less formal
proofs.
Chapter 4 Proof techniques (section 2.1)
4.1 Theorems and Informal proofs
What we have seen so far:

Argument
 P_{1}/\···/\ P_{n} > Q
 Syntax
 how it's written
 Semantic
 meaning in a given interpretation
 Valid argument

True for all interpretations
True because of its very structure
Proofs of valid argument are purely based on syntactical rewriting
rules.
Only arguments that are true for all interpretations can be proved.
We are now interested to work in a particular subject (say integer
arithmetic).
Définition 26
A theorem is an argument that is valid in this particular subject.
Exemple 41
If x is even and y is even then xy is even.
E(x): x is even
(for allx ) (for ally ) [E(x)/\ E(y) > E(xy)]
This argument is true in this context, but not universally true.
How to prove it ?
Implicitly, we add new hypothesis which reflects basic facts of this
subject.
For example, we add the following hypothesis:
Using those new hypothesis, it's now possible to write a formal logic
proof of the theorem.
See [1, Example 4 p. 87]
4.1.2 Formal and informal proofs:
Remember that the goal of a proof is to be read by humans (in particular
yourself) in order to convince those humans.
The formal proof above is convincing in the sense that you can check
that all the steps are valid. However, it's very difficult to extract
the meaning of the proof:

Why does it work ?
 How could I reuse it for a similar problem ?
The problem is that the keys of the proofs are buried under layers
of insignificant details.
So, starting from now, we will write informal proofs:
Définition 27
An informal proof is a narrative descriptions, of the ideas
and of the important steps of the proof, without extraneous details.
Exemple 42
A proof of the theorem above could be written as follow:
Proof.
Let x and y be two even integers. We can take n and m
such that x=2n and y=2m.
Then, xy=2(2nm). Since 2nm is an integer, xy is an even number.
Let see which details we omitted:

Let x and y be two even integers.
Implicit universal instantiation
 We can take n and m such that x=2n and y=2m.
Implicit universal instantiation for x and y of the definition
of an even number
Implicit existential instantiation for n and m
 Then, xy=2(2nm)
Implicit use of rules of arithmetic
 Since 2nm is an integer, xy is an even number:
Implicit universal instantiation of the definition of an even number
Implicit universal generalization to get the final result
Problem 1
Which details can we omit, and which not ?
A reader will be convinced by the proof if he can check that he could
translate each step of the proof into one or several steps of a formal
proof.
So, this all depends on WHO reads the proof!
You should not write your proof for your instructor, but for yourself,
and for everybody else in the class.
A good rule of thumb is to imagine yourself rereading the proof in
a few month, and to check that even then, you could possibly translate
the proof into a formal one.
The difference between formal and informal proofs is very similar
to the difference between assembly and, say, C++ or Java. A C++ program
is not usable by itself. It's usable because it's possible to translate
it into assembly language.
However, there does not exists, and most likely will never exists,
a compiler that transforms informal proofs into formal proofs. English
is much to rich a language for this.
4.1.3 Formal and informal theorems:
Most of the time, we also won't write theorems as a formal argument,
but rather with an English sentence that could be translated into
a formal argument:
Théorème 2
Let x and y be two integers. [Some definitions: interpretation]
Assume x and y are even. [Some hypothesis: P_{1}/\···/\ P_{n}]
Then, xy is even. [Consequent: Q]
4.1.4 To Prove or not to prove
In textbooks, you can be asked: prove that ...

you know in advance it's true;
 you just have to figure out how to prove it.
Usually, in real life, you first have to find what to prove.

you don't even know in advance if it's true.
Two jobs:

Find the good questions
 Prove or disprove those questions
Définition 28
A conjecture is a statement that you guess is true, but that
you have not proved or disproved yet.
Classical steps:

Explore some examples
 Try to see some pattern emerging
 Formulate a conjecture
 Try to prove (or disprove it)
Steps 13 are inductive reasoning, whereas step 4 is deductive reasoning.
Finding the good questions is as important as solving them !
Exemple 43
Fermat's conjecture.
4.1.5 Some ``research'' around the factorial
Définition 29
Let n be an integer. The factorial of n is the number n!:=n(n1)···1.
For example, 1!=1 and 4!=4·3·2·1=24.
Problem 2
How big is n! ?
4.2 Proof techniques
We want to prove some argument of the form P > Q.
4.2.1 Disproof by counter example
Conjecture 1
If n is a positive integer, then n!<n^{3}.
4.2.2 Direct proof
Définition 30
Direct proof

Assume P
 Deduce Q
Exemple 44
Prove that if x and y are even, then xy is even.
Exercice 17
Prove that if x and y are even, then x+y is even.
4.2.3 Proof by contraposition
Exemple 45
Prove that if n^{2} is odd, then n is odd.
Hint: prove instead that if n is even, then n2 is
even.
Définition 31
Proof by contraposition:

Assume Q' (the consequent is false)
 Prove P' (the antecedent is also false)
This technique relies on the fact that P > Q is equivalent to
Q' > P'.
Exercice 18
Prove that xy is odd if and only if x and y are odd.
4.2.4 Exhaustive proof
Problem 1
Can a proof by example be valid ?
Yes, if there is a finite number of cases to be treated.
Exemple 46
Propositional logic formula (the truth table is finite)
Définition 32
Proof by exhaustion means that all cases have been exhausted.
(and so are you ...).
Exemple 47
Drawing a figure without lifting the pencil and without retracing
a line.
4.2.5 Some ``research'' around rational numbers
Définition 33
A number x is rational if it can be written as p/q,
where p and q are integers.
Exemple 48
5, 7/5, 1/3, 14/4, 7/2,
... are rational.
Problem 2
Properties of rational numbers ?
Remarque 7
If x is rational, it's always possible to choose p and q
so that:

q is positive
 p and q are relatively prime
I.e., the biggest common divisor of p and q is 1.
Problem 3
Are all numbers rational ?
Problem 4
Assume x and y are rational.

Is x+y rational ?
 Is xy rational ?
 Is x/y rational ?
Problem 5
Is the square root of an integer a rational number ?
Problem 6
Prove that the square root of 2 is irrational.
4.2.6 Proof by contradiction
Théorème 3
The square root of 2 is irrational.
Proof.
Let's assume the square root of 2 is rational.
Let p and q be two integers such that (2)^{1/2}=p/q
and p and q are relatively prime.
Then, we have 2=(p/q)^{2}, and so 2q^{2}=p^{2}.
Therefore p^{2} is even, and we have seen that this implies that
p is also even.
Let k be the integer such that p=2k.
Then, we have 2q^{2}=p^{2}=(2k)^{2}=4k^{2}, and so q^{2}=2k^{2}.
It follows that q^{2} is even, and so q is also even.
Conclusion: p and q are both even.
That's a contradiction, since p and q are relatively prime!
Définition 34
Proof by contradiction

Assume the contrary
 Deduce a contradiction
This technique relies on the fact that:

Q/\ Q' is always false
 if P>0 is true, then P is false.
4.2.7 Serendipity
Exemple 49
The chess board problem.
4.3 Summary
Goal 
Technique 
Name 
P > Q 
Assume P ; deduce Q. 
Direct proof/Deduction method 
P'> Q' 
Prove Q> P. 
Proof by contraposition 
Q' 
Assume Q; deduce a contradiction. 
Proof by contradiction 
P<> Q 
Prove P> Q; prove Q> P. 

P/\ Q 
Prove P ; prove Q. 

P\/ Q 
Prove P'> Q 

(P_{1}\/ P_{2})> Q 
Prove P_{1}> Q; prove P_{2}> Q 
Proof by cases 
(for allx ) Q(x) 
Let x; prove Q(x) 
ui / ug 
(there existsx ) Q(x) 
Construct a such that Q(a) 
eg 

Be smart 
Serendipity 
Chapter 5 Induction  Recursion
5.1 Introduction
Définition 35
A sequence
S is a list of objects, enumerated in some order:
S(1),S(2),S(3),...,S(k),...
A sequence can be defined by giving the value of S(k), for all
k.
Exemple 50
S(k):=2^{k} defines the sequence: 2,4,8,16,32,...
Imagine instead that I give you the following recipe:

(a) S(1):=1.
 (b) If k>1, then S(k):=S(k1)+k.
Can you compute S(2)? S(3) ? S(4) ?
Could you compute any S(k), where k>0.
Proposition 1
The sequence S(k) is fully and uniquely defined by (a) and (b).
Définition 36
We say that S is defined recursively
by (a) and (b).

(a) is the base case
 (b) is the induction step
Exemple 51
The stair and the baby.
This is the idea of recursion (or induction), which is very useful
in maths and computer science.
It allows to reduce an infinite process to a finite (small) number
of steps.
Exemple 52
Define S(k) by S(1):=4.
Problem: how to compute S(2) ?
Exemple 53
Define S(k) by S(k):=2S(k1)
Problem: both the following sequences satisfies this definition!

1,2,4,8,16,...
 3,6,12,24,48,...
Exemple 54
Proof that any integer is even.
Pitfall: Both base case and induction step are necessary !
You need to know how to start, and you need to know how to continue.
5.2 Proofs by induction
Problem 1
Let S be the sequence defined by:

S(1)=1
 S(k):=S(k1)+2^{k1}
Goal: compute S(60)

S(1) 
= 
1 
= 
S(2) 
= 
1+2 
= 
S(3) 
= 

= 
S(4) 
= 

= 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
S(k) 
= 
1+···+2^{k} 
= 

Conjecture 2
S(60)=
The formula S(k)=2^{k}1 is called a closed form formula
for S(k).
It allows to compute S(k) easily, without computing all the previous
values S(1),S(2),... S(k1).
So, how to prove this conjecture ?

Introduce some notation

Let P(k) be the predicate: S(k)=2^{k}1.
For any positive integer k, P(k) is either true or false.
 Look at examples

Exercice 19
Try the following:

Prove P(1),P(2),P(3)
 Assume that P(27) is true. Can you prove that P(28) is true
?
Now, can you prove P(60)?
5.2.1 First principle of mathematical induction:
Théorème 4
First principle of mathematical induction
Let P(k) be a predicate. If:
(a) P(1) is true
(b) For any k>1, P(k1) true implies P(k) true
Then: P(k) is true for all k>=1.
Définition 37
P is the inductive hypothesis.
(a) is the base case.
(b) is the induction step.
Exemple 55
Consider the sequence S defined as above by:
(a) S(1):=1.
(b) S(k):=S(k)+2^{k1}.
Let's prove that for all k, S(k)=2^{k}1.
Proof.
Let P(k) be the predicate S(k)=2^{k}1.

Base case:
S(1)=1=2^{1}1. So, P(1) is true.
 Induction step:
Let k>1 be an integer, and assume P(k1) is true: S(k1)=2^{k1}1.
Then, S(k)=S(k1)+2^{k1}=2^{k1}1+2^{k1}=2^{k}1.
So, P(k) is also true.
Conclusion: by the first principle of mathematical induction,
P(k) is true for all k>=1, i.e. S(k)=2^{k1}.
Exercice 20
Let S(k):=1+3+5+···+(2k1). Find a closed form formula for S(k).
Exercice 21
Prove that k!>2^{k} for any k>=4.
Exercice 22
Prove that for any integer k, 2^{2k}1 is divisible by 3.
Exercice 23
Prove that a+ar+ar^{2}+···+ar^{n}=aar^{n}/1r.
Exercice 24
Find a closed form formula for S(k):=1^{4}+2^{4}+··· k^{4}.
Hint: the difficulty is to find a suggestion. What tool could you
use for this?
Exercice 25
The chess board problem.
5.2.2 Other forms of induction:
Exemple 56
Define the sequence F(n) by:
F(1):=1
F(2):=1
F(k)=F(k1)+F(k2), for all k>2.
Exercice 26
Can you compute F(3),F(4),F(5)?
This sequence is the famous and useful Fibonacci sequence.
Problem 2
To compute F(k), you not only need F(k1), but also F(k2).
So that does not fit into the previous induction scheme.
That's fine, because to compute F(k), you only need to know the
values of F(r) for r<k.
Théorème 5
Second principle of Mathematical induction:
Let P(k) be a predicate. If
(a) P(1) is true,
(b) For any k>1, [ P(r) true for any r, 1<= r<k ]
implies P(k) true,
then: P(k) is true for all k>=1.
Exemple 57
Any integer is a product of prime integers
Exemple 58
The coin problem
We did not prove that the first (or the second) principle of induction
were valid.
Let's just give an idea of the proof:
Théorème 6
The three following principles are in fact equivalent:

the first principle of induction
 (b) the second principle of induction
 (c) the principle of well ordering:
every non empty collection of positive integers has a smallest member.
Problem 3
Is the principle of well ordering valid for Z ? R ? C
?
5.3 Other uses of induction
Induction is a much more general problem solving principle.
If you have a family of problems such that:

There is some measure of the size of a problem
 You can solve the smallest problems
 You can breakup each bigger problems in one (or several) strictly
smaller cases, and build from it a solution for the big problem.
Then, you can solve by induction any problem in your family.
Exemple 59
Problem: computation of F(k).
Measure of the problem: k.
Value of F(1) is known.
F(k) can be computed from the values of F(k1) and F(k2)
Exemple 60
Inductive definition of well formed formulas from formal logic
<basic proposition>: A  B  C  ...
<binary connective>: /\  \/  >  <
 <>
<wff>: <basic proposition>  <wff>'  (<wff>)  <wff> <binary
connective> <wff>
This kind of inductive description is called *BNF* (Backus Naur
form).
Exemple 61
Recursive proof of a property of wff
Théorème 7
If a wff contains n propositions (counted with repetition: in A\/ A,
there are two propositions), then it contains n1 binary connectives.
Exemple 62
Recursive algorithm for the computation of the value of a wff:

bool function Evaluation(P, C)
// Precondition: P is a wff, C is the context, i.e.
// the truth values of all the basic propositions.
// Postcondition: returns the truth value of P in the // context C
Begin
If P is a proposition (say A) Then
Return the truth value of A;
ElseIf P is of the form (Q) Then
Return Evaluation(Q, C);
ElseIf P is of the form Q' Then
Return not Evaluation(Q,C);
ElseIf P is of the form Q_{1}/\ Q_{2} Then
Return Evaluation(Q_{1}, C)and Evaluation(Q_{2}, C);
... // Other binary operators
Else
Error P is not a wff;
End;
Remarque 8
There are programming languages that are particularly well adapted
to write this kind of algorithms. Actually, the real program would
almost look like the pseudocode. See for example CAML
ftp://ftp.inria.fr/lang/camllight/.
Exemple 63
Bijection between words of parenthesis and forests
5.4 Conclusion
Induction is a fundamental technique for solving problems, in particular
in computer science.
It's the mathematical formalization of the divide and conquer approach.
Chapter 6 Proof Of Correctness (Sections 1.6, 2.3)
6.1 Introduction
``Beware of bugs in the above code; I have only proved it correct,
not tried it.''
Donald E. Knuth http://wwwcsfaculty.stanford.edu/~knuth/
How to check if a program is correct ?
Why checking if a program is correct ?
What is a correct program ?
Définition 38
Specification: precise description of how a program should
behave.
A program is correct, if it behaves in accordance with its
specifications.
How to check if a program is correct?

Testing
Designing good test data sets
You can prove there is a bug. No proof of no bug!
 Proof of Correctness
6.2 Correctness
6.2.1 Assertions

...
x=sqrt(r); // r should be >= 0 !
x=a[i]; // i should be in the correct range
...
Définition 39
Assertion: condition on the variables of a program that should
be verified at some given step, if the program is running correctly.
Using assert in C/C++:

#include <assert.h>
...
assert(r>=0);
x=sqrt(r);
assert((0<=i) && (i<MAX));
x=a[i];
...
Each assertion is automatically checked:

If the assertion is not verified, the program stops immediately with
a message like:

Assertion failed, line 35 of file foo.c

That's more informative for the user than a segmentation fault at
the following line.
 This can be deactivated by adding a #define NDEBUG preprocessor directive.
With C++, java and other languages, you can use exceptions
instead of assert:

More informative error messages
 Error recovery mechanisms
Static type checking also makes some kind of assertion checking.
One way or the other, check assertions in your programs.
6.2.2 Specification of a function:

int factorial(int n);
// Preconditions: n is a nonnegative integer
// Postcondition: returns n!
...
int factorial(int n) {
assert(n>=0);
if (n==0) return 1;
return n*factorial(n1);
}
The specification of a program can be formalized as follow:

P: program
 X: input
 P(X): output
 Q: precondition: predicate Q(X) on the input
 R: postcondition: predicate R(X,P(X)) on the input and the output
Exemple 64
Program to compute of a square root

P: y:=sqrt(x);
 X: x
 P(x): y
 Q(x): x>=0
 R(x,y): y^{2}=x
Définition 40
P is correct if (for all X) Q(X)> R(X,P(X)).
A Hoare triple is just a short hand notation: {Q} P
{R}
Exemple 65
{x>=0} y:=sqrt(x);; {y^{2}=x}
6.3 Proof of Correctness
6.3.1 Divide and Conquer
Let P be a big program:

s_{0}
s_{1}
s_{2}
...
s_{n1}
To prove {Q} P {R}, we break P into elementary steps,
and insert assertions that describes the state of the program at each
step:

{Q}
s_{0}
{R_{1}}
s_{1}
{R_{1}}
s_{2}
...
s_{n1}
{R}
P is correct, if each step is correct, i.e. the following Hoare
triples hold:

{Q} s_{0} {R_{1}},
 {R_{1}} s_{1} {R_{2}},
 ...
 {R_{n1}} s_{n1} {R}
To prove that the elementary steps are correct, we will use some syntactic
rules, exactly as we did in formal logic.
6.3.2 Assignment rule:
Consider the following Hoare triple: {Q} x:=e {R}
Théorème 8
If from Q you can derive R with x substituted everywhere
by e,
then the Hoare triple is valid.
Exemple 66
{x=2} y:=x+1 {y=3}
When substituting, y=3 becomes x+1=3.
From x=2, you can deduce x+1=3.
So this Hoare triple holds.
Exemple 67
{x>0} x:=x+1 {x>1}
Here it can be confusing.
The same name x stands for both the value of x before and after
the assignments.
If you get confused, just rename the variable:
{x_{0}>0} x_{1}:=x_{0}+1 {x_{1}>1}
When substituting, x_{1}>1 becomes x_{0}+1>1, which you can
deduce from x_{0}>0.
6.3.3 Conditional Rule
Consider the following Hoare triple:

{Q}
if condition B then
P_{1}
else
P_{2}
end if
{R}
Théorème 9
If the Hoare triples {Q/\ B} P_{1}{R} and {Q/\ B'}
P_{2} {R} hold,
then the Hoare triple above holds.
Exemple 68
[1, Exercise 11 p. 78]
Consider the following Hoare triple:

{Q}
while condition B do
P
end_while
{R}
{Q} describes the state of the program before the loop.
We need to have a predicate which describes the state of the program
DURING the loop:
The loop invariant {S}
Most of the time, {Q} will do the job, but not always.
Théorème 10
If the Hoare triple {S/\ B} P {S} holds, then the
following Hoare triple will hold:

{S}
while condition B do
P
end_while
{S/\ B'}
6.3.5 A simple example
Consider this little program:

Product(x,y)
// Precondition: x and y are nonnegative integers
// Postcondition: returns the product of x and y
begin
i:=0;
j:=0;
while ( (i=x)' ) do
j=j+y;
i:=i+1;
end while
// Assertion: { j = x*y }
return(j);
end;
It's pretty clear that this algorithm it's correct.
Let's check it anyway to see how it works.
Proof.
We need a loop invariant, which will describe the state of the program
after k iterations.
A reasonable guess is S : j=iy
Let i_{k}, j_{k} and S_{k} be the value of i, j and
S after k iterations.
Base case: using the assignment rule, i_{0}=0 and j_{0}=0 so
S_{0} holds.
Induction step:
Assume the invariant holds after k1 iterations: j_{k1}=i_{k1}y
We need to prove that the invariant is preserved by the k th iteration.
Otherwise said, we have to check that the following Hoare triple holds:

{S_{k1}/\(i=x)'}
j_{k}:=j_{k1}+y;
i_{k}:=i_{k1}+1;
{S_{k}}
Let's applying the assignment rule.
By substituting i_{k} and j_{k} by the expressions they have
been assigned with, we get:

S_{k} 
<=> 
j_{k} 
= 
i_{k}y 


<=> 
j_{k1}+y 
= 
(i_{k1}+1)y 
(By the substitution rule) 

<=> 
i_{k1}y+y 
= 
(i_{k1}+1)y 
(By S_{k1}) 

<=> 
i_{k1}y+y 
= 
i_{k1}y+y 

So, S_{k} indeed holds.
By induction, after any number k of iterations, S_{k} still
hold.
At the end, both S_{k} and i=x hold, so, as expected: j=iy=xy
Remarque 9
We have proved that after the end of the execution the result is correct.
But what if the program does not terminate and loop forever?
This is not usually consider a proper behavior.
We only have spoken about partialcorrectness.
A full proof of correctness needs also a proof of termination!
6.3.6 A more sophisticated example: The Euclidean algorithm

GCD(a,b)
Begin
// Precondition: a and b are nonnegative integers
// with a>=b.
// Postcondition:
// returns gcd(a,b), the greater common divisor of
// a and b.
Local variables: i, j, q, r
i:=a;
j:=b;
while j>0 do
// Assertion:
// r is the rest of the integer division i/j
r:= i mod j;
i:= j;
j:= r;
end while
// Assertion: i is the gcd of a and b.
return(i);
end;
Exercice 27
Use the Euclidean algorithm to compute:
GCD(8,8), GCD(14,4), GCD(31,17).
Here the proof of correctness of the algorithm is nontrivial.
Proof.
Let i_{k} and j_{k} be the value of i and j after k
iterations.
We need to find an invariant which describes the state of the program
after each iteration.
Take S_{k}: gcd(i_{k}, j_{k})=gcd(a,b).

Base case:
Before the loop, i_{0}=a and j_{0}=b.
So the invariant S_{0} holds: gcd(i_{0}, j_{0})=gcd(a,b);
 Induction step:
Assume S_{k1} holds, that is gcd(i_{k1}, j_{k1})=gcd(a,b).
We just need to prove that gcd(i_{k}, j_{k})=gcd(i_{k1}, j_{k1}).
By the assignment rule, we have:

r is the rest of the division of i_{k1} by j_{k1},
 i_{k}=j_{k1},
 j_{k}=r.
So, by the definition of the integer division, we have:
i_{k1}=qj_{k1}+j_{k}
for some integer q.

Assume x divides both i_{k} and j_{k}.
Then, of course j_{k1}, since j_{k1}=i_{k}.
By the equation above, x also divides i_{k1}.
 Assume x divides both i_{k1} and j_{k1}.
Then of course x divides i_{k}.
Once again, using the equation above, x also divides j_{k}
Therefore, gcd(i_{k}, j_{k})=gcd(i_{k1}, j_{k1})=gcd(a,b),
as wanted.
 At the end:
By induction, the loop invariant S still holds: gcd(i, j)=gcd(a,b)
Moreover, the loop condition failed: j=0.
So, gcd(a,b)=gcd(i, j)=gcd(i, 0)=i,
as expected.
6.4 Conclusion
6.4.1 A note on automatic program proving:

There is no mechanical way to decide even if a program terminates
or not.
 Traditional imperative languages:

Automatic proving is very difficult because of side effects
 side effects needs to be considered as some form of output
 Functional languages (lisp, scheme, ML, caml):

Designed to make it easier to prove correctness
 Fun to program. Try them!
 Two difficulties:

Finding the correct assertions
 Checking that the implications holds
 Mixed approach:

The programmer gives the main assertions
 The prover derives the other assertions
6.4.2 Check assertions in your programs!

Use static type checking / assert / exceptions
 When debugging, you can ask for all assertions to be checked
 Documentation
 Help for the proof
Chapter 7 Analysis of Algorithm (Section 2.5)
Exemple 69
Finding an element in a list.
Computing the nth Fibonacci number F(n).
Problem 1
How long will my program take to execute ?
How much memory do I need ?
How efficient is my algorithm ?
Is it the best one ?
Which algorithm should I choose ?
Goal: "measure" the quality of an algorithm:

Time
 Memory
 Easiness of writing
7.1 Complexity of an algorithm: Measuring time
Définition 41
Decide which operations are basic (i.e. multiplications).
Complexity of the algorithm: how many basic operations are
needed.
Exemple 70
Sequential search

SequentialSearch(x,l)
// Preconditions:
//  l is a list ["bonjour","car","hello","juggler"]
//  x is a string
// Postcondition: return true iff x is in the list
Begin
// compare successively x with l[1], l[2], ..., l[n]
For i From 1 To n Do
If (x=l[i]) Then Return TRUE; EndIf
EndFor;
Return FALSE;
End;
Basic operations: comparison of two strings.
How many operations ?

x = "unicycle":
 x = "juggler":
 x = "hello":
 x = "car":
Worst case: n basic operations
Average case: depends on the probability for x to be in the list.
Difficult !
Résumé 1
n: size of the data (here, the length of the list)
Worst case: maximum number of operations for a data of size n.
Average case: average number of operations for a data of size n
(difficult)
Exemple 71
Binary search
A non sorted dictionary is useless!!!

BinarySearch(x, l)
// Preconditions:
//  l is sorted ["bonjour","car","hello","juggler"]
//  x is a string
// Postcondition: return true iff x is in the list
Begin
If n=1 Then
Return x=l[1];
ElseIf x <= l[n/2] Then
Return BinarySearch(x, [l[1], ..., l[n/2]]);
Else
Return BinarySearch(x, [l[n/2]+1, ..., l[n]]);
End if
End
Basic operation: comparison of two strings.
How many operations ?
For n, there is at most k operations, where k is the smallest
integer such that 2^{k1}>= n.
log(n): logarithm of n other base 2
Number of operations: log(n)
7.1.1 How precise ?
The estimate above does not give a precise measure of the time needed.
It depends in particular on:

What computer you are using
 How long are the strings
 The computer language:

How fast are strings
 How fast are for loops
 How fast are function calls
A precise measure of the time is often hard to obtain.
Most of the time, to decide between to algorithms, a rough order of
magnitude is enough.
Draw the graphs to get an idea if there are big differences

4n operations / 7n operations?
 4n operations / 7n operations ?
 n operations / n+3 operations ?
 log(n) operations / n operations ?
 n operations / n^{2} operations ?
 n operations / exp(n) operations ?
Définition 42
An algorithm is of complexity O(n^{5}) if it needs less than Cn^{5}+K
operations to execute, where C and K are some constant.

O(1): constant (random access to an element of an array)
 O(log(n)) : logarithmic (random access in a sorted list
by binary search)
 O(n): linear (random access in a linked list; sequential search)
 O(nlog(n)): (sorting a list by a good algorithm)
 O(n^{2}): quadratic (sorting a list by bubble sort)
 O(n^{k}): polynomial
 O(exp(n)): exponential
Worst case analysis is usually easier. It is important if your need
a warranty on the time needed for each data (e.g.: real time data
acquisition).
Average case analysis is more difficult. It is important when running
over a large sample of data.
7.1.2 Case study: the Fibonacci sequence
Here is a program written in MuPAD www.mupad.de (free Computer
Algebra system).
It present different implementations of the computation of the Fibonacci
sequence.

/*
fibonacci_rec (n: Dom::Integer);
fibonacci_rec_remember (n: Dom::Integer);
fibonacci_loop_array (n: Dom::Integer);
fibonacci_loop (n: Dom::Integer);
Precondition: n non negative integer
Postcondition: returns the nth fibonacci number
Also increments the global variable operations.
*/
fibonacci_rec:=
proc(n: Dom::Integer)
begin
if n=0 then return(1); end_if;
if n=1 then return(1); end_if;
operations:=operations+1; // Count operations
return(fibonacci_rec(n1)+fibonacci_rec(n2));
end_proc:
fibonacci_array:=
proc(n:Dom::Integer)
local i, F;
begin
if n=0 then return(0); end_if;
if n=1 then return(1); end_if;
F[0]:=0;
F[1]:=1;
for i from 2 to n do
operations := operations+1; // Count operations
F[i] := F[i1] + F[i2];
end_for;
return(tableau(n));
end_proc:
fibonacci_loop:=
proc(n:Dom::Integer)
option remember;
local i, value, previous, previousprevious;
begin
if n=0 then return(0); end_if;
if n=1 then return(1); end_if;
previous:=0;
value:=1;
for i from 2 to n do
operations := operations+1; // Count operations
previousprevious:= previous;
previous := value;
value := previousprevious+previous;
end_for;
return(value);
end_proc:
fibonacci_rec_remember:=
proc(n: Dom::Integer)
option remember;
begin
if n=0 then return(0); end_if;
if n=1 then return(1); end_if;
operations := operations+1; // Count operations
return(fibonacci_rec_remember(n1)+
fibonacci_rec_remember(n2) );
end_proc:
printentry:=
proc(x)
begin
userinfo(0, NoNL, x);
userinfo(0, NoNL, "t");
end_proc:
for n from 0 to 30 do
printentry(n);
operations:=0;
printentry(fibonacci_rec(n));
printentry(operations);
operations:=0;
fibonacci_array(n);
printentry(operations);
operations:=0;
fibonacci_loop(n);
printentry(operations);
operations:=0;
fibonacci_rec_remember(n);
printentry(operations);
// Clear remember table;
fibonacci_rec_remember:=
subsop(fibonacci_rec_remember,5=NIL);
print();
end_for:
Figure 7.1: Measurement of the complexity of the different implementations of
the Fibonacci sequence
Problem 2
Why is Fibonacci rec so slow ? (Trace its execution)
How to avoid this?
Don't throw away the results !
The remember option.
Time consumption:

Fibonacci rec: exponential
 Fibonacci rec_remember: linear
 Fibonacci array: linear
 Fibonacci loop: linear
Memory consumption:

Fibonacci rec: exponential
 Fibonacci rec_remember: linear
 Fibonacci array: linear
 Fibonacci loop: constant
Easiness to write:

Fibonacci rec: straight forward
 Fibonacci rec_remember: straight forward
 Fibonacci array: easy
 Fibonacci loop: harder
7.2 The P=NP conjecture
7.2.1 Case study: Satisfying a well formed formula of propositional logic
We consider a well formed formulas of propositional logic.
Définition 43
A choice of truth values for A, B, C, ... satisfies
P if P evaluates to true.
Exemple 72
A true and B true satisfies A/\ B.
A true and B false do not satisfy A/\ B.
Problem 1
Given the truth values of A, B, C, ... check if they satisfy
P.
Algorithm ?
Complexity ?
Définition 44
A wff P is satisfiable if at least one choice of truth values
for A, B, C, ... satisfies P.
I.e. P is not a contradiction.
Exemple 73
A/\ B is satisfiable, whereas A/\ A' is not.
Problem 2
Testing if a wff P is satisfiable.
Algorithm ?
Complexity ?
Is this the best algorithm ?
Does there exist a polynomial algorithm ?
That's an instance of a problem where:

checking a solution is easy (polynomial)
 finding the solution seems to be hard (the best known algorithm is
exponential)
Exemple 74
Checking if the solution of a crossword puzzle is correct is easy.
Solving the crossword puzzle is not !
Exemple 75
Backpack problem with size=23, and objects=5,7,7,20,17,5,11,5,19,14.
7.2.2 Polynomial and Non Deterministic Polynomial problems
Définition 45
Problems as above are called NP (Nondeterministic Polynomial)
P: collection of all polynomial problems
NP: collection of all NP problems
Problem 3
P = NP ?
In other words:
If it's easy to check a potential solution of a problem,
is there necessarily an easy way to find the solution ?
Intuitively, no. This cannot be true.
Well, despite a lot of research, we still have no proof!
This is the main problem of theoretical computer science since 30
years.
So, this is YOUR job to find the solution !
Part III
Sets and Combinatorics
Chapter 8 Sets (section 3.1 )
8.1 Introduction
Problem p 161
Goal:

Introduce the formalism of set theory
 Use it to solve counting problems
8.2 Sets and basic manipulations of sets
8.2.1 Definition
Définition 46
A set A is a collection of objects.
Defining property: you can ask whether an object a belongs to A
or not.
Objects or elements: a,b,c,....
Sets: A,B,C,....
Predicates:

ain S: a belongs to S, or a is an element of S, or
S contains a;
 anot in S: a does not belong to S.
Two sets A and B are equal (A=B) if they contain the
same elements:
(for all x)(xin A)<=>(xin B)
8.2.2 How to define a set S?
You have to describe which elements are in the set S and which
are not.

List the elements of the set S:
S:={ a,b,c}
ain S is true, whereas din S is false.
The order and the repetition are irrelevant:
{ a,b,c}={ a,b,b,b,c}={ c,b,a}
Note that a and { a} are two different things. The first is
the object a, whereas the second one is the set containing a,
and no other elements.
 Provide a predicate which determines which element are in S:
S:={ x  P(x)}
S:={ x  x is an integer and 3<x<=7}
5in S is true, whereas 3not inS, and no car belongs to A.
 Use a recursive definition, as for the set of well formed formulas:
S contains basic propositions A,B,C,...
If P and Q belong to S, then P', P/\ Q, P\/ Q,
P> Q, P<> Q also belong to S.
Some usual sets:

The empty set:
Ø:={ x  false}
 Numbers:
N:={ x  x is a non negative integer }
Z:={ x  x is an integer }
Q:={ x  x is a rational number }
R:={ x  x is a real number }
C:={ x  x is a complex number }
S^{*}:={ x  xin S and x!=0}
S^{+}:={ x  xin S and x>=0}
8.2.3 Relationships between sets
Exercice 28
[1, Exercise 10 p. 117]
R:={1,3,pi,4,1,9,10}, S:={{1},3,9,10}
T:={1,3,pi}, U:={{1,3,pi},1}

Sincluded in or equal to R?
 1in R?
 1in S?
 1included in or equal to U?
 {1}included in or equal to T?
 {1}included in or equal to S?
 Tincluded in R?
 {1}in S?
 Øincluded in or equal to S?
 Tincluded in or equal to U?
 Tin U?
 Tnot in R?
 Tincluded in or equal to R?
 Sincluded in or equal to{1,3,9,10}?
8.2.4 The powerset of a set
The powerset (S) of a set S is the set of all subsets of
S.
Exemple 76
What are the elements of the following powersets ?

({1,2})=
 ({1})=
 (Ø)=
Exercice 29
What is the size of (S) if S has n elements ?
8.2.5 Binary and unary operations
Définition 47
¤ is a unary operation on a set S if for every
element x of S, ¤ x exists, is unique and is a member
of S.
¤ is a binary operation on a set S if for every
ordered pair (x,y) of elements of S, x¤ y exists, is
unique and is a member of S.
Examples:

is +a binary operation on N ?
 is +a binary operation on {0,1,2}?
 is ×a binary operation on {0,1}
 is a binary operation on N ?
 is /a binary operation on Z ?
 are +,, and ×binary operations on Q ?
 are +,, ×and /binary operations on Q^{*}?
 is (x)^{1/2} a unary operation on R^{+} ?
 are /\, \/, >, ', ... operations
on { true,false}? on wff ?
A binary operation can be defined by its multiplication table.
Exemple 77
({ true,false},/\)
/\ 
false 
true 
false 
false 
false 
true 
false 
true 
An operation does not necessarily need to have a particular meaning.
Exemple 78
({2,5,9},¤)
¤ 
2 
5 
9 
2 
2 
5 
2 
5 
2 
9 
5 
9 
9 
2 
5 
5¤9=?
8.3 Operations on sets
8.3.1 The universe of discourse
We have seen operations on numbers, or other objects.
We can also define operations on sets (union, intersection, ...).
We need a set that contains all the sets we want to deal with.
It would be tempting to consider the set of all sets.
Problem 1
Let S be the set of all sets that does not contain themselves.
Does S contain itself ?

if S does not contain itself, then it contains itself !
 if S contains itself, then it does not contain itself !
There is a strange loop here which yields a contradiction: S cannot
exist.
For a similar reason, the set of all sets cannot exist.
See Gödel Esher Bach[2] for a lot of similar
fun stuff with strange loops.
That's been a major source of trouble and work a century ago, when
mathematicians tried to define a strong basis for set theory.
To be safe, we always work inside a set big enough to contain all
the sets we need, but small enough to avoid the strange loop above.
Définition 48
This set is the universal set, or the universe of discourse.
We define operations on the powerset of the universal set.
8.3.2 Union, intersection, complement and Cartesian product
We define operations on subsets of the universe of discourse.
Définition 49
The union
of A and B is the set:
Aunion B:={ x  xin A or xin B}
Définition 50
The intersection
of A and B is the set:
Aintersect B:={ x  xin A and xin B}
Définition 51
The complement
of A is the set:
A':={ x  xnot in A}
Définition 52
The set difference
of A and B is the set:
AB:={ x (xin A) and (xnot in B)}
Définition 53
The Cartesian product
(or cross product
) of A and
B is the set:
A× B:={(x,y)  xin A,yin B}
Notation 2
A^{2}=A× A is the set of all ordered pairs of elements of
A;
A^{n}=A×···× A is the set of all nuples of elements
of A.
Définition 54
Venn diagrams.
Exercice 30
Let
A:={1,2} and
B:={2,3,4}.

Aunion B=,
 Aintersect B=,
 AB=,
 A'= (assuming that the universe of discourse is {1,2,3,4}),
 A× B=,
 A^{3}=.
Exercice 31
Let A be a set of size n, and B a set of size m.

what is the size of Aunion B:
 what is the size of Aintersect B:
 what is the size of A' ?
 what is the size of A× B :
8.3.3 Set identities
Proposition 2
Let A and B be to sets. Then,
Aintersect B=Bintersect A (commutative property)
Proof.
xin Aintersect B
<=>(xin A)/\(xin B) by definition of intersect
<=>(xin B)/\(xin A) by commutativity of /\
<=> xin Bintersect A by definition of intersect
The commutativity of intersectderives directly from the commutativity
of /\.
All identities on logic operators induce identities on set operators.
Proposition 3
Let A, B, and C be two sets. Then,
Aunion B=Bunion A (commutative property)
Aintersect B=Bintersect A (commutative property)
(Aunion B)union C=Aunion(Bunion C)=Aunion Bunion C (associative property)
(Aintersect B)intersect C=Aintersect(Bintersect C)=Aintersect Bintersect C (associative property)
Aunion A'=S (complement, S is the universe of discourse)
Aintersect A'=Ø (complement)
(Aunion B)'=A'intersect B' (de Morgan's)
(Aintersect B)'=A'union B' (de Morgan's)
AB=Aintersect B'
8.4 Countable and uncountable sets
8.4.1 Enumerations
If a set S has k elements, these can be listed one after the
other:
s_{1},s_{2},...,s_{k}
Définition 55
s_{1},s_{2},...,s_{k} is called an enumeration of the
set S.
Exemple 79
1,5,4,3, 5,1,3,4, and 4,1,3,5 are enumerations of the set
{1,5,4,3}.
When a set S is infinite, you can not enumerate all of it's elements
at once.
However, some times, you can still pickup a first element s_{1},
then a second s_{2}, and so on forever. This process is an enumeration
of the elements of S if:

Any element of S gets enumerated eventually;
 No element is enumerated twice.
Exemple 80
0,1,2,3,4,... is an enumeration of the nonnegative integers.
8.4.2 Denumerable and Countable sets
Définition 56
A infinite set S is denumerable
if it has an enumeration
s_{1},s_{2},...,s_{n},...
A set S is countable
if it's either finite or denumerable.
Exemple 81
Are the following sets countable ?

The set of all students in this class.
 The set N of the nonnegative integers (enumeration: 0,1,2,3,4,...);
Actually, the order does not need to be logical.
This is a perfectly legal enumeration: 0,10,5,11,4,2,17,2,1,28,3
 The set of the even integers (enumeration: ;
 The set of the prime integers (enumeration: ;
 The set Z of the integers (enumeration: );
 The set N×N
(e.g. all points in the plane with nonnegative integer coordinates):
 Enumeration: (0,0),(0,1),(1,0),(2,0),(1,1),(0,2),(3,0),...
 The set Z×Z of all points in the plane with integer coordinates:
Enumeration: (0,0),(0,1),(1,0),(0,1),(1,0),(2,0),...
 The set Q of rational numbers:
Théorème 11
The following sets are countable:

Any subset Aincluded in B of a countable set B.
 The union Aunion B of two countable sets A and B is countable.
 The union A_{1}union A_{2}union···union A_{i} of a finite number
i of countable sets A_{1},...,A_{i}.
 The union A_{1}union A_{2}union···union A_{i}union···of a countable
number of countable sets.
 The cross product A× B of two countable sets A and B
is countable.
 The cross product A_{1}×···× A_{i} of a finite number
of countable sets A_{1},...,A_{i}.
Proof.
Enumerations of those sets can be constructed in the following way:

Enumerate the elements of B, and ignore those that are not in A.
 Enumerate the elements of A and B alternatively (a_{1},b_{1},a_{2},b_{2},...).
 Induction, using 2.
 As for N×N: a_{1,1},a_{2,1},a_{1,2},a_{3,1},a_{2,2},a_{1,3},...
(a_{i,j} is the jth element of A_{i}).
 As for N×N: (a_{1},b_{1}),(a_{1},b_{2}),(a_{2},b_{1}),(a_{1},b_{3}),(a_{2},b_{2}),(a_{3},b_{1}),....
 Induction, using 5.
8.4.3 Uncountable sets ?
Remarque 10
All the sets we have seen so far are countable.
Problem 1
Are there any uncountable sets ?
Théorème 12
(Cantor) R is uncountable.
Lemme 1
Any real number
xin[0,1) can be written uniquely in decimal form:
x=0.d_{1}d_{2}d_{3}d_{4}··· d_{i}···,
where the
d_{i} are digits between 0 and 9, and the sequence
does not end by an infinite number of 9.
Exemple 82
Here is how some classical numbers can be written:

0=0.0000···
 0.5=0.50000···
 0.5=0.49999··· (Same number as above, we don't use this form)
 1/3=0.333333···
 pi=3.141592653···
Let's jump to the proof of the theorem, without detailing the proof
of this lemma.
Proof.
(Of the theorem): Cantor diagonalization method.
It's sufficient to prove that [0,1) is not countable.
Let's do this by contradiction: assume [0,1) is countable.
Let s_{1},s_{2},...,s_{i},... be an enumeration of [0,1).
Goal: construct an element which is not in this enumeration.
Using the lemma, we can write the s_{i} as follow:

s_{1} 
= 
0, 

d_{1,2} 
d_{1,3} 
d_{1,4} 
d_{1,5} 
··· 
s_{2} 
= 
0, 
d_{2,1} 

d_{2,3} 
d_{2,4} 
d_{2,5} 
··· 
s_{3} 
= 
0, 
d_{3,1} 
d_{3,2} 

d_{3,4} 
d_{3,5} 
··· 
s_{4} 
= 
0, 
d_{4,1} 
d_{4,2} 
d_{4,3} 

d_{4,5} 
··· 
s_{5} 
= 
0, 
d_{5,1} 
d_{5,2} 
d_{5,3} 
d_{5,4} 

··· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 

We want to construct an element x which is not in this enumeration.
Let's look at an example:

s_{1} 
= 
0, 

5 
7 
3 
5··· 
s_{2} 
= 
0, 
0 

4 
7 
3··· 
s_{3} 
= 
0, 
4 
5 

0 
3··· 
s_{4} 
= 
0, 
9 
7 
3 

7··· 
s_{5} 
= 
0, 
4 
3 
8 
1 

·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
· 
·
·
··
·
· 








x 
= 
0, 
0 
1 
0 
0 
1··· 

The trick : let x_{i}=1 if d_{i,i}=0 and x_{i}=1 else.
Then, x!= s_{1}. But also, x!= s_{2}, x!= s_{3}, ...
Therefore x is indeed never enumerated! Contradiction!
Conclusion: [0,1) is not countable.
Remarque 11
We can deduce that many sets like C, or [5.3,10) are uncountable.
A similar proof can be used to prove that many other sets are uncountable.
This includes the set of all infinite strings, or the set (N)
of all subsets of N.
Résumé 2
We have a hierarchy of bigger and bigger sets:

The empty set
 Finite sets (the empty set, sets with 1 element, sets with 2
elements, ...)
 Denumerable sets: N, Z, Q, ...
 Uncountable sets: R, C, ...
 ...
Problem 3
Is there any set bigger than N and smaller that R? Nobody
knows!
8.5 Conclusion
We are now done with basic properties of sets. We have seen enough
formalism
to go on to the next section, Combinatorics:
How to count the objects in a finite set ?
Chapter 9 Combinatorics (section 3.2...3.5)
9.1 Introduction
Goal: answer ``How many ?'' questions:

How many ways to climb a stair ?
 How many rows in a truth table ?
 How many trees ?
Applications:

Analysis of algorithms
 Statistics
 Size of a database
9.2 Counting
9.2.1 Preliminaries
Définition 57
The size of a finite set is the number of elements in this
set.
The size of S is denoted by S (or, often in the
literature, # S).
Note 1
In the sequel, we only consider finite sets !
A``how many ?'' question is usually formalized as follow:

Define the set S of the objects to be counted
 What is the size of S ?
Techniques to answer this question:

breakup S as some combination of simpler sets
 transform S into a set, the size of which is already known
9.2.2 The addition principle
Exemple 83
A small company wants to know how many computers it owns, given that
it owns 5 Mac's, and 3 UNIX stations.
Let:

C be the sets of computers,
 M be the set of Mac's (M=5),
 U be the set of UNIX stations (U=3).
Then, C=Munion U.
What is the size of C ?

 
C 
 
= 
 
M 
 
+ 
 
U 
 
=5+3=8 
Théorème 13
(Addition principle) Let A and B be disjoint sets (e.g. Aintersect B=Ø).
Then:

 
Aunion B 
 
= 
 
A 
 
+ 
 
B 
 
. 
Exemple 84
Imagine 2 of the 5 Mac's are actually running Linux.
What is the size of C ?
Well, C=6<8.
So, the disjoint hypothesis is important !
What to do when the sets are not disjoint ?
Théorème 14
Let A and B be two sets. Then:

 
Aunion B 
 
= 
 
A 
 
+ 
 
B 
 
 
 
Aintersect B 
 
. 
Proof.
We just have to split A, B, and Aunion B into unions of disjoint
subsets:
A=(AB)union(Aintersect B)
B=(BA)union(Aintersect B)
Aunion B=(AB)union(Aintersect B)union(BA).
(Practice: draw and prove those set identities!).
Then,
 
A 
 
+ 
 
B 
 
 
 
Aintersect B 
 

= 
( 
 
AB 
 
+ 
 
Aintersect B 
 
)+( 
 
BA 
 
+ 
 
Aintersect B 
 
) 
 
Aintersect B 
 


= 
 
AB 
 
+ 
 
Aintersect B 
 
+ 
 
BA 
 


= 

Exemple 85
In the example above, C=M+UMintersect U=5+32=6.
Exercice 32
There are 24 students in a class, all of them are math majors or
computer science major. 10 are math majors, and 20 are computer
science majors. How many doublemajor are there in this class ?
9.2.3 The multiplication principle
Exemple 86
You are a car dealer. For each car you sell, you propose several options
(color, 2WD/4WD, airbag, ABS, radio, ...). Of course, you would
like to always have all possible combinations of options in stock.
On the other hand, you cannot afford to store too many cars in your
parking lot. So you need to know how many possible combinations of
options there are.

2WD / 4WD
 Black / Yellow / Green
How many combinations are there ?
First solution: draw a decision tree
.
The choices are independent!
The number of subbranches at each level is constant.
Therefore, we have 2·3=6 choices.
Let's formalize this.
Let C be the set of all choices. Each choice can be represented
by a couple.
For example a yellow 2WD car with black seats can be represented as:
(2,Y)
So, C can be viewed as the cross product:
{2,4}×{ B,Y,G}.
Théorème 15
Let A and B be two sets. Then,

 
A× B 
 
= 
 
A 
 
· 
 
B 
 
. 
Exemple 87
In the example above, C={2,4}×{ B,Y,G}={2,4}·{ B,Y,G}=2·3=6.
Exercice 33
How many possible different phone numbers are there ?
Strings
Strings are commonly used objects in computer science.
``how many'' questions often be translate into questions about
sets of strings.
Définition 58
Let S be a set, that we call alphabet.
Elements of S are called letters.
A string over S is a sequence (s_{1},s_{2},...,s_{n})
of elements of S.
lambda:=() is the empty string.
The set of all strings of length n is S^{n}=S×···× S.
The set of all finite strings over S is denoted S^{*}.
A language is a set of strings.
Exemple 88
Let S:={0,1}. A string over S, like 001001 is called
a binary string.
Let S:={ a,b,...,z}. Then bonjour is a string over
S of length 7.
A phone number 3032733462 is a string over {0,1,2,3,4,5,6,7,8,9}.
The set of all English words is a language.
Exemple 89
Count the following:

binary strings of length 4?
 strings of length n over an alphabet containing p letters ?
9.2.4 Decision trees
Exemple 90
A kid is allowed to pick 3 candies (one after another), out of
one jelly bean, one lollipop and 2 gummy bears (one blue, one yellow).
In how many ways can he do this ?
Draw a decision tree.
The choices are not independent
The multiplication principle does not apply.
However, the number of choices at each step does not change!
So the answer is: 5·4·3=60.
Exercice 34
Count the following:

Strings of length 4 over {1,2,3,4,5} with no repetitions.
 Strings of length n over {1,...,n} with no repetitions.
Exercice 35
How many ways to toss a coin five times without two heads in a row
?
Note: the decision tree is irregular.
9.2.5 Combining those principles together
Exemple 91
Count the following:
Exercise 1819 p. 195

Strings of length n over {1,...,n} with repetitions.
 Strings of length n over {0,1} with two consecutive 1.
 How many fourdigits numbers begin with 4 or 5?
 How many ways to toss a coin seven times with two heads in a row ?
9.2.6 Principle of Inclusion and Exclusion
Exemple 92
All the guests at a dinner party drink coffee, chocolate or tea; 11drink
coffee, 9 drink tea, 10 drink chocolate; 3 drink coffee and
tea, 5 drink tea and chocolate; 6 drink coffee and chocolate
Formalization: what is the size of Aunion Bunion C?
Generalization ?
Théorème 16
(Principle of Inclusion and exclusion)
Let A_{1},...,A_{n} be n sets. Then,
 
A_{1}union···union A_{n} 
 

= 
 
A_{1} 
 
+···+ 
 
A_{n} 
 



 
 
A_{1}intersect A_{2} 
 
··· 
 
A_{n1}intersect A_{n} 
 



+ 
 
A_{1}intersect A_{2}intersect A_{3} 
 
+··· 
 
A_{n2}intersect A_{n1}intersect A_{n} 
 



+(1)^{n+1} 
 
A_{1}intersect···intersect A_{n} 
 


= 


\
/
 
Iincluded in{1,···,n} 

(1) 




 

A_{i} 



 

Proof.
Induction.
9.2.7 Pigeon Hole principle
Exercice 36
It's early in the morning, and you don't want to wake up your significant
whatever by turning on the light. You know you only have black and
white socks in your drawer. Is it possible to exit the room with at
least two socks with the same color ?
Théorème 17
(Pigeon hole principle)
If more than k items are placed in k bins, then at least one
bin has more than one item inside.
Exemple 93
How many people do you need to have in a room to ensure at least two
of them have the same birthday ?
9.2.8 Permutations: sequences without repetitions
Exemple 94
How many ways to choose a committee with 1 president and 1 vicepresident
out of a group of n persons.
This amount to find the number of strings of length 2 over a set
of size n, without repetitions. This is n(n1).
Définition 59
We denote by P(n,k) the number of strings of length k over a
set of size n without repetitions.
Exemple 95
P(n,0)= , P(n,1)= , P(n,n)= ,
P(1,2)= .
Théorème 18
P(n,k)=n(n1)···(nk+1)=n!/(nk)!.
9.2.9 Combinations
Exemple 96
How many ways to choose 3 voluntaries out of a group of 5 persons?
Définition 60
We denote by C(n,k) the number of subsets of size k of a set
of size n.
Exemple 97
C(n,0)= , C(n,1)= , C(n,n)= ,
C(1,2)= .
Théorème 19
C(n,k)=n(n1)···(nk+1)/k(k1)···1=n!/k!(nk)!.
Exemple 98
Ex. 51 p. 210:
A committee of 8 students is to be selected from a class consisting
of 19 freshmen and 34 sophomores.

In how many ways can 3 freshmen and 5 sophomores be selected?
 In how many ways can a committee with exactly 1 freshman be selected?
 In how many ways can a committee with at most 1 freshman be selected?
 In how many ways can a committee with at least 1 freshman be selected?
Exemple 99
What is the total number of subsets of S ?
Exemple 100
What is the number of paths between Stratton Hall and Golden Tourism
Information center ?
9.2.10 Combinations with repetitions
Exemple 101
In how many ways can you distribute 10 lollipops between 4 kids.
Théorème 20
The number of combinations with repetitions of k elements of a
set S of size n is C(n+k1,n).
Exemple 102
In how many ways can you distribute 10 lollipops between 4 kids,
so that every kid get at least one lollipop ?
Addition principle (A and B disjoint) 
Aunion B=A+B 
Principle of InclusionExclusion 
Aunion B=A+BAintersect BA_{1}union···union A_{n}=··· 
Multiplication principle 
A× B=A·BA^{k}=A^{k} 
Sequences of k elements of S (S=n) 
n^{k} 
Permutations: 
P(n,k)=n!/(nk)! 
Sequences of k elements of S, w/o repetitions 

Sequences of k elements of {0,1}, w/o successive 1 
Fibonacci: F(k) 
Combinations: subsets of size k of S 
C(n,k)=n!/k!(nk)! 
All subsets of S 
2^{n} 
Combinations with repetitions 
C(n+k1,k) 
9.3 Properties of binomial coefficients; Combinatorial proofs
Exemple 103
How many subsets of size 14 of a set of size 15 ?
Théorème 21
C(n,k)=C(n,nk)
Proof.

Algebraic proof
 Combinatorial proof
Définition 61
Combinatorial proof:
Goal: prove that two quantities a and b are equal
Technique:

construct a set A of size a;
 construct a set B of size b;
 construct a bijection between A and B
Problem 1
Recursive computation of C(n,k) ?
Théorème 22
C(n,k)=C(n1,k)+C(n1,k1)
Proof.

Algebraic proof
 Combinatorial proof
Exemple 104
Compute C(7,4), without calculator.
Définition 62
Pascal triangle:
nk 
0 
1 
2 
3 
4 
5 
6 
7 
0 
1 







1 
1 
1 






2 
1 
2 
1 





3 
1 
3 
3 
1 




4 
1 
4 
6 
4 
1 



5 
1 
5 
10 
10 
5 
1 


6 
1 
6 
15 
20 
15 
6 
1 

7 
1 
7 
21 
35 
35 
21 
7 
1 
Théorème 23
C(n,0)+···+C(n,n)=2^{n}
Proof.

Algebraic proof ?
 Combinatorial proof
9.3.1 The binomial theorem
Problem 2
Compute (x+y)^{n} .
Théorème 24
There is a formula for expanding (x+y)^{n}:
(x+y)^{n} 
= 


= 
C(n,0)x^{n}+C(n,1)x^{n1}y+C(n,2)x^{n2}y^{2}+···+C(n,n)y^{n} 
9.4 Conclusion
Part IV
Relations Functions and Matrices
Chapter 10 Relations (section 4.1, 4.2 and 4.4)
10.1 Introduction
A mere set of words would not make a good dictionary.
It would be a pain to find a particular word!
A usable dictionary has some structure: the words are sorted.
In general, the more structure a set have, the more useful
it is.
A way to bring structure into this set is to describe the relations
between its elements, or between its elements and the elements of
another set.
In this section we will see how we can formalize and study relations.
Exemple 105
Imagine you want to build a house.
Figure 10.1 shows the tasks that need to be completed.
Figure 10.1: Tasks to build a house
Let S:={ F,W,E,I,O,R} be the set of all tasks.
Problem: can we do the tasks in any order ?
For example, it would be better to build the walls AFTER the foundations!
S in itself does not contain enough information to choose a correct
order.
Figure 10.2: Constraints between the tasks to build a house
Set of constraints: rho:={(F,W),(W,O),(W,E),(R,E),(R,I)}.
This set of constraints gives some structure to S, and makes it
useful.
10.2 Relations
Définition 63
A binary relation on a set S is a subset rho of S× S.
Let x and y be two elements of S.
Then x is in relation with y (denoted x rho y)
iff (x,y)inrho.
Exemple 106
Let rho be the relation ``is a prerequisite for'':
rho:={(F,W),(W,O),(W,E),(R,E),(R,I)}
Then, (F,W)inrho, but (I,O)not inrho.
So, F rho W is true, whereas I rho O
is false.
A relation can be defined by a property.
Exercice 37
Let S:={1,2,3,4,5}. Draw the relations defined by:

x rho_{1} y iff x=y;
 x rho_{2} y iff x<= y;
 x rho_{3} y iff x divides y;
 x rho_{4} y iff xy is even.
Définition 64
A binary relation between two sets S and T is a subset
rho of S× T.
A nary relation between n sets S_{1},...,S_{n} is
a subset rho of S_{1}×···× S_{n}.
Exemple 107
Let M be a set of male and F a set of female students.
We can define the relation ``is married to''.
10.2.2 Basic properties of relations
Is there anything particular about the relation ``is married to''
?
Définition 65
Let rho be a relation between S and T.

S is onetoone if any element of S and T appears
at most once in rho;

S is onetomany if any element of T appears at most
once in rho;

S is manytoone if any element of S appears at most
once in rho;

S is manytomany in all other cases.
Exemple 108
Is x lower or equal to itself ?
Définition 66
A relation rho on a set S is reflexive
iff
(for all xin S) x rho x.
Exemple 109
Assume x is equal to y. Is y equal to x ?
Définition 67
A relation rho on a set S is symmetric
iff
(for all xin S)(for all yin S) (x rho y)<>(y rho x).
Exemple 110
Assume x<= y and y<= x. What can you say about x and
y?
Définition 68
A relation rho on a set S is antisymmetric
iff
(for all xin S)(for all yin S)[(x rho y) and (y rho x)]>(x=y).
Exemple 111
Assume x<y and y<z. What can you say about x and z?
Définition 69
A relation rho on a set S is transitive
iff
(for all xin S)(for all yin S)(for all zin S)[(x rho y) and (y rho z)]>(x rho z).
Exemple 112
What are the properties of the following relations

<, <=, =;
 ``is married to'';
 ``is a friend of'';
 ``has the same color than''.
10.2.3 Operations on relations
A relation is basically a set. So, we can use all usual set operations
on relations.
Remarque 12
Let S and T be two sets.
The powerset (S× T) is the set of all binary relations
between S and T.
Définition 70
Let rho and sigma be two relations between S and T.
We can construct the following new relations:

union: rhounionsigma,
 intersection: rhointersectsigma,
 complement: rho',
 ...
Exemple 113
Let S:=N.

What is the union of <and =?
 What is the intersection of < and >?
 What is the union of < and >?
 Is <a sub relation of <= ?
10.2.4 Closure of a relation
Exemple 114
Lets come back to the example of the house building.
Figure 10.1: Constraints between the tasks to build a
house
Do you have to do the electricity after the foundations ?
Foundations is not a direct prerequisite for electricity.
However, it still needs to be done before electricity.
The relation ``has to be done before'' is transitive, and contains
the relation
``is a prerequisite for''. It's the transitive closure
of ``is a prerequisite for''.
Définition 71
Let rho be a relation.
The transitive closure of rho is the smallest transitive
relation rho^{*}containing rho.
Exercice 38
Draw the transitive closure of ``is a prerequisite for''.
Problem 1
Is there an algorithm for computing the transivite relation of a relation?
Algorithme 2
Construction of the transitive closure rho^{*}.

rho^{*}:=rho;
 Find x, y, z such that (x rho^{*}y), (y rho^{*}z),
and (x ¬ rho^{*}z);
 If no such triple exists, rho^{*}is transitive; Exit;
 rho^{*}:=rho^{*}union{(x,z)};
 Repeat at step 2.
Remarque 13
The order in which you add the pairs to rho is irrelevant.
Exemple 115
Draw the relation ``has to be done before'' (Figure 10.1).
Définition 72
Let rho be a relation.
The reflexive closure of rho is the smallest reflexive
relation containing rho.
The symmetric closure of rho is the smallest symmetric
relation containing rho.
Exercice 39
Consider the following relation:

Draw its reflexive closure
 Draw its symmetric closure
 Draw its transitive closure
 Draw its antisymmetric closure
Remarque 14
The antisymmetric closure of a relation cannot be uniquely defined!
10.3 Equivalence relations
Exemple 116
Consider a set of cars, and the relation ``has the same color as''.
What are the properties of this relation?
Définition 73
An equivalence relation
is a relation which is

Reflexive
 Symmetric
 Transitive
Définition 74
Let rho be a relation on a set S, and x be an element of
S.
The equivalence class of x is the set [x]of all elements
y such that x rho y.
Exemple 117
The equivalence class of a car is the set of all cars having same
color.
Remarque 15
If x rho y, then [x]=[y].
Exemple 118
Consider the relation same as_{3} on N defined by xsame as_{3}y
iff 3 divides xy.
(we also say that x and y are congruent modulo 3:
xsame as y (mod 3))
What are the equivalence classes?
Définition 75
A partition of a set S is a collection of subsets { A_{1},...,A_{k}}
of S such that:

A_{1}union···union A_{k}=S
 A_{i}intersect A_{j}=Ø for any i, j.
Exemple 119
Consider a set S of car, whose color are either red, blue, or green.
Define the following subsets of S:
R (red cars), B (blue cars), and G (green cars) .
Then, { R,B,G} is a partition of S.
The equivalence relation ``has the same color'' and the partition
of the cars by color are closely related.
Théorème 25
In general:

An equivalence relation on S defines a unique partition of S.
 A partition of S defines a unique equivalence relation on S.
Proof.
Left as exercise:

Construct the collection { A_{1},...,A_{k}}, and prove it
is indeed a partition of S.
 Construct the relation rho, and prove it's indeed an equivalence
relation.
Définition 76
Let S be a set, and rho be a relation on S.
The set of all the equivalence classes is called the quotient
of S by rho.
Remarque 16
This method is used a lot in set theory:
Construction of Z from N.
Construction of Z/nZ (integers modulo n) from Z.
Construction of Q from Z.
10.4 Partially Ordered Sets
Exemple 120
What are the properties of the relation ``has to be done before''
on the tasks to build a house ?
Définition 77
A partially ordered set
(or poset
) is a set S with
a relation rho which is

Reflexive
 Antisymmetric
 Transitive
Exemple 121
Which of the following are posets ?

({1,2,3,4},<=);
 ({1,2,3,4},<);
 ({1,2,3,4},=);
 ({1,2,3,4},rho), with x rho y iff x
divides y.
 (({1,2,3,4}),included in or equal to).
10.4.1 Drawing posets; Hasse diagrams
Définition 78
Some names:

Node (or vertex)
 Predecessor
 Immediate predecessor
 Maximal element
 Minimal element
 Least element
 Greatest element
Définition 79
The Hasse diagram
of a poset is a drawing of this poset such
that:

If x rho y, then y is higher than x;
 The loops are not drawn;
 There is a segment linking x and y iff x is an immediate
predecessor of y.
Exemple 122
Draw all posets on 1, 2, 3, 4 indistinct nodes.
How many such posets are there on 10 nodes ?
Scheduling a set of tasks consist in allocating tasks to resources,
subject to certain constraints.
Exemple 123
Consider the house building problem, with the following assumptions:

All tasks take the one day to complete;
 One worker can only work on one task every day;
 It does not save time to have two workers working on the same task.
Figure 10.1: Constraints between the tasks to build a
house
Schedule the tasks between 2 workers to optimize the completion time:

Day 1 
Day 2 
Day 3 
Day 4 
Day 5 
Day 6 
Day 7 
Worker 1 







Worker 2 







What if there is one worker ?

Day 1 
Day 2 
Day 3 
Day 4 
Day 5 
Day 6 
Day 7 
Worker 1 







What if there are three workers ?

Day 1 
Day 2 
Day 3 
Day 4 
Day 5 
Day 6 
Day7 
Worker 1 







Worker 2 







Worker 3 







Proposition 4
Consider the problem of scheduling n tasks on p processors, subject
to precedence constraints, with the same assumptions and goal as above.

1 or 2 processors: algorithms in O(n^{2});
 3 processors: complexity unknown;
 p processors: NPcomplete
(basically the only known algorithm is exhaustive search);
 oo processors: O(n).
This is typical from scheduling problems, where a very small change
in the constraints can make huge differences in the complexity of
the problems.
Also, usually the complexity is as follow:

No constraints: low complexity;
 Some constraints: higher and higher complexity;
 More constraints: NPcomplete. The only known algorithm is exhaustive
search through all plausible cases (branchandcut);
 Even more constraints: still NPcomplete, but easier. Indeed more
and more cases can be thrown away very early.
Having very good scheduling algorithms is vital in industry, because
a 1% difference in completion time (for example) can save thousands
of dollars. Consequently, research in this topic is well financed!
10.5 Conclusion

The more structure a set has, the more useful it is.
 Defining a relation is a way to add structure to a set.
 Relations, posets, equivalence relations, ... are just abstract
models of natural notions commonly used in real life.
Chapter 11 Functions (section 4.4)
11.1 Introduction
The notion of function is pretty natural, and used in many domains
(programming languages, calculus, ...). So far, the intuitive
notion of function was sufficient, but we now need a more formal definition,
and we will use relations for this.
We will also see how certain functions, called bijections, provide
powerful counting methods.
Exemple 124
Let S be a set of cars and T be the set of all colors.
Consider the relation ``is of color''.
What properties does this relation have ?"
You can speak about THE color of the car: uniquely defined.
11.2 Formal definition of functions
11.2.1 Functions with one variable
Définition 80
A function
f: S > T
between two sets S and T is a relation such that any element
s of S is in relation with a unique element of T.
This unique element, denoted f(s) is the image
of s.
If f(s)=y, then s is a preimage
of y.
S is the domain
of f;
T is the codomain
of f;
The set { f(s) , sin S} is the range
of f.
Exemple 125
The identity on a set S if the function id_{S}:S> S
defined by id_{S}(x):=x.
One way to define a function is to provide an equation, or any other
mean that allows to compute the image of an element of the domain.
Exemple 126
f(n):=n+2;
f(L):=sort(L);
f(x):=[x].
Définition 81
Graph of a function from R to R (or any subset of them):
Exercice 40
Draw the graphs of the functions f(x)=x^{2}, g(x)=(x)^{1/2},
and h(x):=x.
11.2.2 Functions with several variables
The argument of a function does not need to be a simple number.
Exemple 127
f(x,y)=x^{2}+2xy
Définition 82
A function with n variables is a function whose argument is a nuple:
f : S_{1}×···× S_{n} > T.
Exemple 128
Here are a few functions with several variables:

f :{ students}×{ test1,test2,final} > {1,2,...,100}
f(Jason,test2):=···
 f: { True,False}^{3} > { True,False}
f(A,B,C)=(A/\ B)'/\ C
11.2.3 Equality of functions
Exemple 129
Are the functions f(x):=x and g(x):=x equal ?
Définition 83
Two functions f and g are equal iff they have same domain
S, same codomain T, and f(x)=g(x) for all x in S.
11.3 Properties of functions
11.3.1 Injective, surjective and bijective functions
Exemple 130
Let P be the set of all residents in America;
let N be the set of all assigned Social Security Number (SSN);
let SSN: P> N be the function which assigns
to each person x its SSN SSN(x).
A Social Security Number is useful because it uniquely describes a
person!
That is, given an assigned SSN, you can find the person having this
SSN.
Définition 84
The function SSN is invertible.
The inverse of the function SSN is the function SSN^{1}: N> P
such that:
SSN^{1}(n) is the person having n as SSN.
What properties does a function need to be invertible?
To be invertible, a function needs to have the two following properties:
Définition 85
A function f :S > T is injective
(onetoone
) iff
(for all xin S)(for all yin S) (f(x)=f(y))>(x=y)
``If x and y have the same SSN, then x is the same person as y''
Définition 86
A function f :S > T is surjective
(onto) iff
(for all yin T)(there exists xin S) f(x)=y
``For any assigned SSN, there is a person which has this SSN.''
Définition 87
A function f :S > T is bijective
iff it's injective and surjective.
Exercice 41
Which one of the following functions from R to R are bijective?

f(x):=x^{2};
 f(x):=x^{3};
 f(x):=1/1+x^{2};
 f(x):=exp(x);
 f(x):=log(x);
11.3.2 Composition of functions
Définition 88
Let f:S> T and g:T> U be two functions.
The composition function
g¤ f is the function from
S to U defined by
g¤ f(x):=g(f(x)).
Remarque 17
The standard symbol for composition is an empty circle! In those notes
a full circle is used instead due to a bug in lyx ...
Exercice 42
Define f:R>R by f(x):=x+1, and g:R>R
by g(x):=x^{2}.
What is the value of g¤ f(1)? What is g¤ f?
What is the value of f¤ g(1)? What is f¤ g?
Théorème 26
Let f:S> T and g:T> U be two functions.

If f and g are injective, then g¤ f is injective.
 If f and g are surjective, then g¤ f is surjective.
 If f and g are bijective, then g¤ f is bijective.
Proof.
3. is a direct consequence of 1 and 2.

Assume f and g are injective. We have :
(for all x)(for all y) (f(x)=f(y)) > (x=y)
(for all x)(for all y) (g(x)=g(y)) > (x=y)
We want to prove that :
(for all x)(for all y) (g¤ f(x)=g¤ f(y)) > (x=y)
Take x and y such that g¤ f(x)=g¤ f(y)
By definition g(f(x))=g(f(y)).
Then, since g is injective, f(x)=f(y).
Since f is also injective x=y.
We are done!
 Left as exercise.
11.3.3 Inverse of function
Exemple 131
If you take the SSN of a person, and then lookup the person having
this SSN, you will get back the original person. That is, composing
the function SSN and the function SSN^{1} yields the identity.
Définition 89
Let f:S> T be a function.
If there exists a function g:T> S such that g¤ f=id_{S}
and f¤ g=id_{T}, then g is called the inverse function
of f, and is denoted f^{1}.
Théorème 27
A function f is bijective iff its inverse f^{1} exists.
Exercice 43
Give the inverses (if they exists!) of the following functions:

f(x):=x1;
 f(x):=x^{2};
 f(x):=x^{3};
 f(x):=exp(x).
11.4 Functions and counting
11.4.1 Injections, surjections, bijections and cardinality of sets
Exemple 132
You distribute n different cakes between k child.
Such a distribution can be formalized by a function f:
f(4)=5 means that the 4th cake is given to the 5th child.

Suppose f is injective.

What does it mean ?
 What can you say about n and k ?k>= n
 Suppose f is surjective.

What does it mean ?
 What can you say about n and k ?
 Suppose f is bijective.

What does it mean ?
 What can you say about n and k ?
Théorème 28
Let S and T be two sets.
If there exists an injective function between S and T, then
S<=T.
If there exists a surjective function between S and T, then
S>=T.
If there exists a bijective function between S and T, then S=T.
Note that we have never given a formal definition of the size of a
set.
We just relied on the intuitive notion.
In set theory, the theorem above is actually the definition of cardinality:

Two sets have the same cardinality (size), iff they are in bijection.
 A set is of size n iff it's in bijection with {1,...,n}.
11.4.2 Examples of bijections
Finite and countable sets
Do you remember how we proved that Z was countable ?

A set S is finite if there is an enumeration s_{1},...,s_{k}
of S.
This enumeration is actually a bijection from {1,...,k} to
S!
 A set S is denumerable if there is an enumeration s_{1},...,s_{k},...
of S.
This enumeration is actually a bijection from N to S!
Y/B buildings, 0/1 strings and pairs of rabbits
Problem 1
Counting:

buildings with yellow and blue floors, without consecutive yellow
floors.
 strings of 0 and 1 without two consecutive 1.
The answer in both case is the Fibonacci sequence.
Those two problems are fundamentally the same:
There is a bijection between solutions of the first and solutions
to the second!
So if you solve the one of them, you solve both of them.
Problem 2
Counting pairs of rabbits after n generations (ex 29 p. 140).
Here also the solution is the Fibonacci sequence.
Can you find the bijection ?
Strings of wellbalanced parenthesis and Dyck paths
A string of wellbalanced parenthesis is a string such as (()()),
where each open parenthesis is closed and vice versa.
Définition 90
A Dyck path of size n, is a path in N×N from (0,0)
to (2n,0) such that each step is either (1,1) or (1,1).
The important fact is that such a path never goes under the xaxis.
Exercice 44
Find all strings of wellbalanced parenthesis of length 0, 2,
4,6, 8, 10.
Find all Dyck paths of size 0, 1, 2, 3, 4.
Do you notice something ?
Conjecture 3
There are as many Dyck paths of size n than strings of well balanced
parenthesis of length 2n.
How to prove this conjecture?
Let D be the set of Dyck paths of size n.
Let S be the set of strings of well balanced parenthesis of length
2n.
We just have to construct a bijection f:S> D.
Take s=s_{1}··· s_{2n}in S, and build a path f(s) in the
following way:
read s from left to right; if s_{i} is a (, go right
and up, else go right and down.
Exercice 45
Draw f(s) for the following strings: (), ()(),
((()(())())()).
Why is f(s) indeed a Dyck path of size n?

There are 2n steps to the right
 In s, there are as many '(' as ')', so the final position
is (2n,0).
 At any position i in s, there are more '(' than ')' before
i.
How to prove that f a bijection ?
Lets try to construct an inverse g for f:
Take a path p, and build a string g(p) in the following way:
go through p from left two right. For each up step, add a '('
to g(p), and for each down step, add a ')'.
Exercice 46
Apply g to the paths obtained in the previous exercises.
For the same reasons as above, the resulting string g(p) is a string
of well balanced parenthesis.
Moreover, by construction f(g(p))=p for any pin D and g(f(s))=s,
for any sin S.
So g is indeed an inverse for f.
Therefore, f is a bijection and consequently S=D,
as wanted.
Ordered trees and strings of wellbalanced parenthesis
Définition 91
The set of all ordered trees can be defined recursively as follow:

A single node is an ordered tree
 If t_{1},...,t_{k} are k ordered trees, the structure obtained
by adding a common father to t_{1},...,t_{k} is an ordered tree.
Exercice 47
Draw all ordered trees on 1,2,3,4,5 nodes.
Exemple 133
Prove that any ordered tree on n nodes has n1 edges.
Ordered trees are defined recursively, so using recursion seems reasonable.
The property is true on a tree with one node.
Let t be a tree on n>1 nodes.
Assume the property is true for any tree of size <n.
By construction, t is build by adding a common father to k trees
t_{1},...,t_{k}.
Let n_{i} be the number of nodes of t_{i}.
We have n=n_{1}+···+n_{k}+1.
Clearly n_{i}<n, so by induction each t_{i} has n_{i}1 edges.
Conclusion: the number of edges of t is :
(n_{1}1)+···+(n_{k}1)+k=n_{1}+···+n_{k}=n1.
Théorème 29
There are as many trees with n nodes as strings of well balanced
parenthesis of length 2(n1).
Let's define recursively a bijection f between trees and well balanced
parenthesis:

If t has a single node, then f(t) is the empty string;
 If t is build from k subtrees t_{1},...,t_{k}, then f(t):=(f(t_{1}))···(f(t_{k})).
Exercice 48
Apply f to a few trees.
Here also, it's easy to construct the inverse of f, so f is
a bijection.
Since f maps trees on n nodes on strings of length 2(n1),
and viceversa, we are done with the proof of the theorem.
Catalan Numbers
We have seen that many different kind of objects (trees, strings of
well balanced parenthesis, Dyck paths) are counted by the same sequence
of numbers:
1,1,2,5,14,...
This sequence is called the Catalan sequence C(0),C(1),C(2),C(3),....
Problem 3
What is the value C(n) ?
It's enough to count, for example, the number of Dyck paths of length
2n.
There is a beautiful trick to do this. It's called André's inversion
principle.
Will you find it ?
11.5 Conclusion

There are often connections between apparently unrelated objects (trees,
strings of well balanced parenthesis, Dyck paths, ...).
 Those connections, formalized by bijections, provide powerful methods
for counting objects.
 The Sloane is very handy to suggest connections!
Part V
Modeling, Arithmetic, Computations and Languages
Chapter 12 Groups, and other algebraic structures (section 8.1)
12.1 Introduction
A mathematical structure is a formal model which capture common properties
or behavior of different sets.
It consists of an abstract set, together with operations and relations
which obey certain rules.
For example the set of tasks to build a house, and the set of all
courses available at CSM have something in common: the notion of prerequisite.
This notion can be formalized using a structure of partially ordered
set.
We will introduce here certain algebraic structures, which are useful
to model arithmetic, as well as to the study of symmetries.
12.2 Permutations
Définition 92
Let A be a set. A permutation of A is a bijection from
A to A.
Let S_{A} be the set of all permutations of A.
Let S_{n} be the set of all permutations of {1,... n}.
A permutation sigma of S_{n} can be written in array
form as follow:
meaning that sigma(1)=2, sigma(2)=1, sigma(3)=4, and so
on.
Exercice 49
Write all permutations in S_{1}, S_{2}, and S_{3}.
What is the size of S_{n}?
12.2.1 Operation on S_{n}
S_{n} as a set is not very interesting.
Can we define operations on S_{n}?
Proposition 5
The composition ¤is an operation of S_{n}.
Indeed, the composition of two permutations is still a permutation.
We also call this operation product of permutations.
Exercice 50
Compute the following:
The permutation above produces a cycle: 1>2>3>4>5>1
Définition 93
By (a,b,c,d), we denote the permutation sigma such that sigma(a)=b,
sigma(b)=c, sigma(c)=d, sigma(d)=a, and sigma(x)=x
for any other element x.
Such a permutation is called a cycle.
A cycle (i,j) of length two is called a transposition.
A transposition (i,i+1) is called elementary transposition.
Exercice 51
Write (3,4,2,1) and (2,1,3,4) in array notation.
Write the following permutations in array notation:
Proposition 6
The product of two disjoint cycles is commutative.
This is not the case for nondisjoint cycles.
Exercice 52
Are the following permutations cycles?
Problem 1
So there are permutations that are not cycles.
Would it be possible to write them using only cycles?
Théorème 30
Any permutation can be written as a product of disjoint cycles.
This product is unique, up to the order.
Exercice 53
Compute the following permutations

(
( 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
5 
10 
2 
1 
6 
4 
8 
3 
7 
9 

)
) 

= 
( 

) 
Writting a permutation in cycle notation is a good way to compute
powers of this permutation.
12.2.2 Generators of S_{n}
If you have a list, you can use a permutation to change its order.
Well, S_{n} is pretty big, so this makes quite a lot of possible
operations.
However, some operations can be obtained by combining other operations.
Could we try to reduce the size of S_{n}?
Définition 94
A subset G of S_{n} generates S_{n} if any permutation
can be written as a product of elements of G.
Problem 2
Find a set of generators of S_{n} as small as possible.
Problem 3
Does the set of all cycles generate S_{n}?
Problem 4
How many cycles are there ?
Can you do better?
Problem 5
Does the set of all transpositions generate S_{n}?
How many transpositions are there?
Can you do better?
Problem 6
Does the set of all elementary transposition generate S_{n}?
How many elementary transpositions are there ?
By the way, how many elementary transpositions at most do you need
to express a permutation?
Can you do better than n1 generators?
12.3.1 Definitions and examples
The notation [A,¤] denote a set A together with an operation
¤."
Problem 1
Is there anything in common between [S_{n},¤] and [Z +].
Définition 95
Let A be a set with an operation ¤
What are the properties of [Z,+]?

+ associative;
 +is commutative;
 0 is an identity element;
 If x is an integer, thenx is an inverse for x.
What are the properties of [R^{*},·]?
Those structures have many common properties.
Définition 96
[G,¤] is a group
if G is a non empty set, and ¤is
a binary operation on G such that:

¤ is associative
 an identity element exists (in G).
 each element of G has an inverse (in G) with respect to ¤.
If the operation ¤is commutative, G is called a commutative
group
.
Exercice 54
Which of the following are groups ?

[Z,+]
 [N,+]
 [Z,]
 [{ true,false},/\}
 [R,+]
 [R,·]
 [S_{n},¤]
 [nZ,+]
What's the point ?

Many structures, coming in particular from arithmetics, are groups.
 If you prove a property for groups in general, you have proved this
property for all of those structures.
 Group theory is a vast field, and there are hundreds of theorems about
groups waiting for you to apply them on your favorite structure.
 Basic tool to study the Rubicks Cube.
12.3.2 Just some examples of results and questions about groups
Théorème 31
The identity of a group is unique.
The inverse of an element is unique.
The equations a¤ x=b and x¤ a=b have a unique solution.
The inverse of a product can be computed by (x¤ y)^{1}=y^{1}¤ x^{1}.
Définition 97
Let [G,¤] be a group, and Aincluded in or equal to G.
Then [A,¤] is a subgroup of [G,¤] if [A,¤]
itself is a group.
Théorème 32
The size of a subgroup divides the size of the group.
Problem 2
Common problems about a group:

Find a minimal set of generators
 Express an element of the group as product of generators
 Efficient computations
 Find all subgroups
 Are two groups structurally the same (are they isomorphic?)
For playing with groups, there is a very nice program:
http://wwwhistory.mcs.stand.ac.uk/~gap/
12.3.3 Groups and Symmetries
One common use of groups if for the study of problems with symmetries.
Exemple 134
Computing the temperature in a house with a heater in the middle.

Case 1: the house has a random shape;
 Case 2: the house is round;
 Case 3: the house is square.
Most of the time, a problem with symmetry has a symmetric solution.
Using this can save a LOT of time.
Problem: finding the orbits
The set of all symmetries is a group, so group theory can help you.
12.3.4 Groups and isomorphy of graph
Définition 98
The two following structures are labelled graphs:
If you exchange the nodes 3 and 4 of the first graph, you get
the second.
The group S_{n} acts on graphs on n nodes.
Structurally, except for the numbering, there is no difference between
them:
They are isomorphic.
Problem 3
Testing if two graphs are isomorphic.
This is an important but very difficult problem.
Définition 99
The following structure is an unlabelled graph
:
It can be defined as an equivalence class of labelled graphs.
Problem 4
How many labelled graphs on n nodes ?
How many unlabelled graphs on n nodes ?
Group theory provides a tool (Pólya enumeration) for this.
12.4 Conclusion

Groups are an abstract model for many common structures
 Results about groups apply to all of those structures
 Groups are a powerful tool to study problems with symmetries
 Groups often occurs when dealing with unlabeled structures
References
 [1]

Judith L. Gersting.
Mathematical Structures for Computer Science.
W. H. Freeman and Company, 4 edition, 1998.
 [2]

Douglas R. Hofstadter.
Gödel, Escher, Bach: an eternal golden braid.
Basic Books Inc. Publishers, New York, 1979.
 [3]

G. Pólya.
How to solve it.
Princeton University Press, Princeton, NJ, second edition, 1988.
A new aspect of mathematical method.
Class notes
Introduction
Algebraic Structures /
CSM MACS358 /
Nicolas M. Thiéry
Last modified: Fri Jan 19 13:13:02 2007