Automaten und Formale Sprachen“ alias ” Theoretische Informatik

Transcription

Automaten und Formale Sprachen“
”
alias
Theoretische Informatik“
”
Sommersemester 2012
Dr. Sander Bruggink
Übungsleitung: Jan Stückrath
Sander Bruggink
Automaten und Formale Sprachen
1
Organisatorisches und Einführung
Who are we?
Teacher: Dr. Sander Bruggink
Roomm LF 265
E-Mail: [email protected]
Teaching assistent: Jan Stückrath
Room LF 265
E-Mail: [email protected]
Tutors: Dennis Nolte / Martin Kutscher
Dennis Nolte: [email protected]
Martin Kutscher: [email protected]
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Introduction
Who are you?
BAI
ISE
Others
Website
www.ti.inf.uni-due.de/teaching/ss2012/afs/
Moodle-Site
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Appointments
Lecture:
Tuesday, 12pm–2pm, room LB 131
Exercise groups:
Group ISE: Tuesday, 8am–10am, room LC 137 (English)
Jan Stückrath
Group BAI-1: Wednesday, 12pm–2pm, room LE 120 (German)
Martin Kutscher
Group BAI-2: Thursday, 2pm–4pm, room LF 125 (German)
Dennis Nolte
Group BAI-3: Thursday, 4pm–6pm, room LF 125
Dennis Nolte
Group BAI-4: Friday, 10am–12pm, raum LE 120
Dennis Nolte
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Advice about the exercises
Please try to split evenly among the exercise groups.
Visit the exercise groups and do the homework. The material of this
lecture can only be mastered by frequent practice. Memorizing
doesn’t help much.
The exercise groups start in the third week of the semester.
Thus, the first exercise group take place from 24 to 27 April.
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Advice about the exercises
The exercise sheet is put online every week on Tuesday at the latest.
The written exercises must be handed in on Monday, 4pm of the
following week. In this week the exercise sheet is discussed in the
exercise groups.
Handing in:
in the letter box adjacent to room LF 259.
online through Moodle.
Plase write clearly your name, student number and group number on
your exercise. Also write down the name of the lecture.
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Exam
BAI: Oral exam in the module “Theoretische Informatik”
(“Automaten und formale Sprachen” together with “Berechenbarkeit
und Komplexität”).
Students, who have started in the summer semester, can take an
exam in both lectures separately.
ISE: Written exam after the semester.
Others: Written exam after the semester.
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Bonus points
Bonus points:
During the semester the will be 12 (or 11) exercise sheets of 20 points
each.
If you receive 50% of the points, you will recieve one grade level
higher (for example 2,0 instead of 2,3) for your exam.
For the oral exam of the module “ Theoretische Informatik” you must
obtain the bonus in both “Automaten und Formale Sprachen” and in
“Berechenbarkeit und Komplexität”.
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Literature
We use the following book:
Uwe Schöning: Theoretische Informatik – kurzgefaßt. Spektrum,
2008. (5. Auflage)
Other relevant literature:
Neuauflage eines alten Klassikers:
Hopcroft, Motwani, Ullman: Introduction to Automata Theory,
Languages, and Computation. Addison-Wesley, 2001.
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Literatur
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Adventure-Problem
10
1
4
2
6
5
7
11
12
9
3
13
8
14
15
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Adventure-Problem
Adventure Problem (Level 1)
Rules of the Adventure Problem:
The Treasure Rule
You must find at least two treasures.
The Door Rule
You can only go through a door, when you found a key before. (The key
can be used arbitrarily many times.)
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Adventure-Problem
The Dragon Rule
Immediately after the encounter with a dragon, you must jump into a river,
because the dragon will otherwise ignite you. This is not the case anymore,
if you have previously found a sword, because then you can kill the dragon.
Remark: Dragons, treasures and keys are “refilled” after you left the
according field.
We are look for a path from a start to an end state, which satisfies all of
the above conditions.
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Adventure-Problem
Question (Level 1)
Is there a solution in the example?
Adventure
Yes! The shortest solution is:
1, 2, 3, 1, 2, 4, 10, 4, 5, 6, 4, 5, 6, 4, 11, 12 (length 16).
Is there a general solving procedure which – given an adventure in the
form of a graph – can always determine whether there is a solution?
Yes! We will see this procedure in the lecture.
In order to be able to implement this procedure, we need also formal
description of the rules (door rule, dragon rule, treasure rule).
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Adventure-Problem
New Door Rule
The keys are magical and disappear immediately after being used to open
a door. As soon as you go through a door, the door is locked again.
However, you can carry more than one key.
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Adventure-Problem
Questions (Level 2)
Adventure
Yes! The shortest solution is: 1, 2, 3, 1, 2, 4, 10, 4, 7, 8, 9, 4, 7, 8,
9, 4, 11, 12. (length 18)
Is there a general solving procedure?
Yes! We will see this procedure in the lecture.
Why is the new problem harder?
We have to “count” the keys.
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Adventure-Problem
Adventure Problem (Expert Level)
New Dragon Rule
Swords become unusable by the dragon’s blood, as soon as one has killed
a dragon. However, dragons are replaced after being killed.
Key Regel
A magic gate can only be passed, when you don’t own a key.
Sword Rule
A river can only be passed, when you don’t have a sword (otherwise, you’ll
drown).
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Adventure-Problem
Adventure Problem (Expert Level)
Questions (Expert Level)
Adventure
Yes! The shortest solution is:
Ja! Die kürzeste Lösung ist 1, 2,
3, 1, 2, 4, 10, 4, 7, 8, 9, 4, 10, 4, 5, 6, 4, 11, 12. (Länge 19)
Is there a general solving procedure?
No! It is a so-called undecidable problem.
This is not discussed in this lecture, but in “Berechenbarkeit und
Komplexität”.
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Formal Languages
Adventure Problem and Formal Languages
(Formal) Languages
Language = Set of words
Languages contain in general infinitely may words.
Thus: We need finite descriptions of infinite languages.
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Formal Languages
Adventure Problem und Formale Languages
Questions
Typical questions here are:
Is a language L empty or does it contain (at least) one word? L = ∅?
Is a word w in the language? w ∈ L?
Are two languages included in one another? L1 ⊆ L2 ?
Depending on the language (or languages) these question are either
decidable (there is a general procedure to solve the problem) or
undecidable
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Formal Languages
Adventure Problem and Formal Languages
The single levels of the adventure belong to the following language classes:
Level 1 → regular languages
Level 2 → context free languages
Expert level → Chomsky-0 languages (semi-decidable languages)
These are discussed in “Berechenbarkeit & Komplexität”.
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Contents of the Lecture
For theoretical computer science
How can infinite structures be represented by finite descriptions
(automata, grammars)?
There are numerous applications – for example in the following areas:
searching in texts (regular expressions)
syntax of (programming) languages and compiler construction
modelling system behaviour
verification of systems
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Contents of the Lecture
Contents of the lecture
Automata and formal languages
Mathematical foundations and formal proofs
Languages, grammars and automata
Chomsky Hierarchy (different language classes)
Regular languages and context free languages
How can we show that a language is not of a certain class (pumping
lemma)
Decision procedures
Closure properties (is the intersection of two regular languages also
regular?)
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Mathematical Foundations and Formal Proofs
Sets
Set
A set M of elements is denoted as enumerations
M = {0, 2, 4, 6, 8, . . . }
or a a set of elements with a certain property
M = {n | n ∈ N and n even}
General format:
M = {x | P(x)}
(M is the set of all elements x, which satisfy property P.)
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Sets
Remarks:
The elements of a set a unordered, that is, their order is not
important. For example:
{1, 2, 3} = {1, 3, 2} = {2, 1, 3} = {2, 3, 1} = {3, 1, 2} = {3, 2, 1}
An element cannot occur in a set more than once. It is either in the
set, or not. For example:
{1, 2, 3} =
6 {1, 2, 3, 4} = {1, 2, 3, 4, 4}
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Stes
Element of a set
We write a ∈ M, when an element a is contained in the set M.
Number of elements of a set
For a set M the number of elements of M is denoted by |M|.
Subset
We write A ⊆ B when every element of A is also an element of B. A is
then called a subset of B. The relation ⊆ is also called inclusion.
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Sets
Example:
2 ∈ {1, 2, 3}? 3
2 ⊆ {1, 2, 3}? 7
{1, 2} ∈ {1, 2, 3}? 7
{1, 2} ⊆ {1, 2, 3}? 3
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Venn-Diagrams
Venn-Diagrams are graphical representation of sets and the relationships
between them.
• A
B
B
A
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Set operations
Union: A ∪ B = {e | e ∈ A oder e ∈ B}
Intersection: A ∩ B = {e | e ∈ A und e ∈ B}
Difference: A \ B = {e | e ∈ A und e ∈
/ B}
A∪B
A
A∩B
B
A
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A\B
B
A
B
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Power set
Power set
Let M be a set. The set P(M) is the set of all subsets of M.
P(M) = {A | A ⊆ M}
We have: |P(M)| = 2|M| (for a finite set M).
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Tupel
Tupel
Besides sets we also use tuples, which are written with (round)
parenthesis: (a1 , . . . , an )
In a tuple the elements are ordered. For example:
(1, 2, 3) 6= (1, 3, 2)
An element can occur multiple times in a tuple. Tuples of different
size are always unequal. For example
(1, 2, 3, 4) 6= (1, 2, 3, 4, 4)
A tuple (a1 , . . . , an ) consisting of n elements is called n-tuple. A
2-tupel is also called a pair.
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Cross product
Cross product (or cartesian product)
Let A, B be two sets. The set A × B is the set of all pairs (a, b), where a is
an element of A and b an element of B.
A × B = {(a, b) | a ∈ A, b ∈ B}
We have: |A × B| = |A| · |B| (for finite sets A, B).
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Relations
Binary relation
Let A, B be two sets. A binary relation between A and B is a set of pairs
R ⊆ A × B.
Relation
Let A1 , . . . , An be sets. An (n-ary) relation is a set of tuples
R ⊆ A1 × · · · × An .
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Properties of Relations
Let R ⊆ A × A be a relation from A to A.
R is called reflexive, when for all x ∈ A it holds, that x R x.
R is called symmetric, when for all x, y ∈ A it holds, that when x R y ,
then y R x.
R is called antisymmetric, when for all x, y ∈ A it holds, that when
x R y and y R x, then x = y .
R is called transitive, when for all x, y , z ∈ A it holds, that when
x R y and y R z, then x R z.
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Properties of Relations (Continued)
Reflexive:
a
Symmetric:
a
Antisymmetric:
a
Transitive:
a
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b
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b
b
wobei a 6= b
c
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Speical Relations
A quasi-order (or pre-order) is a reflexive, transitive relation.
A order is a reflexive, transitive und antisymmetric relation.
An equivalence relation is a reflexive, transitive and symmetric
relation.
Quasi-order:
a
b
order:
c
a
d
b
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equivalence relation:
c
a
d
b
c
d
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functions
Function
f:A→B
a 7→ f (a)
The function f maps an element a ∈ A to an element f (a) ∈ B. Here, A is
the domain and B the codomain.
Formally: a function f : A → B is a total and unambiguous relation
between A and B.
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Mathematical Statements
In mathematics, statement are either true or false.
Statements can be composed from smaller statements:
P and Q – When P, then Q – For all P, Q – P doesn’t hold
There are:
basic statements of mathematics (axioms)
statements which are already proved from the axioms (theorems)
statements which are assumed to be true (hypotheses, premises,
assumptions)
assertions that we want to prove or refute.
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Formal Proofs
Example of a theorem
For all n > 0 it holds, that 1 + 2 + · · · + n =
n · (n + 1)
.
2
Premise: n is a number greater than 0.
Conclusion: 1 + 2 + · · · + n =
n · (n + 1)
.
2
If the premises are true, then the conclusion is also true.
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Implication
Implication (“if, then”)
Wenn P, dann Q“ (P → Q) ist wahr, wenn Q aus P folgt.
”
Using: If P is known, and P → Q is known, then Q is known. (Modus
Ponens).
Proving: To prove P → Q, assume P, and show under this assumption,
that Q is true.
Refuting: To refute P → Q, show that P is true, but Q isn’t.
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Konjunktion
Conjunction (“and”)
P and Q“ (P ∧ Q) is true, when P and Q are both true.
”
Using: When P ∧ Q ist known, then P and Q are also known separately
(and can be used as premises).
Proving: To prove P ∧ Q, we have to prove P and prove Q.
Refuting: To refute P ∧ Q, we have to refute P or refute Q.
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Disjunction
Disjunction (“or”)
P or Q“ (P ∨ Q) is true, when P is true or Q is true.
”
Refuting: To refute P ∨ Q, one must refute P or refute Q.
Using: To prove R from P ∨ Q:
Assume P, and show under that assumption, that R is true.
Assume Q, and show under that assumption, that R is true.
Since R follows from both P and Q, and P or Q holds, R must also hold.
Prove: To prove P ∨ Q, you have to prove P or you have to prove Q. (In
most cases, this doesn’t work, however.)
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Negation
Negation (“not”)
“Not P” (¬P) is true, when P is not true, and false, when P is true.
Using: Negations can be used to prove contradictions.
Proving: You prove ¬P by refuting P.
(In many cases you need a proof by contradiction.
Refuting: You refute ¬P by proving P.
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Universal quantifier
Universal quantifier (“for all”)
“For all x from P it holds that Q“ (∀x ∈ P : Q) is true, when Q holds for
all objects x from P.
Using: When ∀x ∈ P : Q is known, and you have an object a from P hat,
you know that Q holds for a.
Proving: Assume, that x is an element of P and show under this
assumption that Q holds for x. You cannot assume any other things about
x!
Wiederlegen: Suche einen Gegenbeispiel, das heißt ein Objekt aus P, für
das Q nicht gilt.
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Existential quantifier
Existential quantifier (“there is a”)
“There is a x from P, such that Q” (∃x ∈ P : Q) is true, when an object
x ∈ P exists, such that Q holds for x.
Using: When ∃x ∈ P : Q is known, you may introduce an object from P
(with an arbitrary name) for which Q holds.
You cannot assume any other things about Q.
Proving: Suche ein Beispiel: ein Objekt x ∈ P für das Q gilt.
Refuting: Nehme an, dass x ∈ P, und beweise, dass Q nicht für x gilt.
Sonst darf man nichts über x annehmen.
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Example
Prove the following theorem:
Theorem
Let A be a set, and ⊆ A × A a quasi-order on A. Define the relation ≈
as follows:
x ≈ y whenever x y and y x.
Then ≈ is an equivalence relation.
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Automaten und Formale Sprachen“ alias ” Theoretische Informatik

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