cd5560 faber formal languages, automata and models of computation lecture 0 - intro
DESCRIPTION
CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro Mälardalen University 2005. Content Adminstrivia Mathematical Preliminaries Countable Sets (Uppräkneliga mängder) Uncountable sets (Överuppräkneliga mängder). Lecturer & E xami ner - PowerPoint PPT PresentationTRANSCRIPT
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CD5560
FABER
Formal Languages, Automata and Models of Computation
Lecture 0 - Intro
Mälardalen University
2005
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Content
AdminstriviaMathematical PreliminariesCountable Sets (Uppräkneliga mängder)
Uncountable sets (Överuppräkneliga mängder)
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Lecturer & Examiner
Gordana Dodig-Crnkovic
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Teaching Assistent
Andreas Ermedahl
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http://www.idt.mdh.se/kurser/cd5560/05_04
visit home page regularly!
Course Home Page
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How Much Work?
20 hours a week for this type of course (norm)
4 hours lectures2 hours exercises
14 hours own work a week!
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Mathematical Preliminaries
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• Sets
• Functions
• Relations
• Proof Techniques
• Languages, Alphabets and Strings
• Strings & String Operations
• Languages & Language Operations
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}3,2,1{AA set is a collection of elements
SETS
},,,{ airplanebicyclebustrainB
We write
A1
Bship
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Set Representations
C = { a, b, c, d, e, f, g, h, i, j, k }
C = { a, b, …, k }
S = { 2, 4, 6, … }
S = { j : j > 0, and j = 2k for some k>0 }
S = { j : j is nonnegative and even }
finite set
infinite set
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A = { 1, 2, 3, 4, 5 }
Universal Set: All possible elements
U = { 1 , … , 10 }
1 2 34 5
A
U
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78
910
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Set OperationsA = { 1, 2, 3 } B = { 2, 3, 4, 5}
• Union
A U B = { 1, 2, 3, 4, 5 }
• Intersection
A B = { 2, 3 }
• Difference
A - B = { 1 }
B - A = { 4, 5 }
U
A B
A-B
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• Complement
Universal set = {1, …, 7}
A = { 1, 2, 3 } A = { 4, 5, 6, 7}
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3
4
5
6
7
A A
A = A
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{ even integers } = { odd integers }
024
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1
3
5
7
even
odd
Integers
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DeMorgan’s Laws
A U B = A BU
A B = A U B
U
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Empty, Null Set:
= { }
S U = S
S =
S - = S
- S =
U = Universal Set
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SubsetA = { 1, 2, 3} B = { 1, 2, 3, 4, 5 }
A B
U
Proper Subset: A B
U
A
B
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Disjoint Sets
A = { 1, 2, 3 } B = { 5, 6} A B =
U
A B
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Set Cardinality
For finite sets
A = { 2, 5, 7 }
|A| = 3
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Powersets
A powerset is a set of sets
Powerset of S = the set of all the subsets of S
S = { a, b, c }
2S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }
Observation: | 2S | = 2|S| ( 8 = 23 )
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Cartesian Product
A = { 2, 4 } B = { 2, 3, 5 }
A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 5) }
|A X B| = |A| |B|
Generalizes to more than two sets
A X B X … X Z
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PROOF TECHNIQUES
• Proof by construction
• Proof by induction
• Proof by contradiction
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Construction
We define a graph to be k-regular
if every node in the graph has degree k.
Theorem. For each even number n > 2 there exists
3-regular graph with n nodes.
1
2
4
3
0
5
1 2
0
3n = 4 n = 6
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Construct a graph G = (V, E) with n > 2 nodes.
V= { 0, 1, …, n-1 }
E = { {i, i+1} for 0 i n-2} {{n-1,0}} (*)
{{i, i+n/2 for 0 i n/2 –1} (**)
The nodes of this graph can be written consecutively around the circle.
(*) edges between adjacent pairs of nodes
(**) edges between nodes on opposite sides
Proof by Construction
END OF PROOF
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Induction
We have statements P1, P2, P3, …
If we know
• for some k that P1, P2, …, Pk are true
• for any n k that
P1, P2, …, Pn imply Pn+1
Then
Every Pi is true
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Proof by Induction
• Inductive basis
Find P1, P2, …, Pk which are true
• Inductive hypothesis
Let’s assume P1, P2, …, Pn are true,
for any n k
• Inductive step
Show that Pn+1 is true
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Example
Theorem A binary tree of height n
has at most 2n leaves.
Proof
let L(i) be the number of leaves at level i
L(0) = 1
L(3) = 8
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We want to show: L(i) 2i
• Inductive basis
L(0) = 1 (the root node)
• Inductive hypothesis
Let’s assume L(i) 2i for all i = 0, 1, …, n
• Induction step
we need to show that L(n + 1) 2n+1
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Induction Step
hypothesis: L(n) 2n
Leveln
n+1
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hypothesis: L(n) 2n
Leveln
n+1
L(n+1) 2 * L(n) 2 * 2n = 2n+1
Induction Step
END OF PROOF
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Inductionsbevis: Potensmängdens kardinalitet
Påstående: En mängd med n element har 2n delmängder
Kontroll
• Tomma mängden {} (med noll element) har bara en delmängd: {}.
• Mängden {a} (med ett element) har två delmängder: {} och {a}
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Påstående: En mängd med n element har 2n delmängder
Kontroll (forts.)
• Mängden {a, b} (med två element) har fyra delmängder:
{}, {a}, {b} och {a,b}
• Mängden {a, b, c} (med tre element) har åtta delmängder: {}, {a}, {b}, {c} och {a,b}, {a,c}, {b,c}, {a,b,c}
Påstående stämmer så här långt.
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Bassteg
Enklaste fallet är en mängd med noll element (det finns bara en sådan), som har 20 = 1 delmängder.
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Induktionssteg
Antag att påståendet gäller för alla mängder med k element, dvs antag att varje mängd med k element har 2k delmängder.
Visa att påståendet i så fall också gäller för alla mängder med k+1 element, dvs visa att varje mängd med k+1 element har 2k+1 delmängder.
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Vi betraktar en godtycklig mängd med k+1 element. Delmängderna till mängden kan delas upp i två sorter:
Delmängder som inte innehåller element nr k+1: En sådan delmängd är en delmängd till mängden med de k första elementen, och delmängder till en mängd med k element finns det (enligt antagandet) 2k stycken.
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Delmängder som innehåller element nr k+1: En sådan delmängd kan man skapa genom att ta en delmängd som inte innehåller element nr k+1 och lägga till detta element. Eftersom det finns 2k delmängder utan element nr k+1 kan man även skapa 2k
delmängder med detta element.
Totalt har man 2k + 2k = 2. 2k= 2k+1 delmängder till den betraktade mängden.
END OF PROOF
(Exempel från boken: Diskret matematik och diskreta modeller, K Eriksson, H. Gavel)
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Proof by Contradiction
We want to prove that a statement P is true
• we assume that P is false
• then we arrive at a conclusion that contradicts our assumptions
• therefore, statement P must be true
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Example
Theorem is not rational
Proof
Assume by contradiction that it is rational
= n/m
n and m have no common factors
We will show that this is impossible
2
2
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Therefore, n2 is evenn is even
n = 2 k
2 m2 = 4k2 m2 = 2k2m is even
m = 2 p
Thus, m and n have common factor 2
Contradiction!
= n/m 2 m2 = n2 2
END OF PROOF
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Languages, Alphabets and
Strings
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defined over an alphabet:
Languages
zcba ,,,,
A language is a set of strings
A String is a sequence of letters
An alphabet is a set of symbols
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Alphabets and Strings
We will use small alphabets:
abbawbbbaaavabu
baaabbbaabababaabbaaba
Strings
ba,
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Operations on Strings
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String Operations
m
n
bbbvaaaw
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21
y bbbaaax abba
mn bbbaaawv 2121
Concatenation (sammanfogning)
xy abbabbbaaa
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12aaaw nR
naaaw 21 ababaaabbb
Reverse (reversering)
bbbaaababa
Example:Longest odd length palindrome in a natural language:
saippuakauppias (Finnish: soap sailsman)
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String Length
naaaw 21
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4
aaaabba
nw Length:
Examples:
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Empty String
A string with no letters: (Also denoted as )
Observations:
}{{}
0
abbaabbaabbawww
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Substring (delsträng)Substring of string:
a subsequence of consecutive characters
String Substring
bbabbabbaab
abbababbababbababbab
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Prefix and Suffix
Suffixesabbab
abbababbaabbaba
babbabbbababbab uvw
prefix
suffix
Prefixes
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Repetition
Example:
Definition:
n
n www... w
abbaabbaabba 2
0w
0abba
}
(String repeated n times)w
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The * (Kleene star) Operation
the set of all possible strings from alphabet
*
,,,,,,,,,*
,aabaaabbbaabaaba
ba
[Kleene is pronounced "clay-knee“]
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The + Operation
: the set of all possible strings from alphabet except
,ba ,,,,,,,,,* aabaaabbbaabaaba
*
,,,,,,,, aabaaabbbaabaaba