CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 8 Mälardalen University
CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 1 - Intro
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Transcript of CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 1 - Intro
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CDT314
FABER
Formal Languages, Automata and Models of Computation
Lecture 1 - Intro
School of Innovation, Design and Engineering Mälardalen University
2011
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Content
Adminstrivia
Mathematical Preliminaries
Countable Sets (Uppräkneliga mängder)
Uncountable sets (Överuppräkneliga mängder)
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Lecturer & Examiner
Gordana Dodig-Crnkovic
Lessons: Leo HatvaniAssistant: Svetlana Girs
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http://www.idt.mdh.se/kurser/cd5560/11_11/
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 lectures
2 hours exercises
14 hours own work a week!
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Why Theory of Computation?
1. A real computer can be modelled by a mathematical object: a theoretical computer.
2. A formal language is a set of strings, and can represent a computational problem.
3. A formal language can be described in many different ways that ultimately prove to be identical.
4. Simulation: the relative power of computing models can be based on the ease with which one model can simulate another.
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5. Robustness of a general computational model.
6. The Church-Turing thesis: "everything algorithmically computable is computable by a Turing machine."
7. Study of non-determinism: languages can be described by the existence or no-nexistence of computational paths.
8. Understanding unsolvability: for some computational problems there is no corresponding algorithm that will unerringly solve them.
Why Theory of Computation?
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Practical Applications
1. Efficient compilation of computer languages
2. String search
3. Investigation of the limits of computation,recognizing difficult/unsolvable problems
4. Applications to other areas:– circuit verification– economics and game theory (finite automata as
strategy models in decision-making); – theoretical biology (L-systems as models of
organism growth) – computer graphics (L-systems) – linguistics (modelling by grammars)
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History
• Euclid's attempt to axiomatize geometry
(Archimedes realized, during his own efforts to define the area of a planar figure, that Euclid's attempt had failed and that additional postulates were needed. )
• Leibniz's (1646 - 1716) dream reasoning as calculus - "Characteristica Universalis" aiming at:
"a general method in which all truths of the reason would be reduced to a kind of calculation. At the same time this would be a sort of universal language or script, but infinitely different from all those projected hitherto; for the symbols and even the words in it would direct reason; and errors, except those of fact, would be mere mistakes in calculation.“
• de Morgan, Boole, Frege, Russell, Whitehead:
Mathematics as a branch of symbolic logic!
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1900 Hilberts program for axiomatization of mathematics, redefined "proof" to become a completely rigorous notion, totally different from the psycho/sociological "A proof is something that convinces other mathematicians.“ He confirms the prediction Leibniz made, that "the symbols would direct reason"
1880 -1936 first programming languages
1931 Gödels incompleteness theorems
1936 Turing maschine (showed to be equivalent with recursive functions). Commonly accepted: TM as ultimate computer
1950 automata
1956 language/automata hierarchy
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Every mathematical truth expressed in a formal language is consisting of
• a fixed alphabet of admissible symbols, and
• explicit rules of syntax for combining those symbols into meaningful words and sentences
Gödel's two incompleteness theorems of mathematical logic show limitations of all but the most trivial axiomatic systems. The theorems are widely interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all of mathematics is impossible, thus giving a negative answer to Hilbert's second problem.
Formalization of Mathematics
The main goal of Hilbert's program was to provide secure foundations for all mathematics. In particular this should include:
A formalization of all mathematics; in other words all mathematical statements should be written in a precise formal language, and manipulated according to well defined rules.
Completeness: a proof that all true mathematical statements can be proved in the formalism.
Consistency: a proof that no contradiction can be obtained in the formalism of mathematics. This consistency proof should preferably use only "finitistic" reasoning about finite mathematical objects.
Decidability: there should be an algorithm for deciding the truth or falsity of any mathematical statement.
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Statement of Hilbert’s program
Gödel showed that most of the goals of Hilbert's program were impossible to achieve. His second incompleteness theorem stated that any consistent theory powerful enough to encode addition and multiplication of integers cannot prove its own consistency. This wipes out most of Hilbert's program as follows:
It is not possible to formalize all of mathematics, as any attempt at such a formalism will omit some true mathematical statements.
The most interesting mathematical theories are not complete.
A theory such as Peano arithmetic cannot even prove its own consistency.
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Statement of Hilbert’s program
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Turing used a Universal Turing machine (UTM) to prove an incompleteness theorem even more powerful than Gödel’s because it destroyed not one but two of Hilbert's dreams:
1. Finding a finite list of axioms from which all
mathematical truths can be deduced
2. Solving the entscheidungsproblem, ("decision
problem“) by producing a "fully automatic procedure"
for deciding whether a given proposition (sentence) is
true or false.
Turing’s Contribution
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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 3
4 5
A
U
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7
8
910
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Set Operations
A = { 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
AA
A = A
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{ even integers } = { odd integers }
02
4
6
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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Subset
A = { 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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Proof by 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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Inductive Proof
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
Level
n
n+1
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hypothesis: L(n) 2n
Level
n
n+1
L(n+1) 2 * L(n) 2 * 2n = 2n+1
Induction Step
END OF PROOF
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Proof by induction: Cardinality of a power set
Let S be a finite set with N elements. Then the powerset of S (that is the set of all subsets of S ) contains 2^N elements. In other words, S has 2^N subsets. This statement can be proved by induction. It's true for N=0,1,2,3 as can be shown by examination.
The notation 2^N means 2 to the power N, i.e., the product of N factors all of which equal 2. 2^0 is defined to be 1
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For the induction step suppose that the statement is true for a set with N-1 elements, and let S be a set with N elements. Remove on element x from S to obtain a set T with N-1 elements. There are two types of subsets of S: those that contain x and those that do not contain x. The latter are subsets of T, of which there are 2^ N-1.
Every subset P of S that does contain x can be obtained from a subset Q of T by adding x. The set Q is simply the set P with x removed. Clearly there is a unique and distinct set Q for each set P and every subset Q of T gives rise to a unique and distinct subset P of S. There are thus also 2^ (N-1) subsets of S that contain x, for a total of 2^ (N-1) + 2^ (N-1) = 2^ N subsets of S.
The size of a finite power set
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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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Countable Sets
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Infinite sets are either
Countable or Uncountable
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Countable set
There is a one to one correspondence
between elements of the set
and natural numbers
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We started with the natural numbers, then• add infinitely many negative whole numbers to get the integers, • then add infinitely many rational fractions to get the rationals, • then added infinitely many irrational fractions to get the reals.
Each infinite addition seem to increase cardinality: |N| < |Z| < |Q| < |R|
But is this true? NO!
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Example
Integers: ,2,2,1,1,0
The set of integers is countable
Correspondence:
Natural numbers: ,4,3,2,1,0
oddnnevennnnf 2/)1(;2/)( {
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ExampleThe set of rational numbers
is countable
Positive
Rational numbers:,
87
,43
,21
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Naive Idea
Rational numbers: ,31
,21
,11
Natural numbers:
Correspondence:
,3,2,1
Doesn’t work!
we will never count
numbers with nominator 2:,
32
,22
,12
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Better Approach
11
21
31
41
12
22
32
13
23
14
...
...
...
...
Rows: constant numerator (täljare)
Columns: constant denominator
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11
21
31
41
12
22
32
13
23
14
...
...
...
...
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We proved:
the set of rational numbers is countable
by describing an enumeration procedure
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Definition
An enumeration procedure for is an
algorithm that generates
all strings of one by one
Let be a set of strings S
S
S
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A set is countable if there is an
enumeration procedure for it
Observation
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Example
The set of all finite strings
is countable
},,{ cba
We will describe the enumeration procedure
Proof
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Naive procedure:
Produce the strings in lexicographic order:
aaaaaa
......Doesn’t work!
Strings starting with will never be produced b
aaaa
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Better procedure
1. Produce all strings of length 1
2. Produce all strings of length 2
3. Produce all strings of length 3
4. Produce all strings of length 4
..........
Proper Order
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Produce strings in
Proper Order
aaabacbabbbccacbcc
aaaaabaac......
length 2
length 3
length 1abc
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Theorem
The set of all finite strings is countable
Proof
Find an enumeration procedure
for the set of finite strings
Any finite string can be encoded
with a binary string of 0’s and 1’s
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Produce strings in Proper Order
length 2
length 3
length 10
1
00
01
10
11
000
001
….
0
1
2
3
4
5
6
7
….
String = program Natural number
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PROGRAM = STRING (syntactic way)
PROGRAM = FUNCTION (semantic way)
PROGRAMstring string
PROGRAMnatural number
n
natural number
n
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Uncountable Sets
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A set is uncountable if it is not countable
Definition
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Theorem
The set of all infinite strings is uncountable
We assume we have
an enumeration procedure
for the set of infinite strings
Proof (by contradiction)
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Infinite string Encoding
0w
1w
2w
...
...
...
...
00b
10b
20b
01b
11b
21b
02b
12b
22b
=
=
=
Cantor’s diagonal argument
... ... ... ...
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Cantor’s diagonal argument
We can construct a new string that is missing in our enumeration!
w
The set of all infinite strings is uncountable!
Conclusion
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There are some integer functions that
that cannot be described by finite strings (programs/algorithms).
Conclusion
An infinite string can be seen as FUNCTION (n:th output is n:th bit in the string)
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Theorem
Let be an infinite countable set
The powerset of is uncountable S2 S
S
Example of uncountable infinite sets
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Proof
Since is countable, we can write S
},,,{ 321 sssS
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Elements of the powerset have the form:
},{ 31 ss
},,,{ 10975 ssss
……
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We encode each element of the power set
with a binary string of 0’s and 1’s
1s 2s 3s 4s
1 0 0 0}{ 1s
Powerset
element
Encoding
0 1 1 0},{ 32 ss
1 0 1 1},,{ 431 sss
...
...
...
...
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Let’s assume (by contradiction)
that the powerset is countable.
we can enumerate
the elements of the powerset
Then:
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1 0 0 0 0
1 1 0 0 0
1 1 0 1 0
1 1 0 0 1
Powerset
elementEncoding
1p
2p
3p
4p
...
...
...
...
...
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Take the powerset element
whose bits are the complements
in the diagonal
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1 0 0 0 0
1 1 0 0 0
1 1 0 1 0
1 1 0 0 1
New element: 0011
(binary complement of diagonal)
...
...
...
...
1p
2p
3p
4p
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The new element must be some
of the powerset ip
However, that’s impossible:
the i-th bit of must be
the complement of itself
from definition of
Contradiction!
ip
ip
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Since we have a contradiction:
The powerset of is uncountable S2 S
END OF PROOF
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Example Alphabet : },{ ba
The set of all finite strings:
},,,,,,,,,{},{ * aabaaabbbaabaababaS
infinite and countable
uncountable infinite
}},,,}{,{},{},{{2 aababaabaaS 1L 2L 3L 4L
The powerset of contains all languages:S
An Application: Languages
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Finite strings (algorithms): countable
Languages (power set of strings): uncountable
There are infinitely many more languages
than finite strings.
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There are some languages
that cannot be described by finite strings (algorithms).
Conclusion
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Cardinality - Kardinaltal
Kardinaltal är mått på storleken av mängder. Kardinaltalet för en ändlig mängd är helt enkelt antalet element i mängden.
Två mängder är lika mäktiga om man kan para ihop elementen i den ena mängden med elementen i den andra på ett uttömmande sätt, dvs det finns en bijektion mellan dem.
Numbers
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http://teachers.henrico.k12.va.us/math/ms/c30708/pics/1_4realnumbers.jpg
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Cardinality - Kardinaltal
Kardinaltal kan generaliseras till oändliga mängder. Till exempel är mängden av positiva heltal och mängden av heltal lika mäktiga (har samma kardinaltal).
Däremot kan man inte para ihop alla reella tal med heltalen på detta sätt. Mängden av reella tal har större mäktighet än mängden av heltal.
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Cardinality - Kardinaltal
Man kan införa kardinaltal på ett sådant sätt att två mängder har samma kardinaltal om och endast om de har samma mäktighet. T ex kallas kardinaltalet som hör till de hela talen för 0 (alef 0, alef är den första bokstaven i det hebreiska alfabetet).
Dessa oändliga kardinaltal kallas transfinita kardinaltal.
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Georg Cantor utvecklade i slutet av 1800-talet matematikens logiska grund, mängdläran.
Den enklaste, "minsta", oändligheten kallade han 0.
Det är den uppräkningsbara oändliga mängdens (exempelvis mängden av alla heltal) kardinaltalet.
Kardinaltalet av mängden punkter på en linje, och även punkterna på ett plan och i en kropp, kallade Cantor 1.
Fanns det större oändligheter?
Mer om oändligheter…More about Infinty
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Ja! Cantor kunde visa att antalet funktioner på en linje var ”ännu oändligare” än punkterna på linjen,
och han kallade den mängden 2.
Cantor fann att det gick att räkna med kardinaltalen precis som med vanliga tal, men räknereglerna blev något enahanda..
0 + 1= 0 0 + 0 = 0 0 · 0 = 0.
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Men vid exponering hände det något:
0 0 (0 upphöjt till 0) = 1.
Mer generellt visade det sig att
2 n (2 upphöjt till n) = n+1
Det innebar att det fanns oändligt många oändligheter, den ena mäktigare än den andra!
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Men var det verkligen säkert att det inte fanns någon oändlighet mellan den uppräkningsbara och punkterna på linjen? Cantor försökte bevisa den så kallade kontinuumhypotesen.
Cantor: two different infinities 0 and 1 http://www.ii.com/math/ch/#cardinals
Continuum Hypothesis: 0 < 1 = 2 0
Se även:http://www.nyteknik.se/pub/ipsart.asp?art_id=26484