formal languages, automata and models of computation · formal languages, automata and models of...
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CDT314
FABER
Formal Languages, Automata
and Models of Computation
Lecture 1 - Intro
Mälardalen University
2010
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Content
Adminstrivia
Mathematical Preliminaries
Countable Sets (Uppräkneliga mängder)
Uncountable sets (Överuppräkneliga mängder)
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Examiner
Gordana Dodig-Crnkovic
Lecturer
Stefan Bygde
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http://www.idt.mdh.se/kurser/cd5560/11_10
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 computational model.
6. The Church-Turing thesis: anything that can
be computed can be computed by a Turing
machine.
7. Nondeterminism: languages can be
described by the existence or nonexistence
of computational paths.
8. Unsolvability: for some computational
problems there is no corresponding
algorithm that will unerringly solve them.
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Practical Applications
1. Efficient compilation of computer languages
2. String searching
3. Identifying the limits; Recognizing difficult 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 (modeling 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 dream of a symbolic logic
• de Morgan, Boole, Frege, Russell, Whitehead:
Mathematics as branch of symbolic logic!
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1900 Hilberts program
1880 -1936 first programming languages
1931 Gödels incompleteness theorem
1936 Turing machine (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
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Turing used a Universal Turing machine (UTM) to prove an even more powerful incompleteness theorem 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.
Examination
• Three midterms
• Regular Languages
• Context Free Languages
• Restriction Free Languages
• OR:
• Final Exam in three parts
• If you have finished the midterm of one
time of language you don’t have to do
corresponding part of the final exam
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Form and Resources
• Lectures
• Exercises
• Labs (optional)
• Midterms
• Webpage (many links, news etc)
• Literature:
• Swe: Lennart Salling. Formella språk,
automater och beräkningar
• Eng: Linz, An Introduction to Formal
Languages and Automata
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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{A
A 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 of induction: The cardinality of the powerset
Claim: A set of n elements
has 2n subsets
Check:
• The empty set {} has only one subset: {}.
• The set {a} (a set with exactly one element) has two
subsets: {} och {a}
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Påstående: En mängd med n element
har 2n delmängder
Check (cont.)
• The set of two elements {a, b} has four subsets:
{}, {a}, {b} and {a,b}
• The set {a, b, c} has eight subsets:
{}, {a}, {b}, {c} och {a,b}, {a,c}, {b,c}, {a,b,c}
The claim holds for these basic cases.
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Base Case
The simplest case is a set with no elements
(there is just one such set), which has 20 = 1
subsets.
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Induction step
Suppose that the claim holds for all sets with k
elements, i.e., suppose that every set with k
elements has 2k subsets.
Show that the claim in this case holds also for all
sets with k+1 elements, i.e., show that every
set with k+1 elements has 2k+1 subsets.
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Consider an arbitrary set with k+1 elements. The subsets
of this set can be divided into two groups:
Subsets which does not contain the k+1:th element: Such
a subset is a subset to the set of the k original elements,
and the cardinality of subsets to a set of k elements is
(according to the inductive hypothesis) 2k.
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Subsets containing the k+1:th element: Such a subset can be
constructed by taking any set not containing the k+1th element and
add the k+1th element. Since there are 2k subsets without the
k+1th element (by hypothesis), it is also possible to create 2k
subsets including this element.
In total there are 2k + 2k = 2. 2k= 2k+1 subsets to the considered set.
END OF PROOF
(Exemple taken from the book: Diskret matematik och diskreta modeller,
K Eriksson, H. Gavel)
45
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 = n22
END OF PROOF
48
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
51
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!
52
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/)( {
53
ExampleThe set of rational numbers
is countable
Positive
Rational numbers:,
8
7,
4
3,
2
1
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Naive Idea
Rational numbers: ,3
1,
2
1,
1
1
Natural numbers:
Correspondence:
,3,2,1
Doesn’t work!
we will never count
numbers with nominator 2:,
3
2,
2
2,
1
2
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Better Approach
1
1
2
1
3
1
4
1
1
2
2
2
3
2
1
3
2
3
1
4
...
...
...
...
Rows: constant numerator (täljare)
Columns: constant denominator
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1
1
2
1
3
1
4
1
1
2
2
2
3
2
1
3
2
3
1
4
...
...
...
...
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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
61
Naive procedure:
Produce the strings in lexicographic order:
a
aaaaa
......
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
aa
abacba
bb
bcca
cbcc
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
PROGRAM
natural
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 uncountableS2 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
83
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
84
Finite strings (algorithms): countable
Languages (power set of strings): uncountable
There are infinitely many more languages
than finite strings.
85
There are some languages
that cannot be described by finite strings
(algorithms).
Conclusion
86
Cardinality - Kardinaltal
A Cardinal number is a measure of the size of a set. The cardinality for a finite set is simply the number of elements in the set.
Two sets has equal size if it is possible to pair the elements one by one in the two sets. That is, if there exists a bijection between the two sets.
This concept can be generalised to infinite sets. For instance, the set of positive integers and the set of integers have equal size.
On the contrary, it is not possible to pair the real numbers with the integers in this fashion. The set of real numbers is larger than the set of integers.
Cardinal numbers can be defined as in the following way: two sets have the same cardinal number if and only if they have the same size (i.e., can be paired together one and one). As an example, the cardinal number for the integers is 0 (alef-0, alef is the first letter in the hebrew alphabet).
These infinite cardinal numbers are called transfinite cardinal numbers.
87
Georg Cantor developed, in the end of the 19th century, the logical
foundation of mathematics: set theory.
Cantor introduced the notion transfinite cardinal numbers.
The simplest, ”smallest", infinity, he named 0.
This is the countable infinite sets’ (for instance, the integers) cardinal
number.
The cardinality of the set of points on a line, or the points in a plane or
field, Cantor named 1.
Are there ”larger” infinities?
More about Infinty
88
Yes! Cantor showed that the set of functions on a line
was ”even more infinite” than the points on the line,
and he called that cardinality 2.
Cantor discovered that it was possible to use
artithmetic on cardinal numbers just as with finite
numbers. But the arithmetic rules became quite
simplistic:
0 + 1= 0 0 + 0 = 0 0 · 0 = 0.
89
However, with powers something happened:
0 0 ( 0 to the power of 0) = 1.
In general:
2 n (2 to the power of n) = n+1
This implies that there are infinitly many infinities, in
strictly increasing order.
90
But is it really certain that there is no cardinality in between the countable infinity and the points on the line? Cantor tried to prove the so-called continuum hypothesis.
Cantor: two different infinities 0 and 1
http://www.ii.com/math/ch/#cardinals
Continuum Hypothesis: 0 < 1 = 2 0
See also:
http://www.nyteknik.se/pub/ipsart.asp?art_id=26484
CD5560CD5560
FABER
Formal Languages, Automata and Models of Computationand Models of Computation
Lecture 2
Mälardalen University2010
1
ContentLanguages, Alphabets and Stringsg g , p g
Strings & String OperationsLanguages & Language OperationsLanguages & Language Operations
Regular ExpressionsFinite Automata, FADeterministic Finite Automata, DFA
2
Languages,Languages, Alphabets and
Strings
3
Languages
A language is a set of strings
d fi d l h bA String is a sequence of letters
defined over an alphabet:
{ }{ }zcba ,,,, K=Σ
An alphabet is a set of symbols
4
Alphabets and Strings
We will use small alphabets: { }ba,=Σp
Strings
{ },
ba
Strings
bbbaaavabu =ab
abbawbbbaaav==
b babba
abbaw =
baaabbbaabababa
5baaabbbaaba
Operations on StringsOperations on Strings
6
String Operations
naaaw L21= x = abbam
nbbbv L21
21= y = bbbaaa
x abba
Concatenation (sammanfogning)
bbbaaawv =
Concatenation (sammanfogning)
mn bbbaaawv LL 2121=
bb bbbxy = abbabbbaaa
7
Reverse (reversering)Reverse (reversering)
naaaw L21= ababaaabbb
12aaaw nR L= bbbaaababa12n bbbaaababa
Example:Longest odd length palindrome in a natural language:g g p g g
saippuakauppias (Finnish: soap sailsman)
8
(Finnish: soap sailsman)
String LengthString Length
naaaw L21=nw =Length: nw =Length:
Examples:
4=abbaExamples:
2=aa1=a
9
Recursive Definition of LengthRecursive Definition of Length
F l tt 1For any letter: 1=a
For any string : 1+= wwawa
Example: 1+= abbabba
11111 +++=++= aab41111 =+++=
10
Length of Concatenation
vuuv +=
3== uaabuExample:
53,== vabaabv
uaabuExample:
5, vabaabv
8538== aababaabuv
853 =+=+= vuuv
11
Proof of Concatenation Length
Claim: vuuv +=Claim: vuuv +=
Proof: By induction on the length v
Induction basis: 1=vFrom definition of length:
vuuuv +=+= 1 vuuuv +=+= 1
12
I d i h h i +Inductive hypothesis:
≤
vuuv +=
f nv ≤for
Inductive step: we will prove
vuuv +=
1+= nvfor
13
Inductive StepWrite , wherewav = 1, == anw
From definition of length: 1+== uwuwauv1+= wwa
From inductive hypothesis: wuuw +=
1Thus: vuwauwuuv +=+=++= 1
14END OF PROOF
Empty StringEmpty String
A string with no letters: (Also denoted as ε)
λ( )
Observations: 0λObservations: 0
λλ
λ
==
=
wwwλλ
∅== abbaabbaabba
}{{} λ≠=∅
15
Substring (delsträng)Substring of a string:
a subsequence of consecutive charactersa subsequence of consecutive characters
String SubstringString Substring
ababbab
babbaabbab
bbabb
bb babbab
16bbababbab
Prefix and Suffix
Suffixesabbab
Prefixes Suffixes
λ abbab uvw =Prefixes
aλ
bbabfi
ab babprefix
suffixabb ab
suffix
abbaλb
17abbab λ
Repetitionn www... w =
n (String repeated n times)w
Example: ( ) abbaabbaabba =2p
Definition:
( )
λ0Definition: λ=0w
( )0( ) λ=0abba
18
Th (Kl * t ) O tiThe (Kleene* star) Operation
the set of all possible strings from alphabet *Σ Σ
{ }bΣ { }{ }*
,aabaaabbbaabaaba
baλ=Σ
=Σ{ }K,,,,,,,,,* aabaaabbbaabaabaλ=Σ
[* Kleene is pronounced "clay-knee“]
19http://en.wikipedia.org/wiki/Kleene_star
The + (Kleene plus) Operation( p ) p
: the set of all possible strings from the+Σ galphabet except Σ λ
{ },ba=Σ{ }K,,,,,,,,,* aabaaabbbaabaabaλ=Σ
λ−Σ=Σ+ *
{ }K,,,,,,,, aabaaabbbaabaaba=Σ+
20
Example
{ }, fyoj=Σ , usch
{}
* λ, oj, fy, usch, ojoj, =Σ}Kfyfy,uschusch, ojfy, ojusch
λ−Σ=Σ+ *{
}oj, fy, usch, ojoj, fyfy,uschusch,
f h=Σ+
}Kojfy, ojusch
21
Operations on LanguagesOperations on Languages
22
LanguageA language is any subset of *Σ
Example: { },ba=Σ { }{ }K,,,,,,,,*
,aaabbbaabaabaλ=Σ
Languages:{ }λ{ }{ }aabaaaλ{ }
}{,,
aaaaaaabaababaabbaaabaaa
λ23
},,,,,{ aaaaaaabaababaabbaλ
Example
An infinite language }0:{ ≥= nbaL nnAn infinite language }0:{ ≥= nbaL
λ
Labb∉ab L∈
bbbbbaabbaaaaabbbbb
24
Operations on LanguagesThe usual set operations
{ } { }{ } { } bbbbb
aaaabbabaabbbaaaaaba}{
},,,{,,, =∪{ } { }{ } { } { }aaaaaabbbaaaaaba
ababbbaaaaaba,,,,
}{,,,=−=∩
{ } { } { },,,,
LL −Σ= *Complement:
{ }K,,,,,,,,,* aabaaabbbaabaabaλ=ΣLL Σp
{ } { }K,,,,,,, aaabbabaabbaa λ=25
Reverse
}:{ LwwL RR ∈=Definition:
}:{ LwwL ∈=
Examples:
{ } { }ababbaabababaaabab R ,,,, =Examples:
{ } { }
}0:{ ≥= nbaL nn
}0:{}{
≥= nabL nnR }0:{ ≥nabL
26
ConcatenationCo cate at o
Definition: { }2121 ,: LyLxxyLL ∈∈=
Example
{ }{ }aabbaaba ,,,Example
{ }{ }{ }baaabababaaabbaaaab
aabbaaba,,,,,
,,,= { },,,,,
27
RepeatDefinition: 321Ln LLLL =
n
{ } { }{ }{ }bbbb 3{ } { }{ }{ }{ }bbbbbababbaaabbabaaabaaa
babababa ,,,, 3 ==
Special case:
{ }bbbbbababbaaabbabaaabaaa ,,,,,,,Special case: { }λ=0L
{ } { }λ=0,, aaabbaa28
Example
nn }0:{ ≥= nbaL nn
}0{2 bb mmnn }0,:{2 ≥= mnbabaL mmnn
2Lbb bbb 2Laabbaaabbb∈
29
Star-Closure (Kleene *)
Definition:Definition:L210* LLLL ∪∪=
Example:Example:
⎪⎫
⎪⎧ ,λ
{ } ⎪⎪⎬
⎪⎪⎨=
,,*,
bbbbbbbbbba
bba{ }
⎪⎪⎭
⎬
⎪⎪⎩
⎨ ,,,,abbbbabbaaabbaaa
bbbbbbaabbaa
30
⎪⎭⎪⎩ K,,,, abbbbabbaaabbaaa
Positive ClosureDefinition
+ 21
{ }λ−= *LL=+ 21 LLL UU
{ }λ= *L
⎪⎫
⎪⎧ ,,bba
{ }⎪
⎪⎬
⎪
⎪⎨=
+ ,,,,, bbbbbbaabbaabba⎪⎭⎪⎩ K,,,, abbbbabbaaabbaaa
31
Regular ExpressionsRegular Expressions
32
Regular Expressions: R i D fi itiRecursive Definition
λ∅ αλ,,∅Primitive regular expressions:
2rGiven regular expressions and1r
21 rr +2rGiven regular expressions and 1r
21
21rrrr
⋅+
1
21*r
are Regular Expressions
( )1r33
Examples
( ) )(* ∅+⋅⋅+ ccbaA regular expression: ( ) )( ∅+⋅⋅+ ccbaA regular expression:
( )++ baNot a regular expression:
34
Building Regular Expressions
{ }cba=ΣBuilding Regular Expressions
Zero or more
{ }cba ,,=Σ
Zero or more. a* means "zero or more a's."
To say "zero or more ab's," that is, y , ,{, ab, abab, ababab, ...}, you need to say (ab)*.
ab* denotes {a, ab, abb, abbb, abbbb, ...}.
35
Building Regular Expressions{ }cba ,,=Σ
Building Regular Expressions
One or more
{ },,
One or more. Since a* means "zero or more a's", you can use aa* (or equivalently, a*a) to mean "one or more a's.“
Similarly, to describe "one or more ab's," that is, {ab abab ababab } you can use ab(ab)*{ab, abab, ababab, ...}, you can use ab(ab) .
36
Building Regular Expressions
{ }cba ,,=Σ
Any string at all. To describe any string at all (with = {a b c}) you canTo describe any string at all (with = {a, b, c}), you can use (a+b+c)*.
Any nonempty string.This can be written as any character from followed by any string at all: (a+b+c)(a+b+c)*.
37
Building Regular Expressions{ }cba ,,=Σ
g g p
Any string not containing....To describe any string at all that doesn't contain an aTo describe any string at all that doesn t contain an a (with = {a, b, c}), you can use (b+c)*.
Any string containing exactly one...To describe any string that contains exactly one a, put y g y , p"any string not containing an a," on either side of the a, like this: (b+c)*a(b+c)*. , ( ) ( )
38
Languages of Regular Expressions
( )( )rL rlanguage of regular expression
Example
( ) { },...,,,,,*)( bcaabcaabcacbaL λ=⋅+
39
DefinitionDefinition
For primitive regular expressions:
( )L ∅∅( )L ∅=∅
( ) { }L = λλ( ) { }
( ) { }aaL =
40
Definition (continued)e t o (co t ued)
For regular expressions and1r 2r
( ) ( ) ( )2121 rLrLrrL ∪=+
( ) ( ) ( )( ) ( ) ( )2121 rLrLrrL =⋅
( ) ( )( )( ) ( )( )** 11 rLrL =
( )( ) ( )( )( ) ( )11 rLrL =
41
Example( ) *b+( ) *aba ⋅+Regular expression:
( )( )*abaL ⋅+ ( )( ) ( )*aLbaL +=
( ) ( )( ) ( )*aLbaL +=
( ) ( )( ) ( )( )*LbLL( ) ( )( ) ( )( )*aLbLaL ∪=
{ } { }( ) { }( )*b∪{ } { }( ) { }( )*aba ∪=
{ }{ }aaaaaaba λ= { }{ },...,,,, aaaaaaba λ=
{ }baababaaaaaa=42
{ },...,,,...,,, baababaaaaaa=
Example
Regular expression
( ) ( )bbabar ++= *
( ) { }bbbbaabbaabbarL =( ) { },...,,,,, bbbbaabbaabbarL =
43
Example
Regular expressionRegular expression
( ) ( ) bbbaar **= ( ) ( ) bbbaar =
( ) }0,:{ 22 ≥= mnbbarL mn( ) },{
44
Examplep
{ }1,0=Σ { },
)(rL = { all strings with at leasttwo consecutive 0 }two consecutive 0 }
Regular expression
*)10(00*)10( *)10(00*)10( ++=r
45
Example
{ }1,0=Σ
)(rL = { all strings without
{ }1,0Σ
)(rL = { all strings withouttwo consecutive 0 }
Regular expression
)0(*)011(1 λ++=r(consists of repeating 1’s and 01’s).
46
ExampleL = { all strings without
t ti 0 }two consecutive 0 }
Equivalent solution:
)0(*1)0(**)011*1(2 λλ +++=r
Equivalent solution:
)0(1)0()0111(2 λλ +++=r
( d 00 i i f h 0 h b 1(In order not to get 00 in a string, after each 0 there must be an 1, which means that strings of the form 1....101....1are repeated That is the first parenthesis To take into accountare repeated. That is the first parenthesis. To take into account strings that end with 0, and those consisting of 1’s solely, the rest of the expression is added.)
47
p )
Equivalent Regular Expressions
Definition:
Regular expressions andr rRegular expressions and1r 2r
)()( 21 rLrL =are equivalent if
48
Example
L = { all strings withouttwo consecutive 0 }
)(*)(**)*( λλ)0(*)011(1 λ++=r
)0(*1)0(**)011*1(2 λλ +++=r
LrLrL == )()( 21 1r 2randLrLrL )()( 21 1 2are equivalentregular expressions
49regular expressions.