1. Theory of Automata Chapter 1
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2. Grading Policy Assignments (10%) Quizzes/Tests (10%)
Due at the beginning of the class
10% penalty for each day of delay, up to 3 days
Some programming component
Quizzes/Tests (10%)
Surprise
Announced
Mid-term exam (30%)
Closed books, closed notes
Final exam (50%)
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3. Recommended Book(s) Text Book(s): Reference Book(s):
Nasir S.F.B and P.K. Srimani, “A Textbook on Automata Theory”, Cambridge University Press, India, 2008
Sikander H. Khiyal, “Theory of Automata and Computation”, National Book Foundation, 2004
Reference Book(s):
John E. Hopcroft, Rajeev Motvani, and Jeffrey D. Ullman, “Introduction to Automata theory, Languages and Computation”, Second Edition, Addison-Wesley, New York, 2001.
Daniel I. A. Cohen, “Introduction to Computer Theory”, Second Edition.
K.L.P. Mishra, N. Chandrasekaran, “Theory of Computer Science (Automata, Languages and Computation)”, Prentice-Hall of India, 2002.
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4. Computation
Computation can be defined as finding a solution to a problem from given inputs by means of an algorithm.
Model of Computation
A formal or an abstract computer
Types of Computation Models
Serial Models
Parallel Models
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5. Computation Devices
In this study we consider a mathematical model of computing.
Such a model strips the computational machinery down to the bare minimum, so that it’s easy to manipulate these theoretical machines (there are several such models, for different purposes, as you’ll soon see) mathematically to prove things about their capabilities.
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6. Background
Theory of computation is the study of abstract computing devices, or “machines”.
Allan Turing in the 1930’s.
In the 1940’s and 1950’s, simpler kinds of machines, which we today call “finite automata”, were studied by a number of researchers.
These automata, originally proposed to model brain function.
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7. Background
In the late 1950’s, the linguist Noam Chomsky begun the study of formal “grammars”.
In 1969, S. Cook extended Turing’s study of what could and what could not be computed.
The later class of problems is called “intractable”, or NP-hard.
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8. What is a Language?
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9. Formal languages
By formal we don't mean dressed up, we mean languages with a form defined by rules that say which strings are in the language and which are not. We are only interested in syntax here, not semantics. We are interested in the form of a string of symbols, not the meaning of the string.
Semantic languages
Syntactic languages
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10. Types of Formal Languages
Regular languages (type-3)
Context-free languages (type-2)
Context-sensitive languages (type-1)
Phrase structure languages (type-0)
Language Generators
Language Recognition
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11. Phrase-Structure Languages Context –Sensitive Languages
Context-Free Languages
Regular Languages
Type-3
DFA
NFA
ε-NFA
RE RG
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Type-2
Language generating devices
Language recognition devices
PDA
CFG
Type-1
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LBA
CSG
Type-0 TM
PSG

12. Lets Play a Game!
In this game I have a rule in mind. You guess strings of letters, using only the letters A and B. I’ll tell you whether each string follows the rule or not. Your job is to guess the rule.
e.g., suppose you have done these experiments:
accepted rejected
ABBA BA
GUES The Language
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13. Applications
Finite state machines are a useful model for many important kinds of hardware and software. e.g.
Software for designing digital circuits
Lexical analyzer of a compiler
Searching for keywords in a file or on the Web
Software for verifying finite state systems, such as communication protocols
Operating System (UNIX grep)
Text Editors
Markup Languages (HTML, XML)
Natural Language Processing
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14. Dr. Abdus Salam
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Central concepts

15. Central Concepts Symbol
A symbol is a single object and is an abstract entity that has no meaning by itself.
It can be alphabet, character or special character
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16. Central Concepts Alphabet
An alphabet is a finite, nonempty set of symbols, Σ (sigma) is used for an alphabet.
For example
{0, 1} is an alphabet with two symbols
{a, b} is another alphabet with two symbols, and
English alphabet is also an alphabet
Σ = {0,1}, or Σ = the binary alphabet
Σ = {a,b,…..z}, or Σ=the set of all lower-case letters
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17. Central Concepts String
A string is a finite sequence of symbols chosen from some alphabet. e.g is a string from the binary alphabet Σ = {0,1}.
The number of symbols in a string is called the length of a string. e.g has length 5 and the number of symbols are 2. The standard notation for the length of a string w is |w|. e.g. |011| = 3 and | ε | = 0
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18. Central Concepts Strings
The empty string (also called null string) is the string with length zero. That is, it has no symbols. It is denoted with ε (epsilon), that may be chosen from any alphabet whatsoever. Thus | ε | = 0.
Powers of an Alphabet: If Σ is an alphabet, we define Σk to be the set of strings of length k, each of whose symbols is in Σ. e.g. Σ0 = {ε}. If Σ = {0,1}, then
Σ1 = {0,1}
Σ2 = {00,01,10,11}
Σ3 = {000,001,010,011,100,101,110,111}
The set of all strings over an alphabet is denoted as Σ*. For instance, {0,1}* = {ε,0,1,00,01,10,11,000,….}. Put another way,
Σ* = Σ0 U Σ1 U Σ2 U Σ3 ….
Σ+ = Σ1 U Σ2 U Σ3 ….
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19. Central Concepts Strings
Concatenation of Strings: Let x and y be strings. Then xy denotes the string obtained by concatenating x with y, that is, xy is the string obtained by appending the sequence os symbols of y to that of x. e.g. if x = aab and y = bbab, then xy = aabbbab. Note that xy ≠ yx.
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20. Central Concepts Languages
A language is a set of strings over an alphabet. Thus {a, ab, baa} is a language (over alphabet {a, b}).
A set of strings all of which are chosen from some Σ*, where Σ is a particular alphabet, is called a language. If Σ is an alphabet, and L Σ* , then L is a language over Σ.
An example is English, where the collection of legal English words is a set of strings over the alphabet that consists of all the letters.
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21. Examples of Languages
The language of all strings consisting of n 0’s followed by n 1’s, for some n ≥ 0:
{ε, 01, 0011, ,…..}
The set of strings of 0’s and 1’s with an equal number of each:
{ε, 01,10,0011,0101,1001,…..}
The set of binary numbers whose value is a prime:
{10, 11, 101,111,1011…..}
Note: The only important constraint on what can be a language is that all alphabets are finite, although they can have an infinite number of strings.
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22. Graphs
A graph, denoted by G = (V, E), consists of a finite set of vertices V and a set of pairs of vertices E called edges. An example of graph is shown below:
Here V={1,2,3,4,5} and E={(n,m)|n+m=4, or n+m=7}
Directed graph?
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23. Trees A tree is a digraph with the following properties:
There is one vertex, called the root, that has no predecessors and from which there is path to every vertex.
Each vertex other than root has exactly one predecessor.
The successors of each vertex are ordered “from the left”.
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24. Set Notation
It is common to describe a language using a “set-notation”:
{w | something about w }
It is read “the set of strings w such that (whatever is said about w to the right of the vertical bar)”
Examples:
{ w | w consists of an equal number of 0’s and 1’s }
{ w | w is a binary integer that is prime }
{ w | w is a syntactically correct C++ program }
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25. Set-Notation as a way to Define Languages
It is also common to replace w by some expression with parameters and describe the strings in the language by stating conditions on the parameters. For example:
{ 0n1n | n ≥ 1}
{0i1j | 0 ≤ i ≤ j }
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26. Some Special Languages
The empty set Ø is a language which has no strings.
The set { ε } is a language which has one string, namely ε . Though ε has no symbols, this set has an object in it. So it is not empty.
For any alphabet Σ, the set of all strings over Σ is denoted by Σ*. Thus a language over alphabet Σ is a subset of Σ*.
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27. Operations on Languages
Since languages are sets, all the set operations can be applied to languages.
Thus the union, intersection and difference of two languages over an alphabet Σ are languages over Σ. The complement of a language L over an alphabet Σ is Σ* - L and it is also a language.
Another operation on a language is concatenation. Let L1 and L2 be languages. Then the concatenation of L1 with L2 is denoted as L1L2 and it is defined as L1L2 = ( xy | x Є L1 and y Є L2}
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28. Theory of Automata Chapter 2
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Theory of Automata Chapter # 2

29. Finite Automata Definition of an Automaton
“An automaton is defined as a system where energy, materials and information are transformed, transmitted and used for performing some functions without direct participation of man.”
Examples are
Automatic machine tools, automatic packing machines, and automatic photo printing machines.
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Theory of Automata Chapter # 2

30. Automaton I1 Automaton q1, q2, …..qn O1 I2 O2 I3 O3 Ip Oq
In Computer Science the term ‘automaton’ means “discrete automaton” and is defined in a more abstract way as shown in the figure:
Dr. Abdus Salam I1
Automaton
q1, q2, …..qn O1 I2 O2
Theory of Automata Chapter # 2 I3 O3 Ip Oq
The model of a discrete automaton

31. Characteristics of Discrete Automaton
Input
At each discrete instance of time t1, t2,…. Input values I1, I2,…. Each of which can take a finite number of fixed values from the input alphabet Σ, are applied to the input side of the model.
Output
O1, O2,…. Are the outputs of the model, each of which can take finite number of fixed values from an output O.
States
At any instant of time the automaton can be in one of the states q1, q2, q3….
State relation
The next state of an automaton at any instant of time is determined by the present state and the present input.
Output relation
Output is related to either state only or to both the input and the state. It should be noted that at any instant of time the automaton is in some state.
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Theory of Automata Chapter # 2

32. Example 1
A vending machine dispense piece of candy that cost Rs. 20 each. The machine accepts coins of Rs. 5 and Rs. 10 only and does not return change.
As soon as the amount deposited equal or exceeds Rs. 20, the machine releases a piece of candy.
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Theory of Automata Chapter # 2

33. Example 1 (contd:) Each circle represents a state of the machine
Rs. 15 deposited
Rs. 5 deposited 10
Dr. Abdus Salam 5 5
Rs. 0 deposited 5 10 5 5 10
Rs. 20 or more deposited
Rs. 10 deposited 10
Theory of Automata Chapter # 2 10
Each circle represents a state of the machine
Unlabelled arrow indicates the initial state of the machine
Double circle indicates that candy is released, called accepting state.

34. Example 1 (contd:) Next State Table Inputs State 5 10 
Rs. 0 deposited
Rs. 5 deposited
Rs. 10 deposited
Rs. 15 deposited
Rs. 20 or more deposited
Rs. 20 deposited
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Theory of Automata Chapter # 2

35. Example 2
A story of man with wolf, goat and cabbage to cross the river.
Dr. Abdus Salam G M
MWGC - Φ
WC-MG
MWC-G G M C W C W
C-MWG
W-MCG
Theory of Automata Chapter # 2 G G G G
MGC-W
MWG-C C W C W G M
Φ - MWGC
MG-WC
G-MWC G M

36. Deterministic Finite Automata
A deterministic finite automaton is a simple language recognition device. It is called deterministic because their operation is completely determined by their input.
Strings are fed into the device by means of an input tape, which is divided into squares, with one symbol inscribed in each tape square.
Input tape
Dr. Abdus Salam a b a a b b a b
Reading head q0
Theory of Automata Chapter # 2 q1 q5
Finite control q2 q4 q3

37. Formal Definition of Deterministic Finite Automata (DFA)
A deterministic finite automaton is quintuple
A= (Q, Σ, δ, s, F) where
Q is a finite set of states,
Σ is an alphabet,
s Є Q is the initial state,
F Q is the set of final states, and
δ the transition function, is a function from Q X Σ to Q . If the symbol read from tape is σ Є Σ, then δ(q, σ) Є Q.
Theory of Automata Chapter # 2
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38. A DFA M = (Q, Σ, δ, s, F), where Q = {q0, q1} Σ = {a, b} s = q0
F = {q0} q σ
δ(q,σ) q0 a b q1
Theory of Automata Chapter # 2
Dr. Abdus Salam

39. The Language of DFA
M passes from state q0 to q1 or from q1 back to q0 when a b is read, but M essentially ignores a’s, always remaining in its current state when an a is read. Thus M accepts a string if and only if the number of b’s is even.
Then L(M) is the set of all strings in {a,b}* that have an even number of b’s.
L(M) = { w | w Є { a, b }* and w contains even number of b’s }
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Theory of Automata Chapter # 2

40. Transition Diagram
A transition diagram for a DFA M = (Q, Σ, δ, s, F) is a graph defined as follows:
For each state in Q there is a node (vertex).
For each state q in Q and each input symbol a in Σ, let δ(q, a) = p. Then the transition diagram has an arc (edge) from node q to node p, labeled a.
There is an arrow into the start state q0, labeled start.
Nodes corresponding to accepting states (those in F) are marked by a double circle.
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Theory of Automata Chapter # 2 a b q0 q1 b

41. How a DFA Processes Strings?
Let the input string is aabba. δ(q0, a) = q0 δ(q0, b) = q1 δ(q1, b) = q0 Since q0 belong to F therefore the string is accepted.
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Theory of Automata Chapter # 2

42. Example-2
Let us design a DFA M that accepts the language L(M) = { w | w Є {0,1}* and w is of the form x01y for some strings x and y consisting of 0’s and 1’s only}
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To decide whether 01 is a substring of the input, M needs to remember:
1. Has it already seen 01? If so, then it accepts every sequence of further inputs.
2. Has it never seen 01, but its most recent input was 0, so if it now sees a 1, it will have seen 01.
3. Has it never seen 01, but its last input was either nonexistent or it last saw a 1?
Theory of Automata Chapter # 2

43. Example (continued)
These three conditions can each be represented by three states q0, q1, q2. The transition functions are:
δ(q0, 0) = q2
δ(q0, 1) = q0
δ(q2, 0) = q2
δ(q2, 1) = q1
δ(q1, 0) = q1
δ(q1, 1) = q1
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0,1 1 q0 q2 q1
Theory of Automata Chapter # 2

44. Transition Tables
A transition table is a conventional, tabular representation of a function like δ that takes two arguments and returns a value.
The rows of the table correspond to the states, and the columns correspond to the inputs. 1
 q0 q2 q0
*q1 q1
Transition table for the DFA
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45. Quiz No. 1
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Consider the finite state automaton M shown on the white board. - What are the states of M? - What are the input symbols of M? - What is the initial state of M? - What are the accepting states of M? - Find δ(q0, ). - Find the next state table for M. - Describe informally the language accepted by the DFA.
Theory of Automata Chapter # 2

46. Quiz No. 2 Design a DFA to accept the language:
L = { w | w Є {English language letters}* and w has “the” as a substring}
- What are the states of M?
- What are the input symbols of M?
- What is the initial state of M?
- What are the accepting states of M?
- Find δ(q1, brothers).
- Find the transition state table for M.
- Draw the transition diagram.
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Theory of Automata Chapter # 2

47. Extending the Transition Function to String
Design a DFA to accept the language:
L = { w | w Є {0, 1}* and w has an even number of 0’s and even number of 1’s}
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*->q0 q2 q1 q3 q0
Theory of Automata Chapter # 2

48. Extending the Transition Function to String
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Theory of Automata Chapter # 2 ^

49. Delta hat(q0, )=?
δ(q0, ε) = q0.
δ(q0, 1) = (δ(δ(q0, ε ), 1) = δ(q0,1)=q1.
δ(q0, 11) = (δ(δ(q0, 1), 1) = δ(q1,1)=q0.
δ(q0, 110) = (δ(δ(q0, 11), 0) = δ(q0,0)=q2.
δ(q0, 1101) = (δ(δ(q0, 110), 1) = δ(q2,1)=q3.
δ(q0, 11010) = (δ(δ(q0, 1101), 0) = δ(q3,0)=q1.
δ(q0, ) = (δ(δ(q0, 11010), 1) = δ(q1,1)=q0. ^
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Theory of Automata Chapter # 2 ^ ^ ^ ^

50. Assignment No. 1
Construct deterministic finite automata accepting each of the following languages.
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Theory of Automata Chapter # 2

51. Using theoretical computer simulators for Formal Languages and Automata Theory
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Multipurpose FLAT Software
DEM (Deus ex Machina)
JFLAP (Java Formal Languages & Automata Packages)
JCT (Java Computability Toolkit)
FSME (Finite State Machine Explorer)
Theory of Automata Chapter # 2

52. Nondeterministic Finite Automata
OUTLINE
Example of an NFA
Difference between DFA and NFA
Definition of nondeterministic finite automata
Transition diagram an transition table
Extended transition function
The language of an NFA
Proving that an NFA accepts a language
Equivalence of DFA and NFA
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Theory of Automata Chapter # 2

53. Example of an NFA
A “nondeterministic” finite automata (NFA) has the power to be in several states at once. Nondeterminism is the ability to change states in a way that is only partially determined by the current state and input symbol.
A transition diagram for NFA is shown below:
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0, 1
0, 1
Theory of Automata Chapter # 2 q0 q3 q4 1 1 q1 q2
0, 1

54. Example of an NFA
An input sequence a1,a2,….,an is accepted by a NFA if there exists a sequence of transitions corresponding to the input sequence, that leads from the initial state to some final state.
e.g is accepted by the NFA because there is a sequence of transition through the states q0, q0, q0, q3, q4, q4 labeled 0, 1, 0, 0, 1.
This NFA accepts all strings with either two consecutive 0’s or two consecutive 1’s.
DFA is a special case of NFA.
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Theory of Automata Chapter # 2

55. The states an NFA during the processing of an input sequence
Dr. Abdus Salam 1 1 q0 q0 q0 q0 q0 q0 1 1 q3 q1 q3 q3 q1
Theory of Automata Chapter # 2 q4 1 q4

56. Formal Definition of NFA
An NFA is represented like a DFA:
A = ( Q, Σ, δ, q0, F)
Where:
Q is a finite set of states.
Σ is a finite set of input symbols.
q0 , a member of Q, is the start state.
F, a subset of Q, is the set of final (or accepting) states.
δ, the transition function is a function that takes a state in Q and an input symbol in Σ as arguments and returns a subset of Q (δ:Qx Σ→2Q). Notice that the only difference between an NFA and DFA is in the type of value that δ returns: a set of states in the case of an NFA and a single state in the case of a DFA.
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Theory of Automata Chapter # 2

57. The Transition Table for an NFA
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Input
State 1 q0
{q0, q3 }
{q0, q1 } q1 Ø
{ q2 } q2 q3
{ q4 } q4
Theory of Automata Chapter # 2

58. The difference between DFA and NFA
The difference between DFA and NFA is in the type of δ.
For the NFA, δ is a function that takes a state and input symbol as arguments (like the DFA’s transition function), but returns a set of zero, one, or more states (rather than returning exactly one state, as the DFA must).
In DFA, for a given input string w and state q, there will be exactly one path labeled w starting at q. To determine if a string is accepted by DFA it suffices to check this one path.
For NFA there may be many paths labeled w, and all must be checked to see whether one or more terminate at a final state. The NFA proliferation of states of the transitions diagram with input is
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Theory of Automata Chapter # 2

59. The Extended Transition Function^
The previous figure suggest that δ(q0, 010) = {q0, q3}. Formally, we define δ for an NFA’s transition function δ by:
BASIS: δ(q, ε) = {q} That is, without reading any input symbols, we are only in the state we begin in.
INDUCTION: Suppose w is of the form w = xa, where a is the final symbol of w and x is the rest of w. Then
δ(q, w) = U δ(pi, a) ^
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Theory of Automata Chapter # 2 k ^
i =1

60. Example (Extended Transition Function)
Let us use δ to consider the processing of input by the NFA. A summary of the steps is:
δ (q0, ε) = {q0}
δ (q0, 0) = {q0, q3}
δ (q0, 01) = δ(δ(q0, 0), 1)) = δ({q0, q3},1) = δ(q0, 1) U δ(q3, 1) = {q0, q1} U Ø = {q0, q1}
δ (q0, 010) = δ(δ(q0, 01), 0)) = δ({q0, q1},0) = δ(q0, 0) U δ(q1, 0) = {q0, q3} U Ø = {q0, q3}
δ (q0, 0100) = δ(δ(q0, 010), 0)) = δ({q0, q3},0) = δ(q0, 0) U δ(q3, 0) = {q0, q3} U{q4} = {q0, q3, q4}
δ (q0, 01001) = δ(δ(q0, 0100), 1)) = δ({q0, q3, q},1) = δ(q0, 1) U δ(q3, 1) U δ(q4, 1) = {q0, q1} U Ø U {q4} = {q0, q3, q4}
As F = {q2, q4} thus {q0, q3, q4} ∩ {q2, q4} = {q4} string is accepted. ^
Dr. Abdus Salam ^ ^ ^ ^ ^ ^ ^
Theory of Automata Chapter # 2 ^ ^

61. The Language of an NFA
An NFA accepts a string w if it is possible to make any sequence of choices of next state, while reading the characters of w, and go from the start state to any accepting state.
The language of an NFA is the set of strings w such that δ(q0, w) contains at least one accepting state.
If A = (Q, Σ, δ, q0, F) is an NFA, then
L(A) = { w | δ(q0, w) ∩ F ≠ Ø }
i.e. L(A) is the set of strings w in Σ* such that δ(q0, w) contains at least one accepting state.
Dr. Abdus Salam ^
Theory of Automata Chapter # 2 ^ ^

62. An NFA Example Input State a b
Let M = (Q, Σ, δ, q0, F), where Q = {q0, q1, q2, q3}, Σ = {a, b}, F = {q3} and δ is given below:
Draw the transition diagram for M.
What is the language of M?
Find δ(q0,aaa)=? ^
Input
State a b q0
{q0, q1}
{q0} q1
{q2} q2
{q3} q3 Ø
Theory of Automata Chapter # 2
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63. Equivalence of DFA and NFA
Every language that can be described by some NFA can also be described by some DFA
We can prove this statement using subset construction
Subset construction involve constructing an automaton in terms of the states and transitions of another automaton
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Theory of Automata Chapter # 2

64. Subset Construction
Given an NFA N = (QN, Σ, δN, q0, FN) we will construct a DFA D = (QD, Σ, δD, {q0}, FD) such that L(D) = L(N)
Steps of the subset construction
QD = { S | S QN}
FD = {S QN} | S ∩ FN ≠ Ø }
For every S QN and a Є Σ,
δD(S, a) = U δN (p, a)
Theory of Automata Chapter # 2
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65. Example: Subset Construction
Convert the following NFA to a DFA
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Theory of Automata Chapter # 2

66. The Complete Subset Construction1 Ø
─> {q0}
{q0, q1}
{q0}
{q1}
{q2}
*{q2}
{q0,q1}
{q0,q2}
*{q0, q2}
*{q1, q2}
*{q0, q1, q2}
{q0, q2}
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Theory of Automata Chapter # 2

67. The DFA Constructed from NFA
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{q0}
{q0, q1}
{q0, q2} 1
Theory of Automata Chapter # 2 1

68. Assignment # 2 Write a program to convert an NFA to DFA.
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Theory of Automata Chapter # 2

69. Theorem 1
If D = (QD, Σ, δD, {q0}, FD) is the DFA constructed from NFA N = (QN, Σ, δN, {q0}, FN) by the subset construction, then L(D)=L(N).
PROOF: We prove by induction on |w|, is that
δD({q0}, w) = δN(q0, w)
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Theory of Automata Chapter # 2

70. The inductive part of the definition of δ for NFA’s tell us
BASIS: Let |w| =0; that is, w = ε by the basis definition of δ for DFA’s and NFA’s, both δD({q0}, ε) and δN(q0, ε) are {q0}.
INDUCTION: Let w be of length n + 1, and assume the statement for length n. Break w up as w = xa, where a is the final symbol of w. By the induction hypothesis, δD({q0}, x) = δN({q0}, x). Let both these sets of N’s states be {p1, p2,…,pk}.
The inductive part of the definition of δ for NFA’s tell us
δN(q0, w) = δ(q0, xa)
δN(q0, x) = {p1, p2,…,pk}
δN(q0, w) = UδN(pi, a) ………(1)
The subset construction, on the other hand, tells us that
δD({p1, p2,…,pk}, a) = UδN(pi, a) ……..(2)
Now, let us use (2) and the fact that δD({q0}, x) = {p1, p2,…,pk} in the inductive part of the definition of δ for DFA’s:
δD({q0}, w) = δD(δD({q0}, x), a) = δD({p1, p2,…,pk}, a = UδN(pi, a)
Thus the above equations demonstrate that δD({q0}, w) = δN(q0, w).
When we observe that D and N both accept w if and only if δD({q0}, w) or δN(q0, w), respectively, contain a state in FN, we have a complete proof that L(D) = L(N).
Dr. Abdus Salam ^ ^ ^ ^ ^
Theory of Automata Chapter # 2 ^ ^ ^ ^ ^ ^ ^

71. Convert to a DFA the following NFA1 →p
{p, q}
{p} q
{r} r
{s} Ø *s 1
─> {p}
{p, q}
{p}
{p, q, s}
{p, r}
{p, q, r, s}
*{p, q, s}
{p, r, s}
*{p, q, r, s}
*{p, r, s}
{p, s}
*{p, s}
Dr. Abdus Salam
Theory of Automata Chapter # 2

72. Which of the following strings are accepted by this nondeterministic finite automata?
Dr. Abdus Salam aa
aba
abb ab
abab a b a b e a b b
Theory of Automata Chapter # 2

73. Design an NFA to recognize the following set of stringsb c d
->q0
{q0,q1,q4,q7}
{q0} q1 {}
{q2} q2
{q3}
*q3 q4
{q5} q5
{q6}
*q6 q7
{q8} q8
{q9} q9
{q10}
*q10
Exercise 2.4.1
a) abc, abd, and aacd. Assume the alphabet is {a, b, c, d}
Theory of Automata Chapter # 2
Dr. Abdus Salam

74. Finite Automata with Epsilon-Transitions
The new “feature” is that a transition is allowed on ε, the empty string.
It does not expand the class of languages that can be accepted by finite automata.
It gives us some added “programming convenience”.
NFA’s with ε-transitions are called ε-NFA’s
Dr. Abdus Salam
Theory of Automata Chapter # 2

75. A ε-NFA accepting decimal numbers
An optional + or – sing,
A string of digits,
A decimal point
Another string of digits. Either this string, or the strings (2) can be empty.
Dr. Abdus Salam
Theory of Automata Chapter # 2

76. The Formal Notation for an ε-NFA
An ε-NFA is a quintuple (Q, Σ, δ, q0, F) where δ is a function from Q X Σ U {ε} to the power set of Q.
The ε-NFA from the previous slide is represented formally as
E = ({q0, q1,…,q5}, {.,+,-,0,1,2,…9}, δ, q0, {q5})
where the transition table for δ is
Dr. Abdus Salam
Theory of Automata Chapter # 2

77. Epsilon-Closures
We ε-close a state q by following all transitions out of q that are labeled ε.
When we get to other states by following ε, we follow the ε-transitions out of those states, and so on, eventually finding every state that can be reached from q along any path whose arcs are labeled ε.
Formal Definition
BASIS: State q is in ECLOSE(q).
INDUCTION: If state p is in ECLOSE(q), and there is transition from state p to state r labeled ε, then r is in ECLOSE(q).
Dr. Abdus Salam
Theory of Automata Chapter # 2

78. Ε-Close Example
For the automaton of previous slide, each state is its own ε-closure, with two exceptions: ECLOSE(q0) = {q0, q1} and ECLOSE(q3) = {q3, q5}.
Another Example:
Dr. Abdus Salam
Theory of Automata Chapter # 2
ECLOSE(1) = {1, 2, 3, 4, 6}

79. Extended Transitions and Languages for ε-NFA’s
BASIS:
δ(q, ε) = ECLOSE(q).
INDUCTION:
δ(q, xa) = U ECLOSE(p)
p Є δ(δ(q, x), a)
Example
Lets compute δ(q0, 5.6) for ε-NFA of previous slide on the board.
Dr. Abdus Salam
Theory of Automata Chapter # 2

80. Eliminating ε-Transitions
Given an ε-NFA
E = (QE, Σ, δE, q0, FE)
We will construct a DFA
D = (QD, Σ, δD, q0, FD)
Such that
L(D) = L(E)
Details of the construction:
QD = {S | S QE and S = ECLOSE(S)}
qD = ECLOSE(q0)
FD = { S | S Є QD and S ∩ FE ≠ Ø }
δD(S, a) =U{ ECLOSE(p) | p Є δ(t, a) for some t Є S}
Theory of Automata Chapter # 2
Dr. Abdus Salam

81. Eliminating ε-Transitions
Dr. Abdus Salam
Theory of Automata Chapter # 2
δ(q0, ε) = ECLOSE(q0) = {q0, q1}
δ({q0,q1}, +-) = ECLOSE(δ({q0, q1}, +-))

82. Dr. Abdus Salam
Theory of Automata Chapter # 2

83. Transducers Finite automata with output Moore Machine Mealy Machine
Dr. Abdus Salam
Theory of Automata Chapter # 2

84. Moore Machine
A Moore machine is like a finite automata except the following differences.
There are two alphabets (input alphabet & output alphabet)
No accept states
Not a language recognizer
It is an output producer
Each state produces a one-character output immediately upon the machines entry into that state.
Dr. Abdus Salam
Theory of Automata Chapter # 2

85. Moore Machine Formal definition
A Moore machine is a sixtuple M=(Q,Σ,∆,δ,Λ,q0)
Q is a finite set of states
Σ is an input alphabet
∆ is an output alphabet
δ transitions from one state to another
Λ is output table. Λ is a mapping from Q -> ∆ i.e. Λ(Q) = ∆
q0 is starting state
Dr. Abdus Salam
Theory of Automata Chapter # 2

86. Moore Machine Example Σ = {a, b} ∆ = {0, 1} Q={q0, q1, q2, q3}
Run the string abab through the machine which will produce the output
Input
States
Output a b q0 1 q1 q3 q2
q0/1 a
q1/0 b b a a b
q3/1 a
q2/0
Theory of Automata Chapter # 2 b
Dr. Abdus Salam

87. Exercise Counting machine
Construct a Moore machine that produces a 1 for each occurrence of the substring aab found in the input string.
Dr. Abdus Salam
Theory of Automata Chapter # 2

88. Mealy Machine
A Mealy machine produces output on a transition instead of on entry into a state.
Transitions are labeled i/o.
No accept states
Not a language recognizer, it is an output producer.
Its output will be the same length as its input.
Dr. Abdus Salam
Theory of Automata Chapter # 2

89. Mealy Machine Formal definition
A Mealy machine is a sixtuple M=(Q,Σ,∆,δ,Λ,q0)
Q is a finite set of states
Σ is an input alphabet
∆ is an output alphabet
δ transitions from one state to another
Λ is output table. Λ is a mapping from Q x Σ -> ∆ i.e. Λ(Q) = ∆
q0 is starting state
Dr. Abdus Salam
Theory of Automata Chapter # 2

90. Mealy Machine Example q0 q1/0 q3/0 q1 q3/1 q2/1 q2 q3 q0/1
Input
States a b q0
q1/0
q3/0 q1
q3/1
q2/1 q2 q3
q0/1
Σ = {a, b} ∆ = {0, 1} Q={q0, q1, q2, q3}
Run the string aaabb through the machine which will produce the output q0
a/0 q1
a/1
b/0
b/1
b/1
a/1 q2 q3
Theory of Automata Chapter # 2
b/1, a/0
Dr. Abdus Salam

91. Exercise
Construct a Mealy machine that takes the 1’s complement of its binary input.
Produce a Mealy machine that adds two integers using their binary expansions.
Construct a Mealy machine that accept its input in a binary number and prints out the binary number that is one larger.
Dr. Abdus Salam
Theory of Automata Chapter # 2

92. Theory of Automata Chapter 3-Regular Expressionsby
Dr. Abdus Salam

93. Regular Expressions
The set of strings accepted by a finite automaton is referred to as the language accepted by the finite automaton. We might describe a finite automaton as a language recognizer whereas a regular expression is a language generator.
For each finite automaton there is a regular expression that defines the same language.
Dr. Abdus Salam
Theory of Automata

94. Defining Languages L1 = {xn | n = 1, 2, 3, …}
L4 = {ε x xx xxx …} Let S = {x}. Then L4 = S*
We can also write L4 = {x}*
x* = ε x1 x2 x3… = xn
The language L4 can also be defined by any of the expressions below:
xx* x+ xx*x* x*xx* x+x* x*x+
Dr. Abdus Salam
Theory of Automata

95. Examples The Language defined by the expression ab*a
Is the set of all strings of a’s and b’s that have at least two letters, that begin and end with a’s and that have nothing but b’s inside (if any thing at all).
Language(ab*a)={aa aba abba abbba…}
Dr. Abdus Salam
Theory of Automata

96. More Examples
a*b*
Language(a*b*)={ε a b aa ab bb aaa aab…} Note a*b* ≠ (ab)*
L2 = { xodd}
can be defined as x(xx)* or (xx)*x
01* + 10*
Denotes the language consisting of all strings that are either a single 0 followed by any number of 1’s or a single 1 followed by any number of 0’s.
Dr. Abdus Salam
Theory of Automata

97. Operations on languages
Union
Concatenation
Closure (or star, Kleene closure)
Dr. Abdus Salam
Theory of Automata

98. Languages and Regular Expression
S.No.
Languages
Regular Expression 1
{ε} ε 2
{0} 3
{001} i.e. {0}{0}{1}
001 4
{0,1}
0+1 5
{1, ε}{001}
(1+ ε)001 6
{110}*{0,1}
(110)*(0+1) 7
{1}*{10}
1*10 8
{10, 111, 11010}
( )*
Theory of Automata
Dr. Abdus Salam

99. Formal Definition of RE
A RE over the alphabet Σ, and the corresponding language are defined as follows:
Every letter of Σ can be made into a regular expression; ε itself is a regular expression.
If r1 and r2 are regular expressions, then so are:
(r1)
r1r2
r1 + r2
r1*
Nothing else is a regular expression
Dr. Abdus Salam
Theory of Automata

100. Example Regex for: L = { w Є {0, 1}* | 0 and 1 alternate in w}
(01)* + (10)* + 0(10)* + 1(01)*
Or, equivalently,
(ε + 1)(01)*(ε + 0)
Order of precedence for operators:
Star
Dot
Plus
Example: 01* + 1 is grouped (0(1)*) + 1
Dr. Abdus Salam
Theory of Automata

101. Example-1 Write regular expression for the following languages:
The set of strings over alphabet {a, b, c} containing at least one a and at least one b. Ans: The simplest approach is to consider those strings in which the first a precedes the first b separately from those where the opposite occurs. The expression: c*a(a+c)*b(a+b+c)* + c*b(b+c)*a(a+b+c)*
Dr. Abdus Salam
Theory of Automata

102. Example-2
The language of all words that have at least two a’s can be described by the expression
(a + b)*a (a + b)*a (a + b)*
(some beginning) (the 1st important a) (some middle) (the 2nd important a) (some end)
Dr. Abdus Salam
Theory of Automata

103. Example-3
Even-Even = {ε aa bb aaaa aabb abab abba baab baba bbaa bbbb aaaaaa aaaabb aaabab}
E = [aa+bb+(ab+ba)(aa+bb)*(ab+ba)]*
type1 = aa
type2 = bb
type3 = (ab + ba)(aa + bb)*(ab + ba)
E = [type1 + type2 + type3]*
Dr. Abdus Salam
Theory of Automata

104. Equivalence of FA’s and regex’s
We have already shown that DFA’s, NFA’s, and ε-NFA all are equivalent.
To show FA’s equivalent to regex’s we need to establish that
For every DFA A we can find (construct, in this case) a regex R, such that L(R) = L(A).
For every regex R there is a ε-NFA A, such that L(A) = L(R).
Dr. Abdus Salam
Theory of Automata

105. Simplification Rules
We will be needing the following simplification rules:
(ε + R)* = R*
R + RS* = RS*
ØR = R Ø = Ø (Annihilation)
Ø + R = R + Ø = R (Identity)
Dr. Abdus Salam
Theory of Automata

106. Convert DFA to regex Rij = Rij + Rik (Rkk )*Rkj
L(A) = {x0y | x Є {1}* and y Є {0,1} }
Dr. Abdus Salam
Theory of Automata

107. Convert DFA to regex (con’t)
Dr. Abdus Salam
Theory of Automata

108. Convert DFA to regex (con’t)
Dr. Abdus Salam
Theory of Automata

109. Convert DFA to regex (con’t)
Dr. Abdus Salam
Theory of Automata

110. Observations There are n3 expressions for an n-state automaton
Each inductive step grows the expression 4-fold
R(n) could have size 4n
For all {i, j} < {1,….,n}, R(k) uses R(k-1)
So we have to write n2 times the regex R(k-1)
We need a more efficient approach:
The State Elimination Technique
Dr. Abdus Salam ij ij kk kk
Theory of Automata

111. The State Elimination Technique
When state S is eliminated, all the paths that went through s no longer exist in the automaton.
To not to change the language of automaton, add an arc from q to p.
How to label that arc? Use a Regular Expression.
The language of the automaton is the union over all paths from the start state to an accepting state of the language formed by concatenating the REs along the path.
Dr. Abdus Salam
Theory of Automata

112. The State Elimination Technique
What happens when we eliminate state s.
For each accepting state q eliminate from the original automaton all states except q0 and q.
To compensate, we introduce, for each predecessor qi of s and each successor pj of s, a RE that represents all the paths that start at qi and finally go to pj.
The expression for these paths is QiS*Pj.
Add this expression to the arc from qi to pj.
Dr. Abdus Salam
Theory of Automata

113. Constructing a RE from a FA
For each accepting state q, apply the previous reduction process to produce an equivalent automaton with RE labels on the arcs. Eliminate all states except q and the start state q0.
If q ≠ q0, then we shall be left with a two-state automaton that looks like;
The RE for the accepted strings can be described as (R+SU*T)*SU
Dr. Abdus Salam
Theory of Automata

114. The Strategy for Constructing a RE from a FA
If the start state is also an accepting state, then we are left with a one-state automaton that looks like;
The RE denoting the strings that it accepts is R*.
The desired RE is the sum (union) of all the expressions derived from the reduced automata for each accepting state, by rules (2) and (3).
Dr. Abdus Salam
Theory of Automata

115. Example 3.6
Dr. Abdus Salam
First step is to convert it to an automaton with regular expression labels.
Theory of Automata

116. Example (con’t) Lets eliminate state B.
State B has one predecessor, A, and one successor, C. Thus:
Q1 = 1, P1 = 0 + 1, R11 = Ø (Since the arc from A to C does not exist) and S = Ø (because there is no loop at state B).
The resultant expression is Ø + 1Ø*(0 + 1).
To simplify;
initial Ø may be ignored in a union.
L(Ø*) = {Є} U L(Ø) U L(Ø) …….
Thus Ø + 1Ø*(0 + 1) is equivalent to 1(0 + 1).
Dr. Abdus Salam
Theory of Automata

117. Example (con’t) Lets eliminate state C and obtain AD.
The mechanics is similar to those performed to eliminate state B and the resulting automaton is shown as follows:
Dr. Abdus Salam
Theory of Automata
The REs are R = 0 + 1, S = 1(0 + 1)(0 + 1), T = Ø, U = Ø.
The generic expression (R + SU*T)*SU* thus simplifies in this case to R*S, or (0 + 1)*1(0 + 1)(0 + 1).

118. Example (con’t) with regex (0 + 1)*1(0 + 1)
We can eliminate D to obtain AC.
Dr. Abdus Salam
with regex (0 + 1)*1(0 + 1)
The final expression is the sum of previous two regex’s:
(0 + 1)*1(0 + 1)(0 + 1) + (0 + 1)*1(0 + 1)
Theory of Automata

119. From regex’s to ε-NFA’s
Theorem 3.7: For every regex R we can construct an ε-NFA A, s.t. L(A) = L(R).
Proof: By structural induction:
Basis: Automata for ε, Ø, and a.
Dr. Abdus Salam
Theory of Automata

120. From regex’s to ε-NFA’s
Induction: Automata for R + S, RS, and R*.
Dr. Abdus Salam
Theory of Automata

121. Example 3.8 Let us convert the regular expression (0 + 1)*1(0 + 1)
Dr. Abdus Salam
Theory of Automata

122. Example 3.8 (cont.)
Dr. Abdus Salam
Theory of Automata

123. Application of Regular Expressions
Regular Expression in UNIX
Most real applications deal with the ASCII character set
UNIX regular expression allow us to write character classes to represent large sets of characters. The rules are:
The symbol . (dot) stands for “any character”
The sequence [a1,a2…ak] stands for the regular expression
a1 + a2 + … + ak
A range of the form x-y mean all the characters form x to y in the ASCII sequence. e.g. digits can be expressed [0-9]
Dr. Abdus Salam
Theory of Automata

124. Application of Regular Expressions
There are special notations for several of the most common classes of characters. e.g.
[:digit;] is the set of ten digits, the same as [0-9].
[:alpha:] stands for any alphabetic character, as does [A-Za-z]
[:alnum:] stands for the digits and letters, as does {A-Za-z0-9]
grep stands for “Global (search for) Regular Expression and Print”
Lexical Analysis
Finding Patterns in Text
Dr. Abdus Salam
Theory of Automata

125. Automata Theory Chapter 4 Properties of Regular Languagesby
Dr. Abdus Salam

126. Algebraic Laws for Languages
Commutativity
is the property of an operator that says we can switch the order of its operands and get the same result.
L + M = M + L. The commutative law of union, says that we may take the union of two languages in either order.
Associativity
is the property of an operator that allows us to regroup the operands when the operator is applied twice.
(L + M) + N = L + (M + N). The associative law for union, says that we may take the union of three languages either by taking the union of the first two initially, or taking the union of the last two initially.
Dr. Abdus Salam
Theory of Automata

127. Algebraic Laws for Languages
(LM)N = L(MN). The associative law for concatenation, says that we can concatenate three languages by concatenating either the first two or the last two initially.
Missing from this list is the “law” LM = ML, which would say that concatenation is commutative. However, this law is false.
Dr. Abdus Salam
Theory of Automata

128. Algebraic Laws for regexp’s
Identity
An identity for an operator is a value such that when the operator is applied to the identity and some other value, the result is the other value. e.g. 0 is the identity for addition.
Annihilator
An annihilator for an operator is a value such that when the operator is applied to the annihilator and some other value, the result is the annihilator. e.g. 0 is an annihilator for multiplication.
Dr. Abdus Salam
Theory of Automata

129. Algebraic Laws for regexp’s
There are three laws for regular expressions involving these concepts
Ø + L = L + Ø = L. This law asserts that Ø is the identity for union.
ЄL = LЄ = L. This law asserts that Є is the identity for concatenation.
ØL = LØ = Ø. This law asserts that Ø id the annihilator for concatenation.
Dr. Abdus Salam
Theory of Automata

130. Distributive Laws
A distributive law involves two operators, and asserts that one operator can be pushed down to be applied to each argument of the operator individually.
L(G(E + F)) =L(EG + FG). This law, is the left distributive law of concatenation over union.
e.g. L(( )11) = L( )
L((E + F)G) =L(EG + FG). This law, is the right distributive law of concatenation over union.
Dr. Abdus Salam
Theory of Automata

131. The Idempotent Law
An operator is said to be idempotent if the result of applying it to two of the same values as arguments is that value. e.g = 0.
L + L = L. This law, the idempotence law for union, states that if we take the union of two identical expressions, we can replace them by one copy of the expression
Dr. Abdus Salam
Theory of Automata

132. Laws Involving Closures
Dr. Abdus Salam
(L*)* = L*
Closing an expression that is already closed does not change the language.
Ø* = ε
The closure of Ø contains only the string ε.
ε* = ε
The only string that can be formed by concatenating any number of copies of the empty string is the empty string itself.
L+ = LL*=L*L
L+ is defined to be L+LL+LLL+….
L* = ε + L + LL + LLL + ….
Theory of Automata

133. Properties of Regular Languages
Dr. Abdus Salam
Pumping Lemma. Every regular language satisfies the pumping lemma. If somebody presents you with fake regular language, use a pumping lemma to show a contradiction.
Closure properties. Building automata from components through operations, e.g. given L and M we can build an automaton for L ∩ M.
Decision properties. Computational analysis of a automata, e.g. are two automata equivalent.
Minimization techniques. We can save money since we can build smaller machines.
Theory of Automata

134. Proving Languages not to be Regular
The class of languages known as the regular languages has four different descriptions. They are the languages accepted by:
DFA’s
NFA’s
ε-NFA’s RE
Not every language is a regular language.
Dr. Abdus Salam
Theory of Automata

135. Nonregular Languages Consider the language L = {0n1n | n ≥ 0}
If we attempt to find a DFA that recognizes L we discover that such a machine needs to remember how many 0s have been seen so far as it reads the input
Because the number of 0s isn’t limited, the machine needs to keep track of an unlimited number of possibilities
This cannot be done with any finite number of states
Dr. Abdus Salam
Theory of Automata

136. Intuition may fail us
Just because a language appears to require unbounded memory in order to recognized, it doesn’t mean that it is necessarily so
Example
C = {w |w has an equal number of 0s and 1s} is not regular
D = {w |w has an equal number of 01 and 10 as substrings} is regular
Dr. Abdus Salam
Theory of Automata

137. Language nonregularity
The technique for proving nonregularity of some languages is provided by a theorem about regular languages called pumping lemma
Pumping lemma states that all regular languages have a special property
If we can show that a language does not have this property we are guaranteed that it is not regular
Dr. Abdus Salam
Theory of Automata

138. Pumping Property
All strings in the language can be “pumped" if they are at least as long as a certain special value, called the pumping length
Meaning: each such string in the language contains a section that can be repeated any number of times with the resulting string remaining in the language.
Dr. Abdus Salam
Theory of Automata

139. The Substring Cycle Let us consider the NFA given below.
Dr. Abdus Salam
Theory of Automata

140. The Substring Cycle
This NFA accepts among others some strings of length greater than 5 such as abbabbb, abbabbabbb etc. Those strings which are accepted by this NFA and whose length is greater than 5 have a substring which can be repeated any number of times without being rejected by the NFA.
For example the string abbabbb is accepted by the NFA and if one of its substrings bba is repeated any number of times in abbabbb, the resultant strings such as abbb (bba repeated 0 times), abbabbabbb, abbabbabbabbb etc. are also accepted by the NFA. In general if a string w (such as abbabbb in the example above) is accepted by an NFA with n states and if its length is longer than n, then there must be a cycle in the NFA along some path from the initial state to some accepting state (such as the cycle in the above example). Then the substring representing that cycle (bba in the example) can be repeated any number of times within the string w without being rejected by the NFA.
Dr. Abdus Salam
Theory of Automata

141. The Pumping Lemma for RL
Let L be a regular language.
Then n, w Є L : |w| ≥ n => w = xyz such that:
y ≠ ε
|xy| ≤ n
k ≥ 0, xykz Є L
That is we can always find a nonempty string y not too far from the beginning of w that can be “pumped”; that is, repeating any number of times, or deleting it (the case k = 0), keeps the resulting string in the language L.
Theory of Automata
Dr. Abdus Salam

142. The Pumping Lemma for RL
Proof
Suppose L is regular. Then L = L(A) for some DFA A. Suppose A has n states.
Now, consider any string w of length n or more, say w = a1a2…am, where m ≥ n and each ai is an input symbol.
For i = 0, 1, . . ., n define state pi to be δ(q0, a1a2…ai), where δ is the transition function of A, and q0 is the start state of A. That is, pi is the state A is in after reading the first i symbols of w. Note that p0 = q0.
By the pigeonhole principle, it is not possible for the n+1 different pi’s for i = 0,1,…,n to be distinct, since there are only n different states. Thus we can find two different integers i and j, with 0 ≤ i ≤ j ≤ n, such that pi = pj. Now, we can bring w = xyz as follows:
Dr. Abdus Salam
Theory of Automata

143. The Pumping Lemma for RL
x = a1a2…ai
y = ai+1ai+2…aj
z = aj+1aj+2…am
That is, x takes us to pi once; y takes us from pi back to pi (since pi is also pj), and z is the balance of w. The relationships among the strings and states are suggested by the following figure. Note that x may be empty, in the case that i = 0. Also, z may be empty if j = n = m. However, y cannot be empty, since i is strictly less than j.
Dr. Abdus Salam
Theory of Automata

144. The Pumping Lemma for RL
Now, consider what happens if the automaton A receives the input xykz for any k ≥ 0. if k = 0, then the automaton goes from the start state q0 (which is also p0) to pi on input x. Since pi is also pj, it must be that A goes from pi to the accepting state shown in figure on input z. Thus, A accepts xz.
If k > 0, then A goes from q0 to pi input x, circles from pi to pi k times on input yk, and then goes to the accepting state on input z. Thus, for any k >= 0, zykz is also accepted by A; that is, xykz is in L.
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Theory of Automata

145. Example Leq = { 0n1n | n ≥ 0 }. We wish to show that L is not regular.
Suppose Leq is regular. Then there is a +ve integer n that satisfies the conditions of pumping lemma. We show that no matter what n is, we may find w, with |w| ≥ n, that produces a contradiction.
Consider a string w = 0n1n for that n.
Then there must be strings x, y, and z such that w=xyz
|xy| ≤ n, |y| ≥ n and xykz Є Leq for every k ≥ 0.
w =
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Theory of Automata
X y z

146. Since |y| > 0 , y has at least one symbol
Since |y| > 0 , y has at least one symbol. In string w = 0n1n, y can be inside 0n , inside 1n or straddle 0n and 1n . Hence there are the following three possibilities for y:         1. y = 0p, p > 0 ,         2. y = 1p, p > 0     or         3. y = 0p1q , p, q > 0 . Let us now consider the string xykz for k = 2 for each of these three cases. In case 1, xy2z = 0n-p02p1n = 0n+p1n . Since p > 0 , n + p ≠ n . Hence 0n+p1n can not be in the language L represented by 0n1n . Similarly in case 2, since xy2z = 0n 1n + p , it can not be in L. In case 3, xy2z = 0n-p(0p1q)21n-q = 0n-p(0p1q0p1q)1n-q = 1n 1q0p1n , which can not be in L. Thus in all three cases xy2z is not in L. This violates the condition that for every k >= 0, xykz Є L. Hence L is not a regular language.
Dr. Abdus Salam
Theory of Automata

147. Equivalence and Minimization of Automata
Let A = (Q, Σ, δ, q0, F) be DFA, and {p, q} Q. We define;
p ≡ q  Aw Є Σ* : δ(p, w) Є F iff δ(q, w) Є F
If p ≡ q we say that p and q are equivalent.
If p ≡ q we say that p and q are distinguishable.
In other words p and q are distinguishable iff
≡‌ w : δ(p, w) Є F and δ(q, w) Є F, or vice versa. / /
Theory of Automata
Dr. Abdus Salam

148. Example
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Theory of Automata
δ(C, ε) Є F, δ (G, ε) Є F => C ≡ G δ(A, 01) = C Є F, δ(G, 01) = E Є F => A ≡ G / / / /

149. What about A and E?
Dr. Abdus Salam / /
δ(A, ε) = A Є F, δ (E, ε) = E Є F δ(A, 1) = F = δ (E, 1) Therefore δ(A, 1x) = δ(E, 1x) = δ (F, x) δ(A, 00) = G = δ(E, 00) δ(A, 01) = C = δ(E, 01) Conclusion A≡ E.
Theory of Automata

150. Table Filling Algorithm
We can compute distinguishable pairs with the following inductive table filling algorithm:
Basis: If p Є F and q Є F, then p ≡ q.
Induction: If a Є Σ : δ(p, a) ≡ δ(q, a),
then p ≡ q.
Example: Applying the table filing algo to A: / / / /
Theory of Automata
Dr. Abdus Salam

151. Applying the Table Filling Algo.
Step-1
Step-2
Step-3
Dr. Abdus Salam B B B C x C x x C D x x D x D x x x E E x E x x x F F x F x x x x G G x G x x x x x x H H x H x x x x x x A B C D E F G A B C D E F G
Theory of Automata A B C D E F G
The three remaining pairs, which are therefore equivalent pairs, are {A, E}, {B, H} and {D, F}.
Thus, the table filling algorithm stops with the table shown in Step-3, which is the correct determination of equivalent and distinguishable states.

152. Testing Equivalence of Regular Languages
Let L and M be regular languages (each given in some form).
To test if L = M
Convert both L and M to DFA’s.
Imagine the DFA that is the union of the two DFA’s (never mind there are two start states).
If TF-Algo says that the two start states are distinguishable, then L ≠ M, otherwise L = M.
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Theory of Automata

153. Example
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Theory of Automata

154. Minimization of DFA’s We can minimize the DFA to obtain
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We can minimize the DFA
to obtain
Theory of Automata

155. The End
Thank You