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FORMAL LANGUAGES AND AUTOMATA THEORY
10CS56
FORMAL LANGUAGES AND AUTOMATA THEORY
Subject Code: 10CS56 Hours/Week : 04 Total Hours : 52 I.A. Marks : 25 Exam Hours: 03 Exam Marks: 100
P ART - A UNIT 1 7 Hours Introduction to Finite Automata: Introduction to Finite Automata; The central concepts of Automata theory; Deterministic finite automata; Nondeterministic finite automata UNIT 2 7 Hours Finite Automata, Regular Expressions: An application of finite automata; Finite automata with Epsilon-transitions; Regular expressions; Finite Automata and Regular Expressions; Applications of Regular Expressions UNIT 3 6 Hours Regular Languages, Properties of Regular Languages: Regular languages; Proving languages not to be regular languages; Closure properties of regular languages; Decision properties of regular languages; Equivalence and minimization of automata UNIT 4 6 Hours Context-Free Grammars And Languages : Context free grammars; Parse trees; Applications; Ambiguity in grammars and Languages . P ART B UNIT 5 7 Hours Pushdown Automata: Definition of the Pushdown automata; the languages of a PDA; Equivalence of PDA's and CFG's; Deterministic Pushdown Automata UNIT 6 6 Hours Properties of Context-Free Languages: Normal forms for CFGs; The pumping lemma for CFGs; Closure properties of CFLs UNIT 7 7 Hours Introduction To Turing Machine: Problems that Computers cannot solve; The turning machine; Programming techniques for Turning Machines; Extensions to the basic Turning Machines; Turing Machine and Computers. UNIT 8 6 Hours Undecidability: A that is not recursively enumerable; An Undecidable problem that is RE; Post's Correspondence problem; Other undecidable problems. Dept of ISE,SJBIT 1
FORMAL LANGUAGES AND AUTOMATA THEORY
10CS56
Text Books: 1. John E. Hopcroft, Rajeev Motwani, Jeffrey D.Ullman: Introduction to Automata Theory, Languages and Computation, 3rd Edition, Pearson Education, 2007. (Chapters: 1.1, 1.5, 2.2 to 2.5, 3.1 to 3.3, 4, 5, 6, 7, 8.1 to 8.4, 8.6, 9.1, 9.2, 9.4.1, 9.5) Reference Books: 1. K.L.P. Mishra: Theory of Computer Science, Automata, Languages, and Computation, 3rd Edition, PHI, 2007. 2. Raymond Greenlaw, H.James Hoover: Fundamentals of the Theory of Computation, Principles and Practice, Morgan Kaufmann, 1998.
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Table Of Contents
UNIT-1:INTRODUCTION TO FINITE AUTOMATA: 1.1: Introduction to finite Automata 1.2 : Central concepts of automata theory 1.3: Deterministic finite automata 1.4:Non deterministic finite auto mata UNIT-2:FINITE AUTOMATA, REGULAR EXPRESSIONS 2.1 An application of finite auto mata 2.2 Finite automata with Epsilon transitions 2.3 Regular expressions 2.4 Finite automata and regular expressions 2.5Applications of Regular expressions UNIT- 3: PROPERTIES OF REGULAR LANGUAGES 3.1 Regular languages 3.2 proving languages not to be regular languages 3.3 closure properties of regular languages 3.4 decision properties of regular languages 3.5 equivalence and minimization of automata UNIT-4:Context Free Grammar and languages 4.1 Context free grammars 4.2 parse trees 4.3 Applications 4.4 ambiguities in grammars and languages
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UNIT-5: PUSH DOWN AUTOMATA 5.1: Definition of the pushdown automata 5.2: The languages of a PDA 5.3: Equivalence of PDA and CFG 5.4: Deterministic pushdown automata Unit-6: PROPERTIES OF CONTEXT FREE LANGUAGES 6.1 Normal forms for CFGS 6.2The pumping lemma for CFGS 6.3closure properties of CFLS UNIT -7: INTRODUCTION TO TURING MACHINES 7.1 problems that computers cannot solve 7.2 The Turing machine 7.3 Programming techniques for turing machines 7.4 Extensions to the basic turing machines 7.5 Turing machines and computers Unit-8: Undesirability 8.1: A language that is not recursively enumerable 8.2: a un-decidable problem that is RE 8.3: Posts correspondence problem 8.4: Other undecidable problem
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74
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FORMAL LANGUAGES AND AUTOMATA THEORY
UNIT-1:INTRODUCTION TO FINITE AUTOMATA: 1.1: Introduction to finite Automata 1.2 : Central concepts of automata theory 1.3: Deterministic finite automata 1.4:Non deterministic finite automata
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INTRODUCTION TO FINITE AUTOMATA
1.1:introduction to finite automata
In this chapter we are going to study a class of machines called finite automata. Finite automata are computing devices that accept/recognize regular languages and are used to model operations of many systems we find in practice. Their operations can be simulated by a very simple computer program. A kind of systems finite automnata can model and a computer program to simulate their operations are discussed.
Formal definition
Automaton An automaton is represented formally by a 5-tuple (Q,,,q0,F), where: Q is a finite set of states. is a finite set of symbols, called the alphabet of the automaton. is the transition function, that is, : Q × Q. q0 is the start state, that is, the state of the automaton before any input has been processed, where q0 Q. F is a set of states of Q (i.e. F Q) called accept states.
Input word An automaton reads a finite string of symbols a1,a2,...., an , where ai , which is called an input word. The set of all words is denoted by *. Run A run of the automaton on an input word w = a1,a2,...., an *, is a sequence of states q0,q1,q2,...., qn, where qi Q such that q0 is the start state and qi = (qi-1,ai) for 0 < i n. In words, at first the automaton is at the start state q0, and then the automaton reads symbols of the input word in sequence. When the automaton reads symbol ai it jumps to state qi = (qi-1,ai). qn is said to be the final state of the run. Accepting word A word w * is accepted by the automaton if qn F. Recognized language An automaton can recognize a formal language. The language L * recognized by an automaton is the set of all the words that are accepted by the automaton. Recognizable languages The recognizable languages are the set of languages that are recognized by some automaton. For the above definition of automata the recognizable languages are regular languages. For different definitions of automata, the recognizable languages are different.
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1.2:concepts of automata theory
Automata theory is a subject matter that studies properties of various types of automata. For example, the following questions are studied about a given type of automata. Which class of formal languages is recognizable by some type of automata? (Recognizable languages) Are certain automata closed under union, intersection, or complementation of formal languages? (Closure properties) How much is a type of automata expressive in terms of recognizing class of formal languages? And, their relative expressive power? (Language Hierarchy) Automata theory also studies if there exist any effective algorithm or not to solve problems similar to the following list. Does an automaton accept any input word? (emptiness checking) Is it possible to transform a given non-deterministic automaton into deterministic automaton without changing the recognizable language? (Determinization) For a given formal language, what is the smallest automaton that recognizes it? (Minimization).
Classes of automata
The following is an incomplete list of types of automata. Automata Deterministic finite automata(DFA) Nondeterministic finite automata(NFA) Nondeterministic finite automata with -transitions (FND- or -NFA) Pushdown automata (PDA) Linear bounded automata (LBA) Turing machines Timed automata Deterministic Büchi automata Nondeterministic Büchi automata Nondeterministic/Deterministic Rabin automata Nondeterministic/Deterministic Streett automata Nondeterministic/Deterministic parity automata Nondeterministic/Deterministic Muller automata .1.3:Deterministic finite automata
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Recognizable language regular languages regular languages regular languages context-free languages context-sensitive language recursively enumerable languages -limit languages -regular languages -regular languages -regular languages -regular languages -regular languages
FORMAL LANGUAGES AND AUTOMATA THEORY
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.
Definition: A DFA is 5-tuple or quintuple M = (Q, , , q0, A) where Q is non-empty, finite set of states. is non-empty, finite set of input alphabets. is transition function, which is a mapping from Q x q0 A Q is the start state. Q is set of accepting or final states. to Q.
Note: For each input symbol a, from a given state there is exactly one transition (there can be no transitions from a state also) and we are sure (or can determine) to which state the machine enters. So, the machine is called Deterministic machine. Since it has finite number of states the machine is called Deterministic finite machine or Deterministic Finite Automaton or Finite State Machine (FSM). The language accepted by DFA is L(M) = { w | w * and *(q0, w) A} The non-acceptance of the string w by an FA or DFA can be defined in formal notation as: L(M) = { w | w * and *(q0, w) A}
Obtain a DFA to accept strings of a's and b's starting with the string ab
q b0 q
a a
q
1
b
a,b q
2
3 a,b
Fig.1.1 Transition diagram to accept string ab(a+b)* So, the DFA which accepts strings of a's and b's starting with the string ab is given by M = (Q, , , q0, A) where Q = {q0, q1, q2, q3} = {a, b} q0 is the start state A = {q2}. is shown the transition table 2.4. a q0 States q1 q2 q3
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q1 q3 q3 q2 q2 q2 q3 q3
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Draw a DFA to accept string of 0's and 1's ending with the string 011.
1 q0
0
0 q1
1 0 1
q2 0
1
q3
Obtain a DFA to accept strings of a's and b's having a sub string aa
b q0
a b
q1
a
a,b q2
Obtain a DFA to accept strings of a's and b's except those containing the substring aab.
b q0
a b
q1
a
a q2
b
a,b q3
Obtain DFAs to accept strings of a's and b's having exactly one a, b q0 b q1 b q0 b q0 b q1 b q2 a,b q2 a, b q1 b q3 a, b a q4
a
a
a
a
a
a
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Obtain a DFA to accept strings of a's and b's having even number of a's and b's
The machine to accept even number of a's and b's is shown in fig.2.22. q b
0
a b a a
a
q b
1
b
q
2
q
3
Fig.2.22 DFA to accept even no. of a's and b's a q0 b q2 q0 b q2 b b a a aa a a a a q0 b q2 b a a a b q3 q1 b b q3 b q3 q1 b q1 b
Regular language Definition: Let M = (Q, , , q0, A) be a DFA. The language L is regular if there exists a machine M such that L = L(M).
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* Applications of Finite Automata * String matching/processing Compiler Construction The various compilers such as C/C++, Pascal, Fortran or any other compiler is designed using the finite automata. The DFAs are extensively used in the building the various phases of compiler such as Lexical analysis (To identify the tokens, identifiers, to strip of the comments etc.) Syntax analysis (To check the syntax of each statement or control statement used in the program) Code optimization (To remove the un wanted code) Code generation (To generate the machine code) Other applications- The concept of finite automata is used in wide applications. It is not possible to list all the applications as there are infinite number of applications. This section lists some applications: 1. Large natural vocabularies can be described using finite automaton which includes the applications such as spelling checkers and advisers, multi-language dictionaries, to indent the documents, in calculators to evaluate complex expressions based on the priority of an operator etc. to name a few. Any editor that we use uses finite automaton for implementation. 2. Finite automaton is very useful in recognizing difficult problems i.e., sometimes it is very essential to solve an un-decidable problem. Even though there is no general solution exists for the specified problem, using theory of computation, we can find the approximate solutions. 3. Finite automaton is very useful in hardware design such as circuit verification, in design of the hardware board (mother board or any other hardware unit), automatic traffic signals, radio controlled toys, elevators, automatic sensors, remote sensing or controller etc. In game theory and games wherein we use some control characters to fight against a monster, economics, computer graphics, linguistics etc., finite automaton plays a very important role
1.4 : Non deterministic finite automata(NFA)
Definition: An NFA is a 5-tuple or quintuple M = (Q, , , q0, A) where Q is non empty, finite set of states.
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is non empty, finite set of input alphabets. is transition function which is a mapping from Q x { U } to subsets of 2Q. This function shows the change of state from one state to a set of states based on the input symbol. q0 A Q is the start state. Q is set of final states.
Acceptance of language Definition: Let M = (Q, , , q0, A) be a DFA where Q is set of finite states, is set of input alphabets (from which a string can be formed), is transition function from Q x { U } to 2Q, q0 is the start state and A is the final or accepting state. The string (also called language) w accepted by an NFA can be defined in formal notation as: L(M) = { w | w *and *(q0, w) = Q with atleast one Component of Q in A}
Obtain an NFA to accept the following language L = {w | w The machine to accept either ababn or aban where n ababn or aban where n 0}
0 is shown below:
q1 q0 q5
a
q2
b
q3 a q7
a
b
q4
a
q6
b
Conversion from NFA to DFA Let MN = (QN, N, N, q0, AN) be an NFA and accepts the language L(MN). There should be an equivalent DFA MD = (QD, D, D, q0, AD) such that L(MD) = L(MN). The procedure to convert an NFA to its equivalent DFA is shown below:
Step1: The start state of NFA MN is the start state of DFA MD. So, add q0(which is the start state of NFA) to QD and find the transitions from this state. The way to obtain different transitions is shown in step2. Step2: For each state [qi, qj,....qk] in QD, the transitions for each input symbol in be obtained as shown below: 1.
D([qi,
can
qj,....qk], a) =
N(qi,
a) U
N(qj,
a) U ...... N(qk, a)
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= [ql, qm,....qn] say. 2. Add the state [ql, qm,....qn] to QD, if it is not already in QD. 3. Add the transition from [qi, qj,....qk] to [ql, qm,....qn] on the input symbol a iff the state [ql, qm,....qn] is added to QD in the previous step. Step3: The state [qa, qb,....qc] QD is the final state, if at least one of the state in qa, qb, ..... qc AN i.e., at least one of the component in [qa, qb,....qc] should be the final state of NFA. Step4: If epsilon ( ) is accepted by NFA, then start state q0 of DFA is made the final state.
Convert the following NFA into an equivalent DFA.
0 q0
0,1 q 0, 1 q 1 2
1
Step1: q0 is the start of DFA (see step1 in the conversion procedure). So, QD = {[q0]} (2.7)
Step2: Find the new states from each state in QD and obtain the corresponding transitions. Consider the state [q0]: When a = 0 D([q0], 0) = N([q0], 0) = [q0, q1] (2.8) = N([q0], 1) = [q1] (2.9)
When a = 1 D([q0], 1)
Since the states obtained in (2.8) and (2.9) are not in QD(2.7), add these two states to QD so that QD = {[q0], [q0, q1], [q1] } (2.10)
The corresponding transitions on a = 0 and a = 1 are shown below.
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[q0] Q [q0, q1] [q1] Consider the state [q0, q1]:
0 [q0, q1]
[q1]
1
When a = 0 D([q0, q1], = N([q0, q1], 0) = N(q0, 0) U N(q1, 0) 0) = {q0, q1} U {q2} = [q0, q1, q2] (2.11) When a = 1 D([q0, q1], = N([q0, q1], 1) = N(q0, 1) U N(q1, 1) 1) = {q1} U {q2} = [q1, q2] D (2.12) F Since the states obtained in (2.11) and (2.12) are the not defined in QD(see 2.10), add A these two states to QD so that QD = {[q0], [q0, q1], [q1], [q0, q1, q2], [q1, q2] } and add the transitions on a = 0 and a = 1 as shown below: 0 [q0, q1] [q0, q1, q2] 1 [q1] [q1, q2] (2.13)
Q
[q0] [q0, q1] [q1] [ q0 , q1, q2 ] [q1, q2]
Consider the D ate [q1]: st
F When a = 0 A D([q1], 0)
When a = 1 D F A Dept of ISE,SJBIT
D([q1],
= N([q1], 0) = [q2] D (2.14) F A = =
N([q1],
1)
1)
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[q2] (2.15) Since the states obtained in (2.14) and (2.15) are same and the state q2 is not in QD(see 2.13), add the state q2 to QD so that QD = {[q0], [q0, q1], [q1], [q0, q1, q2], [q1, q2], [q2]} (2.16) and add the transitions on a = 0 and a = 1 as shown below: 0 [q0] [q0, q1] [q0, q1] [q0, q1, q2] [q1] [q2] [ q0 , q1, q2 ] [q1, q2] [q2] 1 [q1] [q1, q2] [q2]
Q
Consider the state [q0,q1,q2]: When a = 0 0)
D([q0,q1,q2],
= N([q0,q1,q2], 0) = N(q0, 0DU N(q1, 0) U ) = {q0,q1} U {q2} U { } F = [q0,q1,q2] A (2.17)
N(q2,
0)
D F WhenA = 1 a ([q0,q1,q2], D 1)
= N([q0,q1,q2], 1) = N(q0, 1) U N(q1, 1) U = {q1} U {q2} U {q2} = [q1, q2] (2.18)
N(q2,
1)
Since the states obtained in (2.17) and (2.18) are not new states (are already in QD, see 2.16), do not add these two states to QD. But, the transitions on a = 0 and a = 1 should be added to the transitional table as shown below: 0 1 D [q0, q1] [q1] F [q0, q1, q2] [q1, q2] A
15
[q0] [q0, q1]
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Q
[q1] [q2] [ q0 , q1, [q0,q1,q2] q2 ] [q1, q2] [q2]
[q2] [q1, q2]
Consider the state [q1,q2]: When a = 0 D([q1,q2], 0) = N([q1,q2], 0) = N(q1, 0) U N(q2, 0) = {q2} U { } = [q2] (2.19) = N([q1,q2], 1) = N(q1, 1) U N(q2, 1) = {q2} U {q2} = [q2] (2.20)
When a = 1 D([q1,q2], 1) D F A
Since the states obtained in (2.19) and (2.20) are not new states (are already in QD see 2.16), do not add these two states to QD. But, the transitions on a = 0 and a = 1 should be added to the transitional table as shown below: 0 [q0, q1] [q0, q1, q2] [q2] [q0,q1,q2] [q2] 1 [q1] [q1, q2] [q2] [q1, q2] [q2]
Q
[q0] [q0, q1] [q1] [ q0 , q1, q2 ] [q1, q2] [q2]
D Consider the state [q2]: F A When a = 0 D([q2], 0) When a = 1 D([q2], 1) D Dept of ISE,SJBIT F A
= N([q2], 0) = { } (2.21) D F = N([q2], A) 1 = [q2]
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(2.22) Since the states obtained in (2.21) and (2.22) are not new states (are already in QD, see 2.16), do not add these two states to QD. But, the transitions on a = 0 and a = 1 should be added to the transitional table. The final transitional table is shown in table 2.14. and final DFA is shown in figure 2.35. [q0] [q0,q1] [q1] [q0,q1,q2] [q1,q2] [q2] 0 [q0, q1] [q0, q1, q2] [q2] [q0,q1,q2] [q2] 1 [q1] [q1, q2] [q2] [q1, q2] [q2] [q2]
[q 0 ]
0
1
[q 0 , q 1 ]
[q 1 ]
0 1
1 0, 1
0, 1
[q 0 , q 1 , q 2 ]
[q 1 , q 2 ]
[q 2 ]
0
1
Fig.2.35 The DFA
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Convert the following NFA to its equivalent DFA.
4 0 a 1 b 2 3 6
a
5 8 9
b
7
Let QD = {0} Consider the state [A]: When input is a: (A, a) When input is b: ( A, b) Consider the state [B]: When input is a: (B, a) When input is b: ( B, b) = N(0, a) = {1} (B) = N(0, b) = { }
(A)
= N(1, a) = { } = N(1, b) = {2} = {2,3,4,6,9}
(C)
This is because, in state 2, due to -transitions (or without giving any input) there can be transition to states 3,4,6,9 also. So, all these states are reachable from state 2. Therefore, (B, b) = {2,3,4,6,9} = C Consider the state [C]: When input is a: (C, a) = N({2,3,4,6,9}, a) = {5} = {5, 8, 9, 3, 4, 6} = {3, 4, 5, 6, 8, 9} order) (D)
(ascending
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This is because, in state 5 due to -transitions, the states reachable are {8, 9, 3, 4, 6}. Therefore, (C, a) = {3, 4, 5, 6, 8, 9} = D When input is b: = N({2, 3, 4, 6, 9}, b) ( C, b) = {7} = {7, 8, 9, 3, 4, 6} = {3, 4, 6, 7, 8, 9}(ascending order) (E) This is because, from state 7 the states that are reachable without any input (i.e., -transition) are {8, 9, 3, 4, 6}. Therefore, (C, b) = {3, 4, 6, 7, 8, 9} = E Consider the state [D]: When input is a: (D, a)
= N({3,4,5,6,8,9}, a) = {5} = {5, 8, 9, 3, 4, 6} = {3, 4, 5, 6, 8, 9} order) (D) When input is b: = N({3,4,5,6,8,9}, b) = {7} = {7, 8, 9, 3, 4, 6} = {3, 4, 6, 7, 8, 9} order) (E)
(ascending
(D, b)
(ascending
Consider the state [E]: When input is a: (E, a)
When input is b: (E, b)
= N({3,4,6,7,8,9}, a) = {5} = {5, 8, 9, 3, 4, 6} = {3, 4, 5, 6, 8, 9}(ascending order) (D)
= N({3,4,6,7,8,9}, b) = {7} = {7, 8, 9, 3, 4, 6} = {3, 4, 6, 7, 8, 9}(ascending order) (E) Since there are no new states, we can stop at this point and the transition table for the DFA is shown in table 2.15.
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Q
A B C D E
a B D D D
b C E E E
Table 2.15 Transitional table The states C,D and E are final states, since 9 (final state of NFA) is present in C, D and E. The final transition diagram of DFA is shown in figure 2.36 A a B b C a a D
D F A
aD b F b E A b
Fig. 2.36 The DFA
D F A D F A
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Unit 1:Assignment questions:
1. Obtain a DFA to accept strings of a's and b's starting with the string ab 2. Draw a DFA to accept string of 0's and 1's ending with the string 011. 3. Obtain a DFA to accept strings of a's and b's having a sub string aa
4. Obtain a DFA to accept strings of a's and b's except those containing the substring aab. 5. Obtain DFAs to accept strings of a's and b's having exactly one a,
6. Obtain a DFA to accept strings of a's and b's having even number of a's and b's 7. Give Applications of Finite Automata * 8. Define DFA, NFA & Language? n m n m 9. (i) Write Regular expression for the following L = { a b : m, n are even} L = { a , b (ii) Write DFA to accept strings of 0's, 1's & 2's beginning with a 0 followed by odd number of 1's and ending with a 2. 10. Design a DFA to accept string of 0's & 1's when interpreted as binary numbers would be multiple of 3. 11. Find closure of each state and give the set of all strings of length 3 or less accepted by automaton.
: m>=2, n>=2}
p q *r {p,q}
12. Convert above automaton to a DFA 13. Write a note on Application of automaton.
a { r} {q} {p} { r}
b {p,r} {p}
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UNIT-2:FINITE AUTOMATA, REGULAR EXPRESSIONS 2.1 An application of finite automata
2.2 Finite automata with Epsilon transitions 2.3 Regular expressions 2.4 Finite automata and regular expressions 2.5 Applications of Regular expressions
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2.1 An application of finite automata
Applications of finite automata includes String matching algorithms, network protocols and lexical analyzers String Processing Consider finding all occurrences of a short string (pattern string) within a Long string (text string).This can be done by processing the text through a DFA: the DFA for all strings that end with the pattern string. Each time the accept state is reached, the current position in the text is output Example: Finding 1001 To find all occurrences of pattern 1001, construct the DFA for all strings ending in 1001.
Finite-State Machines A finite-state machine is an FA together with actions on the arcs. A trivial example for a communication link
:
Example FSM: Bot Behavior
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A bot is a computer-generated character in a video game
.
State charts
State charts model tasks as a set of states and actions. They extend FA diagrams Here is a simplified state chart for a stopwatch
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Lexical Analysis In compiling a program, the first step is lexi-cal analysis. This isolates keywords,identifiersetc., while eliminating irrelevant symbols.A token is a category, for example "identifier","relation operator" or specific keyword. For example, token RE keyword then then variable name [a-zA-Z][a-zA-Z0-9]* where latter RE says it is any string of alphanumeric characters starting with a letter. A lexical analyzer takes source code as a string,and outputs sequence of tokens. For example, for i = 1 to max do x[i] = 0; might have token sequence for id = num to id do id [ id ] = num sep As a token is identified, there may be an action. For example, when a number is identified, itsvalue is calculated 2.2 Finite automata with Epsilon transitions
We can extend an NFA by introducing a "feature" that allows us to make a transition on , the empty string. All the transition lets us do is spontaneously make a transition, without receiving an input symbol. This is another mechanism that allows our NFA to be
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in multiple states at once. Whenever we take an edge, we must fork off a new "thread" for the NFA starting in the destination state. Just as nondeterminism made NFA's more convenient to represent some problems than DFA's but were not more powerful, the same applies to NFA's. While more expressive, anything we can represent with an NFA we can represent with a DFA that has no transitions.
Epsilon Closure Epsilon Closure of a state is simply the set of all states we can reach by following the transition function from the given state that are labeled . Generally speaking, a collection of objects is closed under some operation if applying that operation to members of the collection returns an object still in the collection. In the above example: (q) = { q } (r) = { r, s} let us define the extended transition function for an NFA. For a regular, NFA we said for the induction step: Let ^(q,w) = {p1, p2, ... pk} (pi,a) = Sifor i=1,2,...k Then ^(q, wa) = S1,S2... Sk For an -NFA, we change for ^(q, wa): Union[ (Each state in S1, S2, ... Sk)] This includes the original set S1,S2... Sk as well as any states we can reach via . When coupled with the basis that ^(q, ) = (q) lets us inductively define an extended transition function for a NFA.
Eliminating Transitions Transitions are a convenience in some cases, but do not increase the power of the NFA. To eliminate them we can convert a NFA into an equivalent DFA, which is quite similar to the steps we took for converting a normal NFA to a DFA, except we must now follow all Transitions and add those to our set of states. 1. Compute for the current state, resulting in a set of states S. 2. (S,a) is computed for all a in by a. Let S = {p1, p2, ... pk} b. Compute I=1k (pi,a) and call this set {r1, r2, r3... rm}. This set is achieved by following input a, not by following any transitions
c. Add the transitions in by computing (S,a)= I=1 m (r1) 3. Make a state an accepting state if it includes any final states in the -NFA.
Note :The (epsilon) transition refers to a transition from one state to another without the reading of an input symbol (ie without the tape containing the input string moving). Epsilon transitions can be inserted between
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any states. There is also a conversion algorithm from a NFA with epsilon transitions to a NFA without epsilon transitions. a b C q0 {q0} {q1} q1 {q2} {q2} Consider the NFA-epsilon move machine M = { Q, , , q0, F} q2 {q2} Q = { q0, q1, q2 } = { a, b, c } and moves q0 = q0 F = { q2 }
Note: add an arc from qz to qz labeled "c" to figure above. The language accepted by the above NFA with epsilon moves is the set of strings over {a,b,c} including the null string and all strings with any number of a's followed by any number of b's followed by any number of c's. Now convert the NFA with epsilon moves to a NFA M = ( Q', , ', q0', F') First determine the states of the new machine, Q' = the epsilon closure of the states in the NFA with epsilon moves. There will be the same number of states but the names can be constructed by writing the state name as the set of states in the epsilon closure. The epsilon closure is the initial state and all states that can be reached by one or more epsilon moves. Thus q0 in the NFA-epsilon becomes {q0,q1,q2} because the machine can move from q0 to q1 by an epsilon move, then check q1 and find that it can move from q1 to q2 by an epsilon move. q1 in the NFA-epsilon becomes {q1,q2} because the machine can move from q1 to q2 by an epsilon move.
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q2 in the NFA-epsilon becomes {q2} just to keep the notation the same. q2 can go nowhere except q2, that is what phi means, on an epsilon move. We do not show the epsilon transition of a state to itself here, but, beware, we will take into account the state to itself epsilon transition when converting NFA's to regular expressions. The initial state of our new machine is {q0,q1,q2} the epsilon closure of q0 The final state(s) of our new machine is the new state(s) that contain a state symbol that was a final state in the original machine. The new machine accepts the same language as the old machine, thus same sigma. So far we have for out new NFA Q' = { {q0,q1,q2}, {q1,q2}, {q2} } or renamed { qx, qy, qz } = { a, b, c } F' = { {q0,q1,q2}, {q1,q2}, {q2} } or renamed { qx, qy, qz } q0 = {q0,q1,q2} or renamed qx inputs qx or{q0,q1,q2} qy or{q1,q2} qz or{q2} a b c
Now we fill in the transitions. Remember that a NFA has transition entries that are sets. Further, the names in the transition entry sets must be only the state names from Q'. Very carefully consider each old machine transitions in the first row. You can ignore any entries and ignore the column. In the old machine (q0,a)=q0 thus in the new machine '({q0,q1,q2},a)={q0,q1,q2} this is just because the new machine accepts the same language as the old machine and must at least have the the same transitions for the new state names. inputs a b c qx or{q0,q1,q2} {qx} or{{q0,q1,q2}} qy or{q1,q2} qz or{q2} No more entries go under input a in the first row because old (q1,a)= , (q2,a)= Now consider the input b in the first row, (q0,b)= , (q1,b)={q2} and (q2,b)= . The reason we considered q0, q1 and q2 in the old
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machine was because out new state has symbols q0, q1 and q2 in the new state name from the epsilon closure. Since q1 is in {q0,q1,q2} and (q1,b)=q1 then '({q0,q1,q2},b)={q1,q2}. WHY {q1,q2} ?, because {q1,q2} is the new machines name for the old machines name q1. Just compare the zeroth column of to '. So we have inputs a b c qx or{q0,q1,q2} {qx} or{{q0,q1,q2}} {qy} or{{q1,q2}} qy or{q1,q2} qz or{q2} Now, because our new qx state has a symbol q2 in its name and (q2,c)=q2 is in the old machine, the new name for the old q2, which is qz or {q2} is put into the input c transition in row 1. Inputs a b c qx or{q0,q1,q2} {qx} or{{q0,q1,q2}} {qy} or{{q1,q2}} {qz} or{{q2}} qy or{q1,q2} qz or{q2} Now, tediously, move on to row two, ... . You are considering all transitions in the old machine, delta, for all old machine state symbols in the name of the new machines states. Fine the old machine state that results from an input and translate the old machine state to the corresponding new machine state name and put the new machine state name in the set in delta'. Below are the "long new state names" and the renamed state names in delta'. Inputs a b qx or{q0,q1,q2} {qx} or{{q0,q1,q2}} {qy} or{{q1,q2}} qy or{q1,q2} {qy} or{{q1,q2}} qz or{q2} inputs a b c qx {qx} {qy} {qz} qy {qy} {qz} qz {qz} c {qz} or{{q2}} {qz} or{{q2}} {qz} or{{q2}}
\ /
\ Q /
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The figure above labeled NFA shows this state transition table. It seems rather trivial to add the column for epsilon transitions, but we will make good use of this in converting regular expressions to machines. regular-expression -> NFA-epsilon -> NFA -> DFA. 2.3 :Regular expression Definition: A regular expression is recursively defined as follows. 1. 2. is a regular expression denoting an empty language. -(epsilon) is a regular expression indicates the language containing an empty string. 3. a is a regular expression which indicates the language containing only {a} 4. If R is a regular expression denoting the language LR and S is a regular expression denoting the language LS, then a. R+S is a regular expression corresponding to the language LRULS. b. R.S is a regular expression corresponding to the language LR.LS.. c. R* is a regular expression corresponding to the language LR*. 5. The expressions obtained by applying any of the rules from 1-4 are regular expressions.
The table 3.1 shows some examples of regular expressions and the language corresponding to these regular expressions.
Regular expressions (a+b)* (a+b)*abb ab(a+b)* (a+b)*aa(a+b) * a*b*c*
Meaning Set of strings of a's and b's of any length including the NULL string. Set of strings of a's and b's ending with the string abb Set of strings of a's and b's starting with the string ab. Set of strings of a's and b's having a sub string aa. Set of string consisting of any number of a's(may be empty string also) followed by any number of b's(may include empty string) followed by any number of c's(may include empty string). Set of string consisting of at least one `a' followed by string consisting of at least one `b' followed by string consisting of at least one `c'. Set of string consisting of at least one `a'
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a+b+c+ aa*bb*cc*
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followed by string consisting of at least one `b' followed by string consisting of at least one `c'. (a+b)* (a + Set of strings of a's and b's ending with either a bb) or bb (aa)*(bb)*b Set of strings consisting of even number of a's followed by odd number of b's (0+1)*000 Set of strings of 0's and 1's ending with three consecutive zeros(or ending with 000) (11)* Set consisting of even number of 1's Table 3.1 Meaning of regular expressions
Obtain a regular expression to accept a language consisting of strings of a's and b's of even length.
String of a's and b's of even length can be obtained by the combination of the strings aa, ab, ba and bb. The language may even consist of an empty string denoted by . So, the regular expression can be of the form (aa + ab + ba + bb)* The * closure includes the empty string. Note: This regular expression can also be represented using set notation as L(R) = {(aa + ab + ba + bb)n | n 0}
Obtain a regular expression to accept a language consisting of strings of a's and b's of odd length.
String of a's and b's of odd length can be obtained by the combination of the strings aa, ab, ba and bb followed by either a or b. So, the regular expression can be of the form (aa + ab + ba + bb)* (a+b) String of a's and b's of odd length can also be obtained by the combination of the strings aa, ab, ba and bb preceded by either a or b. So, the regular expression can also be represented as (a+b) (aa + ab + ba + bb)* Note: Even though these two expression are seems to be different, the language corresponding to those two expression is same. So, a variety of regular expressions can be obtained for a language and all are equivalent.
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2.4 :finite automata and regular expressions
Obtain NFA from the regular expression Theorem: Let R be a regular expression. Then there exists a finite automaton M = (Q, , , q0, A) which accepts L(R). Proof: By definition, , and a are regular expressions. So, the corresponding machines to recognize these expressions are shown in figure 3.1.a, 3.1.b and 3.1.c respectively. q0 (a) qf q0 (b) qf q0 a (c) qf
Fig 3.1 NFAs to accept , and a The schematic representation of a regular expression R to accept the language L(R) is shown in figure 3.2. where q is the start state and f is the final state of machine M. L(R) q M f
Fig 3.2 Schematic representation of FA accepting L(R) In the definition of a regular expression it is clear that if R and S are regular expression, then R+S and R.S and R* are regular expressions which clearly uses three operators `+', `-` and `.'. Let us take each case separately and construct equivalent machine. Let M1 = (Q1, 1, 1, q1, f1) be a machine which accepts the language L(R1) corresponding to the regular expression R1. Let M2 = (Q2, 2, 2, q2, f2) be a machine which accepts the language L(R2) corresponding to the regular expression R2. Case 1: R = R1 + R2. We can construct an NFA which accepts either L(R1) or L(R2) which can be represented as L(R1 + R2) as shown in figure 3.3. L(R1) q0 q1 f1 q2 f2 M1 qf M2 L(R2)
Fig. 3.3 To accept the language L(R1 + R2)
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It is clear from figure 3.3 that the machine can either accept L(R1) or L(R2). Here, q0 is the start state of the combined machine and qf is the final state of combined machine M. Case 2: R = R1 . R2. We can construct an NFA which accepts L(R1) followed by L(R2) which can be represented as L(R1 . R2) as shown in figure 3.4. L(R1) L(R2) q1 M1 q2 M2 f1 f2 Fig. 3.4To accept the language L(R1 . R2) It is clear from figure 3.4 that the machine after accepting L(R1) moves from state q1 to f1. Since there is a -transition, without any input there will be a transition from state f1 to state q2. In state q2, upon accepting L(R2), the machine moves to f2 which is the final state. Thus, q1 which is the start state of machine M1 becomes the start state of the combined machine M and f2 which is the final state of machine M2, becomes the final state of machine M and accepts the language L(R1.R2). Case 3: R = (R1)*. We can construct an NFA which accepts either L(R1)*) as shown in figure 3.5.a. It can also be represented as shown in figure 3.5.b.
q0
q1 f1
M1 L(R1) (a)
qf
q0
q1 f1
M1
qf
(b) Fig. 3.5 To accept the language L(R1)* It is clear from figure 3.5 that the machine can either accept or any number of L(R1)s thus accepting the language L(R1)*. Here, q0 is the start state qf is the final state.
Obtain an NFA which accepts strings of a's and b's starting with the string ab.
The regular expression corresponding to this language is ab(a+b)*.
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Step 1: The machine to accept `a' is shown below. 4 a 5
Step 2: The machine to accept `b' is shown below. 6 b 7
Step 3: The machine to accept (a + b) is shown below. 4 3 6 b 7 a 5 8
Step 4: The machine to accept (a+b)* is shown below. a
4 2 3 6
5 8 9
b
7
Step 5: The machine to accept ab is shown below. 0 a 1 b 2
Step 6: The machine to accept ab(a+b)* is shown below.
0
a
1
b
4 2 3 6
a
5 8 9
b
7
Fig. 3.6 To accept the language L(ab(a+b)*)
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Obtain the regular expression from FA Theorem: Let M = (Q, , , q0, A) be an FA recognizing the language L. Then there exists an equivalent regular expression R for the regular language L such that L = L(R). The general procedure to obtain a regular expression from FA is shown below. Consider the generalized graph r1 q0 r
2
r q1 4
r Fig. 3.9 Generalized transition graph where r1, r2, r3 and r4 are the regular expressions and correspond to the labels for the edges. The regular expression for this can take the form: r = r1*r2 (r4 + r3r1*r2)* (3.1)
3
Note: 1. Any graph can be reduced to the graph shown in figure 3.9. Then substitute the regular expressions appropriately in the equation 3.1 and obtain the final regular expression. 2. If r3 is not there in figure 3.9, the regular expression can be of the form r = r1*r2 r4* (3.2) 3. If q0 and q1 are the final states then the regular expression can be of the form r = r1* + r1*r2 r4* (3.3)
Obtain a regular expression for the FA shown below:
0 q0 0 q2 1 1 q3 q1 0 0,1
1
The figure can be reduced as shown below: 01 q0 10
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It is clear from this figure that the machine accepts strings of 01's and 10's of any length and the regular expression can be of the form (01 + 10)*
What is the language accepted by the following FA
0 q0 1 q1
1
0
0, q2 1
Since, state q2 is the dead state, it can be removed and the following FA is obtained. 0 q0 1 q1 1
The state q0 is the final state and at this point it can accept any number of 0's which can be represented using notation as 0* q1 is also the final state. So, to reach q1 one can input any number of 0's followed by 1 and followed by any number of 1's and can be represented as 0*11* So, the final regular expression is obtained by adding 0* and 0*11*. So, the regular expression is R.E = 0* + 0*11* = 0* ( + 11*) = 0* ( + 1+) = 0* (1*) = 0*1* It is clear from the regular expression that language consists of any number of 0's (possibly ) followed by any number of 1's(possibly ).
2.5:Applications of Regular Expressions
Pattern Matching refers to a set of objects with some common properties. We can match an identifier or a decimal number or we can search for a string in the text. In UNIX operating system, we can use the editor ed to search for a specific pattern in the text. For example, if the command specified is /acb*c/
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An application of regular expression in UNIX editor ed.
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then the editor searches for a string which starts with ac followed by zero or more b's and followed by the symbol c. Note that the editor ed accepts the regular expression and searches for that particular pattern in the text. As the input can vary dynamically, it is challenging to write programs for string patters of these kinds.
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Assignment questions:
1. Obtain an NFA to accept the following language L = {w | w 2. Convert the following NFA into an equivalent DFA. ababn or aban where n 0}
0 q0
0,1 q 0, 1 q 1 2
1
3. Convert the following NFA to its equivalent DFA.
4 0 a 1 b 2 3 6
a
5 8 9
b
7
4. P.T. Let R be a regular expression. Then there exists a finite automaton M = (Q, , , q0, A) which accepts L(R).
5. Obtain an NFA which accepts strings of a's and b's starting with the string ab. 6. Define grammar? Explain Chomsky Hierarchy? Give an example 7. (a) Obtain grammar to generate string consisting of any number of a's and b's with at least one b. Obtain a grammar to generate the following language: L ={WWR where W {a, b}*} 8. (a) Obtain a grammar to generate the following language: L = { 0m 1m2n | m>= 1 and n>=0} Obtain a grammar to generate the set of all strings with no more than three a's when = {a, b} 9. Obtain a grammar to generate the following language: ( i ) L = { w | n a ( w) > n b ( w ) } (ii) L = { an bm ck | n+2m = k for n>=0, m>=0} 10. Define derivation , types of derivation , Derivation tree & ambiguous grammar. Give example for each. 11. Is the following grammar ambiguous? S aB | bA A aS | bAA |a B bS | aBB | b 12. Define PDA. Obtain PDA to accept the language L = {an bn | n>=1} by a final state. 13. write a short note on application of context free grammar.
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UNIT 3: PROPERTIES OF REGULAR LANGUAGES
3.1 Regular languages 3.2 proving languages not to be regular languages 3.3 closure properties of regular languages 3.4 decision properties of regular languages 3.5 equivalence and minimization of automata
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3.1:Regular languages
In theoretical computer science and formal language theory, a regular language is a formal language that can be expressed using a regular expression. Note that the "regular expression" features provided with many programming languages are augmented with features that make them capable of recognizing languages that can not be expressed by the formal regular expressions (as formally defined below). In the Chomsky hierarchy, regular languages are defined to be the languages that are generated by Type-3 grammars (regular grammars). Regular languages are very useful in input parsing and programming language design.
Formal definition
The collection of regular languages over an alphabet is defined recursively as follows: The empty language Ø is a regular language. For each a (a belongs to ), the singleton language {a} is a regular language. If A and B are regular languages, then A B (union), A · B (concatenation), and A* (Kleene star) are regular languages. No other languages over are regular. See regular expression for its syntax and semantics. Note that the above cases are in effect the defining rules of regular expression Examples All finite languages are regular; in particular the empty string language {} = Ø* is regular. Other typical examples include the language consisting of all strings over the alphabet {a, b} which contain an even number of as, or the language consisting of all strings of the form: several as followed by several bs. A simple example of a language that is not regular is the set of strings . Intuitively, it cannot be recognized with a finite automaton, since a finite automaton has finite memory and it cannot remember the exact number of a's. Techniques to prove this fact rigorously are given below.
proving languages not to be regular languages
· · Pumping Lemma Used to prove certain languages like L = {0n1n | n 1} are not regular. Closure properties of regular languages Used to build recognizers for languages that are constructed from other languages by certain operations. Ex. Automata for intersection of two regular languages
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·
Decision properties of regular languages Used to find whether two automata define the same language Used to minimize the states of DFA eg. Design of switching circuits.
Pumping Lemma for regular languages ( Explanation)
Let L = {0n1n | n 1} There is no regular expression to define L. 00*11* is not the regular expression defining L. Let L= {0212}
0 1 1 2
0 3 1 6 0,1 0
1 4 0
1 5
0,1
State 6 is a trap state, state 3 remembers that two 0's have come and from there state 5 remembers that two 1's are accepted. This implies DFA has no memory to remember arbitrary `n'. In other words if we have to remember n, which varies from 1 to we have to have infinite states, which is not possible with a finite state machine, which has finite number of states.
Pumping Lemma (PL) for Regular Languages
Theorem:
Let L be a regular language. Then there exists a constant `n' (which depends on L) such that for every string w in L such that |w| n, we can break w into three strings, w=xyz, such that: 1. |y| > 0 2. |xy| n 3. For all k 0, the string xykz is also in L. PROOF:
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Let L be regular defined by an FA having `n' states. Let w= a1,a2 ,a3----an and is in L. |w| = n n. Let the start state be P1. Let w = xyz where x= a1,a2 ,a3 -----an-1 , y=an and z = .
Therefore xykz = a1 ------ an-1 (an)k k=0 k=1 k=2 a1 ------ an-1 is accepted a1 ------ an is accepted a1 ------ an+1 is accepted
k=10 a1 ------ an+9 is accepted and so on. Uses of Pumping Lemma: - This is to be used to show that, certain languages are not regular. It should never be used to show that some language is regular. If you want to show that language is regular, write separate expression, DFA or NFA. General Method of proof: (i) (ii) Select w such that |w| Select y such that |y| n 1 n
(iii) Select x such that |xy| (v)
(iv) Assign remaining string to z Select k suitably to show that, resulting string is not in L. To prove that L={w|w anbn, where n 1} is not regular
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Example 1.
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Proof: Let L be regular. Let n is the constant (PL Definition). Consider a word w in L. Let w = anbn, such that |w|=2n. Since 2n > n and L is regular it must satisfy PL.
xy contain only a's. (Because |xy| n). Let |y|=l, where l > 0 (Because |y| > 0). Then, the break up of x. y and z can be as follows
from the definition of PL , w=xykz, where k=0,1,2,------ , should belong to L. That is an-l (al)k bn L, for all k=0,1,2,------ Put k=0. we get an-l bn Example 2. To prove that L={w|w is a palindrome on {a,b}*} is not regular. i.e., L={aabaa, aba, abbbba,...} Proof: Let L be regular. Let n is the constant (PL Definition). Consider a word w in L. Let w = anban, such that |w|=2n+1. Since 2n+1 > n and L is regular it must satisfy PL. L. Contradiction. Hence the Language is not regular.
xy contain only a's. (Because |xy| n). Let |y|=l, where l > 0 (Because |y| > 0). That is, the break up of x. y and z can be as follows
from the definition of PL w=xykz, where k=0,1,2,------ , should belong to L. That is an-l (al)k ban L, for all k=0,1,2,------ .
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Put k=0. we get an-l b an L, because, it is not a palindrome. Contradiction, hence the language is not regular . Example 3. To prove that L={ all strings of 1's whose length is prime} is not regular. i.e., L={12, 13 ,15 ,17 ,111 ,----}
Proof: Let L be regular. Let w = 1p where p is prime and | p| = n +2
Let y = m. by PL xykz L | xykz | = | xz | + | yk | = (p-m) + m (p-m) = (p-m) (1+m) ----- this can not be prime if p-m 2 or 1+m 2 1. 2. Example 4. To prove that L={ 0i2 | i is integer and i >0} is not regular. i.e., L={02, 04 ,09 ,016 ,025 ,----} Proof: Let L be regular. Let w = 0n2 where |w| = n2 n by PL xykz L, for all k = 0,1,--Select k = 2 | xy2z | = | xyz | + | y | = n2 + Min 1 and Max n Therefore n2 < | xy2z | n2 + n is n2 < | xy2z | < n2 + n + 1+n n2 < | xy2z | < (n + 1)2 adding 1 + n ( Note that less than or equal to replaced by less than sign) (1+m) 2 because m 1 Limiting case p=n+2 (p-m) 2 since m n Let k = p-m
Say n = 5 this implies that string can have length > 25 and < 36 which is not of the form 0i2. a) Show that following languages are not regular
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3.3:closure properties of regular languages
1. The union of two regular languages is regular. 2. The intersection of two regular languages is regular. 3. The complement of a regular language is regular. 4. The difference of two regular languages is regular. 5. The reversal of a regular language is regular. 6. The closure (star) of a regular language is regular. 7. The concatenation of regular languages is regular. 8. A homomorphism (substitution of strings for symbols) of a regular language is regular. 9. The inverse homomorphism of a regular language is regular
Closure under Union
Theorem: If L and M are regular languages, then so is L Ex1. L1={a,a3,a5,-----} L2={a2,a4,a6,-----} L1 L2 = {a,a2,a3,a4,----} RE=a(a)* Ex2. L1={ab, a2 b2, a3b3, a4b4,-----} L2={ab,a3 b3,a5b5,-----} L1 L2 = {ab,a2b2, a3b3, a4b4, a5b5----} RE=ab(ab)* M.
Closure Under Complementation
Theorem : If L is a regular language over alphabet S, then L = language. Ex1. L1={a,a3,a5,-----} * -L1={e,a2,a4,a6,-----} RE=(aa)* Ex2.
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- L is also a regular
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Consider a DFA, A that accepts all and only the strings of 0's and 1's that end in 01. That is L(A) = (0+1)*01. The complement of L(A) is therefore all string of 0's and 1's that do not end in 01
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regular language
Theorem: - If L is a regular language over alphabet
, then, L =
*
- L is also a
Proof: - Let L =L(A) for some DFA. A=(Q, , , q0, F). Then L = L(B), where B is the DFA (Q, , , q0, Q-F). That is, B is exactly like A, but the accepting states of A have become non-accepting states of B, and vice versa, then w is in L(B) if and only if ^ ( q0, w) is in Q-F, which occurs if and only if w is not in L(A).
Closure Under Intersection
Theorem : If L and M are regular languages, then so is L M. Ex1. L1={a,a2,a3,a4,a5,a6,-----} L2={a2,a4,a6,-----} L1L2 = {a2,a4,a6,----} RE=aa(aa)* Ex2 L1={ab,a3b3,a5b5,a7b7-----} L2={a2 b2, a4b4, a6b6,-----} L1 L2 = RE= Ex3. Consider a DFA that accepts all those strings that have a 0.
Consider a DFA that accepts all those strings that have a 1.
The product of above two automata is given below.
This automaton accepts the intersection of the first two languages: Those languages that have both a 0 and a 1. Then pr represents only the initial condition, in which we have
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seen neither 0 nor 1. Then state qr means that we have seen only once 0's, while state ps represents the condition that we have seen only 1's. The accepting state qs represents the condition where we have seen both 0's and 1's. Ex 4 (on intersection) Write a DFA to accept the intersection of L1=(a+b)*a and L2=(a+b)*b that is for L1 L2.
DFA for L1
L2 =
(as no string has reached to final state (2,4))
Ex5 (on intersection) Find the DFA to accept the intersection of L1=(a+b)*ab (a+b)* and L2=(a+b)*ba (a+b)* that is for L1 L2
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DFA for L1
L2
Closure Under Difference
Theorem : If L and M are regular languages, then so is L M. Ex. L1={a,a3,a5,a7,-----} L2={a2,a4,a6,-----} L1-L2 = {a,a3,a5,a7----} RE=a(a)*
Reversal
Theorem : If L is a regular language, so is LR Ex. L={001,10,111,01} LR={100,01,111,10} To prove that regular languages are closed under reversal. Let L = {001, 10, 111}, be a language over ={0,1}. LR is a language consisting of the reversals of the strings of L. That is LR = {100,01,111}. If L is regular we can show that LR is also regular. Proof. As L is regular it can be defined by an FA, M = (Q, , , q0, F), having only one final state. If there are more than one final states, we can use - transitions from the final states going to a common final state. Let FA, MR = (QR,
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R, R,q R,FR) 0
defines the language LR,
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Where QR = Q,
R=
, q0R=F,FR=q0,
and
R (p,a)->
q, iff (q,a) -> p
Since MR is derivable from M, LR is also regular. The proof implies the following method 1. Reverse all the transitions. 2. Swap initial and final states. 3. Create a new start state p0 with transition on to all the accepting states of original DFA Example Let r=(a+b)* ab define a language L. That is L = {ab, aab, bab,aaab, -----}. The FA is as given below
The FA for LR can be derived from FA for L by swapping initial and final states and changing the direction of each edge. It is shown in the following figure.
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Homomorphism
A string homomorphism is a function on strings that works by substituting a particular string for each symbol. Theorem : If L is a regular language over alphabet , and h is a homomorphism on , then h (L) is also regular. Ex. The function h defined by h(0)=ab h(1)=c is a homomorphism. h applied to the string 00110 is ababccab L1= (a+b)* a (a+b)*
h : {a, b}
{0, 1}*
Resulting : h1(L) = (01 + 11)* 01 (01 + 11)* h2(L) = (101 + 010)* 101 (101 + 010)* h3(L) = (01 + 101)* 01 (01 + 101)*
Inverse Homomorphism
Theorem : If h is a homomorphism from alphabet S to alphabet T, and L is a regular language over T, then h-1 (L) is also a regular language. Ex.Let L be the language of regular expression (00+1)*. Let h be the homomorphism defined by h(a)=01 and h(b)=10. Then h-1(L) is the language of regular expression (ba)*.
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3.4: decision properties of regular languages
1. is the language described empty? 2. Is a particular string w in the described language? 3. Do two descriptions of a language actually describe the same language? This question is often called "equivalence" of languages.
Converting Among Representations
Converting NFA's to DFA's Time taken for either an NFA or -NFA to DFA can be exponential in the number of states of the NFA. Computing -Closure of n states takes O(n3) time. Computation of DFA takes O(n3) time where number of states of DFA can be 2n. The running time of NFA to DFA conversion including transition is O(n3 2n). Therefore the bound on the running time is O(n3s) where s is the number of states the DFA actually has. DFA to NFA Conversion Conversion takes O(n) time for an n state DFA. Automaton to Regular Expression Conversion For DFA where n is the number of states, conversion takes O(n34n) by substitution method and by state elimination method conversion takes O(n3) time. If we convert an NFA to DFA and then convert the DFA to a regular expression it takes the time O(n34n 2n) Regular Expression to Automaton Conversion Regular expression to -NFA takes linear time O(n) on a regular expression of length n. Conversion from -NFA to NFA takes O(n3) time. Testing Emptiness of Regular Languages Suppose R is regular expression, then empty. 1. R = R1 + R2. Then L(R) is empty if and only if both L(R1) and L(R2) are 2. R= R1R2. Then L(R) is empty if and only if either L(R1) or L(R2) is empty. 3. R=R1* Then L(R) is not empty. It always includes at least 4. R=(R1) Then L(R) is empty if and only if L(R1) is empty since they are the same language. Testing Emptiness of Regular Languages Suppose R is regular expression, then empty. 1. R = R1 + R2. Then L(R) is empty if and only if both L(R1) and L(R2) are 2. R= R1R2. Then L(R) is empty if and only if either L(R1) or L(R2) is empty.
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3. R=(R1)* Then L(R) is not empty. It always includes at least 4. R=(R1) Then L(R) is empty if and only if L(R1) is empty since they are the same language. Testing Membership in a Regular Language Given a string w and a Regular Language L, is w in L. If L is represented by a DFA, simulate the DFA processing the string of input symbol w, beginning in start state. If DFA ends in accepting state the answer is `Yes' , else it is `no'. This test takes O(n) time If the representation is NFA, if w is of length n, NFA has s states, running time of this algorithm is O(ns2) If the representation is - NFA, - closure has to be computed, then processing of each input symbol , a , has 2 stages, each of which requires O(s2) time. If the representation of L is a Regular Expression of size s, we can convert to an -NFA with almost 2s states, in O(s) time. Simulation of the above takes O(ns2) time on an input w of length n
3.5:Minimization of Automata ( Method 1) Let p and q are two states in DFA. Our goal is to understand when p and q (p q) can be replaced by a single state. Two states p and q are said to be distinguishable, if there is at least one string, w, such that one of ^ (p,w) and ^ (q,w) is accepting and the other is not accepting. Algorithm 1: List all unordered pair of states (p,q) for which p q. Make a sequence of passes through these pairs. On first pass, mark each pair of which exactly one element is in F. On each subsequent pass, mark any pair (r,s) if there is an a for which (r,a) = p, (s,a) = q, and (p,q) is already marked. After a pass in which no new pairs are marked, stop. The marked pair (p,q) are distinguishable. Examples: 1. Let L = { , a2, a4, a6, ....} be a regular language over shown in Fig 1. = {a,b}. The FA is
Fig 2. gives the list of all unordered pairs of states (p,q) with p q.
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The boxes (1,2) and (2,3) are marked in the first pass according to the algorithm 1. In pass 2 no boxes are marked because, (1,a) and (3,a) 2. That is (1,3) where and 3 are non final states. ( ,2),
(1,b) and (3,b) . That is (1,3) ( , ), where is a non-final state. This implies that (1,3) are equivalent and can replaced by a single state A.
Fig 3. Minimal Automata corresponding to FA in Fig 1 Minimization of Automata (Method 2)
Consider set {1,3}. (1,3) (2,2) and (1,3) ( , ). This implies state 1 and 3 are equivalent and can not be divided further. This gives us two states 2,A. The resultant FA is shown is Fig 3. Example 2. (Method1): Let r= (0+1)*10, then L(r) = {10,010,00010,110, ---}. The FA is given below
Following fig shows all unordered pairs (p,q) with p q
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The pairs marked 1 are those of which exactly one element is in F; They are marked on pass 1. The pairs marked 2 are those marked on the second pass. For example (5,2) is one of these, since (5,2) (6,4), and the pair (6,4) was marked on pass 1. From this we can make out that 1, 2, and 4 can be replaced by a single state 124 and states 3, 5, and 7 can be replaced by the single state 357. The resultant minimal FA is shown in Fig. 6
The transitions of fig 4 are mapped to fig 6 as shown below
Example 2. (Method1):
(2,3) (4,6) this implies that 2 and 3 belongs to different group hence they are split in level 2. similarly it can be easily shown for the pairs (4,5) (1,7) and (2,5) and so on.
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Assignment questions 1. Let M = (Q, , , q0, A) be an FA recognizing the language L. Then there exists an equivalent regular expression R for the regular language L such that L = L(R).
2. Obtain a regular expression for the FA shown below:
0 q0 0 q2 1 1 q3 q1 0 0,1
1
3. What is the language accepted by the following FA
0 q0 1 q1
1
0
0,1 q2
4. Write short note on Applications of Regular Expressions
5. Obtain a DFA to accept strings of a's and b's starting with the string ab
q0 b q3 a,b
a a
q1
b
a,b q2
6. Prove pumping lemma? 7. prove that L={w|w is a palindrome on {a,b}*} is not regular. i.e., L={aabaa, aba, abbbba,...} 8. prove that L={ all strings of 1's whose length is prime} is not regular. i.e., L={12, 13 ,15 ,17 ,111 ,----} 9. Show that following languages are not regular
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L={anbm | n, m 0 and nm } L={anbmcmdn | n, m 1 } L={an | n is a perfect square } L={an | n is a perfect cube } 10. Apply pumping lemma to following languages and understand why we cannot complete proof L={anaba | n 0 } L={anbm | n, m 0 } 11. P.T. If L and M are regular languages, then so is L M.
*
12. P.T. If L is a regular language over alphabet S, then L = language. 13. P.T. - If L is a regular language over alphabet regular language , then, L
- L is also a regular =
*
- L is also a
14. Write a DFA to accept the intersection of L1=(a+b)*a and L2=(a+b)*b that is for L1 L2. 15. Find the DFA to accept the intersection of L1=(a+b)*ab (a+b)* and L2=(a+b)*ba (a+b)* that is for L1 L2 16. P.T. If L and M are regular languages, then so is L M.
17.
P.T. If L is a regular language, so is LR
18. If L is a regular language over alphabet , and h is a homomorphism on , then h (L) is also regular. 19. If h is a homomorphism from alphabet S to alphabet T, and L is a regular language over T, then h-1 (L) is also a regular language. 20. Design context-free grammar for the following cases a) L={ 0n1n | nl } b) L={aibjck| ij or jk} 21. Generate grammar for RE 0*1(0+1)*
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4.1 Context free grammars 4.2 parse trees 4.3 Applications 4.4 ambiguities in grammars and languages
UNIT 4:Context Free Grammar and languages
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4.1: Context free grammar
Context Free grammar or CGF, G is represented by four components that is G=(V,T,P,S), where V is the set of variables, T the terminals, P the set of productions and S the start symbol. Example: The grammar Gpal for palindromes is represented by Gpal = ({P},{0,1}, A, P) where A represents the set of five productions 1. P 2. P0 3. P1 4. P0P0 5. P1P1 Derivation using Grammar
Parse trees are trees labeled by symbols of a particular CFG. Leaves: labeled by a terminal or . Interior nodes: labeled by a variable. Children are labeled by the right side of a production for the parent. Root: must be labeled by the start symbol.
4.2: parse trees
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Example: Parse Tree
S -> SS | (S) | ()
Example 1: Leftmost Derivation The inference that a * (a+b00) is in the language of variable E can be reflected in a derivation of that string, starting with the string E. Here is one such derivation: E E * E I * E a * E a * (E) a * (E + E) a * (I + E) a * (a + E) a * (a + I) a * (a + I0) a * (a + I00) a * (a + b00) Leftmost Derivation - Tree
Example 2: Rightmost Derivations The derivation of Example 1 was actually a leftmost derivation. Thus, we can describe the same derivation by: E E * E E *(E) E * (E + E) E * (E + I) E * (E +I0) E * (E + I00) E * (E + b00) E * (I + b00) E * (a +b00) I * (a + b00) a * (a + b00) We can also summarize the leftmost derivation by saying E a * (a + b00), or express several steps of the derivation by expressions such as E * E a * (E).
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Rightmost Derivation - Tree
There is a rightmost derivation that uses the same replacements for each variable, although it makes the replacements in different order. This rightmost derivation is: E E *