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CSE2243 Theory of Computation and Compilers
Lecture 1:Regular Languages
Finite Automata
CSE2243 Lecture 1 2
Areas in the Theory of Computation
● What are the fundamental capabilities and limitations of computers?
● Areas:● Automata● Computability● Complexity
CSE2243 Lecture 1 3
Complexity
● How computationally hard a problem is
● Problem 1: Arranging a list of numbers in ascending order● Even a small computer can sort a million numbers rather quickly
● Problem 2: Finding a schedule of classes for the entire university to satisfy some reasonable constraints like no two classes can be scheduled in the same room at the same time● Much harder! If there are thousand classes, finding the best schedule may take
centuries, even on a supercomputer
● What makes some problems computationally hard and others easy?
● Classify problems as easy or hard
CSE2243 Lecture 1 4
Computability
● Certain basic problems cannot be solved by computers● Example: determining whether a mathematical
statement is true or false – no computer algorithm can perform this task!
● Classify problems as those which can be solved and those which cannot
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Automata
● Deals with the definitions and properties of mathematical models of computation
● Examples of models:● Finite automaton – used in text processing,
compilers and hardware design● Contextfree grammar – used in programming
languages and artificial intelligence
● Practice with formal definitions of computation
CSE2243 Lecture 1 6
Regular Languages
● What is a computer?● Use an idealized computer called a
computational model● We will use several different computational
models depending on the features we want to focus on
● We begin with the simplest model: finite state machine or finite automaton
CSE2243 Lecture 1 7
Finite Automata
● Models computers with an extremely limited amount of memory (finite memory)
● Example: Controller for an automatic door● Pads on each side of the door● Door opens when someone steps on the front pad,
if no one is standing on the rear pad
CSE2243 Lecture 1 8
Automatic Door Example
● Two states of the controller representing the condition of the door:● OPEN● CLOSED
● Four possible input conditions● FRONT – someone is standing on the front pad● REAR – someone is standing on the rear pad● BOTH – people are standing on both pads● NEITHER – no one is standing on either pad
CSE2243 Lecture 1 9
Automatic Door – State Diagram● State Diagram:
● State Transition Table:
● This controller is a computer that has just a single bit of memory, capable of recording which of the two states the controller is in.
● Other examples of computers with limited memories: elevators, dishwashers, electronic thermostats, parts of digital watches and calculators
CSE2243 Lecture 1 10
DFA M1 (1)
● State Diagram:
● 3 States: q1, q
2 and q
3
● Start state: q1 (indicated by an arrow pointing at it from nowhere)
● Accept state: q2 (indicated by a double circle)
● Transitions: the arrows going from one state to another labelled by inputs
CSE2243 Lecture 1 11
DFA M1 (2)
● How does the FA compute?● It is given and input string● It processes the string● It produces an output
● Processing:● Start in the start state● Read the input string, left to right
– After reading a symbol, M1 moves from current state to a new state by
following the transition labeled with the input symbol
● After it processes the last symbol, M1 produces an output
– Accept if M1 is now in an accept state, otherwise reject
CSE2243 Lecture 1 12
DFA M1 (3)
● How does M1 compute?
● On 1101?– Answer: It visits (in order) q
1, q
2, q
2, q
3, q
2 Accept
● On 01?
● On 101000?
● Does M1 accept or reject: 0? 1? 100? 01? 10? Any string
that ends in 1?
CSE2243 Lecture 1 13
Formal Definition of DFA (1)
● Definition of a DFA by a state diagram (Previous slide)
● Formal Definition of DFA● Why formal definition
● It is precise● No ambiguity● It provides notation
CSE2243 Lecture 1 14
Formal Definition of DFA (2)
● A DFA is a 5tuple (Q, ∑, δ, q0, F), where
● Q is a finite set called the states● ∑ is a finite set called the alphabet● δ: Q x ∑ Q is the → transition function
– Q x ∑ is the set of pairs (a, b) where a ∈ Q and b ∈ ∑● q
0 ∈ Q is the start state
● F ⊆ Q is the set of accept states (sometimes called the final states)
CSE2243 Lecture 1 15
Formal Definition of M1
● M1 = (Q, ∑, δ, q
1, F )
● Q = {q1, q
2, q
3}
● ∑ = {0, 1}● δ is described by the table
● q1 is the start state
● F = {q2}
CSE2243 Lecture 1 16
Language of a Machine M
● If A is the set of all strings that machine M accepts, then we say:● A is the language of machine M● L(M) = A● M recognizes A ● M accepts A
● A machine may accept several strings, but it always recognizes only one language
● If M accepts no strings, it recognizes one language, namely the empty language, ⌀
CSE2243 Lecture 1 17
Language of M1
● What is the language of M1?
● Answer: L(M1) = A, where
● A = {w | w contains at least one 1 and an even number of 0s follows the last 1}
● M1 recognises A
CSE2243 Lecture 1 18
Example M2
● What is the formal description of M2?
● What is the language of M2?
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Example M3
● What is the language of M3?
CSE2243 Lecture 1 20
Example M4
● Formal Description of M4:
● M4 = ({s,q
1, q
2,r
1, r
2},{a,b}, δ, s, {q
1, r
1})
where is given byδ
● What is the language of M4?
a b
s q1 r1
q1 q1 q2
q2 q1 q2
r1 r2 r1
r2 r2 r1
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Formal Definition of Computation
● Let M = (Q, ∑, δ, q0, F) be a DFA
● Let w = w1 w
2 … w
n be a string over the alphabet ∑
● Then M accepts w if a sequence of states r0, r
1, …
rn exists in Q with the following three conditions:
● r0 = q
0
● δ(ri,w
i+1) = r
i+1
● rn ∈ F
CSE2243 Lecture 1 22
Designing DFAs (1)
● How to design a DFA to do a specific task?● Put yourself in the place of the machine you are designing● Ask: What states are needed? What do I need to remember?● Then do the transitions and the rest of the state diagram
● Exercise● Suppose the alphabet is {0,1} and the language is the set of all
strings with an odd number of 1s.● Design a DFA to recognize this language
CSE2243 Lecture 1 23
Designing DFAs (2)
● Two states:
● qeven
: even number of 1s
● qodd
: odd number of 1s
● Note the counting idea in this example
CSE2243 Lecture 1 24
Designing DFAs (3)● Exercise
● Suppose the alphabet is {0,1} and the language is the set of all strings that contain 001 as the substring– Eg: 0010, 1001, 001, 1111100111110 are in the language– Eg: 11, 1101, 0000 are not in the language
● Design a DFA to recognize this language– How many states?
● Note the substring idea in this example
CSE2243 Lecture 1 25
Regular Languages and Regular Operations
● Definition:● Language L is regular if L = L(M) for some DFA M
● One way to show a language is regular:● Give a DFA for it
● A tool for showing a language is regular: regular operations● If you have two regular languages● And you combine them by a regular operation,● Then the result is another regular language
CSE2243 Lecture 1 26
Regular Operations Context
● In arithmetic:● Basic objects are numbers● Basic tools (addition, multiplication, etc.) operate on
numbers
● In theory of computation:● Basic objects are languages● Basic tools (regular operations) operate on
languages
CSE2243 Lecture 1 27
Regular Operations Definition
● Let A and B be languages● We define regular operations as follows:
● Union: A ∪ B = {x | x ∈ A or x ∈ B}● Concatenation: A ° B = {xy | x ∈ A and y ∈ B}
● Star: A* = {x1x
2...x
k | k ≥ 0 and each x
i ∈ A}
CSE2243 Lecture 1 28
Regular Operations Example
● Let the alphabet ∑ be {a, b, … z}● If A = {good, bad} and B = {boy, girl}, then
● A ∪ B = {good, bad, boy, girl}● A ° B = {goodboy, goodgirl, badboy, badgirl}● A* = {ε, good, bad, goodgood, goodbad, badgood,
badbad, goodgoodgood, goodgoodbad, goodbadgood,...}
CSE2243 Lecture 1 29
Regular Operations Closure
● A collection of objects is closed under operation X if applying operation X to members of the collection returns an object still in the collection
● The natural numbers are closed under multiplication (Meaning for any natural numbers x and y, the product x x y is also a natural number)
● The class of regular languages is closed under each of the regular operations
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Proving the Closure Theorem
● Theorem: the class of regular languages is closed under ● Union● Concatenation● Star
● The 3 proofs are by construction
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Proof v/s Theorem
● A proof of a mathematical statement is an argument that convinces the reader that the statement is true.● Different readers imply different proofs
● A theorem is a mathematical statement that has been proved true.● If we need the statement to prove another statement of greater
interest, we call the first statement a lemma● If the theorem lets us easily conclude other, related statements,
we call these other statements corollaries of the theorem
CSE2243 Lecture 1 32
Types of Proofs
● Proof by contradiction● Assume that the statement is false● Show that this leads to an obviously false consequence,
called a contradiction
● Proof by construction● If the theorem says an object of a certain type exists,
demonstrate how to construct the object
● Proof by induction● Advanced method to show that all elements of an infinite
set have a specified property
CSE2243 Lecture 1 33
Theorem: The class of regular languages is closed under union
● That is:● If A
1 and A
2 are regular languages, then so is A
1 ∪ A
2
● Proof idea– We know that A
1 and A
2 are regular
● So, there is an M1 which recognizes A
1 and an M
2 which recognizes A
2
– We construct an M from M1 and M
2 that recognizes A
1 ∪ A
2
– M simulates both M1 and M
2 and accepts if either accepts
CSE2243 Lecture 1 34
If A1 and A
2 are regular languages,
then so is A1 ∪ A
2● M simulates both M
1 and M
2 and accepts if either accepts
● M does not know (in advance) whether M1 or M
2 will accept
● So, it must remember the state of each, i.e., a pair of states
● So, a state in M is a pair (x,y), where x is the state M1 would be in
and y is the state M2 would be in
● Transitions in M go from pair to pair● The accept states in M are pairs (x,y) where either x is an accept
state of M1 or y is an accept state of M
2
CSE2243 Lecture 1 35
If A1 and A
2 are regular languages,
then so is A1 ∪ A
2● Proof
● Let M1 recognize A
1 where M
1 = (Q
1, ∑, δ
1, q
1, F
1)
● Let M2 recognize A
2 where M
2 = (Q
2, ∑, δ
2, q
2, F
2)
● We construct M that recognizes A1 ∪ A
2 where M = (Q, ∑, δ, q, F) where
– Q = {(r1,r
2)| r
1 ∈ Q
1 and r
2 ∈ Q
2}
● ∑ is the same as the languages for M1 and M
2
● δ is defined as follows: For each (r1,r
2) ∈ Q and each a in ∑, let
δ((r1,r
2),a) = (δ
1(r
1,a),δ
2(r
2,a))
● q = (q1,q
2)
● F = {(r1,r
2)| r
1 ∈ F
1 or r
2 ∈ F
2}
● It is evident from the construction of M that M recognizes A1 ∪ A
2
