THEORY OF COMPUTATION

There are four sorts of men:

He who knows not and knows not he knows not: he is afool - shun him;

He who knows not and knows he knows not: he is simple– teach him;

He who knows and knows not he knows: he is asleep –wake him;

He who knows and knows he knows: he is wise – followshim

Arabian proverb

Chapter 1: Sets, Relations andLanguages

Recommended Readings: Textbook + A Basis for Theo-retical Computer Science by M. A. Arbib, A. J. Kfoury,R N. Moll.

A set is a collection of objects.

Examples of sets are:

– the set of integers, denoted by Z,

– the set of nonnegative integers, also called natu-

ral numbers, denoted by N,

– the set of truth values B = {T, F}– the set of students taking the theory of compu-

tation course

The elements or members of a set are the ob-

jects comprising it. If b is an element of a set L,

then we write b ∈ L.

There are two ways to display a set:

either explicitly listing the elements belonging to

it like in the case of the set of truth values

B = {T, F},

or by specifying the properties that characterize

the elements of this set like: the set of even integers

{x |x ∈ Z and xmod 2 = 0}

Two sets are equal if they have the same ele-

ments

A set without any element is called the empty set

A set with infinitely many elements is said to be

infinite

A set with finitely many elements is said to be fi-

nite

A set X is a subset of a set Y, written X ⊆ Y , if

each element of X is also an element of Y.

X is a proper subset of Y if X is a subset of

Y and X and Y are not equal.

For example, the set of nonnegative integers is a

proper subset of the set of integers, N ⊂ Z.

Basic facts: For any set A, the empty set ∅ is a

subset of A, ∅ ⊆ A, and A is a subset of A, A ⊆ A.

If A ⊆ B and B ⊆ A then A = B.

Set operations: Union, Difference, In-

tersection

The union of two sets A,B is the set of elements

which belongs to at least one of them.

The intersection of two sets A,B is the set of ele-

ments which belongs to both of them.

We say two sets are disjoint if their intersection

is empty.

The set difference A\B of two sets A,B is the set

of those elements in A that are not in B.

Laws for set operations:

Idempotency A ∪ A = A

A ∩ A = A

Commutativity A ∪B = B ∪ AA ∩B = B ∩ A

Associativity (A∪B)∪C = A∪(B∪C)

(A ∩B) ∩ C = A ∩ (B ∩ C)

Suppose now that there is a big set U such that

both A, and B are subsets of U. Let B = U \ B,

B = U \B. Then

A \B = A ∩BA ∩B = A ∪BA ∪B = A ∩B

Exercises Prove the above equalities.

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Distributivity A ∪ (B ∩ C) = (A ∪ B) ∩(A ∪ C)

A∩(B∪C) = (A∩B)∪(A∩C)

Proof

1. To show A∪ (B ∩C) = (A∪B)∩ (A∪C), we

show A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C) and

(A ∪B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C).(a) Let x ∈ A ∪ (B ∩ C).

From x ∈ A ∪ (B ∩ C), it follows: x ∈ A or

x ∈ B ∩ C.If x ∈ A then x ∈ A ∪ B and x ∈ A ∪ C.

Hence x ∈ (A ∪B) ∩ (A ∪ C).If x ∈ B ∩ C then x ∈ B and x ∈ C. If

x ∈ B then x ∈ A ∪ B. If x ∈ C then

x ∈ A ∪ C. Therefore if x ∈ B ∩ C then

x ∈ (A ∪B) ∩ (A ∪ C).

Hence from x ∈ A or x ∈ B ∩ C we can

conclude x ∈ (A ∪ B) ∩ (A ∪ C). Therefore

we have proved that if x ∈ A∪ (B ∩C) then

x ∈ (A ∪B) ∩ (A ∪ C).

(b) Let x ∈ (A ∪ B) ∩ (A ∪ C). To show x ∈A ∪ (B ∩ C).

From x ∈ (A ∪B) ∩ (A ∪ C), it follows: x ∈A ∪ B and x ∈ A ∪ C. Hence (x ∈ A or

x ∈ B) and (x ∈ A or x ∈ C). Hence there

are four cases: (x ∈ A) or (x ∈ A and x ∈ B)

or (x ∈ A and x ∈ C) or (x ∈ B and x ∈ C).

It follows that x ∈ A or (x ∈ B and x ∈ C).

That means x ∈ A ∪ (B ∩ C).

2. Exercise.

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Absorption A ∩ (A ∪B) = A

A ∪ (A ∩B) = A

Proof Exercise.

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DeMorgan’s Laws

A\(B∪C) = (A\B)∩(A\C)

A\(B∩C) = (A\B)∪(A\C)

Proof

1. Suppose there is a set U such that all sets A,B,C

are subsets of U. Hence

A \ (B ∪ C) = A ∩ (B ∪ C) = A ∩ (B ∩ C)

= (A ∩B) ∩ (A ∩ C) = (A \B) ∩ (A \ C)

2. Exercise.

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If S is a collection of sets then⋃

S is the set

whose elements are the elements of the sets in S.

Example: S = {{a, b}, {c}, {a, d}}⋃S = {a, b, c, d}

S = { {a, b}, {c}, {a, {d, a} } }⋃S = {a, b, c, {d, a} }

The power set of a set S, denoted by 2S, is the

set of all subsets of S.

Example: S = {a} , 2S = {∅, {a} }

S = {a, b} , 2S = {∅, {a}, {b}, {a, b} }

A partition of a set S is a set Π of subsets of

S, i.e. Π ⊆ 2S, such that

1. Each element of Π is nonempty

2. Distinct members of Π are disjoint

3.⋃Π = S

Example S = {a, b, c}

Π = { {a}, {b, c} } is a partition of S.

A = { {a}, {a, b}, {c} } is not a partition.

B = { {a}, {b, c}, ∅ } is not a partition.

Functions or Maps

Ordered pairs are written as (a,b) where a is

the first component and b the second.

The Cartesian product of two sets A,B, de-

noted by A×B, is the set of all ordered pairs (a,b)

with a ∈ A, b ∈ B.

Example {a, b} × {a} = {(a, a), (b, a)}

The plane could be represented as a Cartesian

product

R×R

where R is the set of all real numbers.

A map or function from a set A to a set B,

denoted f : A → B, is an assignment to each

element a in A a single element, denoted by f(a),

in B.

A is called the domain of f and B is the codomain

of f.

f(a) is called the image of a under f.

The range of f is denoted by

f (A) = {b | there is a in A such that b = f (a)}

For A′ ⊆ A, f (A′) = {f (a) : a ∈ A′} is called

the image of A′ under f.

Example of functions:

The integer addition +: N×N → N is a func-

tion from N×N to N

Exercises 1) If A is empty, how many functions are

there from A to B ?

2) If B is empty, how many functions are there

from A to B ?

3) If A contains exactly one element, how many

functions are there from A to B ?

A function f : A −→ B is one to one if for

any two distinct elements a, a′ ∈ A, f (a) 6= f (a′).

A function f : A −→ B is onto B ifB = f (A).

A function f : A −→ B is a bijection be-

tween A and B if it is both one-to-one and onto

B.

Let A and B be two finite sets. A and B have the

same number of elements iff there is a bijection be-

tween A and B

Given a function f : A → B and g : B → C, the

composition f ◦ g : A→ C is defined by:

f ◦ g(a) = g(f (a))

Relations

A binary relation on two sets A,B is a subset

of A×B.

An example of binary relation is the greater or

equal relation≥ in the set of integersR = {(n,m) |n ≥m} ⊆ Z× Z.

An ordered n-tuple is written as (a1, . . . , an)

where ai is the ith component of (a1, . . . , an).

The n-fold Cartesian product of the sets

A1, . . . , An, denoted by A1× . . .×An, is the set of

all ordered n-tuples with ai ∈ Ai for i = 1, . . . , n.

An n-ary relation on sets A1, . . . , An is a subset

of A1 × . . .× An.

Property (A function is a special relation)

A function from a set A to a set B is a binary

relation R on A,B such that following property is

satisfied:

– For each a ∈ A there is exactly one ordered pair

in R with first component a.

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A binary relation R ⊆ A × B has an inverse

R−1 ⊆ B × A defined by:

(b, a) ∈ R−1 iff (a, b) ∈ R.

For binary relations Q ⊆ A×B and R ⊆ B×C,

the composition Q ◦R is defined by

Q◦R = {(a, c) : ∃ b ∈ B s.t. (a, b) ∈ Q and (b, c) ∈ R}

Note that for two functions f : A −→ B and

g : B −→ C, f ◦ g : A −→ C.

Exercise Show that if f : A −→ B is a bijec-

tion, then f−1 is also a bijection from B to A.

Special Types of Relations

A relation R ⊆ A × A can be represented as a

directed graph.

A relation R ⊆ A× A is reflexive if for each

a ∈ A, (a, a) ∈ R.

A relation R ⊆ A×A is symmetric if (a, b) ∈R whenever (b, a) ∈ R.

Symmetric relation can be represented by undi-

rected graph.

A relation R ⊆ A × A is transitive if when-

ever (a, b) ∈ R and (b, c) ∈ R then (a, c) ∈ R

A relation R ⊆ A× A is an equivalent rela-

tion if it is reflexive, transitive and symmetric.

Let R be an equivalent relation on a set A. Then

for each a in A, the equivalence class of a with

respect to R is denoted by [a]R and is defined for-

mally by

[a]R = {b | (a, b) ∈ R}

When the context R is clear, we simply write [a]

for [a]R

A representation of an equivalent relation R ⊆A × A as an undirected graph consists of a num-

ber of ”clusters” where within clusters each pair is

connected by a line. The set of nodes in a cluster

is an equivalence class.

Property Let R be an equivalent relation on a

set A. Then for any two elements a,b in A, either

[a]R = [b]R or [a]R, [b]R are disjoint.

Proof Exercise.

LetR be an equivalent relation on a set A. Define

A modulo R to be the set

A/R = {[a]R | a ∈ A}

Theorem 1. Let R ⊆ A × A be an equivalent

relation. Then A modulo R is a partition of A.

Proof Exercise.

Example of Partition and equivalence relation:

Let Z+ be the set of positive integers. Define an

equivalent relation ≡⊆ Z+ × Z+ as follows:

(p, q) ≡ (r, s) iff ps = qr

The ≡ is an equivalent relation characterizing

the rational numbers, i.e.

The set Z+×Z+ modulo ≡ is the set of rational

numbers.

A relation R ⊆ A × A is antisymmetric if

whenever (a, b) ∈ R and a 6= b then (b, a) 6∈ R

A relation that is reflexive, transitive and anti-

symmetric is called a partial order.

A partial order R ⊆ A × A is called a total

order if for all a, b ∈ A, either (a, b) ∈ R or

(b, a) ∈ R.

A path from a to b in a binary relation R ⊆A×A is a sequence (a1, . . . , an), n ≥ 1 such that

a = a1, b = an, for each i = 1, . . . n−1, (ai, ai+1) ∈R.

A path (a1, . . . , an) is a cycle if a1 = an and all

ai’s are distinct.

Two sets A, B are equinumerous if there is a

bijection f : A −→ B.

We say that the cardinality of A is n if A is

equinumeruos with the set {1,...,n}.

A is countably infinite if it is equinumerous

with the set of natural numbers N .

A is uncountable if it is not equinumerous with

the set of natural numbers N .

Exercise Every subset of a finite set is finite.

Exercise Every subset of a countably infinite set is

finite or countably infinite.

Exercise The set N ×N is countably infinite.

Theorem 2. The union of countably infinite col-

lection of countably infinite sets is again count-

able infinite.

Exercixe Is the set of all finite subsets of N count-

ably infinite ?

Three Fundamental ProofTechniques

Principle of Mathematical Induction: Let

A be a set of natural number such that

1. 0 ∈ A, and

2. for each natural number n, if {0, 1, . . . , n} ⊆ A

then n + 1 ∈ A.

Then A = N.

Exercise Prove that for each natural number n

1 + . . . + n =n(n + 1)

2

Exercise Suppose f, g : N −→ (N − {0}) satisfy

following properties:

1. g(0) ≤ f (0), and

2. for each n ≥ 0,

g(n + 1)

g(n)≤ f (n + 1)

f (n)

Then for any n ≥ 0: g(n) ≤ f (n).

The pigeon principle If A, B are finite sets

and |A| > |B|. Then there is no one-to-one func-

tion from A to B.

Exercise Suppose there are 100 new students for

the august semester at AIT and only 97 vacant

rooms. Is it possible to give each student a sepa-

rate room ? Prove it.

Exercise If there are 10 scholarships and 50 appli-

cants. Is it possible that each applicant get a schol-

arship. (Note that scholarship can not be shared).

Exercise Could you try to prove the pigeon prin-

ciple using mathematical induction ?

The Diagonalization Principle Let R be a

binary relation on a set A, and let

–

D = {a | a ∈ A and (a, a) 6∈ R}D is called the diagonal set of R.

– for each a ∈ A,

Ra = {b | b ∈ A and (a, b) ∈ R}

Then for each a ∈ A , D 6= Ra.

Exercise Let A = {1, 2, 3, 4, 5} and

R = {(1, 3), (2, 4), (3, 5), (5, 1), (2, 3), (1, 1), (4, 2), (4, 4)}.Determine the sets D,R1, R2, R3, R4, R5. Check

whethere the diagonal principle holds in this case.

Exercise Would the diagonal principle still holds if

the sets Ra are replaced by the sets Ra = {b | b ∈A and (b, a) ∈ R}

Exercise Would the diagonal principle still holds

if the sets D is replaced by the set C = {a | a ∈A and (a, a) ∈ R}.

Theorem 3. The set 2N is uncountably infinite

Proof (Sketch) Suppose the theorem is not cor-

rect. Therefore, the set 2N is countably infinite, i.e.

there is a bijection (i.e. one-one and onto) function

f : N→ 2N.

Define R = {(n,m) |m ∈ f (n) }. Apply the

diagonal principle on R and show a contradiction

to the assumption that f is a bijection.

�

Exercice Is the set of all infinite subsets of N count-

ably infinite ?

Algorithms

Definition

Let R ⊆ A2 be a directed graph on a set A.

The reflexive transitive closure of R is the

relation

R∗ = {(a, b) | a, b ∈ A and there is a path from a to b in R}

Algorithms

The question we interested in this course is:

What is an algorithm

Example of an algorithm:

Initially S := ∅for i = 1,..., n do

for each i-tuple (b1, . . . , bi) ∈ Ai do

if (b1, . . . , bi) is a path in R then add (b1, bi)

to S

How much time the algorithm needs to terminate ?

The total number of operations can be no more

than

n + n2 + . . . + nn

that belongs to O(nn+1) where n is the number

of elements of A. This algorithm is therefore not

efficient.

An efficient algorithm:

Initially S = R ∪ {(ai, ai) | ai ∈ A}for each j = 1,2,...,n do

for each i = 1,2,...,n and k = 1,2,...,n do

If (ai, aj), (aj, ak) ∈ S but (ai, ak) 6∈ Sthen add (ai, ak) to S.

The algorithm terminate after no more than n3

steps.

Is this algorithm correct, i.e. S = R∗ after ter-

mination of the algorithm. (Could you use mathe-

matical induction to show it correctness ?)

Rate of Growth of Functions

Definition

1. Let f be a function from N into N . The order

of f, denoted by O(f ) is the set of all functions

g : N → N such that there are positive natural

numbers c, d > 0 such that for each n ∈ Ng(n) ≤ c.f (n) + d

2. We write f^_ g iff f ∈ O(g) and g ∈ O(f ).

�

Lemma^_ is a equivalence relation.

Proof Exercise.

Definition The equivalence class of a function

f : N −→ N with respect to^_ is called the

rate of growth of f.

Lemma Let f (n) = a0 + a1.n + . . . + ak.nk be

a polynomial of degree k with nonnegative coeffi-

cients. Then

1. f^_ nk

2. nk ∈ O(nk+1)

3. nk+1 6∈ O(nk)

Proof

1. Let c = a0 + a1 + . . . + ak and d = 0. It is easy

to see: f (n) ≤ c.nk.

2. Obviously nk ≤ nk+1.

3. Suppose nk+1 ∈ O(nk). Hence, there is c, d > 0:

nk+1 ≤ c.nk + d. Let n = c + d. It follows (c +

d).nk = c.nk + d.nk > c.nk + d. Contradiction.

That means that assuming nk+1 ∈ O(nk) leads

to a contradiction. Therefore nk+1 6∈ O(nk).

Lemma 1. Let f, g : N −→ N be monotonic,

i.e. for all m ≥ n: f (m) ≥ f (n) and g(m) ≥g(n). Further let n0 ∈ N be a natural number

such that for each n ≥ n0,

g(n + 1)

g(n)≤ f (n + 1)

f (n)

Then g ∈ O(f )

Proof Let c be a number such that

g(n0) ≤ cf (n0)

From an earlier exercise:

∀n ≥ n0 : g(n) ≤ c.f (n)

Let d = g(n0). We can prove that for each n ∈ N :

g(n) ≤ cf (n) + d

�

Theorem Let r > 1. Then

1. ni ∈ O(rn)

2. rn 6∈ O(ni)

Proof

1. It follows immediately from lemma 1. (Elabo-

rate)

2. Suppose rn ∈ O(ni). Hence ni^_ rn

^_ ni+1.

Contradiction to lemma 1. (Elaborate)

Alphabet and Languages

An alphabet is a finite set of symbols.

A string over an alphabet is a finite sequence

of symbols from this alphabet.

The empty string is denoted by e.

The length of a string is its length as a sequence.

The concatenation of two strings x,y, denoted

by x ◦ y or simply xy, is the string x followed by

string y.

The concatenation is associative, i.e (x ◦ y) ◦z = x ◦ (y ◦ z).

A string v is a substring of a string w if there

are strings x, y such that w = xvy.

For each string w, each natural number i, define

– w0 = e

– wn+1 = wn ◦ w

This is an example of inductive definition.

The reversal of a string is defined inductively

as follows:

– eR = e

– (ua)R = auR

Exercise Prove (w ◦ v)R = vR ◦ wR

The set of all strings over an alphabet Σ is de-

noted by Σ∗.

Any set of strings over an alphabet Σ is a lan-

guage.

Question: Is Σ∗ countably infinite ?

Language Operations

– Complement The complement of a language

A over Σ is Σ∗ − A– Concatenation Let L1, L2 be two languages

over Σ. The concatenation of L1, L2, denoted

by L1 ◦ L2 is defined by:

L1 ◦ L2 = {x ◦ y |x ∈ L1, y ∈ L2}– Denote

1. L0 = {e},2. L1 = L

3. L2 = L ◦ L4. Ln+1 = Ln ◦ LThe Kleen star of a language L, denoted by

L∗, is

L∗ = {w1 ◦ w2 ◦ . . . ◦ wn |wi ∈ L, n ≥ 0}= L0 ∪ L1 ∪ L2 ∪ . . . ∪ Ln ∪ . . .

Notes ∅∗ = {e}

Define

L+ = {w1 ◦ w2 ◦ . . . ◦ wn |wi ∈ L, n ≥ 1}

Regular Languages

The class of regular languages is the least set of

languages satisfying the following properties:

– The emptyset is regular

– For each character a ∈ Σ, the language {a} is

regular

– If A, B are regular languages then A∗, A ∪ B,

A ◦B are also regular.

Finite Representation ofLanguages

If we want to study languages, we need to rep-

resent them. For example, a context-free language

could be represented by a context-free grammar.

Such a representation must be finite even if the

represented language is infinite, otherwise it would

be useless (just imagine an infinite set of grammar

for Thai, Chinese or English. Would you be able to

learn such a set of grammar ?).

On the other handside, different languages should

have different representations.

Could we have enough representations for every

possible language over a nonempty alphabet Σ ?

How many languages there are over an nonempty

alphabet Σ ?

Since Σ∗ is countable infinite, 2Σ∗

is uncountable

infinite.

The representation of languages must be again

some form of language. hence every finite represen-

tation can be viewed as a finite string over some al-

phabet Σ ′. Since Σ ′∗ is countable infinite, we can

not find a finite representation for each langauge

over Σ.

Finte representation of regular languages

The set of regular expresions over Σ is the least

language over Σ ∪ {∗, (, ), ∅,∪, ◦} satisfying the

following properties:

– ∅ a regular expression

– Every symbol in Σ is a regular expression

– If α, β are regular expressions, then so are (αβ),

α∗, α ∪ β

Any regular expression represents a language as

follows:

– L(∅) = empty set

– Every symbol a ∈ Σ, L(a) = {a}– If α, β are regular expressions, then L(αβ) =

L(α)◦L(β), L(α∗) = L(α)∗, L(α∪β) = L(α)∪L(β)

Theorem 4. A language A is regular iff there

exists a regular expression α such that A = L(α)

Proof Exercise.

Exercise Show that the Kleen star, concatenation

and union operators are monotonic, i.e. for lan-

guages L0, L1, L, followings hold:

1. If L0 ⊆ L1 then L∗0 ⊆ L∗12. If L0 ⊆ L1 then L0 ◦ L ⊆ L1 ◦ L3. If L0 ⊆ L1 then L0 ∪ L ⊆ L1 ∪ L

Exercise Check whether the following equations

are correct.

1. ((a ∪ b)∗)∗ = (a ∪ b)∗2. (a ∪ b)∗(a ∪ b)∗ = (a ∪ b)∗3. (a ∪ b)∗ = a∗ ∪ (a∗b)∗

4. (a ∪ b)∗ = ∅∗ ∪ (a∗b)∗

5. (a ∪ b)∗ = (b∗a)∗ ∪ (a∗b)∗

Exercise Exercise 1.8.7 in text book.

Question Given a regular language L, and

a string w, how could we check whether w ∈ L ?

Example Let L = (a ∪ b∗a)∗

w = aababa.

Question: w ∈ L?

Algorithm:

s : x := get-next-symbol;

if x := end-of-file then accept;

else if x := a then goto s;

else if x := b then goto q;

q : x := get-next-symbol;

if x := end-of-file then reject;

else if x := a then goto s;

else if x := b then goto q;

s : x := get-next-symbol;

if x := end-of-file then accept;

else if x := a then goto s;

else if x := b then goto q;

q : x := get-next-symbol;

if x := end-of-file then reject;

else if x := a then goto s;

else if x := b then goto q;

Fig. 1.

Fig. 2.

Fig. 3.

A shortened representation of the algorithm in

above figure is M = (K, Σ, δ, s, F ):

1. K = {s, q}

2. Σ = {a, b}

3. F = {s}

4. δ : K ×Σ −→ K

δ = {(s, a, s), (s, b, q), (q, a, s), (q, b, q)}

Chapter 2: Finite Automata

Chapter 2.1: Deterministic Finite

Automata

Definition 1. A deterministic finite automa-

ton is a quintuple M = (K, Σ, δ, s, F ) where

– K is a finite set of states

– Σ is an alphabet

– s ∈ K is the initial state

– F ⊆ K is the set of final states

– δ, the transition function, is a function from

K ×Σ to K.

�

An example of finite automata is given in figure

3.

A configuration of a finite automaton M =

(K, Σ, δ, s, F ) is a pair (q, w) where q ∈ K and

w ∈ Σ∗

We say that a configuration (q,w) yields an-

other configuration (q′, w′) in one step, denoted

by (q, w) `M (q′, w′), if there is a symbol σ ∈ Σsuch that w = σw′ and q′ = δ(q, σ).

`∗M is the reflexive and transitive closure of `M ,

i.e. we write (q0, w0) `∗M (qn, wn) iff there are

(q1, w1), . . . , (qn1, wn−1) such that

(q0, w0) `M (q1, w1) `M . . . `M (qn1, wn−1) `M(qn, wn)

– A string w is said to be accepted by M =

(K, Σ, δ, s, F ) if and only if there exists a state

q ∈ F such that (s, w) `∗M (q, e)

– The language accepted by M, L(M), is the

set of all strings accepted by M.

Exercise Design an automaton for recognizing

the language a∗ba∗b.

Chapter 2.2: Nondeterministic Finite

Automata

Definition 2. A nondeterministic finite au-

tomaton is a quintuple M = (K, Σ, ∆, s, F )

where

– K is a finite set of states

– Σ is an alphabet

– s ∈ K is the initial state

– F ⊆ K is the set of final states

– ∆, the transition relation, is a finite subset of

K × (Σ ∪ {e})×K.

Examples A nondeterministic machine for

(a ∪ ab)∗.

Fig. 4.

Fig. 5.

M = (K, Σ, δ, s, F ):

– K = {s, q}

– Σ = {a, b}

– F = {s}

– ∆ ⊆ K × (Σ ∪ {e})×K:

∆ = {(s, a, q), (q, e, s), (q, b, s)}

Exercise Design a deterministic machine for (a ∪ab)∗.

Exercise Could you design a nondeterministic ma-

chine for (a ∪ ab)∗ without using e.

A configuration of a nondeterministic finite

automata M = (K, Σ, ∆, s, F ) is a pair (q, w)

where q ∈ K and w ∈ Σ∗

We say that a configuration (q,w) yields an-

other configuration (q′, w′) in one step, denoted

by (q, w) `M (q′, w′), if there is u ∈ Σ ∪ {e} such

that w = uw′ and (q, u, q′) ∈ ∆.

`∗M is the reflexive and transitive closure of `M ,

i.e. we write (q0, w0) `∗M (qn, wn) iff there are

(q1, w1), . . . , (qn1, wn−1) such that

(q0, w0) `M (q1, w1) `M . . . `M (qn1, wn−1) `M(qn, wn)

– A string w is said to be accepted by M =

(K, Σ, δ, s, F ) if and only if there exists a state

q ∈ F such that (s, w) `∗M (q, e)

– The language accepted by M, L(M), is the

set of all strings accepted by M.

Exercise Design a deterministic and a nondeter-

ministic automaton for recognizing the language

(ab ∪ aba)∗.

Chapter 2.3: Equivalence of

Deterministic and Nondeterministic

Finite Automata

Two finite automata M, M ′ are equivalent iff

L(M) = L(M ′).

Theorem 5. For each nondeterministic finite

automaton there exists an equivalent determin-

istic finite automaton.

Proof Let M = (K, Σ, ∆, s, F ) be a nondeter-

ministic automata. For each q ∈ K, define

E(q) = {p ∈ K | (q, e) `∗M (p, e) }

Define M ′ = (K ′, Σ, δ′, s′, F ′).

– K ′ = 2K

– s′ = E(s)

–

δ′ : K ′ ×Σ −→ K ′

δ′(Q, σ) =⋃q∈Q

{E(p) | p ∈ K and (q, σ, p) ∈ ∆}

– F ′ = {Q ⊆ K | Q ∩ F 6= ∅ }

Property For each w ∈ Σ∗:

(q, w) `∗M (p, e) iff (E(q), w) `∗M ′ (P, e)

for some P ⊆ K such that p ∈ P . �

Therefore, it follows that

w ∈ L(M) iff

(s, w) `∗M (p, e) for some p ∈ F iff

(s′, w) `∗M ′ (P, e) and P ∈ F ′ iff

w ∈ L(M ′).

�

Example Let Σ = {a1, . . . , an}, n ≥ 2 and

L = {w | there is one symbol inΣ not appearing in w }

A nondeterministic finite automaton accepting

L:

M = (K,Σ,∆, s, F ) where

– K = {s, q1, . . . , qn}

– F = {q1, . . . , qn}

– ∆ = {(s, e, qi) | 1 ≤ i ≤ n}∪

{(qi, σ, qi) | 1 ≤ i ≤ n, σ 6= ai }

Translate M into an equivalent deterministic fi-

nite automaton.

Chapter 2.4: Properties of Languages

Accepted by Finite Automata

Theorem 6. The class of languages accepted

by finte automata is closed under

1. union

2. concatenation

3. Kleen star

4. complementation

5. intersection

Theorem 7. A language is regular iff it is ac-

cepted by a finite automata

Proof From the definition of regular languages,

it is clear that every regular language is accepted

by a finite automata (exercise).

It remains to prove that every language accepted

by a finite automata is also regular.

M = (K, Σ, ∆, s, F ) be a deterministic finite

automaton.

Let K = {q1, . . . , qn} with s = q1 and Σ =

{a1, . . . , an}.

R(i, j, k) = {σ1 . . . σm ∈ Σ∗ | (qi, σ1 . . . σm) `M

(qk1, σ2 . . . σm) `M . . . `M (qkm−1, σm) `M (qj, e)

and max{k1, . . . , km−1} ≤ k }

It is clear

L(M) =⋃qj∈F

R(1, j, n)

Now we want to show that R(i, j, k) are regular.

We prove this by induction on k.

Base case: k = 0. Obvious from the definition of

R(i, j, 0)

Inductive Step: Follows from the following equation

R(i, j, k) =

R(i, j, k−1)∪R(i, k, k−1)◦R(k, k, k−1)∗◦R(k, j, k−1)

Pumping Theorem

Theorem 8. Let L be a regular language. There

is an integer n > 0 such that any string w ∈ Lwith |w| ≥ n can be rewritten as w = xyz such

that y 6= e, |xy| ≤ n and xykz ∈ L for any

k ≥ 0.

�

Example

L = {aibi | i ≥ 0 } is not regular.

Example

L = {w ∈ {a, b}∗ |w has an equal number of a’s and b’s }

is not regular.

STATE MINIMIZATION

Problem: Given a regular language L, design

a finite automaton M such that L = L(M) and M

has as few states as possible, i.e. for each finite au-

tomaton M ′ that accepts L, the number of states

of M’ is greater or equal the number of states of M

Definition 3. Let M = (K,Σ, δ, s, F ) be a de-

terministic automaton. We say that two strings

x, y ∈ Σ∗ are equivalent wrt M, denoted by

x ∼M y, if there is a state q such that (s, x) `∗M(q, e) and (s, y) `∗M (q, e)

Definition 4. Let L ⊆ Σ∗ and x, y ∈ Σ∗. We

say that x and y are equivalent wrt to L,

denoted by x ≈L y if for all z ∈ Σ∗,

xz ∈ L iff yz ∈ L.

It is obvious that for all x, y ∈ Σ∗ and a ∈ Σ,

x ≈L y implies xa ≈L ya

Theorem 9. For any deterministic finite au-

tomaton M = (K,Σ, δ, s, F ) and for any string

x, y ∈ Σ∗, if x ∼M y then x ≈L(M) y.

Let M = (K,Σ, δ, s, F ) be a deterministic fi-

nite automaton and q ∈ K and Lq be the set of

all strings x such that (s, x) `∗M (q, e). Further for

each x ∈ Σ∗, let [x]M be the equivalent class of x

wrt ∼M .

It is obvious that for each x ∈ Lq following as-

sertions hold:

– Lq = [x]M .

– The mapping µ(q) = Lq is a bijection from K

onto {[x]M | x ∈ Σ∗} such that for all p, q ∈ K,

δ(q, a) = p iff [xa]M = Lp

Therefore M is equivalent to the following au-

tomaton M ′ = (K ′, Σ, δ′, s′, F ′) where

– K ′ = {[x]M |x ∈ Σ∗}– s′ = [e]M– δ′([x]m, a) = [xa]M– F ′ = {[x]M | (s, x) `∗M (q, e), and q ∈ F}

Let L ⊆ Σ∗ be a regular language.

Define a deterministic finite automaton M =

(K,Σ, δ, s, F ) as follows:

– K = {[x]L | x ∈ Σ∗} where [x]L is the equiva-

lent class of x wrt ≈L– s = [e]L– F = {[x]L | x ∈ L}– δ([x]L, a) = [xa]L

Theorem 10.

L = L(M)

Question Given a deterministic finite automa-

ton M = (K,Σ, δ, s, F ), how to construct a deter-

ministic finite automaton M ′ such that L(M) =

L(M ′) and M ′ has a minimal number of states.

– Define

L′p = {w | ∃f ∈ F : (p, w) `∗M (f, e)}

L′p,n = {w |w ∈ L′p, |w| ≤ n}– For p, q ∈ K, define

p ≡ q

iff L′p = L′q

– For p, q ∈ K, define

p ≡n q iff L′p,n = L′q,n

It is obvious that

≡0⊇≡1⊇ . . . ⊇ ≡n⊇ . . . ⊇≡Because K is finite, there is n such that

≡n = ≡n+1 = ≡

Lemma 2. For any two states p,q and any num-

ber n ≥ 0, p ≡n+1 q iff following conditions

hold:

– p ≡n q– for each a ∈ Σ: δ(p, a) ≡n δ(q, a)

ALGORITHMS FOR FINITE

AUTOMATA

Theorem 11. 1. There is an exponential algo-

rithm which given a nondeterministic finite

automaton, constructs an equivalent determin-

istic finite automaton.

2. There is a polynomial algorithm which given

a deterministic finite automaton, constructs

an equivalent deterministic finite automaton

with a minimal number of states.

3. There is a polynomial algorithm which given

two deterministic finite automata, decides whether

they are equivalent.

�

Theorem 12. Given a deterministic finite au-

tomaton M = (K,Σ, δ, s, F ) and a string w,

there is an algorithm to decide whether w is ac-

cepted by M in O(|w|) time.

�

Theorem 13. Given a nondeterministic finite

automaton M = (K,Σ, δ, s, F ) and a string w,

there is a polynomial algorithm to decide whether

w is accepted by M.

Proof The algorithm is given below:

S0 := E(s); n := 0;

repeat the following

{n := n + 1;

σ := the n-th input symbol;

if σ 6= end-of-file then

Sn :=⋃{E(q) | ∃p ∈ Sn−1 : (p, σ, q) ∈ ∆ }

}until σ = end-of-file

if Sn−1 ∩ F 6= ∅ then accept else reject

The algorithm is polynomial.

�

Context Free Languages

Definition 5. A context-free grammar G is

a quadruple (V,Σ,R, S) where

– V is an alphabet

– Σ (the set of terminals) is a subset of V

– R (the set of rules) is a finite subset of (V −Σ)× V ∗

– S (the start symbol) is an element of V −Σ.

�

– The elements of V − Σ are called nontermi-

nals.

– We write AG−→ u for any rule (A, u) ∈ R.

– For any strings w, v ∈ V ∗, we write

wG

=⇒ v

iff there is a rule AG−→ u in R and w = xAy

and v = xuy.

– A sequence

w0G

=⇒ w1G

=⇒ . . .G

=⇒ wn, n ≥ 0

is called a derivation of wn from w0.

– We write

wG

=⇒∗v

iff there is a derivation of v from w.

The language generated by G is

L(G) = {w ∈ Σ∗ |S G=⇒

∗w }

A context-free language is a language gen-

erated by some context free grammar.

Example A grammar for arithmetic expressions:

– Σ = {x,+, ∗, (, )}– V = Σ ∪ {T, F,E}– R = { E −→ E + T

E −→ T

T −→ T ∗ FT −→ F

F −→ (E)

F −→ x }

Give a derivation for (x), (x + x) ∗ x.

Regular Languages are Context-free

Theorem 14. Any regular language is context

free

Proof Let L be a regular language accepted by a

deterministic finite automatonM = (K,Σ, δ, S, F ).

Construct a grammar G = (V,Σ,R, S) as follows:

– V = K ∪Σ– R consists of rules of the forms• P → aQ where δ(P, a) = Q, and• P → e for P ∈ F

We show by induction on n that

(S, a1 . . . an) `M (Q1, a2 . . . an) `M . . . `M (Qn, e)

iff

SG

=⇒ a1Q1G

=⇒ . . .G

=⇒ a1 . . . anQn

– Basic step: n = 0. Obvious.– Inductive Step: Suppose the assertion holds for

n. We show that it holds for n + 1. Let

(S, a1 . . . anan+1) `M (Q1, a2 . . . an) `M . . . `M (Qn+1, e)

be a computation in M. Hence,

(S, a1 . . . an) `M (Q1, a2 . . . an) `M . . . `M (Qn, e)

is also a computation in M. From induction hy-

pothesis,

SG

=⇒ a1Q1G

=⇒ . . .G

=⇒ a1 . . . anQn

is a derivation wrtG. From (Qn, an+1) `M (Qn+1, e),

it follows δ(Qn, an+1) = Qn+1. Hence Qn →an+1Qn+1 is a rule in R. Therefore a1 . . . anQn

G=⇒

a1 . . . anan+1Qn+1.

The other direction holds obviously.

Hence for any string w ∈ Σ∗, w ∈ L iff

(S,w) `∗M (P, e) and P ∈ F iff SG

=⇒∗wP

and P → e in R iff SG

=⇒∗w.

Parse Trees

Let G = (V,Σ,R, S)

1. For each a ∈ Σ, the tree

◦ a

is a parse tree. The root of this tree is a which

is also its only leaf. The yield of this tree is also a.

2. If A −→ e is a rule in G, then

Fig. 6.

is a parse tree whose root is A, whose only leaf

is e and whose yield is e.

3. If

Fig. 7.

are parse trees (n ≥ 1), with roots A1, . . . , An

and yields y1, . . . , yn respectively, and A −→A1 . . . An is a rule in R then

Fig. 8.

is also a parse tree whose root is A, whose leafs

are the leaves of T1, . . . , Tn and whose yield is

y1 . . . yn.

4. Nothing else is a parse tree

�

Examples Construct parse trees yielding (x), (x+

x) ∗x wrt the grammar for arithmetic expressions.

Lemma 3. Let G = (V,Σ,R, S) be a context-

free grammar, and let A ∈ V −Σ, and w ∈ Σ∗.Then the following statements are equivalent

1. A⇒∗ w2. There is a parse tree with root A and yield w.

– We write xL

=⇒ y if the nonterminal symbol

being replaced is the leftmost nonterminal sym-

bol in the string, i.e

x = wAv, y = wuv where w ∈ Σ∗, v ∈ V ∗,

A→ u ∈ R.

A leftmost derivation is of the form

x1L

=⇒ x2L

=⇒ . . .L

=⇒ xn

– We write xR

=⇒ y if the nonterminal symbol be-

ing replaced is the rightmost nonterminal symbol

in the string, i.e

x = wAv, y = wuv where v ∈ Σ∗, w ∈ V ∗,

A→ u ∈ R.

A rightmost derivation is of the form

x1R

=⇒ x2R

=⇒ . . .R

=⇒ xn

Theorem 15. Let G = (V,Σ,R, S) be a context-

free grammar, and let A ∈ V −Σ, and w ∈ Σ∗.

Then the following statements are equivalent:

1. A⇒∗ w

2. There is a parse tree with root A and yield w

3. There is a leftmost derivation AL

=⇒∗w

4. There is a rightmost derivation AR

=⇒∗w

AMBIGUITY

Let Σ = {x,+, ∗, (, )}. Consider the following

two grammars:

– G1 = (V1, Σ,R1, E)

• V1 = Σ ∪ {T, F,E}

• R1 = { E −→ E + T E −→ TT −→ T ∗ F T −→ FF −→ (E) F −→ x }

– G2 = (V2, Σ,R2, E)

• V2 = Σ ∪ {E}

• R2 = { E −→ E + E E −→ E ∗ EE −→ (E) E −→ x }

G1 is unambiguous in the sense that for each

string w ∈ L(G1), there is exactly one parse tree

that yields w. On the contrary, G2 is ambiguous

as there are distinct parse trees yielding the same

string.

Exercise Could we use the following grammar

as the generator of arithmetic expressions in pro-

gramming languages:

– Σ = {x,+, ∗, (, )}– V = Σ ∪ {T, F,E}– R = { E −→ E ∗ T

E −→ T

T −→ T + F

T −→ F

F −→ (E)

F −→ x }

Pushdown Automata

Definition 6. A pushdown automata is a

sixtuple M = (K, Σ, , Γ, ∆, s, F ) where

– K is a finite set of states

– Σ is an alphabet (the input symbols)

– Γ is an alphabet (the stack symbols)

– s ∈ K is the initial state

– F ⊆ K is the set of final states

– ∆, the transition relation, is a finite subset of

(K × (Σ ∪ {e})× Γ ∗)× (K × Γ ∗)

�

For ease of undertsanding, we often write

(p, a, α)→ (q, β)

for ((p, a, α), (q, β)) ∈ ∆

Let L = {wcwR | w ∈ {a, b}∗ }.

L is accepted by pushdown automaton M =

(K,Σ, Γ,∆, s, F )

– K = {s, p, f}, F = {f}

– Σ = {a, b, c}, Γ = {a, b}

– ∆ consists of the following rules

(s, a, e)→ (p, a) (s, b, e)→ (p, b)

(s, c, e)→ (f, e)

(p, a, e)→ (p, a) (p, b, e)→ (p, b)

(p, c, e)→ (f, e)

(f, a, a)→ (f, e) (f, b, b)→ (f, e)

Fig. 9.

A configuration of a pushdown automata is a

triple

(q, w, v)

where q ∈ K, w ∈ Σ∗ and v ∈ Γ ∗

A configuration (q,w,v) yields another configu-

ration (q′, w′, v′) in one step, denoted by

(q, w, v) `M (q′, w′, v′)

if there is a rule (q, a, α)→ (q′, β) in ∆ such that

w = aw′, v = αv0, v′ = βv0.

We write

(q, w, v) `∗M (q′, w′, v′)

if there is a sequence

(q1, w1, v1) `M . . . `M (qn, wn, vn)

such that (q, w, v) = (q1, w1, v1), (qn, wn, vn) =

(q′, w′, v′).

A string w is said to be accepted by M =

(K, Σ, δ, s, F ) if and only if there exists a state

q ∈ F such that (s, w, e) `∗M (q, e, e)

The language accepted by M, L(M), is the

set of all strings accepted by M.

Theorem 16. A language is context-free iff it

is accepted by a pushdown automaton.

Lemma 4. Each context-free language is accepted

by some pushdown automaton.

Proof Let G = (V,Σ,R, S) be a context-free

grammar. The language generated by G is accepted

by the pushdown automatonM = (K,Σ, Γ,∆, s, F ):

– K = {s, q}, F = {q}– Γ = V

– ∆ consists of the following transitions:

• (s, e, e)→ (q, S)

• (q, e, A)→ (q, x) for each rule A→ x in R

• (q, a, a)→ (q, e) for each a ∈ Σ.

�

Exercise Construct a pushdown automaton accept-

ing the language generated by the following gram-

mar G = (V,Σ,R, S) where

– V = {S, a, b}, Σ = {a, b}– R consists of rules

S → aSa S → bSb, S → e

A pushdown automata M = (K,Σ, Γ,∆, s, F ) is

called simple if following conditions are satisfied:

For each ((q, a, β)(p, γ)) ∈ ∆ such that q 6= s:

β ∈ Γ and |γ| ≤ 2

Claim For every pushdown automaton there ex-

ists an equivalent simple pushdown automata.

Proof Let M = (K,Σ, Γ,∆, s, F ) be a push

down automaton. Construct a simple PA M ′ =

(K ′, Σ, Γ ∪ {Z}, ∆′, s′, {f ′}) as follows:

– Add to ∆ the transitions:

((s′, e, e)(s, Z))

((f, e, Z)(f ′, e)) for each f ∈ F– Replace transitions with |β| ≥ 2.

Let ((q, a, β)(p, γ)) be a transition with β =

B1 . . . Bn with n > 1.

Replace this transition by the following transi-

tions

((q, e, B1)(qB1, e))

((qB1, e, B2)(qB1B2, e))

.....

.....

((qB1B2...Bn−2, e, Bn−1)(qB1B2...Bn−1, e))

((qB1B2...Bn−1, a, Bn)(p, γ))

– Get rid of transition with |γ| > 2

Let ((q, a, β)(p, γ)) be a transition with γ =

C1 . . . Cn with m > 1.

Replace this transition by the following transi-

tions

((q, a, β)(r1, Cm))

((r1, e, e)(r2, Cm−1))

....

....

((rm−2, e, e)(rm−1, C2))

((rm−1, e, e)(p, C1))

– Get rid of transition with β = e

Replace all transitions of the form ((q, a, e)(p, γ))

with q 6= s′ by all transitions of the form ((q, a, A)(p, γA))

Lemma 5. Any language accepted by a simple

pushdown automata is context-free.

Proof LetM = (K,Σ, Γ,∆, s, F ) be a pushdown

automata and M ′ be constructed as above.

Define G = (V,Σ,R, S) as follows:

– V = {S} ∪Σ ∪ K × (Γ ∪ {e})×K– R consists of the following rules:

• S −→ 〈s, Z, f ′〉• for each transtition ((q, a, B), (r, C)) ∈ ∆′

and for each p ∈ K ′:〈q, B, p〉 −→ a〈r, C, p〉• for each transition ((q, a, B), (r, C1C2)) ∈ ∆′,

for all p, p′ ∈ K ′〈q, B, p〉 −→ a〈r, C1, p

′〉〈p′, C2, p〉• 〈q, e, q〉 −→ e for each q ∈ K ′

Properties of Context-freeLanguages

Theorem 17. The context-free languages are closed

under union, concatenation and Kleen star.

Proof Let L, L′ be context-free languages gener-

ated by context-free grammars G = (V,Σ,R, S),

G′ = (V ′, Σ,R′, S ′) where we assume that V −Σand V ′ −Σ are disjoint.

Construct new grammars fromG, G′ to generate

L ∪ L′, L ◦ L′, L∗.�

Theorem 18. The intersection of a context-free

language with a regular language is a context-

free language.

Proof Let L be a context-free language accepted

by pushdown automaton M = (K,Σ, Γ,∆, s, F )

and L′ be a regular language accepted by a deter-

ministic finite automaton M ′ = (K ′, Σ, δ, s′, F ′)

where K,K ′ are disjoint.

Construct a new pushdown automaton fromM, M ′

to accept L ∩ L′.

Theorem 19. The intersection of two context

free languages is not always context-free.

The complementation of a context free lan-

guage is not always context-free.

Proof Let L0 = {ambncn |m,n ≥ 0 }, L1 =

{ambmcn |m,n ≥ 0 }. Both L0, L1 are context free

(exercise).

L = L0 ∩ L1 = {anbncn |n ≥ 0 }. We show

below that L is not context-free.

We use the equation L0∩L1 = L0 ∪ L1 to show

that the complementation of a context free lan-

guage is not always context-free (exercise).

�

Theorem 20. (Pumping Theorem)

Let G = (V,Σ,R, S) be a context-free gram-

mar. Then there is a number n such that for

any string w ∈ L(G) of length greater than n

can be rewritten as w = uvxyz in such a way

that either v or y is nonempty and uvkxykz is

in L(G) for any k.

Proof The fanout of a grammarG = (V,Σ,R, S),

denoted by φ(G) is the largest number ofsymbols

on the right handsid of any rule in G. A path in

the parse tree is a sequence of distinct nodes each

connected to the previous one by a line segment;

the first node is the root and the last node is a

leaf. The length of a path is the number of line

segments in it. The height of the parse tree is the

length of the longest path in it.

Lemma 6. The yield of any parse tree of G

with the height h has length at most φ(G)h.

Lemma 7. Let G = (V,Σ,R, S) be a context-

free grammar. Then any string w ∈ L(G) of

length greater than φ(G)|V−Σ| can be rewritten

as w = uvxyz in such a way that either v or y

is nonempty and uvkxykz is in L(G) for any k.

Algorithmic Properties

Definition 7. A context-free grammar G = (V,Σ,R, S)

is said to be in Chomsky normal form if all rules

are of the form

A→ BC

where A,B,C ∈ V

Theorem 21. There is a polynomial algorithm

which given a context-free grammar G and a

string w, decides whether w ∈ L(G) ?

Proof The proof consists of three main steps:

– A polynomial translation of G into an grammar

G′ in Chomsky normal form such that L(G) −(Σ ∪ {e}) = L(G′).

– A polynomial algorithm to decide whether w ∈L(G′) for |w| ≥ 2.

– A polynomial algorithm to decide whether a ∈L(G) for a ∈ Σ ∪ {e}.

Lemma 8. For every context-free grammar G =

(V,Σ,R, S) there is a context-free grammar G′

in Chomsky normal form such that L(G′) =

L(G)− ({e} ∪Σ).

G′ could be constructed in time polynomial to

the size of G.

Proof

– Replace each rule A→ B1B2 . . . Bn by

A→ B1A1

A1 → B2A2

........................

An−2 → Bn−1Bn

where A1, . . . , An−2 are new nonterminals.

– Construct E = {A |A⇒∗ e} as follows:

• E := ∅• While there is a rule A→ α with α ∈ E∗ and

A 6∈ Edo E := E ∪ {A}.

– Removing e-rules:

• Delete all rules of the form A→ e

• For each rule of the form

A→ BC or A→ CB where B ∈ Eadd the rule A→ C

– Construct for each A, D(A) = {B |A ⇒∗ B}as follows:

• D(A) = {A}• While there is a rule B → C with B ∈ D(A)

and C 6∈ D(A)

do D(A) := D(A) ∪ {C}.

– Removing rules containing only one symbol on

the right hand side:

• Delete all rules of the form A→ B

• Replace each rule of the form A→ BC by all

possible rules of the form A → B′C ′ where

B′ ∈ D(B), C ′ ∈ D(C)

• Add the rule S → BC for each rule A→ BC

such that A ∈ D(S)− {S}.�

Lemma 9. Let G be a grammar in Chomski

normal form. There is a polynomial algorithm

to decide whether w ∈ L(G) for |w| ≥ 2.

Proof Let w = x1 . . . xn. Define

N [i, j] = {A |A⇒∗ xi . . . xj}

The sets N [i, j] could be computed by

– N [i, i] = {xi}

– N [i, j] = {A |A→ BC, B ∈ N [i, s],

C ∈ N [s + 1, j] }

w ∈ L(G) iff w ∈ N [1, n].

�

Example Translate the grammar

S → (S) S → SS S → e

into Chomski normal form

S → SS S → ()

S → (S1 S1 → S)

Check whether the string ( ( ) ( ( ) ) ) is gener-

ated by the Chomski normal form grammar.

Fig. 10.

TURING MACHINES

Definition 8. A Turing machine is a 5-tuple

M = (K, Σ, δ, s,H) where

– K is a finite set of states,

– Σ is an alphabet containing the blank symbol

t and the left end symbol ., but not con-

taining the symbols ← and →

– s ∈ K is the initial state

– H ⊆ K is the set of halting states

– δ, the transition function, is a function from

(K−H)×Σ to K× (Σ ∪{←,→}) such that

• for all q ∈ K −H there exists p such that

δ(q, .) = (p,→)

• for all q ∈ K −H, and a ∈ Σ, if δ(q, a) =

(p, b) then b 6= .

Example Design a Turing machine for accepting

L = {w |w ∈ {a, b}∗, |w| is even }.

M = (K, Σ, δ, s, H) with

– K = {s, q0, q1, y, n}, H = {y, n}

– Σ = {a, b}

– δ : (K −H)×Σ −→ K × (Σ ∪ {←,→})

q σ δ(q, r)

s t q0,→q0 t y,tq0 a or b q1,→q1 t n,tq1 a or b q0,→

M accepts aa:

(s, .taa) `M (q0, . t aa) `M (q1, . t aa) `M(q0, . t aat) `M (y, . t aat)

M rejects a:

(s, .ta) `M (q0, . t a) `M (q1, . t at) `M(n, . t at)

A configuration of a Turing machine M =

(K, Σ, δ, s,H) is a member of

K × .Σ∗ × (Σ∗ × (Σ − {t}) ∪ {e})

Definition 9. Let M = (K, Σ, δ, s,H) and let

(q1, w1a1u1) and (q2, w2a2u2) be configurations of

M. Then

(q1, w1a1u1) `M (q2, w2a2u2)

iff for some b ∈ Σ ∪ {←,→}, δ(q1, a1) = (q2, b)

and either

1. b ∈ Σ, w1 = w2, u1 = u2 and a2 = b

or

2. b =←, w1 = w2a2, and either

(a) u2 = a1u1, if a1 6= t or u1 6= e

or

(b) u2 = e, if a1 = t and u1 = e

or

3. b =→, w2 = w1a1, and either

(a) u1 = a2u2or

(b) u1 = u2 = e and a2 = t

A halted configuration is of the form (h,w1aw2)

with h ∈ H .

A computation is a sequence of configurations

C1, C2, . . . , Cn such that C1 `M C2 `M . . . `MCn. We say that the computation has length n and

C1 yields Cn.

M is said to halt on input w iff (s, .tw) yields

some halted configuration.

COMBINING TURING MACHINES

Symbol writing machine: For each symbol

a ∈ Σ ∪ {←,→} − {.}, Ma is a Turing machine

which write the symbol a into the scanned tape

square.

Ma = ({s, h}, Σ, δ, s, {h}):δ(s, b) = (h, a) for each b ∈ Σ − {.}

Notations:

We often write

– a for Ma

– L for M←– R for M→– Lt: Finds the first blank square to the left of the

currently scanned symbol

– Rt: Finds the first blank square to the right of

the currently scanned symbol

An initial configuration is of the form (s, .tw)

Definition 10. Given is a Turing machine M =

(K,Σ, δ, s,H) with H = {y, n}. Any halting

configuration whose state is y is said to be an

accepting state while a halting configuration whose

state is n is said to be a rejecting state.

We say that M accepts an input w ∈ (Σ −{t, .})∗ if (s, .tw) yields an accepting configu-

ration.

We say that M rejects an input w ∈ (Σ −{t, .})∗ if (s, .tw) yields a rejecting configura-

tion.

Let Σ0 ⊆ Σ − {t, .} be an alphabet, called

input alphabet. We say that M decides a lan-

guage L ⊆ Σ∗0 if for any string w ∈ Σ∗0If w ∈ L then M accepts w

If w 6∈ L then M rejects w

A language L is called recursive if it is de-

cided by some Turing machine.

Example Show that

L0 = {anbn |n ≥ 0 } and L1 = {anbncn |n ≥ 0 }

are recursive.

Fig. 11.

Fig. 12.

Example Eeach deterministic finite automaton could

be ”simulated” by a TM.

Let M = (K,Σ, δ, s, F ) be a DFA.

Define a TM M ′ = (K∪{y, n}, Σ ′, δ′, s′, {y, n})with

Σ ′ = Σ ∪ {.,t} and

δ′ : K ×Σ ′ → (K ∪ {y, n})× (Σ ′ ∪ {←,→})

– δ(s′,t) = (s,→)

– δ′(p, a) = (δ(p),→) for p ∈ K, a ∈ Σ– δ′(p,t) = (y,t) for p ∈ F– δ′(p,t) = (n,t) for p 6∈ F

It holds:

∀w ∈ Σ∗ : w ∈ L(M) iff (s′, .tw) `∗M (y, .wt)

∀w ∈ Σ∗ : w 6∈ L(M) iff (s′, .tw) `∗M (n, .wt)

�

Recursive Functions

Definition 11. Let M = (K, Σ, δ, s, {h}). Let

Σ0 ⊆ (Σ − {t, .} and w ∈ Σ∗0 . Suppose

(s, .tw) `∗M (h, .tu)

Then u is called the output of M on input w

and denoted by M(w).

1. We say that M computes function

f : Σ∗0 → Σ∗0

if for all w ∈ Σ∗0 , M(w) = f (w).

2. We say that M computes function

f : Nk → N, (k ≥ 1)

if for all binary numbers w1, . . . , wk,

M(w; . . . ;wk) = f (w1, . . . , wk)

where f (w1, . . . , wk) is also in binary repre-

sentation.

A function f is recursive if there is a Turing

machine that computes f.

Example Show that g(n) = 2n and f (n) = n + 1

are recursive functions.

Recursively Enumerable Languages

Definition 12. Let M = (K,Σ, δ, s,H) be a

Turing machine. Let Σ0 ⊆ Σ − {t, .} be an

alphabet, and L ⊆ Σ∗0 .

We say that M semidecides L if for any

string w ∈ Σ∗0 :

w ∈ L iff M halts on inputw

A language L is called recursively enumer-

able if it is semidecided by some Turing ma-

chine.

Theorem 22. – If a language is recursive then

it is also recursively enumerable.

– If a language is recursive then its complement

is also recursive.

Extensions of Turing Machines

We study two important extensions of Turing

machines:

1. Allowing several tapes instead of one

2. Allowing nondeterministism

Multi-tape Turing Machines

A k-tape Turing machine consists of a finite

control together with k infinite tapes. Each tape

is scanned by a read/write head. The machine can

in one step sense the symbols scanned by all its

heads and then depending on those symbols and

it s current state, rewrite some of those scanned

squares or move some of the heads to the left or

right.

Definition 13. A k-tape Turing machine is

a 5-tuple M = (K, Σ, δ, s, H) where

– K, Σ, s, H are as in the definition of ordinary

Turing machine

– δ, the transition function, is a function

δ : (K −H)×Σk −→ K × (Σ ∪ {←,→})k

Example Design a two-tapes machine to decide

L = {wcw |w ∈ {a, b}∗ } where c 6∈ {a, b}.

| B | t | w | c | w | t | ...

↑

| B | t | t | ...

↑

q r1 r2 δ(q, r1, r2)

s t t p,→,→p t t n,t,tp σ t p′,t, σp′ t σ p,→,→p c t r2,→,←r2 x σ r2, x,←r2 x t f, x,→f σ σ f,→,→f t t y,t,tf x y n, x, y if x 6= y

σ stands for a or b

x stands for a or b or ty stands for a or b or t

Extending Turing machines with more than one

tape does not increase the power of the machines.

Theorem 23. Let k > 1 and M = (K, Σ, δ, s, H)

be a k-tape Turing machine.

Then there is a standard Turing machine M ′ =

(K ′, Σ ′, δ′, s′, H) where Σ ⊆ Σ ′, and such that

for any w ∈ (Σ − {t, .})∗ :

M halts on input w with output y on its first

tape after t steps if and only if

M ′ halts on input x with the same output after

a polynomial number of steps of t and the size

of x.

Nondeterministic Turing machines

Definition 14. A nondeterministic Turing

machine is a quintuple M = (K, Σ, ∆, s, H)

where

– K, Σ, s, H are as for standard Turing ma-

chines, and

– ∆, the transition relation, is a subset of

((K −H)×Σ)× (K × (Σ ∪ {←,→}) )

Definition 15. Let M = (K, Σ, ∆, s) and let

(q1, w1a1u1) and (q2, w2a2u2) be configurations of

M. Then

(q1, w1a1u1) `M (q2, w2a2u2)

iff for some b ∈ Σ∪{←,→}, ((q1, a1), (q2, b)) ∈∆ and

1. either b ∈ Σ, w1 = w2, u1 = u2 and a2 = b

2. or b =←, w1 = w2a2, and

(a) either u2 = a1u1, if a1 6= t or u1 6= e

(b) or u2 = e, if a1 = t and u1 = e

3. or b =→, w2 = w1a1, and

(a) either u1 = a2u2(b) or u1 = u2 = e and a2 = t

Definition 16. Let M = (K,Σ, δ, s,H) be a

nondeterministic Turing machine.

We say that M accepts w ∈ (Σ − {.,t})∗ if

(s, .tw) `∗M (h, uav) for some h ∈ H, a ∈ Σ,

u, v ∈ Σ∗.We say that M semidecides L ⊆ (Σ−{.,t})∗

if for any string w ∈ (Σ − {.,t})∗: w ∈ L iff

M halts on input w.

Definition 17. Let M = (K,Σ, δ, s, {y, n}) be

a nondeterministic Turing machine.

We say that M decides L ⊆ (Σ − {.,t})∗if the following conditions hold for any string

w ∈ (Σ − {.,t})∗:

1. There is a number N depending on M and w

such that no computation starting from (s, .tw)

has a length greater than N

2. w ∈ L iff (s, .tw) `∗M (y, uav) for some

a ∈ Σ, u, v ∈ Σ∗.

We say that M computes a function f : (Σ−{.,t})∗ → (Σ−{.,t})∗ if the following condi-

tions hold for any string w ∈ (Σ − {.,t})∗:

1. There is a number N depending on M and w

such that no computation starting from (s, .tw)

has a length greater than N

2. (s, .tw) `∗M (h, uav) iff ua = .t, v = f (w).

Theorem 24. If a nondeterministic Turing ma-

chine semidecides or decides a language or com-

putes a function then there is a standard one

that semidecides or decides the same language

or computes the same function.

Unrestricted Grammars

Definition 18. A unrestricted grammar G

is a quadruple (V,Σ,R, S) where

– V is an alphabet

– Σ (the set of terminals) is a subset of V

– R (the set of rules) is a finite subset of

V ∗(V −Σ)V ∗ × V ∗– S (the start symbol) is an element of V −Σ.

�

– We write uG−→ v for any rule (u, v) ∈ R.

– For any strings w,w′ ∈ V ∗, we write

wG

=⇒ w′

iff there is a rule uG−→ v and w = xuy and

w′ = xvy.

– A sequence

w0G

=⇒ w1G

=⇒ . . .G

=⇒ wn, n ≥ 0

is called a derivation of wn from w0.

– We write

wG

=⇒∗v

iff there is a derivation of v from w.

The language generated by G is

L(G) = {w ∈ Σ∗ |S G=⇒

∗w }

Example Design an unrestricted grammar for L =

{anbncn |n ≥ 1 }

G = (V,Σ,R, S) where

– V = {S,A,B,C, Ta, Tb, Tc, a, b, c}– Σ = {a, b, c}

– R consists of the following rules:

S → ABCS S → Tc

CA→ AC BA→ AB CB → BC

CTc → Tcc

BTc → Tbb BTb → Tbb

ATb → Taa ATa → Taa

Ta → e

�

Theorem 25. A language is generated by an

unrestricted grammar if and only if it is recur-

sively enumerable.

UNDECIDABILITY

The Church - Turing Thesis

An algorithm is anything that can be viewed

as corresponding to a Turing machine that halts

on all inputs.

Nothing is considered to be an algorithm if it

cannot be rendered as a Turing machine that is

guaranteed to halt on all inputs.

Universal Turing Machine

Let M = (K,Σ, δ, s,H) be a Turing machine. M

is now encoded over the alphabet {a, q, 0, 1} as

follows:

– Let i, j be smallest integer such that 2i ≥ |K|and 2j ≥ |Σ| + 2

– each state in K is represented by a string from

q{0, 1}i– each symbol in Σ is represented by a string from

a{0, 1}j– The representation of the special symbols is as

follows:

t a0j

. a0j−11

← a0j−210

→ a0j−211

– The start state is represented by q0i

The representation of a Turing machine M,

denoted by ”M”, consists of a sequence of strings of

the form (p,b,r,c) with p,r representations of states

and b,c representations of symbols.

Example Let M = (K,Σ, δ, {h}) be a TM where

K = {s, p, q, h}, Σ = {0, 1,t, .} and

δ(s,t) = (p,→),

δ(p, 0) = (p,→), δ(p, 1) = (p,→),

δ(p,t) = (q, 0),

δ(q, 0) = (q,←), δ(q, 1) = (q,←),

δ(q,t) = (h,t).

s q00

p q01

q q10

h q11

t a000

B a001

← a010

→ a011

0 a100

1 a101

The representation ”M” is the following string:

(q00, a000, q01, a011)(q01, a100, q01, a011)(q01, a101, q01, a011)

(q01, a000, q10, a100)(q10, a100, q10, a011)(q10, a101, q10, a011)

(q10, a000, q11, a000)

The universal Turing machine U is a ma-

chine which uses the encodings of other machine to

direct its operations.

Intuitively U takes two arguments: a description

of a Turing machine M, ”M”, and a description of

an input string w, ”w”.

U has the following property:

U halts on ”M””w” iff M halts on w

U is constructed in two steps:

1. A three-tape universal machineU ′ is constructed.

2. Transform U ′ into an equivalent single tape ma-

chine U .

U ′ simulates a TM M as follows:

1. The first tape contains the encoding of the tape

of M

2. The second tape contains the encoding of M

3. The third tape contains the encoding of the state

of M at the current point in the simulated com-

putation

U ′ starts with the string ”M””w” on its first tape

and the other tapes blank.

– ”M” is moved onto the second tape, and

– ”w” is shifted down to the left end of the first

tape, preceding it by .t

– U ′ extracts the coding of the initial state of M

and puts it on the third tape.

The halting problem

Halting Problem:

Given an arbitrary Turing machine M and an

input w, is there an algorithm which can

decide whether M accepts w ?

Let H = {”M””w” | Turing machine M accepts

input string w }.

From the Church-Turing thesis,

”Yes” answer to the halting problem iff the lan-

guage H is recursive.

It is clear that the universal Turing machine ac-

cepts H . Hence H recursively enumerable.

Theorem 26. H is not recursive.

Proof Suppose H is recursive. Let

H1 = {w |w = ”M” for some TM M and M accepts w }

Because H is recursive, H1 is also recursive. Let

ΣU = {a, q, 0, 1, (, ), , } be the alphabet of the uni-

versal TM.

Define R ⊆ Σ∗U ×Σ∗U by

(u,w) ∈ R iff u = ”M” for some TM machine M accepting w

For each string u, Ru = {w | (u,w) ∈ R }

Therefore, for each language L,

L is r.e. iff ∃u = ”M” : L = Ru = L(M).

Let D = {w | (w,w) 6∈ R }

From the diagonalization principle, there is no u

such that D = Ru. Therefore D is not r.e.

It is not difficult (exercise) to see that

H1 = {w | (w,w) ∈ R } = D

Since H1 is recursive and D = H1, D is also

recursive, hence r.e. Contradiction because we have

showed that D is not r.e. Hence, the assumption

that H is recursive leads to a contradiction. H is

hence not recursive. �

Undecidable Problems

Since H is not recursive, there is no algorithm to

decide whether a TM M accepts an input string w.

We say

the halting problem is undecidable.

Definition 19. Let L1, L2 ⊆ Σ∗ be languages.

A reduction from L1 to L2 is a recursive

function τ : Σ∗ → Σ∗ such that x ∈ L1 iff

τ (x) ∈ L2.

Theorem 27. If L1 is not recursive and there

is a reduction from L1 to L2 then L2 is not re-

cursive.

Theorem 28. The following problems are un-

decidable:

1. Given a Turing machine M, does M hold on

the empty tape ?

2. Given a Turing machine M, is the language

accepted by M empty ?

3. Given a Turing machine M, does M accepts

every input ?

4. Given Turing machines M1,M2, do they ac-

cept the same language ?

5. Given a Turing machine M, Is the language

accepted by M regular ? Is it context-free ? Is

it decidable ?

Proof

1. Let He = {”M” | M hold on the empty tape}.We give a reduction τe from H to He. Let w =

a1 . . . an. Define

τe(”M””w”) = ”M ′”

where M ′ = Ra1Ra2 . . . RanLtM .

It is clear that M accepts w iff M ′ accepts the

empty tape. From the reduction theorem, He is

not recursive. Therefore, the problem is undecid-

able.

2. Let Hi = {”M” |M halts on some input}.We reduce He to Hi. Let M be a TM.

Fig. 13.

Let τi(”M”) = ”M ′”. It is clear that if M halts

on empty tape then M ′ halts on all inputs (and

hence some input).

Hi is not recursive.

3. Similar to the previous case.

4. Let Ha = {”M” | M accepts all inputs}. Ha is

not recursive.

Let Heq = {”M”M ′” |M,M ′ accept the same

language}.Let u be a string representing a TM that accepts

all inputs in the first step, i.e. δ(s,t) = (h,t).

Define

τa(”M”) = ”M”u

It is clear that ”M” ∈ Ha iff ”M”u ∈ Heq.

Since Ha is not recursive, Heq is not recursive.

5. See Rice theorem

Theorem 29. (Rice’s Theorem)

Suppose that C is a proper nonempty subset

of the class of all recursively enumerable lan-

guages. The the following question is undecid-

able: Given a Turing machine M, is L(M) ∈ C

Proof Let H ′ = {M |L(M) ∈ C }. We reduce H

to H ′. Let L ∈ C and ML that accepts L.

Without loss of generality, we could assume that

∅ 6∈ C.

Let ”M””w” ∈ H . Let τ (”M””w”) = ”M ′”

where M ′ is a TM rendering the following algo-

rithm:

If U(”M””w”) halts then ML(x)

where x is the input of M ′.

It is clear that if M accepts w then M ′ accepts

a language in C. Hence, τ is a reduction from H to

H ′. Therefore, H ′ is not recursive.

Computational Complexity

Example Consider the problem of a travelling sale

man who has to visit 10 offices in 10 cities. The sale

man has a map with distances between the cities

and he should find an itinerary with the shortest

distance.

A simple algorithm would check all possible itineraries.

The number would be 9! = 362880. If the number

of cities is say 40, no presently forseeable computer

could handle it as it would take billions of years to

finish.

What represents a practically feasible algorithm ?

The Class P

Definition 20. A Turing machine M = (K,Σ, δ, s,H)

is said to be polynomially bounded if there

is a polynomial p(n) such that for any input x,

there is no configuration C such that

(s, .tx) `p(|x|)+1M C

A language is said polynomially decidable

if there is a polynomially bounded Turing ma-

chine that decides it.

The class of all polynomial decidable languages

is denoted by P.

Theorem 30. P is closed under complementa-

tion

Some Well-Known Problems

Reachability Problem

Given a directed graph G ⊆ V × V where V =

{v1, . . . , vn} and two nodes vi, vj, is there a path

from vi to vj ?

Independent Set

Given an undirected graph G and an integer

K ≥ 2, is there a set of nodes C with |C| ≥ K

such that for all vi, vj ∈ C, there is no edge be-

tween vi and vj ?

Boolean Satisfiability

X = {x1, . . . , xn} : a finite set of boolean vari-

ables and

X = {x1, . . . , xn} where xi is the negation of

xi.

The elements of X ∪X are called literals; vari-

ables are positive literals, whereas negations of

variables are negative literals.

A clause is a set of literals. A Boolean for-

mula is a set of clauses.

A truth assignment T is a mapping from X

into {>,⊥} where > and ⊥ stand for true and

false respectively.

We say that a truth assignment T satisfies a

Boolean formula F if for each clause C of F there

is at least one variable xi such that xi ∈ C and

T (xi) = > or xi ∈ C and T (xi) = ⊥.

F is satisfiable if there is a truth assignment

that satisfies F.

Satisfiability: Given a Boolean formula F, is F

satisfiable ?

3-Satisfiability: Given a Boolean formula F whose

clause contains only three or fewer literals, is F sat-

isfiable ?

2-Satisfiability: Given a Boolean formula F whose

clause contains only two or fewer literals, is F sat-

isfiable ?

Reachability is in P .

We have given before an polynomial algorithm

for computing the set of all pairs (v, u) in a graph

such that there is a path from v to u. Hence reach-

ability is in P .

2-satisfiability is in P .

A 2-satisfiability problem could be reduced to

reachability by presenting any clause x ∨ y as a

link from node x to y and from y to x.

An instance of 2-satisfiability is unsatisfiable if

there is a variable x such that there is a path from

x to x and vice versa (exercise: by mathematical

induction).

Are there practical problems that are recursive

but non-polynomially decidable ?

The Class NP

Definition 21. A nondeterministic Turing ma-

chine M = (K,Σ,∆, s,H) is said to be poly-

nomially bounded if there is a polynomial

p(n) such that for any input x, there is no con-

figuration C such that

(s, .tx) `p(|x|)+1M C

A language is said to be nondeterministic

polynomially decidable if there is a nonde-

terministic polynomially bounded Turing machine

that decides it.

The class of all nondeterministic polynomial

decidable languages is denoted by NP.

The Boolean satisfiability, 3-satisfiability, indepen-

dent set problems are all in NP .

Theorem 31. P ⊆ NP

Open Problem P = NP ?

NP - Completeness

Definition 22. – A function τ : Σ∗ → Σ∗ is

said to be polynomial-time computable

if there is a polynomial-bounded Turing ma-

chine that computes it.

– Let L1, L2 ⊆ Σ∗ be languages. A polynomial

reduction from L1 to L2 is a polynomial-

time computable function τ : Σ∗ → Σ∗ such

that x ∈ L1 iff τ (x) ∈ L2.

Definition 23. A language L is said to be NP-

complete if and only if:

1. L ∈ NP2. For every language L′ ∈ NP, there is a polynomial-

time transformation from L′ to L.

Theorem 32. Let L be a NP-complete language.

Then P = NP iff L ∈ P

Theorem 33. Satisfiablity is NP-complete.

Theorem 34. 3-satisfiability is NP-complete.

Proof We give a polynomial reduction from satis-

fiability to 3-satisfiability.

Let C ≡ (x1 ∨ x2 ∨ . . . ∨ xk).Let y1, . . . , yk−3 be new Boolean variables. C is

reduced to following set of 3-clauses:

(x1∨x2∨y1), (y1∨x3∨y2), (y2∨x4∨y3), . . . ,(yk−3 ∨ xk−1 ∨ xk)

τ (F ) is the collections of all 3-clauses obtained

by transforming clauses in F.

F is satisfiable iff τ (F ) is satisfiable.

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Theorem 35. Independent set problem is NP-

complete.

Proof We reduce 3-satisfiability to independent

set problem where K = m where m is the number

of clauses.

For each clause, there are three nodes labelled

by the literals in the clause. Further, two nodes

labelled by complementary literals are linked.

For example:

(x1∨x2∨x3) (x1∨x2∨x3) (x1∨x2∨x3) (x1∨x2 ∨ x3)

Fig. 14.

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