Module 1 TOC.pptx

BCSE304L
Theory of Computation
Dr. WI. Sureshkumar
Associate Professor
School of Computer Science and Engineering (SCOPE)
VIT Chennai
wi.sureshkumar@vit.ac.in
6- DELTA – 2nd Floor

Objective
• Gain a historical perspective of formal languages and automata
theory and its foundations.
• Become familiar with basic principles of Chomsky grammar and its
hierarchy.
• Explain the basic knowledge automata theory.
• Associating formal languages theory with automata theory.
• Discuss Turing machines as an abstract computational model
• The course should in addition clarify the practical view towards the
applications of these ideas in the engineering part of CS.

Text books
• Introduction to Automata Theory, Languages and Computation,
Hopcroft, Motwani, and Ulmann, Pearson publ, 2006.
• An Introduction to Formal Language and Automata,
Peter Linz, Third Edition, Narosa Publishers, New Delhi, 2002.
• Automata and Computability, Dexter C. Kozen, Springer publ, 2007.
• Introduction to Formal Languages, Automata and Computation,
K. Krithivasan and R. Rama, Pearson publ, 2009.
• Introduction of the Theory and Computation, Micheal Sipser,
Thomson Brokecole, 1997.
• Elements of Theory of Computation, H. R. Lewis and C. H.
Papadimitriou, Printice Hall publ, 1981
• Introduction to Languages and Theory of Computation, John C.
Martin, Tata-MaGraw Hill, 2003

Module -1
• Introduction to Languages and Grammars
• Recall on Proof techniques in Mathematics
• Overview of a Computational Models
• Languages and Grammars
• Alphabets
• Strings
• Operations on Languages
• Overview on Automata

Method of Evaluation
QUIZ-1 5 Marks (Moddle (LMS))
DA 10 Marks (Presentation)
CAT - 1 15 Marks (Closed Book)
CAT - 2 15 Marks (Open Book)
FAT 40 Marks
Total 100 Marks

Overview of a computational model
• Computation
Program running in computer
Input x Algorithm f(x) Output
Y
Input w M
N

Alphabet, strings and operations on strings
• Noam Chomsky, 1959 - Formal definition of grammar
• Languages – finite or infinite set of strings
• Machines – Abstract machines used to recognize the string of a language
• An alphabet is a finite, non-empty set of symbols.
• {0,1} is a binary alphabet.
• {A, B, …, Z, a, b, …, z} is an English alphabet.
• A string over an alphabet Σ is a finite sequence of symbols from Σ.
• 0, 1, 11, 00, and 01101 are strings over {0, 1 }.
• Cat, CAT, and compute are strings over the English alphabet.

Empty String
• An empty string (or null string), denoted by ε, is a string
containing no symbol.
ε is a string over any alphabet.

Length of a String
• The length of a string x, denoted by length(x) or |x|, is the
number of symbols in the string.
Let Σ = {a, b, …, z}.
Let w1= abcd and w2 = fgh
Then,
length(w1) = 4 or |w1 | = 4
length(w2) = 3 or |w2 | = 3
length(ε) = 0 or | ε | = 0
• x(i) or xi , denotes the symbol in the i th position of a string x,
for 1 ≤ i ≤ length(x).

String Operations
• Concatenation
• Substring
• Reversal
• Kleene Closure and Positive Closure

Concatenation
• Concatenation of two strings w1 and w2 is defined as the sequence of symbols in w
followed by sequence of symbols in w2
• Let w1= abcd and w2 = fgh
Then
w1w2 = abcdfgh
w2w1 = fghabcd
The concatenation of string x for n times,where n ≥ 0,is denoted by xn
• x0 = ε
• x1 = x
• x2 = x x
• x3 = x x x
• …

Substring
Let x and y be strings over an alphabet Σ
The string x is a substring of y if there exist strings w and z over Σ
such that y = w x z.
• ε is a substring of every string.
• For every string x, x is a substring of x itself.
Example
• ε, comput and computation are substrings of computation.

Reversal
• The reversal of a string is a string with same symbols in reverse order,
denoted by wR
Example
Let w = abcdfgh
Then
wR = hgfdcba

Kleene Closure and Positive Closure
Let ∑ ={ a, b} i.e. ∑1
={ a, b}
∑2 = ∑1 . ∑1 = { a, b}. { a, b} = {aa, ab, ba, bb}
∑3= ∑1 . ∑1 . ∑1 = {a,b}.{a,b}.{a,b}={aa,ab,ba,bb}.
{a,b}
={aaa, aab, aba, abb, baa, bab, bba, bbb}
… etc
∑0 ={ε}
The Kleene closure of ∑, denoted by ∑* is defined as follows:
∑* = ∑0 U ∑1 U ∑2 U ∑3 …….. = {ε, a, b, aa, ab, ba, bb, aaa,….}
That is, ∑* = i=

0 ∑i
The Positive closure of , denoted by ∑+ is defined as follows:
∑+ = ∑* { ε } or ∑+ = i

= 1 ∑i

Operations on Languages
• Complementation
• Union
• Intersection
• Concatenation
• Reversal
• Closure

Complementation
Let L be a language over an alphabet Σ.
The complementation of L, denoted by L ( or L’), is Σ*– L.
Example:
Let Σ = {0, 1} be the alphabet.
L = {Σ* | the number of 1’s in  is even}.
L’ = {Σ* | the number of 1’s in  is not even}.
L’ = {Σ* | the number of 1’s in  is odd}.

Union
Let L1 and L2 be languages over an alphabet Σ.
The union of L1 and L2, denoted by L1L2, is
L1L2 = {x | x is in L1 or L2}.
Example:
{ x {0,1}* | x begins with 0 }  {x{0,1}*|x ends with 0}
= {x  {0,1}*| x begins or ends with 0}

Intersection
The intersection of L1 and L2, denoted by L1L2, is { x | x
is in L1 and L2}.
Example:
{ x{0,1}* | x begins with 0}  { x{0,1}*| x ends with 0}
= { x{0,1}*| x begins and ends with 0}

Concatenation
The concatenation of L1 and L2, denoted by L1L2, is {w1w2| w1 is in
L1 and w2 is in L2}.
Example
L1 = { x  {0,1}*| x begins with 0} L2 ={x  {0,1}*| x ends with 0}
L1L2 = { x  {0,1}*| x begins and ends with 0 and length(x)  2}
L1 = { x  {0,1}*| x ends with 0} L2 = {x  {0,1}*| x begins with 0}
L1L2 = = { x  {0,1}*| x has 00 as a substring}

Reversal
The reversal of L, denoted by LR, is {wR | w is in L}.
Example
L = {x  {0,1}*| x begins with 0}
LR = {x  {0,1}*| x ends with 0}
L = {x  {0,1}*| x has 00 as a substring}
LR = {x  {0,1}*| x has 00 as a substring}

Kleene’s closure
The Kleene’s closure of L, denoted by L*, is {x | for an integer n  0 x =
x1 x2 … xn and x1, x2 , …, xn are in L}.
That is, L* = i

= 0 Li
Example: Let Σ = {0,1} and
Le = {Σ* | the number of 1’s in  is even}
Le* = {Σ* | the number of 1’s in  is even}
(Le)*= {Σ*| the number of 1’s in  is odd}*
= {Σ*| the number of 1’s in  > 0}

Grammars and Languages
Napoleon
S
NP1 VP
PN VER ART ADJ NP2
is a great warrior
Derivation Tree
Syntax Tree
or
or
Parse Tree

Derivation
<S> → <NP1> <VP>
<NP1> → <PN>
<PN> → Napoleon
<VP> → <VER> <ART> <ADV> < NP2>
<VER> → is
<ART> → a
<ADV> → great
< NP2> → warrior
Rules
or
Productions
or
Production rules
→ - rewritten as
- directly derives



Derivation
<S> <NP1> <VP> <PN> <VP> Napoleon <VP>
Napoleon <VER> <ART> <ADV> < NP2>
Napoleon is <ART> <ADV> < NP2>
Napoleon is a <ADV> < NP2>
Napoleon is a great < NP2>
Napoleon is a great warrior
Leftmost Derivation
  






<S> <NP1> <VP>
<NP1><VER> <ART> <ADV> < NP2>
<NP1><VER> <ART> <ADV> warrior
<NP1><VER> <ART> great warrior
<NP1><VER> a great warrior
<NP1> is a great warrior
Napoleon is a great warrior
Rightmost Derivation








Derivation
• Non terminals - N
• Terminals – T
• Total alphabet - V = N  T
are strings in
is obtained from in one step
is obtained from in zero or more steps
reflexive transitive closure of
*
, V


 *
V

   

 *
  
*
 

Grammar
Definition: A grammar is a 4-tuple G = (N, T, P, S), where
• N is a finite set of symbols called non-terminals (uppercase alph)
• T is a finite set of symbols called non-terminals (lowercase alph)
• S N is the start symbol
• P is the set of productions of the form
Note:


 


T
N 

Grammar (Examples)
G = ({S, A}, {a, b}, P, S), where
P: S → aA
A → b
S
a A
b
L(G) = { ab }

Grammar
G = ({S}, {a, b}, P, S), where
P: S → aS
S → b S
b
S
a S
b
S
a S
b
a S
S
a S
b
a S
a S
L(G) = { b, ab, a2b, a3b,. . . }
L(G) = { anb / n ≥ 0 }

Grammar (Examples)
id
id
id
E
id
id
E
E
id
E
id
E
E
E *
*
*
4
4
2
4
1










E
id
G= ({E}, {+, *, (, ), id }, P, E), where
P: E → E + E (rule 1)
E → E * E (rule 2)
E → (E) (rule 3)
E → id (rule 4)
Leftmost derivation
Rightmost derivation
id
E
E E
id
*
id
id
id
id
id
E
id
E
E
E
E
E
E
E
E *
*
*
*
4
4
4
2
1










+
id
E
E E
id

Grammar
Let G = (N, T, P, S) be a grammar.
Then L(G) is called the language generated by G.
L(G) = { }
• Grammars are language generating device
• Automata are language accepting device
w
S
T
w
*
*
/ 


Classification of Grammars
• RE Type – 0 Unrestricted or Phrase structured grammar
• CS Type – 1 Context-sensitive or length increasing grammar
|u|≤|v|
• CF Type – 2 Context-free grammar
• REG Type – 3 Regular grammar
or
v
u  *
*
NV
V
u


 
A
*
, V


 N
A

V



A *
V

 N
A
aB
A  b
A  N
B
A 
, T
a }
{

T
b

Chomsky Hierarchy
Language Grammar Machine Example
Type 3 Regular languages
Regular grammars
• Right-linear grammars
• Left-linear grammars
Finite-state
automata
a*
Type 2 Context-free languages Context-free grammars
Push-down
automata
anbn
Type 1 Context-sensitive languages Context-sensitive grammars
Linear-bound
automata
anbncn
Type 0
Recursive languages
Recursively enumerable
languages
Unrestricted grammars
Turing
machines
any
computable
function

Chomsky Hierarchy
Non-recursively enumerable
Recursively-enumerable
Recursive
Context-sensitive
Context-free
Regular

1) G= ({S}, {a}, P, S), where
P: S → aS (rule 1) S a
S → a (rule 2) S aS aa
S aS aaS aaa
L(G) = { an / n ≥ 1 }
2) G = ({S}, {a, b}, P, S), where
P: S → aS (rule 1) S b
S → b (rule 2) S aS ab
S aS aaS aab
L(G) = { anb/ n ≥ 0 }

 
  

 
  

3) G= ({S}, {a, b }, P, S), where
P: S → aSb (rule 1) S ab
S → ab (rule 2) S aSb aabb
S aSb aaSbb aaabbb
4) G= ({S}, {a, b, c }, P, S), where
P: S → aSa (rule 1) S c
S → bSb (rule 2) S aSa aca
S → c (rule 3) S bSb bcb
S aSa abSba abcba
S bSb baSab bacab
L(G) = { anbn / n ≥ 1 }
L(G) = { wcwR/ w ∈ {a, b}* }

 
  

 
 
  
  

5) G= ({S, B}, {a, b, c }, P, S), where
P: S → aSBc (rule 1) S abc
S → abc (rule 2) S aSBc aabcBc aabBcc aabbcc
cB → Bc (rule 3)
bB →bb (rule 4) S aSBc aaSBcBc aaabcBcBc
aaabBccBc aaabBcBcc aaabBBccc
aaabbBccc aaabbbccc
L(G) = { anbnCn/ n ≥ 1 }
  

 





 

6) G= ({S, A, B}, {a, b}, P, S), where
P: S → AB (rule 1) S AB aAbB aaB aaε = aa
A → aAb (rule 2) S AB aAbB aAbbbB aAbbbε
bB → bbbB (rule 3) aabb
aAb →aa (rule 4) S AB aAbB aAbbbB aAbbbε
B →ε (rule 5) aaAbbbb
aaabbb
L(G) = { an+1bn+k/ n ≥ 1, k = -1, 1, 3, 5, . . .}
   
   

   



7) Grammar for FORTRAN identifier
Length – 6 First symbol is the letter
Letter = {A, B, C, …, Z}
Digit = {0, 1, …,9}
S → (A / B / C /…./ Z) S1
S1 →ε
S1 → (A / B / C /…./ Z / 0 / 1 / 2 / …./ 9) S2
S2 → ε
S2 → (A / B / C /…./ Z / 0 / 1 / 2 / …./ 9) S3

S3 →ε
S3 → (A / B / C /…./ Z / 0 / 1 / 2 / …./ 9) S4
S4 →ε
S4 → (A / B / C /…./ Z / 0 / 1 / 2 / …./ 9) S5
S5→ε
S5 → (A / B / C /…./ Z / 0 / 1 / 2 / …./ 9) S6
S6→ε
RADIUS
RADIUSS
RADIUS
RADIS
RADS
RAS
RS
S







6
5
4
3
2
1
2
2
2
1 A
S
A
AS
S 



Problems
Construct the grammar for the following languages:
1) L(G) = { ancan / n ≥ 0 }
2) L(G) = { anb2n / n ≥ 0 }
3) L(G) = { an+2bn / n ≥ 1 }
4) L(G) = { anbn-3 / n ≥ 3 }
5) L(G) = { anbm / n ≥ 0, m ˃ n }
6) L(G) = { anbn cm / n, m ≥ 1 }
7) L(G) = { anbn cm dm / n, m ≥ 1 }
8) L(G) = { anbm cm dn / n, m ≥ 1 }
9) L(G) = { anbm / n, m ≥ 1 m ≠ n}
10) L(G) = { w {a, b}* / na(w) = nb(w) + 1 }

1) L(G) = { ancan / n ≥ 0 }
G= ({S}, {a, c }, P, S), where
P: S → aSa
S → c
2) L(G) = { anb2n / n ≥ 0 }
G= ({S}, {a, b }, P, S), where
P: S → aSbb
S → 
3) L(G) = { an+2bn / n ≥ 1 }
G= ({S}, {a, b }, P, S), where
P: S → aSb
S → aaab

4) L(G) = { anbn-3 / n ≥ 3 }
G= ({S}, {a, b }, P, S), where
P: S → aSb
S → aaa
5) L(G) = { anbm / n ≥ 0, m ˃ n }
G= ({S, S1}, {a, b }, P, S), where
P: S → aSb / S1
S1 → bS1 / b

6) L(G) = { anbn cm / n, m ≥ 1 }
G= ({S, S1, S2}, {a, b, c }, P, S), where
P: S → S1S2
S1 → aS1b / ab
S2 → cS2 / c
7) L(G) = { anbn cm dm / n, m ≥ 1 }
G= ({S, S1, S2}, {a, b, c, d}, P, S), where
P: S → S1S2
S1 → aS1b / ab
S2 → cS2d/ cd

8) L(G) = { anbm cm dn / n, m ≥ 1 }
G= ({S, S1}, {a, b, c, d}, P, S), where
P: S → aSd / aS1d
S1 → bS1c / bc
9) L(G) = { anbm / n, m ≥ 1 , m ≠ n}
G= ({S, A, B, S1 }, {a, b}, P, S), where
P: S → AS1 / S1B
S1 → aS1b / ab
A → aA / a
B → bB / b

10) L(G) = { w / w =uav, u, v  {a, b}*, ui ≠ vn-i+1 ,1≤ i ≤ |u|, na(w) = nb(w) + 1 }
G= ({S}, {a, b}, P, S), where
P: S → aSb / bSa / a
11) Find grammars for the following languages on Σ = { a }.
a) L ={ w / |w| mod 3 = 0}
b) L ={ w / |w| mod 3 ˃ 0}
c) L ={ w / |w| mod 3 ≠ |w| mod 2}
d) L ={ w / |w| mod 3 ≥ |w| mod 2}

a) L ={ w / |w| mod 3 = 0}
w ε a3 a6 a9 a12 a15 a18
|w| 0 3 6 9 12 15 18
P: S → aaaS /ε
b) L ={ w / |w| mod 3 ˃ 0}
w a a2 a4 a5 a7 a8 a10 a11
|w| 1 2 4 5 7 8 10 11
P: S → a / aa
S → aaaS

c) L ={ w / |w| mod 3 ≠ |w| mod 2}
w ε a a2 a3 a4 a5 a6 a7
|w| mod 3 0 1 2 0 1 2 0 1
|w| mod 2 0 1 0 1 0 1 0 1
L ={a2 , a3 , a4 , a5 , a8, a9 , a10 , a11, a14 , a15 , a16 , a17 , . . . }
P: S → a2
S → a3
S → a4
S → a5
S → a6 S

d) L ={ w / |w| mod 3 ≥ |w| mod 2}
w ε a a2 a3 a4 a5 a6 a7 a8 a9
|w| mod 3 0 1 2 0 1 2 0 1 2 0
|w| mod 2 0 1 0 1 0 1 0 1 0 1
L ={ε , a ,a2 , a4 , a5, a6, a7 , a8 , a10 , a11 , a12 , a13, a14 , a16, a17 , a18 , a19 , a20 , . . .
}
P: S → ε / S1
S → a
S → a2
S1 → a4 / a5 / a6 / a7 / a8
S1 → a6 S1

Proof Techniques
A proof involves a statement of the form p → q
There are several methods for establishing a proof of statements.
Some of them are
1) Direct proof
2) By contradiction
3) By mathematical induction.

Direct proof
If we have to prove that p → q, then a direct proof assumes p is true
and uses this to show that q is true.
Example: 1
Prove for any integer a and b if a and b are odd then ab is odd.
Solution: Any odd integer u, can be written as 2v+1, where v is an
Integer.
Given that a and b are odd integer, by the above statement
a = 2x + 1 and b = 2y + 1 , where x, y are integers.
Use this fact and prove the product ab is also odd.
ab = (2x +1) (2y+1) = 4xy + 2x + 2y + 1 = 2(2xy + x + y) + 1 = 2w+1
where w = 2xy + x + y is an integer.
Hence, ab is an odd number.

Proof by contradiction
If we have to prove that p → q, then the method of contradiction
assumes p does not imply q and then try to derive some contradiction.
Example: 1
Prove for any integer a and b if a and b are odd then ab is odd.
Solution: To prove this, assume the contrary that ab is even.
ab is even implies, ab = 2z , for some integer z.
Given that a and b are odd integer, by the above statement
a = 2x + 1 and b = 2y + 1 , where x, y are integers.
Therefore, ab = 4xy + 2x + 2y + 1 = 2z
z = 2xy + x + y +(½)
z is not an integer and hence a contradiction. ab is an odd number.

Example: 2
A rational number is a number that can be expressed as the ratio of two
integers n and m have no common factor. A real number that is not
rational is said to be irrational. Show that 2 is irrational.
Solution: Assume the contrary that 2 is rational.
2 is rational implies, 2 = n/m ---------------- (1)
where n and m are integers and have no common factor.
From (1), 2m2 = n2 ---------------- (2)
implies n2 is even and hence n is even.
Therefore, we can write n = 2k for some integer k.
From (2), 2m2 = 4k2
m2= 2k2 implies m2 is even and hence m is even.
This is contradiction to the fact that n and m have no common factor.

Proof by mathematical induction
Proof by mathematical induction consists of three basic steps. If the
statement p is to be proved then:
1) Show that p is true for some particular integer n0
- this is called Basis
2) Assume p is true for some particular integer k ≥ n0
- this is called Induction hypothesis
3) Then to prove is true for k+1
- this is called Induction step

Example: 1
Show that for any n ≥ 1, 1 + 2 + . . . + n = n(n+1)/2
Solution:
Let P(n) : 1 + 2 + . . . + n = n(n+1)/2
Basis step: P(1) : 1 = 1(1+1)/2 = 1
P(1) is true.
Induction hypothesis: Assume that P(k) is true for some k.
P(k): 1 + 2 + . . . + k = k(k+1)/2 ------------(1)
Induction step: To prove P(k+1) is also true.
Take the LHS P(k+1) and prove the RHS.
1 + 2 + . . . + k + (k+1) = k(k+1)/2 + (k+1) = (k+1) (k/2 +1) using (1)
= (k+1)(k+2)/2
Therefore, P(k+1) is true and hence P(n) is true for any n.

Use mathematical induction to show that for any integer n,
a) 13 + 33 + 53 + . . . + (2n+1)3 = (n+1)2 (2n2+4n+1)
b)
1
1 . 2
+
1
2 .3
+ ⋯ +
1
𝑛 𝑛+1
=
𝑛
𝑛+1
c) Use induction to prove that n ≥ 0 , the number
42n+1 + 3n+2 is multiple of 13.
d) Show by the principle of mathematical induction that n4 - 4n2 is
divisible by 3 for all n ≥ 0.
e) Use induction to show that 14 + 24 + 34 + . . . + n4 =
n (n + 1) (2n + 1) (3n2 + 3n -1) / 30 .

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