language , grammar and automata

LANGUAGE , GRAMMAR
&
AUTOMATA
Presentation by
Elakkiya.S,
Santhosh.B,
Balamurugan.A,

❖ Outline of the presentation
❖ Language
➢ Concatenation
➢ Properties of concatenation
➢ Regular expression
➢ Regular language
❖ Grammar
➢ Types of grammar
❖ Finite state machine
❖ Automata
➢ Automata and its types.
Maths- seminar presentation- 07.06.2021
1

❖ Language:
❖ Definition:
➢ An alphabet is a non-empty finite set Σ of symbols.
➢ A string or word on the set Σ is a finite sequence of its elements (symbols)
■ If Σ ={a,b} then the sequences u=abba and v=aabbb of letters are strings
of Σ.
➢ The length of a string u is denoted by 丨u丨is the number of symbols in the
string
■ If Σ={0,1,2} then u=01 v=212 and w=01212 are possible strings on A. the
length of these strings is 丨u丨=2 丨v丨=3 丨w丨=5
➢ The string consisting of no symbols or letters taken from an alphabet Σ, we
define Σ⁰={λ}.
2

❖ Operations on strings
❖ The basic operations for strings is concatenation operations.
❖ The concatenation of strings is a string formed by writing the letters
(symbols) of first string followed by letters of second string
❖ Example :
➢ If u =ice and v= cream are the two strings of length 3 and 5 respectively.Then
concatenation of u and v is uv=ice cream is the string of length 3+5=8.
3

❖Properties of concatenation
❖ If Σ is an alphabet and n is a non-negative integer then the power of ∑ is
defined recursively as
➢ ∑1= ∑ and
➢ ∑n+1 = { ab:a∈A and b∈ ∑n}
❖ The concatenation Operation on an alphabet ∑ is associative , since for
every u,v,w ∈ ∑* ... We have u(vw)=(uv)w.
❖ The concatenation operation is not commutative (unless丨 ∑ 丨=1)on an
alphabet ∑ , since uv≠ vu for every u,v ∈ ∑* .
4

❖ The empty string λ is an identity element for concatenation
operation on an alphabet Σ that is λu=uλ=u for every u ∈ Σ*
❖ If u,v ∈ Σ* then we may write u=u1,u2,..........um and
v=v1,V2,......vn. For m,n and
u1,u2,...um;v1,v2..vn two string u,v are said equal.
❖ For every u,v,w ∈ ∑* we have
➢ wu= wv →u=v ( left cancellation)
➢ uw =vw →u=v ( right cancellation)
❖ For u, v ∈∑* we have
➢ │uv│= │u│+│v│
5

❖ Language
❖ Definition:
➢ For a given alphabet Σ , any subset of Σ* is called a language over Σ and is
denoted by L.
➢ The subset Ф also called empty language.
➢ A string in a language L is called a sentence of Σ.
6

❖ Operations on languages
❖ Operations on language :
➢ Let L1 and L2 be two languages over an alphabet Σ. Then the concatenation
of L1 and L2 is denoted by L1L2 and is the language defined as
■ L1L2={ab: a∈L1 and b∈L2}
❖ Examples :
➢ L1={a,ab,c} and L2={λ,b}
■ L1L2={aλ,ab,abλ,abb,cλ,cb}
={a,ab,ab,abb,c,cb}
7

❖ Regular expression
❖ Regular expression such as ( ),*,U,λ that are used to develop expression for
describing languages are called regular expressions.
❖ Thus the regular expression r over a non -empty alphabet Σ is a word which
attaches the letters of Σ with the help of operation( ),*,U and λ.
9

❖ Examples
❖ Each letter a∈Σ is a regular expression.
❖ Empty word λ and empty expression Φ are regular expression.
❖ If r is a regular expression then it's closure expression r* is an regular
expression.
❖ If r1,r2 are regular expression then r1+r2 is a regular expression.
10

❖ Regular language
❖ The language L(r) represented by the regular expression r over an alphabet Σ
is as follows;
(1).If r =Φ then L(r)=Φ
(2).If r=λ then L(λ)={λ} and L(())=ф empty set
(3).If r=a for some a ∈ Σthen L(a)={a}
(4).L(r*)=(L(r))* the kleene closure of L(r)
(5).L(r1 v r2)= L(r1)UL(r2) union of two languages.
11

❖ Grammar Definition
❖ A phrase structure grammar or simply a grammar G consists of four parts;
1. V
2.𝛴
3. S
4.P
12

❖ V,Σ,S,P Definition
❖ V is a finite set,where elements are called variables.
❖ 𝛴 is a finite subset of V whose elements are called terminals.
❖ vn𝛴=ф
❖ S is special variable(non terminal symbol) in the set V called the start symbol
that begins the generation of any sentence in the language.
❖ P is a finite set of whose elements are ordered pair (𝝰,𝜷) usually written as 𝜶
⇾𝜷 where 𝝰 and𝛃 are string(word) on VUΣ and has at least one non-terminal
on its left side elements of P are called productions or production rule.
❖ The grammar denoted by G=(v,Σ,s,p)
13

❖ Language of grammar G
❖ Definition
❖ Let G=(V,Σ,S,P) be a grammar then the language L(G)generated by G is the set
of all strings of terminals which are derivable from the starting state S. That is
➢ L(G)={w ∈ Σ*,s*⇒w}
❖ The elements of L(G) are usually called sentences.
14

problem (1)
1. Find the language L(G) generated by the grammar with v={S,A,B},𝛴={a,b} and
production p ={s→aB, B→b, B→bA , A→a,B}.
Solution;
S→aB
→a bA
→abaB
→ababA
→ababaB
→ababab
L(G)={(ab)ｎ=ababab…….ab:nƐN}.
15

Problem(2)
Find the language L(G) generated by the grammar with V={s,a},Σ={a,b,c}
and production p={S→asb,aS→Aa,Aab→c}.
Solution; S →asb
S→asb S→a asb b
S→a asb b S→aa asb bb
S→a Aabb S→aa Aa bbb
S→acb S→aa c bb
L(G)={an c bn : n≥1}
16

❖ TYPES OF GRAMMAR
➢ Type 0 (OR) Phrase Structure Grammar
➢ Type 1 (OR) Context sensitive Grammar
➢ Type 2 (OR) Context Free Grammar
➢ Type 3 (OR) Regular Grammar
17

❖ Type 0 (or) Phase Structure Grammar
➢ A Grammar G with no restriction on its productions is called A Type 0
Grammar
18

❖ Type 1 (or)Context Sensitive Grammar
➢ A Grammar G said to be of type 1 if every production is of the form
∝→ß where 丨∝丨≤丨β丨or of the form ∝→λ.
➢ Here λ is the null string and ∝ , β are arbitrary string on v.
19

❖ Type 2 (or)Context Free Grammar
➢ A Grammar G is said to be of type 2 if every production is of the form
A→β.
➢ Where the left hand side string is always a single non terminal symbol.
20

❖ Type 3 (or) Regular Grammar
➢ A Grammar G is said to be of type 3 if every production is of the form
A→∝ (or) A→aB.
➢ Where the left hand side string is always a single non terminal symbol and
the right hand side string is either terminal symbol (or) a terminal symbol
Followed by a non-terminal symbol or of the form S→λ.
21

Problems:
1) Determine the type of grammar G in the following:
a)p = { S→aA, A→aAB, B→b , A→a }
b)p = { S→ b, S→aA, A→b , A→bB, A→aA, B→a, B→aB }
c)p = { S→aAB, AB→a, A→b, B→AB }
d)p = { s→b, S→aA, A→b, A→aA, A→bB, B→a, B→aB }
Soln:
a)Type 2 (OR) Context Free Grammar.
b) Type 3 (OR) Regular Grammar.
c) Type 0 (OR) Phrase Structure Grammar.
d) Type 1 (OR) Context sensitive Grammar.
22

❖ Finite state machine
❖ A finite state machine M consists of six parts:
➢ Σ, finite set of input alphabet symbols ai(i=1,2...n) with or without subscripts
are input symbols.
➢ S,a finite set of internal state. The state S₀ is the initial state,
➢ O is a finite set of the output alphabet that is O={x,y,z}.
➢ f,a transition (or next state)function or mapping from s *Σ into S. that is f:s
*Σ→S
➢ g,a output function from S *Σ into O. that is g:S *Σ→0
➢ Such a machine is written as
■ M={Σ,S,S₀ ,o,f,g}
23

❖ The Features of Finite State machine
● At any given time,the machine can be only one of finite number of states.This
States are called the internal States of the machine.
● The machine accepts as input Only a finite number of symbols the finite set
of all possible inputs constitute the input alphabet for the machine.
● An output and a next state are determined by each combination of Inputs and
internal states.The finite set of all possible outputs constitute the output
alphabet for the machine.
● We assume that the machine has an internal clock,an the sequential
processing are synchronised by separate and distinct clock pulses.
24

❖ State Table and State Diagram of a Finite State
machine:
❖ A Finite state machine M can be represented in compact form by means of
➢ A state-table also called transition table of M
➢ A labelled directed graph called the state diagram or transition diagram of M.
➢ The state table combines the state function ‘f’ and the output function ‘g’ for
all pairs of state and inputs
➢ The first column of the table list the (present )state for the machine .
➢ Entries in the second row are the elements of the input alphabet Σ listed
once under f and then again under g.
➢ The six numbers in the last two columns (and last three rows)are elements of
the output alphabets O.
25

❖ State transition table and state diagram
26

❖ Problem(1)
Let m be the finite state machine with state Table Given the following page.
❖ Find the input set Σ ,the state set S,the output set O and the initial state of M.
❖ Draw the state diagram of M.
❖ Find the output string of the input string 01001.
27

❖ Solution
❖ (A).Input set Σ ={0,1}
State set s ={S0,S1,S2,S3}
Output set 0={0,1}
Initial state=S0
28

❖ Finite state automata:
❖ ite state automaton is a special type of finite state machine in which there is
no output , and some of its states are distinguished as accepting states.
❖ There are two types of automatiA finon
Deterministic finite state automation
Non-deterministic finite state automation
29

❖ Deterministic Finite State Automaton (DFA)
❖ A deterministic finite state automaton or simply an automaton M consists of five
parts:
➢ A finite non empty set 𝚺 of inputs, called alphabets
➢ A finite non empty set s of (internal) state.
➢ A next state function ( also called transition function) f from s* 𝚺 into s .
This function or mapping describes the change of states during the
transition. It is usually represented by a state transition table or a state
transition diagram.
➢ A subset F of set S of accepting (final) states.
➢ An initial state s₀𝟄S. Such a finite state automaton is denoted by M=( 𝚺,S,S𝜊
,f).
30

❖ State Transition Diagram
❖ A typical state transition diagram is shown. This diagram is a finite directed
labelled graph so that
➢ Each vertex (node) represents a state,
➢ The directed edges are represented by arrows and indicate the transition
from state si to state sj.The final state is represented by double circle.
➢ The initial state s₀ is represented by a circle with an arrow pointing towards
31

❖ Definition
❖ A string w is said to be accepting by a finite automaton M provided.
➢ There exists a path which originates from some initial state ,goes along the
arrows and terminates at some final state ,and
➢ The path value obtained by concatenation of all edge labels of the path is
equal to w
❖ Trap state
➢ A state from which a dfa can never come out is called a trap state.
32

❖ problem:
❖ Consider the transition diagram shown
(a) Find its state
(b) Find its input symbols
(c) Find its initial state
(d) Find its accepting states
(e) Find f(s2,1)
(f) Write its next state table
33

❖ Solution
(a) State S={s𝜊,s1,s2,s3}
(b) Input symbol 𝚺={0,1}
(c) Initial state=s𝜊
(d) Accepting states=S𝜊
(e) f(s2,1)=s3
(f) Next state table
-------------------------------------------
State f(input)
0 1
S𝜊 S2 S1
S1 S3 S𝜊
S2 S𝜊 S3
S3 S1 S2
34

❖ Non Deterministic Finite State Automaton
❖ A non-deterministic finite state automaton (nfa) allows zero,one or more
transitions from a state on the same input symbol.
❖ A non-deterministic finite state automaton m consists of five parts:
➢ A finite non empty set 𝚺 of input symbols.
➢ A finite non empty set s of state.
➢ A next state function ( or mapping) f from s* 𝚺 into p(s) , which is the power
set of S. that is the set of all subsets of S which are v equal to 2 power s
➢ A subset F of set S of accepting (final) states.
➢ An initial state s₀𝟄S. Such a finite state automaton is denoted by M=( 𝚺,S,S𝜊
,f).
35

❖ problem
❖ For a non deterministic finite state automaton shown..find
(a) Initial state
(b) State set
(c) Input set
(d) State table defining the next state function
❖ solution:
(a) Initial state=S𝜊
(b) State set={S𝜊,S1,S2}
(c) Input set ={0,1}
(d) Next state func
State f
S𝜊 {S1,S2} ф
S1 ф {S𝜊,S2}
S2 ф S𝜊
36

language , grammar and automata

language , grammar and automata

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