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Algorithm To Decide Whether CFL Is Empty

Theory of Automata & Computation

Context Free Language-

 

Before you go through this article, make sure that you have gone through the previous article on Context Free Language.

 

We have discussed-

  • Context free language is generated using a context free grammar.
  • Each context free language is accepted by a Pushdown automaton.

 

In this article, we will discuss a decision algorithm of CFL.

 

Algorithm To Decide Whether CFL Is Empty Or Not-

 

If we can not derive any string of terminals from the given grammar,

then its language is called as an Empty Language.

L(G) = ϕ

 

For a given CFG, there exists an algorithm to decide whether its language is empty L(G) = ϕ or not.

 

Algorithm-

 

  • Remove all the useless symbols from the grammar.
  • A useless symbol is one that does not derive any string of terminals.
  • If the start symbol is found to be useless, then language is empty otherwise not.

 

Remember

The language generated from a CFG is non-empty iff the start symbol is generating.

 

Example-

 

Consider the following grammar-

S → XY

X → AX

X → AA

A → a

Y → BY

Y → BB

B → b

 

Now, let us check whether language generated by this grammar is empty or not.

 

The given grammar can be written as-

S → aabb

X → aX

X → aa

A → a

Y → bY

Y → bb

B → b

 

Clearly,

  • The start symbol generates at least one string (many more are possible).
  • Therefore, start symbol is useful.
  • Thus, language generated by the given grammar is non-empty.

 

L(G) ≠ ϕ

 

NOTE-

 

If L(G) = { ∈ } i.e.

  • If the language generated by a grammar contains only a null string,
  • then it is considered as non-empty.

 

Next Article- Algorithm To Decide Whether CFL Is Finite

 

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Context Free Grammar | Context Free Language

Theory of Automata & Computation

Context Free Grammar-

 

A context Free Grammar (CFG) is a 4-tuple such that-

G = (V , T , P , S)

where-

  • V = Finite non-empty set of variables / non-terminal symbols
  • T = Finite set of terminal symbols
  • P = Finite non-empty set of production rules of the form A → α where A ∈ V and α ∈ (V ∪ T)*
  • S = Start symbol

 

Why Context Free Grammar Is Called So?

 

Context Free Grammar provides no mechanism to restrict the usage of the production rule A → α within some specific context unlike other types of grammars.

That is why it is called as “Context Free” Grammar.

 

Example-01:

 

Consider a grammar G = (V , T , P , S) where-

  • V = { S }
  • T = { a , b }
  • P = { S → aSbS , S → bSaS , S → ∈ }
  • S = { S }

 

  • This grammar is an example of a context free grammar.
  • It generates the strings having equal number of a’s and b’s.

 

Example-02:

 

Consider a grammar G = (V , T , P , S) where-

  • V = { S }
  • T = { ( , ) }
  • P = { S → SS , S → (S) , S → ∈ }
  • S = { S }

 

  • This grammar is an example of a context free grammar.
  • It generates the strings of balanced parenthesis.

 

Applications-

 

Context Free Grammar (CFG) is of great practical importance. It is used for following purposes-

 

  • For defining programming languages
  • For parsing the program by constructing syntax tree
  • For translation of programming languages
  • For describing arithmetic expressions
  • For construction of compilers

 

Context Free Language-

 

The language generated using Context Free Grammar is called as Context Free Language.

 

Properties-

 

  • The context free languages are closed under union.
  • The context free languages are closed under concatenation.
  • The context free languages are closed under kleen closure.
  • The context free languages are not closed under intersection and complement.
  • The family of regular language is a proper subset of the family of context free language.
  • Each Context Free Language is accepted by a Pushdown automaton.

 

Remember

If L1 and L2 are two context free languages, then-

  • L1 ∪ L2 is also a context free language.
  • L1.L2 is also a context free language.
  • L1* and L2* are also context free languages.
  • L1 ∩ L2 is not a context free language.
  • L1′ and L2′ are not context free languages.

 

Ambiguity in Context Free Grammar-

 

A grammar is said to be ambiguous if for a given string generated by the grammar, there exists-

  • more than one leftmost derivation
  • or more than one rightmost derivation
  • or more than one parse tree (or derivation tree).

 

Read More- Grammar Ambiguity

 

To gain better understanding about Context Free Grammar,

Watch this Video Lecture

 

Next Article- Chomsky Normal Form

 

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DFA Solved Examples | How to Construct DFA

Theory of Automata & Computation

Construction Of DFA-

 

Before you go through this article, make sure that you have gone through the previous article on Type-01 Problems.

 

Type-02 Problems-

 

In Type-02 problems, we will discuss the construction of DFA for languages consisting of strings starting with a particular substring.

 

Steps To Construct DFA-

 

Following steps are followed to construct a DFA for Type-02 problems-

 

Step-01:

 

  • Determine the minimum number of states required in the DFA.
  • Draw those states.

 

Use the following rule to determine the minimum number of states-

 

RULE

Calculate the length of substring.

All strings starting with ‘n’ length substring will always require minimum (n+2) states in the DFA.

 

Step-02:

 

  • Decide the strings for which DFA will be constructed.
  • The method for deciding the strings has been discussed in this Video.

 

Step-03:

 

  • Construct a DFA for the strings decided in Step-02.

 

Remember the following rule while constructing the DFA-

 

RULE

While constructing a DFA,

  • Always prefer to use the existing path.
  • Create a new path only when there exists no path to go with.

 

Step-04:

 

  • Send all the left possible combinations to the dead state.
  • Do not send the left possible combinations over the starting state.

 

PRACTICE PROBLEMS BASED ON CONSTRUCTION OF DFA-

 

Problem-01:

 

Draw a DFA for the language accepting strings starting with ‘ab’ over input alphabets ∑ = {a, b}

 

Solution-

 

Regular expression for the given language = ab(a + b)*

 

Step-01:

 

  • All strings of the language starts with substring “ab”.
  • So, length of substring = 2.

 

Thus, Minimum number of states required in the DFA = 2 + 2 = 4.

It suggests that minimized DFA will have 4 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • ab
  • aba
  • abab

 

Step-03:

 

The required DFA is-

 

Problem-02:

 

Draw a DFA for the language accepting strings starting with ‘a’ over input alphabets ∑ = {a, b}

 

Solution-

 

Regular expression for the given language = a(a + b)*

 

Step-01:

 

  • All strings of the language starts with substring “a”.
  • So, length of substring = 1.

 

Thus, Minimum number of states required in the DFA = 1 + 2 = 3.

It suggests that minimized DFA will have 3 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • a
  • aa

 

Step-03:

 

The required DFA is-

 

Problem-03:

 

Draw a DFA for the language accepting strings starting with ‘101’ over input alphabets ∑ = {0, 1}

 

Solution-

 

Regular expression for the given language = 101(0 + 1)*

 

Step-01:

 

  • All strings of the language starts with substring “101”.
  • So, length of substring = 3.

 

Thus, Minimum number of states required in the DFA = 3 + 2 = 5.

It suggests that minimized DFA will have 5 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • 101
  • 1011
  • 10110
  • 101101

 

Step-03:

 

The required DFA is-

 

Problem-04:

 

Draw a DFA that accepts a language L over input alphabets ∑ = {0, 1} such that L is the set of all strings starting with ’00’.

 

Solution-

 

Regular expression for the given language = 00(0 + 1)*

 

Step-01:

 

  • All strings of the language starts with substring “00”.
  • So, length of substring = 2.

 

Thus, Minimum number of states required in the DFA = 2 + 2 = 4.

It suggests that minimized DFA will have 4 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • 00
  • 000
  • 00000

 

Step-03:

 

The required DFA is-

 

Problem-05:

 

Construct a DFA that accepts a language L over input alphabets ∑ = {a, b} such that L is the set of all strings starting with ‘aa’ or ‘bb’.

 

Solution-

 

Regular expression for the given language = (aa + bb)(a + b)*

 

Step-01:

 

Minimum number of states required in the DFA = 5.

It suggests that minimized DFA will have 5 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • aa
  • aaa
  • aaaa
  • bb
  • bbb
  • bbbb

 

Step-03:

 

The required DFA is-

 

Problem-06:

 

Construct a DFA that accepts a language L over input alphabets ∑ = {a, b} such that L is the set of all strings starting with ‘aba’.

 

Solution-

 

Regular expression for the given language = aba(a + b)*

 

Step-01:

 

  • All strings of the language starts with substring “aba”.
  • So, length of substring = 3.

 

Thus, Minimum number of states required in the DFA = 3 + 2 = 5.

It suggests that minimized DFA will have 5 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • aba
  • abaa
  • abaab
  • abaaba

 

Step-03:

 

The required DFA is-

 

Also Read- Converting DFA to Regular Expression

 

To gain better understanding about Construction of DFA,

Watch this Video Lecture

 

Download Handwritten Notes Here-

 

 

Next Article- Minimization of DFA

 

Get more notes and other study material of Theory of Automata and Computation.

Watch video lectures by visiting our YouTube channel LearnVidFun.

Construction of DFA | DFA Solved Examples

Theory of Automata & Computation

Construction Of DFA-

 

In this article, we will learn the construction of DFA.

 

Type-01 Problems-

 

In Type-01 problems, we will discuss the construction of DFA for languages consisting of strings ending with a particular substring.

 

Steps To Construct DFA-

 

Following steps are followed to construct a DFA for Type-01 problems-

 

Step-01:

 

  • Determine the minimum number of states required in the DFA.
  • Draw those states.

 

Use the following rule to determine the minimum number of states-

 

RULE

Calculate the length of substring.

All strings ending with ‘n’ length substring will always require minimum (n+1) states in the DFA.

 

Step-02:

 

  • Decide the strings for which DFA will be constructed.
  • The method for deciding the strings has been discussed in this Video.

 

Step-03:

 

  • Construct a DFA for the strings decided in Step-02.

 

Remember the following rule while constructing the DFA-

 

RULE

While constructing a DFA,

  • Always prefer to use the existing path.
  • Create a new path only when there exists no path to go with.

 

Step-04:

 

  • Send all the left possible combinations to the starting state.
  • Do not send the left possible combinations over the dead state.

 

PRACTICE PROBLEMS BASED ON CONSTRUCTION OF DFA-

 

Problem-01:

 

Draw a DFA for the language accepting strings ending with ’01’ over input alphabets ∑ = {0, 1}

 

Solution-

 

Regular expression for the given language = (0 + 1)*01

 

Step-01:

 

  • All strings of the language ends with substring “01”.
  • So, length of substring = 2.

 

Thus, Minimum number of states required in the DFA = 2 + 1 = 3.

It suggests that minimized DFA will have 3 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • 01
  • 001
  • 0101

 

Step-03:

 

The required DFA is-

 

 

Problem-02:

 

Draw a DFA for the language accepting strings ending with ‘abb’ over input alphabets ∑ = {a, b}

 

Solution-

 

Regular expression for the given language = (a + b)*abb

 

Step-01:

 

  • All strings of the language ends with substring “abb”.
  • So, length of substring = 3.

 

Thus, Minimum number of states required in the DFA = 3 + 1 = 4.

It suggests that minimized DFA will have 4 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • abb
  • aabb
  • ababb
  • abbabb

 

Step-03:

 

The required DFA is-

 

Problem-03:

 

Draw a DFA for the language accepting strings ending with ‘abba’ over input alphabets ∑ = {a, b}

 

Solution-

 

Regular expression for the given language = (a + b)*abba

 

Step-01:

 

  • All strings of the language ends with substring “abba”.
  • So, length of substring = 4.

 

Thus, Minimum number of states required in the DFA = 4 + 1 = 5.

It suggests that minimized DFA will have 5 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • abba
  • aabba
  • ababba
  • abbabba
  • abbaabba

 

Step-03:

 

The required DFA is-

 

 

Problem-04:

 

Draw a DFA for the language accepting strings ending with ‘0011’ over input alphabets ∑ = {0, 1}

 

Solution-

 

Regular expression for the given language = (0 + 1)*0011

 

Step-01:

 

  • All strings of the language ends with substring “0011”.
  • So, length of substring = 4.

 

Thus, Minimum number of states required in the DFA = 4 + 1 = 5.

It suggests that minimized DFA will have 5 states.

 

Step-02:

 

We will construct DFA for the following strings-

  • 0011
  • 00011
  • 000011
  • 0010011
  • 00110011

 

Step-03:

 

The required DFA is-

 

 

Also Read- Converting DFA to Regular Expression

 

To gain better understanding about Construction of DFA,

Watch this Video Lecture

 

Download Handwritten Notes Here-

 

 

Next Article- Construction of DFA | Type-02 Problems

 

Get more notes and other study material of Theory of Automata and Computation.

Watch video lectures by visiting our YouTube channel LearnVidFun.