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Introduction to Computability Theory

Lecture5: Context Free Languages

Prof. Amos Israeli

On the last lecture we completed our study of regular languages. (There is still a lot to learn but our time is limited…).

Introduction and Motivation

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In our study of RL-s we Covered:

1. Motivation and definition of regular languages.

2. DFA-s, NFA-s, RE-s and their equivalence.

3. Non Regular Languages and the Pumping Lemma.

Introduction and Motivation

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In this lecture, we turn to Context Free Grammars and Context Free Languages.

The class of Context Free Languages is an intermediate class between the class of regular languages and the class of Decidable Languages (To be defined).

Introduction and Motivation

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A Context Free Grammar is a “machine” that creates a language.

A language created by a CF grammar is called A Context Free Language.

(We will show that) The class of Context Free Languages Properly Contains the class of Regular Languages.

Introduction and Motivation

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Consider grammar :

A CFL consists of substitution rules called Productions.

The capital letters are the Variables.

The other symbols are the Terminals.

Context Free Grammar - Example

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1G

I

aIbI

Consider grammar :

The grammar generates the language called the language of , denoted by .

Context Free Grammar - Example

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0| nbaB nn

1G

I

aIbI

1G

1G 1GL

Consider grammar :

This is a Derivation of the word by :

On each step, a single rule is activated. This mechanism is nondeterministic.

Context Free Grammar - Example

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aaabbbaaaIbbbaaIbbaIbI

1G

I

aIbI

1Gaaabbb

This is A Parse Tree of the word by :

Context Free Grammar - Example

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1G

aaabbb

I

I

I

bbbaaa

I

Each internal node of the tree is associated with a single production.

Context Free Grammar - Example

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bbbaaa

I

I

I

I

aIbI

I

A Context Free Grammar is a 4-tupple where:

1. is a finite set called the variables.2. is a finite set, disjoint from V called the

terminals.3. is a set of rules, where a rule is a variable

and a string of variables and terminals, and4. is the start variable .

CF Grammar – A Formal Definition

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,

SRV ,,,V

R

VS

A word is a string of terminals.A derivation of a word w from a context Free

Grammar is a sequence of strings ,over , where:

1. is the start variable of G.2. For each , is obtained by

activating a single production (rule) of G on one of the variables of .

A Derivation – A Formal Definition

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SRVG ,,,

li 1Ss 0

wsssS l ...10

V

is

1is

A word w is in the Language of grammar G, denoted by , if there exists a derivation whose rightmost string is w .

Thus,

CF Grammar – A Formal Definition

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GLw

GwwGL from derived be can |

FACTORFACTORTERMTERM |

Grammar :

Rules:1. 2. 3.

Example2: Arithmetical EXPS

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TERMTERMEXPREXPR |

EXPRS

FACTORTERMEXPV ,,

baEXPRFACTOR ||

)(,,,,, ba

2G

Derivation of by Grammar : input

Example2: Arithmetical EXPS

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EXPR

2G aba

Derivation of by Grammar : input rule

Example2: Arithmetical EXPS

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EXPR

2G aba

TERMEXPR

TERMEXPR

Derivation of by Grammar : input output rule

Example2: Arithmetical EXPS

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2G aba

TERMEXPR

TERMEXPR

Derivation of by Grammar : input

Example2: Arithmetical EXPS

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2G aba

Derivation of by Grammar : input rule

TERMEXPR

Example2: Arithmetical EXPS

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2G aba

FACTORTERMTERM

Derivation of by Grammar : input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

FACTORTERMTERM

Derivation of by Grammar : input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

Derivation of by Grammar : input

rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aFACTOR

Derivation of by Grammar : input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aFACTOR

aTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERM

FACTORTERM

aFACTORaTERM

Derivation of by Grammar :

input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

FACTORTERM

aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aFACTORaTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

EXPRFACTOR

aFACTORaTERM

Derivation of by Grammar :

input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

EXPRFACTOR

aEXPR

aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aEXPR

aFACTORaTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aEXPR

TERMEXPREXPR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

TERMEXPREXPR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

TERMEXPR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

TERMEXPR

aTERMTERM

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

FACTORTERM

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

FACTORTERM

aFACTORFACTOR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM aFACTORFACTOR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input

rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

aFACTOR

aFACTORFACTOR

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

output rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

aFACTOR

aFACTORFACTOR

aFACTORa

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM aFACTORFACTOR

aFACTORa

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

input rule

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM

bFACTOR

aFACTORFACTOR

aFACTORa

aTERMEXPRaEXPR aFACTORaTERM

Derivation of by Grammar :

FACTORTERMTERMEXPR

Example2: Arithmetical EXPS

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2G aba

aTERMTERM aFACTORFACTOR

abaaFACTORa

Derivation of by Grammar :

Example2: Arithmetical EXPS

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TERMEXPREXPR

2G

TERMFACTORTERMTERM

FACTORTERMaTERMa

FACTORFACTORa

FACTORba

aba

aba

Note: There is more than one derivation.

To be Demonstrated on the blackboard

Example3: The Language of WF ( )

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We already saw that a word may have more then a single derivation from the same grammar.

A Leftmost Derivation is a derivation in which rules are applied in order left to right.

A grammar is ambiguous if it has Two parse trees.

Ambiguity

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Reminder: Two parse trees are equal if they are equal as trees and if all productions corresponding to inner nodes are also equal .

Ambiguity

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Grammar :

Rules:

Example4: Similar to Arith. EXPS

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,

,

EXPREXPREXPR

EXPRS

FACTORTERMEXPV ,,

)(,,,,, ba

4G

EXPREXPREXPRaEXPR

aba

Example4: 1st Parse Tree for______

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EXPR

EXPR

EXPR EXPR

EXPR

aba

EXPREXPREXPR

aba

Example4: 2nd Parse Tree for_____

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EXPR

EXPR

EXPR EXPR

EXPR

aba

EXPREXPREXPR

Note: Some ambiguous grammars may have an unambiguous equivalent grammar.

But: There exist Inherently Ambiguous Grammars , i.e. an ambiguous grammar that does not have an equivalent unambiguous one.

Ambiguity

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Q: From a computational point of view, how strong are context free languages?

A: Since the language is not regular and it is CF, we conclude that .

Q: Can one prove ?A: Yes.

Discussion

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0| nbaB nn

RLCFL RLCFL

Q: A language is regular if it is recognized by a DFA (or NFA). Does there exist a type of machine that characterizes CFL?

A: Yes, those are the Push-Down Automata (Next Lecture). .

Q: Can one prove a language not to be CFL ?A: Yes, by the Pumping Lemma for CFL-s . For

example: is not CFL.

Discussion

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0| ncbaL nnn