lecture 06 context-free grammars

8/18/2019 Lecture 06 Context-Free Grammars

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TIC 2151 – Theory of

Computation

Lecture 6

Context-Free Grammars (CFG)

TIC 2151 – Theory of Computation


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Lecture 6 – Outline

Introduction Formal description of a CFG

Ambiguous CFGs

Normal forms


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Context-free grammars (CFG)

context-f ree grammars CFG, a more powerful method of describing

languages. Such grammars can describe certain features that have a

recursive structure, which makes them useful in a variety of applications.The collection of languages associated with context-free grammars are called

the context-f ree languages.

A language is context free if and only if some pushdown automaton recognizes

it.Languages that can be accepted by some PDA or generated by some CFG are

context-free.

Many useful languages are context-free, including most programming

languages, query languages, and markup languages.

We can establish a relationship between the regular languages and the context-

free languages. Because every regular language is recognized by a finite

automaton and every finite automaton is automatically a pushdown automaton

that simply ignores its stack, we now know that every regular language is also a

context-free language.


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The following is an example of a context-free grammar, which we call G.

Grammar G:

A => 0A1

A => B

B => #

derivation of a word

You may also represent the same information pictorially with a parse tree.

See next slide



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parse tr ees for the word 00#11

Formal description of a CFG

Formally written we have the grammar G:G = (V, S, R, S)

with V = {A,B}

S = {0,1,#}

R = {[A, 0A1], [A, B], [B, #]}

S = A



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G = ({S}, {a, b},R, S). The set of rules, R, is S => aSb | SS |

This grammar G generates strings such as abab, aaabbb, and aababb.

Give a parse trees for aaabbb and abab

Some

exercises

Give the formal definition for the following grammar an describe the

produced language: S => aSb | SS |



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Some

exercises

Give the formal definition for the following grammar and give parse trees for

a + a x a and (a + a) x a

Formal Definition



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Some

exercises

Give the formal definition for the following grammar and give parse trees for

a + a x a and (a + a) x a

Formal Definition



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Ambiguity

o Sometimes a grammar can generate the same string in several different

ways. Such a string will have several different parse trees and thus

several different meanings.

o This result may be undesirable for certain applications, such as programming languages, where a program should have a unique

interpretation.

o A grammar is called ambiguous if there is at least one word having two

different parse trees.

AmbiguityContext-free grammars (CFG)


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Ambiguity

Consider the following rules of a grammar and give two different parse trees for

a + a x a

This grammar generates the string a + a x a ambiguously.

Sometimes when we have an ambiguous grammar we can find an

unambiguous grammar that generates the same language.

A grammar is called inherently ambiguous if there is no equivalent

unambiguous grammar to generates it. {aib jck | i=j or j=k} is inherently

ambiguous!

AmbiguityContext-free grammars (CFG)


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When working with context-free grammars CFG, it is often convenient to

have them in simplified form. One of the simplest and most useful forms

is called the Chomsky normal form.

Remark:

any context-free grammar may be transformed to a CNF grammar expressing the

same language

Chomsky Normal Form (CNF)


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Original CF grammar S => ASA | aB

A => B | S

B => b | e

1. Insert a new start symbol

S0 => S

S => ASA | aB

A => B | S

B => b | e

2. Remove the rule B => e

S0 => S

S => ASA | aB | a

A => B | S |

B => b

Converting a CFG to CNF

Chomsky Normal Form (CNF) CFG to CNF

Step 1. Add a new start variable S0 and the rule S0S,

where S was the original start variable. This change

guarantees that the start variable doesn’t occur on the

right-hand side of a rule

Step 2.we take care of all e-rules. We remove an e-ruleC , where C is not the start variable. Then for each

occurrence of an C on the right-hand side of a rule, we

add a new rule with that occurrence deleted. if R uCv

is a rule in which u and v are strings of variables and

terminals, we add rule R uv. The rule R uCvCw

causes us to add R

uvCw, R

uCvw, and R

uvw.If we have the rule R C, we add R unless we had

previously removed the rule R e .

We repeat these steps until we eliminate all e -rules not

involving the start variable.

Try to follow every step! Don t go to the next step, if you

still have problems with the one before!


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3. Remove the rule A => e

S0 => S

S => ASA | aB | a | SA | AS | S

A => B | S

B => b

4. Remove the rule S => S

S0 => S

S => ASA | aB | a | SA | AS

A => B | S

B => b

5. Remove the rule S0 => SS0 => ASA | aB | a | SA | AS

S => ASA | aB | a | SA | AS

A => B | S

B => b


Step 3 .Eliminate all unit rules of the form

CD. We repeat these steps until we

eliminate all unit rules.




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6. Remove the rule A => B

S0 => ASA | aB | a | SA | ASS => ASA | aB | a | SA | AS

A => S | b

B => b

7. Remove the rule A => S

S0 => ASA | aB | a | SA | ASS => ASA | aB | a | SA | AS

A => b | ASA | aB | a | SA | AS

B => b

8. Insert new symbols for the last incorrect rules

S0 => AA1 | UB | a | SA | AS

S => AA1 | UB | a | SA | AS

A => b | AA1 | UB | a | SA | AS

B => b

A1 => SA

U => a




Step 4. Take care of rules with more

than two terminals or variables.

convert all remaining rules into the

proper form.


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Covert the following CFG into Chomsky Normal Form (CNF)

SOLUTION

1. elimination of e, we obtain: 2. elimination of the unit production we obtain


3. convert all remaining rules into the

proper form. replace a by A, b by B

and c by C, we obtain


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Our next problem:

I f we have a word w, how can we determine if it can be

generated by a context-free grammars CFG ?

CFG


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S

C D

w1w2. . . . . . wk wk+1. . . wn

The Cocke-Younger-Kasami

Algorithm (CYK)

CYK – The Concept

For each word, there must be a rule of the form A BC, whereB produces the left part of the word and C the right part.

We replace the occurrence of BC with A.

This process is repeated, and should leave us with the start

variable S.


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Given the following grammar

S => AB

A => ab | aAb

B => c | cB

and in Chomsky normal form

S => ABA => CD | CF

B => c | EB

C => a

D => b

E => c

F => AD

we build a table for aaabbbcc:

a a a b b b c c

C C C D D D E,B E,B

A B

F

A

F

A

S

S

If we find S here,

we knowaaabbbcc

is part of the language.

CYK – In Action Example (1)


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a a a b b b c c

C C C D D D E,B E,BA B

F

A

F

A

S

S

This (darkgray) partis produced by B

As there is a rule S => AB,

we can fill S here.

This (gray) part

is produced by this A



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a a a b b b c c

A F

B E

C D

?

If there is one of the following

rules, so write the left side X in the box:

X => AD

X => BE

X => CF

If there are more than one, write all.

CYK – In Action


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For the grammar:

we prove the membership of 00100111 and 01011100:


S => SS | BA | BC

A => SCB => 0

C => 1

0 0 1 0 0 1 1 1

B B C B B C C C

S S

A

S

A

S

A

S

0 1 0 1 1 1 0 0

B C B C C C B B

S S

S

YES NO


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Learning Outcomes

After completion of this lecture, you should be able to:

Draw parse trees for words generated from CFG.

Given the definition of a CFL, design a grammar (i.e. give the

grammar rules).

Convert a CFG into Chomsky Normal Form.

Apply the CYK algorithm to determine if a word is generated

form a CFG.

lecture 06 context-free grammars

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