TIC 2151 – Theory of Computation Lecture 6 Context-Free Grammars (CFG)
TIC 2151 – Theory of Computation
Lecture 6 – Outline
Introduction
Formal description of a CFG
Ambi biguo guous us CF CFGs Gs Am Normal forms
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Context-free grammars (CFG) context-f r ee gr ammar s CF G, 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 r ee 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 contextfree 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. 3
Context-free grammars (CFG)
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Context-free grammars (CFG) 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 par se tr ee. See next slide
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Context-free grammars (CFG)
par se 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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Context-free grammars (CFG)
Some exercises
Give the formal definition for the following grammar an describe the produced language: S => aSb | SS | 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
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Context-free grammars (CFG)
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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Context-free grammars (CFG)
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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Context-free grammars (CFG)
Ambiguity
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.
This result may be undesirable for certain applications, such as programming languages, where a program should have a unique interpretation.
o
o
A grammar is called ambiguous if there is at least one word having two different parse trees.
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Context-free grammars (CFG)
Ambiguity
Ambiguity Consider the following rules of a grammar and give two different parse trees for
a+axa
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! 11
Chomsky Normal Form (CNF) 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 12
Chomsky Normal Form (CNF) Converting a CFG to CNF 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
CFG to CNF
Try to follow every step! Don t go to the next step, if you still have problems with the one before!
Step 1. Add a new start variable S0 and the rule S0 S, 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-rule , where C is not the start variable. Then for each C 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. 13
Chomsky Normal Form (CNF)
CFG to CNF
Try to follow every step! Don t go to the next step, if you still have problems with the one before!
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
Step 3 .Eliminate all unit rules of the form CD. We repeat these steps until we eliminate all unit rules.
5. Remove the rule S0 => S S0 => ASA | aB | a | SA | AS S => ASA | aB | a | SA | AS A => B | S B => b 14
Chomsky Normal Form (CNF) 6. Remove the rule A => B S0 => ASA | aB | a | SA | AS S => ASA | aB | a | SA | AS A => S | b B => b
CFG to CNF
Try to follow every step! Don t go to the next step, if you still have problems with the one before!
7. Remove the rule A => S S0 => ASA | aB | a | SA | AS S => 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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Chomsky Normal Form (CNF)
CFG to CNF
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:
CFG
I f we have a word w, how can we deter mine if it can be gener ated by a context-f r ee grammar s CF G ?
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The Cocke-Younger-Kasami Algorithm (CYK) CYK – The Concept S
C
D
w1w2. . . . . . w k wk+1. . . wn
For each word, there must be a rule of the form A BC, where B 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. 18
Example (1)
CYK – In Action
Given the following grammar S => AB A => ab | aAb B => c | cB and in Chomsky normal form S => AB A => 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 know aaabbbcc is part of the language.
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Example (1)
CYK – In Action a
a
a
b
b
b
c
c
C
C
C
D
D
D
E,B
E,B
A
B
This (darkgray) part is produced by B
F A F
This (gray) part is produced by this A
A
S S
As there is a rule S => AB, we can fill S here.
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CYK – In Action a
a
a
b
b
b
A
c
F
B C
c
E 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.
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Example (2)
CYK – In Action For the grammar:
S => SS | BA | BC A => SC B => 0 C => 1
we prove the membership of 00100111 and 01011100: 0
0
1
0
0
1
1
1
0
1
0
1
1
1
0
0
B
B
C
B
B
C
C
C
B
C
B
C
C
C
B
B
S
S
S
S
A S
S
A S A S
YES
NO 22
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.
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