review of cfgs and parsing - university of waterloo

Winter 2014 Costas Busch - RPI 1

Review of CFGs and Parsing – I Context-free Languages and Grammars


Regular Languages

}0:{ ≥nba nn }{ Rww

**ba *)( ba +

Context-Free Languages


Context-Free Languages

Pushdown Automata

Context-Free Grammars

stack

automaton


Context-Free Grammars


Formal Definitions

( )PSTVG ,,,=

Set of Variables/ Non-terminal Set of

terminal symbols

Start Symbol

Set of productions

Grammar:


Context-Free Grammar:

All productions in are of the form

sA →

String of Non-terminal and terminals

),,,( PSTVG =

P

Non-terminal


λ|aSbS →

( )PSTVG ,,,=

}{SV =},{ baT =

},{ λ→→= SaSbSP

variables terminals

productions

start variable

Example of Context-Free Grammar


For a grammar with start variable G S

String of terminals or

*},:{)(*

TwwSwGL ∈⇒=

λ

Language of a Grammar:


Example

λ→

→

SaSbSGrammar:

Non-terminals

Sequence of terminals and non-terminals

The right side may be λ


Grammar: Derivation of string :

λ→

→

SaSbS

abaSbS ⇒⇒

ab

aSbS→ λ→S


Grammar: Derivation of string :

aabbaaSbbaSbS ⇒⇒⇒

aSbS→ λ→S

aabb

λ→

→

SaSbS


λ→

→

SaSbS

}0:{ ≥= nbaL nn

Grammar:

Language of the grammar:


We write: Instead of:

aaabbbS*⇒

aaabbbaaaSbbbaaSbbaSbS ⇒⇒⇒⇒

for zero or more derivation steps

A Convenient Notation


nww*

1 ⇒

nwwww ⇒⇒⇒⇒ 321

in zero or more derivation steps

In general we write:

If:

ww*⇒Trivially:


A language is context-free if there is a context-free grammar with

LG

)(GLL =

Context-Free Language:


since context-free grammar :

}0:{ ≥= nbaL nn

G

Example:

is a context-free language

generates LGL =)(

λ|aSbS →


λ||bSbaSaS →

abbaabSbaaSaS ⇒⇒⇒

Context-free grammar : G

Example derivations:

abaabaabaSabaabSbaaSaS ⇒⇒⇒⇒

=)(GL }*},{:{ bawwwR ∈

Palindromes of even length

Another Example


λ|| SSaSbS→

ababSaSbSSSS ⇒⇒⇒⇒

Context-free grammar : G

Example derivations:

abababaSbabSaSbSSSS ⇒⇒⇒⇒⇒

}prefixanyin )()( and

),()(:{

vvnvn

wnwnw

ba

ba≥

==)(GL

() ((( ))) (( ))

Describes matched parentheses: )b (, ==a

Another Example


Derivation Order and

Derivation Trees


Derivation Order Consider the following example grammar with 5 productions:

ABS→.1λ→

→

AaaAA

.3

.2λ→

→

BBbB

.5

.4


aabaaBbaaBaaABABS54321⇒⇒⇒⇒⇒

Leftmost derivation order of string : aab

At each step, we substitute the leftmost variable

ABS→.1λ→

→

AaaAA

.3

.2λ→

→

BBbB

.5

.4


aabaaAbAbABbABS32541⇒⇒⇒⇒⇒

Rightmost derivation order of string : aab

At each step, we substitute the rightmost variable

ABS→.1λ→

→

AaaAA

.3

.2λ→

→

BBbB

.5

.4


aabaaAbAbABbABS32541⇒⇒⇒⇒⇒

Rightmost derivation of : aab

aabaaBbaaBaaABABS54321⇒⇒⇒⇒⇒

Leftmost derivation of : aab

ABS→.1λ→

→

AaaAA

.3

.2λ→

→

BBbB

.5

.4


Derivation Trees Consider the same example grammar:

aabaaBbaaABbaaABABS ⇒⇒⇒⇒⇒

And a derivation of : aab

ABS→ λ|aaAA→ λ|BbB→


ABS⇒S

BA


yield AB


aaABABS ⇒⇒

a a A

S

BA


yield aaAB


aaABbaaABABS ⇒⇒⇒S

BA

a a A B b


yield aaABb


aaBbaaABbaaABABS ⇒⇒⇒⇒S

BA

a a A B b

λ


yield aaBbBbaa =λ


aabaaBbaaABbaaABABS ⇒⇒⇒⇒⇒

yield aabbaa =λλ

S

BA

a a A B b

λ λ

Derivation Tree


(parse tree)


aabaaBbaaBaaABABS ⇒⇒⇒⇒⇒

aabaaAbAbABbABS ⇒⇒⇒⇒⇒S

BA

a a A B b

λ λ

Give same derivation tree

Sometimes, derivation order doesn’t matter Leftmost derivation:

Rightmost derivation:


Ambiguity


Grammar for mathematical expressions

))(()( aaaaaaa +∗++∗+

Example strings:

Denotes any number

aEEEEEE |)(|| ∗+→


A leftmost derivation for aaa ∗+

E

EE

EE

+

a

a a

∗

aaaEaaEEaEaEEE*+⇒∗+⇒

∗+⇒+⇒+⇒



E

EE

+

a a

∗

EE a

aaaEaaEEaEEEEEE

∗+⇒∗+⇒

∗+⇒∗+⇒∗⇒

Another leftmost derivation for


aaa ∗+


aaa ∗+ E

EE

+

a a

∗

EE a

E

EE

EE

+

a

a a

∗

Two derivation trees for



E

EE

+

∗

EE

E

EE

EE

+

∗2

2 2 2 2

2

222 ∗+=∗+ aaa

take 2=a


E

EE

+

∗

EE

E

EE

EE

+

∗

6222 =∗+

2

2 2 2 2

2

8222 =∗+

4

2 2

2

6

2 2

24

8

Good Tree Bad Tree

Compute expression result using the tree


Two different derivation trees may cause problems in applications which use the derivation trees:

•  Evaluating expressions

•  In general, in compilers for programming languages


Ambiguous Grammar:

A context-free grammar is ambiguous if there is a string which has:

two different derivation trees or two leftmost derivations

G)(GLw∈

(Two different derivation trees give two different leftmost derivations and vice-versa)


E

EE

+

a a

∗

EE a

E

EE

EE

+

a

a a

∗

string aaa ∗+ has two derivation trees


this grammar is ambiguous since

Example:


string aaa ∗+ has two leftmost derivations

aaaEaaEEaEEEEEE

∗+⇒∗+⇒

∗+⇒∗+⇒∗⇒

aaaEaaEEaEaEEE*+⇒∗+⇒

∗+⇒+⇒+⇒


this grammar is ambiguous also because


IF_STMT if EXPR then STMT →| if EXPR then STMT else STMT

Another ambiguous grammar:

Variables Terminals

Very common piece of grammar in programming languages


If expr1 then if expr2 then stmt1 else stmt2

IF_STMT

expr1 then

else if expr2 then

STMT

stmt1

if

IF_STMT

expr1 then else

if expr2 then

STMT stmt2 if

stmt1

stmt2

Two derivation trees


In general, ambiguity is bad and we want to remove it

Sometimes it is possible to find a non-ambiguous grammar for a language

But, in general we cannot do so


aEEE

EEEEEE

→

→

∗→

+→

)(aEFFFTTTTEE

|)(||

→

∗→

+→

Ambiguous Grammar Non-Ambiguous

Grammar

Equivalent

generates the same language

A successful example:


aaaFaaFFaFTaTaTFTTTEE

∗+⇒∗+⇒∗+⇒

∗+⇒+⇒+⇒+⇒+⇒

E

E T+

T ∗ F

F

a

T

F

a

a

aEFFFTTTTEE

|)(||

→

∗→

+→

Unique derivation tree for aaa ∗+


}{}{ mmnmnn cbacbaL ∪=

0, ≥mn

An un-successful example:

every grammar that generates this language is ambiguous

L is inherently ambiguous:


}{}{ mmnmnn cbacbaL ∪=

λ||11

aAbAAcSS

→

→

λ||22

bBcBBaSS

→

→21 | SSS→

Example (ambiguous) grammar for : L

Professor Alex Aiken Lecture #5 (Modified by Professor Vijay

Ganesh)

49

Dealing with Ambiguity There are several ways to handle ambiguity Most direct method is to rewrite grammar

unambiguously Enforces precedence of * over +

' '

'

E E E | EE id E | id | (E) E | (E)

→ +

ʹ′ ʹ′→ ∗ ∗

50

Ambiguity No general techniques for handling ambiguity Impossible to convert automatically an

ambiguous grammar to an unambiguous one Used with care, ambiguity can simplify the

grammar. Sometimes allows more natural definitions. However, we still need to eventually eliminate ambiguity. How?

51

Precedence and Associativity Declarations Instead of rewriting the grammar

Use the more natural (ambiguous) grammar Along with disambiguating declarations

Most parser-generators allow precedence

and associativity declarations to disambiguate grammars

52

Associativity Declarations •  Ambiguous •  two parse trees of int + int + int

•  Left associativity declaration: %left +

E

E

E E

E +

int +

int int

E

E

E E

E +

int +

int int

53

Precedence Declarations •  Consider the string int + int * int •  Precedence declarations: %left + %left *

E

E

E E

E +

int *

int int

E

E

E E

E *

int +

int int


Properties of

Context-Free languages


Context-free languages are closed under: Union

1L is context free

2L is context free 21 LL ∪

is context-free

Union


Example

λ|11 baSS →

λ|| 222 bbSaaSS →

Union

}{1nnbaL =

}{2RwwL =

21 | SSS→}{}{ Rnn wwbaL ∪=

Language Grammar


In general:

The grammar of the union has new start variable and additional production 21 | SSS→

For context-free languages with context-free grammars and start variables

21, LL

21, GG21, SS

21 LL ∪S


Context-free languages are closed under: Concatenation

1L is context free

2L is context free 21LL

is context-free

Concatenation


Example

λ|11 baSS →

λ|| 222 bbSaaSS →

Concatenation

}{1nnbaL =

}{2RwwL =

21SSS→}}{{ Rnn wwbaL =

Language Grammar


In general:

The grammar of the concatenation has new start variable and additional production 21SSS→

For context-free languages with context-free grammars and start variables

21, LL

21, GG21, SS

21LLS


Context-free languages are closed under: Star-operation

L is context free *L is context-free

Star Operation


λ|aSbS→}{ nnbaL =

λ|11 SSS →*}{ nnbaL =

Example

Language Grammar

Star Operation


In general:

The grammar of the star operation has new start variable and additional production

For context-free language with context-free grammar and start variable

LGS

*L1S

λ|11 SSS →


Context-free Languages don’t have the following

Properties


Context-free languages are not closed under: intersection

1L is context free

2L is context free

21 LL ∩

not necessarily context-free

Intersection


Example

}{1mnn cbaL =

λ

λ

||

cCCaAbAACS

→

→

→

Context-free:

}{2mmn cbaL =

λ

λ

||

bBcBaAAABS

→

→

→Context-free:

}{21nnn cbaLL =∩ NOT context-free

Intersection


Context-free languages are not closed under: complement

L is context free L not necessarily context-free.

Complement

I.e., the complement of a context-free language may or may not be context-free.


}{2121nnn cbaLLLL =∩=∪

NOT context-free

Example

}{1mnn cbaL =

λ

λ

||

cCCaAbAACS

→

→

→

Context-free:

}{2mmn cbaL =

λ

λ

||

bBcBaAAABS

→

→

→Context-free:

Complement


Intersection of

Context-free languages and

Regular Languages


The intersection of a context-free language and a regular language is a context-free language

1L context free

2L regular 21 LL ∩

context-free


Applications of

Regular Closure


The intersection of a context-free language and a regular language is a context-free language

1L context free

2L regular 21 LL ∩

context-free

Regular Closure


An Application of Regular Closure

Prove that: }0,100:{ ≥≠= nnbaL nn

is context-free


}0:{ ≥nba nn

We know:

is context-free


}{ 1001001 baL = is regular

}{}){( 100100*1 babaL −+= is regular

We also know:


}{}){( 100100*1 babaL −+=

regular }{ nnba

context-free

1}{ Lba nn ∩ context-free

LnnbaLba nnnn =≥≠=∩ }0,100:{}{ 1is context-free

(regular closure)


Another Application of Regular Closure

Prove that: }:{ cba nnnwL ===

is not context-free


}:{ cba nnnwL ===

}{*}**{ nnn cbacbaL =∩

context-free regular context-free

If is context-free

Then

Impossible!!!

Therefore, is not context free L

(regular closure)

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