1

Simplifications of

Context-Free Grammars

2

A Substitution Rule

bB

aAB

abBcA

aaAA

aBS

Substitute

Equivalentgrammar

aAB

abbcabBcA

aaAA

abaBS

|

|

bB

3

A Substitution Rule

EquivalentgrammarabaAcabbcabBcA

aaAA

aaAabaBS

||

||

aAB

abbcabBcA

aaAA

abaBS

|

|

Substitute aAB

4

In general:

1yB

xBzA

Substitute

zxyxBzA 1|equivalentgrammar

1yB

5

Nullable Variables

:production A

Nullable Variable: A

6

Removing Nullable Variables

Example Grammar:

M

aMbM

aMbS

Nullable variable

7

M

M

aMbM

aMbSSubstitute

abM

aMbM

abS

aMbS

Final Grammar

8

Unit-Productions

BAUnit Production:

(a single variable in both sides)

9

Removing Unit Productions

Observation:

AA

Is removed immediately

10

Example Grammar:

bbB

AB

BA

aA

aAS

11

bbB

AB

BA

aA

aAS

SubstituteBA

bbB

BAB

aA

aBaAS

|

|

12

Remove

bbB

BAB

aA

aBaAS

|

|

bbB

AB

aA

aBaAS

|

BB

13

SubstituteAB

bbB

aA

aAaBaAS

||

bbB

AB

aA

aBaAS

|

14

Remove repeated productions

bbB

aA

aBaAS

|

bbB

aA

aAaBaAS

||

Final grammar

15

Useless Productions

aAA

AS

S

aSbS

aAaaaaAaAAS

Some derivations never terminate...

Useless Production

16

bAB

A

aAA

AS

Another grammar:

Not reachable from S

Useless Production

17

In general:

if wxAyS

then variable is usefulA

otherwise, variable is uselessA

)(GLw

contains only terminals

18

A production is useless if any of its variables is useless

xA

DC

CB

aAA

AS

S

aSbS

Productions

useless

useless

useless

useless

Variables

useless

useless

useless

19

Removing Useless Productions

Example Grammar:

aCbC

aaB

aA

CAaSS

||

20

First: find all variables that can producestrings with only terminals

aCbC

aaB

aA

CAaSS

|| },{ BA

AS

},,{ SBA

Round 1:

Round 2:

21

Keep only the variablesthat produce terminal symbols:

aCbC

aaB

aA

CAaSS

||

},,{ SBA

aaB

aA

AaSS

|

(the rest variables are useless)

Remove useless productions

22

Second:Find all variablesreachable from

aaB

aA

AaSS

|

S A B

Use a Dependency Graph

notreachable

S

23

Keep only the variablesreachable from S

aaB

aA

AaSS

|

aA

AaSS

|

Final Grammar

(the rest variables are useless)

Remove useless productions

24

Removing All

Step 1: Remove Nullable Variables

Step 2: Remove Unit-Productions

Step 3: Remove Useless Variables

25

Normal Formsfor

Context-free Grammars

26

Chomsky Normal Form

Each productions has form:

BCA

variable variable

aAor

terminal

27

Examples:

bA

SAA

aS

ASS

Not ChomskyNormal Form

aaA

SAA

AASS

ASS

Chomsky Normal Form

28

Convertion to Chomsky Normal Form

Example:

AcB

aabA

ABaS

Not ChomskyNormal Form

29

AcB

aabA

ABaS

Introduce variables for terminals:

cT

bT

aT

ATB

TTTA

ABTS

c

b

a

c

baa

a

cba TTT ,,

30

Introduce intermediate variable:

cT

bT

aT

ATB

TTTA

ABTS

c

b

a

c

baa

a

cT

bT

aT

ATB

TTTA

BTV

AVS

c

b

a

c

baa

a

1

1

1V

31

Introduce intermediate variable:

cT

bT

aT

ATB

TTV

VTA

BTV

AVS

c

b

a

c

ba

a

a

2

2

1

1

2V

cT

bT

aT

ATB

TTTA

BTV

AVS

c

b

a

c

baa

a

1

1

32

Final grammar in Chomsky Normal Form:

cT

bT

aT

ATB

TTV

VTA

BTV

AVS

c

b

a

c

ba

a

a

2

2

1

1

AcB

aabA

ABaS

Initial grammar

33

From any context-free grammar(which doesn’t produce )not in Chomsky Normal Form

we can obtain: An equivalent grammar in Chomsky Normal Form

In general:

34

The Procedure

First remove:

Nullable variables

Unit productions

35

Then, for every symbol : a

In productions: replace with a aT

Add production aTa

New variable: aT

36

Replace any production nCCCA 21

with

nnn CCV

VCV

VCA

12

221

11

New intermediate variables: 221 ,,, nVVV

37

Theorem:For any context-free grammar(which doesn’t produce )there is an equivalent grammar in Chomsky Normal Form

38

Observations

• Chomsky normal forms are good for parsing and proving theorems

• It is very easy to find the Chomsky normal form for any context-free grammar

39

Greinbach Normal Form

All productions have form:

kVVVaA 21

symbol variables

0k

40

Examples:

bB

bbBaAA

cABS

||

GreinbachNormal Form

aaS

abSbS

Not GreinbachNormal Form

41

aaS

abSbS

Conversion to Greinbach Normal Form:

bT

aT

aTS

STaTS

b

a

a

bb

GreinbachNormal Form

42

Theorem:For any context-free grammar(which doesn’t produce ) there is an equivalent grammarin Greinbach Normal Form

43

Observations

• Greinbach normal forms are very good for parsing

• It is hard to find the Greinbach normal form of any context-free grammar

44

Compilers

45

Compiler

Program

v = 5;if (v>5) x = 12 + v;while (x !=3) { x = x - 3; v = 10;}......

Add v,v,0cmp v,5jmplt ELSETHEN: add x, 12,vELSE:WHILE:cmp x,3...

Machine Code

46

Lexicalanalyzer parser

Compiler

program machinecode

input output

47

A parser knows the grammarof the programming language

48

ParserPROGRAM STMT_LISTSTMT_LIST STMT; STMT_LIST | STMT;STMT EXPR | IF_STMT | WHILE_STMT | { STMT_LIST }

EXPR EXPR + EXPR | EXPR - EXPR | IDIF_STMT if (EXPR) then STMT | if (EXPR) then STMT else STMTWHILE_STMT while (EXPR) do STMT

49

The parser finds the derivation of a particular input

10 + 2 * 5

Parser

E -> E + E | E * E | INT

E => E + E => E + E * E => 10 + E*E => 10 + 2 * E => 10 + 2 * 5

input

derivation

50

10

E

2 5

E => E + E => E + E * E => 10 + E*E => 10 + 2 * E => 10 + 2 * 5

derivation

derivation tree

E E

E E

+

*

51

10

E

2 5

derivation tree

E E

E E

+

*

mult a, 2, 5add b, 10, a

machine code

52

Parsing

53

grammar

Parserinputstring

derivation

54

Example:

Parser

derivation

S

bSaS

aSbS

SSSinput

?aabb

55

Exhaustive Search

||| bSaaSbSSS

Phase 1:

S

bSaS

aSbS

SSSaabb

All possible derivations of length 1

Find derivation of

56

S

bSaS

aSbS

SSS aabb

57

Phase 2

aSbS

SSS

aabb

SSSS

bSaSSSS

aSbSSSS

SSSSSS

Phase 1

abaSbS

abSabaSbS

aaSbbaSbS

aSSbaSbS

||| bSaaSbSSS

58

Phase 2

SSSS

aSbSSSS

SSSSSS

aaSbbaSbS

aSSbaSbS

Phase 3

aabbaaSbbaSbS

||| bSaaSbSSS

aabb

59

Final result of exhaustive search

Parser

derivation

S

bSaS

aSbS

SSSinput

aabb

aabbaaSbbaSbS

(top-down parsing)

60

Time complexity of exhaustive search

Suppose there are no productions of the form

A

BA

Number of phases for string : w ||2 w

61

Time for phase 1: k

k possible derivations

For grammar with rules k

62

Time for phase 2: 2k

possible derivations2k

63

Time for phase : ||2wk

possible derivations||2wk

||2 w

64

Total time needed for string :w

||22 wkkk

Extremely bad!!!

phase 1 phase 2 phase 2|w|

65

There exist faster algorithmsfor specialized grammars

S-grammar: axA

symbol stringof variables

),( aA appears oncePair

66

S-grammar example:

cS

bSSS

aSS

abccabcSabSSaSS

Each string has a unique derivation

67

In the exhaustive search parsingthere is only one choice in each phase

For S-grammars:

Total time for parsing string :w ||w

Time for a phase: 1

68

For general context-free grammars:

There exists a parsing algorithmthat parses a stringin time

||w3||w

(we will show it in the next class)

69

The CYK Parser

70

The CYK Membership Algorithm

Input:

• Grammar in Chomsky Normal Form G

• String

Output:

find if )(GLw

w

71

The Algorithm

• Grammar :G

bB

ABB

aA

BBA

ABS

• String : w aabbb

Input example:

72

a a b b b

aa ab bb bb

aab abb bbb

aabb abbb

aabbb

aabbb

73

aA

aA

bB

bB

bB

aa ab bb bb

aab abb bbb

aabb abbb

aabbb

bB

ABB

aA

BBA

ABS

74

a A

a A

b B

b B

b B

aa

ab S,B

bb A

bb A

aab abb bbb

aabb abbb

aabbb

bB

ABB

aA

BBA

ABS

75

aA

aA

bB

bB

bB

aa abS,B

bbA

bbA

aabS,B

abbA

bbbS,B

aabbA

abbbS,B

aabbbS,B

bB

ABB

aA

BBA

ABS

76

Therefore: )(GLaabbb

Time Complexity:3||w

The CYK algorithm can be easily converted to a parser(bottom up parser)

Observation: