1

Normal Formsfor

Context-free Grammars

Linz 6th, Section 6.2

“Two Important Normal Forms,”

pages 171--178

2

3

Chomsky Normal Form

All productions have form:

BCA

variable variable

aAand

terminal

4

Examples:

bA

SAA

aS

ASS

Not Chomsky

Normal Form

aaA

SAA

AASS

ASS

Chomsky

Normal Form

5

Conversion to Chomsky Normal Form

Example:

AcB

aabA

ABaS

Not Chomsky

Normal Form

6

AcB

aabA

ABaS

Introduce variables for terminals:

cT

bT

aT

ATB

TTTA

ABTS

c

b

a

c

baa

a

cba TTT ,,

7

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

8

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

9

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

10

From any context-free grammar

not in Chomsky Normal Form

we can obtain:

An equivalent grammar

in Chomsky Normal Form

In general:

11

The Procedure

First remove:

Nullable variables

Unit productions

12

For every symbol : a

In productions: replace with a aT

Add production aTa

New variable: aT

13

Replace any productionnCCCA 21

with

nnn CCV

VCV

VCA

12

221

11

New intermediate variables: 221 ,,, nVVV

14

Theorem: For any context-free grammar G such that the empty string is not in L(G), there is an equivalent grammar in Chomsky Normal Form

Linz, 6th, Theorem 6.6, page 172.

15

Observations

• Chomsky normal forms are good

for parsing and proving theorems

• It is very easy to find the Chomsky normal

form of any context-free grammar

16

Greibach Normal Form

All productions have form:

kVVVaA 21

symbol variables

0k

17

Examples:

bB

bbBaAA

cABS

||

Greibach

Normal Form

aaS

abSbS

Not Greibach

Normal Form

18

aaS

abSbS

Conversion to Greibach Normal Form:

bT

aT

aTS

STaTS

b

a

a

bb

Greibach

Normal Form

19

Theorem: For any context-free grammar

G such that the empty string is not in L(G), there is an equivalent grammar

in Greibach Normal Form

Linz, 6th, Theorem 6.7, page 176.Proof not given because it is too

complicated.

20

Observations

• Greibach normal forms are very good

for parsing

• It is hard to find the Greibach normal

form of any context-free grammar

21

An Applicationof

Chomsky Normal Forms

22

The CYK Membership Algorithm

J. Cocke, D. H. Younger, and T. KasamiInput:

• Grammar in Chomsky Normal Form G

• String

Output:

find if )(GLw

w

Considers every possible consecutive subsequence of letters and sets K∈T[i,j] if the sequence of letters starting from i to j can be generated from the non-terminal K.

Once it has considered sequences of length

1, it goes on to sequences of length 2, and so on.

23

For subsequences of length 2 and greater, it considers every possible partition of the subsequence into two halves, and checks to see if there is some production A-> BC such that B matches the first half and C Matches the second half. If so, it records A as matching the whole subsequence.

Once this process is completed, the sentence is recognized by the grammar if the entire string is matched by the start symbol.

24

25

The Algorithm

• Grammar :G

bB

ABB

aA

BBA

ABS

• String : w aabbb

Input example:

26

[0:1]

[1:2] [2:3] [3:4] [4:5]

[0:2] [1:3] [2:4] [3:5]

[0:3] [1:4] [2:5]

[0:4] [1:5]

[0:5]

aabbb

27

a a b b b

aa ab bb bb

aab abb bbb

aabb abbb

aabbb

aabbb

28

a

A

a

A

b

B

b

B

b

B

aa ab bb bb

aab abb bbb

aabb abbb

aabbb

bB

ABB

aA

BBA

ABS

29

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

30

a

A

a

A

b

B

b

B

b

B

aa ab

S,B

bb

A

bb

A

aab

S,B

abb

A

bbb

S,B

aabb

A

abbb

S,B

aabbb

S,B

bB

ABB

aA

BBA

ABS

31

Therefore: )(GLaabbb

Time Complexity:3|| w

The CYK algorithm can be

easily converted to a parser

Observation:

32

Pushdown AutomataPDAs

33

Pushdown Automaton -- PDA

Input String

Stack

States

34

Initial Stack Symbol

Stack

$

Stack

z

bottomspecial symbol

35

The States

q1 q2a, b c

Input

symbol

Pop

symbolPush

symbol

36

q1 q2a, b c

a

b top

input

stack

a

Replaceeh

$eh

$

c

37

q1 q2a, c

a a

Pushb

eh

$eh

$

b

c

top

input

stack

38

q1 q2a, b

a a

Popb

eh

$eh

$

top

input

stack

39

q1 q2a,

a a

No Changeb

eh

$eh

$

btop

input

stack

40

Non-Determinism

q1

q2a, b c

q3a, b c

q1 q2, b c

transition

These are allowed transitions in

a Non-deterministic PDA (NPDA)

41

NPDA: Non-Deterministic PDA

Example:

,

a, a

b, aq0 q1 q2 q3

b, a

, $ $

42

a, a

b, a0q q1 q2 q3

Execution Example:

Input

a a a b b b

current

state

b, a

Time 0

, , $ $

Stack

$

43

a, a

b, aq0 q1 q2 q3

Input

a a a b b b

b, a

Time 1

, , $ $

Stack

$

44

a, a

b, aq0 q1 q2 q3

Input

Stack

a a a b b b

$

a

b, a

Time 2

, , $ $

45

a, a

b, aq0 q1 q2 q3

Input

Stack

a a a b b b

$

a

a

b, a

Time 3

, , $ $

46

a, a

b, aq0 q1 q2 q3

Input

Stack

a a a b b b

$

a

a

a

b, a

Time 4

, , $ $

47

a, a

b, aq0 q1 q2 q3

Input

a a a b b b

Stack

$

a

a

a

b, a

Time 5

, , $ $

48

a, a

b, aq0 q1 q2 q3

Input

a a a b b b

$

a

Stack

b, a

Time 6

, , $ $

a

49

a, a

b, aq0 q1 q2 q3

Input

a a a b b b

$

Stack

b, a

Time 7

, , $ $

a

50

a, a

b, aq0 q1 q2 q3

Input

a a a b b b

b, a

Time 8

accept

, , $ $

$

Stack

51

A string is accepted if there is

a computation such that:

• All the input is consumed

• The last state is a final state

At the end of the computation,

we do not care about the stack contents

52

The input string

is accepted by the NPDA:

aaabbb

a, a

b, aq0 q1 q2 q3

b, a

, , $ $

53

}0:{ nbaL nn

is the language accepted by the NPDA:

a, a

b, aq0 q1 q2 q3

b, a

In general,

, , $ $

54

Another NPDA example

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

NPDA M

}{)( RwwML

55

Execution Example:

Input

Time 0

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a ab b

56

Input

a ab

Time 1

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

ab

57

Input

Time 2

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

aa ab b

b

58

Input

Time 3

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

aa ab b

b

Guess the middle

of string

59

Input

Time 4

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

aa ab b

b

60

Input

Time 5

Stack

$

, $ $1q q2

bb

aa

,

,

, q0

bb

aa

,

,

a ab b a

61

Input

Time 6

Stack

$

, $ $q1

bb

aa

,

,

, q0

bb

aa

,

,

a ab b

accept

q2

62

Rejection Example:

Input

Time 0

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a b b b

63

Input

Time 1

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

aa b b b

64

Input

Time 2

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a

ba b b b

65

Input

Time 3

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a

b

Guess the middle

of string

a b b b

66

Input

Time 4

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a

ba b b b

67

Input

Time 5

Stack

$

, $ $1q q2

bb

aa

,

,

, q0

bb

aa

,

,

aa b b b

There is no possible transition.

Input is not consumed

68

Another computation on same string:

Input Time 0

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a b b b

69

Input

Time 1

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

aa b b b

70

Input

Time 2

Stack

$

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

a

ba b b b

71

Input

Time 3

Stack

$

a

ba b b b

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

b

72

Input

Time 4

Stack

a b b b

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

$

a

b

b

b

73

Input

Time 5

Stack

a b b b

$

a

b

b

b

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

No final state

is reached

74

, $ $q1

q2

bb

aa

,

,

, q0

bb

aa

,

,

There is no computation

that accepts string abbb

)(MLabbb