1 normal forms for context-free grammars. 2 chomsky normal form all productions have form: variable...
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1
Normal Formsfor
Context-free Grammars
2
Chomsky Normal Form
All productions have form:
BCA
variable variable
aAand
terminal
3
Examples:
bA
SAA
aS
ASS
Not ChomskyNormal Form
aaA
SAA
AASS
ASS
Chomsky Normal Form
4
Convertion to Chomsky Normal Form
Example:
AcB
aabA
ABaS
Not ChomskyNormal Form
5
AcB
aabA
ABaS
Introduce variables for terminals:
cT
bT
aT
ATB
TTTA
ABTS
c
b
a
c
baa
a
cba TTT ,,
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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
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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
8
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
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From any context-free grammarnot in Chomsky Normal Form
we can obtain: An equivalent grammar in Chomsky Normal Form
In general:
10
The Procedure
First remove:
Nullable variables
Unit productions
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For every symbol : a
In productions: replace with a aT
Add production aTa
New variable: aT
12
Replace any production nCCCA 21
with
nnn CCV
VCV
VCA
12
221
11
New intermediate variables: 221 ,,, nVVV
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Theorem:For any context-free grammar there is an equivalent grammar in Chomsky Normal Form
14
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
15
Greinbach Normal Form
All productions have form:
kVVVaA 21
symbol variables
0k
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Examples:
bB
bbBaAA
cABS
||
GreinbachNormal Form
aaS
abSbS
Not GreinbachNormal Form
17
aaS
abSbS
Conversion to Greinbach Normal Form:
bT
aT
aTS
STaTS
b
a
a
bb
GreinbachNormal Form
18
Theorem:For any context-free grammarthere is an equivalent grammarin Greinbach Normal Form
19
Observations
• Greinbach normal forms are very good for parsing
• It is hard to find the Greinbach normal form of any context-free grammar
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An Applicationof
Chomsky Normal Forms
21
The CYK Membership Algorithm
Input:
• Grammar in Chomsky Normal Form G
• String
Output:
find if )(GLw
w
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The Algorithm
• Grammar :G
bB
ABB
aA
BBA
ABS
• String : w aabbb
Input example:
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a a b b b
aa ab bb bb
aab abb bbb
aabb abbb
aabbb
aabbb
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aA
aA
bB
bB
bB
aa ab bb bb
aab abb bbb
aabb abbb
aabbb
bB
ABB
aA
BBA
ABS
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aA
aA
bB
bB
bB
aa abS,B
bbA
bbA
aab abb bbb
aabb abbb
aabbb
bB
ABB
aA
BBA
ABS
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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
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Therefore: )(GLaabbb
Time Complexity:3||w
The CYK algorithm can be easily converted to a parser
Observation:
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Pushdown AutomataPDAs
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Pushdown Automaton -- PDA
Input String
Stack
States
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Initial Stack Symbol
Stack
$
Stack
z
bottomspecial symbol
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The States
q1 q2a, b c
Inputsymbol
Popsymbol
Pushsymbol
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q1 q2a, b c
a
b top
input
stack
a
Replaceeh
$eh
$
c
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q1 q2a, c
a a
Pushb
eh
$eh
$
bc
top
input
stack
34
q1 q2a, b
a a
Popb
eh
$eh
$
top
input
stack
35
q1 q2a,
a a
No Changeb
eh
$eh
$
btop
input
stack
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Non-Determinism
q1
q2a, b c
q3a, b c
q1 q2, b c
transition
These are allowed transitions in a Non-deterministic PDA (NPDA)
37
NPDA: Non-Deterministic PDA
Example:
,
a, a
b, a q0 q1 q2 q3
b, a
, $ $
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a, a
b, a 0q q1 q2 q3
Execution Example:
Input
a a a b b b
currentstate
b, a
Time 0
, , $ $
Stack
$
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a, a
b, a q0 q1 q2 q3
Input
a a a b b b
b, a
Time 1
, , $ $
Stack
$
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a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b
$
a
b, a
Time 2
, , $ $
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a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b$
aa
b, a
Time 3
, , $ $
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a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b
$
aaa
b, a
Time 4
, , $ $
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a, a
b, a q0 q1 q2 q3
Input
a a a b b b
Stack
$
aaa
b, a
Time 5
, , $ $
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a, a
b, a q0 q1 q2 q3
Input
a a a b b b$
a
Stack
b, a
Time 6
, , $ $
a
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a, a
b, a q0 q1 q2 q3
Input
a a a b b b$
Stack
b, a
Time 7
, , $ $
a
46
a, a
b, a q0 q1 q2 q3
Input
a a a b b b
b, a
Time 8
accept, , $ $
$
Stack
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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
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The input stringis accepted by the NPDA:
aaabbb
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
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}0:{ nbaL nn
is the language accepted by the NPDA:
a, a
b, a q0 q1 q2 q3
b, a
In general,
, , $ $
50
Another NPDA example
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
NPDA M
}{)( RwwML
51
Execution Example:
Input
Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a ab b
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Input
a ab
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
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Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
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Input
Time 3
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
Guess the middle of string
55
Input
Time 4
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
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Input
Time 5
Stack
$
, $ $1q q2
bb
aa
,
,
, q0
bb
aa
,
,
a ab b a
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Input
Time 6
Stack
$
, $ $q1
bb
aa
,
,
, q0
bb
aa
,
,
a ab b
accept
q2
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Rejection Example:
Input
Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a b b b
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Input
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
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Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
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Input
Time 3
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
Guess the middle of string
a b b b
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Input
Time 4
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
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Input
Time 5
Stack
$
, $ $1q q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
There is no possible transition.
Input is not consumed
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Another computation on same string:
Input Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a b b b
65
Input
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
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Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
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Input
Time 3
Stack
$ab
a b b b
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
b
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Input
Time 4
Stack
a b b b
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
$abbb
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Input
Time 5
Stack
a b b b
$abbb
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
No final state is reached
70
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
There is no computation that accepts string abbb
)(MLabbb