Theory of computation Lecture Slide(Chomsky Normal Form).pptx

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Instructor:
Lecture: # Week: # Semester:
AMERICAN INTERNATIONAL UNIVERSITY-BANGLADESH
CONTEXT-FREE LANGUAGES (CFL)
CSC3113: THEORY OF COMPUTATION
10 5 Spring 2023-2024
Dr. Afroza Nahar
Department of Computer Science, Faculty of Science & Technology.
afroza@aiub.edu

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LECTURE OUTLINE
LECTURE OUTLINE
CSC3113: Theory of Computation
Ambiguity
Chomsky Normal Form (CNF)

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LEARNING OBJECTIVE
LEARNING OBJECTIVE
Concept and Construct of Context Free Grammar (CFG)
Concept of ambiguity and its removal from grammar.
Chomsky Normal From (CNF)

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LEARNING OUTCOME
LEARNING OUTCOME
ALL OUTCOME ARE REPRESENTED WITH EXAMPLES
Understand and practice the construct of CFG
Learn ambiguity in grammar and ways to identify and remove it.
Apply Chomsky Normal From (CNF) on grammar to make it
computable.

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Sample derivations:
S  AB  AAB  aAB  aaB  aabB  aabb
S  AB  AbB  Abb  AAbb  Aabb  aabb
S  A | A B
A  e | a | A b | A A
B  b | b c | B c | b B
S
A B
A
A B
b
a a b
These two derivations are special.
1st derivation is leftmost.
Always picks leftmost variable.
2nd derivation is rightmost.
Always picks rightmost variable.

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DERIVATION TREES
Infinitely
many others
possible.
S
A B
A
A b
a
a
b
A
S
A
A A
A
A b
A
a e
a
b
A
S
A B
A
A B
b
a a b
S  A | A B
A  e | a | A b | A A
B  b | b c | B c | b B
w = aabb
Other derivation
trees for this
string?
?

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DERIVATION TREES
 Observation: Every derivation tree corresponds to one or more derivations.
leftmost: rightmost: mixed:
S => AB S => AB S => AB
=> aAAB => Ab => Ab
=> aaAB => aAAb => aAAb
=> aaaB =>aAab => aaAb
=> aaab => aaab => aaab
 Definition: A derivation is leftmost (rightmost) if at each step in the derivation a
production is applied to the leftmost (rightmost) non-terminal in the sentential
form.
 The first derivation above is leftmost, second is rightmost, the third is neither.
Rules:
S –> AB
A –> aAA
A –> aA
A –> a
B –> bB
B –> b

 Example: Consider the string aaab and the preceding grammar.
S –> AB S => AB S
A –> aAA => aAAB
A –> aA => aaAB A B
A –> a => aaaB
B –> bB => aaab a A A b
B –> b
a a
S => AB S
=> aAB
=> aaAB A B
=> aaaB
=> aaab a A b
a A
a
 The string has two left-most derivations, and therefore has two distinct parse trees.

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EXAMPLE
Draw parse tree and leftmost derivation for the expression
ab#baab using following CFG:
S  CB
C  aCa | bCb | #B
B  AB | 
A  a | b

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AMBIGUITY – PARSE TREE
 If a grammar generates the same string in several different ways, we say that the
string is derived ambiguously in the grammar.
 If a grammar generates some string ambiguously we say that the grammar is
ambiguous.
 Example: Grammar G, EXPREXPR+EXPR|EXPREXPR|(EXPR)|a
 Two parse trees for the string a+aa in G
EXPR
EXPR
EXPR
EXPR
EXPR
a a
a
+ 
EXPR
EXPR
EXPR
EXPR
EXPR
a
a
a 
+

AMBIGUITY AND PARSE TREES
The ambiguity of 01+1 is shown by the two
different parse trees:
S
+
S
 S
1
S
0
S
1
S
 S
+ S
1
S
1
S
0

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AMBIGUITY - DERIVATION
When we say that a grammar generates a string ambiguously, we
mean that the string has two different parse trees, not two different
derivations.
A derivation of string w in a grammar G is a leftmost derivation if at
every step the leftmost remaining variable is the one replaced.
Then we can say, a string w is derived ambiguously in CFG G if it has
two or more different leftmost derivations.
Grammar G is ambiguous if it generates some string ambiguously.
Some CFLs can only be generated by ambiguous grammars. Such
languages are called inherently ambiguous.
Example: {0i1j2k | i=j or j=k}

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AMBIGUITY - DERIVATION
When we say that a grammar generates a string ambiguously, we
mean that the string has two different parse trees, not two different
derivations.
A derivation of string w in a grammar G is a leftmost derivation if at
every step the leftmost remaining variable is the one replaced.
Then we can say, a string w is derived ambiguously in CFG G if it has
two or more different leftmost derivations.
Grammar G is ambiguous if it generates some string ambiguously.
Some CFLs can only be generated by ambiguous grammars. Such
languages are called inherently ambiguous.
Example: {0i1j2k | i=j or j=k}

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Using Parse Tree prove that the following grammar is ambiguous for
the string 00110011.
S → 0A | 1B
A → 0AA | 1S | 1
B → 1BB | 0S | 0

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CHOMSKY NORMAL FORM
It is often convenient to have CFGs in simplified form. One
such form is Chomsky normal form.
A context free grammar is in Chomsky normal form if every
rule is of the form
A  BC
A  a
where a is any terminal and A, B, and C are any variables –
except that B and C may not be the start variable.
In addition S  e is permitted, where S is the start
variable.

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CONVERT ANY GRAMMAR GTO CHOMSKY NORMAL FORM
Add a new start symbol S0 and the new rule S0S,
where S was the original start symbol.
Eliminate all e rules of the form A  e, where A is not
the start symbol.
Add rule Ruv for every rule of the form RuAv,
where u and v are strings of variables and terminals.
Add such rules for every occurrence of A. for example,
add RuvAw, RuAvw, Ruvw for the rule of the
form RuAvAw.
Add Re for the rule of the form RA, unless we have
previously removed the rule Re.

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CONVERT ANY GRAMMAR GTO CHOMSKY NORMAL FORM
Eliminate all unit rules of the form A  B.
Add rule A  u for the rule of the form B  u, unless this was
a unit rule previously removed.
Here u is a string of variables and terminals.
Convert remaining rules into proper form,
RPQ and Ru.
We replace each rule of the form A  u1u2…uk with the rules
Au1A1, A1u2A2, A2u3A3, … , Ak-2uk-1uk.
Here k  3 and each ui is a variable or terminal symbol,
and Ai’s are new variables.
If k  2, replace any terminal ui in the preceding rule(s) with the
new variable Ui and the rule Uiui.
The above procedure converts a Grammar to a Chomsky
normal form. Next, we will go through an example.

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EXAMPLE: CONVERTING TO CNF
Original Grammar
S  ASA | aB
A  B | S
B  b | e
CNF: Rules Format
1. A variable substitute by
two variables
V1  V2V3
a single terminal
V1  t
3. Exception: the start
variable may have an
epsilon rule
S0  e
Conversion Steps:
1. Add a new start variable (substitute by old start variable).
2. Remove all epsilon rules, V1  e.
i. Look for V1 on right side of the arrow of all the rules.
ii. Replace these V1 by e in all possible way, whatever string
comes as a result, add them to the same rule using OR “|”.
3. Remove all unit rules, V1  V2.
i. Look for the rule R for V2.
ii. Replace the V2 of the unit rule by right side of the arrow of R.
4. Convert all rules to CNF rule format, V1  V2V3, V1  t.
i. Introduce new rules with new variables to convert the
existing non-CNF rules to CNF rules.
ii. Try to have these new rules in the CNF format.
CNF Conversion:
Step 1:
Adding new
start variable S0.
Step 2:
S0  S
S  ASA | aB
A  B | S
B  b | e
S0  S
S  ASA | aB | a
A  B | S | e
B  b | e
S0  S
S  ASA | aB | a
A  B | S | e
B  b
S0  S
S  ASA | aB
A  B | S
B  b | e

Remove B  e
Remove A  e
S0  S
S  ASA | aB | a
A  B | S | e
B  b

S0  S
S  ASA | aB | a | SA
A  B | S | e
B  b
S0  S
S  ASA | aB | a | SA | AS
A  B | S | e
B  b
S0  S
S  ASA | aB | a | SA | AS | S
A  B | S | e
B  b
S0  S
A  B | S
B  b
S0  S
S  ASA | aB
A  B | S
B  b | e

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Original Grammar
S  ASA | aB
A  B | S
B  b | e
CNF: Rules Format
two variables
V1  V2V3
a single terminal
V1  t
epsilon rule
S0  e
Conversion Steps:
2. Remove all epsilon rules of the form, V1  e.
3. Remove all unit rules of the form, V1  V2.
CNF Conversion:
Step 4:
Add rule C  SA
Add rule D  a
Find non-CNF rules
B  b
B  b
C  SA
S0  AC | aB | a | SA | AS
S  AC | aB | a | SA | AS
A  b | AC | aB | a | SA | AS
B  b
C  SA
S0  AC | aB | a | SA | AS
A  b | AC | aB | a | SA | AS
B  b
C  SA
S0  AC | aB | a | SA | AS
A  b | AC | aB | a | SA | AS
B  b
C  SA
D  a
S0  AC | DB | a | SA | AS
S  AC | DB | a | SA | AS
A  b | AC | DB | a | SA | AS
B  b
C  SA
D  a
S0  AC | DB | a | SA | AS
S  AC | DB | a | SA | AS
A  b | AC | DB | a | SA | AS
B  b
C  SA
D  a
B  b

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EXAMPLE-2
Original CFG
S  bS | aT | 
T  aT | bR | 
R  bS | 
Add new Start Variable P
P  S
S  bS | aT | 
T  aT | bR | 
R  bS | 
Remove R  
P  S
S  bS | aT | 
T  aT | bR |  | b
R  bS
Remove T  
P  S
S  bS | aT |  | a
T  aT | bR | b | a
R  bS
Remove S  
P  S | 
S  bS | aT | a | b
T  aT | bR | b | a
R  bS | b
Remove P  S
P  bS | aT | a | b | 
S  bS | aT | a | b
T  aT | bR | b | a
R  bS | b
Find rules not in CNF
format
P  bS | aT | a | b | 
S  bS | aT | a | b
T  aT | bR | b | a
R  bS | b
Add new Rules to convert the
rules not in CNF format
P  BS | AT | a | b | 
S  BS | AT | a | b
T  AT | BR | b | a
R  BS | b
A  a
B  b
Finally, in CNF format
P  BS | AT | a | b | 
S  BS | AT | a | b
T  AT | BR | a | b
R  BS | b
A  a
B  b

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EXAMPLE-3
Convert CFG to CNF
Original CFG
S  aXbX
X  aY | bY | 
Y  X | c

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REFERENCES
CONTEXT FREE LANGUAGE & GRAMMAR
Introduction to Theory of Computation, Sipser, (3rd ed),
CFL-2; All Exercises.
Elements of the Theory of Computation, Papadimitriou (2nd ed),
CFL-2.

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