Class7

Context-Free Languages

Derivation tree, parsing, & ambiguity

context free grammar and ambiguity 1

n n R
{a b } {ww }

Regular Languages



Context-Free Pushdown
Grammars Automata

stack

automaton


Context-Free Grammars


n n
L(G ) {a b : n 0} is not regular ….

(((( ))))

we relax the restriction of regular grammar


n n

A context-free grammar G : S aSb
S

The grammar G is linear but not regular.


n n

Regular grammar (e.g., a right linear grammar):
A XB or A  X
where A, B V, and X T*

Context Free
A  X where X (V T)* Grammar

Def. Context-Free Grammars (CFG)

Grammar G (V , T , S , P )

Variables Terminal Start
symbols variable

Productions of the form: A x
x is string of variables and terminals
Note: A regular grammar is a regular grammar.

Def. Context-Free Languages (CFL)

A language L is context-free
if and only if

there is a context free grammar G
with L L (G )

Note: To show L is a CFL find a CFG for L.
Note: A regular language is also a CFL.

n n is a Context-Free
L(G ) {a b : n 0} Language

with context-free grammar G: S aSb
S

A derivation:
S aSb aaSbb aaaSbbb aaabbb


Another Example

A context-free grammar G: S aSa
S bSb
A derivation: S
S aSa abSba abaSaba abaaba

R
L(G ) {ww : w {a, b}*}
is a Context-Free Language

A non Linear Examples

A context-free grammar G: S aSb
S SS
S
What is the L(G)?

A derivation:
S SS aSbS abS abaSb
abaaSbb abaabb () (( ))

S aSb
S SS
S () ((( ))) (( ))
L(G ) {w : na ( w) nb ( w), and
na (v) nb (v) in any prefix v}
How to prove the given grammar
indeed generates the language?
Homework #4 only shows L(G) L.


L(G) L: by induction on the number of steps in deriving a sentential form
Any sentential form obtained from the grammar G: if by one step (rule2 & 3
do not generate any terminal symbols at all, i.e. na(w)=nb(w) = 0, thus is
true; if it is from rule 1: SaSb, then it satisfies the condition of
na(w)=nb(w) and the prefix condition.
Assume any sentential obtained from n steps it is true,
Now assume a sentential form of (n+1) steps: if the last step is by rule 2 or
3, then no new symbols is added, therefore it satisfies the condition since
it satisfies on previous step; it the last step is by rule 1, then as can be
seen that a is added before b is added (from left to right), thus the prefix
condition is also satisfied.

L L(G): by induction on |w|, w L.
By observing the definition of L, w L then |w| is even, let |w|=2n. and if
|w| >= 2, w = a…..b (the first symbol is a and the last symbol is b).
|w|=0: then w= , then w L(G).
Assume |w|=2n, w L(G) is true for n >= 0.
Now |w| = 2n+2, then w=aub with |u|=2n, and u L. by induction
assumption, u L(G), i.e. there exist a derivation s * u, thus S aSb *
aub = w.

Show L={anbmcn: n, m 0 } is CF.

Show L={anbncm: n, m 0 } is CF.


True or False

If L is not regular then L is context
free.

If L is not context free then L is not
regular

From grammar
G: S AB, A aA| , B bB| , thus
L(G) is CF & not regular.

Examples of showing languages are
Context Free

Show { anbm : n = m+3} is CF.

Show { anbm : n m+3} is CF.

Try hw # 7 (a) ~ (d) p.133


{anbm: n m+3}
S1 is for n> m+3 & S2 is for n < m+3 & S  S1
| S2
S1: n m+4, i.e., n = m+t for t=4, 5, 6, …, i.e.
anbm= am+tbm: thus, S1  aaaaB | aS1, B bB | .
S2: n m+2, i.e, n+t =m+2, m = n-2 + t for t=0, 1,
2, …, i.e, anbm= anbn-2+t for t=0,1,2, … which
implies anbm= anbn-2 , anbn-1 , anbn , anbn+1 ,
anbn+2 , ....
(anbn-2 :) D  aa K, K aKb | ; (anbn-1:) E
aK; (anbn , anbn+1 , anbn+2 , .... ) F Fb | K; and
S2  D | E | F

Examples of showing languages are
Context Free

Show { w: na(w) = nb(w)} is CF.

Show { w: na(w) > nb(w)} is CF.

Show { w: na(w) = 2 nb(w)} is CF.

Try hw # 7 (e) ~ (g) p.133


S aSb | bSa |SS| (for na = nb)
S1 for na(w) >nb(w) S1 aS1 | aS | SS1

{ w {a,b,c}*: |w|=3na(w)}
E  aEbEc | aEcEb | bEaEc | bEaEc | cEaEb
| cEbEa | EE | (this is for na = nb = nc)
S  aEKEK | KEaEK | KEKEa
K  b | c.

Derivation Order for w = aab in the CFG:
1. S AB 2. A aaA 4. B Bb
3. A 5. B
Leftmost derivation:
1 2 3 4 5
S AB aaAB aaB aaBb aab
Rightmost derivation:
1 4 5 2 3
S AB ABb Ab aaAb aab

Derivation Trees
Another way of showing derivation,
independent of the order in which
productions are used.


S AB A aaA | B Bb |

S AB
S

A B

Every node to its children is corresponding to a
production rule in the grammar


S AB A aaA | B Bb |

S AB aaAB
S

A B

a a A


S AB A aaA | B Bb |

S AB aaAB aaABb
S

A B

a a A B b


S AB A aaA | B Bb |

S AB aaAB aaABb aaBb
S

A B

a a A B b


S AB A aaA | B Bb |

S AB aaAB aaABb aaBb aab
S
A Derivation
Tree
A B

a a A B b
Every leaf has a
label from T { }

S AB A aaA | B Bb |

S AB aaAB aaABb aaBb aab
Derivation Tree S

A B
yield

a a A B b aa b
aab


derivation order doesn’t matter
Leftmost:
S AB aaAB aaB aaBb aab
Rightmost:
S AB ABb Ab aaAb aab
S
Same derivation tree
A B

a a A B b


S AB A aaA | B Bb |

S Derivation Tree
shows which rules are
A B used in obtaining a w,
but do not give the order
of their applications.
a a A B b
We can find the corresponding
left-most or right-most derivations
from a given derivation tree easily.


Partial Derivation Trees Every leaf has a label
from T { } V
S AB A aaA | B Bb |
The yield of a partial derivation
S AB tree a sentential form of a
derivation

Partial derivation S
tree with root S

A B


sentential
S AB aaAB
form

Partial derivation
S
tree with root S
A B

yield
a a A
aaAB

G: A CFG then Thm. 5.1 on p.132

Every w L(G) there exists a derivation tree
with yield = w. a deriation with the
sentential form w
Conversely,
the yield of any derivation tree w, w L(G)
partial derivation tree tG (with root S), w
Homework for 5.1 (p.133) : 2~5, 7, 8, 11, 13~24
Hand in: 7bef, 8bdg, 13, 15, 17, 18, 21
Prove by induction on the number of steps in the derivation. Check with the
book on p.132.

Ambiguity


E E E | E E | (E) | a
a a a
leftmost derivation

E E E E a E a E E
a a E a a*a
E E

a E E

a a

E E E | E E | (E) | a
a a a leftmost derivation

E E E E E E a E E E
a a E a a a
E E

E E a

a a

E E E | E E | (E) | a
a a a
Two derivation trees
E E

E E E E

a E E E E a

a a a a

The grammarE E E | E E | (E) | a
is ambiguous:
Because a string of L(G)
has two derivation trees

E E
a a a
E E E E

a E E E E a

a a a a

Or we can say

A grammar is ambiguous….


The grammarE E E | E E | (E) | a
is ambiguous:

There are two different leftmost
derivations for a a a

E E E a E a E E
a a E a a*a
E E E E E E a E E
a a E a a a

Definition:
A context-free grammar G is ambiguous

if some string w L(G ) has

two or more derivation trees


In other words:

A context-free grammar G is ambiguous

if some string w L(G ) has:

two or more leftmost derivations
(or rightmost)


Why do we care about ambiguity?
a a a
take a 2
E E

E E E E

a E E E E a

a a a a

2 2 2

E E

E E E E

2 E E E E 2

2 2 2 2

2 2 2 6 2 2 2 8
6 8
E E
2 4 4 2
E E E E
2 2 2 2
2 E E E E 2

2 2 2 2

Correct result: 2 2 2 6

6
E
2 4
E E
2 2
2 E E

2 2

• Ambiguity is bad for programming languages

• We want to remove ambiguity


We fix the ambiguous grammar:
E E E | E E | (E) | a

New non-ambiguous grammar: E E T
E T
T T F
T F
F (E)
context free grammar and ambiguity F a 49

E E T T T F T a T a T F
a F F a a F a a a
E a a a
E E T
E T
E T
T T F T T F
T F
F F a
F (E)
F a a a

Unique derivation tree

E a a a
E T

T T F

F F a

a a

The grammar G: E E T
E T
T T F
T F
F (E)
F a
is non-ambiguous:
i.e. EVERY stringw L (G ) has
a unique derivation tree

Inherent Ambiguity

Some context free languages
have only ambiguous grammars

Example: n n m n m m
L {a b c } {a b c }

S S1 | S2 S1 S1c | A S2 aS2 | B
A aAb | B bBc |

n n n
The string a b c
has two derivation trees

S S

S1 S2

S1 c a S2


Compilers


Machine Code
Program Add v,v,0
v = 5; cmp v,5
if (v>5) jmplt ELSE
x = 12 + v; THEN:
while (x !=3) { Compiler
add x, 12,v
x = x - 3; ELSE:
v = 10; WHILE:
} cmp x,3
...... ...

A parser knows the grammar
of the programming language

Lexical
Compiler parser
analyzer

input output

machine
program
Lexical analyzer : analyze the given input code
is composed of (identifier, or an expression, 57

or a statement, etc)
context free grammar and ambiguity

Parser
PROGRAM STMT_LIST
STMT_LIST STMT; STMT_LIST | STMT;
STMT EXPR | IF_STMT | WHILE_STMT
| { STMT_LIST }

EXPR EXPR + EXPR | EXPR - EXPR | ID
IF_STMT if (EXPR) then STMT
| if (EXPR) then STMT else STMT
WHILE_STMT while (EXPR) do STMT


The parser finds the derivation
of a particular input
derivation
Parser
input E => E + E
E -> E + E
=> E + E * E
10 + 2 * 5 |E*E
=> 10 + E*E
| INT
=> 10 + 2 * E
=> 10 + 2 * 5


derivation tree
derivation
E

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


derivation tree

E machine code

E + E mult a, 2, 5
add b, 10, a
10
E * E

2 5


Parsing


Parser
input
grammar derivation
string


Example:

Parser
S SS derivation
input
S aSb
aabb ?
S bSa
S

Exhaustive Search
S SS | aSb | bSa |

Phase 1: S SS Find derivation of
S aSb aabb
S bSa
S
All possible derivations of length 1

S SS aabb
S aSb
S bSa
S


Phase 2 S SS | aSb | bSa |
S SS SSS
S SS aSbS aabb
Phase 1 S SS bSaS
S SS S SS S
S aSb S aSb aSSb
S aSb aaSbb
S aSb abSab
S aSb ab

Phase 2 S SS | aSb | bSa |
S SS SSS
S SS aSbS aabb
S SS S

S aSb aSSb
S aSb aaSbb
Phase 3
S aSb aaSbb aabb

Final result of exhaustive search
(top-down parsing)
Parser
S SS
input
S aSb
aabb
S bSa
S
What happen if derivation
there is no such
derivation existed?
I.e. what if w L(G)? S aSb aaSbb aabb

Is this searching method an
algorithm?

What if w L(G)?


Time complexity of exhaustive search

Suppose there are no productions of the form
A : - production

A B : unit - production

Number of phases for string w : 2| w|
So that after the execution of every phase, either
(1) the sentential form has additional one or more
terminal symbols; and/or
(2) the length of the sentential form is increased by at
least one.

For grammar with k rules

Time for phase 1: k

k possible derivations


Time for phase 2: k 2

k 2 possible derivations


Time for phase 2 | w |: k 2|w|

2|w| possible derivations
k


Total time needed for string w:

2 2|w|
k k  k

phase 1 phase 2 phase 2|w|

Extremely bad!!!

There exist faster algorithms
for specialized grammars

Simple grammar: A ax

ONE symbol V*: string
from of variables

such that pair ( A, a ) appears at most once


S aS
S-grammar example:
S bSS
S c

Not an S-grammar: S aS
S bSS | bS
S


Simple grammar: A ax

Observe that if G is an S-grammar, then
G has no productions like
A : - production

A B : unit - production

In a derivation process for S-grammar, every
phase will increase one terminal symbol.

S aS
S-grammar example:
S bSS
S c

Find a (left most)derivation for abcc:
S aS abSS abcS abcc
Find a (left most)derivation for bca:
S bSS bcS bcaS
bca L(G) since |bcaS| > | bca |

If G is an S-grammar, then

in the exhaustive search parsing
there is only one choice in each phase
Thus, for any given string w, it
takes a linear time, O(|w|), to decide
either w L(G) by providing a
unique derivation or w L(G).

Not every CF grammar can be converted into an S-grammar.


S-grammar Example

Let G = (V, T, S, P) be an S-grammar:

Give an expression for the maximum
size of P in terms of |V| and |T|.

Is it possible for an s-grammar to be
ambiguous?


For GENERAL context-free grammars:
There exists a parsing algorithm
that parses a string w
in time | w |3

The CYK membership algorithm (on 6.3)

5.2 HW p.144
1~9,11~15,17~19


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