Transcript Languages and Finite Automata

Pumping Lemma
for
Context-free Languages
Prof. Busch - LSU
1
Take an infinite context-free language
Generates an infinite number
of different strings
Example:
S  ABE | bBd
A  Aa | a
B  bSD | cc
D  Dd | d
E  eE | e
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In a derivation of a “long” enough
string, variables are repeated
A possible derivation:
S  ABE  AaBE  aaBE
 aabSDE  aabbBdDE 
 aaabbccdDE  aabbccddE
 aabbccddeE  aabbccddee
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Derivation Tree
S
B
A
A
aabbccddee
a
a
b
E
S
b
D
B
c
e
d
d
E
e
c
Repeated
variable
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Derivation Tree
S
B
A
A
aabbccddee
a
a
b
E
S
b
D
B
c
e
d
d
E
e
c
Repeated
variable
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B  bSD  bbBdD  bbBdd
B
b
S
D
b
B
d
d
*
B  bbBdd
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S  ABE  AaBE  aaBE  aaBeE  aaBee
S
B
A
A
E
e
a
E
e
a
*
S  aaBee
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B
c
c
B  cc
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Putting all together
S
B
A
A
a
a
b
S
b
S  aaBee
e
D
B
c
*
E
d
d
E
e
c
*
B  bbBdd
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B  cc
9
We can remove the middle part
S
B
A
A
a
c
E
e
c
E
e
a
*
0
0
S  aa (bb ) cc (dd ) ee
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*
*
S  aaBee
*
B  bbBdd
*
S  aaBee  aaccee
0
B  cc
0
0
 aa (bb ) cc (dd ) ee
0
aa (bb ) cc (dd ) ee  L(G )
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We can repeated middle part two times
S
B
A
A
a
b
1
a
2
S
e
D
b
B
d
b
S
D
b
B
d
c
*
E
E
e
d
d
c
2
2
S  aa (bb ) cc (dd ) ee
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*
*
S  aaBee
*
B  cc
B  bbBdd
*
S  aaBee  aabbBddee
*
2
2
*
2
2
 aa (bb ) B(dd ) ee  aa (bb ) cc (dd ) ee
2
2
aa (bb ) cc (dd ) ee  L(G )
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We can repeat middle part three times
S
B
A
a
A
1
a
2
3
b
S
D
b
B
d
b
S
D
b
B
d
b
S
D
b
B
d
c
*
E
e
d
E
e
d
d
c
3
3
S  aa (bb ) cc (dd ) ee
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*
S  aaBee
*
*
B  bbBdd
3
B  cc
3
S  aa (bb ) cc (dd ) ee  L(G )
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Repeat middle part i times
S
B
A
a
A
1
a
b
S
b
B
E
e
D
d
E
e
d
B
i
b
S
D
b
B
d
c
*
d
c
i
i
S  aa (bb ) cc (dd ) ee
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*
*
S  aaBee
*
B  bbBdd
i
i
For any
i0
B  cc
S  aa (bb ) cc (dd ) ee  L(G )
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From Grammar
and given string
S  ABE | bBd
A  Aa | a
B  bSD | cc
D  Dd | d
E  eE | e
aabbccddee  L(G)
We inferred that a family of strings is in L(G )
*
i
i
S  aa (bb ) cc (dd ) ee  L(G )
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for any
i0
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Arbitrary Grammars
Consider now an arbitrary infinite
context-free language L
Let
Take
G be the grammar of L  {}
G so that it has no unit-productions
and no
 -productions
(remove them)
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r
Let t
Let
be the number of variables
be the maximum right-hand size
of any production
Example: S  ABE | bBd
A  Aa | a
B  bSD | cc
D  Dd | d
E  eE | e
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r 5
t 3
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Claim:
Take string w  L(G ) with | w | t .
Then in the derivation tree of w
there is a path from the root to a leaf
where a variable of G is repeated
r
Proof:
Proof by contradiction
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Derivation tree of
S
We will show:
some variable
is repeated
w
| w | m
H
H

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First we show that the tree of w
has at least r  2 levels of nodes
Suppose the opposite:
At most
r 1
Levels
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Maximum number of nodes per level
t
Level 0:
1 nodes
Level 1:
t
nodes
nodes
The maximum right-hand side of any production
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Maximum number of nodes per level
t
nodes
t
Level 0:
1 nodes
Level 1:
t
Level 2:
2nodes
t
nodes
nodes
2 nodes
t
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Maximum number of nodes per level
Level 0: 1 nodes
At most
r 1
Level
Levels
i : t i nodes
Level
r: t
r
nodes
Maximum possible string length
r
= max nodes at level r = t
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Therefore,
maximum length of string
w : | w | t
r
| w | t
r
However we took,
Contradiction!!!
Therefore,
the tree must have at least
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r2
levels
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Thus, there is a path from the root
to a leaf with at least r  2 nodes
V1
At least
r2
Levels
V1  S (root)
V2
V3
r  1 Variables
Vr 1

symbol
(leaf)
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r
Since there are at most
some variable is repeated
V1
V2
V3
Pigeonhole
principle
different variables,
S
H
H
Vr 1


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END OF CLAIM PROOF
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Take now a string
w
with
| w | t
r
S
From claim:
some variable H
is repeated
H
H

subtree of H
Take H to be the deepest, so that
only H is repeated in subtree
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We can write
S
w  uvxyz
yield
u
yield
v
H
H
u, v, x, y, z :
Strings of terminals
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x
z
yield
y
yield
yield
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Example:
S
A
B
A
u  aa
v  bb
x  cc
y  dd
z  ee
a
u
b
a
v
E
S
b
B
c
e
D
d
d
y
E
e
z
c
x
B correspond s to H
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Possible derivations

S
S  uHz
u

H
z
H  vHy
v
H
y

Hx
x
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Example:
S
B
A
A
a
a
b
S  uHz

S  aaBee
E
S
b
e
D
B
c

u  aa
v  bb
x  cc
d
d
c

H  vHy

B bbBdd
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y  dd
z  ee
E
e
B correspond s to H

Hx
B  cc
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Remove Middle Part

S
S  uHz
u
z
H

Hx
x
Yield:

*
S uHz uxz
uxz  uv xy z
0
 uv 0xy 0z
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0
 L(G )
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Repeat Middle part two times
S

S  uHz
u
H
z

1
H  vHy
v
H
y

H  vHy
v

Hx
Yield:
H
x
2
y
uvvxyyz  uv xy z
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2
2
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

S  uHz

H  vHy
*

Hx
*
S  uHz  uvHyz  uvvHyyz
*
 uvvxyyz
2
2
 uv xy z  L(G )
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Repeat Middle part

i
times
S
S  uHz
u

H
z
H  vHy
v
1
y
i
H

H  vHy
H
y
v

Hx
H
x
Yield:
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uv xy z
i
i
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

S  uHz
H  vHy




Hx

S  uHz  uvHyz  uvvHyyz




 uv Hy z  uv xy z  L(G)
i
i
i
i
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Therefore,
If we know that:
| w | t
r
w  uvxyz  L(G)
then we also know:
uv xy z  L(G )
i
i
For all
since
L(G)  L  {}
i
i
uv xy z  L
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Observation 1:
S
| vy |  1
Since G has no
unit and
-productions
u

v
z
H
y
H
x
At least one of
v or
y
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is not

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Observation 2:
|vxy |  t
S
r 1
u
since in subtree
only variable H
is repeated
v
H
H
z
y
x
Explanation follows….
subtree of H
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subtree of H
vxy  s1s2 sk
H
T1
s1
Tk
sk
T2
s2
Various yields
| s j | t
r
since no variable is repeated in
k
|vxy | | s j |  k t  t t  t
r
r
Tj
r 1
j 1
Maximum right-hand side of any production
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Thus, if we choose critical length
m t
r 1
t
r
then, we obtain the pumping lemma for
context-free languages
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The Pumping Lemma:
For any infinite context-free language
there exists an integer
for any string
L
m such that
w  L,
| w | m
we can write
w  uvxyz
with lengths
| vxy | m and | vy | 1
and it must be that:
i i
uv xy z  L,
for all i  0
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Applications
of
The Pumping Lemma
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Non-context free languages
{a b c : n  0}
n n n
Context-free languages
{a b : n  0}
n n
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Theorem: The language
n n n
L  {a b c : n  0}
is not context free
Proof:
Use the Pumping Lemma
for context-free languages
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L  {a b c : n  0}
n n n
Assume for contradiction that
L
is context-free
Since L is context-free and infinite
we can apply the pumping lemma
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L  {a b c : n  0}
n n n
Let m be the critical length
of the pumping lemma
Pick any string
We pick:
w L with length | w | m
wa b c
m m m
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L  {a b c : n  0}
n n n
wa b c
m m m
From pumping lemma:
we can write:
w  uvxyz
with lengths
| vxy | m and | vy | 1
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
| vy | 1
Pumping Lemma says:
uv xy z  L
i
i
for all
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i0
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
| vy | 1
We examine all the possible locations
of string vxy in w
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 1:
m
| vxy | m
vxy is in a
| vy | 1
m
m
m
a ...aa...aa...a bbb ...bbb ccc...ccc
u vxy
z
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
v a
| vxy | m
y a
k1
k2
m
| vy | 1
k1  k2  1
m
m
a ...aa...aa...aa...aa...a bbb ...bbb ccc...ccc
u v
x
y
z
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
v a
k1
| vxy | m
y a
k2
| vy | 1
k1  k2  1
m
m
m  k1  k2
a ...aa...aa...aa...aa...a bbb ...bbb ccc...ccc
u v2 x y2
z
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
From Pumping Lemma:
| vy | 1
uv xy z  L
2
2
k1  k2  1
However:
uv xy z  a
2
2
m k1 k2
b c L
m m
Contradiction!!!
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 2:
| vxy | m
| vy | 1
vxy is in b m
Similar to case 1
m
m
m
aaa...aaa b ...bb ...bb ...b ccc...ccc
u
vxy
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z
58
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 3:
| vxy | m
vxy is in c
| vy | 1
m
Similar to case 1
m
m
m
aaa...aaa bbb ...bbb c ...cc...cc...c
u
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vxy z
59
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 4:
| vxy | m
vxy overlaps a
m
| vy | 1
and
b
m
m
m
m
a ...aa...aa b ...bb ...bbb ccc...ccc
u
z
vxy
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Sub-case 1:
| vxy | m
| vy | 1
v contains only a
y contains only b
m
m
m
a ...aa...aa...a b ...bb ...bb ...b ccc...ccc
u v
x
y
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z
61
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
v a
k1
| vxy | m
yb
| vy | 1
k1  k2  1
k2
m
m
m
a ...aa...aa...a b ...bb ...bb ...b ccc...ccc
u v
x
y
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z
62
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
v a
k1
| vxy | m
yb
| vy | 1
k1  k2  1
k2
m  k2
m
m  k1
a ...aa...aa...a b ...bb ...bb ...b ccc...ccc
u v2
x
y
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2
z
63
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
From Pumping Lemma:
| vy | 1
uv xy z  L
2
2
k1  k2  1
However:
uv xy z  a
2
2
m k1 m k2 m
b
c L
Contradiction!!!
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Sub-case 2:
| vxy | m
| vy | 1
v contains a and b
y contains only b
m
m
m
a ...aa...a b ...bb ...bb ...bb ...b ccc...ccc
u
v
x
y
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z
65
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
| vy | 1
By assumption
v a b
k1
m
k2
k1
yb
k1 , k2  1
k3
m
k2
m
k3
a ...aa...a b ...bb ...bb ...bb ...b ccc...ccc
u
v
x
y
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z
66
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
v a b
k1
k2
| vxy | m
yb
k1 , k2  1
k3
m  k3
m
| vy | 1
m
2k3
k1 k2 k1 k2
a ...aa...ab...ba ...ab...bb ...bb ...bb ...b ccc...ccc
u
v
2
x
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y
2
z
67
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
| vxy | m
From Pumping Lemma:
| vy | 1
uv xy z  L
2
2
k1 , k2  1
However:
uv xy z  a b a b
2
2
m
k2
k1
c L
m k3
m
Contradiction!!!
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L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Sub-case 3:
| vxy | m
| vy | 1
v contains only a
y contains a and b
Similar to sub-case 2
m
m
m
a ...aa...aa...aa...a b ...bb ...b ccc...ccc
u
v
x
y
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z
69
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 5:
| vxy | m
vxy overlaps b
m
| vy | 1
and
c
m
Similar to case 4
m
m
m
aaa...aaa bbb...bbb ccc...ccc
z
u
vxy
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70
L  {a b c : n  0}
n n n
wa b c
w  uvxyz
m m m
Case 6:
| vxy | m
vxy overlaps a , b
m
| vy | 1
m and
c
m
Impossible!
m
m
m
aaa...aaa bbb...bbb ccc...ccc
z
u
vxy
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71
In all cases we obtained a contradiction
Therefore: the original assumption that
L  {a b c : n  0}
n n n
is context-free must be wrong
Conclusion:
L is not context-free
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