1. Properties
of
Context-Free
languages

2. Union
Context-free languages
are closed under: Union
L1 is context free
L1 L2
L2 is context free is context-free

3. Example
Language Grammar
n n
L1 {a b } S1 aS1b |
R
L2 {ww } S2 aS2a | bS2b |
Union
n n R S S1 | S2
L {a b } {ww }

4. In general:
For context-free languages L1, L2
with context-free grammars G1, G2
and start variables S1, S2
It is assumed that variable sets of G1 and G2 are
disjoint
The grammar of the union L1 L2
has new start variable S
and additional production S S1 | S2

5. Concatenation
Context-free languages
are closed under: Concatenation
L1 is context free
L1L2
L2 is context free is context-free

6. Example
Language Grammar
n n
L1 {a b } S1 aS1b |
R
L2 {ww } S2 aS2a | bS2b |
Concatenation
n n R S S1S2
L {a b }{ww }

7. In general:
For context-free languages L1, L2
with context-free grammars G1, G2
and start variables S1, S2
The grammar of the concatenation L1L2
has new start variable S
and additional production S S1S2

8. Star Operation
Context-free languages
are closed under: Star-operation
L is context free *
L is context-free

9. Example
Language Grammar
n n
L {a b } S aSb |
Star Operation
n n
L {a b } * S1 SS1 |

10. In general:
For context-free language L
with context-free grammar G
and start variable
S
The grammar of the star operation L *
has new start variable S1
and additional production S1 SS1 |

11. Negative Properties
of
Context-Free
Languages

12. Intersection
Context-free languages
are not closed under: intersection
L1 is context free
L1 L2
L2 is context free not necessarily
context-free

13. Example
n n m n m m
L1 {a b c } L2 {a b c }
Context-free: Context-free:
S AC S AB
A aAb | A aA |
C cC | B bBc |
Intersection
n n n
L1 L2 {a b c } NOT context-free

14. Complement
Context-free languages
are not closed under: complement
L is context free L not necessarily
context-free

15. Example
n n m n m m
L1 {a b c } L2 {a b c }
Context-free: Context-free:
S AC S AB
A aAb | A aA |
C cC | B bBc |
Complement
n n n
L1 L2 L1 L2 {a b c }
NOT context-free

16. Context-free
languages
and
Regular Languages

17. The intersection of
a context-free language and
a regular language
is a context-free language
L1 context free
L1 L2
L2 regular context-free

18. Machine M1 Machine M2
NPDA for L1 DFA for L2
context-free regular
Construct a new NPDA machine M
that accepts L1 L2
M simulates in parallel M1 and M 2

19. NPDA M1 DFA M2
a, b c a
q1 q2 p1 p2
transition transition
NPDA M
q1, p1 a, b c
q2 , p2
transition

20. NPDA M1 DFA M2
q0 p0
initial state initial state
NPDA M
q0 , p0
Initial state

21. Therefore: L ( M1 ) L( M 2 ) is context-free
(since M is NPDA)
L1 L2 is context-free

22. The Pumping Lemma
for
Context-Free
Languages

23. Take an infinite context-free language
Generates an infinite number
of different strings
Example: S AB
A aBb
B Sb
B b

24. S AB
A aBb
B Sb
B b
A derivation: Variables are repeated
S AB aBbB abbB
abbSb abbABb abbaBbBb
abbabbBb abbabbbb

25. Derivation tree string abbabbbb
S
A B
a B b S b
b A B
a B b b
b

26. Derivation tree string abbabbbb
S
A B
a B b S b
b A B
a B b b
repeated b

27. B Sb ABb B
aBbBb aBbbb S b
A B
a B b b
b
B b

28. Repeated Part B
S b
A B
a B b b
B  aBbbb

29. Another possible derivation B
S b
A B
B  aBbbb a B b b
S b
A B
a B b b
B  aBbbb aaBbbbbbb

30. S
A B
a B b S b
b A B
B  aBbbb
a B b b
S  abbaBbbb

31. S
A B
a B b S b
b A B
B  aBbbb
a B b b
S b
A B
a B b b
S  abbaBbbb  abbaaBbbbbbb

32. S
A B
a B b S b
b A B
a B b b
S b
A B
a B b b B b
b
S  abbaaBbbbbbb abbaabbbbbbb

33. S  abbaabbbbbbb
Therefore, the string
abbaabbbbbbb
is also generated by the grammar

34. We know: B b
B  aBbbb
S  abbaBbbb
We also know this string is generated:
S  abbaBbbb
abbaabbbb

35. We know: B b
B  aBbbb
S  abbaBbbb
Therefore, this string is also generated:
S  abbaBbbb
abbaaBbbbbbb
abbaabbbbbbb

36. We know: B b
B  aBbbb
S  abbaBbbb
Therefore, this string is also generated:
S  abbaBbbb
abba(a ) B(bbb)bbb
2 2
abba(a ) B(bbb) bbb
2 2
abba(a ) b(bbb) bbb

37. We know: B b
B  aBbbb
S  abbaBbbb
Therefore, this string is also generated:
S  abbaBbbb

i i
abba(a ) B(bbb) bbb
i i
abba(a ) b(bbb) bbb

38. Therefore, knowing that
abbabbbb
is generated by grammar G,
we also know that
i i
abba(a) b(bbb) bbb
is generated by G

39. In general:
We are given an infinite
context-free grammar G
Assume G has no unit-productions
no -productions

40. Take a string w L (G )
with length bigger than
m > (Number of productions) x
(Largest right side of a production)
Consequence:
Some variable must be repeated
in the derivation of w

41. u , v, x, y, z : strings of terminals
String w uvxyz S
u z
Last repeated variable A
v y
repeated A
x

42. S
Possible
derivations: u z
A
S uAz
v y
A vAy A
A x
x

43. We know:
S uAz A vAy A x
This string is also generated:
*
S uAz uxz
0 0
uv xy z

44. We know:
S uAz A vAy A x
This string is also generated:
* *
S uAz uvAyz uvxyz
1 1
The original w uv xy z

45. We know:
S uAz A vAy A x
This string is also generated:
* * *
S uAz uvAyz uvvAyyz uvvxyyz
2 2
uv xy z

46. We know:
S uAz A vAy A x
This string is also generated:
* *
S uAz uvAyz uvvAyyz
* *
uvvvAyyyz uvvvxyyyz
3 3
uv xy z

47. We know:
S uAz A vAy A x
This string is also generated:
S * uAz * uvAyz * uvvAyyz *
* uvvvAyyyz * 
* uvvvvAy  yyyz *
* uvvvvxy  yyyz
i i
uv xy z

48. Therefore, any string of the form
i i
uv xy z i 0
is generated by the grammar G

49. Therefore,
knowing that uvxyz L(G)
i i
we also know that uv xy z L(G )

50. S
u z
A
v y
A
x
Observation: | vxy | m
Since A is the last repeated variable

51. S
u z
A
v y
A
x
Observation: | vy | 1
Since there are no unit or productions

52. The Pumping Lemma:
For infinite context-free language L
there exists an integer m such that
for any string w L, | w| m
we can write w uvxyz
with lengths | vxy | m and | vy | 1
and it must be:
i i
uv xy z L, for all i 0

53. Applications
of
The Pumping
Lemma

54. Non-context free languages
n n n
{a b c : n 0}
Context-free languages
n n
{a b : n 0}

55. 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

56. n n n
L {a b c : n 0}
Assume for contradiction that L
is context-free
Since L is context-free and infinite
we can apply the pumping lemma

57. n n n
L {a b c : n 0}
Pumping Lemma gives a magic number m
such that:
Pick any string w L with length | w | m
m m m
We pick: w a b c

58. n n n
L {a b c : n 0}
m m m
w a b c
We can write: w uvxyz
with lengths | vxy | m and | vy | 1

59. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Pumping Lemma says:
i i
uv xy z L for all i 0

60. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
We examine all the possible locations
of string vxy in w

61. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
m
Case 1: vxy is within a
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

62. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 1: v and y consist from only a
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

63. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 1: Repeating v and y
k 1
m k m m
aaaaaa...aaaaaa bbb...bbb ccc...ccc
u 2 2 z
v xy

64. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 1: From Pumping Lemma: uv xy z L
k 1
m k m m
aaaaaa...aaaaaa bbb...bbb ccc...ccc
u 2 2 z
v xy

65. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 1: From Pumping Lemma: uv xy z L
k 1
2 2 m k m m
However: uv xy z a b c L
Contradiction!!!

66. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
m
Case 2: vxy is within b
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

67. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 2: Similar analysis with case 1
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

68. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
m
Case 3: vxy is within c
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

69. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 3: Similar analysis with case 1
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

70. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
m m
Case 4: vxy overlaps a and b
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

71. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 1: v contains only a
y contains only b
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

72. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 1: v contains only a
k1 k2 1 y contains only b
m k1 m k2 m
aaa...aaaaaaa bbbbbbb...bbb ccc...ccc
u 2 2 z
v xy

73. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 4: From Pumping Lemma: uv xy z L
k1 k2 1
m k1 m k2 m
aaa...aaaaaaa bbbbbbb...bbb ccc...ccc
u 2 2 z
v xy

74. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 4: From Pumping Lemma: uv xy z L
k1 k2 1
2 2 m k1 m k2 m
However: uv xy z a b c L
Contradiction!!!

75. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 2: v contains a and b
y contains only b
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

76. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 2: v contains a and b
k1 k2 k 1 y contains only b
m k1 k 2 m k m
aaa...aaaaabbaabb bbbbbbb...bbb ccc...ccc
u 2
v xy 2 z

77. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 4: From Pumping Lemma: uv xy z L
k1 k2 k 1
m k1 k 2 m k m
aaa...aaaaabbaabb bbbbbbb...bbb ccc...ccc
u 2
v xy 2 z

78. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
2 2
Case 4: From Pumping Lemma: uv xy z L
However: k1 k2 k 1
2 2 m k1 k2 m k m
uv xy z a b a b c L
Contradiction!!!

79. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 3: v contains only a
y contains a and b
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

80. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 4: Possibility 3: v contains only a
y contains a and b
Similar analysis with Possibility 2

81. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
m m
Case 5: vxy overlaps b and c
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

82. n n n
L {a b c : n 0}
m m m
w a b c
w uvxyz | vxy | m | vy | 1
Case 5: Similar analysis with case 4
m m m
aaa...aaa bbb...bbb ccc...ccc
u vxy z

83. There are no other cases to consider
(since | vxy | m , string vxy cannot
m m m
overlap a , b and c at the same time)

84. In all cases we obtained a contradiction
Therefore: The original assumption that
n n n
L {a b c : n 0}
is context-free must be wrong
Conclusion: L is not context-free