Lecture 18
Properties of Context-free
Languages
COT 4420
Theory of Computation
Section 8.2
Closure properties of Context-free
languages
• If L 1 and L2 are context-free languages, then
we prove that:
Union: L1 ∪ L2
Concatenation: L1L2
Star: L1*
Are Contextfree languages
Union
• L1 is context-free therefore there exists a
context-free grammar G1 = (V1, T1, S1, P1) for it.
• L2 is context-free therefore there exists a
context-free grammar G2 = (V2, T2, S2, P2) for it.
Union
Create a grammar G = (V, T, S, P) such that:
V = V1∪ V2 ∪ S
T = T1∪ T2
P = P1 ∪ P2 U { S  S1 | S2 }
We can show that:
L (G) = L(G1) ∪ L(G2)
Union - Example
S1 → aS1b | λ
S 2 → aS 2 a | bS 2b | λ
L1 = {a b }
n n
L2 = {ww }
R
Union
L = {a b } ∪ {ww }
n n
R
S → S1 | S 2
S1 → aS1b | λ
S 2 → aS 2 a | bS 2b | λ
Concatenation
• L1 is context-free therefore there exists a
context-free grammar G1 = (V1, T1, S1, P1) for it.
• L2 is context-free therefore there exists a
context-free grammar G2 = (V2, T2, S2, P2) for it.
Concatenation
Create a grammar G = (V, T, S, P) such that:
V = V1∪ V2 ∪ S
T = T1∪ T2
P = P1 ∪ P2 U { S  S1S2 }
We can show that:
L (G) = L(G1) . L(G2)
Concatenation- Example
S1 → aS1b | λ
S 2 → aS 2 a | bS 2b | λ
L1 = {a b }
n n
L2 = {ww }
R
Concatenation
L = {a b }{ww }
n n
R
S → S1S 2
S1 → aS1b | λ
S 2 → aS 2 a | bS 2b | λ
Star - Closure
• L1 is context-free therefore there exists a
context-free grammar G1 = (V1, T1, S1, P1) for it.
Star Closure
Create a grammar G = (V, T, S, P) such that:
V = V1 ∪ S
T = T1
P = P1 U { S  S1S | λ }
We can show that:
L (G) = L(G1)*
Star Closure- Example
S1 → aS1b | λ
L1 = {a b }
n n
Star
L = {a b } *
n n
S1 → aS1b | λ
Negative Properties of Contextfree Languages
Negative properties of Context-free
languages
• Context-free languages are not closed under:
Intersection : L1 ∩ L2
Complement:
L1
Intersection - Example
L1 = {a b c }
n n m
Context-free:
S → AC
A → aAb | λ
C → cC | λ
L2 = {a b c }
n m m
Context-free:
S → AB
A → aA | λ
B → bBc | λ
Intersection:
We showed that this is
n n n
L1 ∩ L2 = {a b c }
not context-free
Complement
• De Morgan’s Law:
L1 ∩ L2 = L1 ∪ L2
If the family of context-free languages were closed
under complement, then L1 ∪ L2 would be a
context-free language.  L1 ∩ L2 would be contextfree.
However, we showed that the intersection of two
context-free languages is not necessarily context-free.
Closure under Regular
Intersection
Closure under Regular Intersection
Theorem: If L1 is a context-free language and L2
is a regular language, then L1 ∩ L2 is a contextfree language.
Machine
NPDA for
M1
L1
Machine
M2
DFA for
L2
Closure under Regular Intersection
Proof
Proof: Construct a new NPDA machine M that
accepts L1 ∩ L2. M simulates in parallel M1 and
M2.
NPDA
qi
DFA
M1
a, b  x q
k
transition
NPDA
qi, pj
a
pj
transition
M2
M
a, b  x
transition
qk, pl
pl
NPDA
qi
DFA
M1
λ, b  x q
k
pj
transition
transition
NPDA
qi, pj
M2
M
λ, b  x
transition
qk, pj
NPDA
DFA
M1
M2
q0
p0
initial state
initial state
NPDA
M
q0 , p0
Initial state
NPDA
DFA
M1
p1
q1
final state
M2
p2
final states
NPDA
q1, p1
M
q1, p2
final states
Closure under Regular Intersection
Proof
M accepts w:
((q0, p0), w, z) ⊦* ((qr, ps), λ, x)
qr ∈ F1 and ps ∈ F2
If and only if:
(q0, w, z) ⊦*M1 (qr, λ, x)
And
δ* (p0, w) = ps
Closure under Regular Intersection
Proof
L(M) = L(M1) ∩ L(M2)
M is an NPDA
L(M1) ∩ L(M2) is context-free
Application of Closure under Regular
Intersection
Show that the language L = { anbn : n≥ 0 , n≠100}
is a context-free language.
L1 = {a100b100}
is regular
L1 is finite and therefore regular.
L1
is regular
L = { anbn : n≥ 0 , n≠100}
L1 = {a100b100}
It is easy to see that
L = {anbn : n≥ 0 } ∩ L1
Since {anbn: n≥ 0 } is context-free and L1 is
regular, L is context-free.
Another Example
• Show that the language L = { w: na = nb = nc } is
not context-free.
Suppose L was context-free. Then:
L ∩ { a* b* c*} = { anbncn : n≥ 0}
would also be context-free.
Context-free
Regular
Context-free
But we know that this language is not
context-free. We conclude that L cannot be
contxt-free.
Decision Properties of Contextfree Languages
Membership Problem
Question: Is string w in context-free language L?
Answer: Parsers. Exhaustive search parser /
CYK parsing algorithm.
The Emptiness Problem
Question: Given context-free language L, does it
contain any string at all? Is L empty? (L = ø).
Answer: Use the algorithm to find useless
variables. If start variable S is useless, L is empty.
The Infiniteness Problem
Question: Given context-free language L, Is it finite?
Answer: Assume grammar G for L does not contain
any λ-productions, unit productions or useless
productions.
Create dependency graph for variables.
If there is a loop in the dependency graph then the
language is infinite.
Note: If there is a cycle in the dependency graph it
means that the grammar has a repeating variable.
Non-Decision Properties
• Many questions that can be decided for
regular sets cannot be decided for CFL’s.
• Example: Are two CFL’s the same?
• Example: Are two CFL’s disjoint?
– How would you do that for regular languages?
• At this point we do not have a way of proving
that no algorithm exists.