Download Module 31

Module 31
• Closure Properties for CFL’s
– Kleene Closure
• construction
• examples
• proof of correctness
– Others covered less thoroughly in lecture
• union, concatenation
• CFL’s versus regular languages
– regular languages subset of CFL
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Closure Properties for CFL’s
Kleene Closure
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CFL closed under Kleene Closure
• Let L be an arbitrary CFL
• Let G1 be a CFG s.t. L(G1) = L
– G1 exists by definition of L1 in CFL
•
•
•
•
Construct CFG G2 from CFG G1
Argue L(G2) = L*
There exists CFG G2 s.t. L(G2) = L*
L* is a CFL
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Visualization
• Let L be an arbitrary CFL
• Let G1 be a CFG s.t. L(G1) = L
L
L*
– G1 exists by definition of L1 in CFL
•
•
•
•
Construct CFG G2 from CFG G1
Argue L(G2) = L*
There exists CFG G2 s.t. L(G2) = L*
L* is a CFL
CFL
G1
G2
CFG’s
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Algorithm Specification
• Input
– CFG G1
• Output
– CFG G2 such that L(G2) = (L(G1))*
CFG G1
A
CFG G2
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Construction
• Input
– CFG G1 = (V1, S, S1, P1)
• Output
– CFG G2 = (V2, S, S2, P2)
• V2 = V1 union {T}
– T is a new symbol not in V1 or S
• S2 = T
• P2 = P1 union {T --> TS1 | l}
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Closure Properties for CFL’s
Kleene Closure Examples
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Example 1 *
• Input grammar:
–
–
–
–
V = {S}
S = {a,b}
S=S
P:
V2 = V1 union {T}
T is a new symbol not in V1 or S
S2 = T
P2 = P1 union {T --> TS | l}
• Output grammar
–
–
–
–
V = {S, T}
S = {a,b}
Start symbol is
P:
S --> aa | ab | ba | bb
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Example 2 *
• Input grammar:
–
–
–
–
V = {S, T}
S = {a,b}
Start symbol is T
P:
V2 = V1 union {T}
T is a new symbol not in V1 or S
S2 = T
P2 = P1 union {T --> TS | l}
• Output grammar
–
–
–
–
V = {S, T, U}
S = {a,b}
Start symbol is
P:
T --> TS | l
S --> aa | ab | ba | bb
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Closure Properties for CFL’s
Kleene Closure Proof of Correctness
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Is our construction correct?
• How do we prove our construction is correct?
– Informal
• Test some strings
• Review logic behind construction
– Formal
• First, show every string derived by G2 belongs to (L(G1))*
– That is, show L(G2) is a subset of (L(G1))*
• Second, show every string in (L(G1))* can be derived by G2
– That is, show (L(G1))* is a subset of L(G2)
• Both proofs will be inductive proofs
– Inductive proofs and recursive algorithms go well together
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L(G2) is a subset of (L(G1))*
• We want to prove the following
– If x in L(G2), then x is in (L(G1))*
• This is equivalent to the following
– If T ==>*G2 x, then x is in (L(G1))*
– The two statements are equivalent because
• x in L(G2) means that T ==>*G2 x
• We break the second statement down as follows:
– If T ==>1G2 x, then x is in (L(G1))*
– If T ==>2G2 x, then x is in (L(G1))*
– If T ==>3G2 x, then x is in (L(G1))*
– ...
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L(G2) is a subset of (L(G1))*
*
• Statement to be proven:
– For all n >= 1, if T ==>nG2 x, then x is in (L(G1))*
– Prove this by induction on n
• Base Case:
– n=1
– Examining G2, what is the only string x such that T
==>1G2 x ?
– Prove this string is in (L(G1))*
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Inductive Case *
• Inductive Hypothesis:
– For 1 <= j <= n, if T ==>jG2 x, then x is in (L(G1))*
• Note, this is a “strong” induction hypothesis
• Statement to be Proven in Inductive Case:
– For n above, if T ==>n+1G2 x, then x is in (L(G1))*
• Proving this statement
– Let x be an arbitrary string such that T ==>n+1G2 x
– Examining G2, what are the two possible first derivation steps?
• Case 1: T ==>G2
• Case 2: T ==>G2
==>nG2 x
==>nG2 x
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Case Analysis *
• Case 1: T ==>G2 ==>n x is not possible
– Why not?
• Case 2: T ==>G2 ==>nG2 x
– This means x has the form uv where
• What can we say about u (no IH)?
• What can we say about v (no IH)?
– Applying the inductive hypothesis, what can we conclude?
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Concluding Case 2: *
T ==>G2 ==>nG2 x
– Concluding string v belongs to L(G1)
• Follows from S ==>* G2 v and
• Our construction insures that all strings derived from S in L(G2) are
also in L(G1)
– How do we conclude that x belongs to (L(G1))*
• Wrapping up inductive case
– In all possible derivations of x, we have shown that x belongs to
(L(G1))*
– Thus, we have proven the inductive case
• Conclusion
– By the principle of mathematical induction, we have shown that
L(G2) is a subset of (L(G1))*
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(L(G1))* is a subset of L(G2)
• We want to prove the following
– If x is in (L(G1))*, then x is in L(G2)
• This is equivalent to the following
– If x is in (L(G1))*, then T ==>*G2 x
– The two statements are equivalent because
• x in L(G2) means that T ==>*G2 x
• We break the second statement down as follows:
– If x is in (L(G1))0, then T ==>*G2 x
– If x is in (L(G1))1, then T ==>*G2 x
– If x is in (L(G1))2, then T ==>*G2 x
– ...
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(L(G1))* is a subset of L(G2) *
• Statement to be proven:
– For all n >= 0, if x is in (L(G1))n, then x is in L(G2)
– Prove this by induction on n
• Base Case:
– n=0
– What is the only string x in (L(G1))0?
– Show this string belongs to L(G2)
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Inductive Case *
• Inductive Hypothesis:
– For n>=0, if x is in (L(G1))j, then T ==>*G2 x
• Note, this is a “normal” induction hypothesis
• Statement to be Proven in Inductive Case:
– For n>=0, if x is in (L(G1))n+1, then T ==>*G2 x
• Proving this statement
– Let x be an arbitrary string in (L(G1))n+1
– This means x = uv where
• v in L(G1)
• What can we say about u?
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Deriving x *
– x = uv where
• u is a string in
• v is a string in L(G1)
– Justify all the steps in the following derivation
– T ==> G2 TS ==>* G2 uS ==>* G2 uv = x
• First step:
• Second step:
• Third step:
– Thus T ==>* G2 x
• The inductive case follows
• The result is proven by the principle of mathematical
induction
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Construction for Set Union *
• Input
– CFG G1 = (V1, S, S1, P1)
– CFG G2 = (V2, S, S2, P2)
• Output
– CFG G3 = (V3, S, S3, P3)
• V3 = V1 union V2 union {T}
– Variable renaming to insure no names shared between V1 and V2
– T is a new symbol not in V1 or V2 or S
• S3 = T
• P3 = P1 union P2 union {T -->
}
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Construction for Set
Concatenation *
• Input
– CFG G1 = (V1, S, S1, P1)
– CFG G2 = (V2, S, S2, P2)
• Output
– CFG G3 = (V3, S, S3, P3)
• V3 = V1 union V2 union {T}
– Variable renaming to insure no names shared between V1 and V2
– T is a new symbol not in V1 or V2 or S
• S3 = T
• P3 = P1 union P2 union {T --> }
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CFL’s and regular languages
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CFL Closure Properties
• CFL’s are closed under Kleene closure
– Just proven, proof also in book
• CFL’s are closed under set union
– Proof in book
• CFL’s are closed under set concatenation
– Proof in book
• What can we conclude from these 3 results?
– It follows that regular languages are a subset of CFL’s
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Regular languages subset of CFL
• Recursive definition of regular languages
– Base Case:
• {}, {l}, {a}, {b} are regular languages over {a,b}
• P={}, P={S --> l}, P={S --> a}, P={S --> b}
– Inductive Case:
• If L1 and L2 are are regular languages, then L1*,
L1L2, L1 union L2 are regular languages
• Use previous constructions to see that these
resulting languages are also context-free
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Other CFL Closure Properties
• We will show that CFL’s are NOT closed under
many other set operations
• Examples include
– set complement
– set intersection
– set difference
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Language class hierarchy
?
H
H
Equal
REG
CFL
REC
RE
All languages over alphabet S
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