1

Lecture 32

• 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

2

Closure Properties for CFL’s

Kleene Closure

3

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

4

Visualization

L*L

CFL

CFG’s

• 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 CFLG1 G2

5

Algorithm Specification

• Input– CFG G1

• Output– CFG G2 such that L(G2) = (L(G1))*

CFG G1 CFG G2A

6

Construction

• Input– CFG G1 = (V1, , S1, P1)

• Output– CFG G2 = (V2, , S2, P2)

• V2 = V1 union {T}

– T is a new symbol not in V1 or

• S2 = T

• P2 = P1 union {T --> TS1 | }

7

Closure Properties for CFL’s

Kleene Closure Examples

8

• Input grammar:– V = {S}

– = {a,b}

– S = S

– P: S --> aa | ab | ba | bb

• Output grammar– V = {S, T}

– = {a,b}

– Start symbol is T

– P:T --> TS | S --> aa | ab | ba | bb

Example 1V2 = V1 union {T} T is a new symbol not in V1 or S2 = TP2 = P1 union {T --> TS | }

9

• Input grammar:– V = {S, T}

– = {a,b}

– Start symbol is T

– P:T --> TS | S --> aa | ab | ba | bb

• Output grammar– V = {S, T, U}

– = {a,b}

– Start symbol is U

– P:U --> UT |

T --> TS | S --> aa | ab | ba | bb

Example 2V2 = V1 union {T} T is a new symbol not in V1 or S2 = TP2 = P1 union {T --> TS | }

10

Closure Properties for CFL’s

Kleene Closure Proof of Correctness

11

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

12

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 ==>1

G2 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))*

– ...

13

L(G2) is a subset of (L(G1))*

• Statement to be proven:– For all n >= 1, if T ==>n

G2 x, then x is in (L(G1))*

– Prove this by induction on n

• Base Case:– n = 1

– The only string x such that T ==>1G2

x is the string

• Follows from inspection of G2

– The string belongs to (L(G1))*

• Follows from definition of Kleene Closure

14

Inductive Case

• Inductive Hypothesis:– For 1 <= j <= n, if T ==>j

G2 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+1

G2 x, then x is in (L(G1))*

• Proving this statement– Let x be an arbitrary string such that T ==>n+1

G2 x

– There are two possible first derivation steps

• Case 1: T ==>G2 ==>n

G2 x

• Case 2: T ==>G2 TS ==>n

G2 x

– These 2 cases follow from looking at grammar G2

15

Case Analysis

• Case 1: T ==>G2 is not possible

– n >= 1 which means n+1 >= 2

• Case 2: T ==>G2 TS ==>n

G2 x

– This means x has the form uv where• T ==><n u

• S ==><n v– This follows because in n steps, we go from TS to x

– Thus, T can take at most n-1 steps to generate u

– Likewise for S generating v

– String u belongs to (L(G1))*

• Follows from the inductive hypothesis

16

Concluding Case 2: T ==>G2

TS ==>nG2

x

– 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)

– Conclude x belongs to (L(G1))*

• x = uv where u belongs to (L(G1))* and v belongs to L(G1)

• By definition of Kleene closure, this means x=uv 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))*

17

(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

– ...

18

(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

– The only string x such that x is in (L(G1))0 is the string

• Follows from definition of (L(G1))0

– The string belongs to L(G2)

• Follows from production T -->

19

Inductive Case

• Inductive Hypothesis:– For 0 <= j <= n, if x is in (L(G1))j, then T ==>*

G2 x

• Note, this is a “strong” induction hypothesis

• Statement to be Proven in Inductive Case:– For n above, 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• u is a string in (L(G1))n

• v is a string in L(G1)

– This follows from definition of (L(G1))n+1 and that n>=0

20

Deriving x– x = uv where

• u is a string in (L(G1))n

• v is a string in L(G1)

– T ==> G2 TS ==>*

G2 uS ==>*

G2 uv = x

• The first derivation step is possible from our grammar

• The second sequence of derivation steps follows from T ==>* G2

u

– This follows from applying the induction hypothesis

• Third sequence of derivation steps follows from S ==>* G2

v

– This follows from the fact that v is in L(G1) and

– Our construction includes all productions from G1

– Thus T ==>* G2

x

• The inductive case follows

• The result is proven by the principle of mathematical induction

21

Construction for Set Union

• Input– CFG G1 = (V1, , S1, P1)

– CFG G2 = (V2, , S2, P2)

• Output– CFG G3 = (V3, , 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

• S3 = T

• P3 = P1 union P2 union {T --> S1 | S2}

22

Construction for Set Concatenation

• Input– CFG G1 = (V1, , S1, P1)

– CFG G2 = (V2, , S2, P2)

• Output– CFG G3 = (V3, , 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

• S3 = T

• P3 = P1 union P2 union {T --> S1S2}

23

CFL’s and regular languages

24

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

25

Regular languages subset of CFL

• Recursive definition of regular languages– Base Case:

• {}, {}, {a}, {b} are regular languages over {a,b}

• P={}, P={S --> }, P={S --> a}, P={S --> b}

– Inductive Case:• If L and L’ are are regular languages, then L*, LL’,

L union L’ are regular languages

• Use previous constructions to see that these resulting languages are also context-free

26

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

27

Language class hierarchy

All languages over alphabet

RE

REG

HH

EqualCFLREC

?