closure properties - university of arizona

CSC 473 Automata, Grammars & Languages 10/26/10

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C SC 473 Automata, Grammars & Languages 56

Closure Properties Regular Ops. The CFLs are closed under ∪, ⋅, ∗ Pf: Homework �

Intersection. The CFLs are not closed under intersection. Example: Consider the two CFLs

Then We will later see (CF Pumping Lemma) that this last is

not a CFL. �

However, if is regular and is CF, then is CF.

1 2{ | 0}, { | 0}

n n n nL a b c n L a b n c

! != " # = # "

1 2{ | 0}.

n n nL L a b c n! = "

1L

2L

1 2L L!

!


Closure Properties (contʼd)• Thm: The class of CFLs is closed under intersection with

regular languages. Pf: Assume and

Construction. Construct a “cross-product pda” M asfollows:

where the transition function is defined by:

provided and

Machine M simulates the two given machines “in parallel”, keeping each machine state in one component of the

compound state [ , ]. �

1 1 1 1 1 1 1( ), ( , , , , , )L L P P Q s F!= = " #

2 2 2 2 2 2 2( ), ( , , , , )R L M M Q s F!= = "

1 2 1 2 1, 2 1 2( , , , ,[ ], )M Q Q s s F F!= " # $ # % "

1 2 1 2([ , ], ) ([ , ], , )p p Y q q a X!"

!

1 1 1( , ) ( , , )p Y q a X!" 2 2 2

( , )p q a!=


What is Not Context-Free?• PDA have a limited computing ability. They cannot, for

example, recognize repeated strings like w#w or stringsthat “count” in more than 2 places, such as .

• We will show that some languages are not CF using a CFPumping Lemma, which gives a property that all CFLsmust have. Then, to show that a language L is not CF, wesomehow argue that it lacks this pumping property.

• Closure properties of CFLs can sometimes be used tosimplify non-CFLs and make a pumping argument easier.

n n na b c


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CF Pumping Lemma• Thm [Pumping Lemma for CFLs]. Suppose that L is an

infinite CF language. Then

• For comparison, here is the Regular P.L.:

( )( )[

( , , , , )(

( 0) )]i i

p w w L w p

u v x y z w uvxyz vy

vxy p i uv xy z L

!

" # $ % & '

" = % (

% ) % # & $

( )( )[

( , , )(

( 0) )]i

p w w L w p

x y z w xyz y

xy p i xy z L

!

" # $ % & '

" = % (

% ) % # & $


CF Pumping Lemma (contʼd)• Pf: Let where CFG G is a CFG in

Chomsky Normal Form (Text, Theorem 2.9), i.e. a CFG inwhich all rules are of the (schematic) forms A→BC or A→a(a ≠ ε). If is “sufficiently long”, then anyderivation tree T for w must contain a “long” path—moreprecisely:

• Claim 1: If the derivation tree T for has no path longerthan h then

Pf: Induction on h. Base: h = 1. Only possible tree is and Step: Assume Let T have all paths and be of form (in CNF)

( )w L G!

1| | 2 .

h

w!

"

1.h > h!

{ } ( )L L G!" =

w

0| | 2.a =

S

a


CF Pumping Lemma (contʼd)

• Then have all paths of length By IH, which implies �. Conversely, if a generated string is at least long, then its parse tree must be at least high.

G has variables. Choose If and then Claim 1 ⇒ any parse tree T for w has a path of length at least Such a path has at

least nodes. ∴ some variable appears twice on the path (note the leaf node is a terminal).

2 2| | 2 ,| | 2

h h

s t! !

" "

S

1T

2T

T =

s t w st=

1 2,T T 1.h! "

1| | 2

h

w!

"

2h

1h+

V1

2 .V

p+

= w L!

w p!

2.V +

3V +


3


CF Pumping Lemma (contʼd)• Picture:

1

2V

w p+

! =

T =

R

R

height nodes

variables repeat

2 3

2

V V

V

! + " ! +

" ! + "

R

RR

u v x y zw =

a

ht 2V! +

1

2V

p+

! =

ChooseBottom

variables1V +

vxy p! "

1T

2T


CF Pumping Lemma (contʼd)(1) Center portion is not too long:(2) Pumped portion not empty: cannot both = ε.

1T

2T

vxy p!

,v y

=

R

B C

2T

v x y

x !"

R

B C

2T

v x y

x!"

or

In CNF, no variablegenerates ε


CF Pumping Lemma (contʼd)(3) Pumped strings in L : the following are all parse trees

R

R

( 0)i i

i uv xy z L! " # $ !

R

R

R

R

R

R

R

u v x y z

u vv x yy z

u vvv x yyy z

…

R

u x z

and:


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CF Pumping: Applications• Ex: is not a CFL. Pf: Suppose it is CF. Then the Pumping Lemma ⇒ ∃p ∀ w∈L, |w|≥p ⇒ ∃ uvxyz =w & vy ≠ ε & ∀i≥0 u vi x yi z ∈L. Pick p as the constant guaranteed and choose n ≥ p/3 and Where is Cases: Assume first that

{ | 0}n n n

L a b c n= !

, .n n n

w a b c uvxyz vy != = " ?vxy

.v !"

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

a ab bc c

u v x y z

u v x y z

u v xyz

u v x y z

u v xyz

u v xyz

……… ……… ………


CF Pumping: Applications• In cases 1-3 has an imbalance. In case 4 it has a b before a. In case 5 it has a c before b. In case 6 it has an a after a c. In any case, there is a contradiction to the pumped word being in L. The case where is symmetric. Contradiction. � Cor. The CFLs are not closed under complementation. Pf: is a CFL. But is not a CFL. Therefore cannot be CF. � Ex: is not CF. Proof similar to regular case. � Ex: is not CF. �

y !"

2 2uv xy z

{ | }p q r

L a b c p q q r p r= ! " ! " !

{ | 0}i i i

L a b c a b c i! ! !

" = #

L

{ | prime}i

L a i=

2

{ | 0}i

L a i= !


Pumping: Applications (contʼd) Ex: is not CF. Pf: Intersection with is not a CFL. Therefore cannot be CF. � Ex: is not CF. Pf: By pumping on the word Similar to Text, Example 2.38. � Ex: is not CF. Pf:

Pump on the latter language in a way similar to the previousexample to show it is not CF. �

L

a b c{ { , , } | = = }L w a b c w w w

!= "

a b c! ! !

{ |i 1,j 1}i j i j

L a b c d= ! !n n n n

w a b c d=

{ | {0,1}}L ww w= !

10 10 110 10 1 {10 10 110 10 1| , 1}n m n m

L m n+ + + +

! = "

closure properties - university of arizona

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