Properties of CFL

1
Properties of CFLs

• Reading 8.1a and 8.2

2
Pumping Lemma for CFLs

• Let L be a context-free language. Then there
  exists some positive integer m such that, if s is
  a string of length at least m, then s can be
  written as
• s uvwxy such that
• vwx ltm
• vx gt0
• uviwxiy is also in L for any i gt0

3
Why is it true?

• If the language is finite, then it is regular and
  also context-free.
• If its infinite, it still must have a finite
  number of non-terminals.
• So, at least one non-terminal must be used more
  than once.
• If A is used more than once, at least one
  derivation of it is recursive, and at least one
  is not (or it would be useless.)

4
So

• Suppose the rule is
• A -gt vAx
• A -gt w
• Then if I can generate
• s uvwxy
• I can also generate
• s uv2wx2y and uv3wx3y by using the recursive
  rule for A more times.

5
Using the Pumping Lemma to show a language is
not CF.

• Assume the language is context-free.
• Show it does not satisfy the pumping lemma
• Conclude (by contradiction) the language could
  not have been context-free.

6
Example axbxcx xgt0

• Opponent chooses m.
• I choose s ambmcm.
• Opponent must choose uvwxy so that vwx is no
  more than m but at least one symbol.
• There is no way that vwx contains all three
  characters. It contains either one or two.
• Pumping the string up adds more of one or two
  characters than the third.

7
Example L anbj j n2

• Opponent chooses m.
• I choose string ambm2
• Opponent chooses vwx.
• If vwx contains as only, I can add to the number
  of as while the number of bs stays the same.
  Likewise for bs only.
• If vwx contains as and bs, only interesting
  case is v is some number of as (1ltkltm) and x
  is some number of bs (1ltpltm).
• s am(i-1)k b m2 (i-1)p
• Choose i 0 s am-k b m2-p
• (m-k)2 lt (m-1)2 m2 -2m 1
• m2 -(2m -1)
• lt m2 -p

8
Closure for CFLs

• The family of context-free languages is closed
  under union
• L1 is a CFL with grammar G1, start symbol S1
• L2 is a CFL with grammar G2, start symbol S2
• Let L3 L1UL2
• L3(G3) has start symbol S -gt S1 S2

9
Closure for CFLs

• The family of context-free languages is closed
  under star-closure.
• Let L1 be a context-free language with grammar G1
  and start symbol S1.
• Let L be L1. Then L(G) S -gt S S1 ?

10
Closure for CFLs

• The family of context-free languages is closed
  under concatenation.
• Let L1 be a CFL with grammar G1 and start symbol
  S1.
• Let L2 be a CFL with grammar G2 and start symbol
  S2.
• Let L3 L1 L2.
• Then L3(G3) has start symbol S -gt S1 S2.

11
CFLs are not closed under intersection!

• L1 anbncm n,m gt0
• L2 anbmcm n,m gt0
• L1 n L2 anbncn n gt0

12
CFLs are not closed under complementation!

• Suppose that CFLs are closed under
  complementation.
• Let L1 and L2 be CFLs.
• L1 n L2 L1 U L2
• By closure under union and complementation, we
  get closure under intersection.
• But we just showed that CFLs are not closed
  under intersection!
• Contradiction! So, CFLs are not closed under
  complementation.

13
Regular Intersection

• The intersection of a regular language and a
  context-free language is context-free.
• This is called closure under regular
  intersection.
• Idea of proof Make states equal to the states in
  the PDA x the states in the DFA.

14
Proofs Using Closure

• Prove that the language L w in a,b,c where
  na(w) nb(w) nc(w) is not context-free.
• Assume L is context free.
• LnL(abc) anbncn must be context-free by
  closure under regular intersection.
• But its not, so L must not be context-free.

15
Proofs Using Closure

• Prove that the language L anbn ngt50 is
  context-free.
• Idea Dont want to construct! Use closure.
• Let L1 anbn L1 is context-free.
• Let L2 anbn nlt50 L2 is Regular.
• L L1 ? L2, which is CF by closure under
  regular-intersection

16
Decidable Properties

• A CFL is empty if its start symbol is found to be
  useless.
• A CFL is finite if no variable can ever repeat.
• Draw the dependency graph of the language.
• The language is infinite if and only if the graph
  contains a cycle.