CS 3240: Languages and Computation

Properties of Context-Free Languages

Properties of CFGs

Chomsky Normal Form

In Summary

Conversion to Chomsky Normal Form

Recall: Pumping Lemma for Regular Languages

q1

qk

qj

For every regular language L, there is a finite pumping length p, such that for any string sL and |s|p, we can write s=xyz with:

1) x yi z L for every i{0,1,2,…}2) |y| 13) |xy| p

Pumping Lemma for CFLTheorem: For every context-free language L, there is a pumping length p, such that for any string sL and |s|p, we can write s=uvxyz with

1) u vi x yi z L for every i{0,1,2,…}2) |vy| 13) |vxy| p

Note that 1) implies that uxz L (take i=0), requirement2) says that v,y cannot be the empty strings ε and condition 3) is useful in proving non-CFL.

Pumping a Parse Tree

A

A

u v x y zIf s = uvxyz L is long, then its parse-tree is tall.Hence, there is a path on which a variable A repeats itself. We can pump this A–A part.

S

uvxyz L

A

A

u v x y z

By repeating the A–A part we get…

S

uv2xy2z L

A

u v x y zRA

A

v x y… while removing the A–A gives…

S

uxz L

A

u z

In general: uvixyiz L for all i=0,1,2,…

S

x

Using Pumping Lemma“Proof by contradiction. Assume that L is context free.Let p be the pumping length of L and take sL.

By the pumping lemma it is possible to write s=uvxyz(with |vy| 1 and |vxy| p) such that uvixyiz L for all i0.

However…

Thus the pumping lemma does not hold for s. Contradiction. The language L is not context-free.”

Show that pumping up or down s gives strings that are not in L. Be careful to consider all v,x,y possibilities.

Here you have to be clever in picking the right s.

Example I: Pumping anbncn

Proof by contradiction; assume L = {anbncn | n0} is CF.According to the pumping lemma there is a pumping length p such that

for s = apbpcp L, we can write s = uvxyz = apbpcp, with uvixyiz L for all i0.

Two options for 1 |vxy| p:1) vxy = a*b*, then the string uv2xy2z has not enough letters c, hence

uv2xy2zL2) vxy = b*c*, then the string uv2xy2z has not enough letters a, hence

uv2xy2zL

Contradiction: the pumping lemma does not hold, hence L is not context free.

Example II: (Pumping down)

Prove that C = {aibjck | 0ijk } is not context-free.

Let p be the pumping length, and s = apbpcp C.Pumping lemma: s = uvxyz, such that uv ixyiz C

for every i0. Two options for 1 |vxy| p:1) vxy = a*b*, then the string uv2xy2z has not enough letters c,

hence uv2xy2zC2) vxy = b*c*, then the string uv0xy0z = uxz has too many letters a,

hence uv0xy0zC

Contradiction: The pumping lemma does not hold forthis string apbpcp, hence C is not context-free.

Example III: ww

Prove that the language D = { ww : w{0,1}* } is not CF.

You must be careful when picking the strings sD… Let p be the pumping length, take s=0p1p0p1p.

Options for s=uvxyz with 1 |vxy| p:

• If vxy is to the left of the middle of 0p1p 0p1p,then second half of uv2xy2z starts with a “1”.

• If vxy is to the right of the middle of 0p1p 0p1p, then first half of hence uv2xy2z ends with a “0”.

• If x is in the middle of 0p1p 0p1p, then pumped down uxz equals 0p1i 0j1p D (because i or j < p)

Additional Notes on CFL

Let A1 and A2 be two context-free languages,then the union A1 A2 is also context free.

However, the intersection A1 A2 is not necessarily context free

The complement Ā1= *–A1 are not necessarily context free either

Closure properties of CFLs

Closure under reversal

CFL not closed under intersection!

Decision Problems for CFL

Make a chain of everywhere a variable occurs

From each production send a link to a counter counting how many variables in the prod havenot yet been found to produce

When a variable is found to be generating decrementCounter accordingly

When counter reaches 0 the head variable is generating

Undecidable problems for CFL