CS 461 – Oct. 5
• Pumping lemma #2
– Understanding
– Use to show a language is not a CFL
• Next: Applications of CFLs
– Expression grammars and Compiling
Pumping lemma
L is a CFL implies:
There is a p,
such that for any string s, |s| >= p
We can break up s into 5 parts uvxyz:
| v y | >= 1
| v x y | <= p
and uvixyiz  L.
• How does this differ from the first pumping
lemma?
In other words
• If L is a CFL, then for practically any word w in L,
we should be able to find 2 substrings within w
located near each other that you can
simultaneously pump in order to create more
words in L.
• Contrapositive: If w is a word in L, and we can
always create a word not in L by simultaneously
pumping any two substrings in w that are near
each other, then L is not a CFL.
Gist
• A string can grow in 2 places.
– When we write recursive rules, we can add symbols
to left or right of a variable.
• Ex.
S  #A#
A  ε | 1A2
Look at derivation of: #111…222#
• Even a complex rule like S  S1S1S2 can be
“simplified”, and still only need 2 places to grow the
string.
– This is because any CFG can be written in such a way that there
are no more than 2 symbols on the right side of .
Example
• L = { 1n 2n } should satisfy the pumping lemma.
Let p = 2. Strings in L of length >= 2 are:
{ 12, 1122, 111222, … }. All these words have
both a 1 and a 2, which we can pump.
Just make sure the 2 places to pump are within
p symbols of each other. So let’s say they are in
the middle.
u
v x y
z
11111…1 1 ε 2
22222….2
Could p = 1 work?
Example 2
• { w # wR }. Think about where we can grow.
Let p = 3. All strings of length 3 or more look like
w # wR where | w | >= 1.
As a technicality, would other values of p work?
Non-CFLs
• It’s possible for a language to be non-CFL!
Typically this will be a language that must
simultaneously pump in 3+ places.
• Ex. How could we design a PDA for { 1n2n3n }?
Adversary’s Revenge
• In Game #1, how did we win? Ex. { 0n 1n }.
–
–
–
–
Adversary chose p.
We chose s.
Adversary broke up s = xyz subject to constraints.
We were always able to pump and find words outside
L.
• Game #2 strategy
– Adversary’s constraints looser. The middle 3 parts
have to be within p of each other. Can be anywhere
in the string, not just in first p symbols of word.
{ 1n 2n 3n }
• Let p be any number.
• Choose s = 1p2p3p.
• Let s = uvxyz such that |vxy| <= p and |vy| >= 1.
Where can v and y be?
– All 1’s
– All 2’s
– All 3’s
– Straddling 1’s and 2’s
– Straddling 2’s and 3’s
• In every case, can we find a word not in L?
More examples
• { 1 i 2j 3k : i < j < k }
– What should we chose as s?
– How can the word be split up?
• {ww}
–
–
–
–
Let s = 0p1p0p1p. Where can v & y be?
Same section of 0’s and 1’s.
In neighboring sections
Either v or y straddles a border.