Context Free Pumping Lemma

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Context Free Pumping Lemma

• Zeph Grunschlag

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Agenda

• Context Free Pumping
• Motivation
• Theorem
• Proof
• Proving non-Context Freeness
• Examples on slides
• Examples on blackboard

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Pumping FAs
x
y

• Strings of length 3 or more in DFA above can be
  pumped because such strings correspond to paths
  of length ? 3, so visit ? 4 vertices. Hence,
  pigeonhole principle guarantees some vertex
  visited twice, and hence a pumpable cycle.

z
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Pumping PDAs
(, e ?X ), X?e

• However, regular pumping lemma fails in this
  example.
• Q Give an example of a pattern that cannot be
  pumped.

e , ?e
e , e?
r
s
q
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Pumping PDAs
(, e ?X ), X?e

• A (n )n cant be pumped in the first half.
• However, could pump two substrings at once.
  I.e. could take k left parens if take k right
  parens. I.e. can tandem pump.

e , ?e
e , e?
r
s
q
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Tandem Pumping

• DEF A string s in L is said to be tandem
  pumpable if can break s up into
• s uvxyz
• such that for all i ? 0 we have that
• s uv ixy iz ? L
• with at least one of v,y nonempty.
• Q1 Is 00111 tandem pumpable in 0111 ?
• Q2 Is 00100 tandem pumpable in 0n10n ?
• Q3 Is 00100100 tandem pumpable in
  0n10n10n ?

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Tandem Pumping

• A1 Yes. Any pumpable string is automatically
  tandem pumpable by letting y e. In our case,
  let u e, v 00, x y e, z 111.
• uv ixy iz (00)i111 is indeed in 0111.
• A2 Yes. Let u e, v 00, x 1, y 11 and z
  e.
• uv ixy iz (00)i1(00)i is indeed in 0n10n
• A3 NO! Tandem pumping 00100100 leads either to
  too many 1s, or would increase two of the
  0-streaks, without ability to increase the
  remaining 0-streak.

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Tandem Pumping

• In general, since pumping automatically implies
  tandem pumping, all (infinite) regular languages
  are tandem pumpable. Turns out, that all
  (infinite) context free languages are as well.
  But Q3 can be generalized to show that 0n10n10n
  does not admit tandem pumping of strings which
  are past a certain length. This will end up
  proving that 0n10n10n is not context free

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Context Free Pumping Lemma

• THM Given a context free language L, there is a
  number p (tandem-pumping number) such that any
  string in L of length ? p is tandem-pumpable
  within a substring of length p. In other words,
  for all s ? L with s ? p we we can write
• s uvxyz
• vy ? 1 (some pumpable stuff non-empty)
• vxy ? p (pumping inside length-p portion)
• uv ixy iz ? L for all i ? 0 (tandem-pump v
  and y)

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CFPL Intuition
tk t2 t1 sk s2 s1
-y?

• Intuitively s uvxyz is found as follows Only
  finitely many stack changes possible at cycles in
  the graph of length ? n (the number of states).
  Thus if s is long enough, there will have to be
  some states q,r such that the same string is
  pushed at q as is popped at r and such that the
  path from q to r starts and ends with same stack
  configuration. With these assumption, can then
  pump up v and y in tandem as v pushes same stuff
  that y pops off.

-v?
-x?
-u?
q
r
-z?
s
p
sk s2 s1
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CFPL - Proof

• The previous can actually be formalized and used
  to prove the Context Free Pumping Lemma.
  However, this is actually quite painful compared
  to very simple grammar-theoretic proof
• Proof of CFPL We may assume that the language
  is in CNF. This is not an essential assumption
  but it makes the proof a little easier.
• Consider a derivation tree in which some
  occurring variable node has itself as an ancestor

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CFPL Proof
S
a
A

• Could replace last appearance of A by its first
  appearance. I.e., in tree replace
• A ? and a by
• A ? chuga and a choo
• to get the following

f
o
r
y
o
u
A
c
h
u
g
a
a
n
d
a
c
h
o
o
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CFPL Proof
S
a
A

• And again

f
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r
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o
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A
c
h
u
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a
c
h
o
o
A
c
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a
a
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o
o
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CFPL Proof
S
a
A
f
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r
y
o
u
A
c
h
u
g
a
c
h
o
o
A
c
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a
c
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o
o
A
c
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u
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a
a
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CFPL Proof
S
a
A

• Or could replace
• A ? chuga and a choo by
• A ? and a
• This is called tandem-pumping down

f
o
r
y
o
u
A
c
h
u
g
a
a
n
d
a
c
h
o
o
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CFPL Proof
S
a
A
a
n
d
a
f
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r
y
o
u
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CFPL Proof

• In our particular case, we were able to create
  any string of the form
• a (chuga)i and a (choo)i for you
• In general, any branch down the derivation tree
  with a repeated variable gives rise to strings of
  the form uv ixy iz all of which are in L.
• The end of the proof is just a counting argument
  to see when a repeated variable is guaranteed to
  occur.

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CFPL Proof

• Q If n is the number of variables in the
  grammar, what tree-height guarantees that a
  variable is repeated?
• (Recall the height of the trivial tree just a
  root is 0)

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CFPL Proof

• A If n is the number of variables in the
  grammar, any subtree of height h n1 will have
  a repeated variable. This is because the bottom
  row of a derivation tree is composed of
  terminals, so height n1 ( n2 levels)
  guarantees n1 levels of variables, at least on
  one branch from the root. Pigeonhole principle
  guarantees that some variable will be encountered
  twice!

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CFPL Proof

• Q If the grammar is in CNF, what kind of tree
  is any derivation tree?

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CFPL Proof

• A A binary tree!
• Q What is the maximum number of leaves that a
  binary tree of height n may have?

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CFPL Proof

• A 2n
• Q What the maximum number of leaves that a CNF
  derivation-tree of height n1 may have?

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CFPL Proof

• A Still 2n! This is because the only way to
  get a terminal is through rule of the form A?a so
  there is no branching at the final level.
• Q What string length will guarantee a
  derivation tree of height ? n1 ?

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CFPL Proof

• A 2n. This is because no tree with height lt n1
  could generated this many leaves, or terminal
  letters.
• This leads to setting the tandem-pumping number
  to be p2n.
• The rest of the theorem follows from the above
  considerations. Only fact that need to verify is
  that the pumping can happened within a substring
  of length p. This just follows from finding a
  repeating variable in the last n2 levels of
  tree.