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Pumping Lemma. Problem: Let B be the language of all palindromes over {0,1} containing an equal number of 0s and 1s. Show that B is not CFL Solution: We assume that B is a CFL and obtain a contradiction.

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pumping lemma
Pumping Lemma

Problem:

  • Let B be the language of all palindromes over {0,1} containing an equal number of 0s and 1s. Show that B is not CFL

Solution:

  • We assume that B is a CFL and obtain a contradiction.
  • Let p be the pumping constant of B that is guaranteed to exist by the pumping lemma. Select the string s= 0p1p1p0p.
  • Clearly sis a member of B and of length at least p . We can show that no matter how we divide sinto uvxyz, one of the conditions of the lemma is violated.
    • Pumping lemma, | vxy |<= p , we can only place vxy in the following ways:
slide3
vxy completely falls in the first 0p
    • If we pump v and y , then the new string is no longer a palindrome, and the number of 0s will be greater than the number of 1s, which is a contradiction.
  • vxy falls between 0pand 1p
    • In this case, v will only contain 0s while y will only contain 1s. So if we pump s , the new string is no longer palindrome, which is a contradiction.
  • vxy completely falls in the first 1p
    • Similar to (1), after pumping s , the number of 1s will be greater than the number of 0s, which is a contradiction.
  • vxy falls between 1pand 1p
    • After pumping s , the number of 1s will be greater than the number of 0s, which is a contradiction.
  • vxy completely falls in the second 1p
    • This case is same with (3).
  • vxy falls between 1pand 0p
    • Similar to (2), v will only contain 1s while y will only contain 0s. So if we pump s , the new string is no longer palindrome, which is a contradiction.
  • vxy completely falls in the second 0p
    • This case is same with (1).

Hence, B is not context free.

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Let L = {w  {0; 1}* : w = wR}.
  • (a) Show that L is context-free by giving a context-free grammar for L.
  • (b) Show that L is context-free by giving a pushdown automaton for L.
  • (c) Show that L is not regular.
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CFG
  • L is a symmetric language. Consider a context-free grammar for L : (V, Σ, R, S) , where

i. V = {S}

ii. Σ = {0,1}

iii. Rules:

S  0S0 |1S1| 0 |1|ε

iv. S = S V

pumping lemma1
Pumping Lemma
  • Assume that L regular. Let p be the pumping constant given by the pumping lemma.
  • Let sbe the string 1p-1101p.
  • scan be split into three pieces, s =xyz , where for any i >= 0 the string xyiz is in L .
  • According to the pumping lemma condition, we must have | xy |<= p . If this is the case, then y must consist only of 1’s, so xyyz L . Therefore scannot be pumped.
  • This is a contradiction.
pushdown automaton pda
Pushdown Automaton -- PDA

Input String

Stack

States

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Initial Stack Symbol

Stack

Stack

stack

head

top

bottom

special symbol

Appears at time 0

the states
The States

Pop

symbol

Input

symbol

Push

symbol

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input

stack

top

Replace

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input

stack

top

Push

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input

stack

top

Pop

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input

stack

top

No Change

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Empty Stack

input

stack

empty

Pop

top

The automaton HALTS

No possible transition after

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A Possible Transition

input

stack

Pop

top

non determinism
Non-Determinism

PDAs are non-deterministic

Allowed non-deterministic transitions

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Basic Idea:
  • Push the a’s
  • on the stack

2. Match the b’s on input

with a’s on stack

3. Match

found

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Input

Stack

slide22
Input

Stack

slide23
Input

Stack

slide24
Input

Stack

slide25
Input

Stack

slide26
Input

Stack

slide27
Input

Stack

slide28
Input

Stack

slide29
Input

Stack

accept

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A string is accepted if there is a computation such that:
    • All the input is consumed
    • The last state is an accepting state
    • At the end of the computation,
    • we do not care about the stack contents
    • (the stack can be empty at the last state)
slide40
Push v
  • on stack

3. Match vR on input

with v on stack

2. Guess

middle

of input

4. Match

found

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final state

is not reached

pushing strings
Pushing Strings

Pop

symbol

Input

symbol

Push

string

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Example:

pushed

string

top

Push

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