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## PowerPoint Slideshow about 'Pumping Lemma' - hamlet

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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.
- 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:

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.

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.

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 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.

Example PDA

PDA

Input

Stack

Input

Stack

Input

Stack

Input

Stack

Input

Stack

Input

Stack

Input

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Input

Stack

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)

Guess the middle

of string

Rejection Example:

Input

Guess the middle

of string

There is no possible transition.

Input is not consumed

final state

is not reached

Another PDA example

abbbaa

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