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Pumping Lemma Problem: Solution:
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 s is a member of B and of length at least p . We can show that no matter how we divide s into 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 vxy falls between 0p and 1p
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 0p and 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 1p and 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 1p and 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

PDA

Pumping Lemma Assume that L regular. Let p be the pumping constant given by the pumping lemma. Let s be the string 1p-1101p. s can 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 s cannot be pumped. This is a contradiction.

Pushdown Automaton -- PDA
Input String Stack States

Initial Stack Symbol Stack Stack stack head top bottom special symbol
Appears at time 0

The States Pop symbol Input symbol Push symbol

input stack top Replace

input stack top Push

input stack top Pop

input stack top No Change

Empty Stack input stack empty Pop top The automaton HALTS No possible transition after

A Possible Transition input stack Pop top

Non-Determinism Allowed non-deterministic transitions
PDAs are non-deterministic Allowed non-deterministic transitions

Example PDA PDA

Basic Idea: Push the a’s on the stack 2. Match the b’s on input with a’s on stack 3. Match found

Input Stack

Input Stack

Input Stack

Input Stack

Input Stack

Input Stack

Input Stack

Input Stack

Input Stack accept

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)

is the language accepted by the PDA:

reject

Another PDA example

Push v on stack 3. Match vR on input with v on stack 2. Guess middle of input 4. Match found

Guess the middle of string

accept

Rejection Example: Input

Guess the middle of string

There is no possible transition.
Input is not consumed

Another computation on same string:

final state is not reached

Another PDA example aab abbb (r)

Pushing Strings Pop symbol Input symbol Push string

Example: pushed string top Push

Another PDA example abbbaa

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