Pumping lemma constant, Theory of Computation

a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0.

Since the length of xy ≤n, y consists of all b's Then xy 2 z = anbncn, where the length of of y = j. We know j > 0 so the length of the pumped string contains at as many a's as b's as c's, and is not in L. This is a Contradiction L = {w :| n a (w) = n b (w) = nc(w)}

b)

  • Let n be the pumping lemma constant. Then if L is regular, PL implies that s = anbncm can be decomposed into xyz, |y| > 0, |xy| ≤n, such thatxy i z is in L for all i ≥0.
  • Since the length of xy ≤n, there are three ways to partition s:

1. y consists of all a's

Pumping y will lead to a string with more than n a's -- not in L

2. y consists of all b's

Pumping y will lead to a string with more than m b's, and leave

the number of c's untouched, such that there are no longer 2n more c's than b's -- not in L

3. y consists of a's and b's

Pumping y will lead to a string with b's before a's, -- not in L

  • There is no way to partition anbncm that pumped strings are still in L.
Posted Date: 2/23/2013 12:53:05 AM | Location : United States







Related Discussions:- Pumping lemma constant, Assignment Help, Ask Question on Pumping lemma constant, Get Answer, Expert's Help, Pumping lemma constant Discussions

Write discussion on Pumping lemma constant
Your posts are moderated
Related Questions
The k-local Myhill graphs provide an easy means to generalize the suffix substitution closure property for the strictly k-local languages. Lemma (k-Local Suffix Substitution Clo

Exercise:  Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.

a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0. Since the le


One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included

how many pendulum swings will it take to walk across the classroom?

In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems

Ask question #Minimum 100 words accepte


Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.