In this exercise, we will investigate a technique to prove that a language is notregular. This tool is called the pumping lemma.The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the wordw is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may bedivided into three pieces, w = xyz, satisfying the following conditions:1. For each i ∈ N, xy^i z ∈ L(M ).2. |y| > 0 (i.e., y contains at least one character).3. |xy| ≤ p (i.e., the string xy has at most p characters). Use the pumping lemma to show the following language is not regular (HINT: Use proof by contradictionto assume the language is regular and apply the pumping lemma to the language):L = {0^k1^k | k ∈ N}
In this exercise, we will investigate a technique to prove that a language is notregular. This tool is called the pumping lemma.The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the wordw is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may bedivided into three pieces, w = xyz, satisfying the following conditions:1. For each i ∈ N, xy^i z ∈ L(M ).2. |y| > 0 (i.e., y contains at least one character).3. |xy| ≤ p (i.e., the string xy has at most p characters). Use the pumping lemma to show the following language is not regular (HINT: Use proof by contradictionto assume the language is regular and apply the pumping lemma to the language):L = {0^k1^k | k ∈ N}
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 33EQ
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In this exercise, we will investigate a technique to prove that a language is not
regular. This tool is called the pumping lemma.
The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the word
w is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may be
divided into three pieces, w = xyz, satisfying the following conditions:
1. For each i ∈ N, xy^i z ∈ L(M ).
2. |y| > 0 (i.e., y contains at least one character).
3. |xy| ≤ p (i.e., the string xy has at most p characters).
regular. This tool is called the pumping lemma.
The pumping lemma says that if M = (S, I, f, s0, F ) is a DFA with p states (i.e., p = |S|) and if the word
w is in L(M ) (the language generated by M ) and w has length greater than or equal to p, then w may be
divided into three pieces, w = xyz, satisfying the following conditions:
1. For each i ∈ N, xy^i z ∈ L(M ).
2. |y| > 0 (i.e., y contains at least one character).
3. |xy| ≤ p (i.e., the string xy has at most p characters).
Use the pumping lemma to show the following language is not regular (HINT: Use proof by contradiction
to assume the language is regular and apply the pumping lemma to the language):
L = {0^k1^k | k ∈ N}
to assume the language is regular and apply the pumping lemma to the language):
L = {0^k1^k | k ∈ N}
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