In the theory of formal languages, the pumping lemma for regular languages is a lemma that For example, the following image shows an FSA. An automat
The Pumping Lemma: Examples. Consider the following three languages: The first language is regular, since it contains only a finite numberof strings. The third language is also regular, since it is equivalent to theregular expression (a*)(b*).
We need to show that xy 2 z = 0 p+b 1 p is not in L1. b ≥ 1. So p+b > p. Hence 0 p+b 1 p is not in L. 2. L2 = {xx | x ∈ {0, 1}*} is not regular. We show that the pumping lemma does not hold for L2. Yet more PDA Pumping Lemma Examples . Example 1.
By pumping lemma, let w = xyz, where |xy| ≤ n. Let x = a p, y = a q, and z = a r b n, where p + q + r = n, p ≠ 0, q ≠ 0, r ≠ 0. Thus |y| ≠ 0. Let k = 2. Then xy 2 z = a p a 2q a r b n. Number of as = (p + 2q + r) = (p + q + r) + q = n + q. Hence, xy 2 z = a n+q b n.
Let us assume that L is regular, then by Pumping Lemma the above given rules follow. Now, let x ∈ L and |x| ≥ n.
Pick a particular number k ∈ N and argue that uvkw ∈ L, thus yielding our desired contradiction. What follows are two example proofs using Pumping Lemma.
· N chooses a string s∈L such that |s|≥p. · C chooses strings u,v, that it is necessarily so. Example: ¡ вд £.
Pumping Lemma Example Problems This is an in class exercise. For each problem, choose a partner, and then solve the problem by using the pummping lemma. Write down the proof for the problem on a sheet of paper. We will discuss solutions for each problem, before moving on to the next problem.
One might think that any string of the form wwRw would suffice. This is not correct, however.
L = {ss | s Î {a,b}*}. This language is similar in one respect to even length palindromes. In both cases , to recognize a string in the language , a machine needs to save the first half to compare with the last half. Pumping Lemma Example Problems This is an in class exercise. For each problem, choose a partner, and then solve the problem by using the pummping lemma. Write down the proof for the problem on a sheet of paper.
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The goal is to assume that the language is regular and then derive strings which are not in the language, thereby contradicting the regularity assumption. This language is not regular: Example: The pumping lemma (PL) for ‘a (bc) d’ has the following example: a = x (bc) = y; d = z; The finite automata of this pumping lemma (PL) can be drawn as: Applications for Pumping Lemma (PL) PL can be applied for the confirmation of the following languages are not regular. It should never be used to show a language is regular. Using The Pumping Lemma)In-Class Examples: Using the pumping lemma to show a language L is not regular ¼5 steps for a proof by contradiction: 1.
Partition it according to constraints of pumping lemma in a generic way
The Pumping Lemma: Examples. Consider the following three languages: The first language is regular, since it contains only a finite numberof strings. The third language is also regular, since it is equivalent to theregular expression (a*)(b*). Then there exists some n as in the pumping lemma.
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therefore there must be a DFA with size n, which is the minimal accepting automata. n n n Pick z = a b c obviously (given z=uvw and |uv| <= n and |v|>=1 ) j v = a
B = {0 n 1 n} Assume B is regular, with pumping length p; Let s be 0 p 1 p; s cannot be divided into xyz because ; if y is all 0's then xyyz; if y is all 1's if y is 0 k 1 k then xyyz may have equal number of 0's and 1's, but they will be in the wrong order Se hela listan på liyanxu.blog Once we assume A1 is regular, the lemma provides us with the pumping length, p. We are now free to choose a word s which belongs to A1 and has length ≥ p. If we choose s appropriately, we should be able to “pump up” the size of s in the manner described by the pumping lemma and show the resulting word, s,does not belong to A1. – The Pumping Lemma – Examples – Algorithms that answer questions about FAs. • Reading: Sipser, Sections 1.4, 4.1. •Next: – Computability theory – Readings: • Sipser Chapter 3 • The Nature of Computation, Chapter 8 • GITCS notes, lecture 4 Regular Pumping Lemmas Contents.
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By the pumping lemma, w 2 = xy 2 z must also be in L, but the number of a's before the b in w 2 must be at least p + 1, while the number of a's after b is still p + 1. But this contradicts the condition for w 2 being in L and so the assumption that L is regular is false. Example 2 Using the Pumping Lemma
AMS Euler typeface example. 2008. Public domain Pumping-Lemma xyz.