pumpingstrings1

pumpingstrings1 - p in light of the requirements that | xy...

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Computer Science E-207 A Proof by Contradiction Let’s prove that L = { a k : k = q 2 for some q 0 } is not regular. First, some intuition. Note that L = { ε,a,aaaa,aaaaaaaaa,aaaaaaaaaaaaaaaa,. .. } . Your in- stincts should tell you that a DFA couldn’t possibly accept this language, since its strings’ lengths don’t differ by any multiple of constant factor. Not convinced? Try sketching an NFA that accepts { ε } ; then augment it to accept { ε,a } ; then augment it to accept { ε,a,aaaa } ; then augment it to accept { ε,a,aaaa,aaaaaaaaa } . I don’t imagine your efforts at each step are identical to any previous efforts. In fact, continue in this fashion, and you’ll never converge on a single NFA that handles all of L . Now, our proof. Suppose that L is regular. Then there must exist some pumping length, p , such that any string, s , must be pumpable, provided | s | ≥ p . Well, let’s just see. Let’s choose s = a p 2 , which clearly is in L since s can be written as a k where k = q 2 for q = p . Let’s now consider how we might choose x,y,z Σ * such that s = xyz . Clearly, xy must contain at least one a and no more than
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Unformatted text preview: p , in light of the requirements that | xy | ≤ p and y 6 = ε . Since we’ve supposed that L is regular, s must be pumpable, so it must be that xy 2 z ∈ L . But, wait: what’s the length of xy 2 z ? Well, since | xyz | = p 2 , it must be that | xy 2 z | = p 2 + | y | ≤ p 2 + p , since, if | xy | ≤ p , it must be that y ≤ p . Now wait a minute. The length of s = a p 2 was p 2 . Shouldn’t the length of the next string in L ’s quadratic sequence, a ( p +1) 2 , be ( p +1) 2 = p 2 +2 p +1? Indeed! But note that p 2 < p 2 + p < p 2 + 2 p + 1, the implication of which is that the length of xy 2 z is strictly between two consecutive squares! Hence, it can’t be that xy 2 z = a k , where k = q 2 for some q ≥ 0, in which case xy 2 z isn’t in L , which contradicts the pumping lemma. Since the pumping lemma is provably true, the flaw in our argument must be our assumption that L is regular. Ergo, L is not regular. QED . 1...
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