Science Computer E-207
A Proof by Contradiction
Lets prove that L = {ak : k = q 2 for some q 0} is not regular. First, some intuition. Note that L = {, a, aaaa, aaaaaaaaa, aaaaaaaaaaaaaaaa, . . .}. Your instincts should tell you that a DFA couldnt possibly accept this language, since its strings lengths dont dier by any multiple of constant factor. Not convinced? Try sketching an NFA that accepts {}; then augment it to accept {, a}; then augment it to accept {, a, aaaa}; then augment it to accept {, a, aaaa, aaaaaaaaa}. I dont imagine your eorts at each step are identical to any previous eorts. In fact, continue in this fashion, and youll 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, lets just see. Lets choose 2 s = ap , which clearly is in L since s can be written as ak where k = q 2 for q = p. Lets now consider we how might choose x, y, z such that s = xyz . Clearly, xy must contain at least one a and no more than p, in light of the requirements that |xy | p and y = . Since weve supposed that L is regular, s must be pumpable, so it must be that xy 2 z L. But, wait: whats the length of xy 2 z ? Well, since |xyz | = p2 , it must be that |xy 2 z | = p2 + |y | p2 + p, since, 2 if |xy | p, it must be that y p. Now wait a minute. The length of s = ap was p2 . Shouldnt the 2 length of the next string in Ls quadratic sequence, a(p+1) , be (p + 1)2 = p2 + 2p + 1? Indeed! But note that p2 < p2 + p < p2 + 2p + 1, the implication of which is that the length of xy 2 z is strictly between two consecutive squares! Hence, it cant be that xy 2 z = ak , where k = q 2 for some q 0, in which case xy 2 z isnt in L, which contradicts the pumping lemma. Since the pumping lemma is provably true, the aw in our argument must be our assumption that L is regular. Ergo, L is not regular. QED.
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