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Unformatted text preview: 1.21 b thus, the regular expression is: E U (a U b) a* b ((b U (a(a U b))) a * b)* (E U a) Some good regular expressions to test to see if your regular expressions work are: the empty string, "abbb," and "bbbb." 1.30 In this problem you need to merely explain the error in the "false proof." In the proof a reference is made to example 1.73, however, the logic that for S=0 p 1 p cannot be pumped is flawed. In cases 2 and 3 of example 1.73 the pumping of 0 p 1 p is possible, as a contradiction does not arise. This is because although there are more ones or zeros, the strings generated are still in the language (0*1*). The pumping lemma states only that a sub-divison allowing pumping exists, NOT that all possible sub- divisions can be pumped. 1.46 a There are multiple possible proofs, here is one using the pumping lemma. Suppose L were regular. Let p be the pumping length given by the pumping lemma. Consider the string w p = 0 p 10 p Clearly, |w p | >= p and w p is in L. So, by the pumping lemma, there must be some choice of x,y,z satisfying the conditions of the pumping lemma. But, consider any choice of x,y,z for which w p = xyz, |xy| <= p, and |y| >= 1. Because |xy| <= p, we know that x and y must be 0s, (otherwise |xy|>p ) . This allows us to define y = 0 j and we know that j>0 from point 2 (|y|>=1) of the pumping lemma. It followsand we know that j>0 from point 2 (|y|>=1) of the pumping lemma....
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- Spring '11