l8-handout

l8-handout - Outline Decision problems on regular languages...

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Lecture 8: Finishing Regular expressions David Dill Department of Computer Science 1 Outline Decision problems on regular languages. 2 Closure Properties for Non-Regularity Proof by contradiction: If we can construct a non-regular language using regularity-preserving operations, one of the original languages must have been non-regular. Example: { a i b i | n 0 } is not regular. proof: By contradiction. Consider the homomorphism a 7→ 0 and b 7→ 1 . If a n b n were regular, so would 0 n 1 n be, contradicting a previous theorem. Example: The language L E of strings with equal numbers of 0 s and 1 s is not regular. proof By contradiction. Suppose L E were regular. Then L E L (0 * 1 * ) = { 0 n 1 n | n 0 } would be regular, by closure of regular languages under intersection. But that contradicts a previous theorem. Example: The language L NE , where the number of 0 s is different from the number of 1 s, is not regular. proof
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This note was uploaded on 03/08/2011 for the course CS 154 taught by Professor Motwani,r during the Winter '08 term at Stanford.

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l8-handout - Outline Decision problems on regular languages...

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