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32+Slides--The+Pumping+Lemma

# 32+Slides--The+Pumping+Lemma - CS103 HO#32 Slides-The...

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CS103 HO#32 Slides--The Pumping Lemma 4/27/11 1 More on Regular Expressions All strings over the alphabet {a, b} that have at least 2 a's (some beginning) (1 st impt. a) (some middle) (2 nd impt. a) (some ending) (a b)* a (a b)* a (a b)* More on Regular Expressions All strings over the alphabet {a, b} that have at least 2 a's (some beginning) (1 st impt. a) (some middle) (2 nd impt. a) (some ending) (a b)* a (a b)* a (a b)* (some b's) (1 st a) (some b's) (2 nd a) (some ending) b*ab*a (a b)* (a b)*ab*ab* b*a(a b)*ab* Kleene's Theorem Theorem 1.54 : A language is regular if and only if some regular expression describes it. Lemma 1.55 : If a language is described by a regular expression, then it is regular. Lemma 1.60 : If a language is regular, then it is described by a regular expression. Non-Regular Languages q 1 q 2 q 3 q 4 q 5 a a a a a b b b b b b b b b b a,b Suppose we want to recognize B = { a n b n | n 0 } q 0 q 12 q 13 q 14 q 15 q 9 q 10 q 11 q 7 q 8 q 6 Non-Regular Languages We will show that regular languages have a particular property. To show that a language L is not regular: Assume that L is regular. Thus it has the property. Derive a contradiction. Thus L is not regular. Note that this approach is only useful in proving that L is not regular. Having the property doesn't guarantee that a language is regular. Long Strings Force Repeated States Suppose M = (Q, , , q 1 , F). Suppose s is a string of length at least |Q| that is accepted by M. Then M must visit some state more that once when s is the input. PROOF: M starts in some state, and each time it reads a character, it visits some state. So if |s| = |Q|, M will visit |Q| + 1 states (the initial one plus one for each character in s). Since there more visits than states, the pigeonhole principle tells us that at least one state is visited more than once. This will be true for any accepted string s such that |s| |Q|.

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CS103 HO#32 Slides--The Pumping Lemma 4/27/11 2 b a a c b d a (Assume transitions not shown in this DFA lead to the accepting state.) What is the longest string M can accept without repeating a state? Any accepted string of 6 or more symbols must contain cd one or more times. L(M) = ba(cd)*(ca a)b … strings for missing transitions Examples: baab bacab ba cd ab ba cdcdcd ab ba cdcdcdcd cab M The Pumping Lemma If A is a regular language, then there is a number p (the pumping length) such that, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz satisfying the following three conditions: 1. for each i 0, xy i z A, 2. |y| > 0, and 3. |xy| p. Note that x or z may be , but y may not.
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32+Slides--The+Pumping+Lemma - CS103 HO#32 Slides-The...

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