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Transparency No. 4-1 Formal Language and Automata Theory Chapter 4 Patterns, Regular Expressions and Finite Automata (include lecture 7,8,9) Transparency No. 4-1

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Patterns, regular expression & FAs Transparency No. 4-2 Patterns and their defined languages : a finite alphabet A pattern is a string of symbols representing a set of strings in *. The set of all patterns is defined inductively as follows: 1. atomic patterns: a , , , #, @. 2. compound patterns: if and are patterns, then so are: + , , *, +, ~ and . For each pattern , L( ) is the language represented by and is defined inductively as follows: 1. L(a) = {a}, L( ) = { }, L( )= {}, L(#) = , L(@) = *. 2. If L( ) and L( ) have been defined, then L( + ) = L( ) U L( ), L( ) = L( ) L( ). L( + ) = L( ) +, L( *) = L( )*, L(~ ) = * - L( ), L( ) = L( ) L( ).

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Patterns, regular expression & FAs Transparency No. 4-3 More on patterns We say that a string x matches a pattern iff x L( ). Some examples: * = L(@) = L(#*) 2. L(x) = {x} for any x * 3. for any x 1,…,x n in *, L(x 1 +x 2 +…+x n ) = {x 1,x 2,…,x n }. 4. {x | x contains at least 3 a’s} = L(@a@a@a@} - {a} = # ~a 6. {x | x does not contain a} = (# ~a)* 7. {x | every ‘a’ in x is followed sometime later by a ‘b’ } = = {x | either no ‘a’ in x or ‘b’ in x followed no ‘a’ } = (# ~a)* + @b(# ~a)*

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Patterns, regular expression & FAs Transparency No. 4-4 More on pattern matching Some interesting and important questions: 1. How hard is it to determine if a given input string x matches a given pattern a ? ==> efficient algorithm exists 2. Can every set be represented by a pattern ? ==> no! the set {a n b n | n > 0 } cannot be represented by any pattern. 3. How to determine if two given patterns and are equivalent ? (I.e., L( ) = L( )) --- an exercise ! 4. Which operations are redundant ? = ~(# @) = * ; + = * # = a 1 + a 2 +…+ a n if = {a 1,.., a n } + = ~( ) ; = ~ ( + ) It can be shown that ~ is redundant.

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Patterns, regular expression & FAs Transparency No. 4-5 Equivalence of patterns, regular expr. & FAs Recall that regular expressions are those patterns that can be built from: a , , , +, and *. Notational conventions: + means + ( ) + * means + ( *) Theorem 8: Let A *. Then the following are equivalent: 1. A is regular (I.e., A = L(M) for some FA M ), 2. A = L( ) for some pattern , 3. A = L( ) for some regular expression . pf: Trivial parts: (3) => (2). (2) => (1) to be proved now! (1)=> (3) later.

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Patterns, regular expression & FAs Transparency No. 4-6 (2) => (1) : Every set represented by a pattern is regular Pf: By induction on the structure of pattern . Basis: is atomic: (by construction!) = a : = : = or 4. = #: = @ = #* : Inductive cases: Let M 1 and M 2 be any FAs accepting L( ) and L( ), respectively. = : => L( ) = L(M 1 M 2 ) = * : => L( ) = L(M 1 *) = + = ~ or = : By ind. hyp. and are regular. Hence by closure properties of regular languages, is regular, too. = + = * : Similar to case 8.

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Patterns, regular expression & FAs Transparency No. 4-7 Some examples patterns & their equivalent FAs 1. (aaa)* + (aaaaa)*

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Patterns, regular expression & FAs Transparency No. 4-8 (1)=>(3): Regular languages can be represented by reg. expr. M = (Q, , , S, F) : a NFA; X Q: a set of states; Q : two states X ( ) = def {y * | a path from to labeled y and all intermediate states X }. Note: L(M) = ? X ( ) can be shown to be representable by a regular expr. by induction as follows: Let { a | ,a) } = {a 1,…,a k } ( k 0), then 1. if : ( ) = {a 1, a 2,…,a k }= L(a 1 + a 2 +…+ a k ) if k > 0, = {} = L( ) if k = 0. 2. if = : ( ) = {a 1, a 2,… a k, }=L(a 1 + a 2 +…+ a k + ) if k > 0, = { } = L( ) if k = 0. 3. For nonempty X, let q be any state in X, then, X ( ) = X-{q} ( ) U X-{q} ( q) ( X-{q} (q q))* X-{q} (q ) = L( + ), where by ind. hyp., assume L( ( )) = ( X-{q} ( ), X-{q} ( q), ( X-{q} (q q)), X-{q} (q ) )

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Patterns, regular expression & FAs Transparency No. 4-9 Finally, L(M) = {x | s --x--> f, s S, f F } = s S, f F Q (s,f), is representable by a regular expression. Example (9.3): M : Some examples L(M) = {p,q,r} (p,p) = {p,r} (p,p) + {p,r} (p,q) ( {p,r} (q,q))* {p,q,r} (q,p) {p,r} (p,p) = ? {p,r} (p,q) = ? {p,r} (q,q) = ? {p,r} (q,p) = ? Hence L(M) = ?

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