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LING/C SC/PSYC 438/538 Lecture 11 Sandiway Fong

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Administrivia Homework 3 graded

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Last Time 1.Introduced Regular Languages – can be generated by regular expressions – or Finite State Automata (FSA) – or regular grammars --- not yet introduced 2.Deterministic and non-deterministic FSA 3.DFSA can be easily encoded in Perl: – hash table for the transition function – foreach loop over a string (character by character) – conditional to check for end state 4.NDFSA can be converted into DFSA – example of the set of states construction – Practice: ungraded homework exercise

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Ungraded Homework Exercise do not submit, do the following exercise to check your understanding – apply the set-of-states construction technique to the two machines on the ε- transition slide (repeated below) – self-check your answer: verify in each case that the machine produced is deterministic and accurately simulates its ε- transition counterpart a ε b > a ε b >

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Ungraded Homework Exercise Review Converting a NDFSA into a DFSA 1 a ε 23 b > {1,3} {2} a b {3} > Note: this machine with an ε-transition is non-deterministic Note: this machine is deterministic

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Ungraded Homework Exercise Review Converting a NDFSA into a DFSA 1 a ε 23 b > {1,2} {2} a b {3} b Note: this machine with an ε-transition is non-deterministic Note: this machine is deterministic >

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Last Time Regular Languages Three formalisms – All formally equivalent (no difference in expressive power) – i.e. if you can encode it using a RE, you can do it using a FSA or regular grammar, and so on … Regular Grammars FSA Regular Expressions Regular Languages talk more about formal equivalence later today… Perl regular expressions stuff out here

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Perl Regular Expressions Perl regex can include backreferences to groupings (i.e. \1, etc.) – backreferences give Perl regexs expressive power beyond regular languages: the set of prime numbers is not a regular language L prime = {2, 3, 5, 7, 11, 13, 17, 19, 23,.. } can be proved using the Pumping Lemma for regular languages (later) can have regular Perl code inside a regex

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Backreferences and FSA Deep question: – why are backreferences impossible in FSA? sx y a a b b > Example: Suppose you wanted a machine that accepted /(a+b+)\1/ One idea: link two copies of the machine together x2 y2 a a b b y Doesn’t work! Why? Perl implementation: – how to modify it get the backreference effect?

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Regular Languages and FSA Formal (constructive) set-theoretic definition of a regular language Correspondence between REs and Regular Languages concatenation (juxtaposition) union( | also [ ] ) Kleene closure( * )= (x + = xx*) Note: backreferences are memory devices and thus are too powerful e.g. L = {ww} and prime number testing (earlier slides)

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Regular Languages and FSA Other closure properties: Not true higher up: e.g. context-free grammars as we’ll see later

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Equivalence: FSA and Regexs Textbook gives one direction only Case by case: a)Empty string b)Empty set c)Any character from the alphabet

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Equivalence: FSA and Regexs Concatenation: – Link final state of FSA 1 to initial state of FSA 2 using an empty transition Note: empty transition can be eliminated using the set of states construction (see earlier slides in this lecture)

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Equivalence: FSA and Regexs Kleene closure: – repetition operator: zero or more times – use empty transitions for loopback and bypass

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Equivalence: FSA and Regexs Union: aka disjunction – Non-deterministically run both FSAs at the same time, accept if either one accepts

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Regular Languages and FSA Other closure properties: Let’s consider building the FSA machinery for each of these guys in turn…

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Regular Languages and FSA Other closure properties:

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Regular Languages and FSA Other closure properties:

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Regular Languages and FSA Other closure properties:

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Regular Languages and FSA Other closure properties:

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Regular Expressions from FSA Textbook Exercise: find a RE for Examples (* denotes string not in the language): *ab *ba babbab λ (empty string) bb *baba*baba bababbabab

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Regular Expressions from FSA Draw a FSA and convert it to a RE: 1 1 2 2 3 3 4 4 > b ab b b ε b*ab+( )+ [Powerpoint Animation] = b+(ab+)*| ε b

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Regular Expressions from FSA Perl implementation: $s = "ab ba bab bb baba babab"; while ($s =~ /\b(b+(ab+)*)\b/g) { print " match!\n"; } Output: perl test.perl match! Note: doesn’t include the empty string case Note: /../g global flag for multiple matches

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