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CSE 311 Foundations of Computing I

Learn about the limitations of FSMs and pattern matching in computing. Explore CFG languages like binary palindromes and balanced parentheses. Discover efficient pattern matching algorithms using DFAs.

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CSE 311 Foundations of Computing I

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  1. CSE 311 Foundations of Computing I Lecture 24 FSM Limits, Pattern Matching Autumn 2011 CSE 311

  2. Announcements • Reading assignments • 7thEdition, Section 13.4 • 6th Edition, Section 12.4 • 5th Edition, Section 11.4 CSE 311

  3. Last lecture highlights • NFAs from Regular Expressions (01 1)*0 0  1     0     1 CSE 311

  4. Last lecture highlights 0,1 0 • “Subset construction”: NFA to DFA  a,b a 1 0 1  1 1 b c 0 1 b c 0 0 0 0,1 1 b,c a,b,c 0 NFA DFA CSE 311

  5. What can Finite State Machines do? • We’ve seen how we can get DFAs to recognize all regular languages • What about some other languages we can generate with CFGs? • { 0n1n : n≥0 }? • Binary Palindromes? • Strings of Balanced Parentheses? CSE 311

  6. A={0n1n: n≥0} cannot be recognized by any DFA Consider the infinite set of strings S={, 0, 00, 000, 0000, ...} Claim: No two strings in S can end at the same state of any DFA for A, so no such DFA can exist Proof: Suppose nm and 0n and 0m end at the same state p. Since 0n1n is in A, following 1nafter state p must lead to a final state. But then the DFA would accept 0m1n which is a contradiction CSE 311

  7. The set B of binary palindromes cannot be recognized by any DFA Consider the infinite set of strings S={, 0, 00, 000, 0000, ...} Claim: No two strings in S can end at the same state of any DFA for B, so no such DFA can exist Proof: Suppose nm and 0n and 0m end at the same state p. Since 0n10n is in B, following 10n after state p must lead to a final state. But then the DFA would accept 0m10n which is a contradiction CSE 311

  8. The set P of strings of balanced parentheses cannot be recognized by any DFA CSE 311

  9. Pattern Matching • Given • a string, s, of n characters • a pattern, p, of m characters • usually m<<n • Find • all occurrences of the pattern p in the string s • Obvious algorithm: • try to see if p matches at each of the positions in s • stop at a failed match and try the next position

  10. String s = x y x x y x y x y y x y x y x y y x y x y x x Pattern p = x y x y y x y x y x x

  11. String s = x y xxy x y x y y x y x y x y y x y x y x x x y xy y x y x y x x

  12. String s =xyx x y x y x y y x y x y x y y x y x y x x x y x y xy x y y x y x y x x

  13. String s = x yxxy x y x y y x y x y x y y x y x y x x x y x y x x y x y y x y x y x x

  14. String s = x y xx y x yxy y x y x y x y y x y x y x x x y x y x x y x y x yyx y x y x x

  15. String s = x y x xyx y x y y x y x y x y y x y x y x x x y x y x x y x y x y y xy x y y x y x y x x

  16. String s = x y x x yx y x y y x y x y xy y x y x y x x x y x y x x y x y x y y x x y x y y x y x y xx

  17. String s =x y x x y xyx y y x y x y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x xy x y y x y x y x x

  18. String s = x y x x y x yx yyx y x y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x x y xy y x y x y x x

  19. String s = x y x x y x y xyy x y x y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x x y x xy x y y x y x y x x

  20. String s = x y x x y x y x yyx y x y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x x y x x xy x y y x y x y x x

  21. String s = x y x x y x y x y yx y x yxy y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x x y x x x x y x yyx y x y x x

  22. String s = x y x x y x y x y y xyx y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x x y x x x x y x y y xy x y y x y x y x x

  23. String s = x y x x y x y x y y x yx y x y y x y x y x x x y x y x x y x y x y y x x y x y y x y x y x x x Worst-case time O(mn) x y x x x x y x y y x x y x y y x y x y x x

  24. String s = x y x x y x y x y y x yx y x y y x y x y x x x y xy x Lots of wasted work x y xy x yy x x yx y y x y x y xx x x y x x x x y x y y x x y x y y x y x y x x

  25. Better Pattern Matching via Finite Automata • Build a DFA for the pattern (preprocessing) of size O(m) • Keep track of the ‘longest match currently active’ • The DFA will have only m+1 states • Run the DFA on the string nsteps • Obvious construction method for DFA will be O(m2) but can be done in O(m) time. • Total O(m+n) time

  26. x y y x x x x x y y y Building a DFA for the pattern Patternp=x y x y y x y x y x x

  27. x y y x x x x x y y y Preprocessing the pattern Patternp=x y x y y x y x y x x y

  28. x y y x x x x x y y y Preprocessing the pattern Patternp=x y x y y x y x y x x x y x x y

  29. x y y x x x x x y y y Preprocessing the pattern Patternp=x y x y y x y x y x x x x x y x x y y y

  30. x y x y y x x x x x y y y Preprocessing the pattern Patternp=x y x y y x y x y x x x x y x x x y y y y y

  31. Generalizing • Can search for arbitrary combinations of patterns • Not just a single pattern • Build NFA for pattern then convert to DFA ‘on the fly’. • Compare DFA constructed above with subset construction for the obvious NFA.

  32. A Quick Note... • On how to convert NFAs and DFAs to equivalent regular expressions... • We’ve already seen • DFAs and NFAs recognize the same languages • NFAs (and therefore DFAs) recognize any language given by a regular expression • This completes the equivalence of DFAs and regular expressions CSE 311

  33. Generalized NFAs • Like NFAs but allow • Parallel edges • Regular Expressions as edge labels • NFAs already have edges labeled  or a • An edge labeled by A can be followed by reading a string of input chars that is in the language represented by A • A string x is accepted iff there is a path from start to final state labeled by a regular expression whose language contains x CSE 311

  34. Starting from NFA • Add new start state and final state • Then eliminate original states one by one, keeping the same language, until it looks like: • Final regular expression will be A λ A λ λ CSE 311

  35. Only two simplification rules: • Rule 1: For any two states q1 and q2 with parallel edges (possibly q1=q2), replace • Rule 2: Eliminate non-start/final state q3 by replacing all for every pair of states q1, q2 (even if q1=q2) A A⋃B q2 by q2 q1 q1 q2 q1 B B A C AB*C q2 q3 q1 by CSE 311

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