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LING 388: Language and Computers

LING 388: Language and Computers. Sandiway Fong 9/27 Lecture 10. Adminstrivia. Reminder Homework 4 due Wednesday. Today’s Topic. Finite State Automata (FSA) equivalent to the regular expressions we’ve been studying. Regular Expressions: Example. .... from lecture 8

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LING 388: Language and Computers

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  1. LING 388: Language and Computers Sandiway Fong 9/27 Lecture 10

  2. Adminstrivia • Reminder • Homework 4 due Wednesday

  3. Today’s Topic • Finite State Automata (FSA) • equivalent to the regular expressions we’ve been studying

  4. Regular Expressions: Example .... from lecture 8 • example (sheeptalk) • baa! • baaa! • baaaa! • … • regular expression • baaa*! • baa+!

  5. Regular Expressions: Example .... from lecture 8 • example (sheeptalk) • baa! • baaa! • baaaa! • … • regular expression • baaa*! • baa+! a s b w a x a y > ! z

  6. Regular Expressions: Example • step-by-step • regular expression • baaa*! s > Start state: s

  7. Regular Expressions: Example • step-by-step • regular expression • baaa*! • b • from s, • see ‘b’, • move tow s b w >

  8. Regular Expressions: Example • step-by-step • regular expression • baaa*! • ba • From w, • see an ‘a’, • move tox s b w a x >

  9. Regular Expressions: Example • step-by-step • regular expression • baaa*! • baa • From x, • see an ‘a’, • move toy s b w a x a y >

  10. Regular Expressions: Example • step-by-step • regular expression • baaa*! • baaa* • baa • baaa • baaaa • baaaaa... • from y, • see an ‘a’, • move to ? s b w a x a y > a y’ but machine must have a finite number of states! a y” a ...

  11. Regular Expressions: Example • step-by-step • regular expression • baaa*! • baaa* • baa • Baaa • baaaa • baaaaa... • from y, • see an ‘a’, • “loop” aka return to state y a s b w a x a y >

  12. Regular Expressions: Example • step-by-step • regular expression • baaa*! • baaa*! • from y, • see an ‘!’, • move to final state z (indicated in red) a s b w a x a y > ! z Note: machine cannot finish (i.e. reach the end of the input string) in states s, x or y

  13. Finite State Automata (FSA) • construction • the step-by-step FSA construction method we just used • works for any regular expression • conclusion • anything we can encode with a regular expression, we can build a FSA for it • an important step in showing that FSA and REs are equivalent

  14. a b c d x y e ... ... z etc. a etc. one loop for each character y Microsoft Word Wildcards • basic wildcards • ? and * • ? any single character • e.g. p?t put, pit, pat, pet • * zero or more characters

  15. a x y a a e i o x y u Microsoft Word Wildcards • basic wildcards • @ • one or more of the preceding character • e.g. a@ • [ ] • range of characters • e.g. [aeiou]

  16. see anything but ‘<‘ see anything but ‘>‘ x x y y < > Microsoft Word Wildcards • basic wildcards • < > • < • beginning of a word • can think of there being a special symbol/invisible character marking the beginning of each word • > • end of a word • suppose there is an invisible character marking the end of each word

  17. see anything but ‘>‘ x y > see anything but ‘m‘ x m y z > Microsoft Word Wildcards • basic wildcards • < > • > • end of a word • Note • the see-anything-but loop is implicit • m> • “word that ends in m” • example: • mom is...

  18. Finite State Automata (FSA) • more formally • (Q,s,f,Σ,) • set of states (Q): {s,w,x,y,z} 5 states must be a finite set • start state (s): s • end state(s) (f): z • alphabet (Σ): {a, b, !} • transition function : signature: character × state → state • (b,s)=w • (a,w)=x • (a,x)=y • (a,y)=y • (!,y)=z a s b w a x a y > ! z

  19. Finite State Automata (FSA) • in Prolog • define one predicate for each state • taking one argument (the input list L) • consume input character (take the head of the list) • call next state with the tail of the list • rule • fsa(L) :- s(L). i.e. call start state s

  20. Finite State Automata (FSA) • state s: (start state) • s([b|L]) :-w(L). match input string beginning with b and call statewwith remainder of input • state w: • w([a|L]) :- x(L). • state x: • x([a|L]) :- y(L). • state y: • y([a|L]) :- y(L). • y([!|L]) :- z(L). • state z: (end state) • z([]). a s b w a x a y > ! z

  21. Finite State Automata (FSA) • query • ?-s([b,a,a,a,!]). a s b w a x a y > [a,a,a,!] [a,a,!] [a,!] [b,a,a,a,!] ! [!] Database s([b|L]) :-w(L). w([a|L]) :- x(L). x([a|L]) :- y(L). y([a|L]) :- y(L). y([!|L]) :- z(L). z([]). z []

  22. Finite State Automata (FSA) • In which state does query • ?-s([b,a,b,a,!]). fail? a s b w a x a y > [a,b,a,!] [b,a,b,a,!] [b,a,!] ! Database s([b|L]) :-w(L). w([a|L]) :- x(L). x([a|L]) :- y(L). y([a|L]) :- y(L). y([!|L]) :- z(L). z([]). z

  23. FSA • Finite State Automata (FSA) have a limited amount of expressive power • Let’s look at a modification to FSA and its effect on its power

  24. b a b b abb String Transitions • so far... • all machines have had just a single character label on the arc • so if we allow strings to label arcs • do they endow the FSA with any more power? • Answer: No • because we can always convert a machine with string-transitions into one without

  25. Finite State Automata (FSA) • equivalent a a s b w a x a y > s baa y > ! ! z machine with 5 states z

  26. Finite State Automata (FSA) • equivalent a a s b w a x a y > s baa y > ! ! Database s([b|L]) :-w(L). w([a|L]) :-x(L). x([a|L]) :- y(L). y([a|L]) :- y(L). y(['!'|L]) :- z(L). z([]). Database s([b,a,a|L]) :- y(L). y([a|L]) :- y(L). y(['!'|L]) :- z(L). z([]). z z

  27. b ε x y Empty Transitions • so far... • how about allowing the empty character? • i.e. go from x to y without seeing a input character • does this endow the FSA with any more power? • Answer: No • because we can always convert a machine with empty transitions into one without

  28. a b b Empty Transitions • example • (ab)|b a b > > ε

  29. a b a b > ε Empty Transitions • example • (ab)|(empty string) = final state

  30. s x a a b y a b b 1 2 3 ε NDFSA • Basic FSA • deterministic • it’s clear which state we’re always in, or • deterministic = no choice point • NDFSA • ND = non-deterministic • i.e. we could be in more than one state • non-deterministic  choice point • example: • initially, either in state 1 or 2 > >

  31. a b 1 2 3 a NDFSA • more generally • non-determinism can be had not just withε-transitionsbut with any symbol • example: • given a, we can proceed to either state 2 or 3 >

  32. a b a b 1 2 3 2,3 2,3 2 1 3 a NDFSA • NDFSA • are they more powerful than FSA? • similar question asked earlier forε-transitions • Answer: No • We can always convert a NDFSA into a FSA • example • (set of states) > >

  33. > > > > a a {1} {1} {1} {1} {2,3} {2,3} b {3} a b a b > 1 2 3 2,3 2,3 2 1 3 a b {2,3} {3} a > NDFSA • example • (set of states) • construct new machine with states = set of possible states of the old machine • Essential trick: • i.e. simulate the old (non-deterministic) machine with the new machine

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