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

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Adminstrivia Reminder –Homework 2 due tonight –Need Help?

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Administrivia some lectures are now available on the course hompage in audio format –as MP3 files –courtesy of Josh Harrison

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Last Time Expressing regular expressions as FSA

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Last Time Expressing regular expressions as FSA Microsoft Word Wildcards one or more of the preceding character e.g. –[ ] range of characters e.g. [aeiou] xy a a xy o a e i u

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Last Time Prolog FSA implementation –regular expression: ba + ! –query ?- s([b,a,a,’!’]). –code one predicate per state (but each predicate may have more than one rule) –s([b|L]) :- x(L). sx z b ! y a a > –y([a|L]) :- y(L). –y([‘!’|L]) :- z(L). –z([]). –x([a|L]) :- y(L).

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Today’s Topic More on the power of FSA …

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FSA Finite State Automata (FSA) have a limited amount of expressive power Let’s look at some modifications to FSA and their effects on its power

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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? b Answer: No –because we can always convert a machine with string-transitions into one without abb abb

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Empty Transitions –so far... all machines have had just a single character label on the arc how about allowing the empty string? –i.e. go from x to y without seeing a input character –does this endow the FSA with any more power? b Answer: No –because we can always convert a machine with empty transitions into one without xy

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Empty Transitions example –(ab)|b a b a b b

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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 sx y a a b b 12 a 3 b > >

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NDFSA more generally –non-determinism can be had not just with -transitions but with any symbol example: –given a, we can proceed to either state 2 or 3 12 a a 3 b >

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NDFSA –are they more powerful than FSA? –similar question asked earlier for -transitions –Answer: No –We can always convert a NDFSA into a FSA example –(set of states) 12 a a 3 b 1 2,3 a 3 b 2 > >

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NDFSA example –(set of states) trick: –construct new machine with states = set of possible states of the old machine 12 a a 3 b > {1} > a > {2,3} {3}{3} ba {1} > {2,3} {3} ba {1} > {2,3} 1 2,3 a 3 b 2 >

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Equivalence FSANDFSA FSA with -transitions same expressive power

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NDFSA and Prolog we have already seen how to encode a regular FSA in Prolog –(using one predicate per state) s([b|L]) :- x(L). x([a|L]) :- y(L). y([a|L]) :- y(L). y([‘!’|L]) :- z(L). z([]). s([b|L]) :- x(L). x([a|L]) :- y(L). y([a|L]) :- y(L). y([‘!’|L]) :- z(L). z([]). sx z b ! y a a >

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NDFSA and Prolog we can do the same for NDFSA –without the set-of-states conversion to FSA –let Prolog ’ s computation rule keep track of the possible states and choice points for us s([a|L]) :- x(L). s([a|L]) :- y(L). x([b|L]) :- y(L). y([]). s([a|L]) :- x(L). s([a|L]) :- y(L). x([b|L]) :- y(L). y([]). sx a a y b >

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NDFSA and Prolog computation tree –?- s([a]).L=[] rule 1 ?- x([]). –No ?- y([]).L=[] rule 2 –Yes rule 4 1.s([a|L]) :- x(L). 2.s([a|L]) :- y(L). 3.x([b|L]) :- y(L). 4.y([]). 1.s([a|L]) :- x(L). 2.s([a|L]) :- y(L). 3.x([b|L]) :- y(L). 4.y([]). sx a a y b >

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-Transitions and Prolog earlier example s([a|L]):- x(L). s(L):- x(L). x([b|L]):- y(L). y([]). s([a|L]):- x(L). s(L):- x(L). x([b|L]):- y(L). y([]). sx a y b >

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-Transitions and Prolog Computation tree: –?- s([b]). rule 2 ?- x([b]).L=[] rule 3 –?- y([]). rule 4 »Yes 1.s([a|L]):- x(L). 2.s(L):- x(L). 3.x([b|L]):- y(L). 4.y([]). 1.s([a|L]):- x(L). 2.s(L):- x(L). 3.x([b|L]):- y(L). 4.y([]). sx a y b >

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