LING/C SC/PSYC 438/538 Computational Linguistics

1. LING/C SC/PSYC 438/538Computational Linguistics Sandiway Fong Lecture 13: 10/9

2. Administrivia • Homework 2 • grade (to be) sent by email today • Upcoming midterm • this Thursday

3. s x a b a b z y a a b b NDFSA → (D)FSA: Recap • set of states construction • new machine simulates in parallel all possible situations for the original machine • procedure must terminate • 4 states • 24-1=15 possible different states in the new machine

4. Practice Question • convert the NDFSA into a deterministic FSA • implement both the NDFSA and the equivalent FSA in Prolog using the “one predicate per state” encoding • run both machines on the strings abab and abaaba, • how many steps (transitions + final stop) does each machine make? from figure 2.27 in the textbook

5. Today’s Topics • FSA to regexp • FSA and complementation • Regexp and complementation

6. FSA to Regexp • Give a Perl regexp for this FSA

7. FSA to Regexp • Let’s define Ei to be the possible strings that can lead to an accepting computation from state i • Then: • E1 = aE2 • E2 = bE3 | bE4 • E3 = aE2 | λ • E4 = aE3

8. FSA to Regexp • Solve simultaneous equations • E1 = aE2 • E2 = bE3 | bE4 • E3 = aE2 | λ • E4 = aE3

9. FSA to Regexp • Solve simultaneous equations • E1 = aE2 • E2 = bE3 | bE4 (eliminate) • E3 = aE2 | λ • E4 = aE3 • Then • E1 = abE3 | abE4 • E3 = abE3 | abE4 | λ

10. FSA to Regexp • Solve reduced equations • E1 = abE3 | abE4 • E3 = abE3 | abE4 | λ • E4 = aE3

11. FSA to Regexp • Solve reduced equations • E1 = abE3 | abE4 • E3 = abE3 | abE4 | λ • E4 = aE3 (eliminate E4) • Then • E1 = abE3 | abaE3 • E3 = abE3 | abaE3 | λ

12. E3 ab|aba FSA to Regexp • Solve reduced equations • E1 = abE3 | abaE3 • E3 = abE3 | abaE3 | λ • Solve recursive definition for E3 rewrite E3 as • E3 = (ab|aba)E3 |λ

13. E3 ab|aba FSA to Regexp • Then • E1 = abE3 | abaE3 • E3 = (ab|aba)* • Also • E1 = (ab|aba)E3 • Substituting for E3 • E1 = (ab|aba)(ab|aba)*

14. FSA to Regexp • E1 = (ab|aba)(ab|aba)* • We know • e+ = ee* = e*e • Hence • E1 = (ab|aba)+ Note: the order of variable elimination matters

15. b a b reject b a,b FSA and Complementation • FSA are closed under complementation • i.e. make a new FSA’ accept strings rejected by FSA, and reject strings accepted by FSA over the alphabet • ∑= {a,b}

16. b a b reject b a,b FSA and Complementation • A Simple Idea: • make old accepting state(s) non-final states • make non-final and reject states accepting states • Nearly correct, but why does this not work? • hint: consider aba 3 1 2 4 5

17. Prolog one([a|L]) :- two(L). two([b|L]) :- three(L). two([b|L]) :- four(L). three([]). three([a|L]) :- two(L). four([a|L]) :- three(L). To construct the complement FSA in Prolog: add a single line fsab(L) :- \+ one(L). \+ is the Prolog operator for negation as failure to prove i.e. \+ one(L) is true if one(L) cannot be true \+ one(L) is false if one(L) can be true FSA and Complementation weakness is that \+ can’t be used to generate

18. Regexps and Complementation Not part of basic regexps • Formally, we can define: • complement of regexp e = ∑*– e • where • ∑* = set of all possible strings over alphabet ∑ • i.e. zero or more occurrences ... • Class Exercise • Let ∑= {a,b} • Give a Perl regexp equivalent to the complement regexp for (ab|aba)+