LINGC SCPSYC 438538 Computational Linguistics

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LING/C SC/PSYC 438/538Computational Linguistics

• Sandiway Fong
• Lecture 13 10/9

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Administrivia

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

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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
• 24-115 possible different states in the new
  machine

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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
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Todays Topics

• FSA to regexp
• FSA and complementation
• Regexp and complementation

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FSA to Regexp

• Give a Perl regexp for this FSA

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FSA to Regexp

• Lets 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 ?
• E4 aE3

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FSA to Regexp

• Solve simultaneous equations
• E1 aE2
• E2 bE3 bE4
• E3 aE2 ?
• E4 aE3

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FSA to Regexp

• Solve simultaneous equations
• E1 aE2
• E2 bE3 bE4 (eliminate)
• E3 aE2 ?
• E4 aE3
• Then
• E1 abE3 abE4
• E3 abE3 abE4 ?

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FSA to Regexp

• Solve reduced equations
• E1 abE3 abE4
• E3 abE3 abE4 ?
• E4 aE3

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FSA to Regexp

• Solve reduced equations
• E1 abE3 abE4
• E3 abE3 abE4 ?
• E4 aE3 (eliminate E4)
• Then
• E1 abE3 abaE3
• E3 abE3 abaE3 ?

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FSA to Regexp

• Solve reduced equations
• E1 abE3 abaE3
• E3 abE3 abaE3 ?
• Solve recursive definition for E3
• rewrite E3 as
• E3 (ababa)E3 ?

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FSA to Regexp

• Then
• E1 abE3 abaE3
• E3 (ababa)

• Also
• E1 (ababa)E3

• Substituting for E3
• E1 (ababa)(ababa)

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FSA to Regexp

• E1 (ababa)(ababa)
• We know
• e ee ee

• Hence
• E1 (ababa)

Note the order of variable elimination matters
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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

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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

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1
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4
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FSA and Complementation

• Prolog
• one(aL) - two(L).
• two(bL) - three(L).
• two(bL) - four(L).
• three().
• three(aL) - two(L).
• four(aL) - 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

weakness is that \ cant be used to generate
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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 (ababa)