LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.
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Transcript of LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.
![Page 1: LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.](https://reader036.fdocuments.us/reader036/viewer/2022062421/56649d555503460f94a32bb4/html5/thumbnails/1.jpg)
LING/C SC/PSYC 438/538Computational Linguistics
Sandiway Fong
Lecture 16: 10/23
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Midterm
• graded and returned
• +1 scaling applied
– Final will be on material covered after midterm only
– There will be one or two graded homeworks as well to come
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Question 6
[Correction to lecture 15 slides] • Give the deterministic FSA corresponding to:
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Answer 6
• [Correction to lecture 15 slides] • Deterministic machine
1
5
2a
3c 4
c
a
b b
c
6a
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Recap
• Following Chapter 3 of the textbook
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Stage 1: Lexical Intermediate Levels
• Figure 3.17 (top half):tape view of input/output pairs
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The Mapping Problem
• Example: (3.11)
• (Context-Sensitive) Spelling Rule: (3.5) e / {x,s,z}^__ s#
rewrites to letter e in left context x^ or s^ or z^ and right context s#
• i.e. insert e after the ^ when you see x^s# or s^s# or z^s#
• in particular, we have x^s# x^es#
![Page 8: LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.](https://reader036.fdocuments.us/reader036/viewer/2022062421/56649d555503460f94a32bb4/html5/thumbnails/8.jpg)
Stage 2: Intermediate Surface Levels
• also can be implemented using a FSTimportant!machine is designed to pass input not matching the rule through unmodified (rather than fail)
implements context-sensitive ruleq0 to q2 : left contextq3 to q0 : right context
![Page 9: LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.](https://reader036.fdocuments.us/reader036/viewer/2022062421/56649d555503460f94a32bb4/html5/thumbnails/9.jpg)
Stage 2: Intermediate Surface Levels
• Example (3.17)
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Stage 2: Intermediate Surface Levels
• Transition table for FST in 3.14
pg.79
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Stage 2: Intermediate Surface Levels
• Class Exercise– Let’s build this
FST in Prolog together
– state by state
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Stage 2: Intermediate Surface Levels
• A Note on Prolog– it will be useful to employ the cut
predicate for the implementation of “other”
• ! is known as the “cut” predicate– it affects how Prolog backtracks for
another solution– it means “cut” the backtracking off– Prolog will not try any other possible
matching rule on backtracking
![Page 13: LING/C SC/PSYC 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/23.](https://reader036.fdocuments.us/reader036/viewer/2022062421/56649d555503460f94a32bb4/html5/thumbnails/13.jpg)
FST 3.14 in Prolog
• % To run: ?- q0([f,o,x,^,s,#],L). L = [f,o,x,e,s,#]
• % q0• q0([],[]). % final state case• q0([z|L1],[z|L2]) :- !, q1(L1,L2). % q0 - z -> q1• q0([s|L1],[s|L2]) :- !, q1(L1,L2). % q0 - s -> q1• q0([x|L1],[x|L2]) :- !, q1(L1,L2). % q0 - x -> q1• q0([#|L1],[#|L2]) :- !, q0(L1,L2). % q0 - # -> q0• q0([^|L1],L2) :- !, q0(L1,L2). % q0 - ^:ep -> q0• q0([X|L1],[X|L2]) :- q0(L1,L2). % q0 - other -> q0
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FST 3.14 in Prolog
• % q1• q1([],[]). % final state case• q1([z|L1],[z|L2]) :- !, q1(L1,L2). % q1 - z -> q1• q1([s|L1],[s|L2]) :- !, q1(L1,L2). % q1 - s -> q1• q1([x|L1],[x|L2]) :- !, q1(L1,L2). % q1 - x -> q1• q1([^|L1],L2) :- !, q2(L1,L2). % q1 - ^:ep -> q2• q1([#|L1],[#|L2]) :- !, q0(L1,L2). % q1 - # -> q0• q1([X|L1],[X|L2]) :- q0(L1,L2). % q1 - other -> q0
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FST 3.14 in Prolog
• % q2• q2([],[]). % final state case• q2(L1,[e|L2]) :- q3(L1,L2). % q2 - ep:e -> q3• q2([s|L1],[s|L2]) :- !, q5(L1,L2). % q2 - s -> q5• q2([z|L1],[z|L2]) :- !, q1(L1,L2). % q2 - z -> q1• q2([x|L1],[x|L2]) :- !, q1(L1,L2). % q2 - x -> q1• q2([#|L1],[#|L2]) :- !, q0(L1,L2). % q2 - # -> q0• q2([X|L1],[X|L2]) :- \+ X = ^, q0(L1,L2). % q2 - other -
> q0
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FST 3.14 in Prolog
• % q3• q3([s|L1],[s|L2]) :- q4(L1,L2). % q3 - s -> q4
• % q4• q4([#|L1],[#|L2]) :- q0(L1,L2). % q4 - # -> q0
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FST 3.14 in Prolog
• % q5• q5([z|L1],[z|L2]) :- !, q1(L1,L2). % q5 - z -> q1• q5([s|L1],[s|L2]) :- !, q1(L1,L2). % q5 - s -> q1• q5([x|L1],[x|L2]) :- !, q1(L1,L2). % q5 - x -> q1• q5([^|L1],L2) :- !, q2(L1,L2). % q5 - ^:e -> q2• q5([X|L1],[X|L2]) :- \+ X = #, q0(L1,L2). % q5 -
other -> q0