1. LING 581: Advanced Computational Linguistics
Lecture Notes
April 24th

2. Administrivia
Thanks for submitting homework 9 – I've commented on a few of your submissions

3. From last time
psg2.pl 

4. From last time
psg2.pl 

5. Evaluation Check our computer implementation on… fjpSlides4.pdf

6. Evaluation Load psg2.pl into SWI Prolog; Syntax: (S) Proposition (P):
?- s(P,S,[jack,is,hungry],[]).
P =  (findall(_1820, is_hungry(_1820), _1796), member(jack, _1796)),
S = s(n(jack), vp(v(v(is), ap(hungry)))) ;
false.
Syntax:
(S)
Proposition (P):
find all X s.t. is_hungry(X) is true.
Name that list S
Jack must be a member of S

7. Evaluation James likes Jack: Syntax: (S) Proposition (P):
?- s(P,S,[james,likes,jack],[]).
P =  (findall(_90, likes(_90, jack), _62), member(james, _62)),
S = s(n(james), vp(vt(likes), n(jack))) ;
false.
Syntax:
(S)
Proposition (P):
find all X s.t. likes(X,jack) is true.
Name that list S
James must be a member of S

8. Evaluation Proposition (P): Syntax:
It s not the case that James likes Jack:
?- s(P,S,[it,is,not,the,case,that,james,likes,jack],[]).
P =  (\+ (findall(_102, likes(_102, jack), _74), member(james, _74))),
S = s(neg, s(n(james), vp(vt(likes), n(jack)))) ;
false.
Proposition (P):
find all X s.t. likes(X,jack) is true.
Name that list S
James must NOT be a member of S
Syntax:
(S)

9. Evaluation Syntax (S):
Jack is hungry, and it s not the case that James likes Jack:
?- s(P,S,[jack,is,hungry,and,it,is,not,the,case,that,james,likes,jack],[]).
P =  ((findall(_178, is_hungry(_178), _154), member(jack, _154)), \+ (findall(_288, likes(_288, jack), _250), member(james, _250))),
S = s(s(n(jack), vp(v(v(is), ap(hungry)))), conj(and), s(neg, s(n(james), vp(vt(likes), n(jack))))) ;
false.
Syntax (S):

10. Evaluation Proposition (P1): find all X s.t. is_hungry(X) is true.
Jack is hungry, and it s not the case that James likes Jack:
?- s(P,S,[jack,is,hungry,and,it,is,not,the,case,that,james,likes,jack],[]).
P =  ((findall(_178, is_hungry(_178), _154), member(jack, _154)), \+ (findall(_288, likes(_288, jack), _250), member(james, _250))),
S = s(s(n(jack), vp(v(v(is), ap(hungry)))), conj(and), s(neg, s(n(james), vp(vt(likes), n(jack))))) ;
false.
Proposition (P1):
find all X s.t. is_hungry(X) is true.
Name that list S
Jack must be a member of S
Proposition (P2):
find all X s.t. likes(X,jack) is true.
Name that list S
James must NOT be a member of S
P1 and P2 true

11. Evaluation Situations:
Assert facts into database
Closed world assumption: is fact is not in database, false.
Prolog requires use to declare predicates (to guard against programming errors) using dynamic predicate/arity (= # of arguments)

12. Evaluation
Situation V': Jack is hungry, and it is not the case that James likes Jack
Assume:
?- assert(is_hungry(jack)).
true.
?- dynamic likes/2.
i.e.:
?- likes(jack, james).
false.
?- s(P,_,[jack,is,hungry,and,it,is,not,the,case,that,james,likes,jack],[]), call(P).
P =  ((findall(_2668, is_hungry(_2668), [james, jack]), member(jack, [james, jack])), \+ (findall(_2778, likes(_2778, jack), _2740), member(james, _2740))) ;
false.
True!

13. Evaluation
Situation V'': Jack is hungry, and it is not the case that James likes Jack
Assume:
?- assert(is_hungry(jack)).
true.
?- assert(likes(jack, james)).
?- s(P,_,[jack,is,hungry,and,it,is,not,the,case,that,james,likes,jack],[]), call(P).
false.
False!

14. Evaluation
Consider now (16) below (note the bracketing!):

15. Evaluation
Situation V''': It is not the case that Jack is hungry or Sophia is boring
?- s(P,S,[it,is,not,the,case,that,jack,is,hungry,or,sophia,is,boring],[]).
P =  (\+ (findall(_18234, is_hungry(_18234), _18210), member(jack, _18210));findall(_18314, is_boring(_18314), _18290), member(sophia, _18290)),
S = s(s(neg, s(n(jack), vp(v(v(is), ap(hungry))))), conj(or), s(n(sophia), vp(v(v(is), ap(boring))))) ;
P =  (\+ (findall(_18162, is_hungry(_18162), _18138), member(jack, _18138);findall(_18242, is_boring(_18242), _18218), member(sophia, _18218))),
S = s(neg, s(s(n(jack), vp(v(v(is), ap(hungry)))), conj(or), s(n(sophia), vp(v(v(is), ap(boring)))))) ;
false.
1st parse:
Narrow
scope for
negation
2nd parse:
Wide
scope for
negation

16. Evaluation
Situation V''': It is not the case that Jack is hungry or Sophia is boring
?- assert(is_hungry(jack)).
true.
?- assert(is_boring(sophia)).
Cut (!) preserves only the first reading (narrow scope for neg):
?- s(P,_,[it,is,not,the,case,that,jack,is,hungry,or,sophia,is,boring], []), !, call(P).
P =  (\+ (findall(_2828, is_hungry(_2828), _2804), member(jack, _2804));findall(_2908, is_boring(_2908), [sophia]), member(sophia, [sophia])).
True!

17. Evaluation
Suppose Jack is not hungry, does this change the proposition's truth value?
?- retract(is_hungry(jack)).
true.
?- is_hungry(jack).
false.
Still true!
?- s(P,_,[it,is,not,the,case,that,jack,is,hungry,or,sophia,is,boring],[]), !, call(P).
P =  (\+ (findall(_2838, is_hungry(_2838), _2814), member(jack, _2814));findall(_2918, is_boring(_2918), _2894), member(sophia, _2894)) .

18. Evaluation
Almost situation V'''': It is not the case that Jack is hungry or Sophia is boring
?- retract(is_boring(sophia)).
true.
?- is_boring(sophia).
false.
Since Jack is not hungry, proposition remains true!
?- s(P,_,[it,is,not,the,case,that,jack,is,hungry,or,sophia,is,boring],[]) , !, call(P).
P =  (\+ (findall(_2844, is_hungry(_2844), _2820), member(jack, _2820));findall(_2924, is_boring(_2924), _2900), member(sophia, _2900)) .

19. Evaluation Proposition now false!
Now situation V'''': It is not the case that Jack is hungry or Sophia is boring
?- assert(is_hungry(jack)).
true.
Proposition now false!
?- s(P,_,[it,is,not,the,case,that,jack,is,hungry,or,sophia,i s,boring],[]), !, call(P).
false.

20. Quantifiers nobody has seen a unicorn means roughly (Prolog-style):
Not all noun phrases (NPs) are (by nature) directly referential like names
Quantifiers: “something to do with indicating the quantity of something”
Examples:
every child
nobody
two dogs
several animals
most people
nobody has seen a unicorn
means roughly (Prolog-style):
?- setof(X,(person(X), seen(X,Y), unicorn(Y)),Set),cardinality(Set,0).

21. Quantifiers Database setof vs. findall (recall last lecture) Fix:
nobody has seen a unicorn
means roughly (Prolog-style):
?- setof(X,(person(X), seen(X,Y), unicorn(Y)),Set),cardinality(Set,0).
Database
setof vs. findall (recall last lecture)
Fix:

22. Quantifiers Semantic compositionality: Example:
elements of a sentence combine in piecewise fashion to form an overall (propositional) meaning for the sentence
Example:
(4) Every baby cried
Word Meaning
cried cried(X).
baby baby(X).
every ?
every baby cried proposition (True/False)
that can be evaluated for a given situation

23. Quantifiers Scenario (Possible World):
suppose there are three babies...
baby(noah).
baby(merrill).
baby(dani).
all three cried
cried(noah).
cried(merrill).
cried(dani).
only Dani jumped
jumped(dani).
Noah and Dani swam
swam(noah).
swam(dani).
(6)
every baby
exactly one baby
most babies
cried ✓
jumped
swam
think of quantifiers as “properties-of-properties”
every_baby(P) is a proposition
P: property
every_baby(P) true for P=cried
every_baby(P) false for P=jumped and P=swam