1. LING 388: Language and Computers
Sandiway Fong
10/15
Lecture 15

2. Adminstrivia Reminder Extra credit homework out today
Homework 4 due tonight…
Extra credit homework out today
due Thursday
easy way to pick up extra points…

3. Last Time Applied the transformation: to NP and VP rules:
np(np(DT,NN)) --> dt(DT), nn(NN).
np(np(NP,PP)) --> np(NP), pp(PP).
vp(vp(VBD,NP)) --> vbd(VBD), np(NP).
vp(vp(VP,PP)) --> vp(VP), pp(PP).
x(X) --> [z], w(X,x(z)).
x(x(z)) --> [z].
w(W,X) --> [y], w(W,x(X,y)).
w(x(X,y),X) --> [y].
x(x(X,y)) --> x(X), [y].
x(x(z)) --> [z].
[z]
[y] x x
[z]
[y] x x

4. Last Time

5. Last Time
2nd page:

6. Last Time Parses for: Stanford Parser:
I saw a boy with a telescope with a limp with no money
Stanford Parser:

7. Last Time
Parses for:
I saw a boy with a telescope with a limp with no money
3 PPs
with a telescope
with a limp
with no money
14 parses …

8. saw
a boy
a telescope
a limp
no money

9. saw
a boy
a telescope
a limp
no money

10. saw
a boy
a telescope
a limp
no money

11. saw
a boy
a telescope
a limp
no money

12. saw
a boy
a telescope
a limp
no money

13. saw
a telescope
a limp
no money
a boy

14. saw
a telescope
a limp
no money
a boy

15. saw
a telescope
a limp
no money
a boy

16. saw
a telescope
a limp
no money
a boy

17. saw
a telescope
a limp
no money
a boy

18. saw
no money
a boy
a telescope
a limp

19. saw
a limp
no money
a boy
a telescope

20. saw
a limp
no money
a boy
a telescope

21. saw
no money
a boy
a telescope
a limp

22. Why 14 parses?
Parses for:
I saw a boy with a telescope with a limp with no money
Two attachment points VP (saw a boy) and NP (a boy) for the three PPs (1. with a telescope, 2. with a limp, 3. with no money)
Four cases:
1. NP (1,2,3) VP
2. NP (1,2) VP (3)
3. NP (1) VP (2,3)
4. NP VP (1,2,3)
Two items to attach: two possibilities
XP XP 1
Total number of parses: = 14
Three items to attach: five possibilities
XP XP XP XP 1 XP 1

23. General recursive formula for # parses
Let dk = configuration at depth k (k is depth of last PP)
Formula:
dk  d dk + dk+1 (k=1,2,3,…)
Start:
d1 (case: one PP, only one place to attach it)
d1+ d2
(d1+ d2 ) + (d1+ d2 + d3 ) = 2 d1+ 2 d2 + 1 d3
2 (d1+ d2 ) + 2 (d1+ d2 + d3 ) + (d1+ d2 + d3 + d4 ) = 5 d1+ 5 d2 + 3 d3 + 1 d4
XP XP XP 1
Depth:
Total for one PP: 1
Total for two PPs: 2
Total for 3 PPs: 5
Total for 4 PPs: 14

24. General recursive formula for # parses
In principle, the formula allows to extrapolate the number of syntactically valid parses to an arbitrary number of PPs in a row…
perhaps of academic interest only…

25. General recursive formula for # parses
Use the formula in conjunction with the possible attachments at the top level:
Case: just one PP phrase, e.g. [PP1 with a telescope]
NP (1) VP 1 d1
NP VP (1) 1 d1 = 2 (total)
Case: two PP phrases, e.g. [PP1with a telescope] [PP2 with a limp]
NP (1,2) VP d1+ d2
NP (1) VP (2) 1 d1 x 1 d1
NP VP (1,2) d1+ d2 = 5 (total)

26. General recursive formula for # parses
Case: three PP phrases, e.g. [PP1with a telescope] [PP2 with a limp] [PP3 with no money]
NP (1,2,3) VP 2 d1+ 2 d2 + d3
NP (1,2) VP (3) (d1+ d2 ) x 1 d1
NP (1) VP (2,3) 1 d1 x (d1+ d2 )
NP VP (1,2,3) 2 d1+ 2 d2 + d3 = 14 (total)

27. General recursive formula for # parses
Let’s see if our formula correctly predicts the number of parses for four PPs:
Case: three PP phrases, e.g. [PP1with a telescope] [PP2 with a limp] [PP3 with no money] ] [PP4 with a smile]
NP (1,2,3,4) VP 5 d1+ 5 d2 + 3 d3 + d4
NP (1,2,3) VP (4) (2 d1+ 2 d2 + d3 ) x 1 d1
NP (1,2) VP (3,4) (d1+ d2 ) x (d1+ d2 )
NP (1) VP (2,3,4) 1 d1 x (2 d1+ 2 d2 + d3 )
NP VP (1,2,3,4) 5 d1+ 5 d2 + 3 d3 + d4= 42 (total)

28. Extended grammar

29. Counting Parses To count the number of parses, run the command: Note:
findall(Parse,s(Parse,[i,saw,a,boy,with,a,telescope,with,a,limp,with,no,money,with,a,smile],[]),List), length(List,Number).
Note:
findall/3 finds all solutions to the s/3 query, each time it finds a solution it puts it into a list (called List here)
we evaluate the length of that List = Number

30. Counting Parses

31. Counting Parses

32. Counting Parses

33. Counting Parses

34. Counting Parses

35. Counting Parses
Number of parses increases exponentially..

36. Homework 5 Extra Credit (optional)
We know there are 5 attachment possibilities for 2 PPs following V NP
Show that all 5 are possible in natural language
i.e. construct sentences where each of the 5 possibilities would be the most likely parse
(you may change the words for each example)
e.g. I saw a boy with a telescope on a heavy tripod