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

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Presentation on theme: "LING 388: Language and Computers Sandiway Fong 10/15 Lecture 15."— Presentation transcript:

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

2 Adminstrivia Reminder – 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: 1.np(np(DT,NN)) --> dt(DT), nn(NN). 2.np(np(NP,PP)) --> np(NP), pp(PP). 3.vp(vp(VBD,NP)) --> vbd(VBD), np(NP). 4.vp(vp(VP,PP)) --> vp(VP), pp(PP). x(x(X,y)) --> x(X), [y]. x(x(z)) --> [z]. x(x(X,y)) --> x(X), [y]. x(x(z)) --> [z]. [z] [y] x x 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) --> [z], w(X,x(z)). x(x(z)) --> [z]. w(W,X) --> [y], w(W,x(X,y)). w(x(X,y),X) --> [y]. [z] [y] x x

4 Last Time

5 2 nd page:

6 Last Time Parses for: – 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 … 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 1 2 2 Two items to attach: two possibilities XP 1 2 2 Three items to attach: five possibilities XP 1 XP 1XP 1XP1XP 1 2 2 2 22 3 3 333 Three items to attach: five possibilities XP 1 XP 1XP 1XP1XP 1 2 2 2 22 3 3 333 Total number of parses: 5 + 2 + 2 + 5 = 14

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

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) VP1 d 1 NP VP (1)1 d 1 = 2 (total) – Case: two PP phrases, e.g. [ PP1 with a telescope] [ PP2 with a limp] NP (1,2) VPd 1 + d 2 NP (1) VP (2)1 d 1 x 1 d 1 NP VP (1,2)d 1 + d 2 = 5 (total)

26 General recursive formula for # parses – Case: three PP phrases, e.g. [ PP1 with a telescope] [ PP2 with a limp] [ PP3 with no money] NP (1,2,3) VP2 d 1 + 2 d 2 + d 3 NP (1,2) VP (3)(d 1 + d 2 ) x 1 d 1 NP (1) VP (2,3)1 d 1 x (d 1 + d 2 ) NP VP (1,2,3)2 d 1 + 2 d 2 + d 3 = 14 (total)

27 General recursive formula for # parses Lets see if our formula correctly predicts the number of parses for four PPs: – Case: three PP phrases, e.g. [ PP1 with a telescope] [ PP2 with a limp] [ PP3 with no money] ] [ PP4 with a smile] NP (1,2,3,4) VP5 d 1 + 5 d 2 + 3 d 3 + d 4 NP (1,2,3) VP (4)(2 d 1 + 2 d 2 + d 3 ) x 1 d 1 NP (1,2) VP (3,4)(d 1 + d 2 ) x (d 1 + d 2 ) NP (1) VP (2,3,4)1 d 1 x (2 d 1 + 2 d 2 + d 3 ) NP VP (1,2,3,4)5 d 1 + 5 d 2 + 3 d 3 + d 4 = 42 (total)

28 Extended grammar

29 Counting Parses To count the number of parses, run the command: –findall(Parse,s(Parse,[i,saw,a,boy,with,a,telescope,with,a,limp,with,no,money,with,a,s mile],[]),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

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35 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


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