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**LING 388: Language and Computers**

Sandiway Fong 10/15 Lecture 15

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**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…

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**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

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Last Time

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Last Time 2nd page:

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**Last Time Parses for: Stanford Parser:**

I saw a boy with a telescope with a limp with no money Stanford Parser:

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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 …

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saw a boy a telescope a limp no money

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saw a boy a telescope a limp no money

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saw a boy a telescope a limp no money

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saw a boy a telescope a limp no money

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saw a boy a telescope a limp no money

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saw a telescope a limp no money a boy

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saw a telescope a limp no money a boy

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saw a telescope a limp no money a boy

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saw a telescope a limp no money a boy

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saw a telescope a limp no money a boy

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saw no money a boy a telescope a limp

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saw a limp no money a boy a telescope

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saw a limp no money a boy a telescope

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saw no money a boy a telescope a limp

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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

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**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

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**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…

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**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)

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**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)

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**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)

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Extended grammar

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**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

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Counting Parses

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Counting Parses

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Counting Parses

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Counting Parses

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Counting Parses

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Counting Parses Number of parses increases exponentially..

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**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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