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LING 388: Language and Computers

LING 388: Language and Computers. Sandiway Fong 10/15 Lecture 15. Adminstrivia. Reminder Homework 4 due tonight … Extra credit homework out today d ue Thursday easy way to pick up extra points…. Last Time. Applied the transformation: to NP and VP rules:

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LING 388: Language and Computers

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  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: • 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: • 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 1 XP 1 2 2 Total number of parses: 5 + 2 + 2 + 5 = 14 Three items to attach: five possibilities XP 1 XP 1 XP 1 XP 1 XP 1 2 2 2 2 2 3 3 3 3 3

  23. General recursive formula for # parses • Let dk = configuration at depth k (k is depth of last PP) • Formula: • dk d1 + ..+ 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+ 2d2 + 1 d3 • 2 (d1+ d2 ) + 2 (d1+ d2 + d3 ) + (d1+ d2 + d3 + d4 ) = 5 d1+ 5d2 + 3d3 + 1 d4 XP 1 XP 1 XP 1 2 2 2 3 3 3 Depth: 112 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. [PP1with 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] [PP2with a limp] [PP3with 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] [PP2with a limp] [PP3with no money] ] [PP4with 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: • 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

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