LING 388: Language and Computers Sandiway Fong 10/15 Lecture 15.
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Transcript of LING 388: Language and Computers Sandiway Fong 10/15 Lecture 15.
LING 388: Language and Computers
Sandiway Fong10/15
Lecture 15
Adminstrivia
• Reminder– Homework 4 due tonight…
• Extra credit homework out today– due Thursday– easy way to pick up extra points…
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].
[z]
[y]xx
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]xx
Last Time
Last Time
• 2nd page:
Last Time
• Parses for:– I saw a boy with a telescope with a limp with no money
• Stanford Parser:
Last Time
• Parses for:– I saw a boy with a telescope with a limp with no money
3 PPswith a telescopewith a limpwith no money
14 parses …
sawa boy
a telescopea limpno money
sawa boy
a telescopea limp
no money
sawa boy
a telescopea limp
no money
sawa boy
a telescopea limpno money
sawa boy
a telescopea limp
no money
sawa telescopea limpno money
a boy
sawa telescopea limp
no moneya boy
sawa telescope
a limpno money
a boy
sawa telescope
a limpno money
a boy
sawa telescope
a limpno money
a boy
sawno money
a boya telescopea limp
sawa limpno money
a boya telescope
sawa limp
no moneya boy
a telescope
sawno money
a boya telescope
a limp
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 possibilitiesXP 1 XP 1 2 2
Three items to attach: five possibilitiesXP 1 XP 1 XP 1 XP 1
XP 1 2 2 2
2 2 3 3 3 3
3
Total number of parses: 5 + 2 + 2 + 5 = 14
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+ 2 d2 + 1 d3
– 2 (d1+ d2 ) + 2 (d1+ d2 + d3 ) + (d1+ d2 + d3 + d4 ) = 5 d1+ 5 d2 + 3 d3 + 1 d4
Total for one PP: 1
Total for two PPs: 2Total for 3 PPs: 5
Total for 4 PPs: 14
XP 1 XP 1 XP 1
2 2 2 3 3 3Depth: 1 1
2
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…
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)
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)
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)
Extended grammar
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
Counting Parses
Counting Parses
Counting Parses
Counting Parses
Counting Parses
Counting Parses
Number of parses increases exponentially..
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