22 Why 14 parses?Parses for:I saw a boy with a telescope with a limp with no moneyTwo 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) VP2. NP (1,2) VP (3)3. NP (1) VP (2,3)4. NP VP (1,2,3)Two items to attach: two possibilitiesXP XP 1Total number of parses: = 14Three items to attach: five possibilitiesXP 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 d32 (d1+ d2 ) + 2 (d1+ d2 + d3 ) + (d1+ d2 + d3 + d4 ) = 5 d1+ 5 d2 + 3 d3 + 1 d4XP XP XP 1Depth:Total for one PP: 1Total for two PPs: 2Total for 3 PPs: 5Total 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 d1NP VP (1) 1 d1 = 2 (total)Case: two PP phrases, e.g. [PP1with a telescope] [PP2 with a limp]NP (1,2) VP d1+ d2NP (1) VP (2) 1 d1 x 1 d1NP 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 + d3NP (1,2) VP (3) (d1+ d2 ) x 1 d1NP (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 + d4NP (1,2,3) VP (4) (2 d1+ 2 d2 + d3 ) x 1 d1NP (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)
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
35 Counting ParsesNumber of parses increases exponentially..
36 Homework 5 Extra Credit (optional) We know there are 5 attachment possibilities for 2 PPs following V NPShow that all 5 are possible in natural languagei.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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