LING 438/538 Computational Linguistics Sandiway Fong Lecture 16: 10/19

Administrivia review homework #3 new homework #4 –out today –usual rules apply - due next Thursday

Last Time Spelling errors and correction Error Correction –correct Bayesian Probability –Minimum Edit Distance Computation Dynamic Programming

Minimum Edit Distance example –assuming insert =1 delete=1 substitution=2 (or 0 for substituting the same character) recursive formula –incrementally computed from minimum edit distances of shorter strings intent execut intent execu inten execut inten execu one edit operation away L DB min(L+1,D+0,B+1) cost: =8

Minimum Edit Distance Computation one formula Microsoft Excel implementation $ in a cell reference means don’t change when copied from cell to cell e.g. in C$1, 1 stays the same in $A3,A stays the same (not 3) min(C2+1,B3+1,B2+if(C$1=$A3,0,2))min(D2+1,C3+1,C2+if(D$1=$A3,0,2)) min(C3+1,B4+1,B3+if(C$1=$A4,0,2)) inc col inc row row columnprotected

Minimum Edit Distance Computation demo example pairs –intention, intent: –intention, intentional: –intention, ten: –intention, ton: –intention, teen: min edit distance (assuming substitution cost 2)

Homework 3 Review

Question 1 438/538 (4pts) Give the minimum size regular expression for the FSA below (2pt) Minimum size regular expression for the FSA: –a + b* not minimum size in terms of number of symbols: –aa*b* –(aa*)|(aa*b*) s xy a a b ε

Question 1 438/538 (4pts) Give an equivalent FSA without the ε-transition (2pts) –answer in the form of a diagram or formal definition or Prolog definition are all ok Equivalent ε-free FSA s xy a a b ε sab ab ab How to arrive at this answer? by inspection or by consideration of a + b* b* = ε | b + sa a a sb b b

Question 1 438/538 (4pts) Give an equivalent FSA without the ε-transition (2pts) –answer in the form of a diagram or formal definition or Prolog definition are all ok Set-of-States Construction method: s xy a a b ε {s}{x,y}{y} ab aba sab ab ab

Question 2 438/538 (8pts) convert the NDFSA into a deterministic FSA (3pts) figure 2.27 in the textbook {1} a {2} b {3,4} a {2,3} b a {1} a {2} b {3,4} a {2,3} b a set-of-states construction:

Question 2 438/538 (8pts) implement both the NDFSA and the equivalent FSA in Prolog using the “one predicate per state” encoding Prolog code: one([a|L]) :- two(L). two([b|L]) :- three(L). two([b|L]) :- four(L). three([]). three([a|L]) :- two(L). four([a|L]) :- three(L). strings abab and abaaba, how many steps (transitions + final stop)?

Question 2 438/538 (8pts) implement both the NDFSA and the equivalent FSA in Prolog using the “one predicate per state” encoding Prolog code: s1([a|L]) :- s2(L). s2([b|L]) :- s34(L). s34([]). s34([a|L]) :- s23(L). s23([]). s23([b|L]) :- s34(L). s23([a|L]) :- s2(L). {1} a {2} b {3,4} a {2,3} b a strings abab and abaaba, how many steps (transitions + final stop)?

Question 3 438/538 (8pts) (5pts) Give a FSA in Prolog that accepts a binary string (made up of 0’s and 1’s) if and only if it begins with a 1 and contains exactly one 0 –examples: – –10 –* FSA:

Question 3 438/538 (8pts) (5pts) Give a FSA in Prolog that accepts a binary string (made up of 0’s and 1’s) if and only if it begins with a 1 and contains exactly one 0 (3pts) Given the regular expression equivalent of the FSA Regular Expression: –11*01*

Homework #4

Question 1 438/538 (8pts) Implement the e-insertion rule (Context-Sensitive) Spelling Rule: (3.5) –   e / { x, s, z } ^ __ s# –as a FST in Prolog Goals: –pass through non-matching cases unchanged –implement rule exactly –no deletion of boundaries ^ and #

Question 2 438/538 (6pts) What does the Porter Stemmer output for the following words: –(2 pts) availability –(2 pts) shipping –(2pts) unbelievable Show the steps (stages) in your answer

Question 2 438/538 (6pts) –the Porter Stemmer handles -ement for cases like replacement  replac(e) –it doesn’t handle statement  stat(e) i.e. it outputs statement –Why? Explain (2pts) –Modify the Porter rule responsible to allow for statement  stat(e) Submit your rule (2pts) Give 2 examples where the modified rule would be too liberal, i.e. it overstems (2pts)

Summary Q1: 8pts Q2: 6+6=12pts Total: 20 pts