LING 388 Language and Computers Lecture 4 9/11/03 Sandiway FONG

Administrivia Homework 1 Homework 1  account for submissions   Need help with Homework 1?  Ask Charles after class today  I’ll also be in Douglass 309 this afternoon  Homework hint  Number of inference steps?  Count the number of calls in the debugger

Administrivia Question from Tuesday’s computer lab class Question from Tuesday’s computer lab class  How to make SWI-Prolog print the entire list?  Answer:  ?- set_prolog_flag(toplevel_print_options,[max_depth(0)]).

Strings and Languages String: String:  sequence of characters, possibly empty  constructor operation: concatenation  Examples: a aaabc  (empty string) Alphabet: Alphabet:  set of possible characters  Examples: {a,b} {0,1,2,3,4,5,6,7,8,9} Language: Language:  set of strings defined over an alphabet  Examples: {,a,aa,aaa,aaaa} {}

Regular Expressions Regular Expressions Regular Expressions shorthand for describing sets of strings  Examples:  string + - set of one or more occurrences of string a + = {a, aa, aaa, aaaa, aaaaa, …}a + = {a, aa, aaa, aaaa, aaaaa, …} (abc) + = {abc, abcabc, abcabcabc, …}(abc) + = {abc, abcabc, abcabcabc, …}  string * - set of zero or more occurrences of string a * = {, a, aa, aaa, aaaa, …}a * = {, a, aa, aaa, aaaa, …} (abc) * = {, abc, abcabc, …}(abc) * = {, abc, abcabc, …}  Note: a a * = a +a a * = a + a {, a, aa, aaa, aaaa, …}a {, a, aa, aaa, aaaa, …} = {a, aa, aaa, aaaa, aaaaa, …}

Regular Expressions Regular Expressions Regular Expressions  Examples: contd.  string n - exactly n occurrences of string a 4 b 3 = { aaaabbb }a 4 b 3 = { aaaabbb }  (string 1 | string 2 ) - either string 1 or string 2 (a|b) = {a,b}(a|b|c) = {a, b, c}(a|b) = {a,b}(a|b|c) = {a, b, c}

Regular Expressions Regular Expressions contd. Regular Expressions contd.  can’t describe all possible languages  formally equivalent to regular grammars/finite state automata How to show this?How to show this? –Proof by construction… as we’ll see  How to show this?  Examples: {a n b n | n>=0}is not regular{a n b n | n>=0}is not regular {ww R | w  {a,b} + } is not regular{ww R | w  {a,b} + } is not regular –R = reverse, e.g. abc R = cba Proof by Pumping LemmaProof by Pumping Lemma

Regular Expressions Regular Expressions contd. Regular Expressions contd.  popularized by  string matching utilities - grep in Unix  Perl  typically augmented with  meta-characters denoting classes of characters e.g. [:digit:] = [0-9],. = any single charactere.g. [:digit:] = [0-9],. = any single character  special operators negation: e.g. [^0-9]negation: e.g. [^0-9] optionality: e.g. 1?optionality: e.g. 1? for dealing with lines: e.g. ^ (beginning) $ (end)for dealing with lines: e.g. ^ (beginning) $ (end)

Finite State Automata (FSA) Example: Example:  Language: L = {a + b + } “one or more a’s followed by one or more b’s”  regular language described by a regular expressiondescribed by a regular expression  Note: infinite set of strings belonging to language Linfinite set of strings belonging to language L –e.g. abbb, aaaab, aabb, *abab, * –e.g. abbb, aaaab, aabb, *abab, *

Finite State Automata (FSA) sx y a a b b L = {a + b + }

Finite State Automata (FSA) FSA shown on previous slide: FSA shown on previous slide:  acceptor - no output  cf. transducer - input/output pairs  deterministic- no ambiguity i.e. at any given state and input character, the next state is uniquely determined  cf. non-deterministic FSA (NDFSA)  Note: NDFSA are not more powerful than FSANDFSA are not more powerful than FSA Proof: by constructionProof: by construction

Finite State Automata (FSA) More formally: More formally:  Set of states: {s,x,y}  Start state: s, end state: y  Alphabet: {a, b}  Transition function  : signature: character x state -> state   (a,s)=x   (a,x)=x   (b,x)=y   (b,y)=y

Finite State Automata (FSA) A possible Prolog encoding strategy: A possible Prolog encoding strategy:  Define one predicate for each state  taking one argument (the input string)  consume input character  call next state with remaining input string  Initially:  fsa(L) :- s(L). call start state s

Finite State Automata (FSA)  State s: (start state)  s([a|L]) :- x(L). match input string beginning with a and call state x with remainder of input  State x:  x([a|L]) :- x(L).  x([b|L]) :- y(L).  State y: (end state)  y([]).  y([b|L]) :- y(L).

Finite State Automata (FSA)  Example: 1. ?- fsa([a,a,b]). 2. s([a,a,b). 3. x([a,b]). 4. x([b]). 5. y([]). succeeds

Finite State Automata (FSA)  Example: 1. ?- fsa([a,b,a]). 2. s([a,b,a]). 3. x([b,a]). 4. y([a]). fails

Finite State Automata (FSA) Another possible Prolog encoding strategy: Another possible Prolog encoding strategy:  fsa(S,L) :- L = [C|M], transition(S,C,T),fsa(T,M).  fsa(y,[]).% End state  transition(s,a,x).% Encodes transition  transition(x,a,x).% function   transition(x,b,y).  transition(y,b,y).

Finite State Automata (FSA)  Example: 1. ?- fsa(s,[a,a,b]). 2. transition(s,a,T). T=x 3. fsa(x,[a,b]). 4. transition(x,a,T’). T’=x 5. fsa(x,[b]). 6. transition(x,b,T”). T”=y 7. fsa(y,[]). succeeds