1. LING 388 Language and Computers Lecture 9 9/30/03 Sandiway FONG

2. Administrivia Reminder Reminder  Homework 2 due this Thursday  No late submissions please  Need help?  Talk to Charles or me after the class to set up an appointment  We’re here to help you understand the concepts

3. Recap Regular Grammars FSA Regular Expressions DCG

4. Recap Extensions to regular grammars Extensions to regular grammars 1. Use left and right recursive rules 2. Use multiple terminal/non-terminals on the right (Not a problem for DCG) => Last time, a DCG for the non-regular (context-free) language L n ={a n b n | n>=0} using 1. Now, an alternative DCG for L n using 2.

5. Multiple Non-Terminals/Terminals on the Right Free from the single non-terminal/terminal symbol restriction imposed by regular grammars… Free from the single non-terminal/terminal symbol restriction imposed by regular grammars…  we can encode L n ={a n b n | n>=0 } using just 2 context-free rules as follows:  S -> s --> [].  S -> aSbs --> [a],s,[b].

6. Multiple Non-Terminals/Terminals on the Right Reminder: Reminder:  This rather simple-looking grammar cannot be encoded with a FSA  Requires “working memory” unavailable to the FSA  Intuitively, given … s --> [a],s,[b].s --> [a],s,[b]. recall the regular grammar -> FSA construction idea S a How to keep track of the bs as we read the as?

7. Multiple Non-Terminals/Terminals on the Right However, G n can be simulated using a Push-Down Automata (PDA) However, G n can be simulated using a Push-Down Automata (PDA)  i.e. a FSA plus a stack bb…bbb…b S a / push b B b / pop b Stack

8. Multiple Non-Terminals/Terminals on the Right DCG rule mapping to Prolog clauses: DCG rule mapping to Prolog clauses:  s --> [].  s(L,L).  s --> [a],s,[b].  s(L1,L4) :- ‘C’(L1,a,L2), s(L2,L3), ‘C’(L3,b,L4). L1 L2 L3 L4 Prolog query ?- s(X,[]).

9. Multiple Non-Terminals/Terminals on the Right Informally, we have the following Prolog sentential forms… Informally, we have the following Prolog sentential forms…  ?- s([a,a,a,b,b,b],[]).  a s([a,a,b,b,b],X) b a a s([a,b,b,b],X’) b ba a s([a,b,b,b],X’) b b –a a a s([b,b,b],X”) b b b –a a a s([b,b,b],[b,b,b]) b b b –Note: [b,b,b] - [b,b,b] = [] a a s([a,b,b,b],[b,b]) b ba a s([a,b,b,b],[b,b]) b b Note: [a,b,b,b] – [b,b] = [a,b]Note: [a,b,b,b] – [b,b] = [a,b]  a s([a,a,b,b,b],[b]) b  Note: [a,a,b,b,b] – [b] = [a,a,b,b]

10. So far… We have encoded L n = { a n b n | n >= 0 } using two different DCGs containing: We have encoded L n = { a n b n | n >= 0 } using two different DCGs containing: 1. Left and right recursive rules 2. Three symbols on the right  Now let’s examine one more technique:  Modifying regular grammar G ab where L(G ab ) = { a + b + } without changing the rule format  i.e. without using either 1. or 2.

11. More on the capabilities of DCG rules To do this, we’ll need two more things … To do this, we’ll need two more things …  The ability to call Prolog predicates from DCG rules, and  The capability for non-terminals to be more than just simple names  general structures like: a(X) b(X,Y)a(X) b(X,Y)

12. Calling Prolog Predicates We can insert calls to any Prolog predicate, either built-in or user- defined, anywhere using the notation { … } We can insert calls to any Prolog predicate, either built-in or user- defined, anywhere using the notation { … } Example: Example:  s --> [a], { playChess }, [b], { playGo }.  Perhaps a slightly extreme example … playChess/0 and playGo/0 are not SWI-Prolog built-insplayChess/0 and playGo/0 are not SWI-Prolog built-ins  but the point is that this capability gives the DCG formalism the expressive power of Turing Machines (Turing, 1949) aka “general computing power”  Somewhat more down-to-earth and relevant for L n …  we can use { … } to do arithmetic

13. Non-terminals as structures In general, non-terminals need not be limited to names only In general, non-terminals need not be limited to names only Some instances where we’ll find non-terminals as structures useful: Some instances where we’ll find non-terminals as structures useful:  Subject/Inflection agreement  John likes Mary[ IP [ NP John ][ I’ [ Infl ] …]]  I like Mary[ IP [ NP I ][ I’ [ Infl ] …]]  John is tired  We are tired  IP -> NP(A) I’(A) A a variable controlling agreement between the subject and inflectionA a variable controlling agreement between the subject and inflection

14. Non-terminals as structures In Prolog In Prolog  IP -> NP(A) I’(A) =>  ip --> np(A), ibar(A). A a variable controlling agreement between the subject and inflectionA a variable controlling agreement between the subject and inflection  Simple implementation: np(sg) --> [john]. np(pl) --> [we].np(sg) --> [john]. np(pl) --> [we]. propagates subject number (sg/pl) into non- terminal ibarpropagates subject number (sg/pl) into non- terminal ibar

15. Non-terminals as structures Another use, to build parse trees as Prolog structures… Another use, to build parse trees as Prolog structures…  s(s(NP,VP)) --> np(NP), vp(VP).  np(np(N)) --> pronoun(N).  np(np(det(the),n(ball))) --> [the,ball].  pronoun(i) --> [i].pronoun(we) --> [we].  vp(vp(V)) --> unergative(V).  vp(vp(V,NP)) --> transitive(V), np(NP).  unergative(v(ran)) --> [ran].  transitive(v(hit)) --> [hit]. Prolog query Prolog query  ?- s(X,[i,hit,the,ball],[]).  Note: extra parameter precedes difference list Answer Answer  X = s(np(i),vp(v(hit),np(det(the),n(ball))))

16. From a + b + to a n b n Define G ab such that L(G ab ) = a + b + Define G ab such that L(G ab ) = a + b + We have seen this grammar before as: We have seen this grammar before as:  s --> [a], b.  b --> [a], b.  b --> [b], c.  b --> [b].  c --> [b], c.  c --> [b].

17. From a + b + to a n b n Idea: Idea:  Add a parameter as a counter to the non-terminals to count the number of as and bs  Input stringCounter Value  a1  aa2  aaa3  aaab 2  aaabb 1  aaabbb 0  So:  every time we see an a, we add 1 to the counter  every time we see a b, we subtract 1 from the counter  This way we can be sure the numbers of as and bs match

18. From a + b + to a n b n Idea: Idea:  Add a parameter to the non-terminals to count the number of as and bs  s --> [a], b(1).1 means we have seen one a  b(N) --> [a], b(M).where M is N+1  b(N) --> [b], c(M).where M is N-1  b(N) --> [b].N must be 1  c(N) --> [b], c(M).where M is N-1  c(N) --> [b].N must be 1

19. From a + b + to a n b n DCG rules: DCG rules:  s --> [a], b.s --> [a], b(1).  b --> [a], b.b(N) --> [a], {M is N+1}, b(M).  b --> [b], c.b(N) --> [b], {M is N-1}, c(M).  b --> [b].b(1) --> [b].  c --> [b], c.c(N) --> [b], {M is N-1}, c(M).  c --> [b].c(1) --> [b].

20. Context-Free Grammars Upgrading from regular to context-free grammars: Upgrading from regular to context-free grammars:  By removing all restrictions from the right side of the rule  X ->   { V N u V T } *  How powerful are context-free grammars?

21. Context-Free Grammars The Expressive Power of Formal Grammars The Expressive Power of Formal Grammars Chomsky Hierarchy: Chomsky Hierarchy:  Type-0 General rewrite rules  Type-1 Context-sensitive rules  a n b n c n Type-2 Context-free rulesType-2 Context-free rules a n b n a n b n –Type-3 Regular grammar rules – a + b +

22. DCG and Chomsky Hierarchy Regular Grammars = Type-3 FSA Regular Expressions DCG = Type-0 Type-2 Type-1

23. DCG is more powerful than Context-Free Next time… Next time…  We can write a DCG for the context- sensitive language L abc = { a n b n c n | n >= 0 }  But: No context-free grammar can encode L abcNo context-free grammar can encode L abc