1 / 18

LING 388: Language and Computers

LING 388: Language and Computers. Sandiway Fong Lecture 4: 9/1. Homework 1 given out in the lab class last Wednesday reminder: due tonight ( midnight in my mailbox ) We’ll go through the solution to the homework next time Homework 2 to be given out today

mikkel
Télécharger la présentation

LING 388: Language and Computers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. LING 388: Language and Computers Sandiway Fong Lecture 4: 9/1

  2. Homework 1 given out in the lab class last Wednesday reminder: due tonight (midnight in my mailbox) We’ll go through the solution to the homework next time Homework 2 to be given out today Due by next Wednesday midnight Reminder Next Monday is Labor Day (no class) Administrivia

  3. Last Time • Prolog lists: • comma-separated notation[1,2,3,4,5] • head-tail notation [1,2|[3]] • Recursion: • defining something in terms of itself • but reducing the problem on each iteration • example: • lastinlist/2 finding the last element of a list • Today: • more on recursion...

  4. SWI-Prolog Built-in Predicates • Actually… last/2 is a built-in in SWI-Prolog http://www.swi-prolog.org/pldoc/doc_for?object=section%282%2c%27F.1%27%2cswi%28%27%2fdoc%2fManual%2fpredsummary.html%27%29%29

  5. Definition lastinlist([X],X). (base case) lastinlist([X|L],Y) :- lastinlist(L,Y). (recursive case) Notes: There are two cases We use the head/tail notation in the recursive case we list the base case first and the recursive case last because Prolog accesses its database in first-to-last order Captures intuition Base Case: i.e. simplest case last element of a list containing just one element Recursive Case: we define last element of a nontrivial list in terms of the last element of its tail captures idea that last [1,2,3] = last tail [1,2,3] tail [1,2,3] = [2,3] last [2,3] = last tail [2,3] tail [2,3] = [3] last [3] is 3 (by the case case) so last [1,2,3] is 3 Re-cap of lastinlist/2 Definition

  6. Simple Procedure Given a query, e.g. ?- q(A1,A2). match query against the database in first-to-last order Fact a fact matches the query q(B1,B2). only if the predicate names match (here: q) the number of arguments (arity) match (here: 2) and each corresponding argument matches (here: A1 must match B1, and A2 must match B2) Rule a rule matches the query if the head of the rule matches the query q(C1,C2) :- p(C1), r(C2). if the head of the rule matches the query (like Fact-matching) ?- q(A1,A2). q(C1,C2) :- p(C1), r(C2). the body of the rule creates new sub-queries that must all be true in Prolog’s world for the entire rule to be true ?- p(C1). ?- r(C2). Prolog Computation Summary

  7. Prolog Lists lastinlist([X],X). (base case) lastinlist([X|L],Y) :-lastinlist(L,Y).(recursive case) • Query ?- lastinlist([1,2,3],Z). Z=3 • Computation procedure • ?- lastinlist([1,2,3],Z). • match factlastinlist([X],X). NO • match head of rulelastinlist([X|L],Y) :-lastinlist(L,Y).YES • match names:lastinlist • match arity: 2 • match corresponding 1st arguments: [1,2,3] = [X|L] • X=1 L=[2,3] • match corresponding 2nd arguments: • Y=Z • create new sub-query: • ?- lastinlist([2,3],Y).

  8. Prolog Lists lastinlist([X],X). (base case) lastinlist([X|L],Y) :-lastinlist(L,Y).(recursive case) • Sub-Query ?- lastinlist([2,3],Y). • Computation procedure (2nd time around) • ?- lastinlist([2,3],Y). • match factlastinlist([X],X). NO • match head of rulelastinlist([X’|L’],Y’) :-lastinlist(L’,Y’).YES • match names:lastinlist • match arity: 2 • match corresponding 1st arguments: [2,3] = [X’|L’] • X’=2 L’=[3] • match corresponding 2nd arguments: • Y=Y’ • create new sub-query: • ?- lastinlist([3],Y’).

  9. Prolog Lists lastinlist([X],X). (base case) lastinlist([X|L],Y) :-lastinlist(L,Y).(recursive case) • Sub-Query ?- lastinlist([3],Y’). • Computation procedure (3rd time around) • ?- lastinlist([3],Y’). • match factlastinlist([X”],X”). YES • match names:lastinlist • match arity: 2 • match corresponding 1st arguments: [3] = [X”] • X”=3 • match corresponding 2nd arguments: • X”=Y’ • no new sub-query • We’re done! • X”=3, X”=Y’, Y=Y’, Y=Z • Original query was ?- lastinlist([1,2,3],Z). Z=3 YES

  10. commands ?- trace. switch tracing on [trace] ?- prompt ?- notrace. switch tracing off in trace mode [return] creep (one step forward) trace mode display Call <query> about to evaluate <query> Exit <query> <query> succeeded Redo <query> see if query can be satisfied another way Fail <query> <query> can’t be matched Let’s run through our example using the SWI-Prolog trace mechanism Prolog tracing

  11. Prolog Lists lastinlist([X],X). (base case) lastinlist([X|L],Y) :-lastinlist(L,Y).(recursive case) • Query ?-lastinlist([],Z). No • Computation tree • ?-lastinlist([],Z). (Neither case matches!) • No

  12. Prolog Lists lastinlist([X],X). (base case) lastinlist([X|L],Y) :-lastinlist(L,Y).(recursive case) • What happens to the computation tree for this query? • ?-lastinlist(W,3). W=[3] (base case) W=[X] X=3 • ; • ?- lastinlist(W,3). (recursive case)W=[X|L] Y=3 • ?- lastinlist(L,3). (base case) L=[3] • W = [X,3] • ; • ?- lastinlist(L,3). (recursive case)L=[X’|L’] Y’=3 • ?- lastinlist(L’,3). (base case) L’=[3] • W = [X, X’,3] • ;and so on… all lists with last element = 3 [3] [ _ ,3] [ _ , _ ,3] [ _ , _ , _ ,3] and so on...

  13. Computation tree • Definition • lastinlist([X],X). (base case) • lastinlist([X|L],Y) :-lastinlist(L,Y). (recursive case) • Query • ?-lastinlist([mary,likes,john],Y). View of recursive matching procedure as a computation tree lastinlist([mary,likes,john],Y). lastinlist([X],X). [X] ≠ [mary,likes,john] [X|L] = [mary,likes,john] Y = Y’ Y = john lastinlist([likes,john],Y). lastinlist([X’],X’). [X’] ≠ [likes,john] [X’|L’] = [likes,john] lastinlist([john],Y’). X” = Y’ Y’ = john lastinlist([X”],X”). [X”] = [john] X” = john

  14. Computation tree • Definition • lastinlist([X],X). (base case) • lastinlist([X|L],Y) :-lastinlist(L,Y). (recursive case) Answer: Y = john did we completely explore the search space? NO Ask for more answers (;) what happens? lastinlist([mary,likes,john],Y). lastinlist([X],X). [X] ≠ [mary,likes,john] [X|L] = [mary,likes,john] lastinlist([likes,john],Y). lastinlist([X’],X’). [X’] ≠ [likes,john] [X’|L’] = [likes,john] lastinlist([john],Y’). lastinlist([X”|L”],Y”). lastinlist([X”],X”). [X”] = [john] [X”|L”] = [john] lastinlist([],Y”’). X” = john no match

  15. Length of a List • Usage • ?- len(L,N). • L a list, N the length of L • Definition • len([],0). (base case) • len([X|L],N) :- len(L,M), N is M+1. (recursive case) • Notes • recursive definition (similar in style to lastinlist/2) • Prolog builtin predicate is/2 evaluates arithmetic expressions on the right-hand-side (RHS) • Examples • ?-len([john,[saw,mary]],X). X=2 • ?-len([john,saw,mary], 4). No

  16. Length of a List • Usage • ?- len(L,N). • L a list, N the length of L • Definition • len([],0). (base case) • len([X|L],N) :- len(L,M), N is M+1. (recursive case)

  17. Length of a List • Usage • ?- len(L,N). • L a list, N the length of L • Definition • len([],0). (base case) • len([X|L],N) :- len(L,M), N is M+1. (recursive case) • Example • ?- len(L,3). what happens? First answer: [ _ , _ , _ ] what happens after you hit semicolon ? (try it for yourself in Prolog and see)

  18. Homework 2 • Submit your programs and examples in one file! • Exercise 1 (4pts): • write a recursive definition lastbut1/2 to find the 2nd last element of a list • e.g. ?- lastbut1([a,b,c,d,e],X). X=d • Exercise 2 (6pts): • Write a recursive definition midlist/2 to pick out the middle element of a list • (if there is no middle element, it should fail to return an answer) • e.g. ?-midlist([a,b,c,d,e],X). X=c • ?-midlist([a,b,c,d],X). No

More Related