n Bottom-up (LR) parsing n Characteristic Finite State Machine (CFSM) n SLR(1) parsing table n Conflicts in SLR(1) n LR parsing variants

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1 Aoucemets Gradig HW1 Should be doe by the ed of today. Grades are set to release toight You have 1 week to issue re-grade request Raibow grades comig up over the weeked Last Class Bottom-up (LR) parsig Characteristic Fiite State Machie (CFSM) SLR(1) parsig table Coflicts i SLR(1) LR parsig variats Dowload SWI Prolog! HW3 is posted, due o Sept 28 th Logic Programmig Cocepts Laguage costructs: facts, queries, rules Idividual assigmet Fall 18 CSCI 4430, A Milaova 1 Fall 18 CSCI 4430, A Milaova 2 Today s Lecture Outlie Laguage costructs: facts, rules, queries Search tree, uificatio, backtrackig, backward chaiig Lists Arithmetic Logic Programmig ad Prolog Read: Scott, Chapter 12 Fall 18 CSCI 4430, A Milaova 3 4 Prolog Dowload ad istall SWI Prolog o laptop! Write your Prolog program ad save i.pl file, e.g., family.pl Ru swipl (Prolog iterpreter) o commad lie Load your file:? [family]. Issue query at prompt:? female(c). J.R.Fisher s Prolog Tutorial: Logic Programmig Logic programmig Logic programmig is declarative programmig Logic program states what (logic), ot how (cotrol) Programmer declares axioms I, facts ad rules Programmer states a theorem, or a goal (the what) I, a query Laguage implemetatio determies how to use the axioms to prove the goal Fall 18 CSCI 4430, A Milaova 5 Fall 18 CSCI 4430, A Milaova 6 1

2 Logic Programmig Logic programmig style is characterized by Database of facts ad rules that represet logical relatios. Computatio is modeled as search (queries) over this database Maipulatio of lists ad use of recursio, i which it is very similar to the fuctioal programmig style Logic Programmig Cocepts A Hor Clause is: H B 1,B 2,,B Atecedets (B s): cojuctio of zero or more terms i predicate calculus; this is the body of the hor clause Cosequet (H): a term i predicate calculus Resolutio priciple: if two Hor clauses A B 1,B 2,B 3,, B m C D 1,D 2,D 3,, D are such that A matches D 1, the we ca replace D 1 with B 1,B 2,B 3,, B m C B 1,B 2,B 3,, B m,d 2,D 3,, D Fall 18 CSCI 4430, A Milaova 7 Fall 18 CSCI 4430, A Milaova 8 Hor Clauses i I, a Hor clause is writte h :- b 1,...,b. Hor Clause is called clause Cosequet is called goal or head Atecedets are called subgoals or tail Hor Clause with o tail is a fact E.g., raiy(seattle). Depeds o o other coditios Hor Clause with a tail is a rule sowy(x) :- raiy(x),cold(x). Fall 18 CSCI 4430, A Milaova 9 Hor Clauses i Clause is composed of terms Costats Number, e.g., 123, etc. Atoms e.g., seattle, rochester, raiy, foo I, atoms begi with a lower-case letter! Variables X, Foo, My_var, Seattle, Rochester, etc. I, variables must begi with upper-case letter! Structures cosists of a atom, called a fuctor ad a list of argumets raiy(seattle), sowy(x) 10 A Prolog Program raiy(seattle). raiy(rochester). sowy(x) :- raiy(x),cold(x).?- [sowy].?- raiy(c).?- sowy(c). Today s Lecture Outlie Laguage costructs: facts, rules, queries Search tree, uificatio, backtrackig, backward chaiig Lists Arithmetic Fall 18 CSCI 4430, A Milaova 11 Fall 18 CSCI 4430, A Milaova 12 2

3 Logical Sematics program cosists of facts ad rules raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x). Rules like sowy(x):- raiy(x),cold(x). correspod to logical formulas X[sowy(X) raily(x) ^ cold(x)] /* For every X, X is sowy, if X is raiy ad X is cold */ Fall 18 CSCI 4430, A Milaova 13 Logical Sematics A query such as?- raiy(c). triggers resolutio. Logical sematics does ot impose restrictio i the order of applicatio of resolutio rules C = seattle C = rochester Fall 18 CSCI 4430, A Milaova raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x). C = rochester C = seattle 14 Procedural Sematics?- sowy(c). raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x). sowy(x) :- raiy(x),cold(x). First, fid the first clause i the database whose head matches the query, i our case this is clause sowy(x) :- raiy(x),cold(x) The, fid a bidig for X to make raiy(x) true; the, check if cold(x) is true with that bidig If yes, report bidig as successful Otherwise, backtrack to the bidig of X, ubid ad cosider the ext bidig s computatio is well-defied procedurally by search tree, rule orderig, uificatio ad backtrackig Questio raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x). sowy(troy). What does this query yield??- sowy(c). Fall 18 CSCI 4430, A Milaova 15 Fall 18 CSCI 4430, A Milaova 16 Procedural Sematics raiy(seattle). raiy(rochester). sowy(x) :- raiy(x),cold(x). raiy(x) X = seattle OR sowy(c) _C = _X success sowy(x) AND X = rochester cold(seattle) fails; backtrack. cold(x) raiy(seattle) raiy(rochester) Prolog Cocepts: Search Tree OR levels: paret: goal (e.g., raiy(x)) childre: heads-of-clauses (raiy( )) ORDER: from left to right AND levels: paret: goal (e.g., sowy(x)) childre: subgoals (raiy(x), cold(x)) ORDER: from left to right raiy(x) OR sowy(c) sowy(x) AND raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x).?- sowy(c). cold(x) raiy(seattle) raiy(rochester) Fall 18 CSCI 4430, A Milaova

4 Prolog Cocepts: Uificatio Prolog Cocepts: Uificatio At OR levels Prolog performs uificatio Uifies paret (goal), with child (head-of-clause) E.g., sowy(c) = sowy(x) success, _C = _X raiy(x) = raiy(seattle) success, X = seattle parets(alice,m,f) = parets(edward,victoria,albert) fail parets(alice,m,f) = parets(alice,victoria,albert) success, M = victoria, F = albert I, = deotes uificatio, ot assigmet! A costat uifies oly with itself E.g., alice=alice, but alice=edward fails Two structures uify if ad oly if (i) they have the same fuctor, (ii) they have the same umber of argumets, ad (iii) their argumets uify recursively E.g., raiy(x) = raiy(seattle) A variable uifies with aythig. If the other thig has a value, the variable is boud to that value. If the other thig is a uboud variable, the the two variables are associated ad if either oe gets a value, both do Prolog Cocepts: Backtrackig If at some poit, a goal fails, Prolog backtracks to the last goal (i.e., last uificatio poit) where there is a utried bidig, udoes curret bidig ad tries ew bidig (a alterative OR brach), etc. raiy(x) X = seattle OR?- sowy(c). sowy(c) _C = _X sowy(x) AND raiy(seattle). raiy(rochester). sowy(x):-raiy(x),cold(x). cold(seattle) fails; backtrack. cold(x) raiy(seattle) raiy(rochester) Prolog Cocepts: Backward Chaiig Backward chaiig: starts from goal, towards facts? sowy(rochester). sowy(rochester):- raiy(rochester), raiy(rochester) sowy(rochester): sowy(rochester). Forward chaiig: starts from facts towards goal? sowy(rochester). raiy(rochester) sowy(rochester):- raiy(rochester), sowy(rochester): sowy(rochester). Fall 18 CSCI 4430, A Milaova 21 Fall 18 CSCI 4430, A Milaova 22 Exercise takes(jae, his). takes(jae, cs). takes(ajit, art). takes(ajit, cs). classmates(x,y):-takes(x,z),takes(y,z).?- classmates(jae,c). Draw search tree for query. What are the bidigs for C? Fall 18 CSCI 4430, A Milaova 23 Ruig Prolog Programs Eter database of facts ad rules ito a file. E.g., classmates.pl At the iterpreter prompt, load the file the ru queries. E.g.,?- [classmates]. true.?- takes(ajit,c). C = art ; C = cs. 24 4

5 Today s Lecture Outlie Laguage costructs: facts, rules, queries Search tree, uificatio, backtrackig, backward chaiig Lists Arithmetic Fall 18 CSCI 4430, A Milaova 25 Lists list head tail [a,b,c] a [b,c] [X,[cat],Y] X [[cat],y] [a,[b,c],d] a [[b,c],d] [X Y] X Y Fall 18 CSCI 4430, A Milaova a b a c b c [ ] [ ] [ ] d 26 Lists: Uificatio Questio [ H1 T1 ] = [ H2 T2 ] Head H1 uifies with H2, possibly recursively Tail T1 uifies with T2, possibly recursively E.g., [ a [b, c] ] = [ X Y ] X = a Y = [b, c] NOTE: I, = deotes uificatio, ot assigmet! [X,Y,Z] = [joh, likes, fish] X = joh, Y = likes, Z = fish [cat] = [X Y] X = cat, Y = [ ] [[the, Y] Z] = [[X, hare] [is,here]] X = the, Y = hare, Z = [is, here] Fall 18 CSCI 4430, A Milaova 27 Fall 18 CSCI 4430, A Milaova 28 Lists: Uificatio Lists Uificatio Sequece of comma separated terms, or [ first term rest_of_list ] [ [the Y] Z] = [ [X, hare] [is, here] ] look at the trees to see how this works! [ a, b, c ] = [ X Y ] X = a, Y = [b,c]. the Y Z X hare [ ] is here [ ] [a Z ] =? [ X Y ] X = a, Y = Z. Fall 18 CSCI 4430, A Milaova 29 Fall 18 CSCI 4430, A Milaova 30 5

6 Improper ad Proper Lists [1 2] versus [1, 2] Questio. Ca we uify these lists? [abc, Y] =? [ abc Y ] [ ] abc Y [ ] abc Y Aswer: No. There is o value bidig for Y that makes these two trees isomorphic Fall 18 CSCI 4430, A Milaova Member_of Procedure?- member(a,[a,b]). true.?- member(a,[b,c]). false.?- member(x,[a,b,c]). X = a ; 1. member(a, [A B]). X = b ; 2. member(a, [B C]) :- member (A, C). X = c.?- member(a,[b,c,x]). X = a ; false. Fall 18 CSCI 4430, A Milaova 33 Prolog Search Tree (OR levels oly) member(x,[a,b,c]) A=X=a,B=[b,c] X=a success A =X=b,B =[c] X=b success A=X,B=a,C=[b,c] member(x,[b,c]) A =X,B =b,c =[c] member(x,[c]) A =X=c,B =[ ] A =X B =c, C =[ ] X=c member(x,[ ]) success fail 1. member(a, [A B]). 2. member(a, [B C]) :- member (A, C). fail 34 Member_of Procedure member(a, [A B]). member(a, [B C]) :- member(a,c). logical sematics: For every value assigmet of A, B ad C, we have member(a,[b C]) if member(a,c); procedural sematics: Head of clause is procedure etry. Procedure body cosists of calls withi this procedure. Procedural Iterpretatio member(a, [A B]). member(a, [B C]) :- member(a,c). member is a recursive procedure member(a, [A B]). is the base case. Procedure exits with true if the elemet we are lookig for, A, is the first elemet i the list. It exits with false if we have reached the ed of the list member(a, [B C]) :- member(a,c). is the recursive case. If elemet A is ot the first elemet i the list, call member recursively with argumets A ad tail C Fall 18 CSCI 4430, A Milaova 35 Fall 18 CSCI 4430, A Milaova 36 6

7 Questio Questio 1. member(a, [A B]). 2. member(a, [B C]) :- member(a, C). 1. member(a, [A B]). 2. member(a, [B C]) :- member(a, C). Give all aswers to the followig query:?- member(a,[b, a, X]). Give all aswers to the followig query:?- member(a, [b a]). Aswer: Fall 18 CSCI 4430, A Milaova 37 Fall 18 CSCI 4430, A Milaova 38 Apped Procedure apped([ ], A, A). apped([a B], C, [A D]) :- apped(b,c,d). Build a list?- apped([a],[b],y). Y = [ a,b ]?- apped([a,b,c],[d,e],y). Y = [ a,b,c,d,e ] More Apped apped([ ], A, A). apped([a B], C, [A D]) :- apped(b,c,d). Break a list ito costituet parts?- apped(x,[b],[a,b]). X = [ a ]?- apped([a],y,[a,b]). Y = [ b ] Fall 18 CSCI 4430, A Milaova 39 Fall 18 CSCI 4430, A Milaova 40 More Apped More Apped? - apped(x,y,[a,b]). Geeratig a ubouded umber of lists?- apped(x,[b],y). X = [ ] Y = [b] ; X = [ _G604] Y = [ _G604, b] ; X = [ _G604, _G610] Y = [ _G604, _G610, b] ; Etc. Be careful whe usig apped with 2 ubouded argumets!!! Fall 18 CSCI 4430, A Milaova 41 Fall 18 CSCI 4430, A Milaova 42 7

8 Questio What does this procedure do: p([],[]). p([a B],[[A] Rest]) :- p(b,rest). Puts brackets aroud each elemet i the list?- p([a,b,c],y). Y = [ [a],[b],[c] ]. It ca also flatte a list:?- p(x,[[a],[b],[c]]). X = [ a,b,c ]. Fall 18 CSCI 4430, A Milaova 43 Commo Structure Processig a list: proc([],[]). proc([h T],[H1 T1]) :- f(h,h1),proc(t,t1). Base case: we have reached the ed of list. I our case, the result for [ ] is [ ]. Recursive case: result is [H1 T1]. H1 was obtaied by callig f(h,h1) --- processes elemet H ito result H1. T1 is the result of recursive call of proc o T. Fall 18 CSCI 4430, A Milaova 44 Today s Lecture Outlie Arithmetic Laguage costructs: facts, rules, queries Search tree, uificatio, backtrackig, backward chaiig Lists Arithmetic Fall 18 CSCI 4430, A Milaova 45 has all arithmetic operators Built-i predicate is is(x, 1+3) or more commoly we write X is 1+3 is forces evaluatio of 1+3:?- X is 1+3 X = 4 = is uificatio ot assigmet!?- X = 4-1 X = 4-1 % uifies X with 4-1!!! Fall 18 CSCI 4430, A Milaova 46 Arithmetic: Commo Pitfalls is is ot ivertible! That is, argumets o the right caot be uboud! 3 is 3 X. ERROR: is/2: Argumets are ot sufficietly istatiated This does t work either:?- X is 4, X = X+1. false. Why? What is goig o here? Exercise Write sum, which takes a list of itegers ad computes the sum of the itegers. E.g., sum([1,2,3],r).?- R = 6. How about if the itegers are arbitrarily ested? E.g., sum([[1],[[[2]],3]],r).?- R = 6. Fall 18 CSCI 4430, A Milaova 47 Fall 18 CSCI 4430, A Milaova 48 8

9 Exercise Write plus10, which takes a list of itegers ad computes aother list, where all itegers are shifted +10. E.g., plus10([1,2,3],r).?- R = [11,12,13]. Write le, which takes a list ad computes the legth of the list. E.g., le([1,[2],3],r).?- R = 3. Fall 18 CSCI 4430, A Milaova 49 Exercise Write atoms, which takes a list ad computes the umber of atoms i the list. E.g., atoms([a,[b,[[c]]]],r).?- R = 3. Hit: built-i predicate atom(x) yields true if X is a atom (i.e., symbolic costat such as x, abc, tom). Fall 18 CSCI 4430, A Milaova 50 9