Logic Programming. fac(n-1,m) fac(n,n m) fac(1,1) ex) factorial program query. cf) in procedural programming. ?- fac(5,120). yes?- fac(5,x).

Transcription

1 Logic Programming

2 Logic Programming = ex) factorial program query fac(1,1) fac(n-1,m) fac(n,n m)?- fac(5,120). yes?- fac(5,x). X=120 cf) in procedural programming x=1; for (i=1;i<=n;i++) x=x*i;

3 예 ) 타입시스템 Γ n : ι Γ(x) =τ Γ x : τ Γ + x : τ 1 E : τ 2 Γ λx.e : τ 1 τ 2 Γ E 1 : τ 1 τ 2 Γ E 2 : τ 1 Γ E 1 E 2 : τ 2 Γ E 1 : ι Γ E 2 : ι Γ E 1 +E 2 : ι

4 Introduction to Prolog Programming in logic The first and most popular logic programming language First-order predicate logic: Turing complete Apps) natural language processing, theorem proving, expert systems, games, web programming, etc

5 Defining Relations by Facts pam tom ann bob pat liz parent(tom,bob). parent(pam,bob). parent(tom,liz). parent(bob,ann). parent(bob,pat). parent(pat,jim). jim

6 Defining Relations by Facts pam tom bob liz?- parent(bob,pat). yes?- parent(liz,pat). no ann pat?- parent(tom,ben). no jim

7 Defining Relations by Facts pam tom?- parent(x,pat). X=tom bob liz?- parent(bob,x). X=ann X=pat ann jim pat?- parent(x,y). X=pam Y=bob X=tom Y=bob...

8 Defining Relations by Facts pam tom?- parent(y,jim), parent(x,y). X=bob Y=pat bob liz?- parent(x,y), parent(y,jim). X=bob Y=pat ann pat?- parent(tom,x), parent(x,y). X=bob Y=ann jim?- parent(x,ann), parent(x,pat). X=bob

9 Defining Relations by Rules male pam tom ann male bob pat liz (pam). male(tom). male(bob). (liz). (pat). (ann). male(jim). male jim

10 Defining Relations by Rules male pam tom inference rule parent(x,y) male offspring(y,x) bob liz in prolog offspring(y,x):-parent(x,y). ann pat?- offspring(liz,tom). yes male jim offspring(liz,tom):-parent(tom,liz).

11 Defining Relations by Rules male pam ann male bob male tom pat liz inference rule in prolog parent(x,y) parent(y,z) grandparent(x,z) grandparent(x,z):- parent(x,y), parent(y,z). jim

12 Defining Relations by Rules male pam tom inference rule male bob liz parent(z,x) parent(z,y) (X) sister(x,y) in prolog ann male pat sister(x,y):- parent(z,x), parent(z,y), (X). jim

13 Defining Relations by Rules male pam male tom sister(x,y):- parent(z,x), parent(z,y), (X). bob liz?- sister(ann,pat). yes ann male pat?- sister(x,pat). X=ann X=pat jim

14 Defining Relations by Rules male pam male bob tom liz sister(x,y):- parent(z,x), parent(z,y), (X), different(x,y).?- sister(ann,pat). yes ann male pat?- sister(x,pat). X=ann jim

15 Recursive Rules inference rule parent(x,z) predecessor(x,z) parent(x,y) predecessor(y,z) predecessor(x,z) in prolog predecessor(x,z):- parent(x,z). predecessor(x,z):- parent(x,y), predecessor(y,z).?- predecessor(pam,x). X=bob X=ann X=pat X=jim

16 How prolog answers parent(tom,bob). parent(pam,bob). parent(tom,liz). parent(bob,ann). parent(bob,pat). parent(pat,jim). predecessor(x,z):- parent(x,z). predecessor(x,z):- parent(x,y), predecessor(y,z). parent(tom,pat) no predecessor(tom,pat) parent(tom,y) predecessor(y,pat) Y=bob predecessor(bob,pat) parent(bob,pat) yes

17 Declarative/Procedural Meanings Declarative meanings Procedural meanings What is the output of the program? How is this output obtained?

18 Data objects in Prolog data objects (=terms) simple objects structures constants variables X,Y,Z,... atoms numbers tom,bob,nil,x_,x25,... 1, 1313, 0, -97, 3.14,...

19 Structured Objects Structured object = tree Objects that have several components. date date(1, may, 2011) 1 may 2011 functor arguments point(1,1) seg(point(1,1),point(2,3)) triangle(point(4,2),point(6,4),point(7,1)) (a+b)*(c-5) = *(+(a,b), -(c,5))

20 Matching Two terms match if they are unifiable Prolog finds the most general unifier?- date(d, M, 2011) = date(d1, may, Y1). D = D1 M = may Y1 = 2001 D = 1 D = 1 M = may Y1 = 2001?- date(d, M, 2011) = date(d1, may, 2144) fail

21 Prolog and Logic Syntax: first-order predicate logic, clause form Procedural meaning: resolution principle for mechanical theorem proving (Robinson 65) Matching = unification in logic occur check?- X = f(x). X = f(x) or X = f(f(f(f(f(f(f(f(f(...

22 Lists [ann, tennis, tom, skiing] =.(ann,.(tennis,.(tom,.(skiing, [])))) [ann] =.(ann,[])?- Hobbies1 =.(tennis,.(music, [])), Hobbies2 = [skiing, food], L = [ann, Hobbies1, tom, Hobbies2]. Hobbies1 = [tennis, music] Hobbies2 = [skiing, food] L = [ann,[tennis,music],tom,[skiing,food]] [a,b,c] = [a [b,c]] = [a,b, [c]] = [a,b,c []]

23 List operations member(x,l) : X is a member of list L conc(l1,l2,l3) : L3 is the concatenation of L1 and L2 add(x,l1,l2) : add(x,l,[x L) del(x,l,l1) : L1 is equal to L with X removed sublist(s,l) : S is a sublist of L permutation (L,P) : P is a permutation of L

24 Membership member(x,l) is true if X occurs in L member(x,[x Tail]). member(x,[head Tail]) :- member(x,tail).?- member(b,[a,b,c]). yes?- member(b,[a,[b,c]]). no?- member([b,c],[a,[b,c]]). yes

25 Concatenation conc(l1,l2,l3) is true if L1 and L2 are two lists and L3 is their concatenation conc([],l,l). conc([x L1],L2,[X L3]) :- conc(l1,l2,l3). X L1 L2?- conc([a,b],[c,d],[a,b,c,d]). yes L3 [X L3]?- conc([a,b,c],[1,2,3],l). L = [a,b,c,1,2,3]

26 Other uses of conc decompose a given list into two lists?- conc(l1,l2,[a,b,c]). L1 = [] L2 = [a,b,c]; L1 = [a] L2 = [b,c]; L1 = [a,b] L2 = [c]; L1 = [a,b,c] L2 = []

27 Other uses of conc look for a certain pattern in a list?- conc(before,[may After], [jan,feb,mar,apr,may,jun,jul,aug,sep,oct,nov,dec]). Before = [jan,feb,mar,apr] After = [jun,jul,aug,sep,oct,nov,dec]?- conc(_,[pred,may,succ _], [jan,feb,mar,apr,may,jun,jul,aug,sep,oct,nov,dec]). Pred = apr Succ = jun?- L1 = [a,b,z,z,c,z,z,z,d,e], conc(l2,[zzz, _],L1). L2 = [a,b,z,z,c] L1 = [a,b,z,z,c,z,z,z,d,e]

28 Other uses of conc member(x,l) is true if X occurs in L member(x,l) :- conc(_,[x _],L). vs. member(x,[x Tail]). member(x,[head Tail]) :- member(x,tail).

29 Deleting an item del(x,l,l1) is true if L1 is equal to L with the item X removed del(x,[x Tail],Tail). del(x,[y Tail],[Y,Tail]) :- del(x,tail,tail1)?- del(a,[a,b,a,a],l). L = [b,a,a]; L = [a,b,a]; L = [a,b,a];?- del(a,l,[1,2,3]). L = [a,1,2,3]; L = [1,a,2,3]; L = [1,2,a,3]; L = [1,2,3,a];

30 (Quiz1) Inserting an item Define insert(x,l,l1) : insert(x,l,l1) is true if L1 is equal to L with the item X inserted (at any place)?- insert(a,[b,c],[a,b,c]). yes?- insert(a,[b,c],l). L = [a,b,c]; L = [b,a,c]; L = [b,c,a];?- insert(a,l,[a,b,c]). L = [b,c]

31 Sublist sublist(s,l) is true if S is a sublist of L?- sublist([c,d,e],[a,b,c,d,e]). yes?- sublist([c,e],[a,b,c,d,e,f]). no L L1 S L2 sublist(s,l):- sublist(l1,l3,l), sublist(s,l2,l3). L3

32 Sublist?- sublist (S,[a,b,c]). S = []; S = [a]; S = [a,b]; S = [a,b,c]; S = []; S = [b]; S = [b,c]; S = []; S = [c]; S = [];

33 Permutations permutation(l,p) is true if P is a permutation of L permutation([],[]). permutation([x L],P):- permutation(l,p1), insert(x,p1,p).?- permutation ([a,b,c],p). P = [a,b,c]; P = [b,c,a]; P = [b,c,a]; P = [a,c,b]; P = [c,a,b];

34 (Quiz2). Reverse Define the relation that reverses list. reverse(list,reversedlist)?- reverse([a,b,c,d],[d,c,b,a]). yes

35 (Quiz3). Palindrome Define the predicate palindrome(list) that is true if List is a palindrome. A list is palindrome if it reads the same in the forward and in the backward direction.?- palindrome([m,a,d,a,m]). yes?- permutation([a,a,b,b,c],l), palindrome(l). L = [a,b,c,b,a]; L = [b,a,c,a,b];...

36 The eight queens problem

37 Board template X/Y: a queen sitting at (X,Y) [1/2, 2/4, 3/6, 4/8, 5/3, 6/1, [1/Y1, 2/Y2, 3/Y3, 4/Y4, 5/Y5, 6/Y6, /7, 8/5] 7/Y7, 8/Y8]

38 Solution solution(s) is true if position S is safe solution([x/y Others]):- solution(others), member(y,[1,2,3,4,5,6,7,8]), noattack(x/y,others). noattack(_,[]). noattack(x/y,[x1/y1 Others]):- Y =\= Y1, Y1-Y =\= X1-X, Y1-Y =\= X-X1, noattack(x/y,others). template([1/y1,2/y2,3/y3,4/y4,5/y5,6/y6,7/y7,8/y8]).?- template(s), solution(s). S = [1/4,2/2,3/7,4/3,5/6,6/8,7/5,8/1]...