Logic Programming

Logic Programming

fac(1,1) fac(n,n×m)

fac(n-1,m)

ex) factorial program query

?- fac(5,120).yes?- fac(5,X).X=120

cf) in procedural programming

x=1;for (i=1;i<=n;i++) x=x*i;

프로그램 = 추론규칙

구하고자 하는 것이 무엇인지를 명시

어떻게 구하는지를 명시

예) 타입시스템

•타입시스템 (추론규칙)

•올바른 타입을 가지는 프로그램이란 무엇인가?

•타입 유추 알고리즘

•입력 프로그램의 타입을 어떻게 찾는가?

Γ � E : τ

Γ � n : ι

Γ(x) = τ

Γ � x : τ

Γ + x : τ1 � E : τ2

Γ � λx.E : τ1 → τ2

Γ � E1 : τ1 → τ2 Γ � E2 : τ1

Γ � E1 E2 : τ2

Γ � E1 : ι Γ � E2 : ιΓ � E1+E2 : ι

E → n | x | λx.E | E E | E + E

v → n | λx.EC → [] | C E | v C | C + E | v + C

∅ � ff1 f2

f1 ∧ f2

T· · ·

let fac = rec f λn.if n = 0 then 1 else n ∗ (f (n− 1)) in (fac 2)

σ � 1 ⇒ 1· · ·

σ{x �→ 1} � x + 2 ⇒ 3σ � let x = 1 in x + 2 ⇒ 3

σ � λx.x ⇒ �λx.x, σ� σ � 1 ⇒ 1 σ{x �→ 1} � x ⇒ 1σ � (λx.x) 1 ⇒ 1

σ � λx.z x ⇒ �λx.z x, σ� σ � λy.y ⇒ �λy.y,σ�

σ{x �→ �λy.y,σ�} � z ⇒ �λk.k, σ��σ{x �→ �λy.y,σ�} � x ⇒ �λy.y,σ�σ�{k �→ �λy.y,σ� � k ⇒ �λy.y,σ�}σ{x �→ �λy.y,σ�} � z x ⇒ �λy.y,σ�

σ(= {z �→ �λk.k, σ��}) � (λx.z x) (λy.y) ⇒ �λy.y,σ�

σ � λx.E ⇒ λx.E

σ � E1 ⇒ λx.E� σ � E2 ⇒ v σ{x �→ v} � E� ⇒ v�

σ � E1 E2 ⇒ v�

λx.E

E1 E2

S = {x1 �→ Y1, . . . , xn �→ Yn}S E

1

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

Defining Relations by Facts

parent(tom,bob).parent(pam,bob).parent(tom,liz).parent(bob,ann).parent(bob,pat).parent(pat,jim).

pam tom

bob

ann pat

jim

liz

Defining Relations by Facts

pam tom

bob

ann pat

jim

liz

?- parent(bob,pat).yes

?- parent(liz,pat).no

?- parent(tom,ben).no

Defining Relations by Facts

pam tom

bob

ann pat

jim

liz

?- parent(X,pat).X=tom

?- parent(bob,X).X=annX=pat

?- parent(X,Y).X=pam Y=bobX=tom Y=bob...

Defining Relations by Facts

pam tom

bob

ann pat

jim

liz

?- parent(Y,jim), parent(X,Y).X=bob Y=pat

?- parent(X,Y), parent(Y,jim).X=bob Y=pat

?- parent(tom,X), parent(X,Y).X=bob Y=ann

?- parent(X,ann), parent(X,pat).X=bob

Defining Relations by Rules

female(pam).male(tom).male(bob).female(liz).female(pat).female(ann).male(jim).

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

Defining Relations by Rules

offspring(Y,X):-parent(X,Y).

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

offspring(Y,X)

parent(X,Y)

inference rule

in prolog

?- offspring(liz,tom).yes

offspring(liz,tom):-parent(tom,liz).

Defining Relations by Rules

grandparent(X,Z):- parent(X,Y), parent(Y,Z).

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

grandparent(X,Z)

parent(X,Y) parent(Y,Z)

inference rule

in prolog

Defining Relations by Rules

sister(X,Y):- parent(Z,X), parent(Z,Y), female(X).

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

sister(X,Y)

parent(Z,X) parent(Z,Y) female(X)

inference rule

in prolog

Defining Relations by Rules

?- sister(ann,pat).yes

?- sister(X,pat).X=annX=pat

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

sister(X,Y):- parent(Z,X), parent(Z,Y), female(X).

Defining Relations by Rules

?- sister(ann,pat).yes

?- sister(X,pat).X=ann

pam tom

bob

ann pat

jim

liz

female male

female female

femalemale

male

sister(X,Y):- parent(Z,X), parent(Z,Y), female(X), different(X,Y).

Recursive Rules

predecessor(X,Z):- parent(X,Z).

predecessor(X,Z):- parent(X,Y), predecessor(Y,Z).

predecessor(X,Z)

parent(X,Y) predecessor(Y,Z)

inference rule

in prolog

predecessor(X,Z)

parent(X,Z)

?- predecessor(pam,X).X=bobX=annX=patX=jim

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).

predecessor(tom,pat)

parent(tom,pat)

no

parent(tom,Y)predecessor(Y,pat)

predecessor(bob,pat)

Y=bob

parent(bob,pat)

yes

Declarative/Procedural Meanings

• Declarative meanings

• What is the output of the program?

• Procedural meanings

• How is this output obtained?

Data objects in Prolog

data objects (=terms)

simple objects structures

constants

atoms numbers

variables

tom,bob,nil,x_,x25,... 1, 1313, 0, -97, 3.14,...

X,Y,Z,...

Structured Objects

• Structured object = tree

• Objects that have several components.

date(1, may, 2011)

functor arguments

date

1 may 2011

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))

Matching• Two terms match if they are unifiable

• Prolog finds the most general unifier

?- date(D, M, 2011) = date(D1, may, Y1).D = D1M = mayY1 = 2001

?- date(D, M, 2011) = date(D1, may, 2144)fail

D = 1D = 1M = mayY1 = 2001

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(...

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|[]]

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

Membership

member(X,L) is true if X occurs in L

?- member(b,[a,b,c]).yes

?- member(b,[a,[b,c]]).no

?- member([b,c],[a,[b,c]]).yes

member(X,[X|Tail]).member(X,[Head|Tail]) :- member(X,Tail).

Concatenationconc(L1,L2,L3) is true if L1 and L2

are two lists and L3 is their concatenation

?- conc([a,b],[c,d],[a,b,c,d]).yes

?- conc([a,b,c],[1,2,3],L).L = [a,b,c,1,2,3]

conc([],L,L).conc([X|L1],L2,[X|L3]) :- conc(L1,L2,L3).

X L1 L2

L3

[X|L3]

Other uses of conc

?- 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 = []

decompose a given list into two lists

Other uses of conc

?- 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 = aprSucc = 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]

look for a certain pattern in a list

Other uses of conc

member(X,L) is true if X occurs in L

member(X,[X|Tail]).member(X,[Head|Tail]) :- member(X,Tail).

member(X,L) :- conc(_,[X|_],L).

vs.

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];

(Quiz1) Inserting an itemDefine 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]

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

L1 L2

L

Ssublist(S,L):- sublist(L1,L3,L), sublist(S,L2,L3).

L3

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 = [];

Permutations

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];

permutation(L,P) is true if P is a permutation of L

(Quiz2). Reverse

Define the relation

reverse(List,ReversedList)

that reverses list.

?- reverse([a,b,c,d],[d,c,b,a]).yes

(Quiz3). PalindromeDefine 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];...

The eight queens problem

1 2 3 4 5 6 7 8

1 2

3 4

5 6

7 8

Board templateX/Y: a queen sitting at (X,Y)

1 2 3 4 5 6 7 8

1 2

3 4

5 6

7 8

[1/2, 2/4, 3/6,4/8,5/3,6/1,7/7,8/5]

[1/Y1, 2/Y2, 3/Y3,4/Y4,5/Y5,6/Y6,7/Y7,8/Y8]

Solutionsolution(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]...