George F Luger ARTIFICIAL INTELLIGENCE 5th edition Structures and Strategies for Complex Problem Solving Programming in Logic (PROLOG) Luger: Artificial.

George F Luger

ARTIFICIAL INTELLIGENCE 5th editionStructures and Strategies for Complex Problem Solving

Programming in Logic (PROLOG)

Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005

1

PROLOG

• What is True

• What needs to be Proved

Not

• How to do it like in other Programming languages

2

PROLOG

• Edinburgh Syntax

• Programming in PROLOG– Clocksin and Mellish, 1984

3

PROLOG EXAMPLES

4

EXAMPLE 1Set of facts:

parent(ali,ahmad).

parent(ahmad,salem).

parent(ali,fatema).

parent(fatema,osama).

male(ali).

male(ahmad).

male(salem).

male(osama).

female(fatema).

5

EXAMPLE 1• Every sentence ends with a period “.”

• Names starts with lower case, constants

6

EXAMPLE 1? parent(ali, ahmad).

By unifying this statement with the axioms (assertions) in order.

The above query (goal) also ends with a period

Ans:

yes

?parent(ali,X).

The above statement asks about all ali’s child

Semicolon asks for more answers

Ans:

X=ahmad; Note: by unifying X with ahmad

X=fatema;

no 7

EXAMPLE 1?parent(X,Y).

X=ali

Y=ahmad;

X=ahmad

Y=salem;

X=ali

Y=fatema;

X=fatema

Y=osama;

no

8

EXAMPLE 1?parent(X,ahmad), male(X).

Comma “,” means and

Who is ahmad’s father

It first unifies the first part, i.e parent(X,ahmad).

It finds that X could take the value of ali.

It then attempts to prove male(ali)

Ans:

X=ali;

No

Backtracking on two levels to find other answers9

EXAMPLE 1Add the following rule (considered rule not fact):

father(X,Y):- parent(X,Y) , male(X).

father(X,Y) called head

parent(X,Y) , male(X) called body

?father(X,ahmad).

Ans:

X=ali;

No

?father(fatema,Y).

no

Because fatema is unified with X and there is no way to prove that male(fatema)

10

EXAMPLE 1sibling(X,Y):- parent(Z,X) , parent(Z,Y).

?sibling(X,Y)

X=ahmad,

Y=ahmad;

X=ahmad,

Y=fatema;

X=salem,

Y=salem;

X=fatema,

Y=ahmad;

X=fatema,

Y=fatema;

X=osama,

Y=osama;

no11

EXAMPLE 1sibling(X,Y):- parent(Z,X) , parent(Z,Y), X\

==Y.

?sibling(X,Y)

X=ahmad,

Y=fatema;

X=fatema,

Y=ahmad;

no12

EXAMPLE 1Define uncle relation, uncle(X,Y).

uncle(X,Y):-

parent(Z,Y),

parent(G,Z),

parent(G,X),

X\==Z. 13

HW: Define the following rules:• mother

• son

• daughter

• sister

• sibling

• grandfather(X,Y):-parent(X,Z),parent(Z,Y), male(X).

• grandmother

• cousin(X,Y)14

Built-in mathematical predicates• X=:=Y true when X equals Y

• X=\=Y true when X does not equal to Y

• X<Y true when X is less than Y

• X>Y true when X is greater than Y

• X=<Y true when X is less than or equal to Y

• X>=Y true when X is greater than or equal to Y

15

Built-in Mathematical predicates?6 = : = 4 +2.

yes

?8 = : = 4*2.

yes

?7= : = 4+6.

no

?6 =:= 6.

yes 16

Built-in logical predicates• X= =Y true when X can be unified with

Y

• X\= =Y true when X is not the same as Y

• X=Y attempts to unify X with Y

17

Built-in logical predicates?5= =5.

yes

?X= = Y.

no

?5+1 == 5+1.

yes

?5+1 = = 1+5.

no 18

Built-in logical predicates?X= ali.

X=ali

?X= 1/2.

X=1/2

? f(X)= f(ali).

X=ali

?f = g.

no 19

Built-in logical predicates?5 \= 4.

yes

?ahmad @<basem.

yes

? 'Ali' @< 'Basem'.

yes

? X=1+2.

X=1+2 20

Built-in logical predicates? X is 1+2.

X=3

? X is 4*2.

X=8

?X is 7//3.

X=2

? X is 5, X is 3+3.

no 21

Mathematical operations+ Addition

- Subtraction

* Multiplication

/ Real division

// Integer division

mod modulus

^ Exponent

22

Negation as failurehome(X):-not out(X).

out(ali).

?home(ali).

no

?home(X).

no

?home(zaki).

yes

?out(zaki).

no23

Horn ClausesP1 ^ P2 ^ P3 ^ P4 R

R:- P1, P2, P3, P4.

24

Recursion

A program that calls itself

Assume the following relations and attempt to find predecessor

parent(ali,ahmad).

parent(ahmad,fatema).

parent(fatema,osama).

25

Recursion• A non recursion version

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

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

predecessor(X,Y):- parent (X,Z1), parent (Z1,Z2), parent(Z2,Y).

• The above program is applicable to a limited number of generations.

26

Recursion• Recursion version

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

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

27

Recursion? predecessor(ali,ahmad).

yes

? predecessor(ali,fatema).

yes

? predecessor(ali,osama).

yes.

? predecessor(ali,X).

X=ahmad 28

Recursion• Factorial

29

0N if1)!-(N*N

0 1!

NifN

RecursionFactorial

• fact(0,1).

• fact(N,X):-

N>0, N1 is N-1,

fact(N1,X1),

X is N*X1.

?fact(0,X).

X=1.

?fact(1,X).

X=130

Recursion• HW:

• Write a program to calculate the sum of the numbers:

1, 2, 3, ..., N

31

List Processing• Empty lists are denoted as []

• [H | T] list of :– head H representing the first element.

– tail T represents the rest of the elements.

– [5, 2, 7, 10] H is 5, T is [2, 7, 10]

32

List Processing?[H | T]=[5, 2, 7, 10].

H=5

T=[2, 7, 10]

?[a,b,c,d]=[X,Y | T].

X=a

Y=b

T=[c, d]

33

List Processing?[H | T]=[ali].

H=ali

T=[]

?[H | T]=[].

no

34

List Processing - membership• Test for a membership in a list

member(X,[X | T]).

member(X,[ _ | T]):-

member(X,T).

?member(5,[5, 7]).

yes

?member(5,[1, 5]).

yes 35

List Processing - membershipTo get all members of a list you can use:

?member(X,[a,b,c,d]).

X=a;

X=b;

X=c;

X=d;

no

36

List Processing - count• To count how many members in a list

count([],0).

count([X | T],C):-

count(T,C1),

C is C1+1.

?count([2,4,7],C).

C=3

37

List Processing - sum• To calculate the sum of the elements in a list

sum([],0).

sum([X | T],S):-

sum(T, S1),

S is S1+X.

?sum([2,4,7],S).

S=13

38

List Processing - append• To append two lists and produce a list containing the

elements of the first list then the elements of the second list

append([], L, L).

append([H | T1], L, [H | T]):

append(T1, L, T).

39

List Processing - append?append([r,t],[a,b,c],L).

L=[r,t,a,b,c].

?append(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=[];

no 40

List Processing - split• To divide a list into two parts positive and negative

positive([],[],[]).

positive([X|Y],[X|L1], L2) :-

X >= 0,

positive(Y,L1, L2) .

positive([X|Y],L1, [X|L2]) :-

positive(Y,L1, L2).

41

List Processing – write elements• To print the elements of a list

write_a_list([]).

write_a_list([H|T]):-

write(H), nl, write_a_list(T).

? write_a_list([5, 6, 7]).

5

6

7 42

Cut operator• To find a maximum of two numbers, you might

write:

max(X,Y,X):-X>=Y.

max(X,Y,Y).

?max(5,3,M).

M=5;

M=3;

no43

Cut operator• To find a maximum of two numbers, you might

write:

max(X,Y,X):-X>=Y,!.

max(X,Y,Y).

?max(5,3,M).

M=5;

no

44

Cut operator• Add an element to a list only if that element does

not exist already in the list

add(X, L, L):-

member(X,L), !.

add(X, L, [X|L]).

45

Self test list exercises1. Calculate the max/min value in a list.

? min_in_list([5,7,3,7,9], A).

A=3

? max_in_list([8,4,9,4],X).

X=9

46

Self test list exercises2. Calculate how many times a specific

number appears in a list.

3. Take a list as input and return its reverse in another list.

4. Take two lists as input and return their union, intersection, and difference in another list.

5. Delete one number from a list.

47

Homework1. Write a program to calculate the absolute

value of X, absval(X,Y) where X is the input and Y is the output.

2. Write a program to define the even(X) which is true only if X is even.

3. Write a program write_reverse to print the elements of a list in a reverse order

4. Write a program that simulates repeat assuming repeat does not exist.

48

PROLOG Self test examples

?concatenate([2, 3, 5], [7, 9, 5], B).

B=[2, 3, 5, 7, 9, 5]

?order([2, 8, 3, 5], B).

B=[2, 3, 5, 8]

49

The fail predicatecountry(jordan).

country(egypt).

country(england).

print_countries:-

country(X),

write(X),

nl,

fail.

print_countries.50

The fail predicate

?print_countries.

jordan

egypt

england

yes

51

The repeat command

do:-

repeat, read(X), square(X).

do.

square(stop):-!.

square(X):-

Y is X*X, write(Y), nl, fail.

52

The repeat command

?do.

5.

25

6.

36

stop.

yes53

The repeat commanddo:-

read(X), square(X).

square(stop).

square(X):-

X\==stop,

Y is X*X,

write(Y), nl,

do.

54

Input / output procedures - write

?X=5, write(X).

5

yes

?write([a,b,5,6]).

[a,b,5,6]

yes55

Input / output procedures - write

?write(date(16,11,2008)).

date(16,11,2008)

yes

56

Input / output procedures - read

?read(X).

a.

X=a.

?read(stop).

test

no57

Dynamic programs

• Programs that are able to modify themselves.

• Modification means the ability to add/ remove facts at runtime.

• Such procedures include:– assert(C)– retract(C)– abolish(C)

58

Dynamic programs - assert

• assert(C)

?parent(zaid, faris).

no

?assert(parent(zaid, faris)).

yes


yes59

Dynamic programs - assert

• asserta(C) adds the new statement C to the beginning of the program.

• assertz(C) adds the new statement C to the end of the program.

60

Dynamic programs - retract

• retract removes a statement from the program

?retract(parent(zaid, faris)).

yes


no

61

Dynamic programs - abolish

?abolish(parent,2).

yes

removes all statements defined as parent with arity 2.

62

Structuresemployee(Name,House_no,Street_name,City_name, Day,Month,Year).

employee(ali,5,salt_street,amman,12,10,1980).

?employee(ali,No,S,C,D,M,Y).

No=5

S=salt_street

C=amman

D=12

M=10

Y=1980

63

Structuresemployee(ali,address(5,salt_street,amman),date(12,10,1980)).

employee(hasan,address(12,university_street,zarqa),date(7,3,1985)).

employee(samer,address(9,madina_street,amman),date(2,9,1987)).

?employee(hasan,A,D).

A=address(12,university_street,zarqa)

D=date(7,3,1985)

?employee(Name,address(No,St,amman),_).

Name=ali

No=5

St=salt_street;

Name=samer

No=9

St=madina_street;

false

no

64

Structures?employee(Name,A,date(_,_,1985)).

Name=hasan

A=address(12,university_street,zarqa);

no

?employee(ali,A,date(Day, Month, Year)), 2011- Year>25.

A=address(5,salt_street,amman)

Day=12

Month=10

Year=1980;

no 65

Structures – Binary Trees

66

2

5

10

7 9

3

b

a

c

a

Example 3

Example 2

Example 1


• Binary trees when empty are called nil.

• It may contain 3 components:– Left subtree

– Root

– Right subtree

• It can be represented using the following structure:bin_tree(Left_subtree, Root, Right_subtree).

67


bin_tree(nil,a,nil).

68

a

Example 1


bin_tree(bin_tree(nil,b,nil),a, bin_tree(nil,c,nil)).

69

Example 2

b

a

c


bin_tree(bin_tree(nil,2,nil),5, bin_tree(bin_tree(nil,7,nil),10, bin_tree(nil,9, bin_tree(nil,3,nil)))). 70

Example 3

2

5

10

7 9

3

Structures – Binary Trees - Count

count(nil,0).

count(bin_tree(Left,Root,Right),C):-

count(Left,C1),

count(Right,C2),

C is C1+C2+1.

71

Structures – Binary Trees – Count V2

count(nil,0).

count(bin_tree(Left,_,Right),C):-

count(Left,C1),

count(Right,C2),

C is C1+C2+1.

72

Structures – Binary Trees - Sum

sum(nil,0).

sum(bin_tree(Left,Root,Right),S):-

sum(Left,S1),

sum(Right,S2),

S is S1+S2+Root.

73

Structures – Binary Trees - depth

depth(nil,0).

depth(bin_tree(Left,Root,Right),D):-

depth(Left,D1),

depth(Right,D2),

D1>=D2,

D is D1+1.


depth(Left,D1),

depth(Right,D2),

D2>D1,

D is D2+1. 74

Structures – Binary Trees – depth V2

depth(nil,0).


depth(Left,D1),

depth(Right,D2),

max(D1,D2,M),

D is M+1.

max(X,Y,X):- X>=Y,!.

max(X,Y,Y):- X<Y.

75

Binary Tree - Hw• Write a PROLOG program max_tree to find

the maximum value in a binary tree (bin_tree) assuming all values are numeric.

• Write a PROLOG program to add a new value to an ordered binary tree.

• Write a PROLOG program to display the values stored in an ordered binary tree in ascending order.

76

PROLOG Resources

http://en.wikibooks.org/wiki/Prolog/Lists

http://www.csupomona.edu/~jrfisher/www/prolog_tutorial/2_7.html

77

George F Luger ARTIFICIAL INTELLIGENCE 5th edition Structures and Strategies for Complex Problem Solving Programming in Logic (PROLOG) Luger: Artificial.

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