1

member(X,[Y| _ ] ) :- X = Y.member(X, [ _ | Y]) :- member(X, Y).

It would be easier to write this as:

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

?- member(1, [3,4,5,8,1,9]).Yes?- member(X, [prolog, c, ada, haskell]).X= prolog;X= cX= ada;X= haskell;No

Recursion and Lists

2

Other Examples

change(you, i).change(are, [am, not]).change(french, australian).change(do, no).change(X, X). /* catchall */alter([ ], [ ]).alter([H|T], [X|Y]) :- change(H, X), alter(T,Y).

?- alter([you,are,a,computer],R).

R = [i, [am, not], a, computer]

Yes

?- alter([you,are,french],R).

R = [i, [am, not], australian]

Yes

?-

3

asserta/1.assertz/1assert/1retract/1retractall/1

Example:

assertz(fib(N,F)).

4

:-dynamic fibfib(1,1).fib(2,1).fib(N,F) :-

N > 2,N1 is N-1, fib(N1,F1),N2 is N-2, fib(N2,F2),F is F1 + F2,asserta(fib(N,F)).

?- fib(8, F).

F = 21

?- fib(6,F).

F = 8

5

F(6)+

f(5) f(4)+

f(3) f(2)

f(2) f(1)

+

1 1

+

f(4) f(3)+

f(3) f(2) 1

+

f(2) f(1)

+

f(2) f(1)

1 1 1 1 1

6

asserta/1.assertz/1assert/1retract/1retractall/1

Example:

assertz(fib(N,F)).

7

:-dynamic animal/1. % A directive.

animal(tiger).

animal(lion).

animal(monkey).

animal(X):-mamal(X),

asserta(animal(X)).

mamal(cat).

mamal(dog).

?- animal(dog).Yes?- listing(animal).:- dynamic animal/1.animal(dog).animal(tiger).animal(lion).animal(monkey).animal(A) :- mamal(A), asserta(animal(A)).

8

leng([ ], 0).leng([H|T], Len) :- leng(T, Len1),

Len is Len1 + 1.

Finding the length of a list

9

[trace] ?- leng([1,2,3,4,5],Ln).

Call: (6) leng([1, 2, 3, 4, 5], _G405)?creep

Call: (7) leng([2, 3, 4, 5], _G474) ? creep

Call: (8) leng([3, 4, 5], _G474) ? creep

Call: (9) leng([4, 5], _G474) ? creep

Call: (10) leng([5], _G474) ? creep

Call: (11) leng([], _G474) ? creep

Exit: (11) leng([], 0) ? creep

^ Call: (11) _G479 is 0+1 ? creep

^ Exit: (11) 1 is 0+1 ? creep

Exit: (10) leng([5], 1) ? creep

^ Call: (10) _G482 is 1+1 ? creep

^ Exit: (10) 2 is 1+1 ? creep

Exit: (9) leng([4, 5], 2) ? creep

^ Call: (9) _G485 is 2+1 ? creep

^ Exit: (9) 3 is 2+1 ? creep

Exit: (8) leng([3, 4, 5], 3) ? creep

^ Call: (8) _G488 is 3+1 ? creep

^ Exit: (8) 4 is 3+1 ? creep

Exit: (7) leng([2, 3, 4, 5], 4) ? creep

^ Call: (7) _G405 is 4+1 ? creep

^ Exit: (7) 5 is 4+1 ? creep

Exit: (6) leng([1, 2, 3, 4, 5], 5) ? creep

Ln = 5

Yes

leng([ ], 0).leng([H|T], Len) :- leng(T, Len1),

Len is Len1 + 1.

10

remove(X, [ ], [ ]).remove(X, [X|T], C):-

remove(X,T,C).remove(X, [H|T], [H|T1]):- X \= H,

remove(X,T,T1).

?- remove(1,[4,9,8,7,1,9,7,5,1],R).

R = [4, 9, 8, 7, 9, 7, 5]

Removing an element from a list

11

isort([], []).isort([H|T],R):- isort(T,Q),

ins(H,Q,R).ins(X, [], [X]).ins(X, [H|T], [H|R]):- X>H, ins(X,T,R).ins(X, [H|T], [X,H|T]):- X=<H.

?- isort([4,1,9,5,8,3,2],S).

S = [1, 2, 3, 4, 5, 8, 9]

?- isort([a,b,c,d],S).

ERROR: Arithmetic: `c/0' is not a function

Insertion Sort

12

I/O

13

A full-stop must follow a term to be read

?- read(X).|: massey.X = massey Yes?- read(X).|: Y.X = _G185 Yes?- read(X).|: likes(john,mary).X = likes(john, mary) Yes?- read(X), name(X,List).|: massey.X = masseyList = [109, 97, 115, 115, 101, 121]

14

?- write('hello world').hello worldYes?- write('hello world'),nl,nl,tab(9),write('I am here :)').hello world

I am here :)Yes?- write("abc").[97, 98, 99]Yes?- write('abc').abcYes

15

printlst([ ]) :- put('.'), nl.

printlst([H|T]) :- write(H), tab(1), printlst(T).?- write([‘This’,is,a,list]).

[This, is, a, list]

?- printlst([‘This’,is, a,list]).

This is a list .

?- read(X),nl,write(‘Here is what was entered: ‘),write(X).|: massey.

Here is what was entered: masseyX = massey

A predicate for printing a list

16

male(andrew).male(john). male(george). male(greg). male(adam). female(mary). female(jennifer). female(eve). parents(john,george,mary).parents(greg,adam,eve).parents(jennifer, adam,eve).parents(andrew, adam,eve).is_brother_of(X,Y):-male(X),parents(X, Father, Mother),parents(Y, Father,Mother),

X\=Y,write(X),tab(1),write('is'),tab(1),write(Y),write('s sister.').is_sister_of(X,Y):-female(X),parents(X, Father, Mother),parents(Y, Father, Mother),

X\=Y,write(X),tab(1),write('is'),tab(1),write(Y),write('s sister.').

?- is_sister_of(X,Y).jennifer is greg`s sister.X = jenniferY = greg ;jennifer is andrew`s sister.X = jenniferY = andrew

17

Built-in predicates “fail” and “true”

Fail always fails and true always succeeds

?- likes(mary,X).X = john Yes?- likes(mary,X),fail.No?- likes(mary,X),write(X),nl,fail.johnfruitbookNo?- likes(mary,X),write(X),nl,fail ; true.johnFruit ORbookX = _G354 Yes

likes(tom,jerry).likes(mary,john).likes(tom,mouse).likes(tom,jerry).likes(jerry,cheeze).likes(mary,fruit).likes(john,book).likes(mary,book).likes(tom,john).

Output predicates do not re-succeed on backtracking

18

likes(tom,jerry).likes(mary,john).likes(tom,mouse).likes(tom,jerry).likes(jerry,cheeze).likes(mary,fruit).likes(john,book).likes(mary,book).likes(tom,john).?- likes(mary,X), write('Mary likes '), write(X),put('.'), nl, fail.Mary likes john.Mary likes fruit.Mary likes book.No

?- likes(mary,X),write(X),nl,fail;write('reached the OR part').johnfruitbookreached the OR partX = _G471 Yes

19

write('Enter a number X to be raised to the power of 5 '), read(N),

P5 is N**5,write(N),write(‘ to the power of 5 is: '),write(P5).

?- write('Enter a number X to be raised to the power of 5: '), read(N),| P5 is N**5,write(N),write(' to the power of 5 is: '),write(P5).Enter a number X to be raised to the power of 5:|: 2.2 to the power of 5 is: 32N = 2P5 = 32 Yes?-

?- repeat,write('Enter a number X to be raised to the power of 5: '),

| read(N), P5 is N**5,write(N),write(' to the power of 5 is: '),

| write(P5),fail.

Built-in predicate “repeat”

20

File I/O

21

Prolog

accepts input from an input stream

send output to an output stream

Streams can be files or I/O devices

Active streams

are the keyboard and the screen by default (user)

We may only have

one stream open for input and

one open for output

22

tell(Filename) Opens the file named by Filename for writing and makes it the current output stream. It creates the file if it does not exist.

telling(X) Retrieves the current input stream.

told Closes the file opened by the built-in predicate tell.

see(Filename) Opens the file named by Filename for input and makes it the current input straeam.

seeing(X) Retrieves the current input stream.

seen Closes the file opened by the built-in predicate see.

23

?- tell('D:/test.txt').Yes?- telling(X).X = '$stream'(1996248) Yes?- write('testing tell').Yes?- told.Yes

testing tellend_of_file

Contents of D:/test.txt

?- telling(Old),tell(‘D:/test.txt’),write(‘testing’),told,tell(Old).

24

?- tell(‘lksout.txt’),likes(mary,X), write(‘Mary likes ‘),write(X), put(‘.’),nl, fail; told.X = _G662 Yes

likes(tom,jerry).likes(mary,john).likes(tom,mouse).likes(tom,jerry).likes(jerry,cheeze).likes(mary,fruit).likes(john,book).likes(mary,book).likes(tom,john).

Mary likes john.

Mary likes fruit.

Mary likes book.

Contents of the file “lks.txt”

25

Cut: !Eliminates choicesAlways succeeds but stops backtracking

a:-b,c,!,d.a:-e,f.

max(X,Y,Y) :- Y>X.max(X,Y,X). ?- max(1,2,X).X = 2 ;X = 1 ;No?-

max(X,Y,Y) :- Y>X, !. max(X,Y,X). ?- max(1,2,X).X = 2 ;No?-

26

Cut makes programs less readable.

Using cut can make your programs more efficient.

You have to be careful when using cut. Cut, may alter the way your program behaves. When you use cut in a rule, you have to be certain of how your rule will be used. If you are not careful your program may behave strangely.

Built-in predicate cut, “!”

27

Unexpected results: Using member with cut we only get one value for X, which is not what we expect.

 

member1(X,[X|_]):-!.

member1(X,[_|T]):-member1(X,T).

 

?- trace.

 

Yes

[trace] ?- member1(X,[1,2,3,4]).

Call: (6) member1(_G401, [1, 2, 3, 4]) ? creep

Exit: (6) member1(1, [1, 2, 3, 4]) ? creep

 

X = 1 ;

Fail: (6) member1(1, [1, 2, 3, 4]) ? creep

 

No

X = 1 ;

 

X = 2 ;

 

X = 3 ;

 

X = 4 ;

 

No

Member without cut 

membernocut(X,[X|_]).

membernocut(X,[_|T]):-membernocut(X,T).

 

?- membernocut(X,[1,2,3,4]).

 

28

Efficiency:

Using member without cut here, we are wasting time trying out the second part of the predicate when we know that it will fail.

Call: (7) membernocut(1, [1, 2, 3, 4]) ? creep

Exit: (7) membernocut(1, [1, 2, 3, 4]) ? creep

member succeeded

Redo: (7) membernocut(1, [1, 2, 3, 4]) ? creep

Call: (8) membernocut(1, [2, 3, 4]) ? creep

Call: (9) membernocut(1, [3, 4]) ? creep

Call: (10) membernocut(1, [4]) ? creep

Call: (11) membernocut(1, []) ? creep

Fail: (11) membernocut(1, []) ? creep

Fail: (10) membernocut(1, [4]) ? creep

Fail: (9) membernocut(1, [3, 4]) ? creep

Fail: (8) membernocut(1, [2, 3, 4]) ? creep

Fail: (7) membernocut(1, [1, 2, 3, 4]) ? creep

No

membernocut(X,[X|_]).

membernocut(X,[_|T]):-membernocut(X,T).

?- trace.Yes[trace] ?- membernocut(1,[1,2,3,4]),write('member succeeded'),fail.

29

  Using member with cut: 

?- trace. Yes[trace] ?- member1(1,[1,2,3,4]),write('member succeeded'),fail. Call: (7) member1(1, [1, 2, 3, 4]) ? creep Exit: (7) member1(1, [1, 2, 3, 4]) ? creepmember succeeded

member1(X,[X|_]):-!.

member1(X,[_|T]):-member1(X,T).

Fail: (7) member1(1, [1, 2, 3, 4]) ? creep

 

No

?-

30

member(X, [X|T]).

member(X, [H|T]) :- not(X=H), member(X, T).

Using not instead of cut makes your program more clear

member1(X,[X|_]):-!.

member1(X,[_|T]):-member1(X,T).

31

male(andrew).male(john). male(george). male(greg). male(adam). female(mary). female(jennifer). female(eve). parents(john,george,mary).parents(greg,adam,eve).parents(jennifer, adam,eve).parents(andrew, adam,eve).is_brother_of(X,Y):-male(X),parents(X, Father, Mother),parents(Y, Father,Mother), X\=Y.is_sister_of(X,Y):-female(X),parents(X, Father, Mother),parents(Y, Father, Mother),X\=Y.

?- male(X).X = andrew ;X = john ;X = george ;X = greg ;X = adam ;No?-

?- male(X), ! .X = andrew ;No

32

male(andrew).

male(john).

male(george).

male(greg).

male(adam).

female(mary).

female(jennifer).

female(eve).

parents(john,george,mary).

parents(greg,adam,eve).

parents(jennifer, adam,eve).

parents(andrew, adam,eve).

is_brother_of(X,Y):-male(X),!,parents(X, Father, Mother),parents(Y, Father,Mother), X\=Y.

is_sister_of(X,Y):-female(X),!,parents(X, Father, Mother),parents(Y, Father, Mother),X\=Y.

33

apend([],L,L):-!.apend([H|T1],L2,[H|T2]):- apend(T1,L2,T2).

[trace] ?- apend([],[1,2,3,4],L). Call: (6) apend([], [1, 2, 3, 4], _G409) ? creep Exit: (6) apend([], [1, 2, 3, 4], [1, 2, 3, 4]) ? creepL = [1, 2, 3, 4] ; Fail: (6) apend([], [1, 2, 3, 4], [1, 2, 3, 4]) ? creepNo

apend([],L,L).apend([H|T1],L2,[H|T2]):-apend(T1,L2,T2).

[trace] ?- apend([],[1,2,3,4],L). Call: (6) apend([], [1, 2, 3, 4], _G409) ? creep Exit: (6) apend([], [1, 2, 3, 4], [1, 2, 3, 4]) ? creepL = [1, 2, 3, 4] ; Fail: (6) apend([], [1, 2, 3, 4], _G409) ? creepNo

34

?- tell(‘fam.pl’), listing(male/1), told.Yes

male(andrew).

male(john).

male(george).

male(greg).

male(adam).

Contents of the file fam

consult(Filename) :- see(Filename), repeat,

read(X), assert(X),

X=end_of_file, !, seen.

Definition of “consult”

H

35

male(andrew).male(john). male(george). male(greg). male(adam). female(mary). female(jennifer). female(eve). parents(john,george,mary).parents(greg,adam,eve).parents(jennifer, adam,eve).parents(andrew, adam,eve).

?- male(X).X= andrew;X= john;X= george;X= greg;X= adam;

?- female(X).X= mary;X= jennifer;X= eve;

?- parents(X, adam, eve).X= greg;X= jennifer;X= andrew;

?- findall(X, male(X), List).List= [andrew, john, george, greg, adam]?- findall(X, female(X), List).List= [mary, jennifer, eve]?- findall(X, parents(X,adam,eve), List).List= [greg, jennifer, andrew]

findall(X,Term,List).

36

Other examples

?- findall(t(X,Y),parents(X,Y,Z),List).X = _G353Y = _G354Z = _G358List = [t(john, george), t(greg, adam), t(jennifer, adam), t(andrew, adam)] Yes?-

?- findall(t(X,Y),append(X,Y,[a,b]),L).

X = _68

Y = _69

L = [t([],[a,b]),t([a],[b]),t([a,b],[])]

yes

male(andrew).male(john). male(george). male(greg). male(adam). female(mary). female(jennifer). female(eve). parents(john,george,mary).parents(greg,adam,eve).parents(jennifer, adam,eve).parents(andrew, adam,eve).

37

call(Goal) Invoke Goal as a goal.?- call(a=a).

Yes

?- [user].

|: likes(mary,john).

|:

% user compiled 0.02 sec, 152 bytes

Yes

?- call(likes(mary,john)).

Yes

?- call(likes(mary,X)).

X = john

Yes

?- call(likes(X,Y)).

X = mary

Y = john

Yes

?- call(likes(X,mary)).

No

?-

38

?- not(likes(mary,john)).

No

?- not(likes(mary,X)).

No

?- not(likes(mary,george)).

Yes

?-

not(Goal) Succeeds when Goal fails.

Implementation of not using callnot(P):-call(P),!,fail.

not(P).

39

In Prolog, programs and data are made of terms. A term may be: a constant (an atom or a number) 10, ‘abc’, likes, mary a variable X, Y a compound term likes(mary,john), likes(X,Y).

There are several built-in predicates that succeed or fail depending on the type of the term their argument is.

atomic(X) Succeeds if X is bound to an atom, a string or anumber (integer or floating point).

atom(X) Succeeds if X is bound to an atom.

number(X) Succeeds if X is bound to a number (integer or float)

float(X) Succeeds if X is bound to a floating point number

40

?- atom(a).Yes?- atom(X).No?- X=a,atom(X).X = a Yes?- X=a,atomic(X).X = a Yes?- atomic(a).Yes?- atomic(likes).Yes?- atom(likes).Yes

?- atom(1).No?- atom(1.2).No?- atomic(1.2).Yes?- number(5).Yes?- number(1.2).Yes?- float(1).No?- float(1.1).Yes

41

An exampleCounting the number of occurances of an atomcount(Atom,List,N).

count( _ , [ ] , 0 ).count(Atom, [Atom | Tail] , N):- count(Atom,Tail,N1), N is N1 + 1.count(Atom,[ _ | Tail], N):- count(Atom,Tail,N).

?- count(1,[1,2,3,1],N).N = 2 Yes?- count(1,[1,X,Y,Z],N).X = 1Y = 1Z = 1N = 4 % incorrectYes?-

?- L= [1,2,3,X,2],write(L),count(2,L,N).

[1, 2, 3, _G368, 2]

L = [1, 2, 3, 2, 2]

X = 2

N = 3 % incorrect

Yes

?-

42

Solutioncount( _ , [ ] , 0 ).count(Atom,[Head | Tail], N):- atomic(Head),

Atom=Head, count(Atom,Tail,N1), N is N1 + 1;count(Atom,Tail,N).

?- count(1,[1,2,3,1],N).

N = 2

Yes

?- count(1,[1,X,Y,Z],N).

X = _G269

Y = _G272

Z = _G275

N = 1

Yes

?-

?- L= [1,2,3,X,2],write(L),count(2,L,N).

[1, 2, 3, _G371, 2]

L = [1, 2, 3, _G371, 2]

X = _G371

N = 2

Yes

?-