1. 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= c X= ada; X= haskell; No Recursion and Lists

2. 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. 3 asserta/1. assertz/1 assert/1 retract/1 retractall/1 Example: assertz(fib(N,F)).

4. 4 :-dynamic fib fib(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. 5 F(6) + f(5)f(4) + f(3) f(2) f(2) f(1) + 1 + f(4) f(3) + f(3) f(2) 1 + f(2) f(1) + 1 1 1 1 1

6. 6 asserta/1. assertz/1 assert/1 retract/1 retractall/1 Example: assertz(fib(N,F)).

7. 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. 8 leng([ ], 0). leng([H|T], Len) :- leng(T, Len1), Len is Len1 + 1. Finding the length of a list

9. 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. 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. 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. 12 I/O

13. 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 = massey List = [109, 97, 115, 115, 101, 121]

14. 14 ?- write('hello world'). hello world Yes ?- 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'). abc Yes

15. 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: massey X = massey A predicate for printing a list

16. 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 = jennifer Y = greg ; jennifer is andrew`s sister. X = jennifer Y = andrew

17. 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. john fruit book No ?- likes(mary,X),write(X),nl,fail ; true. john Fruit OR book X = _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. 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'). john fruit book reached the OR part X = _G471 Yes

19. 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: 32 N = 2 P5 = 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. 20 File I/O

21. 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. 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. 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. 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. 25 Cut: ! Eliminates choices Always 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. 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. 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. 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. 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]) ? creep member succeeded member1(X,[X|_]):-!. member1(X,[_|T]):-member1(X,T). Fail: (7) member1(1, [1, 2, 3, 4]) ? creep No ?-

30. 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. 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. 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. 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]) ? creep L = [1, 2, 3, 4] ; Fail: (6) apend([], [1, 2, 3, 4], [1, 2, 3, 4]) ? creep No 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]) ? creep L = [1, 2, 3, 4] ; Fail: (6) apend([], [1, 2, 3, 4], _G409) ? creep No

34. 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. 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. 36 Other examples ?- findall(t(X,Y),parents(X,Y,Z),List). X = _G353 Y = _G354 Z = _G358 List = [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. 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. 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 call not(P):-call(P),!,fail. not(P).

39. 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 a number (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. 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. 41 An example Counting the number of occurances of an atom count(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 = 1 Y = 1 Z = 1 N = 4 % incorrect Yes ?- ?- 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. 42 Solution count( _, [ ], 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 ?-