1. Week 3
PROLOG

2. OBJECTIVES FOR THIS SESSION
The Lists in Prolog
Tracing

3. SIMPLE LISTS IN PROLOG

4. HOW TO MAKE A LIST IN PROLOG
pam tom
bob liz
ann pat
jim

5. HOW TO MAKE A LIST IN PROLOG
parent(pam, bob).
pam tom
parent(tom, bob).
parent(tom, liz).
parent(bob, pat).
liz parent(pat, jim).
bob
female(pam).
female(liz).
female(pat).
ann pat male(tom).
male(bob).
male(jim).
jim

6. HOW TO MAKE A LIST IN PROLOG
pam tom
bob liz
ann pat
jim

7. HOW FINDALL WORKS
findall(X,predecessor(pam,X),Preds).
Output list
The query variable

8. EXAMPLE1
The following conversation with Prolog:
?- Hobbies1 = [tennis, music],
Hobbies2 = [skiing, food],
L = [ann, Hobbies1, tom, Hobbies2].
Hobbies1 = [tennis, music]
Hobbies2 = [skiing, food]
L = [ann, [tennis, music], tom, [skiing, food] ]
8

9. EXAMPLE2
Treat the whole tail as a single object:
L = [a, b, c]
Tail = [b, c]
L = .(a, Tail)
L = [a | Tail]
[a, b, c] = [a | [b, c] ] = [a, b | [c] ] = [a, b, c | [ ] ]

10. EXAMPLE2
Treat the whole tail as a single object:
L = [a, b, c]
Tail = [b, c]
L = .(a, Tail)
L = [a | Tail]
[a, b, c] = [a | [b, c] ] = [a, b | [c] ] = [a, b, c | [ ] ]

11. MEMBERSHIP
 X is a member of L either:
1) X is the head of L, or
2) X is a member of the tail of L.
member(X, L)
 Can be written in two clauses:
member(X, [X | Tail] ). %simple fact
member(X, [Head | Tail] ) :-
member(X, Tail). %rule
 Example:
member(b, [a,b,c] ). %true
member(b, [a,[b,c]] ). %false, the tail is [b,c]
member([b,c],[a,[b,c]] ). %true

12. MEMBERSHIP
Simple other tests:
findall(X,parent(X,_),Parents).
Parents = [pam, tom, tom, bob, pat].
Is pam a member of Parents????
member(pam, [pam, tom,bob, pat]).

13. MEMBERSHIP
Parents = [pam, tom, tom, bob, pat].
Is pam a member of HEAD of Parents????
member(pam,[pam,Parents]).

14. HEAD AND TAIL
 A non-empty list can be thought of as
consisting of two parts
 The head
 The tail
 The head is the first item in the list
 The tail is everything else
 The tail is the list that remains when we take the
first element away
 The tail of a list is always a list

15. HEAD AND TAIL EXAMPLE 1
 [mia, vincent, jules, yolanda]
Head:
Tail:

16. HEAD AND TAIL EXAMPLE 1
 [mia, vincent, jules, yolanda]
Head: mia
Tail:

17. HEAD AND TAIL EXAMPLE 1
 [mia, vincent, jules, yolanda]
Head: mia
Tail: [vincent, jules, yolanda]

18. HEAD AND TAIL EXAMPLE 2
 [[ ], dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]
Head:
Tail:

19. HEAD AND TAIL EXAMPLE 2
 [[ ], dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]
Head: [ ]
Tail:

20. HEAD AND TAIL EXAMPLE 2
 [[ ], dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]
Head: [ ]
Tail: [dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]

21. HEAD AND TAIL EXAMPLE 3
 [dead(z)]
Head:
Tail:

22. HEAD AND TAIL EXAMPLE 3
 [dead(z)]
Head: dead(z)
Tail:

23. HEAD AND TAIL EXAMPLE 3
 [dead(z)]
Head: dead(z)
Tail: [ ]

24. THE BUILT-IN OPERATOR |
?- [Head|Tail] = [mia, vincent, jules, yolanda].
Head = mia
Tail = [vincent,jules,yolanda]
yes
?-

25. THE BUILT-IN OPERATOR |
?- [X|Y] = [mia, vincent, jules, yolanda].
X = mia
Y = [vincent,jules,yolanda]
yes
?-

26. THE BUILT-IN OPERATOR |
?- [X|Y] = [ ].
no
?-

27. THE BUILT-IN OPERATOR |
?- [X,Y|Tail] = [[ ], dead(z), [2, [b,c]], [], Z, [2, [b,c]]] .
X=[]
Y = dead(z)
Z = _4543
Tail = [[2, [b,c]], [ ], Z, [2, [b,c]]]
yes
?-

28. ANONYMOUS VARIABLE
 Suppose we are interested in the
second and fourth element of a list
?- [X1,X2,X3,X4|Tail] = [mia, vincent, marsellus, jody, yolanda].
X1 = mia
X2 = vincent
X3 = marsellus
X4 = jody
Tail = [yolanda]
yes
?-

29. ANONYMOUS VARIABLES
?- [ _,X2, _,X4|_ ] = [mia, vincent, marsellus, jody, yolanda].
X2 = vincent
X4 = jody
yes
?-
 There is a simpler way of obtaining only
the information we want:
The underscore is the anonymous
variable

30. THE ANONYMOUS VARIABLE
 Is used when you need to use a variable, but
you are not interested in what Prolog
instantiates it to
 Each occurrence of the anonymous variable
is independent, i.e. can be bound to
something different

31. MEMBER
 One of the most basic things we would like to
know is whether something is an element of
a list or not
 So let`s write a predicate that when given a
term X and a list L, tells us whether or not X
belongs to L
 This predicate is usually called member/2

32. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?-

33. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(yolanda,[yolanda,trudy,vincent,jules]).

34. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(yolanda,[yolanda,trudy,vincent,jules]).
yes
?-

35. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(vincent,[yolanda,trudy,vincent,jules]).

36. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(vincent,[yolanda,trudy,vincent,jules]).
yes
?-

37. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(zed,[yolanda,trudy,vincent,jules]).

38. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(zed,[yolanda,trudy,vincent,jules]).
no
?-

39. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(X,[yolanda,trudy,vincent,jules]).

40. MEMBER/2
member(X,[X|T]).
member(X,[H|T]):- member(X,T).
?- member(X,[yolanda,trudy,vincent,jules]).
X = yolanda;
X = trudy;
X = vincent;
X = jules;
no

41. CONCATENATION OF LISTS
For concatenation we first define our rule.
Step 1: if the first list is empty
conc([],L,L).
Step2: if first list is not empty
conc(L1,L2,L3)=> conc([X|L1],L2,[X|L3]):-
conc(L1,L2,L3).

42. [X|L1]
X L1 L2
L3
X L3
L3

43. SAMPLE OF CONCATENATION
conc([a],[c,d],L).
L = [a, c, d].

44. LENGTH OF A LIST IN PROLOG
len([],0). %% not a Prolog native predicate!
len([_|L],N):-
len(L,X),
N is X + 1.
?- len([a,b,c,d,e,[a,x],t],X).
X=7
yes
?-

45. SIMPLE TRACE

46. SIMPLE TRACE

47. SIMPLE TRACE

48. COME OUT OF TRACE

49. GUI TRACER

50. GUI TRACER