prolog -cpt114 - week3

PROLOGWeek 3

OBJECTIVES FOR THIS SESSION

The Lists in Prolog

Tracing

SIMPLE LISTS IN PROLOG

HOW TO MAKE A LIST IN PROLOG

pam tom

bob liz

pat

jim

ann

HOW TO MAKE A LIST IN PROLOG

pam tom

bob liz

pat

jim

ann

parent(pam, bob).parent(tom, bob).parent(tom, liz).parent(bob, pat).parent(pat, jim).female(pam).female(liz).female(pat).male(tom).male(bob).male(jim).

HOW TO MAKE A LIST IN PROLOGpam tom

bob liz

pat

jim

ann

HOW FINDALL WORKS

findall(X,predecessor(pam,X),Preds).

Output listThe 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] ]

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 | [ ] ]

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 | [ ] ]

MEMBERSHIP

X is a member of L either:1) X is the head of L, or2) 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

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]).

MEMBERSHIP

Parents = [pam, tom, tom, bob, pat].

Is pam a member of HEAD of Parents????

member(pam,[pam,Parents]).

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

HEAD AND TAIL EXAMPLE 1

[mia, vincent, jules, yolanda]

Head: Tail:

HEAD AND TAIL EXAMPLE 1

[mia, vincent, jules, yolanda]

Head: miaTail:

HEAD AND TAIL EXAMPLE 1

[mia, vincent, jules, yolanda]

Head: miaTail: [vincent, jules, yolanda]

HEAD AND TAIL EXAMPLE 2

[[ ], dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]

Head: Tail:

HEAD AND TAIL EXAMPLE 2

[[ ], dead(z), [2, [b,c]], [ ], Z, [2, [b,c]]]

Head: [ ]Tail:

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]]]

HEAD AND TAIL EXAMPLE 3

[dead(z)]

Head: Tail:

HEAD AND TAIL EXAMPLE 3

[dead(z)]

Head: dead(z) Tail:

HEAD AND TAIL EXAMPLE 3

[dead(z)]

Head: dead(z) Tail: [ ]

THE BUILT-IN OPERATOR |

?- [Head|Tail] = [mia, vincent, jules, yolanda].

Head = miaTail = [vincent,jules,yolanda]yes

?-

THE BUILT-IN OPERATOR |

?- [X|Y] = [mia, vincent, jules, yolanda].

X = miaY = [vincent,jules,yolanda]yes

?-

THE BUILT-IN OPERATOR |

?- [X|Y] = [ ].

no

?-

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

?-

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 = miaX2 = vincentX3 = marsellusX4 = jodyTail = [yolanda]yes

?-

ANONYMOUS VARIABLES

There is a simpler way of obtaining only the information we want:

?- [ _,X2, _,X4|_ ] = [mia, vincent, marsellus, jody, yolanda].X2 = vincentX4 = jodyyes

?-

The underscore is the anonymous variable

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

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

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?-

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(yolanda,[yolanda,trudy,vincent,jules]).

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(yolanda,[yolanda,trudy,vincent,jules]).yes?-

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(vincent,[yolanda,trudy,vincent,jules]).

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(vincent,[yolanda,trudy,vincent,jules]).yes?-

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(zed,[yolanda,trudy,vincent,jules]).

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(zed,[yolanda,trudy,vincent,jules]).no?-

MEMBER/2

member(X,[X|T]).member(X,[H|T]):- member(X,T).

?- member(X,[yolanda,trudy,vincent,jules]).

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

CONCATENATION OF LISTS

For concatenation we first define our rule.

Step 1: if the first list is emptyconc([],L,L).Step2: if first list is not emptyconc(L1,L2,L3)=> conc([X|L1],L2,[X|

L3]):-

conc(L1,L2,L3).

L1X L2

[X|L1]

L3

L3X

L3

SAMPLE OF CONCATENATION

conc([a],[c,d],L).L = [a, c, d].

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=7yes?-

SIMPLE TRACE

SIMPLE TRACE

SIMPLE TRACE

COME OUT OF TRACE

GUI TRACER

GUI TRACER