Reputation: 815
I have two problems that have bugged me for hours. connected/2
is supposed to judge whether two people are connected or not; distance/3
is supposed to measure the kinship. But:
true
s infinitely for the query connected(x,y)
;N
for distance(x,y,N)
query. Any suggestions?Here are my facts:
male(ted).
male(barney).
male(ranjit).
male(marshall).
male(tony).
male(swarley).
male(steve).
male(chuck).
male(john).
male(devon).
male(morgan).
female(robin).
female(lily).
female(wendy).
female(stellar).
female(abby).
female(victoria).
female(carina).
female(sarah).
female(ellie).
married(ted, robin).
married(marshall, lily).
married(ranjit, wendy).
married(stellar, tony).
married(steve, carina).
married(sarah, chuck).
married(ellie, devon).
father(ted, barney).
father(ted, ranjit).
father(marshall, wendy).
father(ranjit, stellar).
father(tony, abby).
father(tony, swarley).
father(tony, victoria).
father(steve, chuck).
father(steve, ellie).
father(chuck, john).
father(devon, morgan).
mother(robin, barney).
mother(robin, ranjit).
mother(lily, wendy).
mother(wendy, stellar).
mother(stellar, abby).
mother(stellar, swarley).
mother(stellar, victoria).
mother(carina, chuck).
mother(carina, ellie).
mother(sarah, john).
mother(ellie, morgan).
Now, my predicates:
parent(X,Y) :- father(X,Y).
parent(X,Y) :- mother(X,Y).
son(X,Y) :-
male(X),
parent(Y,X).
daughter(X,Y) :-
female(X),
parent(Y,X).
sibling(X,Y) :-
parent(Z,X),
parent(Z,Y).
cousin(X,Y) :-
parent(Z,X),
parent(W,Y),
parent(G,Z),
parent(G,W).
ancestor(X,Y) :-
parent(X,Z),
ancestor(Z,Y).
ancestor(X,Y) :- parent(X,Y).
notmember(X,[]).
notmember(X,[H|T]) :-
X \= H,
notmember(X,T).
connected(X,Y,_) :- X == Y.
connected(X,Y,Visited) :-
ancestor(X,Z),
notmember(Z,Visited),
connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :-
ancestor(Z,X),
notmember(Z,Visited),
connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :-
sibling(X,Z),
notmember(Z,Visited),
connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :-
married(X,Z),
notmember(Z,Visited),
connected(Z,Y,[Z|Visited]).
connected(X,Y) :- connected(X,Y,[X]).
minimum(X,[X]).
minimum(X,[M,H|T]) :-
M =< H,
minimum(X,[M|T]).
minimum(X,[M,H|T]) :-
M > H,
minimum(X,[H|T]).
distance(X,X,_,0).
distance(X,Y,Visited,N) :-
parent(X,Z),
notmember(Z,Visited),
distance(Z,Y,[Z|Visited],N1),
N is N1+1.
distance(X,Y,Visited,N) :-
parent(Z,X),
notmember(Z,Visited),
distance(Z,Y,[Z|Visited],N1),
N is N1+1.
distance(X,Y,N) :- distance(X,Y,[],N).
EDIT: Thank you, i think i've managed to solve half way through the problems now. Taking @twinterer's advice, I have fixed the predicates like this
connected(X,Y,_) :- X == Y.
connected(X,Y,V) :-
married(X,Z),
notmember(Z,V),
connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
sibling(X,Z),
notmember(Z,V),
connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
parent(X,Z),
notmember(Z,V),
connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
parent(Z,X),
notmember(Z,V),
connected(Z,Y,[Z|V]),!.
connected(X,Y) :- connected(X,Y,[X]).
minimum(X,[X]).
minimum(X,[M,H|T]) :-
M =< H,
minimum(X,[M|T]).
minimum(X,[M,H|T]) :-
M > H,
minimum(X,[H|T]).
count(X,[],0).
count(X,[X|T],N) :-
count(X,T,N1),
N is N1+1.
count(X,[H|T],N) :-
X \== H,
count(X,T,N1),
N is N1.
distance(X,X,Visited,0) :-
count(X,Visited,N),
N =< 1, !.
distance(X,Y,Visited,N) :-
parent(X,Z),
(notmember(Z,Visited)->
distance(Z,Y,[Z|Visited],N1),
N is N1+1
;
fail
),!.
distance(X,Y,Visited,N) :-
parent(Z,X),
(notmember(Z,Visited)->
distance(Z,Y,[Z|Visited],N1),
N is N1+1
;
fail
),!.
distance(X,Y,N) :-
findall(N1,distance(X,Y,[X],N1),L),!,
minimum(N,L),!.
But there is a new set of problems now
distance(X,y,n)
connected(X,y)
return duplicate resultsI think removing duplicate results can be achieved by using that findall/3
predicate,
but I am clueless as to how I can actually implement it.
Upvotes: 4
Views: 2387
Reputation: 2436
1) I don't think that you end in an infinite loop, but you let your program explore all ways that two people are connected, which will be a very large number. Since you probably are only interested in whether they are connected at all, you should add cuts at the end of the connected/3
clauses that will prevent backtracking once you have successfully determined that two people are connected, e.g.:
connected(X,Y,_) :- X == Y,!.
connected(X,Y,Visited) :-
ancestor(X,Z),
notmember(Z,Visited),
connected(Z,Y,[Z|Visited]),!.
...
2) I didn't get infinitely increasing values for N when I tested your code, but still the distance/3
predicate would determine different paths how two people are connected. Depending on the people, the minimum distance would not be the first to be computed. I would change the definition of distance/3
to something like this:
distance(X,Y,N) :-
findall(N0, distance(X,Y,[],N0), Ns), !,
minimum(N, Ns).
This is reusing your minimum/2
predicate. Note that you should add cuts to the first two clauses of this predicate in order to avoid spurious choicepoints.
Regarding your additional questions:
3) you have to distinguish between looking for the smallest N and looking for people matching degree N:
distance(X,Y,N) :-
nonground(N),!,
findall(N0, distance(X,Y,[],N0), Ns), !,
minimum(N, Ns).
distance(X,Y,N) :-
distance(X,Y,[X],N).
Also, you need to remove the cuts that you added to distance/4
. This is different from the cuts added to connected/3
!
There's probably a better way that avoids distinguishing between these two modes, but all I can think of at the moment is using some kind of breadth-first-search (in order to guarantee the minimum degree)...
4) I don't get duplicate answers for queries like ?- connected(X,victoria).
Do you have an example?
Upvotes: 5