:- use_module(library(lists)).
friendof(mark, frank).
friendof(mark, sylvie).
friendof(john, sylvie).
friendof(marie, chris).
symmetrical(X, Y) :- friendof(X, Y).
symmetrical(X, Y) :- friendof(Y, X).
:- dynamic(friendlist/1).
friends(Friend, LL) :-
asserta(friendlist([])),
friends2(Friend)
retract(friendlist(LL)).
friends2(F) :-
What I have to do is, using friends as it is (so, using friends2/1), get all the friends of mark as such:
?-friends (mark, X). would give this list, (take note of the order in which they appear):
X = [sylvie, frank]
I was told many things, of which I should be doing
symmetrical(..),retract(..),assert(..),fail.
But it's just not working to put the list back in friendlist from friends2/1.
Can someone show me how it's done?
It's really important!
Thanks!
EDIT1:
These are some attempts at getting it to work.
friends2(F) :- findall(B,symmetrical(A,B),B).
symmetrical(F,Y),retract(friendlis([Y:_])),assert(...
%symmetrical(..),retract(..),assert(..),fail.
Maybe you're doing it much more complex that actually is.
Most importantly: assert/retract should not be used for computing temporary relations in DB. Use them only to change knowledge - that is, well structured and meaningful relations that need to be persistent for the duration of the program.
For your task, try
friends_of(Person, Friends) :-
setof(Friend, (friendof(Person, Friend) ; friendof(Friend, Person)), Friends).
edit Of course, I don't understand why you need to do it so much more complicated... maybe you need to show you are able to issue modification to DB ? I fear you will discover that such problem (keeping a coherent db state) is much more difficult than foreseen, and learning to assert/retract useless relations will not help you much. Anyway...
friends(Friend, LL) :-
asserta(friendlist([])),
friends2(Friend),
retract(friendlist(LL)), !.
friends2(F) :-
( friendof(F, P) ; friendof(P, F) ),
retract(friendlist(L)),
assert(friendlist([P|L])),
fail.
friends2(_).
could work for you...
Related
I'm unable to figure out why my code isn't working despite looking through answers to similar questions. I'm too new at Prolog to properly name things, but I hope you can see what I'm trying to get at.
I am defining a timetable roughly based on this program and am struggling with getting a list of the Classes that Mike teaches for a given result (Next step will be to declare that only results where both Mike and Phil teach 2 should be returned, but I want to work through it so that I can see and understand what's going on).
I imagine this should be simple but any combinations of the addToList(List,C) predicate never work. I know there is the append predicate but I hear it's inefficient, and would like to learn the 'raw' way. I don't know how many variations I've attempted and can't get my head around the way Prolog works in this regard and don't know on what level I'm going wrong - it's all a bit of a black box mystery working with it.
var program =
:- use_module(library(lists)).
prefers(may,a).
prefers(may,b).
prefers(may,c).
prefers(may,d).
prefers(bob,a).
prefers(bob,b).
prefers(bob,c).
prefers(pete,a).
prefers(pete,b).
prefers(pete,c).
prefers(pete,d).
prefers(tom,a).
prefers(tom,b).
prefers(tom,c).
prefers(tom,d).
teacher_pref(mike,a).
teacher_pref(mike,b).
teacher_pref(mike,c).
teacher_pref(phil,b).
teacher_pref(phil,c).
teacher_pref(phil,d).
addToList([C|List],C):- addToList(List,C).
timetable([a,[C1,S1,T1],b,[C2,S2,T2],c,[C3,S3,T3],d,[C4,S4,T4]],List1):-
teacher_pref(T1,C1),
teacher_pref(T2,C2),
teacher_pref(T3,C3),
teacher_pref(T4,C4),
prefers(S1,C1),
prefers(S2,C2),
S1\\=S2,
prefers(S3,C3),
S1\\=S3,
S2\\=S3,
prefers(S4,C4),
S1\\=S4,
S2\\=S4,
S3\\=S4,
addToList(List1,C):-
teacher_pref(mike,C).
session.consult( program );
session.query('timetable([C1,[a,S1,T1],C2,[b,S2,T2],C3,[c,S3,T3,L3],C4,[d,S4,T4]],List1).')
If I understand correctly you have:
teacher_pref(mike,a).
teacher_pref(mike,b).
teacher_pref(mike,c).
And you want to get a list of these classes, which would be:
[a, b, c]
In Prolog we have some higher-order predicates that are for occasions like this:
% (What to find, the goal to call, all the results)
?- findall(Class, teacher_pref(mike, Class), Classes).
Classes = [a, b, c].
In the Tau-Prolog docs they're under All Solutions, in SWI-Prolog there's a couple more.
To make this into a more generic predicate:
teacher_prefs(Teacher, Prefs) :-
findall(Pref, teacher_pref(Teacher, Pref), Prefs).
I have an array L of some type, I'm trying to extract the data to an array, for example:
L=[day(sunday),day(monday)]
to
Target=[sunday,monday]
Tried using forall and searched for related questions on Prolog lists.
extract_data_to_list(L,Target) :-
member(day(Day),L),
length(L, L1),
length(Target, L1),
member(Day,Target).
Current output:
?- extract_data_to_list([day(sunday),day(monday)],Target).
Target = [sunday, _5448] ;
Target = [_5442, sunday] ;
Target = [monday, _5448] ;
Target = [_5442, monday].
Desired output:
?- extract_data_to_list([day(sunday),day(monday)],Target).
Target=[sunday,monday]
This is an ideal problem for maplist:
day_name(day(DayName), DayName).
dates_daylist(Dates, DayList) :-
maplist(day_name, Dates, DayList).
Maplist applies day_name to each corresponding pair of elements in Dates and DayList.
This is an ideal problem for library(lambda) for SICStus|SWI:
maplist(\day(N)^N^true, Dates, Daylist).
I have a couple other ways you can do this, just in case you're wondering.
?- findall(D, member(day(D), [day(monday), day(tuesday)]), Days).
Days = [monday, tuesday].
The trick here is that you can use findall/3 to drive a simple loop, if the Goal (argument 2) uses member/2. In this case, we're unifying day(D) with each item in the list; no further work really needs to happen besides the unification, so we're able to "tear off the wrapping" just with member/2 but you could drive a more complex loop here by parenthesizing the arguments. Suppose you wanted to change day to day-of-week, for instance:
?- findall(DoW, (member(day(D),
[day(monday), day(tuesday)]), DoW=day_of_week(D)),
Days).
Days = [day_of_week(monday), day_of_week(tuesday)].
Making the goal more complex works, in other words, as long as you parenthesize it.
The second trick is specific to SWI-Prolog (or Logtalk, if you can use that), which is the new library(yall):
?- maplist([Wrapped,Day]>>(Wrapped=day(Day)),
[day(monday),day(tuesday)], X).
X = [monday, tuesday].
library(yall) enables you to write anonymous predicates. [Wrapped,Day]>>(Wrapped=day(Day)) is sort of like an inline predicate, doing here exactly what #lurker's day_name/2 predicate is doing, except right inside the maplist/3 call itself without needing to be a separate predicate. The general syntax looks something like [Variables...]>>Goal. This sort of thing was previously available as library(lambda) and has been a feature of Logtalk for many years.
I'm struggling with my project in Prolog. My problem is, given a file containing the traffic lines in the form of line(NameOfLine, Type, ListOfStations). For example:
line(m1, train, [a,b,c,d,e])
line(m2, train, [h,e,j,i])
...
where e is the intersecting station of two line (The real line file is very complicated and containing a dozen of lines with hundreds of stations). I have to do the classic graph question like find a routing, calculate the cost, etc.
I knew that I have to build a graph undirected before start, so I've tried something like this:
adjacent(X,Y,[X,Y|_]).
adjacent(X,Y,[_|T]) :- adjacent(X,Y,T).
find_edge(LArret) :- forall(adjacent(X, Y, LArret), assert(edge(X,Y))).
connected(X,Y) :- edge(X,Y) ; edge(Y,X).
graph :-
forall(ligne(_,_,L), find_edge(L)).
But I didn't get all the edges as expected. Can you guys give me some advice for that? Or was I wrong at very first for solving this kind of problem?
Supplementary question:
Thanks for the solution proposed and i've finally succeed to define the edges. Then i tried this algorithm classic to find a path betwwen A and B, but sometimes the search doesn't seem to end and sometimes the program stuck in searching.
connected(X,Y) :- edge(X,Y) ; edge(Y,X).
path(A,B,Path) :-
travel(A,B,[A],Q),
reverse(Q,Path).
travel(A,B,P,[B|P]) :-
connected(A,B).
travel(A,B,Visited,Path) :-
connected(A,C),
C \== B,
\+member(C,Visited),
travel(C,B,[C|Visited],Path).
I think the reason could be a loop in the graph or a infinite loop in seraching, how can I avoid this kind of problem but still find all the path possible?
I was able to use the following code with your line/3 definitions to create your list of edge/2 definitions:
adjacent(X,Y, [X,Y|_]).
adjacent(X,Y, [_|T]) :- adjacent(X,Y, T).
adjacent_lines(X,Y) :- line(_,_,L), adjacent(X,Y, L).
:- forall(adjacent_lines(X, Y), assert(edge(X, Y))).
It created all applicable and valid edge/2 facts as specified by the line/3 facts.
(at least, with SWI-Prolog)
So I am trying to use a recursive method to find a path between two people. Here is the quick background:
I define some facts in(X,Y). That show who is related, ie. in(person1,project1), in(person2,project1), etc etc. Now any two people are related if they were in the same project as each other, or there is a linking path of people between them. For example p1 worked on A p2 worked on A and B and p3 worked on B therefore there is a path from p1 to p3 through p2. These paths can be any length.
I am trying to solve this recursively (don't see any other way), but there is an annoying problem:
related(A,B) :-
in(A,X),
in(B,X),
not(A=B).
chain(A,B) :-
related(A,B).
chain(A,B) :-
related(A,Y),
chain(Y,B).
The issue is that the path can repeat itself. It can go from p1 to p2 back to p1 endless times. A person should not be in the path more than 1 time.
I tried to fix this with a list that I add to. If a person is already in the list, they can't be added again:
related(A,B,L) :-
in(A,X),
in(B,X),not(A=B).
chain(A,B,L) :-
related(A,B,L).
chain(A,B,L) :-
related(A,Y,L),
not(member(Y,L)),
append(L,[Y],Q),
chain(Y,B,Q).
And it sort of worked, but caused a ton of random errors, repeating some people multiple times, some only once, and then failing. Does this approach look right? Am I totally using lists wrong?
Thank You.
First improvement. Are you looking for all the chains of relations or do you want to check if there is one chain of relation? In the first case add a cut.
chain(A,B) :-
related(A,B), !.
chain(A,B) :-
related(A,Y),
chain(Y,B).
In the second case, Prolog does exactly what it's asked to do, that is finding all the possible chains.
Please post a query that causes problems so that we can reason together on it and improve the solution.
Here is an alternative way, maybe less efficient but rather general, based on fixpoint computation.
connected(Found, Connected) :-
collect(Found, [], Ext),
( Ext == Found
-> Connected = Found
; connected(Ext, Connected)
).
collect([], Set, Set).
collect([E|Es], Set, Fix) :-
extend(E, Set, Ext),
collect(Es, Ext, Fix).
extend(E, Set, Ext) :-
directly(E, DirectConn),
ord_union(DirectConn, Set, Ext).
directly(A, DirectConn) :-
setof(B, P^(in(A, P), in(B, P)), DirectConn).
We must call connected(Found, Connected) with a sorted list, and it loops until the set cannot be extended. For instance, with this test data
in(anna, project1).
in(bob, project1).
in(bob, project2).
in(chris, project2).
in(dan, project3).
?- connected([bob],L).
L = [anna, bob, chris].
?- connected([dan],L).
L = [dan].
I allow on purpose directly/2 get identity, i.e.
?- directly(X,Y).
X = anna,
Y = [anna, bob] ;
...
X = dan,
Y = [dan].
I think I was never perfectly clear enough, but I ended up solving this myself. I put the code below.
What really mattered was an effective notchain predicate and then making sure I did the appends correctly. I also created a notsame predicate to replace the not(A=B). The code is below. Most of the answer was making sure that there were [] around what was being appended to the list. Not having the correct [] around what was being appended caused errors.
notchain(X,L) :-
member(X,L),!,fail.
notchain(X,L).
And then:
chain(A,B,L) :-
related(A,B),
append(L,[A],Q),
append(Q,[B],Z),
write(final),writeln(Z).
chain(A,B,L) :-
notchain(A,L),
append(L,[A],Q),
related(A,Y),
chain(Y,B,Q).
This was used in related:
notsame(A,B) :-
(A=B),!,fail.
notsame(A,B).
I have three types of facts:
album(code, artist, title, date).
songs(code, songlist).
musicians(code, list).
Example:
album(123, 'Rolling Stones', 'Beggars Banquet', 1968).
songs(123, ['Sympathy for the Devil', 'Street Fighting Man']).
musicians(123, [[vocals, 'Mick Jagger'], [guitar, 'Keith Richards', 'Brian Jones']].
I need to create these 4 rules:
together(X,Y) This succeeds if X and Y have played on the same album.
artistchain(X,Y) This succeeds if a chain of albums exists from X to Y;
two musicians are linked in the chain by 'together'.
role(X,Y) This succeeds if X had role Y (e.g. guitar) ever.
song(X,Y) This succeeds if artist X recorded song Y.
Any help?
I haven't been able to come up with much but for role(X,Y) I came up with:
role(X,Y) :- prole(X,Y,musicians(_,W)).
prole(X,Y,[[Y|[X|T]]|Z]).
prole(X,Y,[[Y|[H|T]]|Z]) :- prole(X,Y,[[Y|T]|Z]).
prole(X,Y,[A|Z]) :- prole(X,Y,Z).
But that doesn't work. It does work if I manually put in a list instead of musicians(_,W) like [[1,2,3],[4,5,6]].
Is there another way for me to insert the list as a variable?
As for the other rules I'm at a complete loss. Any help would really be appreciated.
You have a misconception about Prolog: Answering a goal in Prolog is not the same as calling a function!
E.g.: You expect that when "role(X,Y) :- prole(X,Y,musicians(_,W))." is executed, "musicians(_,W)" will be evaluated, because it is an argument to "prole". This is not how Prolog works. At each step, it attempts to unify the goal with a stored predicate, and all arguments are treaded either as variables or grounded terms.
The correct way to do it is:
role(X,Y) :- musicians(_, L), prole(X,Y,L).
The first goal unifies L with a list of musicians, and the second goal finds the role (assuming that the rest of your code is correct).
Little Bobby Tables is right, you need to understand the declarative style of Prolog. Your aim is to provide a set of rules that will match against the set of facts in the database.
Very simply, imagine that I have the following database
guitarist(keith).
guitarist(jim).
in_band('Rolling Stones', keith).
in_band('Rolling Stones', mick).
Supposed I want to find out who is both a guitarist and in the Rolling Stones. I could use a rule like this
stones_guitarist(X):-
guitarist(X),
in_band('Rolling Stones', X).
When a variable name is given within a rule (in this case X) it holds its value during the rule, so what we're saying is that the X which is a guitarist must also be the same X that is in a band called 'Rolling Stones'.
There are lots of possible ways for you to arrange the database. For example it might be easier if the names of the musicians were themselves a list (e.g. [guitar,[keith,brian]]).
I hope the following example for song(X,Y) is of some help. I'm using Sicstus Prolog so import the lists library to get 'member', but if you don't have that it's fairly easy to make it yourself.
:- use_module(library(lists)).
song(ARTIST,SONG):-
album(CODE,ARTIST,_,_),
songs(CODE,TRACKS),
member(SONG,TRACKS).