My aim is to break the given list of lists in such a way that I am able to access the individual lists by the same name. I have the following list:-
mylist([[1,2],[2,3],[3,4],[4,6]]).
I want to break the list into [1,2], [2,3], [3,4], [4,6] so that I can access the items (like [1,2]) individually.
For that, can I create a new fact from the separated list elements? I am able to separate the elements into individual lists. But, I want to convert those individual lists into facts. Like:-
mylist([[1,2],[2,3],[3,4],[4,6]]).
should become the following:-
node([1,2]).
node([2,3]).
node([3,4]).
node([4,6]).
And then I should be able to access each and every list using "node".
The other answer is fine (esp. the forall solution), but here is what you could do if you knew your list at compile time, and wanted to add the node/1 facts to the database at compile time.
This code is simplified from the example available at the very bottom of this page:
In your file (I will call it nodes.pl):
term_expansion(nodes_list(NL), Nodes) :-
maplist(to_node, NL, Nodes).
to_node(X, node(X)).
nodes_list([[1,2],[2,3],[3,4],[4,6]]).
When I consult the file, I get:
?- [nodes].
true.
?- listing(node).
node([1, 2]).
node([2, 3]).
node([3, 4]).
node([4, 6]).
true.
Two details:
The expanded predicate, here nodes_list/1, is not going to be in the database.
The clause of term_expansion/2 must come before the definition of nodes_list/1 in the source file.
A great answer (maybe the best) recommended(in the comments) by CapelliC is:
?-forall(member(X,[[1,2],[2,3],[3,4]]),assertz(node(X))).
You could also write:
my_list(L):- member(X,L),assertz(node(X)).
Example:
?- my_list([[1,2],[2,3],[3,4]]).
true ;
true ;
true.
?- node([1,2]).
true ;
false.
Another way thanks to CapelliC's recommendation would be:
?- maplist(X>>assertz(node(X)), [[1,2],[2,3],[3,4]]).
Related
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 am new to prolog and I am trying to create a predicate and am having some trouble.
I have a list of cities that are connected via train. They are connected via my links/2 clause.
links(toronto, ajax).
links(toronto, markham).
links(toronto, brampton).
links(brampton, markham).
links(markham, ajax).
links(brampton, mississauga).
links(mississauga, toronto).
links(mississuaga, oakville).
links(oakville, st.catharines).
links(oakville, hamilton).
links(hamilton, st.catharines).
I am writing a predicate called addnewcities which will take a list of cities and then return a new list containing the original list, plus all the cities that are directly connected to each of the cities in the original list.
Here is a (rough looking) visual representation of the links.
If my input list was [toronto] I want my output to be (order doesnt matter) [ajax,markham,brampton,mississauga,toronto].
If input was [oakville,hamilton] I want the output to be [mississauga,st.catharines,oakville,hamilton].
Here is my predicate so far.
addnewcities([],_).
addnewcities([CitiesH|Tail],Ans):- directer(CitiesH,Ans2), merger(Ans2,[CitiesH],Ans), addnewcities(Tail,Ans).
directer/2 takes a city and saves a list containing all the directly connected cities in the second arg.
merger/3 just merges two lists making sure there are no duplicates in the final list.
When my input is a list with one element ie [toronto] it works!
But when I have a list with multiple elements [toronto,ajax] it says "false" every time.
I'm pretty sure my issue is that when it recurses for the second time, merge is what says its false. I just don't know how to get around this so that my list can keep being updated instead of being checked if true or false.
Any help is appreciated!
this query uses library support to solve the problem:
addcities(Cs, L) :-
setof(D, C^(member(C,Cs), (C=D;link(C,D);link(D,C))), L).
This should work for what you want:
addcities(A,B):-
addcitiesaux(A,[],B).
addcitiesaux([],X,X).
addcitiesaux([X|Xs],L,R):-
link(X,A),
\+ member(A,L),
!,
addcitiesaux([X|Xs],[A|L],R).
addcitiesaux([X|Xs],L,R):-
link(A,X),
\+ member(A,L),
!,
addcitiesaux([X|Xs],[A|L],R).
addcitiesaux([X|Xs],L,R):-
addcitiesaux(Xs,[X|L],R).
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 am attempting to learn the basics of Prolog for a class. I'm running into the seemingly simple problem of not being able to store a list within a rule and retrieve it for usage in other clauses. For example:
% These are the contents of the pl file I want to consult
% Numbers I want to process
inputList([3,2,1,0]).
% Prints out the contents of a list
printList([First | Tail]) :-
write(First),nl,
printList(Tail).
What I want to do is call the following within Prolog:
?- inputList(X).
?- printList(X).
The goal is to avoid constantly entering long lists into the Prolog interpreter and instead store them in the .pl file. However, entering the commands above results in the list not being properly checked against the given clause. How can this be accomplished, preferably using the structure above to store a list {listContents([a,b,c,d]).}?
I think you need to modify your call in Prolog to
?- inputList(X), printList(X).
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).