Trying to implement following predicate, which takes on input
list of lists - one list = one tested graph T (as edges)
graph G itself (as edges)
and tests whether T includes all of the vertices of G. If it does(true) should return it. Basically it's list filtering.
What I have so far is:
covb([],G).
covb([H|R],G) :-
isItCov(G,H), covb(R,G).
isItCov([],H).
isItCov([V-W|R],H) :-
vertex(V,H),
vertex(W,H),
isItCov(R,H).
vertex(V,[V-_|G]).
vertex(V,[_-V|G]).
vertex(V,[_|G]):- vertex(V,G).
For covb([[a-c,c-b,c-d]],[a-b,a-c,a-d,c-d,b-c]) works fine(true). For covb([[a-c]],[a-b,a-c,a-d,c-d,b-c]) works fine too (false). I got an issue while I call it with multiple lists like covb([[a-c,c-b,c-d],[a-c]],[a-b,a-c,a-d,c-d,b-c]). which should work only for the first one.
I actually got two questions -
Why does it work for one list only?
I want to return items of lists of lists which passed the condition and returned true (that's the filtering part). How should I do that?
First of all, your program has a number of singleton variable warnings. Do not ignore singleton variable warnings. They can hide real bugs. Also, since more experienced Prolog users here know that programs with singleton variable warnings are not even worth running, they will (a) just see the warnings and decide that they are no longer interested in trying to help you, or (b) fix the warnings on their side, but then by definition they will be working on a program that is no longer the program you posted!
Now for your questions.
Why does it work for one list only?
It's really not clear what you are asking here, or just above with "covb([[a-c,c-b,c-d],[a-c]],[a-b,a-c,a-d,c-d,b-c]). which should work only for the first one."
This query does fail:
?- covb([[a-c,c-b,c-d],[a-c]],[a-b,a-c,a-d,c-d,b-c]).
false.
This comes down to testing each of the two lists:
?- isItCov([a-b,a-c,a-d,c-d,b-c], [a-c,c-b,c-d]).
true .
?- isItCov([a-b,a-c,a-d,c-d,b-c], [a-c]).
false.
The first list does cover the graph, while the second doesn't. Overall, your definition of covb/2 is written to succeed if all lists cover the graph. This is not the case, so your covb/2 query fails.
Was this what you wanted to know?
I want to return items of lists of lists which passed the condition and returned true (that's the filtering part). How should I do that?
You could see if your Prolog's documentation has something for the word "filter". On SWI-Prolog you can do this:
?- apropos(filter).
true.
This will point you to the include/3 predicate, which seems to do what you want:
?- include(isItCov([a-b,a-c,a-d,c-d,b-c]), [[a-c,c-b,c-d],[a-c]], Covers).
Covers = [[a-c, c-b, c-d]].
If you want to write a filter predicate for your concrete application, it might look something like this:
graph_covers(_Graph, [], []).
graph_covers(Graph, [Nodes|NodesRest], Covers) :-
( isItCov(Graph, Nodes)
-> Covers = [Nodes|CoversRest]
; Covers = CoversRest ),
graph_covers(Graph, NodesRest, CoversRest).
This is similar to your predicate, it just adds an extra argument to collect those node lists for which isItCov/2 succeeded. If it did not succeed, it continues with a list not containing that current node list.
?- graph_covers([a-b,a-c,a-d,c-d,b-c], [[a-c,c-b,c-d],[a-c]], Covers).
Covers = [[a-c, c-b, c-d]] ;
false.
Related
I am writing a program that checks if a list is valid or not valid. Here is what I have so far and I know it should be short I just cannot grasp my mind around prolog
These are my facts
pkg(pkg_name)
pkg(p1)
pkg(p2)
dpnd(p2,p1)
My rule is going to be true_list()
this function will run as follows, if i select true_list([p1,p2]) then it should return true, if i do true_list([p1,p3,p2]) it should return false since p3 is not part of it. I have been looking at this for a while and feel like it should be so simple. I have tried this method
true_list(p1,p2):-package(package_name).
any assistance would be appreciated.
The best solution for this is to have a simple recursive predicate check every single element for existence.
for ?- true_list([p1,p2]).
pkg(pkg_name).
pkg(p1).
pkg(p2).
dpnd(p2,p1).
%your input is a list of elements [Element1, Element2, Element3...ElementN]
true_list([]).
true_list([H|T]) :-
(pkg(H) ->
(true_list(T)) ;
(fail)
).
This code checks every element in your list to see if it exists as a fact pkg and fails if it finds an element that is NOT a pkg. It also succeeds with an empty list.
I defined my knowledge base as:
edge(mammal,isa,animal).
edge(human,isa,mammal).
edge(simba,isa,human).
edge(animal,swim,bybirth).
edge(human,swim,mustlearn).
path(X,Y) :- edge(X,isa,Y).
path(X,Y) :- edge(X,isa,Z), path(Z,Y).
swim(X,Y) :- edge(X,swim,Y).
swim(X,Y) :- path(X,Z), swim(Z,Y).
Now, to use the above knowledge base, I use the following:
?- swim(simba,bybirth).
?- swim(simba,mustlearn).
And for both the queries, Prolog returns true. I want Prolog to check for the property swim locally first, then look at the direct parent, and so on in a hierarchical fashion. And it should stop searching as soon as we know that Simba "mustlearn" to swim, and shouldn't look any further. Thus, it should return false for the first query and true for the second.
I know it has to be done by limiting backtracking. I tried using the cut and not operators, but couldn't succeed. Is there a way to achieve this?
I tried it and ran into a problem too. I thought this might work:
swim(X,Y) :- once((edge(X,swim,Y); path(X,Z), swim(Z,Y))).
It doesn't work, because if Y is already instantiated on the way in, the first step will fail to unify and it will try the second route going through the human intermediate. So even though the query only produces one result, it can be fooled into producing swim(simba, bybirth). The solution is to force Prolog to commit to a binding on another variable and then check that binding after the commitment:
swim(X,Y) :-
once((edge(X,swim,Method); path(X,Z), swim(Z,Method))),
Method = Y.
This tells Prolog, there is only one way to get to this method, so find that method, and then it must be Y. If you find the wrong method, it won't go on a search, it will just fail. Try it!
I am a total beginner at Prolog. I am struggling with creating a rule, which takes a list as parameter and passes the list onto another rule. Here is my code:
combine([], '').
combine([L|List], Total) :-
combine(List, CombinedRest),
atom_concat(L, CombinedRest, Total).
findHeadline([W|Words], Combined) :-
combine(Words, Combined).
findHeadline2([Words], Combined) :-
combine(Words, Combined).
findHeadline works as expected, but findHeadline2 does not. Here is the output:
1 ?- findHeadline([billig, enkeltmand], Combination).
Combination = enkeltmand.
2 ?- findHeadline2([billig, enkeltmand], Combination).
false.
The output I was expecting from findHeadline was:
Combination = billigenkeltmand.
How can it be that this does not work?
I tried to utilize trace in SWI-prolog, but it gave me no clue whatsoever, as the findHeadline rule just exits immediately and does not call the combine rule at all.
It is not very clear what it is exactly that you are after. If you just want to concatenate a list of atoms to get one atom, use atomic_list_concat/2 available in SWI-Prolog:
?- atomic_list_concat([foo, bar, baz], C).
C = foobarbaz.
At the moment, your findHeadline2/2 reads:
"Take a list containing exactly one element, and combine/2 that element."
This is not what you are after, I have the feeling.
Your findHeadline/2, on the other hand, says:
"Take a list of at least one element, and combine/2 all elements except the first".
This is important: never ever ignore compilation warnings. You get code that does something, but you can be almost certain that it does not do what you want it to do, which is bad, or that if someone else reads your code, they will be confused, which is also bad.
I have a predicate "lookupOptions" which returns one by one some lists (Menus).
I'm trying to get it to satisfy the case of multiple inputs. I can return a single set of options as follows, by reading the head of the "list_places" list.
find_options(Restaurant,Town,Menu) :- lookupOptions(Restaurant,H,Menu), list_places(Town,[H|T])
But, I'm not able to get it to iterate.
I have tried a lot of things, these were my best efforts so far.
a) standard enough iteration, but it wont resolve ...
doStuff(X,[],_).
doStuff(Restaurant,[H|T],_):- lookupOptions(Resturant,H,_), doStuff(Restaurant,T,_).
find_options(Restaurant,Town,Menu) :- doStuff(Restaurant,[H|T],Menu), list_places(Town,[H|T]).
b) expanding the goal predicate ...
find_options(_,Town,[H|T],_)
find_options(Restaurant,Town,Menu) :- find_options(Restaurant,Town,[],Menu).
find_options(Restaurant,Town,X,Menu) :- list_places(Town,X).
find_options(Restaurant,Town,[H|T],Menu) :- lookupOptions(Restaurant,[H],Menu), find_options(Restaurant,Town,T,Menu).
Would either of these work ? if the pattern was written correctly. Or if there was an appropriate cut put in place?
Any help most appreciated ...
It's no clear on what you want iterate. Prolog uses backtracking to examine all alternatives, then you should start backtracking if you are after some alternative, or use the all solutions family.
Now I think you want simply declare there could be more find_options(Restaurant,Town,Menu). Then try replacing the head match [H|T] with this:
find_options(Restaurant,Town,Menu) :-
lookupOptions(Restaurant,H,Menu),
list_places(Town, Places),
member(H, Places).
BTW T is a singleton in your original rule. This could be a hint for the need of generalize it.
I have these two lists =
fruits([banana, apple, mangoes, pears]).
foodILike([hamburgers, banana, shakes, fries]).
I want to write a prolog predicate that will return true as soon as it sees 1 items in the foodsILike list in the fruits list. How can I go about doing so?
First, for the plain answer:
fruitsILike(F) :-
fruits(Fs)
member(F, Fs),
foodILike(Ls),
member(F, Ls).
You could avoid the membership check by flattening the fruits and foods lists:
fruit(banana).
fruit(apple).
...
foodILike(hamburger).
foodILike(banana).
...
fruitsILike(F) :-
fruit(F),
foodILike(F).
That said, you seem to try and solve problems in Prolog using imperative idioms, and that won't work. First, predicates do not return anything. When calling a predicate, Prolog unifies its arguments with valid values according to the facts and rules in the program. Therefore, the "returned value" are the assignments to unbound variables. Second, Prolog does not do something "as soon as". It iterates over all possible solutions. You get the first solution, then the second solution, and so on.
member can 1) individually generate all the members of a given list and/or 2) give a yes/no answer as to whether a particular element is in a particular list. I believe you want to use the first form on fruits to generate each of the elements of fruit, and the second form on foodILike to see if any of those is present.