I'm new to Prolog and I'm stuck on a predicate that I'm trying to do. The aim of it is to recurse through a list of quads [X,Y,S,P] with a given P, when the quad has the same P it stores it in a temporary list. When it comes across a new P, it looks to see if the temporary list is greater than length 2, if it is then stores the temporary list in the output list, if less than 2 deletes the quad, and then starts the recursion again the new P.
Heres my code:
deleteUP(_,[],[],[]).
deleteUP(P,[[X,Y,S,P]|Rest],Temp,Output):-
!,
appends([X,Y,S,P],Temp,Temp),
deleteUP(P,[Rest],Temp,Output).
deleteUP(NextP,[[X,Y,S,P]|Rest],Temp,Output):-
NextP =\= P,
listlen(Temp,Z),
Z > 1, !,
appends(Temp,Output,Output),
deleteUP(NextP,[_|Rest],Temp,Output).
listlen([], 0).
listlen([_|T],N) :-
listlen(T,N1),
N is N1 + 1.
appends([],L,L).
appends([H|T],L,[H|Result]):-
appends(T,L,Result).
Thanks for any help!
Your problem description talks about storing, recursing and starting. That is a very imperative, procedural description. Try to focus first on what the relation should describe. Actually, I still have not understood what minimal length of 2 is about.
Consider to use the predefined append/3 and length/2 in place of your own definitions. But actually, both are not needed in your example.
You might want to use a dedicated structure q(X,Y,S,P) in place of the list [X,Y,S,P].
The goal appends([X,Y,S,P],Temp,Temp) shows that you assume that the logical variable Temp can be used like a variable in an imperative language. But this is not the case. By default SWI creates here a very odd structure called an "infinite tree". Forget this for the moment.
?- append([X,Y,S,P],Temp,Temp).
Temp = [X, Y, S, P|Temp].
There is a safe way in SWI to avoid such cases and to detect (some of) such errors automatically. Switch on the occurs check!
?- set_prolog_flag(occurs_check,error).
true.
?- append([X,Y,S,P],Temp,Temp).
sto. % ERROR: lists:append/3: Cannot unify _G392 with [_G395,_G398,_G401,_G404|_G392]: would create an infinite tree
The goal =\=/2 means arithmetical inequality, you might prefer dif/2 instead.
Avoid the ! - it is not needed in this case.
length(L, N), N > 1 is often better expressed as L = [_,_|_].
The major problem, however, is what the third and fourth argument should be. You really need to clarify that first.
Prolog variables can't be 'modified', as you are attempting calling appends: you need a fresh variables to place results. Note this code is untested...
deleteUP(_,[],[],[]).
deleteUP(P,[[X,Y,S,P]|Rest],Temp,Output):-
!,
appends([X,Y,S,P],Temp,Temp1),
deleteUP(P, Rest, Temp1,Output). % was deleteUP(P,[Rest],Temp,Output).
deleteUP(NextP,[[X,Y,S,P]|Rest],Temp,Output1):-
% NextP =\= P, should be useless given the test in clause above
listlen(Temp,Z),
Z > 1, !, % else ?
deleteUP(NextP,[_|Rest],Temp,Output),
appends(Temp,Output,Output1).
Related
Hey so this is my code so far. I am only a begginer in prolog but i need it for school
firstElement([_|_], [Elem1|List1], [Elem2|List2]):-
Elem1 =< Elem2, merge([Elem1] , List1, [Elem2|List2]);
merge([], [Elem2], List2).
merge([Head|Tail], [Elem1|List1], [Elem2|List2]):-
Elem1 =< Elem2,!, add(Elem1,[Head|Tail],[Head|Tail1]),
merge([Head|Tail1], List1, [Elem2|List2]);
add(Elem2,[Head|Tail],[Head|Tail1]),
merge([Head|Tail1], [Elem1|List1], List2).
merge([Head|Tail], [], [Elem2|List2]):-
add(Elem2,[Head|Tail],[Head|Tail1]).
merge([Head|Tail], [Elem1|List1], []):-
add(Elem1,[Head|Tail],[Head|Tail1]).
merge([Head|Tail], [], []).
add(X,[],[X]).
add(X,[Y|Tail],[Y|Tail1]):-
add(X,Tail,Tail1).
I found out that everytime it gets out of a merge it keeps forgetting the last number so it gets back to nothing in the end.
I think you’ve gotten very mixed up here with your code. A complete solution can be had without helpers and with only a few clauses.
First let us discuss the two base cases involving empty lists:
merge(X, [], X).
merge([], X, X).
You don’t quite have these, but I see some sort of recognition that you need to handle empty lists specially in your second and third clauses, but I think you got confused and overcomplicated them. There’s really three scenarios covered by these two clauses. The case where both lists are empty is a freebie covered by both of them, but since that case would work out to merge([], [], []), it’s covered. The big idea here is that if you exhaust either list, because they were sorted, what you have left in the other list is your result. Think about it.
This leaves the interesting case, which is one where we have some items in both lists. Essentially what you want to do is select the smaller of the two, and then recur on the entire other list and the remainder of the one you selected the smaller value from. This is one clause for that:
merge([L|Ls], [R|Rs], [L|Merged]) :-
L #< R,
merge(Ls, [R|Rs], Merged).
Here’s what you should note:
The “result” has L prepended to the recursively constructed remainder.
The recursive call to merge rebuilds the entire second list, using [R|Rs].
It should be possible to build the other clause by looking at this.
As an intermediate Prolog user, I would be naturally a bit suspicious of using two clauses to do this work, because it’s going to create unnecessary choice points. As a beginner, you will be tempted to erase those choice points using cuts, which will go badly for you. A more intermediate approach would be to subsume both of the necessary clauses into one using a conditional operator:
merge([L|Ls], [R|Rs], [N|Ns]) :-
( L #< R ->
N = L, merge(Ls, [R|Rs], Ns)
; —- other case goes here
).
An expert would probably build it using if_/3 instead:
#<(X,Y,true) :- X #< Y.
#<(X,Y,false) :- X #>= Y.
merge([L|Ls], [R|Rs], [N|Ns]) :-
if_(#<(L,R),
(N = L, merge(Ls, [R|Rs], Ns)),
( -- other case here )).
Anyway, I hope this helps illustrate the situation.
This is probably the most trivial implementation of a function that returns the length of a list in Prolog
count([], 0).
count([_|B], T) :- count(B, U), T is U + 1.
one thing about Prolog that I still cannot wrap my head around is the flexibility of using variables as parameters.
So for example I can run count([a, b, c], 3). and get true. I can also run count([a, b], X). and get an answer X = 2.. Oddly (at least for me) is that I can also run count(X, 3). and get at least one result, which looks something like X = [_G4337877, _G4337880, _G4337883] ; before the interpreter disappears into an infinite loop. I can even run something truly "flexible" like count(X, A). and get X = [], A = 0 ; X = [_G4369400], A = 1., which is obviously incomplete but somehow really nice.
Therefore my multifaceted question. Can I somehow explain to Prolog not to look beyond first result when executing count(X, 3).? Can I somehow make Prolog generate any number of solutions for count(X, A).? Is there a limitation of what kind of solutions I can generate? What is it about this specific predicate, that prevents me from generating all solutions for all possible kinds of queries?
This is probably the most trivial implementation
Depends from viewpoint: consider
count(L,C) :- length(L,C).
Shorter and functional. And this one also works for your use case.
edit
library CLP(FD) allows for
:- use_module(library(clpfd)).
count([], 0).
count([_|B], T) :- U #>= 0, T #= U + 1, count(B, U).
?- count(X,3).
X = [_G2327, _G2498, _G2669] ;
false.
(further) answering to comments
It was clearly sarcasm
No, sorry for giving this impression. It was an attempt to give you a synthetic answer to your question. Every details of the implementation of length/2 - indeed much longer than your code - have been carefully weighted to give us a general and efficient building block.
There must be some general concept
I would call (full) Prolog such general concept. From the very start, Prolog requires us to solve computational tasks describing relations among predicate arguments. Once we have described our relations, we can query our 'knowledge database', and Prolog attempts to enumerate all answers, in a specific order.
High level concepts like unification and depth first search (backtracking) are keys in this model.
Now, I think you're looking for second order constructs like var/1, that allow us to reason about our predicates. Such constructs cannot be written in (pure) Prolog, and a growing school of thinking requires to avoid them, because are rather difficult to use. So I posted an alternative using CLP(FD), that effectively shields us in some situation. In this question specific context, it actually give us a simple and elegant solution.
I am not trying to re-implement length
Well, I'm aware of this, but since count/2 aliases length/2, why not study the reference model ? ( see source on SWI-Prolog site )
The answer you get for the query count(X,3) is actually not odd at all. You are asking which lists have a length of 3. And you get a list with 3 elements. The infinite loop appears because the variables B and U in the first goal of your recursive rule are unbound. You don't have anything before that goal that could fail. So it is always possible to follow the recursion. In the version of CapelliC you have 2 goals in the second rule before the recursion that fail if the second argument is smaller than 1. Maybe it becomes clearer if you consider this slightly altered version:
:- use_module(library(clpfd)).
count([], 0).
count([_|B], T) :-
T #> 0,
U #= T - 1,
count(B, U).
Your query
?- count(X,3).
will not match the first rule but the second one and continue recursively until the second argument is 0. At that point the first rule will match and yield the result:
X = [_A,_B,_C] ?
The head of the second rule will also match but its first goal will fail because T=0:
X = [_A,_B,_C] ? ;
no
In your above version however Prolog will try the recursive goal of the second rule because of the unbound variables B and U and hence loop infinitely.
I need to write a predicate longestList/2 such that longestList(L1,L2) is satisfied if L2 is the longest
nested list from the list of lists L1.
?- longestList([[1],[1,2],[1,2,3],[1,2,3,4]],LI).
LI = [[1, 2, 3, 4]] ;
No
?- longestList([[a,b,c],[d,e],[f,g,h]],LI).
LI = [[f, g, h],[a,b,c]];
No
Could someone please help me with intuition to go about solving it?
Here's an outline for a basic, recursive approach. Not quite as crisp as the answer #CapelliC gave, but on the same order of simplicity.
The idea is to traverse the list and keep track of the longest list you've seen so far, and what it's length is. Then you step through the list recursively and update these arguments for the recursion if the conditions indicate so. It's a slight elaboration on the technique used to do a recursive "max list element" predicate. To do this, you set up a call to include more arguments (the current longest list, and its length).
longestList([], []).
longestList([L|Ls], LongestList) :-
length(L, Length),
% Start with the first element (L) being my best choice so far
longestList(Ls, L, Length, LongestList).
Here is the expanded predicate with the new arguments.
longestList([L|Ls], LongestListSoFar, GreatestLengthSoFar, LongestList) :-
% Here, you need to examine L and determine if it should supersede
% the longest list so far and its length. You need to keep in mind that
% if the length of L is the same as the max length so far, then I
% may choose to keep the LongestListSoFar, or choose L. Both are
% valid solutions for this call. This is a good place to use the `;`
% operator, and to be cautious about parenthesizing expressions since
% the comma has higher precedence than the semi-colon.
% Also, you'll need to make a recursive call to longestList(Ls, ??, ??, LongestList).
% The arguments to the recursion will depend upon which way the decision flow goes.
%
% After all that to-do, don't let it scare you: it's about 5 lines of code :)
%
longestList([], LongestListSoFar, ??, ??).
% Fill in the ??. What should they be at list's end ([])?
% Do I even care now what the 3rd argument is?
Hopefully that's enough to give you something to think about to make progress. Or, use #CapelliC's solution and write the member_length/3 predicate. :) Note that, as in his solution, the above solution would generate each maximum list on backtracking if there are more than one. So, you could use findall/3 if you want to get all the solutions in one list.
member/2 will allow you to peek an element (a list for your case) from a list: so, if you have a member_length/3 predicate, you could code
longestList(Lists, Longest) :-
member_length(Lists, Longest, N),
\+ ( member_length(Lists, _Another, M), M > N ).
then to find all longest, you can use findall/3...
Following code, which uses if else, seem to work partially. It has to be called with 0 length and so one cannot get length itself from here.
longest_list([H|T], LongList, LongLen):-
length(H, Len),
(Len > LongLen ->
longest_list(T, [H], Len);
longest_list(T, LongList, LongLen)).
longest_list([],Finallist,_):-writeln(Finallist).
?- longest_list([[1,2,3], [3,4], [4,5,6,7,8], [5,3,4]], Longestlist, 0).
[[4,5,6,7,8]]
true.
However, the variable itself is not coming:
?- writeln(Longestlist).
_G1955
true.
It may give some ideas.
I have the following code for getting the length of a list in prolog, it works recursively.
Is there any other way for getting the length?
len([], 0).
len([H|T], N) :-
len(T, NT), N is NT + 1.
Any suggestions would be appreciated.
You are asking the wrong question :)
But seriously: the only sensible way of finding the length of a list is to use the built-in length/2. How it is implemented is irrelevant -- more important are its semantics:
?- length([a,b], 2).
true.
?- length([a,b], 4).
false.
?- length([a,b,c], Len).
Len = 3.
?- length(List, 3).
List = [_G937, _G940, _G943].
?- length(List, Len).
List = [],
Len = 0 ;
List = [_G949],
Len = 1 ;
List = [_G949, _G952],
Len = 2 . % and so on
Either way, it doesn't get simpler than that. Any other way of finding the length of a list, or checking for the length of a list, or creating a list of a certain length, or enumerating lists of increasing length is going to be less "simple" than using length/2.
And then: learning Prolog means learning how length/2, and the other nicely declarative built-ins can be used.
Repeating an element N times
Splitting a list into segments of some length
Exactly one pair in a list
Rotate a list
I am sure you can think of many other uses of length/2.
Here is an iterative solution that uses repeat/0 predicate:
getlength(L,N) :-
retractall(getlength_res(_)),
assert(getlength_res(0)),
retractall(getlength_list(_)),
assert(getlength_list(L)),
repeat,
(
getlength_list([]), !, getlength_res(N)
;
retract(getlength_res(V)), W is V + 1, assert(getlength_res(W)),
retract(getlength_list([_|T])), assert(getlength_list(T)), fail
).
This solution creates and retracts facts getlength_res/1 and getlength_list/1 as it walks through the list, replacing the old list with a shorter one, and the old number with a number that is greater by one at each iteration of repeat/0. In a sense, the two dynamically asserted/retracted facts behave very much like assignable variables of imperative languages.
Demo.
In general, iterative solutions in Prolog are harder to read than their recursive counterparts. This should come as no surprise, considering that anything that has an effect of an assignment statement of an imperative programming language goes against the grain with Prolog's design philosophy.
Sorry I could not resist to try out this "challenge":
Input=[a,b,b,b,b,b,b,b,b,a,b,c,d,f], between(1,inf,K),findall( _,between(1,K,_),FreeList), ( FreeList=Input,!,true).
findall/3 is doing the behind-the-scenes recursion, code is making unifications of lists FreeList and Input until they unify
I have the following problem. I have a certain number of facts such as:
parent(jane,dick).
parent(michael,dick).
And I want to have a predicate such as:
numberofchildren(michael,X)
so that if I call it like that it shows X=1.
I've searched the web and everyone puts the children into lists, is there a way not to use lists?
Counting number of solutions requires some extra logical tool (it's inherently non monotonic). Here a possible solution:
:- dynamic count_solutions_store/1.
count_solutions(Goal, N) :-
assert(count_solutions_store(0)),
repeat,
( call(Goal),
retract(count_solutions_store(SoFar)),
Updated is SoFar + 1,
assert(count_solutions_store(Updated)),
fail
; retract(count_solutions_store(T))
),
!, N = T.
I can only see two ways to solve this.
The first, which seems easier, is to get all the solutions in a list and count it. I'm not sure why you dislike this option. Are you worried about efficiency or something? Or just an assignment?
The problem is that without using a meta-logical predicate like setof/3 you're going to have to allow Prolog to bind the values the usual way. The only way to loop if you're letting Prolog do that is with failure, as in something like this:
numberofchildren(Person, N) :- parent(Person, _), N is N+1.
This isn't going to work though; first you're going to get arguments not sufficiently instantiated. Then you're going to fix that and get something like this:
?- numberofchildren(michael, N)
N = 1 ;
N = 1 ;
N = 1 ;
no.
The reason is that you need Prolog to backtrack if it's going to go through the facts one by one, and each time it backtracks, it unbinds whatever it bound since the last choice point. The only way I know of to pass data across this barrier is with the dynamic store:
:- dynamic(numberofchildrensofar/1).
numberofchildren(Person, N) :-
asserta(numberofchildrensofar(0)),
numberofchildren1(Person),
numberofchildrensofar(N), !.
numberofchildren1(Person) :-
parent(Person, _),
retract(numberofchildrensofar(N)),
N1 is N + 1,
asserta(numberofchildrensofar(N1),
!, fail.
numberofchildren1(_).
I haven't tested this, because I think it's fairly disgusting, but it could probably be made to work if it doesn't. :)
Anyway, I strongly recommend you take the list option if possible.