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I'm trying to define a predicate "delete(L1, L2, L3)" that is valid when L3 equals L2 minus any of these elements that are contained in L1. E.g. delete([1], [1,2,3], X) => would unify for X = [2,3]. My code is as follows:
isNonElement(_, []).
isNonElement(X, [Y|Z]) :- X \= Y, isNonElement(X,Z).
delete(_, [], []).
delete(Y, [X|W], Z) :- \+(isNonElement(X, Y)), delete(Y, W, Z).
delete(Y, [X|W], [X|Z]) :- isNonElement(X, Y), delete(Y, W, Z).
However it seems not to work for every test case. Can anyone help me out on what could be wrong with my code?
Thanks in advance!
Best regards,
Skyfe.
P.S. I can't tell the cases for which my predicate doesn't work correctly since it's tested by a school system which doesn't tell me which test cases it failed for.
You are very close!
One remaining problem is that you are applying a perfectly sound logical reasoning to predicates that do not admit such a reading. Your code works exactly as intended if you simply apply the following extremely straight-forward changes:
instead of (\=)/2, use dif/2
instead of \+(isNonElement(X, Y)), simply write: member(X, Y).
The first change is always advisable: It typically makes your programs usable in more directions. The second change avoids the use of impure negation by using a pure predicate instead.
In total, we now have:
isNonElement(_, []).
isNonElement(X, [Y|Z]) :- dif(X, Y), isNonElement(X,Z).
delete(_, [], []).
delete(Y, [X|W], Z) :- member(X, Y), delete(Y, W, Z).
delete(Y, [X|W], [X|Z]) :- isNonElement(X, Y), delete(Y, W, Z).
Now check this out: First, your test case:
?- delete([1], [1,2,3], X).
X = [2, 3] ;
false.
Works as expected!
Second, a case with a variable for L2:
?- delete([], L2, []).
L2 = [] ;
false.
This seems also very nice.
Third, another variable:
?- delete([X], [1,2,3], Ls3).
X = 1,
Ls3 = [2, 3] ;
X = 2,
Ls3 = [1, 3] ;
X = 3,
Ls3 = [1, 2] ;
Ls3 = [1, 2, 3],
dif(X, 3),
dif(X, 2),
dif(X, 1) ;
false.
Note now the different possibilities for X, and how dif/2 is used in answers to express that X must be different from certain integers in this case.
The use of impure predicates precludes such more general uses, and your grading system possible tries such cases too.
Note that you can of course easily implement member/2 yourself. It is one of the most straight-forward relations. However, note also that delete/3 is terribly named: An imperative always implies a particular direction of use, but the relation we are considering here also admits many other usage modes!
I am fairly new to prolog and am trying to mess around with lists of lists. I am curious on how to add two lists of lists or subtract them resulting in one list of list. If I have two lists of lists lets say,
SomeList = [[1,2,3,4],[5,6,7,8]]
SomeList2 = [[1,2,3,4],[5,6,7,8]]
How could I add or subtract SomeList and SomeList2 to create a list of lists? Resulting in a sum of say
sumList([[2,4,6,8],[10,12,14,16]])
or vice-versa for subtraction? Any help would be appreciated not looking for code but for insight !
The easiest approach is with maplist:
add(X, Y, Z) :- Z is X + Y.
op_lists(L1, L2, R) :-
maplist(maplist(add), L1, L2, R).
Which gives:
| ?- op_lists([[1,2,3,4],[5,6,7,8]], [[1,2,3,4],[5,6,7,8]], R).
R = [[2,4,6,8],[10,12,14,16]]
yes
| ?-
In the expression:
maplist(maplist(add), L1, L2, R).
maplist(G, L1, L2, R) calls G on each element of L1 and L2, resulting in each element of R. Since each element of L1 and L2 is a list, then G in this case is maplist(add) which calls add on each element of the sublists.
You can obviously modify add(X, Y, Z) to be whatever operation you wish on each pair of elements. You can also make the addition more "relational" by using CLP(FD):
add(X, Y, Z) :- Z #= X + Y.
Then you also get, for example:
| ?- op_lists([[1,2,3,4],[5,6,7,8]], L, [[3,6,9,12],[10,12,14,16]]).
L = [[2,4,6,8],[5,6,7,8]]
yes
| ?-
If you wanted to do this without maplist, you could still use add/3 and use a two-layer approach:
op_lists([], [], []).
op_lists([LX|LXs], [LY|LYs], [LR|LRs]) :-
op_elements(LX, LY, LR),
op_lists(LXs, LYs, LRs).
op_elements([], [], []).
op_elements([X|Xs], [Y|Ys], [R|Rs]) :-
add(X, Y, R),
op_elements(Xs, Ys, Rs).
You can see the simple list processing pattern here, which the use of maplist takes care of for you.
Besides the solutions presented by #lurker (+1), I would also add the possibility to use DCGs, since you are working on lists. For the available operations I suggest to define a slightly more general predicate opfd/4 instead of add/3. Here are exemplary rules for addition and subtraction as asked in your question, you can use these as templates to add other two-place arithmetic operations:
opfd(+,X,Y,Z) :-
Z #= X+Y.
opfd(-,X,Y,Z) :-
Z #= X-Y.
As the desired operation is an argument, you only need one DCG-rule to cover all operations (marked as (1) at the corresponding goal). This way, of course, you have to specify the desired operation as an argument in your relation and pass it on to the DCGs. The structure of these DCGs is very similar to the last solution presented by #lurker, except that the resulting list does not appear as an argument since that is what the DCGs describe. For easier comparison I will stick with the names op_lists//3 and op_elements//3, the calling predicate shall be called lists_op_results/4:
lists_op_results(L1,L2,Op,Rs) :-
phrase(op_lists(Op,L1,L2),Rs).
op_lists(_Op,[],[]) -->
[].
op_lists(Op,[X|Xs],[Y|Ys]) -->
{phrase(op_elements(Op,X,Y),Rs)},
[Rs],
op_lists(Op,Xs,Ys).
op_elements(_Op,[],[]) -->
[].
op_elements(Op,[X|Xs],[Y|Ys]) -->
{opfd(Op,X,Y,R)}, % <-(1)
[R],
op_elements(Op,Xs,Ys).
Example queries:
?- lists_op_results([[1,2,3,4],[5,6,7,8]], [[1,2,3,4],[5,6,7,8]], +, R).
R = [[2,4,6,8],[10,12,14,16]]
?- lists_op_results([[1,2,3,4],[5,6,7,8]], [[1,2,3,4],[5,6,7,8]], -, R).
R = [[0,0,0,0],[0,0,0,0]]
#lurker's example:
?- lists_op_results([[1,2,3,4],[5,6,7,8]], L, +, [[3,6,9,12],[10,12,14,16]]).
L = [[2,4,6,8],[5,6,7,8]]
You can also ask if there is an operation that fits the given lists:
?- lists_op_results([[1,2,3,4],[5,6,7,8]], L, Op, [[3,6,9,12],[10,12,14,16]]).
L = [[2,4,6,8],[5,6,7,8]],
Op = + ? ;
L = [[-2,-4,-6,-8],[-5,-6,-7,-8]],
Op = -
On a sidenote: Since the operation is the first argument of opfd/4 you can also use it with maplist as suggested in #lurker's first solution. You just have to pass it lacking the last three arguments:
?- maplist(maplist(opfd(Op)),[[1,2,3,4],[5,6,7,8]], L, [[3,6,9,12],[10,12,14,16]]).
L = [[2,4,6,8],[5,6,7,8]],
Op = + ? ;
L = [[-2,-4,-6,-8],[-5,-6,-7,-8]],
Op = -
My aim is writing a predicate filter/3. With input list [bar(a,12),bar(b,12),bar(c,13)] and filter criteria bar(A,12) the expected output is [bar(a,12),bar(b,12)].
The code below works but what is the difference between writing \+ \+ Filter = X and Filter = X (for me it is same). I wrote down the program by using 2 versions and it gave the same correct result. But I am sure that they are different?!
filter([],_,[]).
filter([X|XS],Filter,[X|ZS]) :-
\+ \+ Filter=X,
!,
filter(XS,Filter,ZS).
filter([_|XS],Filter,ZS) :-
filter(XS,Filter,ZS).
EDIT:
#lurker you are right, they do not give the same result. ( it was my mistake)
----using \+ \+ Filter = X -----
?- filter([foo(a,12),foo(c,12),foo(b,13)],foo(A,12),Res).
Res = [foo(a, 12), foo(c, 12)].
----using Filter = X -----
?- filter([foo(a,12),foo(c,12),foo(b,13)],foo(A,12),Res).
A = a,
Res = [foo(a, 12)].
?- filter([foo(a,12),foo(a,12),foo(b,13)],foo(A,12),Res).
A = a,
Res = [foo(a, 12), foo(a, 12)].
TL;DR
?- tfilter(\bar(_,S)^(S=12), Xs, Ys).
Now, step-by-step:
There are several issues with your program. The biggest is the actual problem statement which leaves several things open. For example, I assume that you expect that all elements are of the form bar(X, N) and you want to select those with N = 12. What you have implemented is slightly different:
?- filter([bar(a,12),bar(b,12),bar(c,13)], bar(_,12), []).
true.
This anomaly is due to your specific use of the cut. As you can see from the other answers, many versions avoid it. Cut is extremely difficult to use without any surprising effects. #CapelliC's version with cut actually avoids this one problem, but this is a very tricky business.
A further anomaly concerns the way how you might want to generalize your query. What about asking:
?- filter([X], bar(_,12), Xs).
What should a correct answer be? Should Xs include X or not? After all, instances of this query produce different results, too! I will show two of them by adding the goals X = bar(a,12) and X = bar(a,13) in front.
?- X = bar(a,12), filter([X], bar(_,12), Xs).
Xs = [bar(a,12)].
?- X = bar(a,13), filter([X], bar(_,12), Xs).
Xs = [].
So in one case we have an element, and in the other we have not. The general query should thus consequently produce two answers.
Here is an approach which does not have such problems:
State the positive selection criteria.
Let's use a separate predicate for the selection criteria, and call it _true:
snd_bar_true(N, bar(_,N)).
State the negative selection criteria.
snd_bar_false(N, bar(_,S)) :-
dif(N, S).
Now, with both, we can write a clean and correct filter program. Note that N is now just the second argument.
filter([], _N, []).
filter([X|Xs], N, [X|Ys]) :-
snd_bar_true(N, X),
filter(Xs, N, Ys).
filter([X|Xs], N, Ys) :-
snd_bar_false(N, X),
filter(Xs, N, Ys).
?- filter([X], 12, Xs).
X = bar(_A, 12), Xs = [bar(_A, 12)]
; X = bar(_A, _B), Xs = [], dif(_B, 12).
So we get two answers: One selecting the element X provided it is of the form bar(_,12). And the other one, which does not select the element, but ensures that the second element is not 12.
While these answers are all perfect and fine, I'm not very happy with it: It is correct but soo verbose. Here is a way to make it more compact.
Merge the criteria into one "reified" definition
snd_bar_t(N, bar(_,N), true).
snd_bar_t(N, bar(_,S), false) :-
dif(S,N).
There is a more compact and efficient way to express this using (=)/3
snd_bar_t(N, bar(_,S), T) :-
=(S, N, T).
=(X, X, true).
=(X, Y, false) :-
dif(X,Y).
This (=)/3 can be more efficiently implemented as:
=(X, Y, T) :-
( X == Y -> T = true
; X \= Y -> T = false
; T = true, X = Y
; T = false,
dif(X, Y)
).
Now, we can use the generic tfilter/3:
filter(Xs, N, Ys) :-
tfilter(snd_bar_t(N), Xs, Ys).
And then, we can use library(lambda) to avoid the auxiliary definition:
filter(Xs, N, Ys) :-
tfilter(N+\bar(_,S)^(S = N), Xs, Ys).
Note that this (S = N) is not what you probably think! It is effectively not simple equality, but actually, the reified version of it! So it will be called like: call((S = 12), T) and thus =(S, 12, T).
Double negation it's an old 'trick of the trade' often used while writing metainterpreters.
Since variables instantiation due to unification it's undone on backtracking, it has a procedural only semantic of "prove a goal without binding its variables", whatever the meaning of such phrase could be.
1 ?- filter([bar(a,12),bar(b,12),bar(c,13)],bar(_,12),L).
L = [bar(a, 12), bar(b, 12)].
If you comment out (i.e. remove) the double negation, you observe the undue instantiation effect: X has been bound to bar(a,12), and then cannot be matched to bar(b,12).
2 ?- filter([bar(a,12),bar(b,12),bar(c,13)],bar(_,12),L).
L = [bar(a, 12)].
edit for the simple case at hand, an alternative implementation of filter/3 could be
filter([],_,[]).
filter([X|XS],Filter,ZS):-
X \= Filter, !, filter(XS, Filter, ZS).
filter([X|XS],Filter,[X|ZS]):-
filter(XS, Filter, ZS).
or, better
filter([],_,[]).
filter([X|XS],Filter,R):-
(X \= Filter -> R = ZS ; R = [X|ZS]), filter(XS, Filter, ZS).
but if your system implements subsumes_term/2, #Boris' answer is to be preferred
The answer by #CapelliC answers your question.
There is another standard predicate, subsumes_term/2, which can be used to achieve the same effect as the double negation:
filter0([], _, []).
filter0([X|Xs], T, Ys) :-
\+ subsumes_term(T, X),
filter0(Xs, T, Ys).
filter0([X|Xs], T, [X|Ys]) :-
subsumes_term(T, X),
filter0(Xs, T, Ys).
As to how to do the iteration over all elements, instead of a cut, prefer a conditional:
filter1([], _, []).
filter1([X|Xs], T, R) :-
( subsumes_term(T, X)
-> R = [X|Ys]
; R = Ys
),
filter1(Xs, T, Ys).
And if you write this, you can as well use include/3 (which, by the way, is literally a "filter" predicate):
filter(List, Term, Filtered) :-
include(subsumes_term(Term), List, Filtered).
I've started to learn Prolog recently and I can't solve how to make union of three lists.
I was able to make union of 2 lists :
%element
element(X,[X|_]).
element(X,[_|Y]):-
element(X,Y).
%union
union([],M,M).
union([X|Y],L,S) :- element(X,L),union(Y,L,S).
union([X|Y],L,[X|S]) :- (not(element(X,L))),union(Y,L,S).
can anybody help me please ?
union(A, B, C, U) :-
union(A, B, V),
union(C, V, U).
Your definition of union/3 can be improved by replacing
... not(element(X,L)), ...
by
... maplist(dif(X),L), ...
or
... non_member(X, L), ....
non_member(_X, []).
non_member(X, [E|Es]) :-
dif(X, E),
non_member(X, Es).
Here is a case where the difference shows:
?- union([A],[B],[C,D]).
A = C, B = D, dif(C, D).
How must [A] and [B] look like such that their union contains 2 elements?
The answer is: they must be different.
Your original version fails for this query, yet, it succeeds for a specialized instance like:
?- A = 1, B = 2, union([A],[B],[C,D]).
So it succeeds for this, but fails for a generalization of it. Therefore it is not a pure, logical relation.
So is everything fine and perfect with dif/2? Unfortunately not. #TudorBerariu has good reason to go for a cut, since it reflects some of the intention we have about the relation. The cut effectively reflects two key intentions
that the alternative of not being a member is now excluded, which is true for certain modes, like Arg1 and Arg2 being both sufficiently instantiated terms. A safe approximation would be ground terms.
that there is no need to look at further elements in the list Arg2, which again is only true if Arg1 and Arg2 are sufficiently instantiated.
Problems only show when terms are not sufficiently instantiated..
The drawback of OP's definition and the one above, is that both are unnecessarily too general which can be observed with repeated elements in Arg2:
?- union([a,a],[a,a],Zs).
Zs = [a, a]
; Zs = [a, a]
; Zs = [a, a]
; Zs = [a, a]
; false.
In fact, we get |Arg2||Arg1|-1 redundant answers. So the cut had some good reason to be there.
Another reason why union/3 as it stands is not very efficient is that for the (intended) ground case it leaves open unnecessary choice points. Again, #TudorBerariu's solution does not have this problem:
?- union([a],[a],Zs).
Zs = [a]
; false. % <--- Prolog does not know that there is nothing left
Eliminating redundancy
The actual culprit for that many redundant answers is the first rule. element(a,[a,a]) (commonly called member/2) will succeed twice.
union([X|Y],L,S) :- element(X,L), union(Y,L,S).
^^^^^^^^^^^^
Here is an improved definition:
memberd(X, [X|_Ys]).
memberd(X, [Y|Ys]) :-
dif(X,Y), % new!
memberd(X, Ys).
The recursive rule, reading it right-to-left, reads as follows:
Assume memberd(X, Ys) is true already for some X and Ys. Given that, and given that we have a fitting Y which is different from X. Thenwe can conclude that also memberd(X, [Y|Ys]) is true.
So this has eliminated the redundant solutions. But our definition is still not very efficient: it still has to visit Arg2 twice for each element, and then it is unable to conclude that no alternatives are left. In any case: resist to place a cut to remove this.
Introducing determinism via reification.
Compare the definitions of memberd/2 and non_member/2. Although they describe "the opposite" of each other, they look very similar:
non_member(_X, []).
non_member(X, [Y|Ys]) :-
dif(X,Y),
non_member(X, Ys).
memberd(X, [X|_Ys]).
memberd(X, [Y|Ys]) :-
dif(X,Y),
memberd(X, Ys).
The recursive rule is the same! Only the fact is a different one. Let's merge them into one definition - with an additional argument telling whether we mean memberd (true) or non_member (false):
memberd_t(_X, [], false).
memberd_t(X, [X|_Ys], true).
memberd_t(X, [Y|Ys], Truth) :-
dif(X, Y),
memberd_t(X, Ys, Truth).
Now, our definition gets a bit more compact:
unionp([], Ys, Ys).
unionp([X|Xs], Ys, Zs0) :-
if_( memberd_t(X, Ys), Zs0 = Zs, Zs0 = [X|Zs] ),
unionp(Xs, Ys, Zs).
memberd_t(_X, [], false). % see below
memberd_t(X, [Y|Ys], Truth) :-
if_( X = Y, Truth=true, memberd_t(X, Ys, Truth) ).
Note the difference between if_(If_1, Then_0, Else_0) and the if-then-else control construct ( If_0 -> Then_0 ; Else_0 ). While If_1 may succeed several times with different truth values (that is, it can be both true and false), the control construct makes If_0 succeed only once for being true only.
if_(If_1, Then_0, Else_0) :-
call(If_1, T),
( T == true -> call(Then_0)
; T == false -> call(Else_0)
; nonvar(T) -> throw(error(type_error(boolean,T),_))
; /* var(T) */ throw(error(instantiation_error,_))
).
=(X, Y, T) :-
( X == Y -> T = true
; X \= Y -> T = false
; T = true, X = Y
; T = false,
dif(X, Y) % ISO extension
% throw(error(instantiation_error,_)) % ISO strict
).
equal_t(X, Y, T) :-
=(X, Y, T).
To ensure that memberd_t/3 will always profit from first-argument indexing, rather use the following definition (thanks to #WillNess):
memberd_t(E, Xs, T) :-
i_memberd_t(Xs, E, T).
i_memberd_t([], _E, false).
i_memberd_t([X|Xs], E, T) :-
if_( X = E, T = true, i_memberd_t(Xs, E, T) ).
You can make the union of the first two lists and then the union between that result and the third:
union(L1, L2, L3, U):-union(L1, L2, U12), union(U12, L3, U).
You can improve union/3 with a cut operator:
union([],M,M).
union([X|Y],L,S) :- element(X,L), !, union(Y,L,S).
union([X|Y],L,[X|S]) :- union(Y,L,S).
Using only predicates with an extra argument such as memberd_t/3 leads only to weak reification. For strong reification we also need to generate constraints. Strong reification is a further approach to eliminate non-determinism.
But strong reification is difficult, a possible way to archive this is to use a CLP(*) instance which has also reified logical operators. Here is an example if using CLP(FD) for the union problem. Unfortunately this covers only the domain Z:
Strong Reification Code:
member(_, [], 0).
member(X, [Y|Z], B) :-
(X #= Y) #\/ C #<==> B,
member(X, Z, C).
union([], X, X).
union([X|Y], Z, T) :-
freeze(B, (B==1 -> T=R; T=[X|R])),
member(X, Z, B),
union(Y, Z, R).
The above doesn't suffer from unnecessary choice points. Here are some example that show that this isn't happening anymore:
Running a Ground Example:
?- union([1,2],[2,3],X).
X = [1, 2, 3].
Also the above example even doesn't create choice points, if we use variables somewhere. But we might see a lot of constraints:
Running a Non-Ground Example:
?- union([1,X],[X,3],Y).
X#=3#<==>_G316,
1#=X#<==>_G322,
_G316 in 0..1,
freeze(_G322, (_G322==1->Y=[X, 3];Y=[1, X, 3])),
_G322 in 0..1.
?- union([1,X],[X,3],Y), X=2.
X = 2,
Y = [1, 2, 3].
Since we didn't formulate some input invariants, the interpreter isn't able to see that producing constraints in the above case doesn't make any sense. We can use the all_different/1 constraint to help the interpreter a little bit:
Providing Invariants:
?- all_different([1,X]), all_different([X,3]), union([1,X],[X,3],Y).
Y = [1, X, 3],
X in inf..0\/2\/4..sup,
all_different([X, 3]),
all_different([1, X]).
But we shouldn't expect too much from this singular example. Since the CLP(FD) and the freeze/2 is only an incomplete decision procedure for propositions and Z equations, the approach might not work as smooth as here in every situation.
Bye
I am completely new to Prolog and trying some exercises. One of them is:
Write a predicate set(InList,OutList)
which takes as input an arbitrary
list, and returns a list in which each
element of the input list appears only
once.
Here is my solution:
member(X,[X|_]).
member(X,[_|T]) :- member(X,T).
set([],[]).
set([H|T],[H|Out]) :-
not(member(H,T)),
set(T,Out).
set([H|T],Out) :-
member(H,T),
set(T,Out).
I'm not allowed to use any of built-in predicates (It would be better even do not use not/1). The problem is, that set/2 gives multiple same solutions. The more repetitions in the input list, the more solutions will result. What am I doing wrong? Thanks in advance.
You are getting multiple solutions due to Prolog's backtracking. Technically, each solution provided is correct, which is why it is being generated. If you want just one solution to be generated, you are going to have to stop backtracking at some point. This is what the Prolog cut is used for. You might find that reading up on that will help you with this problem.
Update: Right. Your member() predicate is evaluating as true in several different ways if the first variable is in multiple positions in the second variable.
I've used the name mymember() for this predicate, so as not to conflict with GNU Prolog's builtin member() predicate. My knowledge base now looks like this:
mymember(X,[X|_]).
mymember(X,[_|T]) :- mymember(X,T).
not(A) :- \+ call(A).
set([],[]).
set([H|T],[H|Out]) :-
not(mymember(H,T)),
set(T,Out).
set([H|T],Out) :-
mymember(H,T),
set(T,Out).
So, mymember(1, [1, 1, 1]). evaluates as true in three different ways:
| ?- mymember(1, [1, 1, 1]).
true ? a
true
true
no
If you want to have only one answer, you're going to have to use a cut. Changing the first definition of mymember() to this:
mymember(X,[X|_]) :- !.
Solves your problem.
Furthermore, you can avoid not() altogether, if you wish, by defining a notamember() predicate yourself. The choice is yours.
A simpler (and likely faster) solution is to use library predicate sort/2 which remove duplicates in O(n log n). Definitely works in Yap prolog and SWIPL
You are on the right track... Stay pure---it's easy!
Use reified equality predicates =/3 and dif/3 in combination with if_/3, as implemented in Prolog union for A U B U C:
=(X, Y, R) :- X == Y, !, R = true.
=(X, Y, R) :- ?=(X, Y), !, R = false. % syntactically different
=(X, Y, R) :- X \= Y, !, R = false. % semantically different
=(X, Y, R) :- R == true, !, X = Y.
=(X, X, true).
=(X, Y, false) :-
dif(X, Y).
% dif/3 is defined like (=)/3
dif(X, Y, R) :- X == Y, !, R = false.
dif(X, Y, R) :- ?=(X, Y), !, R = true. % syntactically different
dif(X, Y, R) :- X \= Y, !, R = true. % semantically different
dif(X, Y, R) :- R == true, !, X \= Y.
dif(X, Y, true) :- % succeed first!
dif(X, Y).
dif(X, X, false).
if_(C_1, Then_0, Else_0) :-
call(C_1, Truth),
functor(Truth,_,0), % safety check
( Truth == true -> Then_0 ; Truth == false, Else_0 ).
Based on these predicates we build a reified membership predicate list_item_isMember/3. It is semantically equivalent with memberd_truth/3 by #false. We rearrange the argument order so the list is the 1st argument. This enables first-argument indexing which prevents leaving useless choice-points behind as memberd_truth/3 would create.
list_item_isMember([],_,false).
list_item_isMember([X|Xs],E,Truth) :-
if_(E = X, Truth = true, list_item_isMember(Xs,E,Truth)).
list_set([],[]).
list_set([X|Xs],Ys) :-
if_(list_item_isMember(Xs,X), Ys = Ys0, Ys = [X|Ys0]),
list_set(Xs,Ys0).
A simple query shows that all redundant answers have been eliminated and that the goal succeeds without leaving any choice-points behind:
?- list_set([1,2,3,4,1,2,3,4,1,2,3,1,2,1],Xs).
Xs = [4,3,2,1]. % succeeds deterministically
Edit 2015-04-23
I was inspired by #Ludwig's answer of set/2, which goes like this:
set([],[]).
set([H|T],[H|T1]) :- subtract(T,[H],T2), set(T2,T1).
SWI-Prolog's builtin predicate subtract/3 can be non-monotone, which may restrict its use. list_item_subtracted/3 is a monotone variant of it:
list_item_subtracted([],_,[]).
list_item_subtracted([A|As],E,Bs1) :-
if_(dif(A,E), Bs1 = [A|Bs], Bs = Bs1),
list_item_subtracted(As,E,Bs).
list_setB/2 is like set/2, but is based on list_item_subtracted/3---not subtract/3:
list_setB([],[]).
list_setB([X|Xs1],[X|Ys]) :-
list_item_subtracted(Xs1,X,Xs),
list_setB(Xs,Ys).
The following queries compare list_set/2 and list_setB/2:
?- list_set([1,2,3,4,1,2,3,4,1,2,3,1,2,1], Xs).
Xs = [4,3,2,1]. % succeeds deterministically
?- list_setB([1,2,3,4,1,2,3,4,1,2,3,1,2,1],Xs).
Xs = [1,2,3,4]. % succeeds deterministically
?- list_set(Xs,[a,b]).
Xs = [a,b]
; Xs = [a,b,b]
; Xs = [a,b,b,b]
... % does not terminate universally
?- list_setB(Xs,[a,b]).
Xs = [a,b]
; Xs = [a,b,b]
; Xs = [a,b,b,b]
... % does not terminate universally
I think that a better way to do this would be:
set([], []).
set([H|T], [H|T1]) :- subtract(T, [H], T2), set(T2, T1).
So, for example ?- set([1,4,1,1,3,4],S) give you as output:
S = [1, 4, 3]
Adding my answer to this old thread:
notmember(_,[]).
notmember(X,[H|T]):-X\=H,notmember(X,T).
set([],[]).
set([H|T],S):-set(T,S),member(H,S).
set([H|T],[H|S]):-set(T,S),not(member(H,S)).
The only virtue of this solution is that it uses only those predicates that have been introduced by the point where this exercise appears in the original text.
This works without cut, but it needs more lines and another argument.
If I change the [H2|T2] to S on line three, it will produce multiple results. I don't understand why.
setb([],[],_).
setb([H|T],[H|T2],A) :- not(member(H,A)),setb(T,T2,[H|A]).
setb([H|T],[H2|T2],A) :- member(H,A),setb(T,[H2|T2],A).
setb([H|T],[],A) :- member(H,A),setb(T,[],A).
set(L,S) :- setb(L,S,[]).
You just have to stop the backtracking of Prolog.
enter code here
member(X,[X|_]):- !.
member(X,[_|T]) :- member(X,T).
set([],[]).
set([H|T],[H|Out]) :-
not(member(H,T)),
!,
set(T,Out).
set([H|T],Out) :-
member(H,T),
set(T,Out).
Using the support function mymember of Tim, you can do this if the order of elements in the set isn't important:
mymember(X,[X|_]).
mymember(X,[_|T]) :- mymember(X,T).
mkset([],[]).
mkset([T|C], S) :- mymember(T,C),!, mkset(C,S).
mkset([T|C], S) :- mkset(C,Z), S=[T|Z].
So, for example ?- mkset([1,4,1,1,3,4],S) give you as output:
S = [1, 3, 4]
but, if you want a set with the elements ordered like in the list you can use:
mkset2([],[], _).
mkset2([T|C], S, D) :- mkset2(C,Z,[T|D]), ((mymember(T,D), S=Z,!) ; S=[T|Z]).
mkset(L, S) :- mkset2(L,S,[]).
This solution, with the same input of the previous example, give to you:
S = [1, 4, 3]
This time the elements are in the same order as they appear in the input list.
/* Remove duplicates from a list without accumulator */
our_member(A,[A|Rest]).
our_member(A, [_|Rest]):-
our_member(A, Rest).
remove_dup([],[]):-!.
remove_dup([X|Rest],L):-
our_member(X,Rest),!,
remove_dup(Rest,L).
remove_dup([X|Rest],[X|L]):-
remove_dup(Rest,L).