I need to find the combinations in a list of lists. For example, give the following list,
List = [[1, 2], [1, 2, 3]]
These should be the output,
Comb = [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3]]
Another example:
List = [[1,2],[1,2],[1,2,3]]
Comb = [[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3]....etc]
I know how to do it for a list with two sublists but it needs to work for any number of sublists.
I'm new to Prolog, please help.
This answer hunts the bounty offered "for a pure solution that also takes into account for Ess".
Here we generalize this previous
answer like so:
list_crossproduct(Xs, []) :-
member([], Xs).
list_crossproduct(Xs, Ess) :-
Ess = [E0|_],
same_length(E0, Xs),
maplist(maybelonger_than(Ess), Xs),
list_comb(Xs, Ess).
maybelonger_than(Xs, Ys) :-
maybeshorter_than(Ys, Xs).
maybeshorter_than([], _).
maybeshorter_than([_|Xs], [_|Ys]) :-
maybeshorter_than(Xs, Ys).
list_crossproduct/2 gets bidirectional by relating Xs and Ess early.
?- list_comb(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
nontermination % BAD!
?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
Xs = [[1],[2],[3,4,5]] % this now works, too
; false.
Sample query having multiple answers:
?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5],X,Y,Z]).
X = [1,2,_A],
Y = [1,2,_B],
Z = [1,2,_C], Xs = [[1],[2],[3,4,5,_A,_B,_C]]
; X = [1,_A,3],
Y = [1,_A,4],
Z = [1,_A,5], Xs = [[1],[2,_A],[3,4,5]]
; X = [_A,2,3],
Y = [_A,2,4],
Z = [_A,2,5], Xs = [[1,_A],[2],[3,4,5]]
; false.
For completeness, here is the augmented version of my comment-version. Note nilmemberd_t/2 which is inspired by memberd_t/2.
nilmemberd_t([], false).
nilmemberd_t([X|Xs], T) :-
if_(nil_t(X), T = true, nilmemberd_t(Xs, T)).
nil_t([], true).
nil_t([_|_], false).
list_comb(List, []) :-
nilmemberd_t(List, true).
list_comb(List, Ess) :-
bagof(Es, maplist(member,Es,List), Ess).
Above version shows that "only" the first clause was missing in my comment response. Maybe even shorter with:
nilmemberd([[]|_]).
nilmemberd([[_|_]|Nils]) :-
nilmemberd(Nils).
This should work for Prologs without constraints. With constraints, bagof/3 would have to be reconsidered since copying constraints is an ill-defined terrain.
Here's a way to do it using maplist/3 and append/2:
list_comb([], [[]]).
list_comb([Xs|Xss], Ess) :-
Xs = [_|_],
list_comb(Xss, Ess0),
maplist(aux_x_comb(Ess0), Xs, Esss1),
append(Esss1, Ess).
aux_x_comb(Ess0, X, Ess1) :-
maplist(head_tail_list(X), Ess0, Ess1).
head_tail_list(X, Xs, [X|Xs]).
Sample query:
?- list_comb([[a,b],[f,g],[x,y,z]], Ess).
Ess = [[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z],
[b,f,x],[b,f,y],[b,f,z],
[b,g,x],[b,g,y],[b,g,z]].
Here's how it works!
As an example, consider these goals:
list_comb([[a,b],[f,g],[x,y,z]], Ess)
list_comb([ [f,g],[x,y,z]], Ess0)
How can we get from Ess0 to Ess?
We look at the answers to the
latter query:
?- list_comb([[f,g],[x,y,z]], Ess0).
Ess0 = [[f,x],[f,y],[f,z], [g,x],[g,y],[g,z]].
... place a before [f,x], ..., [g,z] ...
?- maplist(head_tail_list(a),
[[f,x],[f,y],[f,z],
[g,x],[g,y],[g,z]], X).
X = [[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z]].
... then do the same for b.
maplist(aux_x_comb) helps us handle all items:
?- maplist(aux_x_comb([[f,x],[f,y],[f,z],
[g,x],[g,y],[g,z]]),
[a,b], X).
X = [[[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z]],
[[b,f,x],[b,f,y],[b,f,z],
[b,g,x],[b,g,y],[b,g,z]]].
To get from a list of lists to a list use append/2.
I hope this explanation was more eludicating than confusing:)
A twist in #false's approach:
%list_comb( ++LL, -Ess)
list_comb( LL, Ess):-
is_list( LL),
maplist( is_list, LL),
findall( Es, maplist( member, Es, LL), Ess).
Testing:
41 ?- list_comb( [[1,2],[1],[1]], X).
X = [[1, 1, 1], [2, 1, 1]].
42 ?- list_comb( [[1,2],[1],[1,2,3]], X).
X = [[1, 1, 1], [1, 1, 2], [1, 1, 3], [2, 1, 1], [2, 1, 2], [2, 1, 3]].
43 ?- list_comb( [[1,2],[],[1,2,3]], X).
X = [].
44 ?- list_comb( [[1,2],t,[1,2,3]], X).
false.
45 ?- list_comb( t, X).
false.
Related
I'm trying to create list2 from list1, but every member should have Z increment.
if Z=1 then: [1,2,3] -> [2,3,4]
addZ(Z,[X | Xs],[Y | Ys]):-
Y is X+Z,
addZ(Z,Xs,Ys).
If I try addZ(1,[1,2,3],X). I just get no as answer.
After adding the base case, as recommended by #DavidTonhofer, you can change the order of the first two arguments of the predicate to avoid spurious choice points (since Prolog apply first argument indexing to quickly narrow down applicable clauses):
addZ([], _, []).
addZ([X|Xs], Z, [Y|Ys]):-
Y is X + Z,
addZ(Xs, Z, Ys).
Example:
?- addZ([1,2,3], 1, Z).
Z = [2, 3, 4].
?-
As always the most fun starts when you use CLP, like
:- use_module(library(clpz)).
addZ(_, [], []).
addZ(I, [X|Xs], [Y|Ys]) :-
Y #= I + X,
addZ(I, Xs, Ys).
And now you can do stuff like:
?- addZ(1, [1, 2, 3], C).
C = [2,3,4].
?- addZ(1, [1, 2, X], [X, Y, 9]).
false.
?- addZ(4, [1, 2, X], [A, Y, 9]).
A = 5, Y = 6, X = 5.
?- addZ(I, [1, 2, X], [5|Ys]).
I = 4, Ys = [6,_A], clpz:(4+X#=_A).
Here's another solution using maplist/3 and library yall:
addZ(Z, Xs, Ys):-
maplist({Z}/[X, Y]>>(Y is X+Z), Xs, Ys).
Sample runs
?- addZ(1, [1,2,3], X).
X = [2, 3, 4].
?- addZ(5, [1,2,3], X).
X = [6, 7, 8].
I'm currently trying to implement an automated theorem prover in prolog and have stumbled across a problem.
If I have a list of lists such as:
[[1,2],[-1,3],[4,5,7],[-2,4]]
How would I get the "set difference" of two compatible list items:
What I mean by compatible is, if the negation of a certain number exists in another list, then replace those two lists with the set difference, ie:
[1,2] and [-1,3] are compatible because -1 is present in the second clause and thus it should return the set difference of [2,3] and the new list should be [[2,3],[4,5,7],[-2,4]].
Currently I have the following step predicate:
memberlist(X,[[X|_]|_]).
memberlist(X,[[_|T1]|T2]) :-
memberlist(X,[T1|T2]).
memberlist(X,[[]|T2]) :-
memberlist(X,T2).
step([]).
step([_|T]) :-
memberlist(neg X,T),
write(X),
nl,
step(T).
step([_|T]) :-
step(T).
So it simply checks each list and checks if the negation of a variable exists and if it does simply write it out. I've already added code which deals with negative numbers, so X will match a -X, X being any integer.
I'm quite stuck at this point, and any help would be greatly appreciated.
Another possible solution:
shrink([L1|R1], [L3|R2]) :-
select(L2, R1, R2),
difference(L1, L2, L3).
shrink([L1|R1], [L1|S]) :-
shrink(R1, S).
difference(L1, L2, L3) :-
select(X, L1, R1),
compatible(X, Y),
select(Y, L2, R2),
union(R1, R2, L3).
compatible(neg(P), P) :- !.
compatible(P, neg(P)).
Some examples:
?- shrink([[1,2], [neg(1),3], [4,5,6], [neg(2),4]], S).
S = [[2, 3], [4, 5, 6], [neg(2), 4]] ;
S = [[1, 4], [neg(1), 3], [4, 5, 6]] ;
false.
?- shrink([[a,neg(b)], [a,b]], S).
S = [[a]] ;
false.
?- shrink([[rainning], [neg(rainning)]], S).
S = [[]] ;
false.
?- shrink([[rainning], [neg(rainning), wet_grass], [neg(wet_grass), green_grass]], S).
S = [[wet_grass], [neg(wet_grass), green_grass]] ;
S = [[rainning], [neg(rainning), green_grass]] ;
false.
?- shrink([[neg(green_grass)], [rainning], [neg(rainning), wet_grass], [neg(wet_grass), green_grass]], A), shrink(A, B), shrink(B, C).
A = [[neg(wet_grass)], [rainning], [neg(rainning), wet_grass]],
B = [[neg(rainning)], [rainning]],
C = [[]] ;
A = [[neg(wet_grass)], [rainning], [neg(rainning), wet_grass]],
B = [[neg(wet_grass)], [wet_grass]],
C = [[]] ;
A = [[neg(green_grass)], [wet_grass], [neg(wet_grass), green_grass]],
B = [[neg(wet_grass)], [wet_grass]],
C = [[]] ;
A = [[neg(green_grass)], [wet_grass], [neg(wet_grass), green_grass]],
B = [[neg(green_grass)], [green_grass]],
C = [[]] ;
A = [[neg(green_grass)], [rainning], [neg(rainning), green_grass]],
B = [[neg(rainning)], [rainning]],
C = [[]] ;
A = [[neg(green_grass)], [rainning], [neg(rainning), green_grass]],
B = [[neg(green_grass)], [green_grass]],
C = [[]] ;
false.
An alternative formulation.
memberlist will give you the triples p(X, Y, Z) where Z and neg(Z) are in X and Y.
collapse will take such a triple and remove X, Y from Xs and add X+Y-Z-neg(Z) to it.
memberlist([X|Xs], p(X, Y, Z)) :-
member(Z, X), member(Y, Xs), member(neg(Z), Y).
memberlist([X|Xs], p(X, Y, Z)) :-
member(neg(Z), X), member(Y, Xs), member(Z, Y).
memberlist([_|Xs], A) :-
memberlist(Xs, A).
collapse(Xs, Ys) :-
memberlist(Xs, p(A, B, I)), % A and B have some I and neg(I) in them
select(A, Xs, XsA), % remove A
select(B, XsA, XsAB), % remove B
append(A, B, AB), select(I, AB, ABI), select(neg(I), ABI, ABII),
Ys = [ABII|XsAB].
Your example
?- collapse([[1, 2], [neg(1), 3], [4, 5, 7], [neg(2), 4]], X).
X = [[2, 3], [4, 5, 7], [neg(2), 4]] ;
X = [[1, 4], [neg(1), 3], [4, 5, 7]] ;
false.
I need to find the combinations in a list of lists. For example, give the following list,
List = [[1, 2], [1, 2, 3]]
These should be the output,
Comb = [[1,1],[1,2],[1,3],[2,1],[2,2],[2,3]]
Another example:
List = [[1,2],[1,2],[1,2,3]]
Comb = [[1,1,1],[1,1,2],[1,1,3],[1,2,1],[1,2,2],[1,2,3]....etc]
I know how to do it for a list with two sublists but it needs to work for any number of sublists.
I'm new to Prolog, please help.
This answer hunts the bounty offered "for a pure solution that also takes into account for Ess".
Here we generalize this previous
answer like so:
list_crossproduct(Xs, []) :-
member([], Xs).
list_crossproduct(Xs, Ess) :-
Ess = [E0|_],
same_length(E0, Xs),
maplist(maybelonger_than(Ess), Xs),
list_comb(Xs, Ess).
maybelonger_than(Xs, Ys) :-
maybeshorter_than(Ys, Xs).
maybeshorter_than([], _).
maybeshorter_than([_|Xs], [_|Ys]) :-
maybeshorter_than(Xs, Ys).
list_crossproduct/2 gets bidirectional by relating Xs and Ess early.
?- list_comb(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
nontermination % BAD!
?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5]]).
Xs = [[1],[2],[3,4,5]] % this now works, too
; false.
Sample query having multiple answers:
?- list_crossproduct(Xs, [[1,2,3],[1,2,4],[1,2,5],X,Y,Z]).
X = [1,2,_A],
Y = [1,2,_B],
Z = [1,2,_C], Xs = [[1],[2],[3,4,5,_A,_B,_C]]
; X = [1,_A,3],
Y = [1,_A,4],
Z = [1,_A,5], Xs = [[1],[2,_A],[3,4,5]]
; X = [_A,2,3],
Y = [_A,2,4],
Z = [_A,2,5], Xs = [[1,_A],[2],[3,4,5]]
; false.
For completeness, here is the augmented version of my comment-version. Note nilmemberd_t/2 which is inspired by memberd_t/2.
nilmemberd_t([], false).
nilmemberd_t([X|Xs], T) :-
if_(nil_t(X), T = true, nilmemberd_t(Xs, T)).
nil_t([], true).
nil_t([_|_], false).
list_comb(List, []) :-
nilmemberd_t(List, true).
list_comb(List, Ess) :-
bagof(Es, maplist(member,Es,List), Ess).
Above version shows that "only" the first clause was missing in my comment response. Maybe even shorter with:
nilmemberd([[]|_]).
nilmemberd([[_|_]|Nils]) :-
nilmemberd(Nils).
This should work for Prologs without constraints. With constraints, bagof/3 would have to be reconsidered since copying constraints is an ill-defined terrain.
Here's a way to do it using maplist/3 and append/2:
list_comb([], [[]]).
list_comb([Xs|Xss], Ess) :-
Xs = [_|_],
list_comb(Xss, Ess0),
maplist(aux_x_comb(Ess0), Xs, Esss1),
append(Esss1, Ess).
aux_x_comb(Ess0, X, Ess1) :-
maplist(head_tail_list(X), Ess0, Ess1).
head_tail_list(X, Xs, [X|Xs]).
Sample query:
?- list_comb([[a,b],[f,g],[x,y,z]], Ess).
Ess = [[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z],
[b,f,x],[b,f,y],[b,f,z],
[b,g,x],[b,g,y],[b,g,z]].
Here's how it works!
As an example, consider these goals:
list_comb([[a,b],[f,g],[x,y,z]], Ess)
list_comb([ [f,g],[x,y,z]], Ess0)
How can we get from Ess0 to Ess?
We look at the answers to the
latter query:
?- list_comb([[f,g],[x,y,z]], Ess0).
Ess0 = [[f,x],[f,y],[f,z], [g,x],[g,y],[g,z]].
... place a before [f,x], ..., [g,z] ...
?- maplist(head_tail_list(a),
[[f,x],[f,y],[f,z],
[g,x],[g,y],[g,z]], X).
X = [[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z]].
... then do the same for b.
maplist(aux_x_comb) helps us handle all items:
?- maplist(aux_x_comb([[f,x],[f,y],[f,z],
[g,x],[g,y],[g,z]]),
[a,b], X).
X = [[[a,f,x],[a,f,y],[a,f,z],
[a,g,x],[a,g,y],[a,g,z]],
[[b,f,x],[b,f,y],[b,f,z],
[b,g,x],[b,g,y],[b,g,z]]].
To get from a list of lists to a list use append/2.
I hope this explanation was more eludicating than confusing:)
A twist in #false's approach:
%list_comb( ++LL, -Ess)
list_comb( LL, Ess):-
is_list( LL),
maplist( is_list, LL),
findall( Es, maplist( member, Es, LL), Ess).
Testing:
41 ?- list_comb( [[1,2],[1],[1]], X).
X = [[1, 1, 1], [2, 1, 1]].
42 ?- list_comb( [[1,2],[1],[1,2,3]], X).
X = [[1, 1, 1], [1, 1, 2], [1, 1, 3], [2, 1, 1], [2, 1, 2], [2, 1, 3]].
43 ?- list_comb( [[1,2],[],[1,2,3]], X).
X = [].
44 ?- list_comb( [[1,2],t,[1,2,3]], X).
false.
45 ?- list_comb( t, X).
false.
i'm starting up learning prolog (i use SWI-prolog) and i did a simple exercise in which i have 2 lists and i want to calculate their intersection and union.
Here is my code that works pretty well but i was asking myself if there is a better way to do it as i don't like to use the CUT operator.
intersectionTR(_, [], []).
intersectionTR([], _, []).
intersectionTR([H1|T1], L2, [H1|L]):-
member(H1, L2),
intersectionTR(T1, L2, L), !.
intersectionTR([_|T1], L2, L):-
intersectionTR(T1, L2, L).
intersection(L1, L2):-
intersectionTR(L1, L2, L),
write(L).
unionTR([], [], []).
unionTR([], [H2|T2], [H2|L]):-
intersectionTR(T2, L, Res),
Res = [],
unionTR([], T2, L),
!.
unionTR([], [_|T2], L):-
unionTR([], T2, L),
!.
unionTR([H1|T1], L2, L):-
intersectionTR([H1], L, Res),
Res \= [],
unionTR(T1, L2, L).
unionTR([H1|T1], L2, [H1|L]):-
unionTR(T1, L2, L).
union(L1, L2):-
unionTR(L1, L2, L),
write(L).
Keep in mind that i want to have just 1 result, not multiple results (even if correct) so running the code with this:
?- intersect([1,3,5,2,4] ,[6,1,2]).
should exit with:
[1,2]
true.
and not with
[1,2]
true ;
[1,2]
true ;
etc...
The same must be valid for union predicate.
As i said my code works pretty well but please suggest better ways to do it.
Thanks
Also, not sure why you're dead against cuts, so long as their removal would not change the declaritive meaning of the code, as per your link. For example:
inter([], _, []).
inter([H1|T1], L2, [H1|Res]) :-
member(H1, L2),
inter(T1, L2, Res).
inter([_|T1], L2, Res) :-
inter(T1, L2, Res).
test(X):-
inter([1,3,5,2,4], [6,1,2], X), !.
test(X).
X = [1, 2].
In the test bit where I call the code, I'm just saying do the intersection but I'm only interested in the first answer. There are no cuts in the predicate definitions themselves.
The following is based on my previous answer to Remove duplicates in list (Prolog);
the basic idea is, in turn, based on #false's answer to Prolog union for A U B U C.
What message do I want to convey to you?
You can describe what you want in Prolog with logical purity.
Using if_/3 and (=)/3 a logically pure implementation can be
both efficient (leaving behind choice points only when needed)
and monotone (logically sound with regard to generalization / specialization).
The implementation of #false's predicates if_/3 and (=)/3 does use meta-logical Prolog features internally, but (from the outside) behaves logically pure.
The following implementation of list_list_intersection/3 and list_list_union/3 uses list_item_isMember/3 and list_item_subtracted/3, defined in a previous answer:
list_list_union([],Bs,Bs).
list_list_union([A|As],Bs1,[A|Cs]) :-
list_item_subtracted(Bs1,A,Bs),
list_list_union(As,Bs,Cs).
list_list_intersection([],_,[]).
list_list_intersection([A|As],Bs,Cs1) :-
if_(list_item_isMember(Bs,A), Cs1 = [A|Cs], Cs1 = Cs),
list_list_intersection(As,Bs,Cs).
Here's the query you posted as part of your question:
?- list_list_intersection([1,3,5,2,4],[6,1,2],Intersection).
Intersection = [1, 2]. % succeeds deterministically
Let's try something else... The following two queries should be logically equivalent:
?- A=1,B=3, list_list_intersection([1,3,5,2,4],[A,B],Intersection).
A = 1,
B = 3,
Intersection = [1, 3].
?- list_list_intersection([1,3,5,2,4],[A,B],Intersection),A=1,B=3.
A = 1,
B = 3,
Intersection = [1, 3] ;
false.
And... the bottom line is?
With pure code it's easy to stay on the side of logical soundness.
Impure code, on the other hand, more often than not acts like "it does what it should" at first sight, but shows all kinds of illogical behaviour with queries like the ones shown above.
Edit 2015-04-23
Neither list_list_union(As,Bs,Cs) nor list_list_intersection(As,Bs,Cs) guarantee that Cs doesn't contain duplicates. If that bothers you, the code needs to be adapted.
Here are some more queries (and answers) with As and/or Bs containing duplicates:
?- list_list_intersection([1,3,5,7,1,3,5,7],[1,2,3,1,2,3],Cs).
Cs = [1, 3, 1, 3].
?- list_list_intersection([1,2,3],[1,1,1,1],Cs).
Cs = [1].
?- list_list_union([1,3,5,1,3,5],[1,2,3,1,2,3],Cs).
Cs = [1, 3, 5, 1, 3, 5, 2, 2].
?- list_list_union([1,2,3],[1,1,1,1],Cs).
Cs = [1, 2, 3].
?- list_list_union([1,1,1,1],[1,2,3],Cs).
Cs = [1, 1, 1, 1, 2, 3].
Edit 2015-04-24
For the sake of completeness, here's how we could enforce that the intersection and the union are sets---that is lists that do not contain any duplicate elements.
The following code is pretty straight-forward:
list_list_intersectionSet([],_,[]).
list_list_intersectionSet([A|As1],Bs,Cs1) :-
if_(list_item_isMember(Bs,A), Cs1 = [A|Cs], Cs1 = Cs),
list_item_subtracted(As1,A,As),
list_list_intersectionSet(As,Bs,Cs).
list_list_unionSet([],Bs1,Bs) :-
list_setB(Bs1,Bs).
list_list_unionSet([A|As1],Bs1,[A|Cs]) :-
list_item_subtracted(As1,A,As),
list_item_subtracted(Bs1,A,Bs),
list_list_unionSet(As,Bs,Cs).
Note that list_list_unionSet/3 is based on list_setB/2, defined here.
Now let's see both list_list_intersectionSet/3 and list_list_unionSet/3 in action:
?- list_list_unionSet([1,2,3,1,2,3,3,2,1],[4,5,6,2,7,7,7],Xs).
Xs = [1, 2, 3, 4, 5, 6, 7].
?- list_list_intersectionSet([1,2,3,1,2,3,3,2,1],[4,5,6,2,7,7,7],Xs).
Xs = [2].
Edit 2019-01-30
Here is an additional query taken from #GuyCoder's comment (plus two variants of it):
?- list_list_unionSet(Xs,[],[a,b]).
Xs = [a,b]
; Xs = [a,b,b]
; Xs = [a,b,b,b]
...
?- list_list_unionSet([],Xs,[a,b]).
Xs = [a,b]
; Xs = [a,b,b]
; Xs = [a,b,b,b]
...
?- list_list_unionSet(Xs,Ys,[a,b]).
Xs = [], Ys = [a,b]
; Xs = [], Ys = [a,b,b]
; Xs = [], Ys = [a,b,b,b]
...
With the old version of list_item_subtracted/3, above queries didn't terminate existentially.
With the new one they do.
As the solution set size is infinite, none of these queries terminate universally.
To cheat slightly less than my first answer, you could use the findall higher order predicate which gets Prolog to do the recursion for you :
4 ?- L1=[1,3,5,2,4], L2=[6,1,2], findall(X, (nth0(N, L1, X), member(X, L2)), Res).
L1 = [1, 3, 5, 2, 4],
L2 = [6, 1, 2],
Res = [1, 2].
If the aim is to just 'get the job done', then swi prolog has built in primitives for exactly this purpose:
[trace] 3 ?- intersection([1,3,5,2,4] ,[6,1,2], X).
intersection([1,3,5,2,4] ,[6,1,2], X).
X = [1, 2].
[trace] 4 ?- union([1,3,5,2,4] ,[6,1,2], X).
X = [3, 5, 4, 6, 1, 2].
Try this, analogue to union/3 here:
:- use_module(library(clpfd)).
member(_, [], 0).
member(X, [Y|Z], B) :-
(X #= Y) #\/ C #<==> B,
member(X, Z, C).
intersect([], _, []).
intersect([X|Y], Z, T) :-
freeze(B, (B==1 -> T=[X|R]; T=R)),
member(X, Z, B),
intersect(Y, Z, R).
It works if the elements are integer, and doesn't leave any choise point:
?- intersect([X,Y],[Y,Z],L).
freeze(_15070, (_15070==1->L=[X, Y];L=[Y])),
_15070 in 0..1,
_15166#\/_15168#<==>_15070,
_15166 in 0..1,
X#=Y#<==>_15166,
X#=Z#<==>_15168,
Y#=Z#<==>_15258,
_15168 in 0..1,
_15258 in 0..1.
?- intersect([X,Y],[Y,Z],L), X=1, Y=2, Z=3.
X = 1,
Y = 2,
Z = 3,
L = [2].
?- intersect([X,Y],[Y,Z],L), X=3, Y=2, Z=3.
X = Z, Z = 3,
Y = 2,
L = [3, 2].
And finally (really), you could use findall to find all the solutions, then use nth0 to extract the first one, which will give you the result you want without cuts, and keeps the predicates nice and clean, without have any additional predicates to trap/stop prolog doing what it does best - backtracking and finding multiple answers.
Edit: It's arguable that putting in extra predicates in the 'core logic' to prevent multiple results being generated, is as ugly/confusing as using the cuts that you are trying to avoid. But perhaps this is an academic exercise to prove that it can be done without using higher order predicates like findall, or the built-ins intersection/union.
inter([], _, []).
inter([H1|T1], L2, [H1|Res]) :-
member(H1, L2),
inter(T1, L2, Res).
inter([_|T1], L2, Res) :-
inter(T1, L2, Res).
test(First):-
findall(Ans, inter([1,3,5,2,4], [6,1,2], Ans), Ansl),
nth0(0, Ansl, First).
% Element X is in list?
pert(X, [ X | _ ]).
pert(X, [ _ | L ]):- pert(X, L).
% Union of two list
union([ ], L, L).
union([ X | L1 ], L2, [ X | L3 ]):- \+pert(X, L2), union(L1, L2, L3).
union([ _ | L1 ], L2, L3):- union(L1, L2, L3).
% Intersection of two list
inter([ ], _, [ ]).
inter([ X | L1 ], L2, [ X | L3 ]):- pert(X, L2), inter(L1, L2, L3).
inter([ _ | L1 ], L2, L3):- inter(L1, L2, L3).
I know this post is very old but I found a solution with minimum coding.
% intersection
intersection([],L1,L2,L3).
intersection([H|T],L2,L3,[H|L4]):-member(H,L2),intersection(T,L3,L3,L4).
% member
member(H,[H|T]).
member(X,[H|T]):-member(X,T).
To test the above code you should not enter L3. Here is an examples.
?- intersection([w,4,g,0,v,45,6],[x,45,d,w,30,0],L).
L = [w, 0, 45].
I want to access list permutation and pass it as argument to other functions.
This is the permutation code:
takeout(X,[X|R],R).
takeout(X,[F|R],[F|S]) :-
takeout(X,R,S),
write(S).
perm([X|Y],Z) :-
perm(Y,W),
takeout(X,Z,W).
perm([],[]).
To start with, let's redefine your predicates so they don't do any unnecessary I/O:
takeout(X,[X|R],R).
takeout(X,[F |R],[F|S]) :- takeout(X,R,S).
perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W).
perm([],[]).
Now you have what could be considered a "pure" permutation function:
?- perm([1,2,3], X).
X = [1, 2, 3] ;
X = [2, 1, 3] ;
X = [2, 3, 1] ;
X = [1, 3, 2] ;
X = [3, 1, 2] ;
X = [3, 2, 1] ;
false.
So, suppose you have a max_heap function that takes a list of values and produces a tree. I'll let you worry about that, so let's just posit that it exists and is called max_heap/2 and let's further posit that you have a way to display this attractively called display_heap/1. To "take" the permutation and "send" it as a parameter to these functions, you're really saying in math-ese: suppose P is a permutation of X, let's make a max_heap with it and display it. Or, suppose P is a permutation of X, H is a max heap made from X, let's display H:
show_heaps(List) :- perm(List, P), max_heap(P, H), display_heap(H).
This says the same thing as my English sentence: suppose P is a permutation of the list, then H is a heap representation of it, then display it. Technically, display_heap/1 is still a predicate which could be true or false for a given heap. In practice, it will always be true, and if you run this you'll still have to hit ; repeatedly to say, give me another solution, unless you use a failure-driven loop or an extralogical predicate like findall/3 to cause all the solutions to be found.
Edit: Let's discuss failure-driven loops and findall/3. First let me add some new predicates, because I don't know exactly what you're doing, but it doesn't matter for our purposes.
double([X|Xs], [Y|Ys]) :- Y is X*2, double(Xs, Ys).
double([],[]).
showlist(Xs) :- print(Xs).
So now I have a predicate double/2 which doubles the values in the list and a predicate showlist/1 that prints the list on standard output. We can try it out like so:
?- perm([1,2,3], X), double(X, Y), showlist(Y).
[2,4,6]
X = [1, 2, 3],
Y = [2, 4, 6] ;
[4,2,6]
X = [2, 1, 3],
Y = [4, 2, 6] ;
[4,6,2]
X = [2, 3, 1],
Y = [4, 6, 2] ;
[2,6,4]
X = [1, 3, 2],
Y = [2, 6, 4] ;
[6,2,4]
X = [3, 1, 2],
Y = [6, 2, 4] ;
[6,4,2]
X = [3, 2, 1],
Y = [6, 4, 2] ;
false.
When you type ; you're saying, "or?" to Prolog. In other words, you're saying "what else?" You're telling Prolog, in effect, this isn't the answer I want, try and find me another answer I like better. You can formalize this process with a failure-driven loop:
?- perm([1,2,3], X), double(X, Y), showlist(Y), fail.
[2,4,6][4,2,6][4,6,2][2,6,4][6,2,4][6,4,2]
false.
So now you see the output from each permutation having gone through double/2 there, and then Prolog reported false. That's what one means by something like this:
show_all_heaps(List) :- perm(List, X), double(X, Y), showlist(Y), nl, fail.
show_all_heaps(_).
Look at how that works:
?- show_all_heaps([1,2,3]).
[2,4,6]
[4,2,6]
[4,6,2]
[2,6,4]
[6,2,4]
[6,4,2]
true.
The other option is using findall/3, which looks more like this:
?- findall(Y, (perm([1,2,3], X), double(X, Y)), Ys).
Ys = [[2, 4, 6], [4, 2, 6], [4, 6, 2], [2, 6, 4], [6, 2, 4], [6, 4, 2]].
Using this to solve your problem is probably beyond the scope of whatever homework it is you're working on though.
We can define list_permutation/2 based on same_length/2 and select/3 like this:
:- use_module(library(lists),[same_length/2,select/3]).
list_permutation(As,Bs) :-
same_length(As,Bs), % redundant goal helps termination
list_permutation_(As,Bs).
list_permutation_([],[]).
list_permutation_([A|As],Bs0) :-
select(A,Bs0,Bs),
list_permutation_(As,Bs).
Thanks to same_length/2, both of the following queries1,2 terminate universally:
?- list_permutation([1,2,3],Ys).
Ys = [1,2,3]
; Ys = [1,3,2]
; Ys = [2,1,3]
; Ys = [3,1,2]
; Ys = [2,3,1]
; Ys = [3,2,1]
; false.
?- list_permutation(Xs,[1,2,3]).
Xs = [1,2,3]
; Xs = [1,3,2]
; Xs = [2,1,3]
; Xs = [2,3,1]
; Xs = [3,1,2]
; Xs = [3,2,1]
; false.
So far, so good. But what does the answer sequence look like if there are duplicate list items?
?- list_permutation([1,1,1],Ys).
Ys = [1,1,1]
; Ys = [1,1,1]
; Ys = [1,1,1]
; Ys = [1,1,1]
; Ys = [1,1,1]
; Ys = [1,1,1]
; false.
5/6 answers are redundant! What can we do? We simply use selectd/3 instead of select/3!
list_permuted(As,Bs) :-
same_length(As,Bs),
list_permuted_(As,Bs).
list_permuted_([],[]).
list_permuted_([A|As],Bs0) :-
selectd(A,Bs0,Bs), % use selectd/3, not select/3
list_permuted_(As,Bs).
Let's re-run above query that gave us 5 redundant solutions before!
?- list_permuted([1,1,1],Ys).
Ys = [1,1,1]
; false.
?- list_permuted(Xs,[1,1,1]).
Xs = [1,1,1]
; false.
Better! All redundant answers are gone.
Let's compare the solution set for some sample case:
?- _Xs = [1,2,1,1,2,1,1,2,1],
setof(Ys,list_permutation(_Xs,Ys),Yss),
setof(Ys,list_permuted(_Xs,Ys),Yss),
length(Yss,N).
N = 84, Yss = [[1,1,1,1,1,1,2,2,2],[1,1,1,1,1,2,1,2,2],[...|...]|...].
OK! How about empirical runtime measurements with a problem of a slightly bigger size?
We use call_time/2 for measuring the runtime in milli-seconds T_ms.
?- call_time(\+ (list_permutation([1,2,1,1,1,2,1,1,1,2,1],_),false),T_ms).
T_ms = 8110.
?- call_time(\+ (list_permuted( [1,2,1,1,1,2,1,1,1,2,1],_),false),T_ms).
T_ms = 140.
OK! And with proper compilation of if_/3 and (=)/3, list_permuted/2 is even faster!
Footnote 1: Using SICStus Prolog version 4.3.2 (x86_64-linux-glibc2.12).
Footnote 2: The answers given by the Prolog toplevel have been post-processed for the sake of readability.
If you just want to explore the permutations without the "False" in the end, this code might be helpful
takeout(X,[F |R],[F|S]) :- F\=X, takeout(X,R,S).
takeout(X,[X|R],R).
perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W).
perm([],[]).
So, the output of perm([a,b],B) would be
B=[a,b]
B=[b,a]