Prolog intersection of three lists - list

please can give some of you advice about interception of three lists in Prolog?
I done interception of two lists:
prunik([], _, []).
prunik([H1|T1], L2, [H1|Res]) :-
member(H1, L2),
prunik(T1, L2, Res).
prunik([_|T1], L2, Res) :-
prunik(T1, L2, Res).
And it works, when I put this question:
prunik([1,3,5,2,4], [6,1,2], X).
I try remake this for three lists, but I really dont know. Any advice, please?


Your code for intersecting two lists is a bit faulty, see the last 3 answers:
?- prunik([1,3,5,2,4], [6,1,2], X).
X = [1, 2] ;
X = [1] ;
X = [2] ;
X = [].
But if you would like to extend your code to work on three lists:
prunik([], _, _, []).
prunik([H1|T1], L2, L3, [H1|Res]) :-
member(H1, L2),
member(H1, L3),
prunik(T1, L2, L3, Res).
prunik([_|T1], L2, L3, Res) :-
prunik(T1, L2, L3, Res).
Sample input/output:
?- prunik([1,2,3], [3,2,4,5,6], [2,3,4], R).
R = [2, 3] ;
R = [2] ;
R = [3] ;
R = [].

Related

SWI Prolog - Combining elements from two lists into pairs

How can I "join" two lists while keeping their elements's positions? Like, I have List1 = [1,2,3] and List2 = [a,b,c]. How can I get the list [[1,a], [2,b], [3,c]]?
I've been trying to look for built-in predicates but all I've found can't reproduce this.

join_2_lists(L1, L2, JoinLst) :-
must_be(list, L1),
must_be(list, L2),
join_2_lists_(L1, L2, JoinLst).
join_2_lists_([], [], []).
join_2_lists_([H1|T1], [H2|T2], [[H1, H2]|JoinLst]) :-
join_2_lists_(T1, T2, JoinLst).
Result in swi-prolog:
?- join_2_lists([1, 2, 3], [a, b, c], J).
J = [[1,a],[2,b],[3,c]].

Predicate to unzip a list

List1=[(x,1),(y,1),(z,1)]
I'm attempting to split this list:
into two lists:
List3=[x,y,z]
List4=[1,1,1]
So I have written this predicate to try to do it:
splt([], [], []).
splt([X|Xs], [Y|Ys], [X,Y|Zs]) :-
splt(Xs,Ys,Zs).
However instead of the desired result, the predicate returns:
1 ?- splt([(x,1),(y,2),(z,3)],L3,L4).
L3 = [_G1760, _G1769, _G1778],
L4 = [ (z, 1), _G1760, (y, 2), _G1769, (z, 3), _G1778].

First, the term you have chosen. This: (a, b), is most definitely not how you would usually represent a "tuple" in Prolog. You almost always use a-b for a "pair", and pairs are used throughout the standard libraries.
So your initial list would look like this: [x-1, y-1, z-1].
This should also explain why you are having your problem. You write (a, b), but your predicate says a, b, and you consume two elements when you expect to get one ,(a,b) term. So, to fix your current predicate you would write:
split([], [], []).
split([X|Xs], [Y|Ys], [(X,Y)|XYs]) :-
split(Xs, Ys, XYs).
?- split(Xs, Ys, [(x,1), (y,1), (z,1)]).
Xs = [x, y, z],
Ys = [1, 1, 1].
But instead, using a more conventional name, term order, and Prolog pairs:
zip([], [], []).
zip([X-Y|XYs], [X|Xs], [Y|Ys]) :-
zip(XYs, Xs, Ys).
?- zip([x-1, y-1, z-1], Xs, Ys).
Xs = [x, y, z],
Ys = [1, 1, 1].
And of course, SWI-Prolog at least has a library(pairs), and it comes with a pairs_keys_values/3:
?- pairs_keys_values([x-1, y-1, z-1], Xs, Ys).
Xs = [x, y, z],
Ys = [1, 1, 1].

I find comfortable using library(yall):
?- maplist([(X,Y),X,Y]>>true, [(x,1),(y,2),(z,3)],L3,L4).
L3 = [x, y, z],
L4 = [1, 2, 3].
or, maybe clearer
?- maplist([A,B,C]>>(A=(B,C)), [(x,1),(y,2),(z,3)],L3,L4).
L3 = [x, y, z],
L4 = [1, 2, 3].

You're matching the tuple as a whole, rather than it's component parts.
You should match on [(X1,Y1)|XS], instead of [X|XS] and [Y|Ys].
splt([],[],[]).
splt([(X1,Y1)|Xs],[X1|T1],[Y1|T2]):-
splt(Xs,T1,T2).
Here the first term is used as input, the second and third as output.
Ideone example, using SWI-Prolog, here.

Decomposing a matrix to a list of first element and sub list of remaining elements in prolog

Test Case
?- decompose([[1,2,8],[3,4],[5,6]], L1, L2).
L1 = [1,3,5], L2 = [[2,8],[4],[6]] ? ;
no
I had tried another implementation however the feedback given was that it was inefficient.
The inefficient implementation
listFirst([], []).
listFirst([H1|T1], [H2|Z]):-
H1 = [H2|_],
listFirst(T1, Z).
listFollowers([], []).
listFollowers([H1|T1], [T2|Z]):-
H1 = [H2|T2],
listFollowers(T1, Z).
decompose(A,L1,L2) :-
listFollowers(A, L2),
listFirst(A, L1).

Following up on #findall's previous answer... How about using meta-predicate maplist/4?
list_head_tail([X|Xs], X, Xs).
decompose(Mss, Hs, Ts) :-
maplist(list_head_tail, Mss, Hs, Ts).
Sample queries:
?- decompose([[a,b,c],[d,e,f]], Heads, Tails).
Heads = [a,d], Tails = [[b,c],[e,f]].
?- decompose([[1,2,8],[3,4],[5,6]], L1, L2).
L1 = [1,3,5], L2 = [[2,8],[4],[6]].

As lurker says, the functions of your listFirst and listFollowers can be combined into a predicate to do those at once. Like this;
decompose([[H|T]|T0], [H|L1], [T|L2]) :- decompose(T0, L1, L2).
decompose([], [], []).

Intersection and union of 2 lists

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].

Working with list of lists in Prolog

Please help me to solve this problem:
I have a list of lists
[[1,2],[3,4]]
How do I get:
[1,3]
[1,4]
[2,3]
[2,4]
Or if I have a list of lists
[[1,2],[3,4],[6,7]]
How do I get:
[1,3,6]
[1,3,7]
[1,4,6]
[1,4,7]
[2,3,6]
[2,3,7]
[2,4,6]
[2,4,7]

The predicate for accessing a single list element is the most basic Prolog building block: member/2.
And you want a list of all lists' elements: maplist/3 does such mapping. Thus we can write
combine(Ls, Rs) :-
maplist(get1, Ls, Rs).
get1(L, E) :-
member(E, L).
note that get1/2 is only required so that we swap the member/2 arguments. But because in (pure) Prolog we are describing relations between arguments, we can swap arguments' order and simplify it even more:
combine(Ls, Rs) :-
maplist(member, Rs, Ls).
Test output:
?- combine( [[1,2],[a,b]], Xs).
Xs = [1, a] ;
Xs = [1, b] ;
Xs = [2, a] ;
Xs = [2, b].
%% this is the same as:
%% maplist( member, Xs, [[1,2],[a,b]]) :-
%% member( X1, [1,2] ),
%% member( X2, [a,b]), Xs = [X1,X2].
edit
A joke: really, my first combine/2 should have been written like
combine(Ls, Rs) :-
maplist(rebmem, Ls, Rs).
rebmem(L, E) :-
member(E, L).

You can do something like this:
lists([], []).
lists([[Head|_]|Lists], [Head|L]):-
lists(Lists, L).
lists([[_,Head|Tail]|Lists], L):-
lists([[Head|Tail]|Lists], L).
That is, take the first element of the first list in your input list and continue recursively with the remaining lists. As a second chance, skip that element and redo with the remaining elements.