Definition of a path/trail/walk - prolog

Many predicates define some kind of an acyclic path built from edges defined via a binary relation, quite similarly to defining transitive closure. A generic definition is thus called for.
Note that the notions defined in graph theory do not readily match what is commonly expected. Most notably, we are not interested in the edges' names.
Worse, also graph theory has changed a bit, introducing the notion of walk, noting
Traditionally, a path referred to what is now usually known as an open walk. Nowadays, when stated without any qualification, a path is usually understood to be simple, meaning that no vertices (and thus no edges) are repeated. (The term chain has also been used to refer to a walk in which all vertices and edges are distinct.)
So my question is: How to name and define this functionality?
What I have done so far is to define:
path(Rel_2, Path, X0,X)
The first argument has to be the continuation of the relation which is an incomplete goal that lacks two further arguments. Then comes either the Path or the pair of vertices.
Example usage
n(a, b).
n(b, c).
n(b, a).
?- path(n,Xs, a,X).
Xs = [a], X = a
; Xs = [a, b], X = b
; Xs = [a, b, c], X = c
; false.
Implementation
:- meta_predicate(path(2,?,?,?)).
:- meta_predicate(path(2,?,?,?,+)).
path(R_2, [X0|Ys], X0,X) :-
path(R_2, Ys, X0,X, [X0]).
path(_R_2, [], X,X, _).
path(R_2, [X1|Ys], X0,X, Xs) :-
call(R_2, X0,X1),
non_member(X1, Xs),
path(R_2, Ys, X1,X, [X1|Xs]).
non_member(_E, []).
non_member(E, [X|Xs]) :-
dif(E,X),
non_member(E, Xs).

How about defining path/4 like this?
path(R_2, Xs, A,Z) :- % A path `Xs` from `A` to `Z` is ...
walk(R_2, Xs, A,Z), % ... a walk `Xs` from `A` to `Z` ...
all_dif(Xs). % ... with no duplicates in `Xs`.
To aid universal termination, we swap the two goals in above conjunction ...
path(R_2, Xs, A,Z) :-
all_dif(Xs), % enforce disequality ASAP
walk(R_2, Xs, A,Z).
... and use the following lazy implementation of all_dif/1:
all_dif(Xs) :- % enforce pairwise term inequality
freeze(Xs, all_dif_aux(Xs,[])). % (may be delayed)
all_dif_aux([], _).
all_dif_aux([E|Es], Vs) :-
maplist(dif(E), Vs), % is never delayed
freeze(Es, all_dif_aux(Es,[E|Vs])). % (may be delayed)
walk/4 is defined like path/4 and path/5 given by the OP:
:- meta_predicate walk(2, ?, ?, ?).
walk(R_2, [X0|Xs], X0,X) :-
walk_from_to_step(Xs, X0,X, R_2).
:- meta_predicate walk_from_to_step(?, ?, ?, 2).
walk_from_to_step([], X,X, _).
walk_from_to_step([X1|Xs], X0,X, R_2) :-
call(R_2, X0,X1),
walk_from_to_step(Xs, X1,X, R_2).
IMO above path/4 is simpler and more approachable, particularly for novices. Would you concur?

I want to focus on naming the predicate.
Unlike maplist/2,
the argument order isn't of primary importance here.
The predicate name should make the meaning of the respective arguments clear.
So far, I like path_from_to_edges best, but it has its pros and cons, too.
path_from_to_edges(Path,From,To,Edges_2) :-
path(Edges_2,Path,From,To).
Let's pick it apart:
pro: path is a noun, it cannot be mis-read a verb. To me, a list of vertices is implied.
pro: from stands for a vertex, and so does to.
con: edges is somewhat vague, but using lambdas here is the most versatile choice.
con: According to Wikipedia, a path is a trail in which all vertices (except possibly the first and last) are distinct. So that would need to be clarified in the description.
Using lambdas for a lists of neighbor vertices Ess:
?- Ess = [a-[b],b-[c,a]],
From = a,
path_from_to_edges(Path,From,To,\X^Y^(member(X-X_neibs,Ess),member(Y,X_neibs))).
Ess = [a-[b],b-[c,a]], From = a, To = a, Path = [a]
; Ess = [a-[b],b-[c,a]], From = a, To = b, Path = [a,b]
; Ess = [a-[b],b-[c,a]], From = a, To = c, Path = [a,b,c]
; false.
Edit 2015-06-02
Another shot at better naming! This leans more on the side of maplist/2...
graph_path_from_to(P_2,Path,From,To) :-
path(P_2,Path,From,To).
Here, graph, of course, is a noun, not a verb.
Regarding the meaning of "path": paths definitely should allow From=To and not exclude that by default (with pairwise term inequalities). It is easy to exclude this with an additional dif(From,To) goal, but not the other way round.

I do not see the reason to define in path/4 the arguments "start node" and "end node". It seems that a simple path/2 with the rule and the list of nodes must be enough.
If the user wants a list starting with some node (by example, 'a'), he can query the statement as: path( some_rule, ['a'|Q] ).
A user could, by example, request for path that have length 10 in the way: length(P,10), path( some_rule, P).
* Addendum 1 *
Some utility goals can be easily added, but they are not the main subject. Example, path/3 with start node is:
path( some_rule, [start|Q], start ) :-
path ( some_rule, [start|Q ] ).
* Addendum 2 *
Addition of last node as argument could give the false idea that this argument drives the algorithm, but it doesn't. Assume by example:
n(a, b).
n(a, c).
n(a, d).
and trace algorithm execution for the query:
[trace] ?- path( n, P, X, d ).
Call: (6) path(n, _G1025, _G1026, d) ? creep
Call: (7) path(n, _G1107, _G1026, d, [_G1026]) ? creep
Exit: (7) path(n, [], d, d, [d]) ? creep
Exit: (6) path(n, [d], d, d) ? creep
P = [d],
X = d ;
Redo: (7) path(n, _G1107, _G1026, d, [_G1026]) ? creep
Call: (8) n(_G1026, _G1112) ? creep
Exit: (8) n(a, b) ? creep
Call: (8) non_member(b, [a]) ? creep
Call: (9) dif:dif(b, a) ? creep
Exit: (9) dif:dif(b, a) ? creep
Call: (9) non_member(b, []) ? creep
Exit: (9) non_member(b, []) ? creep
Exit: (8) non_member(b, [a]) ? creep
Call: (8) path(n, _G1113, b, d, [b, a]) ? creep
Call: (9) n(b, _G1118) ? creep
Fail: (9) n(b, _G1118) ? creep
Fail: (8) path(n, _G1113, b, d, [b, a]) ? creep
Redo: (9) non_member(b, []) ? creep
Fail: (9) non_member(b, []) ? creep
Fail: (8) non_member(b, [a]) ? creep
Redo: (8) n(_G1026, _G1112) ? creep
Exit: (8) n(a, c) ? creep
Call: (8) non_member(c, [a]) ? creep
Call: (9) dif:dif(c, a) ? creep
Exit: (9) dif:dif(c, a) ? creep
Call: (9) non_member(c, []) ? creep
Exit: (9) non_member(c, []) ? creep
Exit: (8) non_member(c, [a]) ? creep
Call: (8) path(n, _G1113, c, d, [c, a]) ? creep
Call: (9) n(c, _G1118) ? creep
Fail: (9) n(c, _G1118) ? creep
Fail: (8) path(n, _G1113, c, d, [c, a]) ? creep
Redo: (9) non_member(c, []) ? creep
Fail: (9) non_member(c, []) ? creep
Fail: (8) non_member(c, [a]) ? creep
Redo: (8) n(_G1026, _G1112) ? creep
Exit: (8) n(a, d) ? creep
Call: (8) non_member(d, [a]) ? creep
Call: (9) dif:dif(d, a) ? creep
Exit: (9) dif:dif(d, a) ? creep
Call: (9) non_member(d, []) ? creep
Exit: (9) non_member(d, []) ? creep
Exit: (8) non_member(d, [a]) ? creep
Call: (8) path(n, _G1113, d, d, [d, a]) ? creep
Exit: (8) path(n, [], d, d, [d, a]) ? creep
Exit: (7) path(n, [d], a, d, [a]) ? creep
Exit: (6) path(n, [a, d], a, d) ? creep
P = [a, d],
X = a .
as you can see, in this case algorithm fails to brute force.
For this reason, if algorithm is not improved, I suggest do not add "end node" as "path" argument.

Related

Why prolog fails to find a route to a vertices (in graph)?

I have a question to you:
I have an exercise that says:
Let there be given edges in the directed graph arc (a, b), arc (b, c), arc (a, d), arc (d, e), arc (d, f), arc (f, a) and arc (f, g). Test go / 2, go1 / 3 predicates.
I need to know why Prolog answers false for the following query:
? - go1 (a, b, X).
I have a graph illustrated below and a code pasted below.
The code is here:
arc(a,b).%these are the edges.
arc(b,c).
arc(a,d).
arc(d,e).
arc(d,f).
arc(f,a).
arc(f,g).
go(X,X).
go(X,Y):-arc(X,Z),go(Z,Y).
% without arc(f,a):
% yes ?- go(a,c).
% no ?- go(d,c).
% no ?- go(f,a).
% yes ?- go(a,g).
%
% with arc(f,a):
% yes ?- go(a,c).
% out of local stack ?- go(a,g).
member1(H,[H|_]).
member1(H,[_|T]):-member1(H,T).
go1(X,X,_).
go1(X,Y,T):-arc(X,Z),not(member1(Z,[X|T])),go1(Z,Y,[X|T]).
% with arc(f,a):
% yes ?- go1(a,c,[]).
% yes ?- go1(a,g,[]).
So guys, please help me to figure out why it happens? Why does Prolog answer false for the following query:
? - go1 (a, b, X).
I've tried to trace the query and I've gotten the answer that is below:
[trace] ?- go1(a,b,X).
Call: (8) go1(a, b, _982) ? creep
Call: (9) arc(a, _1202) ? creep
Exit: (9) arc(a, b) ? creep
^ Call: (9) not(member1(b, [a|_982])) ? creep
Call: (10) member1(b, [a|_982]) ? creep
Call: (11) member1(b, _982) ? creep
Exit: (11) member1(b, [b|_1206]) ? creep
Exit: (10) member1(b, [a, b|_1206]) ? creep
^ Fail: (9) not(user:member1(b, [a|_982])) ? creep
Redo: (9) arc(a, _1202) ? creep
Exit: (9) arc(a, d) ? creep
^ Call: (9) not(member1(d, [a|_982])) ? creep
Call: (10) member1(d, [a|_982]) ? creep
Call: (11) member1(d, _982) ? creep
Exit: (11) member1(d, [d|_1206]) ? creep
Exit: (10) member1(d, [a, d|_1206]) ? creep
^ Fail: (9) not(user:member1(d, [a|_982])) ? creep
Fail: (8) go1(a, b, _982) ? creep
false.
Well, what I don't understand is why we have here the answer below:
Exit: (11) member1(b, [b|_1206]) ? creep
Exit: (10) member1(b, [a, b|_1206]) ? creep
^ Fail: (9) not(user:member1(b, [a|_982])) ? creep
Why does the Prolog not recognize that there is an edge from a to b? Why does it say that b is in the list that there is only [a] after all. Why does this call fail there? What's the reason of the query failure?
You pass it a free variable X, so that means that the:
member(b, [a|X]).
call will succeed, indeed:
?- member(b, [a|X]).
X = [b|_2142] ;
X = [_2140, b|_2148] ;
X = [_2140, _2146, b|_2154]
Now not(member(Z, [X|T])) will succeed, given that the search of member(Z, [X|T]) fails, but here it will always succeed, since a free variable can always be grounded to a list that contains Z, hence it fails.
If you call it with:
go1 (a, b, [])
it will succeed, since then member(b, [a]) will indeed fail.
You can define a predicate that will reconstruct the path with:
go1(A, B, P) :-
go1(A, B, P, []).
go1(A, A, [A], _).
go1(A, C, [A|T], V) :-
arc(A, B),
\+ member(B, [A|V]),
go1(B, C, T, [A|V]).
and thus call it with go1(A, B, P) to obtain the path P.

confusion in how backtracking works given query with member predicate

Suppose i invoke the query member(a, [c,b,a,y]), the output is -
true ;
false.
How come we have "another solution"? Why we dont finish at true.?
Here's the trace:
%% [trace] ?- member(a,[c,b,a,y]).
%% Call: (6) member(a, [c, b, a, y]) ?
%% Call: (7) member(a, [b, a, y]) ?
%% Call: (8) member(a, [a, y]) ?
%% Exit: (8) member(a, [a, y]) ?
%% Exit: (7) member(a, [b, a, y]) ?
%% Exit: (6) member(a, [c, b, a, y]) ?
%% true ;
%% Redo: (8) member(a, [a, y]) ?
%% Call: (9) member(a, [y]) ?
%% Call: (10) member(a, []) ?
%% Fail: (10) member(a, []) ?
%% Fail: (9) member(a, [y]) ?
%% Fail: (8) member(a, [a, y]) ?
%% Fail: (7) member(a, [b, a, y]) ?
%% Fail: (6) member(a, [c, b, a, y]) ?
%% false.
I don't understand how can we backtrack to (8) and why?
member(X,[X|_]).
member(X,[Y|T]) :- member(X,T).
Let's re-write it as the nearly-equivalent one-liner,
membr(X,[A|B]):- ( X = A ; membr(X,B) ).
Tracing it can be a bit clearer:
[trace] 5 ?- trace, membr(a, [c,b,a,y]). X=a
Call: (8) membr(a, [c, b, a, y]) ? creep [A8|B8]=[c,b,a,y]
Call: (9) a=c ? creep X=A8 ?
Fail: (9) a=c ? creep false
Redo: (8) membr(a, [c, b, a, y]) ? creep membr(X,B8) ?
Call: (9) membr(a, [b, a, y]) ? creep [A9|B9]=B8
Call: (10) a=b ? creep X=A9 ?
Fail: (10) a=b ? creep false
Redo: (9) membr(a, [b, a, y]) ? creep membr(X,B9) ?
Call: (10) membr(a, [a, y]) ? creep [A10|B10]=B9
Call: (11) a=a ? creep X=A10 ?
Exit: (11) a=a ? creep true
Exit: (10) membr(a, [a, y]) ? creep success!
Exit: (9) membr(a, [b, a, y]) ? creep success!
Exit: (8) membr(a, [c, b, a, y]) ? creep success!
true ;
Redo: (10) membr(a, [a, y]) ? creep membr(X,B10) ?
Call: (11) membr(a, [y]) ? creep [A11|B11]=B10
Call: (12) a=y ? creep X=A11 ?
Fail: (12) a=y ? creep .........
Redo: (11) membr(a, [y]) ? creep .........
Call: (12) membr(a, []) ? creep
Fail: (12) membr(a, []) ? creep
Fail: (11) membr(a, [y]) ? creep
Fail: (10) membr(a, [a, y]) ? creep
Fail: (9) membr(a, [b, a, y]) ? creep
Fail: (8) membr(a, [c, b, a, y]) ? creep
false.
There's clearly a choice point waiting, after the first success is reached:
(10) membr(X,[A|B]):- ( X = A ; membr(X,B) ).
and after the success is reported, with all the frames still there on the stack, the top -- 10th -- frame is retried from the point where it has stopped, having succeeded:
(10) membr(X,[A|B]):- ( X = A ; membr(X,B) ).
The chain of exits is provisional: we've only traced back up the stack in order to be able to report the found value, as its bits and pieces are stored in the frames leading to the successful 11th frame (with X=a, A10=a, X=A10). The frames were not eliminated, for if they were, we'd only ever could report only the one first found value. But Prolog wants to report all of them.

Prolog - Confusion With Order of Printing in Recursion

Given the following program:
postIt([]).
postIt([c|R]) :- postIt(R), !, nl.
postIt([X|R]) :- postIt(R), write(X).
Query:
?- postIt([a,b,c,d,e]).
Trace:
[trace] ?-
| postIt([a,b,c,d,e]).
Call: (8) postIt([a, b, c, d, e]) ? creep
Call: (9) postIt([b, c, d, e]) ? creep
Call: (10) postIt([c, d, e]) ? creep
Call: (11) postIt([d, e]) ? creep
Call: (12) postIt([e]) ? creep
Call: (13) postIt([]) ? creep
Exit: (13) postIt([]) ? creep
Call: (13) write(e) ? creep
e
Exit: (13) write(e) ? creep
Exit: (12) postIt([e]) ? creep
Call: (12) write(d) ? creep
d
Exit: (12) write(d) ? creep
Exit: (11) postIt([d, e]) ? creep
Call: (11) nl ? creep
Exit: (11) nl ? creep
Exit: (10) postIt([c, d, e]) ? creep
Call: (10) write(b) ? creep
b
Exit: (10) write(b) ? creep
Exit: (9) postIt([b, c, d, e]) ? creep
Call: (9) write(a) ? creep
a
Exit: (9) write(a) ? creep
Exit: (8) postIt([a, b, c, d, e]) ? creep
true.
I don't quite understand why the output is 'ed' 'ba'. I figured that the program would print 'ab' skip 'c' then print 'de'. I am hoping someone may be able to clarify this for me.
Before I address the actual question, one additional point: Please consider using more_readable_names_with_underscores insteadOfStickingItAllTogetherLikeThis. Thus, I suggest for example the name post_it/1.
Next, to make our life simpler, I will only consider the following fragment of your code:
post_it([]).
post_it([X|R]) :- post_it(R), write(X).
That is, I have simply omitted the second clause entirely.
What do you expect now from the query ?- post_it([a,b,c,d]).? Is it abcd?
No, obviously it isn't:
?- post_it([a,b,c,d]).
dcba
Why? In your case, since you are using the impure predicate write/1, the reason for this can only be understood procedurally, that is taking into account the actual execution strategy of Prolog.
The execution strategy of Prolog is called, among other names, depth first search with chronological backtracking. When you have a clause like:
a :- b, c.
then, when a/0 is executed, b/0 is invoked first, and only if b/0 succeeds, c/0 is invoked.
Thus, consider for example the simpler query:
?- post_it([a,b]).
First, post_it([b]) is invoked, and it emits b (Exercise: Why?). Only then, a is emitted, because write(a) occurs as the second goal in post_it/1.
Obviously, in such simple cases, we can still somehow grasp the procedural definition. The major deficit of this is that this quickly becomes way too complex to understand, and so I can only recommend to avoid side-effects: They will invariably make your code too hard to understand, and you are already seeing the first signs of this.
Recommendation: Work in the pure subset of Prolog. In your case, consider the following relation between lists:
without_c(Ls0, Ls) :-
tfilter(dif(c), Ls0, Ls).
This uses tfilter/3 from Ulrich Neumerkel's library(reif) to declaratively describe the relation between two lists, where the second list is the same as the first, but without any occurrence of the atom c.
Example query:
?- without_c([a,b,c,d], Ls).
Ls = [a, b, d].
Much more general cases work too, for example:
?- length(Ls0, _), without_c(Ls0, Ls).
Ls0 = Ls, Ls = [] ;
Ls0 = [c],
Ls = [] ;
Ls0 = Ls, Ls = [_7366],
dif(_7366, c) ;
Ls0 = [c, c],
Ls = [] .
Note that such relations can be easily tested, because the arguments can be reasoned about explicitly, much in contrast to output that only occurs on the terminal. In addition, this program admits a declarative reading, freeing you from many operational considerations.
Note also an additional problem of your original definition: Due to the usage of !/0, it has become extremely hard to predict the effect of removing the second clause. The resulting program is not more specific (as would be expected otherwise), but actually emits solutions that were previously not emitted!

Prolog two list intersection - why does it keep checking?

I need to find elements which exist in both lists S1 and S2 and I need to print out these elements (R).
The problem is that when I type bendri([a,b,c,d],[d,b,e],R), it returns the correct result [b,d], but it doesn't stop. If you press ; symbol, then it keeps on checking again and returns b, after that - d.
Why is this happening? It should only return [b,d] and end its job.
bendri(S1,S2,R) :-
skaiciavimai(S1,S2,R).
skaiciavimai([],_,[]).
skaiciavimai([First|Tail], S2, [First|Rest]) :-
member(First, S2),
skaiciavimai(Tail, S2, Rest).
skaiciavimai([_|Tail], S2, Rest) :-
skaiciavimai(Tail, S2, Rest).
Your problem is that the third clause of skaiciavimai/3 also succeeds on backtracking even if the second clause succeeded. I guess you want to skip the third clause if the second clause succeeds.
To do that you can add a check in the third clause:
skaiciavimai([First|Tail], S2, Rest) :-
\+ (member(First, S2)),
skaiciavimai(Tail, S2, Rest).
so that the third clause fails if the head of the first list is found in S2.
Simple story: ; means that Prolog should look for alternatives, and since bendri(A,B,C) is defined as "C is a list of elements that occur in both A and B" (according to your definition), it will print all possible results. You can see this on the trace below (added below because the trace is rather long).
Resolution
The problem is that you introduce a decision point in your skaiciavimai/3 predicate: indeed given the first argument is a list with at least one element (of the form [_|_]), the Prolog environment can select both the second and third clause: after it has tried the skaiciavimai([First|Tail], S2, [First|Rest]) clause successfully, it can also aim to select the next clause skaiciavimai([_|Tail], S2, Rest). Since there is no predicate that prevents that selection from being successful, it will also find a solution omitting the head element. You can solve this by adding an additional constraint in the last clause:
skaiciavimai([],_,[]).
skaiciavimai([First|Tail], S2, [First|Rest]) :-
member(First, S2),
skaiciavimai(Tail, S2, Rest).
skaiciavimai([First|Tail], S2, Rest) :-
\+ member(First,S2),
skaiciavimai(Tail, S2, Rest).
The \+ means something like the logical not (although one must be careful, because not is a problematic topic in Logic Programming). So now we prevent Prolog from selecting the third clause successfully given First is a member of S2. When you use this code the result of your query is:
?- bendri([a,b,c,d],[d,b,e],R).
R = [b, d] ;
false.
We have thus altered the definition of skaiciavimai/3 it now reads something like: "C is a list of all elements that occur in A that occur in B as well." since in order to omit an element from A (third clause), it should not be a member of B.
Towards a better predicate
In Prolog the aim is to make predicate multidirectional. Indeed, you want to be able to call predicates in different directions. bendri/3 can be implemented such that bendri(A,B,[a,c]) also returns A = [a, c], B = [a, c] ;, etc. (which is the case here). In designing predicates one needs to take into account multiple uses of the predicate.
Trace
?- trace.
true.
[trace] ?- bendri([a,b,c,d],[d,b,e],R).
Call: (6) bendri([a, b, c, d], [d, b, e], _G360) ? creep
Call: (7) skaiciavimai([a, b, c, d], [d, b, e], _G360) ? creep
Call: (8) lists:member(a, [d, b, e]) ? creep
Fail: (8) lists:member(a, [d, b, e]) ? creep
Redo: (7) skaiciavimai([a, b, c, d], [d, b, e], _G360) ? creep
Call: (8) skaiciavimai([b, c, d], [d, b, e], _G360) ? creep
Call: (9) lists:member(b, [d, b, e]) ? creep
Exit: (9) lists:member(b, [d, b, e]) ? creep
Call: (9) skaiciavimai([c, d], [d, b, e], _G448) ? creep
Call: (10) lists:member(c, [d, b, e]) ? creep
Fail: (10) lists:member(c, [d, b, e]) ? creep
Redo: (9) skaiciavimai([c, d], [d, b, e], _G448) ? creep
Call: (10) skaiciavimai([d], [d, b, e], _G448) ? creep
Call: (11) lists:member(d, [d, b, e]) ? creep
Exit: (11) lists:member(d, [d, b, e]) ? creep
Call: (11) skaiciavimai([], [d, b, e], _G451) ? creep
Exit: (11) skaiciavimai([], [d, b, e], []) ? creep
Exit: (10) skaiciavimai([d], [d, b, e], [d]) ? creep
Exit: (9) skaiciavimai([c, d], [d, b, e], [d]) ? creep
Exit: (8) skaiciavimai([b, c, d], [d, b, e], [b, d]) ? creep
Exit: (7) skaiciavimai([a, b, c, d], [d, b, e], [b, d]) ? creep
Exit: (6) bendri([a, b, c, d], [d, b, e], [b, d]) ? creep
R = [b, d] ;
Redo: (11) lists:member(d, [d, b, e]) ? creep
Fail: (11) lists:member(d, [d, b, e]) ? creep
Redo: (10) skaiciavimai([d], [d, b, e], _G448) ? creep
Call: (11) skaiciavimai([], [d, b, e], _G448) ? creep
Exit: (11) skaiciavimai([], [d, b, e], []) ? creep
Exit: (10) skaiciavimai([d], [d, b, e], []) ? creep
Exit: (9) skaiciavimai([c, d], [d, b, e], []) ? creep
Exit: (8) skaiciavimai([b, c, d], [d, b, e], [b]) ? creep
Exit: (7) skaiciavimai([a, b, c, d], [d, b, e], [b]) ? creep
Exit: (6) bendri([a, b, c, d], [d, b, e], [b]) ? creep
R = [b] ;
This answer follows up on #gusbro's answer... Why not preserve logical purity? It's easy!
Simply replace the third clause by:
skaiciavimai([First|Tail], S2, Rest) :-
non_member(First, S2),
skaiciavimai(Tail, S2, Rest).
And define non_member/2 like this:
non_member(X, Es) :-
maplist(dif(X), Es).
This is a follow-up to this logically-pure answer presented earlier.
:- use_module(library(lambda)).
bendri(Es, Fs, Xs) :-
% ^^
% ||
% |+----------------+\
% || ||
% vv vv
tfilter(Fs+\E^memberd_t(E,Fs), Es, Xs).
% ^^
% ||
% \+------------------ Fs has global scope
Sample query given by the OP:
?- bendri([a,b,c,d], [d,b,e], Xs).
Xs = [b,d].

Not sufficiently instantiated for maplist(all_distinct, list)

I'm not able to run this code, what exactly do I have to say about the list to allow maplist/2 to run all_distinct/1?
Solution = [A, B, C, D, E, F, G, H, I],
Solution ins 1..9,
maplist(all_distinct, Solution).
I get ERROR: Arguments are not sufficiently instantiated. I understand that I'm not telling it enough about the list of numbers, but I don't know what I need to tell it. I want a list of 9 different numbers 1 through 9.
Here's a trace when I try to execute:
Call: (7) puzzle(_G548) ? creep
Call: (8) _G548=[_G656, _G659, _G662, _G665, _G668, _G671, _G674, _G677|...] ? creep
Exit: (8) [_G656, _G659, _G662, _G665, _G668, _G671, _G674, _G677|...]=[_G656, _G659, _G662, _G665, _G668, _G671, _G674, _G677|...] ? creep
Call: (8) clpfd: ([_G656, _G659, _G662, _G665, _G668, _G671, _G674|...]ins 1..9) ? creep
Call: (9) error:must_be(list, [_G656, _G659, _G662, _G665, _G668, _G671, _G674|...]) ? creep
Exit: (9) error:must_be(list, [_G656, _G659, _G662, _G665, _G668, _G671, _G674|...]) ? creep
Call: (9) clpfd:'__aux_maplist/2_fd_variable+0'([_G656, _G659, _G662, _G665, _G668, _G671, _G674|...]) ? creep
Call: (10) clpfd:fd_variable(_G656) ? creep
Call: (11) var(_G656) ? creep
Exit: (11) var(_G656) ? creep
Call: (11) true ? creep
Exit: (11) true ? creep
Exit: (10) clpfd:fd_variable(_G656) ? creep
Call: (10) clpfd:'__aux_maplist/2_fd_variable+0'([_G659, _G662, _G665, _G668, _G671, _G674, _G677|...]) ? creep
It looks like ins/2 might be not working and then still passing off to maplist/2? I've got no idea what is happening.
What you are doing is that you are making a list of variables, Solutions, and then Solutions ins 1..9 makes each variable an integer between 1 and 9.
all_distinct/1 expects a list, not an integer.
So, if you want a list of 9 distinct integers:
?- Solutions = [A,B,C,D,E,F,G,H,I],
Solutions ins 1..9,
all_distinct(Solutions).
L = [A, B, C, D, E, F, G, H, I],
A in 1..9,
all_distinct([A, B, C, D, E, F, G, H|...]),
B in 1..9,
C in 1..9,
D in 1..9,
E in 1..9,
F in 1..9,
G in 1..9,
H in 1..9,
I in 1..9.

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