Related
I'm new to Prolog and trying to do this question. We have a list
List = [a,a,a,a,b,c,c,a,a,d,e,e,e,e]
I want to pack it into sub-lists of similar elements.
Pack( [a,a,a,a,b,c,c,a,a,d,e,e,e,e], Sublists)
should give
Sublists = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]]
This is what I have tried so far:
pack([],[],[]).
pack([H],[H],[H]).
pack([H,H1|T],Z,X):- H==H1 , append([H],Z,Z1) , pack([H1|T],Z1,X).
pack([H,H1|T],Z,X):- H=\=H1 , append([H],Z,Z1) ,
append(Z1,X,Xs) , pack([H1|T],Z1,Xs).
Below is the error:
Arithmetic: `a/0' is not a function
In:
[4] a=\=b
[3] pack([a,b|...],[a,a],_1608) at line 13
[2] pack([a,a|...],[a],_1688) at line 13
[1] pack([a,a|...],[],_1762) at line 13
Thanks in advance. I'm trying to solve these problems:
P-99: Ninety-Nine Prolog Problems.
You can approach this kind of problem with simple list processing and using SWI Prolog's dif/2 to provide a general solution:
pack([], []). % packing empty is empty
pack([X], [[X]]). % packing a single element
pack([X,X|T], [[X|PH]|PT]):- % rule for packing when next two terms are the same
pack([X|T], [PH|PT]).
pack([X,Y|T], [[X]|PT]):- % rule for different term
dif(X, Y),
pack([Y|T], PT).
2 ?- pack([a,a,a,a,b,c,c,a,a,d,e,e], L).
L = [[a, a, a, a], [b], [c, c], [a, a], [d], [e, e]] ;
false.
3 ?- pack(L, [[a,a,a], [b,b], [c]]).
L = [a, a, a, b, b, c] ;
false.
4 ?-
Note that lurker's solution still has some performance issues. See the ; false for each solution? This is an indication that Prolog still retains some memory (called a choice point - actually there may be even several such choice points). For many cases however, no such choice point is needed. Here is a solution that overcomes that problem (The name group in place of pack is quite common in the context of Haskell)
group([], []).
group([E|Es], [[E|Gs]|Gss]) :-
igroup(Es, E, Gs, Gss).
igroup([], _, [], []).
igroup([E|Es], F, Gs1, Gss1) :-
( E\=F
-> Gs1=[], Gss1=[[E|Gs2]|Gss2]
; E==F
-> Gs1=[E|Gs2], Gss1=Gss2
; E=F,
Gs1=[E|Gs2], Gss1=Gss2
; dif(E, F),
Gs1=[], Gss1=[[E|Gs2]|Gss2]
),
igroup(Es, E, Gs2, Gss2).
Note how the testing for the equality of E and F is split into four cases:
First E \= F which means that both are definitely different.
Then E == F which means that both are definitely identical.
Then E = F which is the general case of equality, and
dif(E, F) which is the case of general inequality
For the last two cases there is no -> because both may be true.
Since it is quite cumbersome to maintain so many cases, there is library(reif)
for
SICStus and
SWI which permits to write the same more compactly:
igroup([], _, [], []).
igroup([E|Es], F, Gs1, Gss1) :-
if_(E = F
, ( Gs1 = [E|Gs2], Gss1 = Gss2 )
, ( Gs1 = [], Gss1 = [[E|Gs2]| Gss2] )),
igroup(Es, E, Gs2, Gss2).
The error you got is because =\=/2 is true if expr1 is evaluates to a number not equal to expr2 . Instead you can use \=\2 which evaluates \+term1=term2. ==/2 evaluates to term1 equivalent to term2, =:=/ is true if expr1 is a number which is equal to expr2. One more mistake I found in your code is you're not clearing the Intermediate List. You have to flush the values in it after you have added the similar elements list to your Sublists list. I have used cut ! to reduce backtracking. Instead, if you write mutually exclusive predicates, it's better.
I have edited your code:
pack1([],[],[]).
pack1([H],L,[Z]):- append([H],L,Z),!.
pack1([H,H1|T],Z,X):- H == H1 , append([H],Z,Z1) , pack1([H1|T],Z1,X),!.
pack1([H,H1|T],Z,[Z1|Zs]):- H\=H1 ,append([H],Z,Z1) ,pack1([H1|T],[],Zs),!.
Output:
?-pack1([a,a,a,a,b,c,c,a,a,d,e,e,e,e],[],Z).
Z=[[a, a, a, a], [b], [c, c], [a, a], [d], [e, e, e, e]]
?-pack1([a,a,a,a,b,c,1,c,a,a,d,e,e,e,e],[],Z).
Z= [[a, a, a, a], [b], [c], [1], [c], [a, a], [d], [e, e, e, e]]
?-pack1([],[],Z).
Z= []
Hope this helps.
without using dif/2
my_pack([],[[]]).
my_pack([X], [[X]]).
my_pack([X,X|L], [F|R]) :- my_pack([X|L], [F1|R]), append([X], F1, F).
my_pack([X|L], [F|R]) :- my_pack(L, R), append([X], [], F).
This question was asked but there are no answers: here. I read the comments and tried to implement in both ways, but there are more problems that I don't understand.
I first tried the easy way that doesn't keep original order:
list_repeated(L, Ds) :-
msort(L, S),
sorted_repeated(S, Ds).
sorted_repeated([], []).
sorted_repeated([X|Xs], Ds) :-
first(Xs, X, Ds).
first([], _, []).
first([X|Xs], X, [X|Ds]) :-
more(Xs, X, Ds).
first([X|Xs], Y, Ds) :-
dif(X, Y),
first(Xs, X, Ds).
more([], _, []).
more([X|Xs], X, Ds) :-
more(Xs, X, Ds).
more([X|Xs], Y, Ds) :-
dif(X, Y),
first(Xs, X, Ds).
Once the list is sorted without removing duplicates, using first and more I add the element to the second argument if it occurs at least twice and skip all consecutive copies of the element.
This is not working properly because if I have:
?- list_duplicates([b,a,a,a,b,b], Ds).
I get answer [a,b] instead of [b,a] and also I get ; false after the answer.
I also tried another way, but this doesn't work because the accumulator is immutable?
list_duplicates(L, Ds) :-
ld_acc(L, [], Ds).
ld_acc([], _, []).
ld_acc([X|Xs], Acc, Ds) :-
( memberchk(X, Acc)
-> Ds = [X|Ds0],
ld_acc(Xs, Acc, Ds0)
; Acc1 = [X|Acc],
ld_acc(Xs, Acc1, Ds)
).
This cannot work because when I check that an element is member of accumulator I remove only one occurrence of each element: if I have three times the same element in the first argument, I am left with two. If I could change the element in the accumulator then I could maybe put a counter on it? In the first version I used different states, first and more, but here I have to attach state to the elements of the accumulator, is that possible?
A plea for purity
When programming in Prolog, a major attraction is the generality we enjoy from pure relations.
This lets us use our code in multiple directions, and reason declaratively over our programs and answers.
You can enjoy these benefits if you keep your programs pure.
Possible solution
As always when describing lists, also consider using DCG notation. See dcg for more information.
For example, to describe the list of duplicates in a pure way, consider:
list_duplicates([]) --> [].
list_duplicates([L|Ls]) -->
list_duplicates_(Ls, L),
list_duplicates(Ls).
list_duplicates_([], _) --> [].
list_duplicates_([L0|Ls], L) -->
if_(L0=L, [L], []),
list_duplicates_(Ls, L).
This uses if_//3 to retain generality and determinism (if applicable).
Examples
Here are a few example queries and answers. We start with simple ground cases:
?- phrase(list_duplicates([a,b,c]), Ds).
Ds = [].
?- phrase(list_duplicates([a,b,a]), Ds).
Ds = [a].
Even the most impure version will be able to handle these situations correctly. So, slightly more interesting:
?- phrase(list_duplicates([a,b,X]), Ds).
X = a,
Ds = [a] ;
X = b,
Ds = [b] ;
Ds = [],
dif(X, b),
dif(X, a).
Pretty nice, isn't it? The last part says: Ds = [] is a solution if X is different from b and a. Note the pure relation dif/2 automatically appears in these residual goals and retains the relation's generality.
Here is an example with two variables:
?- phrase(list_duplicates([X,Y]), Ds).
X = Y,
Ds = [Y] ;
Ds = [],
dif(Y, X).
Finally, consider using iterative deepening to fairly enumerate answers for lists of arbitrary length:
?- length(Ls, _), phrase(list_duplicates(Ls), Ds).
Ls = Ds, Ds = [] ;
Ls = [_136],
Ds = [] ;
Ls = [_136, _136],
Ds = [_136] ;
Ls = [_156, _162],
Ds = [],
dif(_162, _156) ;
Ls = Ds, Ds = [_42, _42, _42] ;
Ls = [_174, _174, _186],
Ds = [_174],
dif(_186, _174) .
Multiple occurrences
Here is a version that handles arbitrary many occurrences of the same element in such a way that exactly a single occurrence is retained if (and only if) the element occurs at least twice:
list_duplicates(Ls, Ds) :-
phrase(list_duplicates(Ls, []), Ds).
list_duplicates([], _) --> [].
list_duplicates([L|Ls], Ds0) -->
list_duplicates_(Ls, L, Ds0, Ds),
list_duplicates(Ls, Ds).
list_duplicates_([], _, Ds, Ds) --> [].
list_duplicates_([L0|Ls], L, Ds0, Ds) -->
if_(L0=L, new_duplicate(L0, Ds0, Ds1), {Ds0 = Ds1}),
list_duplicates_(Ls, L, Ds1, Ds).
new_duplicate(E, Ds0, Ds) -->
new_duplicate_(Ds0, E, Ds0, Ds).
new_duplicate_([], E, Ds0, [E|Ds0]) --> [E].
new_duplicate_([L|Ls], E, Ds0, Ds) -->
if_(L=E,
{ Ds0 = Ds },
new_duplicate_(Ls, E, Ds0, Ds)).
The query shown by #fatalize in the comments now yields:
?- list_duplicates([a,a,a], Ls).
Ls = [a].
The other examples yield the same results. For instance:
?- list_duplicates([a,b,c], Ds).
Ds = [].
?- list_duplicates([a,b,a], Ds).
Ds = [a].
?- list_duplicates([a,b,X], Ds).
X = a,
Ds = [a] ;
X = b,
Ds = [b] ;
Ds = [],
dif(X, b),
dif(X, a).
?- list_duplicates([X,Y], Ds).
X = Y,
Ds = [Y] ;
Ds = [],
dif(Y, X).
I leave the case ?- list_duplicates(Ls, Ls). as an exercise.
Generality: Multiple directions
Ideally, we want to be able to use a relation in all directions.
For example, our program should be able to answer questions like:
What does a list look like if its duplicates are [a,b]?
With the version shown above, we get:
?- list_duplicates(Ls, [a,b]).
nontermination
Luckily, a very simple change allows as to answer such questions!
One such change is to simply write:
list_duplicates(Ls, Ds) :-
length(Ls, _),
phrase(list_duplicates(Ls, []), Ds).
This is obviously declaratively admissible, because Ls must be a list. Operationally, this helps us to enumerate lists in a fair way.
We now get:
?- list_duplicates(Ls, [a,b]).
Ls = [a, a, b, b] ;
Ls = [a, b, a, b] ;
Ls = [a, b, b, a] ;
Ls = [a, a, a, b, b] ;
Ls = [a, a, b, a, b] ;
Ls = [a, a, b, b, a] ;
Ls = [a, a, b, b, b] ;
Ls = [a, a, b, b, _4632],
dif(_4632, b),
dif(_4632, a) ;
etc.
Here is a simpler case, using only a single element:
?- list_duplicates(Ls, [a]).
Ls = [a, a] ;
Ls = [a, a, a] ;
Ls = [a, a, _3818],
dif(_3818, a) ;
Ls = [a, _3870, a],
dif(_3870, a) ;
Ls = [_4058, a, a],
dif(a, _4058),
dif(a, _4058) ;
Ls = [a, a, a, a] ;
etc.
Maybe even more interesting:
What does a list without duplicates look like?
Our program answers:
?- list_duplicates(Ls, []).
Ls = [] ;
Ls = [_3240] ;
Ls = [_3758, _3764],
dif(_3764, _3758) ;
Ls = [_4164, _4170, _4176],
dif(_4176, _4164),
dif(_4176, _4170),
dif(_4170, _4164) .
Thus, the special case of a list where all elements are distinct naturally exists as a special case of the more general relation we have implemented.
We can use this relation to generate answers (as shown above), and also to test whether a list consists of distinct elements:
?- list_duplicates([a,b,c], []).
true.
?- list_duplicates([b,b], []).
false.
Unfortunately, the following even more general query still yields:
?- list_duplicates([b,b|_], []).
nontermination
On the plus side, if the length of the list is fixed, we get in such cases:
?- length(Ls, L), maplist(=(b), Ls),
( true ; list_duplicates(Ls, []) ).
Ls = [],
L = 0 ;
Ls = [],
L = 0 ;
Ls = [b],
L = 1 ;
Ls = [b],
L = 1 ;
Ls = [b, b],
L = 2 ;
Ls = [b, b, b],
L = 3 ;
Ls = [b, b, b, b],
L = 4 .
This is some indication that the program indeed terminates in such cases. Note that the answers are of course now too general.
Efficiency
It is well known in high-performance computing circles that as long as your program is fast enough, its correctness is barely worth considering.
So, the key question is of course: How can we make this faster?
I leave this is a very easy exercise. One way to make this faster in specific cases is to first check whether the given list is sufficiently instantiated. In that case, you can apply an ad hoc solution that fails terribly in more general cases, but has the extreme benefit that it is fast!
So as far as I can tell, you were on the right track with the accumulator, but this implementation definitely works as you want (assuming you want the duplicates in the order they first appear in the list).
list_duplicates(Input,Output) is just used to wrap and initialise the accumulator.
list_duplicates(Accumulator,[],Accumulator) unifies the accumulator with the output when we have finished processing the input list.
list_duplicates(Accumulator,[H|T],Output) says "if the head (H) of the input list is in the rest of the list (T), and is not in the Accumulator already, put it at the end of the Accumulator (using append), then recurse on the tail of the list".
list_duplicates(Accumulator,[_|T],Output) (which we only get to if either the head is not a duplicate, or is already in the Accumulator) just recurses on the tail of the list.
list_duplicates(Input,Output) :-
once(list_duplicates([],Input,Output)).
list_duplicates(Accumulator,[],Accumulator).
list_duplicates(Accumulator,[H|T],Output) :-
member(H,T),
\+member(H,Accumulator),
append(Accumulator,[H],NewAccumulator),
list_duplicates(NewAccumulator,T,Output).
list_duplicates(Accumulator,[_|T],Output) :-
list_duplicates(Accumulator,T,Output).
You could also recurse in list_duplicates(Accumulator,[H|T],Output) with list_duplicates([H|Accumulator],T,Output) and reverse in the wrapper, looking like this:
list_duplicates(Input,Output) :-
once(list_duplicates([],Input,ReverseOutput)),
reverse(ReverseOutput,Output).
list_duplicates(Accumulator,[],Accumulator).
list_duplicates(Accumulator,[H|T],Output) :-
member(H,T),
\+member(H,Accumulator),
list_duplicates([H|Accumulator],T,Output).
list_duplicates(Accumulator,[_|T],Output) :-
list_duplicates(Accumulator,T,Output).
The once call in the wrapper prevents the false output (or in this case, partial duplicate lists due to a lack of guards on the second rule).
I'm new in Prolog.
I have a problem about predicate prefix but a little bit different.
I want to get a prefix of a list but until an element
The list can have repeat elements.
An example:
prefix(Element, List, Prefix)
prefix(c, [a,b,c,d,e,f], [a, b])
The element is not included.
What I have so far is this
prefix(X, [X|T], []).
prefix(X, [Y|T], [Y|Z]):-
prefix(X, T, Z).
But it does not work.
L = [a,b,c] ? prefix(b, L, Prefix).
no
?-
Thanks
With dif/2 you can explicitly state that for any member X preceding Element, X \== Element:
prefix(Element, [Element|_], []).
prefix(Element, [Head|List], [Head|Prefix]) :-
dif(Element, Head),
prefix(Element, List, Prefix).
or equally, because I wanted to use append/3 in the first iteration of my answer:
prefix(Element, List, Prefix) :-
append(Prefix, [Element|_Suffix], List),
maplist(dif(Element), Prefix).
For the suffix it is basically the same:
suffix(Element, List, Suffix) :-
append(_Prefix, [Element|Suffix], List),
maplist(dif(Element), Suffix).
If you don't want to use maplist(dif(Element), List):
all_dif(_, []).
all_dif(X, [H|T]) :- dif(X, H), all_dif(X, T).
Here is a solution using Definite Clause Grammars dcg and the non-terminal all_seq//2:
prefix(X, Xs, Ys) :-
phrase( ( all_seq(dif(X), Ys), [X], ... ), Xs).
... --> [] | [_], ... .
So the grammar (within phrase/2) reads:
There is
1. an initial sequence Ys with all elements different to X, followed by 2. X, followed by 3. anything.
There is still a downside, which is often the case when using DCGs: The implementation is not as determinate as it could be and thus leaves superfluous choicepoints around.
prefix(X,[X|T],[]).
prefix(X,[Y|T],Z) :- prefix(X,T,M) , Z = [Y|M].
output:
?- L = [a,b,c,d,e,f] , prefix(d,L,G). L = [a, b, c, d, e, f], G = [a,
b, c] .
?- L = [a,b,c,d,e,f] , prefix(e,L,G). L = [a, b, c, d, e, f], G = [a,
b, c, d] .
EDIT #1
the original code is working , use (,) instead of (?) as following.
prefix(X,[X|T],[]).
prefix(X,[Y|T],[Y|Z]) :- prefix(X,T,Z).
output:
?- prefix(d , [a,b,c,d,e] , G). G = [a, b, c]
?- L = [a,b,c] , prefix(b, L, Prefix).
L = [a, b, c],
Prefix = [a] .
EDIT #2
as user false mentioned in comment, I can confirm that you are right, but in my solution, I assume that the list contains unique elements:
prefix(d,[d,d],[d]) succeeds - it should fail ,
I have a list [a, b, a, a, a, c, c]
and I need to add two more occurrences of each element.
The end result should look like this:
[a, a, a, b, b, b, a, a, a, a, a, c, c, c, c]
If I have an item on the list that is the same as the next item, then it keeps going until there is a new item, when it finds the new item, it adds two occurrences of the previous item then moves on.
This is my code so far, but I can't figure out how to add two...
dbl([], []).
dbl([X], [X,X]).
dbl([H|T], [H,H|T], [H,H|R]) :- dbl(T, R).
Your code looks a bit strange because the last rule takes three parameters. You only call the binary version, so no recursion will ever try to derive it.
You already had a good idea to look at the parts of the list, where elements change. So there are 4 cases:
1) Your list is empty.
2) You have exactly one element.
3) Your list starts with two equal elements.
4) Your list starts with two different elements.
Case 1 is not specified, so you might need to find a sensible choice for that. Case 2 is somehow similar to case 4, since the end of the list can be seen as a change in elements, where you need to append two copies, but then you are done. Case 3 is quite simple, we can just keep the element and recurse on the rest. Case 4 is where you need to insert the two copies again.
This means your code will look something like this:
% Case 1
dbl([],[]).
% Case 2
dbl([X],[X,X,X]).
% Case 3
dbl([X,X|Xs], [X|Ys]) :-
% [...] recursion skipping the leading X
% Case 4
dbl([X,Y|Xs], [X,X,X|Ys]) :-
dif(X,Y),
% [...] we inserted the copies, so recursion on [Y|Xs] and Ys
Case 3 should be easy to finish, we just drop the first X from both lists and recurse on dbl([X|Xs],Ys). Note that we implicitly made the first two elements equal (i.e. we unified them) by writing the same variable twice.
If you look at the head of case 4, you can directly imitate the pattern you described: supposed the list starts with X, then Y and they are different (dif(X,Y)), the X is repeated 3 times instead of just copied and we then continue with the recursion on the rest starting with Y: dbl([Y|Xs],Ys).
So let's try out the predicate:
?- dbl([a,b,a,a,a,c,c],[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]).
true ;
false.
Our test case is accepted (true) and we don't find more than one solution (false).
Let's see if we find a wrong solution:
?- dif(Xs,[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]), dbl([a,b,a,a,a,c,c],Xs).
false.
No, that's also good. What happens, if we have variables in our list?
?- dbl([a,X,a],Ys).
X = a,
Ys = [a, a, a, a, a] ;
Ys = [a, a, a, X, X, X, a, a, a],
dif(X, a),
dif(X, a) ;
false.
Either X = a, then Ys is single run of 5 as; or X is not equal to a, then we need to append the copies in all three runs. Looks also fine. (*)
Now lets see, what happens if we only specify the solution:
?- dbl(X,[a,a,a,b,b]).
false.
Right, a list with a run of only two bs can not be a result of our specification. So lets try to add one:
?- dbl(X,[a,a,a,b,b,b]).
X = [a, b] ;
false.
Hooray, it worked! So lets as a last test look what happens, if we just call our predicate with two variables:
?- dbl(Xs,Ys).
Xs = Ys, Ys = [] ;
Xs = [_G15],
Ys = [_G15, _G15, _G15] ;
Xs = [_G15, _G15],
Ys = [_G15, _G15, _G15, _G15] ;
Xs = [_G15, _G15, _G15],
Ys = [_G15, _G15, _G15, _G15, _G15] ;
Xs = [_G15, _G15, _G15, _G15],
Ys = [_G15, _G15, _G15, _G15, _G15, _G15] ;
[...]
It seems we get the correct answers, but we see only cases for a single run. This is a result of prolog's search strategy(which i will not explain in here). But if we look at shorter lists before we generate longer ones, we can see all the solutions:
?- length(Xs,_), dbl(Xs,Ys).
Xs = Ys, Ys = [] ;
Xs = [_G16],
Ys = [_G16, _G16, _G16] ;
Xs = [_G16, _G16],
Ys = [_G16, _G16, _G16, _G16] ;
Xs = [_G86, _G89],
Ys = [_G86, _G86, _G86, _G89, _G89, _G89],
dif(_G86, _G89) ;
Xs = [_G16, _G16, _G16],
Ys = [_G16, _G16, _G16, _G16, _G16] ;
Xs = [_G188, _G188, _G194],
Ys = [_G188, _G188, _G188, _G188, _G194, _G194, _G194],
dif(_G188, _G194) ;
[...]
So it seems we have a working predicate (**), supposed you filled in the missing goals from the text :)
(*) A remark here: this case only works because we are using dif. The first predicates with equality, one usually encounters are =, == and their respective negations \= and \==. The = stands for unifyability (substituting variables in the arguments s.t. they become equal) and the == stands for syntactic equality (terms being exactly equal). E.g.:
?- f(X) = f(a).
X = a.
?- f(X) \= f(a).
false.
?- f(X) == f(a).
false.
?- f(X) \== f(a).
true.
This means, we can make f(X) equal to f(a), if we substitute X by a. This means if we ask if they can not be made equal (\=), we get the answer false. On the other hand, the two terms are not equal, so == returns false, and its negation \== answers true.
What this also means is that X \== Y is always true, so we can not use \== in our code. In contrast to that, dif waits until it can decide wether its arguments are equal or not. If this is still undecided after finding an answer, the "dif(X,a)" statements are printed.
(**) One last remark here: There is also a solution with the if-then-else construct (test -> goals_if_true; goals_if_false, which merges cases 3 and 4. Since i prefer this solution, you might need to look into the other version yourself.
TL;DR:
From a declarative point of view, the code sketched by #lambda.xy.x is perfect.
Its determinacy can be improved without sacrificing logical-purity.
Code variant #0: #lambda.xy.x's code
Here's the code we want to improve:
dbl0([], []).
dbl0([X], [X,X,X]).
dbl0([X,X|Xs], [X|Ys]) :-
dbl0([X|Xs], Ys).
dbl0([X,Y|Xs], [X,X,X|Ys]) :-
dif(X, Y),
dbl0([Y|Xs], Ys).
Consider the following query and the answer SWI-Prolog gives us:
?- dbl0([a],Xs).
Xs = [a,a,a] ;
false.
With ; false the SWI prolog-toplevel
indicates a choicepoint was left when proving the goal.
For the first answer, Prolog did not search the entire proof tree.
Instead, it replied "here's an answer, there may be more".
Then, when asked for more solutions, Prolog traversed the remaining branches of the proof tree but finds no more answers.
In other words: Prolog needs to think twice to prove something we knew all along!
So, how can we give determinacy hints to Prolog?
By utilizing:
control constructs !/0 and / or (->)/2 (potentially impure)
first argument indexing on the principal functor (never impure)
The code presented in the earlier answer by #CapelliC—which is based on !/0, (->)/2, and the meta-logical predicate (\=)/2—runs well if all arguments are sufficiently instantiated. If not, erratic answers may result—as #lambda.xy.x's comment shows.
Code variant #1: indexing
Indexing can improve determinacy without ever rendering the code non-monotonic. While different Prolog processors have distinct advanced indexing capabilities, the "first-argument principal-functor" indexing variant is widely available.
Principal? This is why executing the goal dbl0([a],Xs) leaves a choicepoint behind: Yes, the goal only matches one clause—dbl0([X],[X,X,X]).—but looking no deeper than the principal functor Prolog assumes that any of the last three clauses could eventually get used. Of course, we know better...
To tell Prolog we utilize principal-functor first-argument indexing:
dbl1([], []).
dbl1([E|Es], Xs) :-
dbl1_(Es, Xs, E).
dbl1_([], [E,E,E], E).
dbl1_([E|Es], [E|Xs], E) :-
dbl1_(Es, Xs, E).
dbl1_([E|Es], [E0,E0,E0|Xs], E0) :-
dif(E0, E),
dbl1_(Es, Xs, E).
Better? Somewhat, but determinacy could be better still...
Code variant #2: indexing on reified term equality
To make Prolog see that the two recursive clauses of dbl1_/3 are mutually exclusive (in certain cases), we reify the truth value of
term equality and then index on that value:
This is where reified term equality (=)/3 comes into play:
dbl2([], []).
dbl2([E|Es], Xs) :-
dbl2_(Es, Xs, E).
dbl2_([], [E,E,E], E).
dbl2_([E|Es], Xs, E0) :-
=(E0, E, T),
t_dbl2_(T, Xs, E0, E, Es).
t_dbl2_(true, [E|Xs], _, E, Es) :-
dbl2_(Es, Xs, E).
t_dbl2_(false, [E0,E0,E0|Xs], E0, E, Es) :-
dbl2_(Es, Xs, E).
Sample queries using SWI-Prolog:
?- dbl0([a],Xs).
Xs = [a, a, a] ;
false.
?- dbl1([a],Xs).
Xs = [a, a, a].
?- dbl2([a],Xs).
Xs = [a, a, a].
?- dbl0([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b] ;
false.
?- dbl1([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b] ;
false.
?- dbl2([a,b,b],Xs).
Xs = [a, a, a, b, b, b, b].
To make above code more compact, use control construct if_/3 .
I was just about to throw this version with if_/3 and (=)/3 in the hat when I saw #repeat already suggested it. So this is essentially the more compact version as outlined by #repeat:
list_dbl([],[]).
list_dbl([X],[X,X,X]).
list_dbl([A,B|Xs],DBL) :-
if_(A=B,DBL=[A,B|Ys],DBL=[A,A,A,B|Ys]),
list_dbl([B|Xs],[B|Ys]).
It yields the same results as dbl2/2 by #repeat:
?- list_dbl([a],DBL).
DBL = [a,a,a]
?- list_dbl([a,b,b],DBL).
DBL = [a,a,a,b,b,b,b]
The example query by the OP works as expected:
?- list_dbl([a,b,a,a,a,c,c],DBL).
DBL = [a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]
Plus here are some of the example queries provided by #lambda.xy.x. They yield the same results as #repeat's dbl/2 and #lambda.xy.x's dbl/2:
?- dif(Xs,[a,a,a,b,b,b,a,a,a,a,a,c,c,c,c]), list_dbl([a,b,a,a,a,c,c],Xs).
no
?- list_dbl(X,[a,a,a,b,b]).
no
?- list_dbl(L,[a,a,a,b,b,b]).
L = [a,b] ? ;
no
?- list_dbl(L,DBL).
DBL = L = [] ? ;
DBL = [_A,_A,_A],
L = [_A] ? ;
DBL = [_A,_A,_A,_A],
L = [_A,_A] ? ;
DBL = [_A,_A,_A,_A,_A],
L = [_A,_A,_A] ? ;
...
?- list_dbl([a,X,a],DBL).
DBL = [a,a,a,a,a],
X = a ? ;
DBL = [a,a,a,X,X,X,a,a,a],
dif(X,a),
dif(a,X)
?- length(L,_), list_dbl(L,DBL).
DBL = L = [] ? ;
DBL = [_A,_A,_A],
L = [_A] ? ;
DBL = [_A,_A,_A,_A],
L = [_A,_A] ? ;
DBL = [_A,_A,_A,_B,_B,_B],
L = [_A,_B],
dif(_A,_B) ? ;
DBL = [_A,_A,_A,_A,_A],
L = [_A,_A,_A] ?
dbl([X,Y|T], [X,X,X|R]) :- X \= Y, !, dbl([Y|T], R).
dbl([H|T], R) :-
T = []
-> R = [H,H,H]
; R = [H|Q], dbl(T, Q).
The first clause handles the basic requirement, adding two elements on sequence change.
The second one handles list termination as a sequence change, otherwise, does a plain copy.
When given a list I would like to compute all the possible combinations of pairs in the list.
e.g 2) input is a list (a,b,c) I would like to obtain pairs (a,b) (a,c) (b,c)
e.g 1) input is a list (a,b,c,d) I would like to obtain pairs (a,b) (a,c) (a,d) (b,c) (b,d) and (c,d)
Using select/3 twice (or select/3 once and member/2 once) will allow you to achieve what you want here. I'll let you work out the details and ask for help if it's still troublesome.
BTW, Prolog syntax for list isn't (a, b, c) but [a, b, c] (well, it's syntactic sugar but I'll leave it at that).
edit: as #WillNess pointed out, you're not looking for any pair (X, Y) but only for pairs where X is before Y in the list.
DCGs are a really good fit: as #false described, they can produce a graphically appealing solution:
... --> [] | [_], ... .
pair(L, X-Y) :-
phrase((..., [X], ..., [Y], ...), L).
Alternatively, if you use SWI-Prolog, a call to append/2 does the trick too, in a similar manner, but is less efficient than DCGs:
pair2(L, X-Y) :-
append([_, [X], _, [Y], _], L).
You can do it with a basic recursion, as #WillNess suggested in his comment. I'll leave this part for him to detail if needed!
OK, so to translate the Haskell definition
pairs (x:xs) = [ (x,y) | y <- xs ]
++ pairs xs
pairs [] = []
(which means, pairs in the list with head x and tail xs are all the pairs (x,y) where y is in xs, and also the pairs in xs; and there's no pairs in an empty list)
as a backtracking Prolog predicate, producing the pairs one by one on each redo, it's a straightforward one-to-one re-write of the above,
pair( [X|XS], X-Y) :- member( ... , XS) % fill in
; pair( XS, ... ). % the blanks
%% pair( [], _) :- false.
To get all the possible pairs, use findall:
pairs( L, PS) :- findall( P, pair( L, P), PS).
Consider using bagof if your lists can contain logical variables in them. Controlling bagof's backtracking could be an intricate issue though.
pairs can also be written as a (nearly) deterministic, non-backtracking, recursive definition, constructing its output list through an accumulator parameter as a functional programming language would do -- here in the top-down manner, which is what Prolog so excels in:
pairs( [X|T], PS) :- T = [_|_], pairs( X, T, T, PS, []).
pairs( [_], []).
pairs( [], []).
pairs( _, [], [], Z, Z).
pairs( _, [], [X|T], PS, Z) :- pairs( X, T, T, PS, Z).
pairs( X, [Y|T], R, [X-Y|PS], Z) :- pairs( X, T, R, PS, Z).
Here is a possible way of solving this.
The following predicate combine/3 is true
if the third argument corresponds to a list
contains pairs, where the first element of each pair
is equal to the first argument of combine/3.
The second element of each pair will correspond to an item
of the list in the second argument of the predicate combine/3.
Some examples how combine/3 should work:
?- combine(a,[b],X).
X = [pair(a,b)]
?- combine(a,[b,c,d],X).
X = [pair(a,b), pair(a,c), pair(a,d)]
Possible way of defining combine/3:
combine(A,[B],[pair(A,B)]) :- !.
combine(A,[B|T],C) :-
combine(A,T,C2), % Create pairs for remaining elements in T.
append([pair(A,B)],C2,C). % Append current pair and remaining pairs C2.
% The result of append is C.
Now combine/3 can be used to define pair/2:
pairs([],[]). % Empty list will correspond to empty list of pairs.
pairs([H|T],P) :- % In case there is at least one element.
nonvar([H|T]), % In this case it expected that [H|T] is instantiated.
pairs(H,T,P).
pairs(A,[B],[pair(A,B)]) % If remaining list contains exactly one element,
:- !. % then there will be only one pair(A,B).
pairs(A,[B|T],P) :- % In case there are at least two elements.
combine(A,[B|T],P2), % For each element in [B|T] compute pairs
% where first element of each pair will be A.
pairs(B,T,P3), % Compute all pairs without A recursively.
append(P2,P3,P). % Append results P2 and P3 together.
Sample usage:
?- pairs([a,b,c],X).
X = [pair(a, b), pair(a, c), pair(b, c)].
?- pairs([a,b,c,d],X).
X = [pair(a, b), pair(a, c), pair(a, d), pair(b, c), pair(b, d), pair(c, d)].
You can use append/ to iterate through the list:
?- append(_,[X|R],[a,b,c,d]).
X = a,
R = [b, c, d] ;
X = b,
R = [c, d] ;
X = c,
R = [d] ;
X = d,
R = [] ;
false.
Next, use member/2 to form a pair X-Y, for each Y in R:
?- append(_,[X|R],[a,b,c,d]), member(Y,R), Pair=(X-Y).
X = a,
R = [b, c, d],
Y = b,
Pair = a-b ;
X = a,
R = [b, c, d],
Y = c,
Pair = a-c ;
X = a,
R = [b, c, d],
Y = d,
Pair = a-d ;
X = b,
R = [c, d],
Y = c,
Pair = b-c ;
X = b,
R = [c, d],
Y = d,
Pair = b-d ;
X = c,
R = [d],
Y = d,
Pair = c-d ;
false.
Then, use findall/3 to collect all pairs in a list:
?- findall(X-Y, (append(_,[X|R],[a,b,c,d]), member(Y,R)), Pairs).
Pairs = [a-b, a-c, a-d, b-c, b-d, c-d].
Thus, your final solution can be expressed as:
pairs(List, Pairs) :-
findall(X-Y, (append(_,[X|R],List), member(Y,R)), Pairs).
An example of use is:
?- pairs([a,b,c,d], Pairs).
Pairs = [a-b, a-c, a-d, b-c, b-d, c-d].