This one tickled my interest in theory:
Is it possible to write an inconsistent Prolog program, i.e. a program that answers both false and true depending on how it is queried, using only pure Prolog, the cut, and false?
For example, one could query p(1) and the Prolog Processor would says false. But when one queries p(X) the Prolog Processor would give the set of answers 1, 2, 3.
This can be easily achieved with "computational state examination predicates" like var/1 (really better called fresh/1) + el cut:
p(X) :- nonvar(X),!,member(X,[2,3]).
p(X) :- member(X,[1,2,3]).
Then
?- p(1).
false.
?- p(X).
X = 1 ;
X = 2 ;
X = 3.
"Ouch time" ensues if this is high-assurance software. Naturally, any imperative program has no problem going off the rails like this on every other line.
So. can be done without those "computational state examination predicates"?
P.S.
The above illustrates that all the predicates of Prolog are really carrying a threaded hidden argument of the "computational state":
p(X,StateIn,StateOut).
which can be used to explain the behavour of var/1 and friends. The Prolog program is then "pure" when it only calls predicates that neither consult not modify that State. Well, at least that seems to be a good way to look at what is going on. I think.
Here's a very simple one:
f(X,X) :- !, false.
f(0,1).
Then:
| ?- f(0,1).
yes
| ?- f(X,1).
no
| ?- f(0,Y).
no
So Prolog claims there are no solutions to the queries with variables, although f(0,1) is true and would be a solution to both.
Here is one attempt. The basic idea is that X is a variable iff it can be unified with both a and b. But of course we can't write this as X = a, X = b. So we need a "unifiable" test that succeeds without binding variables like =/2 does.
First, we need to define negation ourselves, since it's impure:
my_not(Goal) :-
call(Goal),
!,
false.
my_not(_Goal).
This is only acceptable if your notion of pure Prolog includes call/1. Let's say that it does :-)
Now we can check for unifiability by using =/2 and the "not not" pattern to preserve success while undoing bindings:
unifiable(X, Y) :-
my_not(my_not(X = Y)).
Now we have the tools to define var/nonvar checks:
my_var(X) :-
unifiable(X, a),
unifiable(X, b).
my_nonvar(X) :-
not(my_var(X)).
Let's check this:
?- my_var(X).
true.
?- my_var(1).
false.
?- my_var(a).
false.
?- my_var(f(X)).
false.
?- my_nonvar(X).
false.
?- my_nonvar(1).
true.
?- my_nonvar(a).
true.
?- my_nonvar(f(X)).
true.
The rest is just your definition:
p(X) :-
my_nonvar(X),
!,
member(X, [2, 3]).
p(X) :-
member(X, [1, 2, 3]).
Which gives:
?- p(X).
X = 1 ;
X = 2 ;
X = 3.
?- p(1).
false.
Edit: The use of call/1 is not essential, and it's interesting to write out the solution without it:
not_unifiable(X, Y) :-
X = Y,
!,
false.
not_unifiable(_X, _Y).
unifiable(X, Y) :-
not_unifiable(X, Y),
!,
false.
unifiable(_X, _Y).
Look at those second clauses of each of these predicates. They are the same! Reading these clauses declaratively, any two terms are not unifiable, but also any two terms are unifiable! Of course you cannot read these clauses declaratively because of the cut. But I find this especially striking as an illustration of how catastrophically impure the cut is.
I'm trying to figure out a way to check if two lists are equal regardless of their order of elements.
My first attempt was:
areq([],[]).
areq([],[_|_]).
areq([H1|T1], L):- member(H1, L), areq(T1, L).
However, this only checks if all elements of the list on the left exist in the list on the right; meaning areq([1,2,3],[1,2,3,4]) => true. At this point, I need to find a way to be able to test thing in a bi-directional sense. My second attempt was the following:
areq([],[]).
areq([],[_|_]).
areq([H1|T1], L):- member(H1, L), areq(T1, L), append([H1], T1, U), areq(U, L).
Where I would try to rebuild the lest on the left and swap lists in the end; but this failed miserably.
My sense of recursion is extremely poor and simply don't know how to improve it, especially with Prolog. Any hints or suggestions would be appreciated at this point.
As a starting point, let's take the second implementation of equal_elements/2 by #CapelliC:
equal_elements([], []).
equal_elements([X|Xs], Ys) :-
select(X, Ys, Zs),
equal_elements(Xs, Zs).
Above implementation leaves useless choicepoints for queries like this one:
?- equal_elements([1,2,3],[3,2,1]).
true ; % succeeds, but leaves choicepoint
false.
What could we do? We could fix the efficiency issue by using
selectchk/3 instead of
select/3, but by doing so we would lose logical-purity! Can we do better?
We can!
Introducing selectd/3, a logically pure predicate that combines the determinism of selectchk/3 and the purity of select/3. selectd/3 is based on
if_/3 and (=)/3:
selectd(E,[A|As],Bs1) :-
if_(A = E, As = Bs1,
(Bs1 = [A|Bs], selectd(E,As,Bs))).
selectd/3 can be used a drop-in replacement for select/3, so putting it to use is easy!
equal_elementsB([], []).
equal_elementsB([X|Xs], Ys) :-
selectd(X, Ys, Zs),
equal_elementsB(Xs, Zs).
Let's see it in action!
?- equal_elementsB([1,2,3],[3,2,1]).
true. % succeeds deterministically
?- equal_elementsB([1,2,3],[A,B,C]), C=3,B=2,A=1.
A = 1, B = 2, C = 3 ; % still logically pure
false.
Edit 2015-05-14
The OP wasn't specific if the predicate
should enforce that items occur on both sides with
the same multiplicities.
equal_elementsB/2 does it like that, as shown by these two queries:
?- equal_elementsB([1,2,3,2,3],[3,3,2,1,2]).
true.
?- equal_elementsB([1,2,3,2,3],[3,3,2,1,2,3]).
false.
If we wanted the second query to succeed, we could relax the definition in a logically pure way by using meta-predicate
tfilter/3 and
reified inequality dif/3:
equal_elementsC([],[]).
equal_elementsC([X|Xs],Ys2) :-
selectd(X,Ys2,Ys1),
tfilter(dif(X),Ys1,Ys0),
tfilter(dif(X),Xs ,Xs0),
equal_elementsC(Xs0,Ys0).
Let's run two queries like the ones above, this time using equal_elementsC/2:
?- equal_elementsC([1,2,3,2,3],[3,3,2,1,2]).
true.
?- equal_elementsC([1,2,3,2,3],[3,3,2,1,2,3]).
true.
Edit 2015-05-17
As it is, equal_elementsB/2 does not universally terminate in cases like the following:
?- equal_elementsB([],Xs), false. % terminates universally
false.
?- equal_elementsB([_],Xs), false. % gives a single answer, but ...
%%% wait forever % ... does not terminate universally
If we flip the first and second argument, however, we get termination!
?- equal_elementsB(Xs,[]), false. % terminates universally
false.
?- equal_elementsB(Xs,[_]), false. % terminates universally
false.
Inspired by an answer given by #AmiTavory, we can improve the implementation of equal_elementsB/2 by "sharpening" the solution set like so:
equal_elementsBB(Xs,Ys) :-
same_length(Xs,Ys),
equal_elementsB(Xs,Ys).
To check if non-termination is gone, we put queries using both predicates head to head:
?- equal_elementsB([_],Xs), false.
%%% wait forever % does not terminate universally
?- equal_elementsBB([_],Xs), false.
false. % terminates universally
Note that the same "trick" does not work with equal_elementsC/2,
because of the size of solution set is infinite (for all but the most trivial instances of interest).
A simple solution using the sort/2 ISO standard built-in predicate, assuming that neither list contains duplicated elements:
equal_elements(List1, List2) :-
sort(List1, Sorted1),
sort(List2, Sorted2),
Sorted1 == Sorted2.
Some sample queries:
| ?- equal_elements([1,2,3],[1,2,3,4]).
no
| ?- equal_elements([1,2,3],[3,1,2]).
yes
| ?- equal_elements([a(X),a(Y),a(Z)],[a(1),a(2),a(3)]).
no
| ?- equal_elements([a(X),a(Y),a(Z)],[a(Z),a(X),a(Y)]).
yes
In Prolog you often can do exactly what you say
areq([],_).
areq([H1|T1], L):- member(H1, L), areq(T1, L).
bi_areq(L1, L2) :- areq(L1, L2), areq(L2, L1).
Rename if necessary.
a compact form:
member_(Ys, X) :- member(X, Ys).
equal_elements(Xs, Xs) :- maplist(member_(Ys), Xs).
but, using member/2 seems inefficient, and leave space to ambiguity about duplicates (on both sides). Instead, I would use select/3
?- [user].
equal_elements([], []).
equal_elements([X|Xs], Ys) :-
select(X, Ys, Zs),
equal_elements(Xs, Zs).
^D here
1 ?- equal_elements(X, [1,2,3]).
X = [1, 2, 3] ;
X = [1, 3, 2] ;
X = [2, 1, 3] ;
X = [2, 3, 1] ;
X = [3, 1, 2] ;
X = [3, 2, 1] ;
false.
2 ?- equal_elements([1,2,3,3], [1,2,3]).
false.
or, better,
equal_elements(Xs, Ys) :- permutation(Xs, Ys).
The other answers are all elegant (way above my own Prolog level), but it struck me that the question stated
efficient for the regular uses.
The accepted answer is O(max(|A| log(|A|), |B|log(|B|)), irrespective of whether the lists are equal (up to permutation) or not.
At the very least, it would pay to check the lengths before bothering to sort, which would decrease the runtime to something linear in the lengths of the lists in the case where they are not of equal length.
Expanding this, it is not difficult to modify the solution so that its runtime is effectively linear in the general case where the lists are not equal (up to permutation), using random digests.
Suppose we define
digest(L, D) :- digest(L, 1, D).
digest([], D, D) :- !.
digest([H|T], Acc, D) :-
term_hash(H, TH),
NewAcc is mod(Acc * TH, 1610612741),
digest(T, NewAcc, D).
This is the Prolog version of the mathematical function Prod_i h(a_i) | p, where h is the hash, and p is a prime. It effectively maps each list to a random (in the hashing sense) value in the range 0, ...., p - 1 (in the above, p is the large prime 1610612741).
We can now check if two lists have the same digest:
same_digests(A, B) :-
digest(A, DA),
digest(B, DB),
DA =:= DB.
If two lists have different digests, they cannot be equal. If two lists have the same digest, then there is a tiny chance that they are unequal, but this still needs to be checked. For this case I shamelessly stole Paulo Moura's excellent answer.
The final code is this:
equal_elements(A, B) :-
same_digests(A, B),
sort(A, SortedA),
sort(B, SortedB),
SortedA == SortedB.
same_digests(A, B) :-
digest(A, DA),
digest(B, DB),
DA =:= DB.
digest(L, D) :- digest(L, 1, D).
digest([], D, D) :- !.
digest([H|T], Acc, D) :-
term_hash(H, TH),
NewAcc is mod(Acc * TH, 1610612741),
digest(T, NewAcc, D).
One possibility, inspired on qsort:
split(_,[],[],[],[]) :- !.
split(X,[H|Q],S,E,G) :-
compare(R,X,H),
split(R,X,[H|Q],S,E,G).
split(<,X,[H|Q],[H|S],E,G) :-
split(X,Q,S,E,G).
split(=,X,[X|Q],S,[X|E],G) :-
split(X,Q,S,E,G).
split(>,X,[H|Q],S,E,[H|G]) :-
split(X,Q,S,E,G).
cmp([],[]).
cmp([H|Q],L2) :-
split(H,Q,S1,E1,G1),
split(H,L2,S2,[H|E1],G2),
cmp(S1,S2),
cmp(G1,G2).
A simple solution using cut.
areq(A,A):-!.
areq([A|B],[C|D]):-areq(A,C,D,E),areq(B,E).
areq(A,A,B,B):-!.
areq(A,B,[C|D],[B|E]):-areq(A,C,D,E).
Some sample queries:
?- areq([],[]).
true.
?- areq([1],[]).
false.
?- areq([],[1]).
false.
?- areq([1,2,3],[3,2,1]).
true.
?- areq([1,1,2,2],[2,1,2,1]).
true.
:- use_module(library(clpfd)).
fact(treated=A) :- A in 0..1.
fact(numYears=B) :- B in 0..sup.
fact(numDrugs=C) :- C in 0..sup.
fact(treated2=D) :- D in 0..1.
fact(cParam=E) :- E in 0..4.
is_differentfact(X,X) :- false.
is_differentfact(Element=_,OtherElement=_) :-
dif(Element,OtherElement).
is_fakt([]).
is_fakt([X|Xs]) :-
fact(X),
maplist(is_differentfact(X),Xs),
is_fakt(Xs).
Why does ?- is_fakt(X) return a list of results answers but after a number of results answers it hangs. I don't know why Prolog cannot return all possible values of X.
You ask:
Why does ?- is_fakt(L) ... but after a number of results answers it hangs.
You say a number. That number is 62 times pressing SPACE to get to that moment of looping. Pretty long isn't it? And your program is tiny. How will you ever get the chance to do the same with a bigger program? Don't worry, there is help. But you need to look at the program from a different angle.
In Prolog understanding the very precise execution of a concrete query is next to impossible. You have two different kinds of control flows interleaved plus strange data structures that do not need to be present, but "come in" later ; sometimes. All that opens up a veritable panoply of possible execution traces that are so full of detail, that your mind will overflow — worse: your mind will still pretend you understand everything but effectively you don't. And the bugs have big party time in your program. Those bugs will bite at some point in time in the future, but only on a bug-to-bite basis. That can be very demoralizing. After all, the program is so small, that should be easy to understand (by the standards of imperative languages). But then, Prolog programs tend to be very compact for problems that are very complex in other languages.
Try to step through with a tracer to see what I mean. You will see all kinds of things happening. And most of them are irrelevant.
Fortunately, there are ways to understand Prolog, but here you have to rely on nice properties of the language itself. For localizing reasons for non-termination, the best is to start to consider a failure-slice. You obtain a failure slice from your program by adding goals false into your program. If the resulting program then still does not terminate, we have a reason why also our original program does not terminate.
Think of it: instead of trying to understand your program we do something humans are much better at: Making an educated guess. That guess can go wrong but we can check that easily. In the beginning you will be pretty awful at guessing. Soon you will see that you can do a lot of things systematically. All code that now becomes irrelevant is stike through.
:- use_module(library(clpfd)).
fact(treated=A) :- A in 0..1.
fact(numYears=B) :- B in 0..sup, false.
fact(numDrugs=C) :- C in 0..sup, false.
fact(treated2=D) :- D in 0..1, false.
fact(cParam=E) :- E in 0..4, false.
is_differentfact(X,X) :- false.
is_differentfact(Element=_,OtherElement=_) :-
dif(Element,OtherElement).
is_fakt([]).
is_fakt([X|Xs]) :-
fact(X),
maplist(is_differentfact(X),Xs),
is_fakt(Xs).
What did we gain? We can narrow down the problem much faster:
?- is_fakt(Xs).
Xs = []
; Xs = [treated=_A], _A in 0..1
; loops.
Before continuing, I try to understand what you mean with is_fakt/1. You probably mean: All the facts by their name, and make sure none is repeated. Now we have only the fact named treated, so we can only produce a list of length 1. And then it loops.
You said:
I don't know why Prolog cannot return all possible values of X.
To be picky, that is not true. Prolog did enumerate all possible values of X. But then it did not terminate.
((Some remarks to consider: Do you really want to get that list in that manner? You will get all permutations! With a list of length n you will get n! different answers. For n = 10 that is 3628800. Is this, what you want? Probably not.))
But let us first stick to identify the precise reason for non-termination.
To better identify the reason, lets "turn off" all answers. So we query is_fakt(L), false instead with:
:- use_module(library(clpfd)).
fact(treated=A) :- A in 0..1.
fact(numYears=B) :- B in 0..sup, false.
fact(numDrugs=C) :- C in 0..sup, false.
fact(treated2=D) :- D in 0..1, false.
fact(cParam=E) :- E in 0..4, false.
is_differentfact(X,X) :- false.
is_differentfact(Element=_,OtherElement=_) :-
dif(Element,OtherElement).
is_fakt([]) :- false.
is_fakt([X|Xs]) :-
fact(X),
maplist(is_differentfact(X),Xs), false,
is_fakt(Xs).
That is a minimal failure-slice. So it is the maplist/2 which does not terminate in the first place. Your idea was to ensure that X has a fact-name that is different to the fact-names in Xs. But if Xs is not bound, that will never terminate. Let's try it:
?- maplist(is_differentfact(X),Xs).
Xs = []
; X = (_A=_B), Xs = [_C=_D], dif(_A,_C)
; X = (_A=_B), Xs = [_C=_D,_E=_F], dif(_A,_C), dif(_A,_E)
; X = (_A=_B), Xs = [_C=_D,_E=_F,_G=_H],
dif(_A,_C), dif(_A,_E), dif(_A,_G)
; X = (_A=_B), Xs = [_C=_D,_E=_F,_G=_H,_I=_J],
dif(_A,_C), dif(_A,_E), dif(_A,_G), dif(_A,_I)
; X = (_A=_B), Xs = [_C=_D,_E=_F,_G=_H,_I=_J,_K=_L],
dif(_A,_C), dif(_A,_E), dif(_A,_G), dif(_A,_I), dif(_A,_K)
; ... .
Not so nice to look at... but we can do it better:
?- maplist(is_differentfact(X),Xs), false.
loops.
So it loops. This is the reason for non-termination. To fix the problem we have to do something in the remaining visible part of the failure slice...
For more, look up other explanations tagged failure-slice
Edited version based on the comments of false.
:- use_module(library(clpfd)).
:- use_module(library(lists)).
fact(treated-X) :- X in 0..1.
fact(numYears-X) :- X in 0..sup.
fact(numDrugs-X) :- X in 0..sup.
fact(treated2-X) :- X in 0..1.
fact(cParam-X) :- X in 0..4.
facts(Facts) :-
findall(X,fact(X),Facts).
is_fact2(_, []).
is_fact2(Facts, [X|Xs]) :-
member(X,Facts),
select(X,Facts,Remaining),
is_fact2(Remaining,Xs).
is_fakt(X) :-
facts(Facts),
is_fact2(Facts,X),
keysort(X,X).
This terminates now.