The predicate if_/3 seems to be fairly popular among the few main contributors in the Prolog part of Stack Overflow.
This predicate is implemented a
In old-fashioned Prolog code, the following pattern arises rather frequently:
predicate([], ...). predicate([L|Ls], ...) :- condition(L), then(Ls, ...). predicate([L|Ls], ...) :- \+ condition(L), else(Ls, ...).
I am using lists here as an example where this occurs (see for example include/3
, exclude/3
etc.), although the pattern of course also occurs elsewhere.
The tragic is the following:
'.'(_, _)
as the primary functor and arity of their first argument.In summary, the existing constructs and language features all fall short in some way to express a pattern that often occurs in practice. Therefore, for decades, it seemed necessary to compromise. And you can make a pretty good guess in which direction the "compromises" usually go in the Prolog community: Almost invariably, correctness is sacrificed for efficiency in case of doubt. After all, who cares about correct results as long as your programs are fast, right? Therefore, until the invention of if_/3
, this was frequently wrongly written as:
predicate([], ...). predicate([L|Ls], ...) :- ( condition(L) -> then(Ls, ...). ; else(Ls, ...). )
The mistake in this is of course that when the elements are not sufficiently instantiated, then this may incorrectly commit to one branch even though both alternatives are logically possible. For this reason, using if-then-else is almost always declaratively wrong, and stands massively in the way of declarative debugging approaches due to its violation of the most elementary properties we expect from pure Prolog programs.
Using if_/3
, you can write this as:
predicate([], ...). predicate([L|Ls], ...) :- if_(condition(L), then(Ls, ...), else(Ls, ...)).
and retain all desirable aspects. This is:
The price of this is rather affordable: As Boris mentioned in the comments, you need to implement a reification. I have now some experience with this and found it rather easy with some practice.
Good news everyone: In many cases, condition
is of the form (=)/2
, or (#=)/2
, and the first even ships with library(reif) for free.
For more information, see Indexing dif/2 by Ulrich Neumerkel and Stefan Kral!
Let's try to solve a simple problem using if_/3
; for example, I will try to partition a list (sorted on a predicate p/2
) in two lists: a prefix in which, for every element X
, we have p(X, true)
, and a rest (in which, if the list was sorted on p/2
, we would have p(X, false)
.
I will use the library reif
as here. So, here is the complete code of my program:
:- use_module(reif).
pred_prefix(Pred_1, List, L_true, L_false) :-
pred_prefix_aux(List, Pred_1, L_true, L_false).
pred_prefix_aux([], _, [], []).
pred_prefix_aux([X|Xs], Pred_1, True, False) :-
if_( call(Pred_1, X),
( True = [X|True0],
pred_prefix_aux(Xs, Pred_1, True0, False)
),
( True = [],
False = [X|Xs]
)
).
The predicate passed to this meta-predicate will take two arguments: the first is the current list element, and the second will be either true
or false
. Ideally, this predicate will always succeed and not leave behind choice points.
In the first argument of if_/2
, the predicate is evaluated with the current list element; the second argument is what happens when true
; the third argument is what happens when false
.
With this, I can split a list in leading a
s and a rest:
?- pred_prefix([X, B]>>(=(a, X, B)), [a,a,b], T, F).
T = [a, a],
F = [b].
?- pred_prefix([X, B]>>(=(a, X, B)), [b,c,d], T, F).
T = [],
F = [b, c, d].
?- pred_prefix([X, B]>>(=(a, X, B)), [b,a], T, F).
T = [],
F = [b, a].
?- pred_prefix([X, B]>>(=(a, X, B)), List, T, F).
List = T, T = F, F = [] ;
List = T, T = [a],
F = [] ;
List = T, T = [a, a],
F = [] ;
List = T, T = [a, a, a],
F = [] .
How can you get rid of leading 0's for example:
?- pred_prefix([X, B]>>(=(0, X, B)), [0,0,1,2,0,3], _, F).
F = [1, 2, 0, 3].
Of course, this could have been written much simpler:
drop_leading_zeros([], []).
drop_leading_zeros([X|Xs], Rest) :-
if_(=(0, X), drop_leading_zeros(Xs, Rest), [X|Xs] = Rest).
Here I have just removed all unnecessary arguments.
If you would have to do this without if_/3
, you would have had to write:
drop_leading_zeros_a([], []).
drop_leading_zeros_a([X|Xs], Rest) :-
=(0, X, T),
( T == true -> drop_leading_zeros_a(Xs, Rest)
; T == false -> [X|Xs] = Rest
).
Here, we assume that =/3
will indeed always succeed without choice points and the T
will always be either true
or false
.
And, if we didn't have =/3
either, you'd write:
drop_leading_zeros_full([], []).
drop_leading_zeros_full([X|Xs], Rest) :-
( X == 0 -> T = true
; X \= 0 -> T = false
; T = true, X = 0
; T = false, dif(0, X)
),
( T == true -> drop_leading_zeros_full(Xs, Rest)
; T == false -> [X|Xs] = Rest
).
which is not ideal. But now at least you can see for yourself, in one single place, what is actually going on.
PS: Please read the code and the top level interaction carefully.