Question mark in SWI-Prolog - nondeterministic functor

Question

I've just tried to implement absolute function in Prolog and I've got some strange behavior. My code was:

absval(X, RESULT) :- X >= 0, RESULT is X.
absval(X, RESULT) :- X < 0, RESULT is -X.

And when I try in SWI-Prolog absval(-2,X). I get

X = 2

yes

as expected. But otherwise when I invoke absval(2,X), I get X = 2 ? and I should insert another input. After pressing enter I get also yes.

What does mean the second one result? What's wrong with my solution?

The behavior you encounter can be explained by the fact that your program is not deterministic. There are two rules for `absval/2`. So, after Prolog satisfies `absval(2,X)` using the first rule, it backtracks in order to resatisfy the goal. The first subgoal `(X < 0)` of the second rule will fail and you will not get a second solution. In order to have a deterministic behavior, add a cut operator after `X>=0` like this: `absval(X, RESULT) :- X >= 0, !, RESULT is X.`. — Tudor Berariu, May 17 '14 at 17:49

score 2 · Accepted Answer · answered May 17 '14 at 17:53

2

what you report doesn't match the behaviour I get here:

?- absval(2,X).
X = 2 ;
false.

that's actually what is expected.

If you need to make it deterministic, use a cut, or better, the 'if/then/else' construct:

absval(X, RESULT) :- X >= 0 -> RESULT is X ; RESULT is -X.

answered May 17 '14 at 17:53

CapelliC

ok, I see. and additional question: is there any way to express this without using neither if/then/else nor cuts? – pt12lol May 17 '14 at 18:30
AFAIK, no. Your code it's already correct, the non determinism is unavoidable without cut... – CapelliC May 17 '14 at 18:32
No, no, no! You have to see the signs! :-) – Paulo Moura May 17 '14 at 18:48

Paulo Moura · Answer 2 · 2014-05-17T19:03:40.590

1

A deterministic version using a single clause without cuts of if-then-else constructs:

absolute_value(Value, AbsoluteValue) :-
    AbsoluteValue is sign(Value) * Value.

Note that sign/1 is a standard built-in arithmetic function.

edited May 17 '14 at 19:03

answered May 17 '14 at 18:47

Paulo Moura

`abs` also. I try to express if/else construction from procedural language using multiple functors. – pt12lol May 17 '14 at 18:52

2 Answers2