Element handling issue in Prolog, but not in a regular list?

Question

I'm trying to write an element handling function in Prolog. It's almost the same as the prolog predicate member/2 but it must do the job in a different way. For being specific; I must say the member/2 predicate function is this:

member(X, [X|_]).
member(X, [_|Tail]) :-
   member(X,Tail).

When you give a query for example: member(X, [1,2,3]).

It gives you X = 1; X = 2; X = 3; in this order for all redo's. I want an element function almost the same. I want the same result with the member function when I give a query like this:

element(X, (1,2,3)).

The difference is just parenthesis instead of bracekts like these : []
In order to do this I tried that:

element(X, (X,_)).
element(X, (_,Tail)) :-
   element(X,Tail).

Which is exactly the same as member/2 predicate function implementation. But this doesn't work because it doesn't giving the last element which is X=3. So I added one more fact that is:

element(X, X).

But this doesn't work either because (obviously) it is giving unnecessary answer with real elements like these:

X=(1,2,3)
X=(2,3)

How can I handle this?

your `member/2` code is off: it should read `[_|Tail]`, not `[_,Tail]` — m09, Apr 22 '13 at 13:56
Yes I wrote it wrong here sorry, I'm making it correct now. By the way I'm using it in the program in the right way as you said. — Hazım Türkkan, Apr 23 '13 at 09:10

score 3 · Accepted Answer · answered Apr 22 '13 at 15:59

3

Seems that a cut can solve your problem:

element(X, (X, _)).
element(X, (_, Tail)) :-
    !, element(X, Tail).
element(X, X).

test:

?- element(X, (1,2,3)).
X = 1 ;
X = 2 ;
X = 3.

?- element(2, (1,2,3,2)).
true ;
true.

answered Apr 22 '13 at 15:59

CapelliC

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Thank you very much, I'm so embarrased that I couldn't see this. I tried to use a cut but because I'm very new to prolog I couldn't place it properly where it's necessary. I'm trying to figure out how prolog works for different situations by the trace option while testing but I couldn't really understand what cut (!) operator does. Can anybody explain it in a proper way? Thanks a lot for the answer again. – Hazım Türkkan Apr 23 '13 at 09:09
I found an explanation like this and it looks good; I could understand now some more but for usage I must get used to it: [link](http://en.wikipedia.org/wiki/Cut_(logic_programming). – Hazım Türkkan Apr 23 '13 at 09:17

score 1 · Answer 2 · edited May 23 '17 at 12:18

1

Terms like (1,2,3) in Prolog have commas as their primary functor. Probably you want to use operator univ, denoted by infix =.., or its close relations functor/3 and arg/3 to pick apart these tuples.

edited May 23 '17 at 12:18

Community

1
1

answered Apr 22 '13 at 15:14

hardmath

8,753
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Element handling issue in Prolog, but not in a regular list?

2 Answers2