Question

I’ve been tasked with the portion of the code that reads through two lists and finds a path between the two lists.

My problem is however with reading and returning a TRUE case if I find the required node in the first list. I’ve been writing my code as below.

%declare the list
circle_line([cc1,cc2,cc3,cc4,cc5,cc6,cc7,cc8,cc9,cc10,cc11,cc12,cc13,cc14,cc15,cc16]).   

%the predicate that finds the station i need
check_for_station(A) :- circle_line(X),
            member(A,X),
            write(A),nl.

Then in the cosole I type: check_for_station(cc9).

But the answer I get is “no.”
I have a feeling that I’m declaring the list wrong, because in the debugger the value of X comes out to be “H135” or something and it clearly does not loop through every element to find the one that I need.

Answer 1

As whythehack found, in Amzi! Prolog the classic (and nondeterministic) member/2 predicate is not a “built-in” but is provided (along with other useful list manipulation predicates like length/2) in the list library. See here for some context about the built-in versus library predicates for lists.

If check_for_station/1 is only to be called with its argument A bound (as the example suggests), then Amzi!’s built-in (and deterministic) predicate is_member/2 will do the job (and a bit more quickly).

The thinking seems to be that nondeterministic member/2 is a two-line user-definition, and providing a faster deterministic version (not permitting backtracking over members of a list) is something the user could not easily provide for themselves.