Question

I am new to prolog and i want to solve this problem. Suppose I have a list say
List i.e. [a,b,c]
now i have some facts say
likes(a,banana).
likes(b,orange).
likes(c,apple).
likes(d,grapes).

---

So if I make a query

?- my_functor(List,X).
X=[banana,orange,apple].

---

Thanks you.

Answer 1

Consider:

?- List=[a,b,c], findall(X, (member(Y, List), likes(Y, X)), Xs).
List = [a, b, c],
Xs = [banana, orange, apple].

Explanation:

findall/3 is called an ‘all-solutions’ predicate which seeks to find all possible values unifiable to the first argument (here, that’s the variable X) to solutions for the seconds argument (here, that’s the conjunction (member(Y, List), likes(Y, X))), and places all values for X into a list, bound to the third argument (here, that’s Xs).

Notice that the inner expression generating the values for X is a statement that backtracks to provide different assignments for X:

?- member(Y, [a,b,c]), likes(Y, X).
Y = a,
X = banana ;
Y = b,
X = orange ;
Y = c,
X = apple ;
false.

Tested with SWI-Prolog.

Note that findall/3 also appears in GNU Prolog amongst most other implementations.