replacing elements between two lists using prolog

Question

My task is to make myReplace(E1,L1,E2,L2) such that the very first occurrence of E1 in L1 gets replaced by E2 and is returned in L2. I have written the below described code and it is working properly.

myReplace(E1,[],E2,[]).
myReplace(E1,[E1|Xs],E2,[E2|Ys]):-
  myReplace(E1,Xs,E1,Ys).
myReplace(E1,[H|Hs],E2,[H|Ts]):-
  E1 \= H,
  myReplace(E1,Hs,E2,Ts).

However, For example myReplace(2,[1,2,3,2,1],5,X) should give X = [1,5,3,2,1] and X = [1,2,3,5,1]. But my code is only giving one solution which is X = [1,5,3,2,1].

Similarly, when myReplace(2,X,5,[1,5,3,5,1]) should backtrack over the solutions X = [1,2,3,5,1] and X = [1,5,3,2,1] only, but my solution gives me one more solution as X = [1,5,3,5,1].

Could you please help me resolve this.

Thank you :)

Answer 1

What about

myReplace(E1,[E1|Xs],E2,[E2|Xs]).

myReplace(E1,[H|Hs],E2,[H|Ts]):-
  myReplace(E1,Hs,E2,Ts).

?

If I'm not wrong, this impose one (and only one) replacement.

Take in count that you have to delete

myReplace(E1,[],E2,[]).

otherwise you get L1 = L2 (no replacement) as a solution.

And observe that, as pointed by Lurker, this isn't "the very first occurrence of E1 in L1 gets replaced".

Answer 2

If I understand your question, you want to get all answers from substituting each one occurrence of E1 in L1. You can get the backtracking for free if you use append/3 for this:

my_replace(X, As, Y, Bs) :-
    append(As_front, [X|As_back], As),
    append(As_front, [Y|As_back], Bs).

With this definition I get:

?- my_replace(2,[1,2,3,2,1],5,X).
X = [1, 5, 3, 2, 1] ;
X = [1, 2, 3, 5, 1] ;
false

See the other solution to see how to see what the problem with your original code was.